Solucionario INCROPERA - Transf de calor

Solucionario PROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal conductivity k and inner temperature, T1. FIND: The outer temperature of the wall, T2. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law,

q cond = q x = q ′′x ⋅ A = -k

T −T dT ⋅ A = kA 1 2 . dx L

Solving for T2 gives

T2 = T1 −

q cond L . kA

Substituting numerical values, find

T2 = 415$ C -

3000W × 0.025m 0.2W / m ⋅ K × 10m2

T2 = 415$ C - 37.5$ C T2 = 378$ C. COMMENTS: Note direction of heat flow and fact that T2 must be less than T1.

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PROBLEM 1.2 KNOWN: Inner surface temperature and thermal conductivity of a concrete wall. FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from -15 to 38°C. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties, (4) Outside wall temperature is that of the ambient air. ANALYSIS: From Fourier’s law, it is evident that the gradient, dT dx = − q′′x k , is a constant, and hence the temperature distribution is linear, if q′′x and k are each constant. The heat flux must be constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 = -15°C are

)

(

25$ C − −15$ C dT T1 − T2 q′′x = − k =k = 1W m ⋅ K = 133.3W m 2 . dx L 0.30 m q x = q′′x × A = 133.3 W m 2 × 20 m 2 = 2667 W .

(1) (2)

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Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of ambient temperature, -15 ≤ T2 ≤ 38°C, with different wall thermal conductivities, k. 3500

Heat loss, qx (W)

2500

1500

500

-500

-1500 -20

-10

0

10

20

30

40

Ambient air temperature, T2 (C) Wall thermal conductivity, k = 1.25 W/m.K k = 1 W/m.K, concrete wall k = 0.75 W/m.K

For the concrete wall, k = 1 W/m⋅K, the heat loss varies linearily from +2667 W to -867 W and is zero when the inside and ambient temperatures are the same. The magnitude of the heat rate increases with increasing thermal conductivity. COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane wall would not be linear.

PROBLEM 1.3 KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency of gas furnace and cost of natural gas. FIND: Daily cost of heat loss. SCHEMATIC:

ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties. ANALYSIS: The rate of heat loss by conduction through the slab is

T −T 7°C q = k ( LW ) 1 2 = 1.4 W / m ⋅ K (11m × 8 m ) = 4312 W t 0.20 m

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The daily cost of natural gas that must be combusted to compensate for the heat loss is

Cd =

q Cg

ηf

( ∆t ) =

4312 W × $0.01/ MJ 0.9 ×106 J / MJ

( 24 h / d × 3600s / h ) = $4.14 / d

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COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation between it and the concrete.

PROBLEM 1.4 KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed thickness. FIND: Thermal conductivity, k, of the wood. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be determined from Fourier’s law, Eq. 1.2. Rearranging, k=q′′x

L W = 40 T1 − T2 m2

k = 0.10 W / m ⋅ K.

0.05m

( 40-20 ) C

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COMMENTS: Note that the °C or K temperature units may be used interchangeably when evaluating a temperature difference.

PROBLEM 1.5 KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions. FIND: Heat loss through window. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: Subject to the foregoing conditions the heat flux may be computed from Fourier’s law, Eq. 1.2. T −T q′′x = k 1 2 L  W (15-5 ) C q′′x = 1.4 m ⋅ K 0.005m q′′x = 2800 W/m 2 . Since the heat flux is uniform over the surface, the heat loss (rate) is q = q ′′x × A q = 2800 W / m2 × 3m2 q = 8400 W. COMMENTS: A linear temperature distribution exists in the glass for the prescribed conditions.

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PROBLEM 1.6 KNOWN: Width, height, thickness and thermal conductivity of a single pane window and the air space of a double pane window. Representative winter surface temperatures of single pane and air space. FIND: Heat loss through single and double pane windows. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction through glass or air, (2) Steady-state conditions, (3) Enclosed air of double pane window is stagnant (negligible buoyancy induced motion). ANALYSIS: From Fourier’s law, the heat losses are

Single Pane:

T1 − T2 35 $C 2 qg = k g A = 1.4 W/m ⋅ K 2m = 19, 600 W L 0.005m

( )

( )

T −T 25 $C Double Pane: qa = k a A 1 2 = 0.024 2m2 = 120 W L 0.010 m COMMENTS: Losses associated with a single pane are unacceptable and would remain excessive, even if the thickness of the glass were doubled to match that of the air space. The principal advantage of the double pane construction resides with the low thermal conductivity of air (~ 60 times smaller than that of glass). For a fixed ambient outside air temperature, use of the double pane construction would also increase the surface temperature of the glass exposed to the room (inside) air.

PROBLEM 1.7 KNOWN: Dimensions of freezer compartment. Inner and outer surface temperatures. FIND: Thickness of styrofoam insulation needed to maintain heat load below prescribed value. SCHEMATIC:

ASSUMPTIONS: (1) Perfectly insulated bottom, (2) One-dimensional conduction through 5 2 walls of area A = 4m , (3) Steady-state conditions, (4) Constant properties. ANALYSIS: Using Fourier’s law, Eq. 1.2, the heat rate is q = q ′′ ⋅ A = k

∆T A total L 2

Solving for L and recognizing that Atotal = 5×W , find 5 k ∆ T W2 L = q



L=

( )

5 × 0.03 W/m ⋅ K 35 - (-10 ) C 4m 2 500 W

L = 0.054m = 54mm.

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COMMENTS: The corners will cause local departures from one-dimensional conduction and a slightly larger heat loss.

PROBLEM 1.8 KNOWN: Dimensions and thermal conductivity of food/beverage container. Inner and outer surface temperatures. FIND: Heat flux through container wall and total heat load. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through bottom wall, (3) Uniform surface temperatures and one-dimensional conduction through remaining walls. ANALYSIS: From Fourier’s law, Eq. 1.2, the heat flux is $ T2 − T1 0.023 W/m ⋅ K ( 20 − 2 ) C ′′ q =k = = 16.6 W/m 2 L 0.025 m

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Since the flux is uniform over each of the five walls through which heat is transferred, the heat load is q = q′′ × A total = q′′  H ( 2W1 + 2W2 ) + W1 × W2  q = 16.6 W/m2  0.6m (1.6m + 1.2m ) + ( 0.8m × 0.6m ) = 35.9 W

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COMMENTS: The corners and edges of the container create local departures from onedimensional conduction, which increase the heat load. However, for H, W1, W2 >> L, the effect is negligible.

PROBLEM 1.9 KNOWN: Masonry wall of known thermal conductivity has a heat rate which is 80% of that through a composite wall of prescribed thermal conductivity and thickness. FIND: Thickness of masonry wall. SCHEMATIC:

ASSUMPTIONS: (1) Both walls subjected to same surface temperatures, (2) Onedimensional conduction, (3) Steady-state conditions, (4) Constant properties. ANALYSIS: For steady-state conditions, the conduction heat flux through a one-dimensional wall follows from Fourier’s law, Eq. 1.2, q ′′ = k

∆T L

where ∆T represents the difference in surface temperatures. Since ∆T is the same for both walls, it follows that L1 = L2

k1 q ′′ ⋅ 2. k2 q1′′

With the heat fluxes related as q1′′ = 0.8 q ′′2 L1 = 100mm

0.75 W / m ⋅ K 1 × = 375mm. 0.25 W / m ⋅ K 0.8

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COMMENTS: Not knowing the temperature difference across the walls, we cannot find the value of the heat rate.

PROBLEM 1.10 KNOWN: Thickness, diameter and inner surface temperature of bottom of pan used to boil water. Rate of heat transfer to the pan. FIND: Outer surface temperature of pan for an aluminum and a copper bottom. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction through bottom of pan. ANALYSIS: From Fourier’s law, the rate of heat transfer by conduction through the bottom of the pan is T −T q = kA 1 2 L Hence, T1 = T2 +

qL kA

2 where A = π D2 / 4 = π (0.2m ) / 4 = 0.0314 m 2 . Aluminum:

T1 = 110 $C +

Copper:

T1 = 110 $C +

600W ( 0.005 m )

(

240 W/m ⋅ K 0.0314 m 2 600W (0.005 m )

(

390 W/m ⋅ K 0.0314 m2

) )

= 110.40 $C

= 110.25 $C

COMMENTS: Although the temperature drop across the bottom is slightly larger for aluminum (due to its smaller thermal conductivity), it is sufficiently small to be negligible for both materials. To a good approximation, the bottom may be considered isothermal at T ≈ 110 °C, which is a desirable feature of pots and pans.

PROBLEM 1.11 KNOWN: Dimensions and thermal conductivity of a chip. Power dissipated on one surface. FIND: Temperature drop across the chip. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform heat dissipation, (4) Negligible heat loss from back and sides, (5) One-dimensional conduction in chip. ANALYSIS: All of the electrical power dissipated at the back surface of the chip is transferred by conduction through the chip. Hence, from Fourier’s law, P = q = kA

∆T t

or ∆T =

t ⋅P kW 2

=

∆T = 1.1$ C.

0.001 m × 4 W

2

150 W/m ⋅ K ( 0.005 m )

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COMMENTS: For fixed P, the temperature drop across the chip decreases with increasing k and W, as well as with decreasing t.

PROBLEM 1.12 KNOWN: Heat flux gage with thin-film thermocouples on upper and lower surfaces; output voltage, calibration constant, thickness and thermal conductivity of gage. FIND: (a) Heat flux, (b) Precaution when sandwiching gage between two materials. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat conduction in gage, (3) Constant properties. ANALYSIS: (a) Fourier’s law applied to the gage can be written as q ′′ = k

∆T ∆x

and the gradient can be expressed as ∆T ∆E / N = ∆x SABt where N is the number of differentially connected thermocouple junctions, SAB is the Seebeck coefficient for type K thermocouples (A-chromel and B-alumel), and ∆x = t is the gage thickness. Hence, q ′′ =

k∆E NSABt

q ′′ =

1.4 W / m ⋅ K × 350 × 10-6 V = 9800 W / m2 . -6 -3 $ 5 × 40 × 10 V / C × 0.25 × 10 m

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(b) The major precaution to be taken with this type of gage is to match its thermal conductivity with that of the material on which it is installed. If the gage is bonded between laminates (see sketch above) and its thermal conductivity is significantly different from that of the laminates, one dimensional heat flow will be disturbed and the gage will read incorrectly. COMMENTS: If the thermal conductivity of the gage is lower than that of the laminates, will it indicate heat fluxes that are systematically high or low?

PROBLEM 1.13 KNOWN: Hand experiencing convection heat transfer with moving air and water. FIND: Determine which condition feels colder. Contrast these results with a heat loss of 30 W/m2 under normal room conditions. SCHEMATIC:

ASSUMPTIONS: (1) Temperature is uniform over the hand’s surface, (2) Convection coefficient is uniform over the hand, and (3) Negligible radiation exchange between hand and surroundings in the case of air flow. ANALYSIS: The hand will feel colder for the condition which results in the larger heat loss. The heat loss can be determined from Newton’s law of cooling, Eq. 1.3a, written as

q′′ = h ( Ts − T∞ ) For the air stream:

q′′air = 40 W m 2 ⋅ K 30 − ( −5) K = 1, 400 W m 2

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For the water stream:

q′′water = 900 W m 2 ⋅ K (30 − 10 ) K = 18,000 W m 2

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COMMENTS: The heat loss for the hand in the water stream is an order of magnitude larger than when in the air stream for the given temperature and convection coefficient conditions. In contrast, the heat loss in a normal room environment is only 30 W/m2 which is a factor of 400 times less than the loss in the air stream. In the room environment, the hand would feel comfortable; in the air and water streams, as you probably know from experience, the hand would feel uncomfortably cold since the heat loss is excessively high.

PROBLEM 1.14 KNOWN: Power required to maintain the surface temperature of a long, 25-mm diameter cylinder with an imbedded electrical heater for different air velocities. FIND: (a) Determine the convection coefficient for each of the air velocity conditions and display the results graphically, and (b) Assuming that the convection coefficient depends upon air velocity as h = CVn, determine the parameters C and n. SCHEMATIC: V(m/s) Pe′ (W/m) h (W/m2⋅K)

1 450 22.0

2 658 32.2

4 983 48.1

8 1507 73.8

12 1963 96.1

ASSUMPTIONS: (1) Temperature is uniform over the cylinder surface, (2) Negligible radiation exchange between the cylinder surface and the surroundings, (3) Steady-state conditions. ANALYSIS: (a) From an overall energy balance on the cylinder, the power dissipated by the electrical heater is transferred by convection to the air stream. Using Newtons law of cooling on a per unit length basis,

Pe′ = h (π D )(Ts − T∞ ) where Pe′ is the electrical power dissipated per unit length of the cylinder. For the V = 1 m/s condition, using the data from the table above, find $ h = 450 W m π × 0.025 m 300 − 40 C = 22.0 W m 2⋅K

(

)

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Repeating the calculations, find the convection coefficients for the remaining conditions which are tabulated above and plotted below. Note that h is not linear with respect to the air velocity. (b) To determine the (C,n) parameters, we plotted h vs. V on log-log coordinates. Choosing C = 22.12 W/m2⋅K(s/m)n, assuring a match at V = 1, we can readily find the exponent n from the slope of the h vs. V curve. From the trials with n = 0.8, 0.6 and 0.5, we recognize that n = 0.6 is a reasonable

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100 80 60 40 20 0

2

4

6

8

10

12

Air velocity, V (m/s) Data, smooth curve, 5-points

Coefficient, h (W/m^2.K)

Coefficient, h (W/m^2.K)

choice. Hence, C = 22.12 and n = 0.6.

100 80 60 40 20 10 1

2

4

6

Air velocity, V (m/s) Data , smooth curve, 5 points h = C * V^n, C = 22.1, n = 0.5 n = 0.6 n = 0.8

8

10

PROBLEM 1.15 KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required to maintain a specified surface temperature for water and air flows. FIND: Convection coefficients for the water and air flow convection processes, hw and ha, respectively. SCHEMATIC:

ASSUMPTIONS: (1) Flow is cross-wise over cylinder which is very long in the direction normal to flow. ANALYSIS: The convection heat rate from the cylinder per unit length of the cylinder has the form q′ = h (π D ) ( Ts − T∞ ) and solving for the heat transfer convection coefficient, find h=

q′ . π D (Ts − T∞ )

Substituting numerical values for the water and air situations: Water

hw =

Air

ha =

28 × 103 W/m

 π × 0.030m (90-25 ) C 400 W/m

π × 0.030m (90-25 ) C

= 4,570 W/m 2 ⋅ K

= 65 W/m 2 ⋅ K.

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COMMENTS: Note that the air velocity is 10 times that of the water flow, yet hw ≈ 70 × ha. These values for the convection coefficient are typical for forced convection heat transfer with liquids and gases. See Table 1.1.

PROBLEM 1.16 KNOWN: Dimensions of a cartridge heater. Heater power. Convection coefficients in air and water at a prescribed temperature. FIND: Heater surface temperatures in water and air. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) All of the electrical power is transferred to the fluid by convection, (3) Negligible heat transfer from ends. ANALYSIS: With P = qconv, Newton’s law of cooling yields P=hA (Ts − T∞ ) = hπ DL ( Ts − T∞ ) P Ts = T∞ + . hπ DL In water, Ts = 20$ C +

2000 W 5000 W / m ⋅ K × π × 0.02 m × 0.200 m 2

Ts = 20$ C + 31.8$ C = 51.8$ C.

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In air, Ts = 20$ C +

2000 W 50 W / m ⋅ K × π × 0.02 m × 0.200 m 2

Ts = 20$ C + 3183$ C = 3203$ C.

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COMMENTS: (1) Air is much less effective than water as a heat transfer fluid. Hence, the cartridge temperature is much higher in air, so high, in fact, that the cartridge would melt. (2) In air, the high cartridge temperature would render radiation significant.

PROBLEM 1.17 KNOWN: Length, diameter and calibration of a hot wire anemometer. Temperature of air stream. Current, voltage drop and surface temperature of wire for a particular application. FIND: Air velocity SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from the wire by natural convection or radiation. ANALYSIS: If all of the electric energy is transferred by convection to the air, the following equality must be satisfied Pelec = EI = hA (Ts − T∞ ) where A = π DL = π ( 0.0005m × 0.02m ) = 3.14 × 10−5 m 2 . Hence, h=

EI 5V × 0.1A = = 318 W/m 2 ⋅ K A (Ts − T∞ ) 3.14 × 10−5m 2 50 $C

(

(

)

V = 6.25 × 10−5 h 2 = 6.25 ×10−5 318 W/m 2 ⋅ K

)

2

= 6.3 m/s

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COMMENTS: The convection coefficient is sufficiently large to render buoyancy (natural convection) and radiation effects negligible.

PROBLEM 1.18 KNOWN: Chip width and maximum allowable temperature. Coolant conditions. FIND: Maximum allowable chip power for air and liquid coolants. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from sides and bottom, (3) Chip is at a uniform temperature (isothermal), (4) Negligible heat transfer by radiation in air. ANALYSIS: All of the electrical power dissipated in the chip is transferred by convection to the coolant. Hence, P=q and from Newton’s law of cooling, 2

P = hA(T - T∞) = h W (T - T∞). In air, 2

2

Pmax = 200 W/m ⋅K(0.005 m) (85 - 15) ° C = 0.35 W.

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In the dielectric liquid 2

2

Pmax = 3000 W/m ⋅K(0.005 m) (85-15) ° C = 5.25 W.

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COMMENTS: Relative to liquids, air is a poor heat transfer fluid. Hence, in air the chip can dissipate far less energy than in the dielectric liquid.

PROBLEM 1.19 KNOWN: Length, diameter and maximum allowable surface temperature of a power transistor. Temperature and convection coefficient for air cooling. FIND: Maximum allowable power dissipation. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through base of transistor, (3) Negligible heat transfer by radiation from surface of transistor. ANALYSIS: Subject to the foregoing assumptions, the power dissipated by the transistor is equivalent to the rate at which heat is transferred by convection to the air. Hence, Pelec = q conv = hA (Ts − T∞ )

(

)

2 where A = π DL + D2 / 4 = π 0.012m × 0.01m + ( 0.012m ) / 4  = 4.90 ×10−4 m 2 .   For a maximum allowable surface temperature of 85°C, the power is

(

Pelec = 100 W/m2 ⋅ K 4.90 × 10−4 m 2

) (85 − 25)$ C = 2.94 W

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COMMENTS: (1) For the prescribed surface temperature and convection coefficient, radiation will be negligible relative to convection. However, conduction through the base could be significant, thereby permitting operation at a larger power. (2) The local convection coefficient varies over the surface, and hot spots could exist if there are locations at which the local value of h is substantially smaller than the prescribed average value.

PROBLEM 1.20 KNOWN: Air jet impingement is an effective means of cooling logic chips. FIND: Procedure for measuring convection coefficients associated with a 10 mm × 10 mm chip. SCHEMATIC:

ASSUMPTIONS: Steady-state conditions. ANALYSIS: One approach would be to use the actual chip-substrate system, Case (a), to perform the measurements. In this case, the electric power dissipated in the chip would be transferred from the chip by radiation and conduction (to the substrate), as well as by convection to the jet. An energy balance for the chip yields q elec = q conv + q cond + q rad . Hence, with q conv = hA ( Ts − T∞ ) , where A = 100 mm2 is the surface area of the chip,

q − q cond − q rad h = elec A (Ts − T∞ )

(1)

While the electric power ( q elec ) and the jet ( T∞ ) and surface ( Ts ) temperatures may be measured, losses from the chip by conduction and radiation would have to be estimated. Unless the losses are negligible (an unlikely condition), the accuracy of the procedure could be compromised by uncertainties associated with determining the conduction and radiation losses. A second approach, Case (b), could involve fabrication of a heater assembly for which the conduction and radiation losses are controlled and minimized. A 10 mm × 10 mm copper block (k ~ 400 W/m⋅K) could be inserted in a poorly conducting substrate (k < 0.1 W/m⋅K) and a patch heater could be applied to the back of the block and insulated from below. If conduction to both the substrate and insulation could thereby be rendered negligible, heat would be transferred almost exclusively through the block. If radiation were rendered negligible by applying a low emissivity coating (ε < 0.1) to the surface of the copper block, virtually all of the heat would be transferred by convection to the jet. Hence, q cond and q rad may be neglected in equation (1), and the expression may be used to accurately determine h from the known (A) and measured ( q elec , Ts , T∞ ) quantities. COMMENTS: Since convection coefficients associated with gas flows are generally small, concurrent heat transfer by radiation and/or conduction must often be considered. However, jet impingement is one of the more effective means of transferring heat by convection and convection coefficients well in excess of 100 W/m2⋅K may be achieved.

PROBLEM 1.21 KNOWN: Upper temperature set point, Tset, of a bimetallic switch and convection heat transfer coefficient between clothes dryer air and exposed surface of switch. FIND: Electrical power for heater to maintain Tset when air temperature is T∞ = 50°C. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Electrical heater is perfectly insulated from dryer wall, (3) Heater and switch are isothermal at Tset, (4) Negligible heat transfer from sides of heater or switch, (5) Switch surface, As, loses heat only by convection. ANALYSIS: Define a control volume around the bimetallic switch which experiences heat input from the heater and convection heat transfer to the dryer air. That is, E in - E out = 0 qelec - hAs ( Tset − T∞ ) = 0. The electrical power required is, qelec = hAs ( Tset − T∞ ) qelec = 25 W/m2 ⋅ K × 30 × 10-6 m2 ( 70 − 50 ) K=15 mW.

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COMMENTS: (1) This type of controller can achieve variable operating air temperatures with a single set-point, inexpensive, bimetallic-thermostatic switch by adjusting power levels to the heater. (2) Will the heater power requirement increase or decrease if the insulation pad is other than perfect?

PROBLEM 1.22 KNOWN: Hot vertical plate suspended in cool, still air. Change in plate temperature with time at the instant when the plate temperature is 225°C. FIND: Convection heat transfer coefficient for this condition. SCHEMATIC:

ASSUMPTIONS: (1) Plate is isothermal and of uniform temperature, (2) Negligible radiation exchange with surroundings, (3) Negligible heat lost through suspension wires. ANALYSIS: As shown in the cooling curve above, the plate temperature decreases with time. The condition of interest is for time to. For a control surface about the plate, the conservation of energy requirement is

E in - E out = E st dT − 2hA s ( Ts − T∞ ) = M c p dt where As is the surface area of one side of the plate. Solving for h, find

h=

h=

Mcp

dT 2As (Ts − T∞ ) dt 3.75 kg × 2770 J/kg ⋅ K 2 × ( 0.3 × 0.3) m 2 ( 225 − 25 ) K

× 0.022 K/s=6.4 W/m 2 ⋅ K

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COMMENTS: (1) Assuming the plate is very highly polished with emissivity of 0.08, determine whether radiation exchange with the surroundings at 25°C is negligible compared to convection. (2) We will later consider the criterion for determining whether the isothermal plate assumption is reasonable. If the thermal conductivity of the present plate were high (such as aluminum or copper), the criterion would be satisfied.

PROBLEM 1.23 KNOWN: Width, input power and efficiency of a transmission. Temperature and convection coefficient associated with air flow over the casing. FIND: Surface temperature of casing. SCHEMATIC:

ASSUMPTIONS: (1) Steady state, (2) Uniform convection coefficient and surface temperature, (3) Negligible radiation. ANALYSIS: From Newton’s law of cooling,

q = hAs ( Ts − T∞ ) = 6 hW 2 ( Ts − T∞ ) where the output power is η Pi and the heat rate is

q = Pi − Po = Pi (1 − η ) = 150 hp × 746 W / hp × 0.07 = 7833 W Hence,

Ts = T∞ +

q 6 hW 2

= 30°C +

7833 W 6 × 200 W / m 2 ⋅ K × (0.3m )

2

= 102.5°C

COMMENTS: There will, in fact, be considerable variability of the local convection coefficient over the transmission case and the prescribed value represents an average over the surface.

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PROBLEM 1.24 KNOWN: Air and wall temperatures of a room. Surface temperature, convection coefficient and emissivity of a person in the room. FIND: Basis for difference in comfort level between summer and winter. SCHEMATIC:

ASSUMPTIONS: (1) Person may be approximated as a small object in a large enclosure. ANALYSIS: Thermal comfort is linked to heat loss from the human body, and a chilled feeling is associated with excessive heat loss. Because the temperature of the room air is fixed, the different summer and winter comfort levels can not be attributed to convection heat transfer from the body. In both cases, the heat flux is Summer and Winter: q′′conv = h ( Ts − T∞ ) = 2 W/m 2 ⋅ K × 12 $C = 24 W/m 2 However, the heat flux due to radiation will differ, with values of Summer:

)

(

)

(

−8 4 4 2 4 4 4 4 2 q ′′rad = εσ Ts − Tsur = 0.9 × 5.67 × 10 W/m ⋅ K 305 − 300 K = 28.3 W/m

(

)

)

(

−8 4 2 4 4 4 4 2 Winter: q ′′rad = εσ Ts4 − Tsur = 0.9 × 5.67 × 10 W/m ⋅ K 305 − 287 K = 95.4 W/m

There is a significant difference between winter and summer radiation fluxes, and the chilled condition is attributable to the effect of the colder walls on radiation. 2

COMMENTS: For a representative surface area of A = 1.5 m , the heat losses are qconv = 36 W, qrad(summer) = 42.5 W and qrad(winter) = 143.1 W. The winter time radiation loss is significant and if maintained over a 24 h period would amount to 2,950 kcal.

PROBLEM 1.25 KNOWN: Diameter and emissivity of spherical interplanetary probe. Power dissipation within probe. FIND: Probe surface temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible radiation incident on the probe. ANALYSIS: Conservation of energy dictates a balance between energy generation within the probe and radiation emission from the probe surface. Hence, at any instant -E out + E g = 0

εA sσTs4 = E g  E g Ts =   επ D2σ 

1/ 4

   

1/ 4

  150W  Ts =   0.8π 0.5 m 2 5.67 × 10−8 W/m2 ⋅ K 4  ( )   Ts = 254.7 K.

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COMMENTS: Incident radiation, as, for example, from the sun, would increase the surface temperature.

PROBLEM 1.26 KNOWN: Spherical shaped instrumentation package with prescribed surface emissivity within a large space-simulation chamber having walls at 77 K. FIND: Acceptable power dissipation for operating the package surface temperature in the range Ts = 40 to 85°C. Show graphically the effect of emissivity variations for 0.2 and 0.3. SCHEMATIC:

ASSUMPTIONS: (1) Uniform surface temperature, (2) Chamber walls are large compared to the spherical package, and (3) Steady-state conditions. ANALYSIS: From an overall energy balance on the package, the internal power dissipation Pe will be transferred by radiation exchange between the package and the chamber walls. From Eq. 1.7, 4 q rad = Pe = ε Asσ Ts4 − Tsur

)

(

For the condition when Ts = 40°C, with As = πD2 the power dissipation will be 4 Pe = 0.25 (π × 0.10 m ) × 5.67 ×10−8 W m 2 ⋅ K 4 × ( 40 + 273) − 77 4  K 4 = 4.3 W  





<

Repeating this calculation for the range 40 ≤ Ts ≤ 85°C, we can obtain the power dissipation as a function of surface temperature for the ε = 0.25 condition. Similarly, with 0.2 or 0.3, the family of curves shown below has been obtained. Power dissipation, Pe (W)

10

8

6

4

2 40

50

60

70

80

90

Surface temperature, Ts (C) Surface emissivity, eps = 0.3 eps = 0.25 eps = 0.2

COMMENTS: (1) As expected, the internal power dissipation increases with increasing emissivity and surface temperature. Because the radiation rate equation is non-linear with respect to temperature, the power dissipation will likewise not be linear with surface temperature. (2) What is the maximum power dissipation that is possible if the surface temperature is not to exceed 85°C? What kind of a coating should be applied to the instrument package in order to approach this limiting condition?

PROBLEM 1.27 KNOWN: Area, emissivity and temperature of a surface placed in a large, evacuated chamber of prescribed temperature. FIND: (a) Rate of surface radiation emission, (b) Net rate of radiation exchange between surface and chamber walls. SCHEMATIC:

ASSUMPTIONS: (1) Area of the enclosed surface is much less than that of chamber walls. ANALYSIS: (a) From Eq. 1.5, the rate at which radiation is emitted by the surface is q emit = E ⋅ A = ε A σ Ts4

)

(

qemit = 0.8 0.5 m 2 5.67 × 10-8 W/m 2 ⋅ K 4 (150 + 273) K 

4

<

q emit = 726 W.

(b) From Eq. 1.7, the net rate at which radiation is transferred from the surface to the chamber walls is

(

4 q = ε A σ Ts4 − Tsur

(

)

)

4 4 q = 0.8 0.5 m2 5.67 × 10-8 W/m 2 ⋅ K 4 ( 423K ) - ( 298K )    q = 547 W.

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COMMENTS: The foregoing result gives the net heat loss from the surface which occurs at the instant the surface is placed in the chamber. The surface would, of course, cool due to this heat loss and its temperature, as well as the heat loss, would decrease with increasing time. Steady-state conditions would eventually be achieved when the temperature of the surface reached that of the surroundings.

PROBLEM 1.28 KNOWN: Length, diameter, surface temperature and emissivity of steam line. Temperature and convection coefficient associated with ambient air. Efficiency and fuel cost for gas fired furnace. FIND: (a) Rate of heat loss, (b) Annual cost of heat loss. SCHEMATIC:

ASSUMPTIONS: (1) Steam line operates continuously throughout year, (2) Net radiation transfer is between small surface (steam line) and large enclosure (plant walls). ANALYSIS: (a) From Eqs. (1.3a) and (1.7), the heat loss is

(

)

4  q = qconv + q rad = A  h ( Ts − T∞ ) + εσ Ts4 − Tsur   where A = π DL = π ( 0.1m × 25m ) = 7.85m 2 . Hence,

(

)

q = 7.85m2 10 W/m2 ⋅ K (150 − 25) K + 0.8 × 5.67 × 10−8 W/m2 ⋅ K 4 4234 − 2984 K 4    q = 7.85m2 (1, 250 + 1, 095) w/m 2 = (9813 + 8592 ) W = 18, 405 W

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(b) The annual energy loss is E = qt = 18, 405 W × 3600 s/h × 24h/d × 365 d/y = 5.80 ×1011 J With a furnace energy consumption of Ef = E/ηf = 6.45 ×1011 J, the annual cost of the loss is C = Cg Ef = 0.01 $/MJ × 6.45 ×105MJ = $6450

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COMMENTS: The heat loss and related costs are unacceptable and should be reduced by insulating the steam line.

PROBLEM 1.29 KNOWN: Exact and approximate expressions for the linearized radiation coefficient, hr and hra, respectively. FIND: (a) Comparison of the coefficients with ε = 0.05 and 0.9 and surface temperatures which may exceed that of the surroundings (Tsur = 25°C) by 10 to 100°C; also comparison with a free convection coefficient correlation, (b) Plot of the relative error (hr - rra)/hr as a function of the furnace temperature associated with a workpiece at Ts = 25°C having ε = 0.05, 0.2 or 0.9. ASSUMPTIONS: (1) Furnace walls are large compared to the workpiece and (2) Steady-state conditions. ANALYSIS: (a) The linearized radiation coefficient, Eq. 1.9, follows from the radiation exchange rate equation, 2 h r = εσ (Ts + Tsur ) Ts2 + Tsur

)

(

If Ts ≈ Tsur, the coefficient may be approximated by the simpler expression h r,a = 4εσ T3 T = ( Ts + Tsur ) 2 For the condition of ε = 0.05, Ts = Tsur + 10 = 35°C = 308 K and Tsur = 25°C = 298 K, find that h r = 0.05 × 5.67 × 10−8 W m 2 ⋅ K 4 (308 + 298 ) 3082 + 2982 K 3 = 0.32 W m 2 ⋅ K

)

(

h r,a = 4 × 0.05 × 5.67 ×10−8 W m 2 ⋅ K 4 ((308 + 298 ) 2 ) K3 = 0.32 W m 2 ⋅ K 3

< <

The free convection coefficient with Ts = 35°C and T∞ = Tsur = 25°C, find that

h = 0.98∆T1/ 3 = 0.98 (Ts − T∞ )

1/ 3

= 0.98 (308 − 298 )

1/ 3

<

= 2.1W m 2 ⋅ K

For the range Ts - Tsur = 10 to 100°C with ε = 0.05 and 0.9, the results for the coefficients are tabulated below. For this range of surface and surroundings temperatures, the radiation and free convection coefficients are of comparable magnitude for moderate values of the emissivity, say ε > 0.2. The approximate expression for the linearized radiation coefficient is valid within 2% for these conditions. (b) The above expressions for the radiation coefficients, hr and hr,a, are used for the workpiece at Ts = 25°C placed inside a furnace with walls which may vary from 100 to 1000°C. The relative error, (hr hra)/hr, will be independent of the surface emissivity and is plotted as a function of Tsur. For Tsur > 150°C, the approximate expression provides estimates which are in error more than 5%. The approximate expression should be used with caution, and only for surface and surrounding temperature differences of 50 to 100°C.

Ts (°C) 35 135

ε 0.05 0.9 0.05 0.9

Coefficients (W/m ⋅K) hr,a h hr 0.32 0.32 2.1 5.7 5.7 0.51 0.50 4.7 9.2 9.0

Relative error, (hr-hra)/hr*100 (%)

30

2

20

10

0 100

300

500

700

Surroundings temperature, Tsur (C)

900

PROBLEM 1.30 KNOWN: Chip width, temperature, and heat loss by convection in air. Chip emissivity and temperature of large surroundings. FIND: Increase in chip power due to radiation. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between small surface and large enclosure. ANALYSIS: Heat transfer from the chip due to net radiation exchange with the surroundings is

(

4 q rad = ε W 2σ T 4 - Tsur

) (

)

q rad = 0.9 ( 0.005 m ) 5.67 × 10−8 W/m 2 ⋅ K 4 3584 - 2884 K 4

2

q rad = 0.0122 W. The percent increase in chip power is therefore q ∆P 0.0122 W × 100 = rad × 100 = × 100 = 35%. . P q conv 0.350 W COMMENTS: For the prescribed conditions, radiation effects are small. Relative to convection, the effect of radiation would increase with increasing chip temperature and decreasing convection coefficient.

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PROBLEM 1.31 KNOWN: Width, surface emissivity and maximum allowable temperature of an electronic chip. Temperature of air and surroundings. Convection coefficient. 2 1/4 FIND: (a) Maximum power dissipation for free convection with h(W/m ⋅K) = 4.2(T - T∞) , (b) 2 Maximum power dissipation for forced convection with h = 250 W/m ⋅K.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between a small surface and a large enclosure, (3) Negligible heat transfer from sides of chip or from back of chip by conduction through the substrate. ANALYSIS: Subject to the foregoing assumptions, electric power dissipation by the chip must be balanced by convection and radiation heat transfer from the chip. Hence, from Eq. (1.10),

(

4 Pelec = q conv + q rad = hA (Ts − T∞ ) + ε Aσ Ts4 − Tsur

)

2 where A = L2 = (0.015m ) = 2.25 × 10−4 m 2 . (a) If heat transfer is by natural convection,

(

)

qconv = C A ( Ts − T∞ )5 / 4 = 4.2 W/m 2 ⋅ K5/4 2.25 × 10−4 m2 ( 60K )5 / 4 = 0.158 W

(

)

(

)

q rad = 0.60 2.25 × 10−4 m2 5.67 × 10−8 W/m2 ⋅ K 4 3584 − 2984 K 4 = 0.065 W

<

Pelec = 0.158 W + 0.065 W = 0.223 W (b) If heat transfer is by forced convection,

(

)

qconv = hA ( Ts − T∞ ) = 250 W/m2 ⋅ K 2.25 × 10−4 m2 ( 60K ) = 3.375 W Pelec = 3.375 W + 0.065 W = 3.44 W

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COMMENTS: Clearly, radiation and natural convection are inefficient mechanisms for transferring 2 heat from the chip. For Ts = 85°C and T∞ = 25°C, the natural convection coefficient is 11.7 W/m ⋅K. 2 Even for forced convection with h = 250 W/m ⋅K, the power dissipation is well below that associated with many of today’s processors. To provide acceptable cooling, it is often necessary to attach the chip to a highly conducting substrate and to thereby provide an additional heat transfer mechanism due to conduction from the back surface.

PROBLEM 1.32 KNOWN: Vacuum enclosure maintained at 77 K by liquid nitrogen shroud while baseplate is maintained at 300 K by an electrical heater. FIND: (a) Electrical power required to maintain baseplate, (b) Liquid nitrogen consumption rate, (c) Effect on consumption rate if aluminum foil (εp = 0.09) is bonded to baseplate surface. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) No heat losses from backside of heater or sides of plate, (3) Vacuum enclosure large compared to baseplate, (4) Enclosure is evacuated with negligible convection, (5) Liquid nitrogen (LN2) is heated only by heat transfer to the shroud, and (6) Foil is intimately bonded to baseplate. PROPERTIES: Heat of vaporization of liquid nitrogen (given): 125 kJ/kg. ANALYSIS: (a) From an energy balance on the baseplate,

E in - E out = 0

q elec - q rad = 0

and using Eq. 1.7 for radiative exchange between the baseplate and shroud,

(

)

qelec = ε p A pσ Tp4 - T 4 .

sh

(

)

Substituting numerical values, with A p = π D 2 p / 4 , find

(

)

2 qelec = 0.25  π ( 0.3 m ) / 4  5.67 × 10−8 W/m2 ⋅ K 4 3004 - 774 K 4 = 8.1 W.  

<

(b) From an energy balance on the enclosure, radiative transfer heats the liquid nitrogen stream causing evaporation,

E in - E out = 0

 LN2 h fg = 0 q rad - m

 LN2 is the liquid nitrogen consumption rate. Hence, where m  LN2 = q rad / h fg = 8.1 W / 125 kJ / kg = 6.48 × 10-5 kg / s = 0.23 kg / h. m

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(c) If aluminum foil (εp = 0.09) were bonded to the upper surface of the baseplate,

(

)

q rad,foil = q rad ε f / ε p = 8.1 W ( 0.09/0.25 ) = 2.9 W

and the liquid nitrogen consumption rate would be reduced by (0.25 - 0.09)/0.25 = 64% to 0.083 kg/h.

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PROBLEM 1.33 KNOWN: Width, input power and efficiency of a transmission. Temperature and convection coefficient for air flow over the casing. Emissivity of casing and temperature of surroundings. FIND: Surface temperature of casing. SCHEMATIC:

ASSUMPTIONS: (1) Steady state, (2) Uniform convection coefficient and surface temperature, (3) Radiation exchange with large surroundings. ANALYSIS: Heat transfer from the case must balance heat dissipation in the transmission, which may be expressed as q = Pi – Po = Pi (1 - η) = 150 hp × 746 W/hp × 0.07 = 7833 W. Heat transfer from the case is by convection and radiation, in which case

(

)

4  q = As  h ( Ts − T∞ ) + εσ Ts4 − Tsur   2

where As = 6 W . Hence,

(

)

2 7833 W = 6 ( 0.30 m )  200 W / m 2 ⋅ K ( Ts − 303K ) + 0.8 × 5.67 × 10−8 W / m 2 ⋅ K 4 Ts4 − 3034 K 4 





A trial-and-error solution yields

Ts ≈ 373K = 100°C

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COMMENTS: (1) For Ts ≈ 373 K, qconv ≈ 7,560 W and qrad ≈ 270 W, in which case heat transfer is dominated by convection, (2) If radiation is neglected, the corresponding surface temperature is Ts = 102.5°C.

PROBLEM 1.34 KNOWN: Resistor connected to a battery operating at a prescribed temperature in air.

 (W ) , FIND: (a) Considering the resistor as the system, determine corresponding values for E in

E g ( W ) , E out ( W ) and E st ( W ) . If a control surface is placed about the entire system, determine  , E , E  . (b) Determine the volumetric heat generation rate within , and E the values for E

g in out st 3 the resistor, q (W/m ), (c) Neglecting radiation from the resistor, determine the convection coefficient. SCHEMATIC:

ASSUMPTIONS: (1) Electrical power is dissipated uniformly within the resistor, (2) Temperature of the resistor is uniform, (3) Negligible electrical power dissipated in the lead wires, (4) Negligible radiation exchange between the resistor and the surroundings, (5) No heat transfer occurs from the battery, (5) Steady-state conditions. ANALYSIS: (a) Referring to Section 1.3.1, the conservation of energy requirement for a control volume at an instant of time, Eq 1.11a, is

E in + E g − E out = E st  , E where E in out correspond to surface inflow and outflow processes, respectively. The energy  is associated with conversion of some other energy form (chemical, electrical, generation term E g  is associated with electromagnetic or nuclear) to thermal energy. The energy storage term E st  , changes in the internal, kinetic and/or potential energies of the matter in the control volume. E g E st are volumetric phenomena. The electrical power delivered by the battery is P = VI = 24V×6A = 144 W. Control volume: Resistor.

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E in = 0

E out = 144 W

E g = 144 W

E st = 0

 term is due to conversion of electrical energy to thermal energy. The term E The E g out is due to convection from the resistor surface to the air. Continued...

PROBLEM 1.34 (Cont.)

Control volume: Battery-Resistor System.

E in = 0 E = 0 g

E out = 144 W E = −144 W

<

st

 term represents the decrease in the chemical energy within the battery. The conversion of The E st chemical energy to electrical energy and its subsequent conversion to thermal energy are processes  or E . The E internal to the system which are not associated with E g st out term is due to convection from the resistor surface to the air. (b) From the energy balance on the resistor with volume, ∀ = (πD2/4)L,

E g = q ∀

(

)

144 W = q π (0.06 m ) / 4 × 0.25 m 2

q = 2.04 ×105 W m3

<

(c) From the energy balance on the resistor and Newton's law of cooling with As = πDL + 2(πD2/4),

E out = qcv = hAs ( Ts − T∞ )

)

(

$ 144 W = h π × 0.06 m × 0.25 m + 2 π × 0.062 m 2 4  (95 − 25 ) C  

144 W = h [0.0471 + 0.0057 ] m 2 (95 − 25 ) C $

h = 39.0 W m 2⋅K COMMENTS: (1) In using the conservation of energy requirement, Eq. 1.11a, it is important to  and E   recognize that E out will always represent surface processes and E g and Est , volumetric in

 is associated with a conversion process from some form of processes. The generation term E g  represents the rate of change of internal energy. energy to thermal energy. The storage term E st (2) From Table 1.1 and the magnitude of the convection coefficient determined from part (c), we conclude that the resistor is experiencing forced, rather than free, convection.

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PROBLEM 1.35 KNOWN: Thickness and initial temperature of an aluminum plate whose thermal environment is changed. FIND: (a) Initial rate of temperature change, (b) Steady-state temperature of plate, (c) Effect of emissivity and absorptivity on steady-state temperature. SCHEMATIC:

ASSUMPTIONS: (1) Negligible end effects, (2) Uniform plate temperature at any instant, (3) Constant properties, (4) Adiabatic bottom surface, (5) Negligible radiation from surroundings, (6) No internal heat generation. ANALYSIS: (a) Applying an energy balance, Eq. 1.11a, at an instant of time to a control volume about the plate, E in − E out = E st , it follows for a unit surface area.

( ) ( )

( )

(

)

αSGS 1m 2 − E 1m 2 − q′′conv 1m 2 = ( d dt )( McT ) = ρ 1m 2 × L c ( dT dt ) . Rearranging and substituting from Eqs. 1.3 and 1.5, we obtain dT dt = (1 ρ Lc ) αSGS − εσ Ti4 − h ( Ti − T∞ ) .



(



dT dt = 2700 kg m3 × 0.004 m × 900 J kg ⋅ K

)

−1

×

0.8 × 900 W m 2 − 0.25 × 5.67 × 10−8 W m 2 ⋅ K 4 ( 298 K )4 − 20 W m 2 ⋅ K ( 25 − 20 )$ C    dT dt = 0.052$ C s . (b) Under steady-state conditions, E st = 0, and the energy balance reduces to

αSGS = εσ T 4 + h ( T − T∞ )

< (2)

0.8 × 900 W m 2 = 0.25 × 5.67 × 10−8 W m 2 ⋅ K 4 × T 4 + 20 W m 2 ⋅ K ( T − 293 K ) The solution yields T = 321.4 K = 48.4°C. (c) Using the IHT First Law Model for an Isothermal Plane Wall, parametric calculations yield the following results. Plate temperature, T (C)

70

60

50

40

30

20 0

0.2

0.4

0.6

0.8

1

Coating emissivity, eps Solar absorptivity, alphaS = 1 alphaS = 0.8 alphaS = 0.5

COMMENTS: The surface radiative properties have a significant effect on the plate temperature, which decreases with increasing ε and decreasing αS. If a low temperature is desired, the plate coating should be characterized by a large value of ε/αS. The temperature also decreases with increasing h.

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PROBLEM 1.36 KNOWN: Surface area of electronic package and power dissipation by the electronics. Surface emissivity and absorptivity to solar radiation. Solar flux. FIND: Surface temperature without and with incident solar radiation. SCHEMATIC:

ASSUMPTIONS: Steady-state conditions. ANALYSIS: Applying conservation of energy to a control surface about the compartment, at any instant E in - E out + E g = 0. It follows that, with the solar input, ′′ − As E + P=0 αSAsqS ′′ − Asεσ Ts4 + P=0 αSAsqS

1/ 4

 α A q′′ + P  Ts =  S s S  Asεσ  

.

′′ = 0 ) , In the shade ( qS

1/ 4

  1000 W Ts =    1 m 2 × 1× 5.67 × 10−8 W/m 2 ⋅ K 4 

= 364 K.

<

In the sun,

1/ 4

 0.25 × 1 m 2 × 750 W/m 2 + 1000 W   Ts =   1 m 2 ×1× 5.67 ×10−8 W/m 2 ⋅ K 4   

= 380 K.

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COMMENTS: In orbit, the space station would be continuously cycling between shade and sunshine, and a steady-state condition would not exist.

PROBLEM 1.37 KNOWN: Daily hot water consumption for a family of four and temperatures associated with ground water and water storage tank. Unit cost of electric power. Heat pump COP. FIND: Annual heating requirement and costs associated with using electric resistance heating or a heat pump. SCHEMATIC:

ASSUMPTIONS: (1) Process may be modelled as one involving heat addition in a closed system, (2) Properties of water are constant. PROPERTIES: Table A-6, Water ( Tave = 308 K): ρ = vf−1 = 993 kg/m3, cp,f = 4.178 kJ/kg⋅K. ANALYSIS: From Eq. 1.11c, the daily heating requirement is Qdaily = ∆U t = Mc∆T

= ρ Vc (Tf − Ti ) . With V = 100 gal/264.17 gal/m3 = 0.379 m3,

(

)

(

)

Qdaily = 993kg / m3 0.379 m3 4.178kJ/kg ⋅ K 40$ C = 62,900 kJ The annual heating requirement is then, Qannual = 365days ( 62,900 kJ/day ) = 2.30 × 107 kJ , or, with 1 kWh = 1 kJ/s (3600 s) = 3600 kJ,

Qannual = 6380 kWh

<

With electric resistance heating, Qannual = Qelec and the associated cost, C, is

C = 6380 kWh ($0.08/kWh ) = $510

<

If a heat pump is used, Qannual = COP ( Welec ). Hence,

Welec = Qannual /( COP ) = 6380kWh/(3) = 2130 kWh The corresponding cost is

C = 2130 kWh ($0.08/kWh ) = $170

<

COMMENTS: Although annual operating costs are significantly lower for a heat pump, corresponding capital costs are much higher. The feasibility of this approach depends on other factors such as geography and seasonal variations in COP, as well as the time value of money.

PROBLEM 1.38 KNOWN: Initial temperature of water and tank volume. Power dissipation, emissivity, length and diameter of submerged heaters. Expressions for convection coefficient associated with natural convection in water and air. FIND: (a) Time to raise temperature of water to prescribed value, (b) Heater temperature shortly after activation and at conclusion of process, (c) Heater temperature if activated in air. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss from tank to surroundings, (2) Water is wellmixed (at a uniform, but time varying temperature) during heating, (3) Negligible changes in thermal energy storage for heaters, (4) Constant properties, (5) Surroundings afforded by tank wall are large relative to heaters. ANALYSIS: (a) Application of conservation of energy to a closed system (the water) at an instant, Eq. (1.11d), yields dU dT dT = Mc = ρ∀c = q = 3q1 dt dt dt Tf t dt ρ c/3q = ∀ 1 ) ∫T dT ∫0 ( i

Hence,

t=

(

)

990 kg/m3 ×10gal 3.79 ×10−3m3 / gal 4180J/kg ⋅ K 3 × 500 W

(335 − 295) K = 4180 s <

(b) From Eq. (1.3a), the heat rate by convection from each heater is 4/3 q1 = Aq1′′ = Ah ( Ts − T ) = (π DL ) 370 ( Ts − T ) Hence, 3/ 4 3/ 4   500 W  q1  Ts = T +  = T+ = ( T + 24 ) K    370π DL   370 W/m 2 ⋅ K 4/3 × π × 0.025 m × 0.250 m  With water temperatures of Ti ≈ 295 K and Tf = 335 K shortly after the start of heating and at the end of heating, respectively, Ts,i = 319 K

<

Ts,f = 359 K Continued …..

PROBLEM 1.38 (Continued) (c) From Eq. (1.10), the heat rate in air is

(

)

4  q1 = π DL  0.70 ( Ts − T∞ )4 / 3 + εσ Ts4 − Tsur   Substituting the prescribed values of q1, D, L, T∞ = Tsur and ε, an iterative solution yields Ts = 830 K

<

COMMENTS: In part (c) it is presumed that the heater can be operated at Ts = 830 K without experiencing burnout. The much larger value of Ts for air is due to the smaller convection coefficient. However, with qconv and qrad equal to 59 W and 441 W, respectively, a significant portion of the heat dissipation is effected by radiation.

PROBLEM 1.39 KNOWN: Power consumption, diameter, and inlet and discharge temperatures of a hair dryer. FIND: (a) Volumetric flow rate and discharge velocity of heated air, (b) Heat loss from case. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Constant air properties, (3) Negligible potential and kinetic energy changes of air flow, (4) Negligible work done by fan, (5) Negligible heat transfer from casing of dryer to ambient air (Part (a)), (6) Radiation exchange between a small surface and a large enclosure (Part (b)). ANALYSIS: (a) For a control surface about the air flow passage through the dryer, conservation of energy for an open system reduces to  ( u + pv ) − m  ( u + pv ) + q = 0 m i o  (ii − io ) = mc  p (Ti − To ) , where u + pv = i and q = Pelec. Hence, with m  p (To − Ti ) = Pelec mc  = m

Pelec 500 W = = 0.0199 kg/s cp ( To − Ti ) 1007 J/kg ⋅ K 25$C

( )

 0.0199 kg/s =m ∀ = = 0.0181 m3 / s ρ 1.10 kg/m3 Vo =

<

  4∀ 4 × 0.0181 m3 / s ∀ = = = 4.7 m/s 2 Ac π D2 π (0.07 m )

<

(b) Heat transfer from the casing is by convection and radiation, and from Eq. (1.10)

(

4 q = hAs ( Ts − T∞ ) + ε Asσ Ts4 − Tsur

) Continued …..

PROBLEM 1.39 (Continued) where As = π DL = π (0.07 m × 0.15 m ) = 0.033 m 2 . Hence,

(

)( )

(

)

q = 4W/m2 ⋅ K 0.033 m 2 20$ C + 0.8 × 0.033 m2 × 5.67 × 10−8 W/m2 ⋅ K 4 3134 − 2934 K 4 q = 2.64 W + 3.33 W = 5.97 W

<

The heat loss is much less than the electrical power, and the assumption of negligible heat loss is justified. COMMENTS: Although the mass flow rate is invariant, the volumetric flow rate increases as the air is heated in its passage through the dryer, causing a reduction in the density.  , is However, for the prescribed temperature rise, the change in ρ, and hence the effect on ∀ small.

PROBLEM 1.40 KNOWN: Speed, width, thickness and initial and final temperatures of 304 stainless steel in an annealing process. Dimensions of annealing oven and temperature, emissivity and convection coefficient of surfaces exposed to ambient air and large surroundings of equivalent temperatures. Thickness of pad on which oven rests and pad surface temperatures. FIND: Oven operating power. SCHEMATIC:

ASSUMPTIONS: (1) steady-state, (2) Constant properties, (3) Negligible changes in kinetic and potential energy.

(

)

PROPERTIES: Table A.1, St.St.304 T = (Ti + To )/2 = 775 K : ρ = 7900 kg/m3, c p = 578 J/kg⋅K; Table A.3, Concrete, T = 300 K: k c = 1.4 W/m⋅K. ANALYSIS: The rate of energy addition to the oven must balance the rate of energy transfer to the

 − E steel sheet and the rate of heat loss from the oven. With E in out

= 0, it follows that

 (ui − uo ) − q = 0 Pelec + m  = ρ Vs ( Ws t s ) , ( u i − u o ) = cp ( Ti − To ) , and where heat is transferred from the oven. With m

(

)

4  +k W L q = ( 2Ho Lo + 2Ho Wo + Wo Lo ) ×  h ( Ts − T∞ ) + ε sσ Ts4 − Tsur c ( o o )(Ts − Tb )/t c ,   it follows that

Pelec = ρ Vs ( Ws t s ) cp ( To − Ti ) + ( 2Ho Lo + 2H o Wo + Wo Lo ) ×

)

(

 h ( T − T ) + ε σ T 4 − T 4  + k ( W L )(T − T )/t s o s s sur  b c   c o o s Pelec = 7900 kg/m3 × 0.01m/s ( 2 m × 0.008 m ) 578J/kg ⋅ K (1250 − 300 ) K + ( 2 × 2m × 25m + 2 × 2m × 2.4m + 2.4m × 25m )[10W/m 2 ⋅ K (350 − 300 ) K

(

)

+0.8 × 5.67 × 10−8 W/m 2 ⋅ K 4 3504 − 3004 K 4 ] + 1.4W/m ⋅ K ( 2.4m × 25m )(350 − 300 ) K/0.5m Continued.….

PROBLEM 1.40 (Cont.) Pelec = 694, 000W + 169.6m 2 (500 + 313 ) W/m2 + 8400W = ( 694, 000 + 84,800 + 53,100 + 8400 ) W = 840kW COMMENTS: Of the total energy input, 83% is transferred to the steel while approximately 10%, 6% and 1% are lost by convection, radiation and conduction from the oven. The convection and radiation losses can both be reduced by adding insulation to the side and top surfaces, which would reduce the corresponding value of Ts .

<

PROBLEM 1.41 KNOWN: Hot plate-type wafer thermal processing tool based upon heat transfer modes by conduction through gas within the gap and by radiation exchange across gap. FIND: (a) Radiative and conduction heat fluxes across gap for specified hot plate and wafer temperatures and gap separation; initial time rate of change in wafer temperature for each mode, and (b) heat fluxes and initial temperature-time change for gap separations of 0.2, 0.5 and 1.0 mm for hot plate temperatures 300 < Th < 1300°C. Comment on the relative importance of the modes and the influence of the gap distance. Under what conditions could a wafer be heated to 900°C in less than 10 seconds? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions for flux calculations, (2) Diameter of hot plate and wafer much larger than gap spacing, approximating plane, infinite planes, (3) One-dimensional conduction through gas, (4) Hot plate and wafer are blackbodies, (5) Negligible heat losses from wafer backside, and (6) Wafer temperature is uniform at the onset of heating. 3

PROPERTIES: Wafer: ρ = 2700 kg/m , c = 875 J/kg⋅K; Gas in gap: k = 0.0436 W/m⋅K. ANALYSIS: (a) The radiative heat flux between the hot plate and wafer for Th = 600°C and Tw = 20° C follows from the rate equation,

(

)

4 q′′rad = σ Th4 − Tw = 5.67 × 10−8 W / m 2 ⋅ K 4

((600 + 273)

4

− ( 20 + 273)

4

)K

4

= 32.5 kW / m2

<

The conduction heat flux through the gas in the gap with L = 0.2 mm follows from Fourier’s law,

(600 − 20) K = 126 kW / m2 T − Tw q′′cond = k h = 0.0436 W / m ⋅ K L 0.0002 m

<

The initial time rate of change of the wafer can be determined from an energy balance on the wafer at the instant of time the heating process begins,

′′ − E ′′out = E ′′st E in

 dT  E ′′st = ρ c d  w   dt i

where E ′′out = 0 and E ′′in = q ′′rad or q ′′cond . Substituting numerical values, find

dTw dt

q′′rad 32.5 × 103 W / m 2  = = = 17.6 K / s  i,rad ρ cd 2700 kg / m3 × 875 J / kg ⋅ K × 0.00078 m

<

dTw dt

q′′  = cond = 68.4 K / s  ρ cd i,cond

< Continued …..

PROBLEM 1.41 (Cont.) (b) Using the foregoing equations, the heat fluxes and initial rate of temperature change for each mode can be calculated for selected gap separations L and range of hot plate temperatures Th with Tw = 20°C.

200 Initial rate of change, dTw/dt (K.s^-1)

400

Heat flux (kW/m^2)

300

200

100

150

100

50

0

0 300

500

700

900

1100

1300

300

500

700

900

1100

Hot plate temperature, Th (C)

Hot plate temperature, Th (C) q''rad q''cond, L = 1.0 mm q''cond, L = 0.5 mm q''cond, L = 0.2 mm

q''rad q''cond, L = 1.0 m m q''cond, L = 0.5 m m q''cond, L = 0.2 m m

In the left-hand graph, the conduction heat flux increases linearly with Th and inversely with L as expected. The radiative heat flux is independent of L and highly non-linear with Th, but does not approach that for the highest conduction heat rate until Th approaches 1200°C. The general trends for the initial temperature-time change, (dTw/dt)i, follow those for the heat fluxes. To reach 900°C in 10 s requires an average temperature-time change rate of 90 K/s. Recognizing that (dTw/dt) will decrease with increasing Tw, this rate could be met only with a very high Th and the smallest L.

1300

PROBLEM 1.42 KNOWN: Silicon wafer, radiantly heated by lamps, experiencing an annealing process with known backside temperature. FIND: Whether temperature difference across the wafer thickness is less than 2°C in order to avoid damaging the wafer. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wafer, (3) Radiation exchange between upper surface of wafer and surroundings is between a small object and a large enclosure, and (4) Vacuum condition in chamber, no convection. PROPERTIES: Wafer: k = 30 W/m⋅K, ε = α " = 0.65. ANALYSIS: Perform a surface energy balance on the upper surface of the wafer to determine Tw,u . The processes include the absorbed radiant flux from the lamps, radiation exchange with the chamber walls, and conduction through the wafer.

E ′′in − E ′′out = 0

α "q′′s − q′′rad − q′′cd = 0

(

)

4 − T4 − k α "q′′s − εσ Tw,u sur

Tw,u − Tw," L

=0

(

4 − 27 + 273 0.65 × 3.0 × 105 W / m 2 − 0.65 × 5.67 × 10−8 W / m 2 ⋅ K 4 Tw,u ( )

4

)K

4

−30W / m ⋅ K Tw,u − (997 + 273) K / 0.00078 m = 0 Tw,u = 1273K = 1000°C

<

COMMENTS: (1) The temperature difference for this steady-state operating condition, Tw,u − Tw,l , is larger than 2°C. Warping of the wafer and inducing slip planes in the crystal structure could occur. (2) The radiation exchange rate equation requires that temperature must be expressed in kelvin units. Why is it permissible to use kelvin or Celsius temperature units in the conduction rate equation? (3) Note how the surface energy balance, Eq. 1.12, is represented schematically. It is essential to show the control surfaces, and then identify the rate processes associated with the surfaces. Make sure the directions (in or out) of the process are consistent with the energy balance equation.

PROBLEM 1.43 KNOWN: Silicon wafer positioned in furnace with top and bottom surfaces exposed to hot and cool zones, respectively. FIND: (a) Initial rate of change of the wafer temperature corresponding to the wafer temperature Tw,i = 300 K, and (b) Steady-state temperature reached if the wafer remains in this position. How significant is convection for this situation? Sketch how you’d expect the wafer temperature to vary as a function of vertical distance. SCHEMATIC:

ASSUMPTIONS: (1) Wafer temperature is uniform, (2) Transient conditions when wafer is initially positioned, (3) Hot and cool zones have uniform temperatures, (3) Radiation exchange is between small surface (wafer) and large enclosure (chamber, hot or cold zone), and (4) Negligible heat losses from wafer to mounting pin holder. ANALYSIS: The energy balance on the wafer illustrated in the schematic above includes convection from the upper (u) and lower (l) surfaces with the ambient gas, radiation exchange with the hot- and cool-zone (chamber) surroundings, and the rate of energy storage term for the transient condition.

E ′′in − E ′′out = E ′′st

q′′rad,h + q′′rad,c − q′′cv,u − q′′cv,l = ρ cd

(

) (

d Tw dt

)

4 4 + εσ T 4 4 εσ Tsur,h − Tw sur,c − Tw − h u ( Tw − T∞ ) − h l ( Tw − T∞ ) = ρ cd

d Tw dt

(a) For the initial condition, the time rate of temperature change of the wafer is determined using the energy balance above with Tw = Tw,i = 300 K,

(

)

(

)

0.65 × 5.67 × 10−8 W / m 2 ⋅ K 4 15004 − 3004 K 4 + 0.65 × 5.67 × 10−8 W / m 2 ⋅ K 4 3304 − 3004 K 4

−8 W / m 2 ⋅ K (300 − 700 ) K − 4 W / m 2 ⋅ K (300 − 700 ) K = 2700 kg / m3 × 875 J / kg ⋅ K ×0.00078 m ( d Tw / dt )i

(d Tw / dt )i = 104 K / s

<

(b) For the steady-state condition, the energy storage term is zero, and the energy balance can be solved for the steady-state wafer temperature, Tw = Tw,ss .

Continued …..

PROBLEM 1.43 (Cont.)

)

(

)

(

4 4 0.65 σ 15004 − Tw,ss K 4 + 0.65 σ 3304 − Tw,ss K4

(

)

(

)

−8 W / m 2 ⋅ K Tw,ss − 700 K − 4 W / m 2 ⋅ K Tw,ss − 700 K = 0 Tw,ss = 1251 K

<

To determine the relative importance of the convection processes, re-solve the energy balance above ignoring those processes to find ( d Tw / dt )i = 101 K / s and Tw,ss = 1262 K. We conclude that the radiation exchange processes control the initial time rate of temperature change and the steady-state temperature. If the wafer were elevated above the present operating position, its temperature would increase, since the lower surface would begin to experience radiant exchange with progressively more of the hot zone chamber. Conversely, by lowering the wafer, the upper surface would experience less radiant exchange with the hot zone chamber, and its temperature would decrease. The temperature-distance trend might appear as shown in the sketch.

PROBLEM 1.44 KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactive wastes. Surface convection conditions. FIND: Total energy generation rate and surface temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible temperature drop across thin container wall. ANALYSIS: The rate of energy generation is

r   o ∫ o 1- ( r/ro )2  2π rLdr E g = ∫ qdV=q  0  E g = 2π Lq o ro2 / 2 − ro2 / 4

(

)

or per unit length,

πq r E ′g = o o . 2 2

<

Performing an energy balance for a control surface about the container yields, at an instant, E g′ − E out ′ =0 and substituting for the convection heat rate per unit length,

π q o ro2 = h ( 2π ro )( Ts − T∞ ) 2 Ts = T∞ +

q o ro . 4h

<

COMMENTS: The temperature within the radioactive wastes increases with decreasing r from Ts at ro to a maximum value at the centerline.

PROBLEM 1.45 KNOWN: Rod of prescribed diameter experiencing electrical dissipation from passage of electrical current and convection under different air velocity conditions. See Example 1.3. FIND: Rod temperature as a function of the electrical current for 0 ≤ I ≤ 10 A with convection 2 coefficients of 50, 100 and 250 W/m ⋅K. Will variations in the surface emissivity have a significant effect on the rod temperature? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform rod temperature, (3) Radiation exchange between the outer surface of the rod and the surroundings is between a small surface and large enclosure. ANALYSIS: The energy balance on the rod for steady-state conditions has the form,

q′conv + q′rad = E ′gen

)

(

4 = I 2R ′ π Dh (T − T∞ ) + π Dεσ T 4 − Tsur e Using this equation in the Workspace of IHT, the rod temperature is calculated and plotted as a function of current for selected convection coefficients. 150

R o d te m p e ra tu re , T (C )

125

100

75

50

25

0 0

2

4

6

8

10

C u rre n t, I (a m p e re s ) h = 5 0 W /m ^2 .K h = 1 0 0 W /m ^2 .K h = 2 5 0 W /m ^2 .K

COMMENTS: (1) For forced convection over the cylinder, the convection heat transfer coefficient is 0.6 dependent upon air velocity approximately as h ~ V . Hence, to achieve a 5-fold change in the 2 convection coefficient (from 50 to 250 W/m ⋅K), the air velocity must be changed by a factor of nearly 15. Continued …..

PROBLEM 1.45 (Cont.) 2

(2) For the condition of I = 4 A with h = 50 W/m ⋅K with T = 63.5°C, the convection and radiation exchange rates per unit length are, respectively, q ′cv = 5.7 W / m and q ′rad = 0.67 W / m. We conclude that convection is the dominate heat transfer mode and that changes in surface emissivity could have 2 only a minor effect. Will this also be the case if h = 100 or 250 W/m ⋅K? (3) What would happen to the rod temperature if there was a “loss of coolant” condition where the air flow would cease? (4) The Workspace for the IHT program to calculate the heat losses and perform the parametric analysis to generate the graph is shown below. It is good practice to provide commentary with the code making your solution logic clear, and to summarize the results. It is also good practice to show plots in customary units, that is, the units used to prescribe the problem. As such the graph of the rod temperature is shown above with Celsius units, even though the calculations require temperatures in kelvins.

// Energy balance; from Ex. 1.3, Comment 1 -q'cv - q'rad + Edot'g = 0 q'cv = pi*D*h*(T - Tinf) q'rad = pi*D*eps*sigma*(T^4 - Tsur^4) sigma = 5.67e-8 // The generation term has the form Edot'g = I^2*R'e qdot = I^2*R'e / (pi*D^2/4) // Input parameters D = 0.001 Tsur = 300 T_C = T – 273 eps = 0.8 Tinf = 300 h = 100 //h = 50 //h = 250 I = 5.2 //I = 4 R'e = 0.4

// Representing temperature in Celsius units using _C subscript

// Values of coefficient for parameter study // For graph, sweep over range from 0 to 10 A // For evaluation of heat rates with h = 50 W/m^2.K

/* Base case results: I = 5.2 A with h = 100 W/m^2.K, find T = 60 C (Comment 2 case). Edot'g T T_C q'cv q'rad qdot D I R'e Tinf Tsur eps h sigma 10.82 332.6 59.55 10.23 0.5886 1.377E7 0.001 5.2 0.4 300 300 0.8 100 5.67E-8 */ /* Results: I = 4 A with h = 50 W/m^2.K, find q'cv = 5.7 W/m and q'rad = 0.67 W/m Edot'g T T_C q'cv q'rad qdot D I R'e Tinf Tsur eps h sigma 6.4 336.5 63.47 5.728 0.6721 8.149E6 0.001 4 0.4 300 300 0.8 50 5.67E-8 */

PROBLEM 1.46 KNOWN: Long bus bar of prescribed diameter and ambient air and surroundings temperatures. Relations for the electrical resistivity and free convection coefficient as a function of temperature. FIND: (a) Current carrying capacity of the bus bar if its surface temperature is not to exceed 65°C; compare relative importance of convection and radiation exchange heat rates, and (b) Show graphically the operating temperature of the bus bar as a function of current for the range 100 ≤ I ≤ 5000 A for bus-bar diameters of 10, 20 and 40 mm. Plot the ratio of the heat transfer by convection to the total heat transfer for these conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Bus bar and conduit are very long in direction normal to page, (3) Uniform bus-bar temperature, (4) Radiation exchange between the outer surface of the bus bar and the conduit is between a small surface and a large enclosure. PROPERTIES: Bus-bar material, ρe = ρe,o [1 + α ( T − To )] , ρ e,o = 0.0171µΩ ⋅ m, To = 25°C,

α = 0.00396 K −1. ANALYSIS: An energy balance on the bus-bar for a unit length as shown in the schematic above has the form

E ′in − E ′out + E ′gen = 0

−q′rad − q′conv + I2R ′e = 0

)

(

4 − hπ D T − T + I 2 ρ / A = 0 −επ Dσ T 4 − Tsur ( ∞) e c where R ′e = ρ e / A c and A c = π D 2 / 4. Using the relations for ρ e ( T ) and h ( T, D ) , and substituting numerical values with T = 65°C, find

(

q′rad = 0.85 π ( 0.020m ) × 5.67 × 10−8 W / m 2 ⋅ K 4 [65 + 273] − [30 + 273] 4

4

)K

4

= 223 W / m

q′conv = 7.83W / m 2 ⋅ K π ( 0.020m )( 65 − 30 ) K = 17.2 W / m

< <

−0.25 where h = 1.21W ⋅ m −1.75 ⋅ K −1.25 ( 0.020m ) (65 − 30 )0.25 = 7.83 W / m2 ⋅ K

(

)

I2 R ′e = I2 198.2 × 10−6 Ω ⋅ m / π (0.020 ) m 2 / 4 = 6.31× 10−5 I2 W / m where

2

ρe = 0.0171× 10−6 Ω ⋅ m 1 + 0.00396 K −1 (65 − 25) K  = 198.2 µΩ ⋅ m  

The maximum allowable current capacity and the ratio of the convection to total heat transfer rate are

I = 1950 A

q′cv / ( q′cv + q′rad ) = q′cv / q′tot = 0.072

<

For this operating condition, convection heat transfer is only 7.2% of the total heat transfer. (b) Using these equations in the Workspace of IHT, the bus-bar operating temperature is calculated and plotted as a function of the current for the range 100 ≤ I ≤ 5000 A for diameters of 10, 20 and 40 mm. Also shown below is the corresponding graph of the ratio (expressed in percentage units) of the heat transfer by convection to the total heat transfer, q ′cv / q ′tot . Continued …..

PROBLEM 1.46 (Cont.) 13 11

Ratio q'cv / q'tot, (%)

Ba r te m p e ra tu re , Ts (C )

100 80 60 40

9 7 5 3

20 0

1000

2000

3000

4000

1

5000

20

40

C u rre n t, I (A)

60

80

Bus bar temperature, T (C)

D = 10 m m D = 20 m m D = 40 m m

D = 10 mm D = 20 mm D = 40 mm

COMMENTS: (1) The trade-off between current-carrying capacity, operating temperature and bar diameter is shown in the first graph. If the surface temperature is not to exceed 65°C, the maximum current capacities for the 10, 20 and 40-mm diameter bus bars are 960, 1950, and 4000 A, respectively. (2) From the second graph with q ′cv / q ′tot vs. T, note that the convection heat transfer rate is always a small fraction of the total heat transfer. That is, radiation is the dominant mode of heat transfer. Note also that the convection contribution increases with increasing diameter. (3) The Workspace for the IHT program to perform the parametric analysis and generate the graphs is shown below. It is good practice to provide commentary with the code making your solution logic clear, and to summarize the results. /* Results: base-case conditions, Part (a) I R'e cvovertot hbar q'cv Tsur_C eps 1950 6.309E-5 7.171 7.826 17.21 30 0.85 */

q'rad

rhoe

D

222.8

1.982E-8 0.02

Tinf_C

Ts_C

30

65

// Energy balance, on a per unit length basis; steady-state conditions // Edot'in - Edot'out + Edot'gen = 0 -q'cv - q'rad + Edot'gen = 0 q'cv = hbar * P * (Ts - Tinf) P = pi * D q'rad = eps * sigma * (Ts^4 - Tsur^4) sigma = 5.67e-8 Edot'gen = I^2 * R'e R'e = rhoe / Ac rhoe = rhoeo * (1 + alpha * (Ts - To) ) To = 25 + 273 Ac = pi * D^2 / 4 // Convection coefficient hbar = 1.21 * (D^-0.25) * (Ts - Tinf)^0.25 // Convection vs. total heat rates cvovertot = q'cv / (q'cv + q'rad) * 100 // Input parameters D = 0.020 // D = 0.010 // D = 0.040 // I = 1950 rhoeo = 0.01711e-6 alpha = 0.00396 Tinf_C = 30 Tinf = Tinf_C + 273 Ts_C = 65 Ts = Ts_C + 273 Tsur_C = 30 Tsur = Tsur_C + 273 eps = 0.85

// Compact convection coeff. correlation

// Values of diameter for parameter study // Base case condition unknown

// Base case condition to determine current

100

PROBLEM 1.47 KNOWN: Elapsed times corresponding to a temperature change from 15 to 14°C for a reference sphere and test sphere of unknown composition suddenly immersed in a stirred water-ice mixture. Mass and specific heat of reference sphere. FIND: Specific heat of the test sphere of known mass. SCHEMATIC:

ASSUMPTIONS: (1) Spheres are of equal diameter, (2) Spheres experience temperature change from 15 to 14°C, (3) Spheres experience same convection heat transfer rate when the time rates of surface temperature are observed, (4) At any time, the temperatures of the spheres are uniform, (5) Negligible heat loss through the thermocouple wires. PROPERTIES: Reference-grade sphere material: cr = 447 J/kg K. ANALYSIS: Apply the conservation of energy requirement at an instant of time, Eq. 1.11a, after a sphere has been immersed in the ice-water mixture at T∞.

E in − E out = E st −q conv = Mc

dT dt

where q conv = h A s ( T − T∞ ). Since the temperatures of the spheres are uniform, the change in energy storage term can be represented with the time rate of temperature change, dT/dt. The convection heat rates are equal at this instant of time, and hence the change in energy storage terms for the reference (r) and test (t) spheres must be equal.

M r cr

dT  dT   = M t ct  dt r dt  t

Approximating the instantaneous differential change, dT/dt, by the difference change over a short period of time, ∆T/∆t, the specific heat of the test sphere can be calculated.

0.515 kg × 447 J / kg ⋅ K c t = 132 J / kg ⋅ K

(15 − 14 ) K 6.35s

= 1.263kg × c t ×

(15 − 14 ) K 4.59s

<

COMMENTS: Why was it important to perform the experiments with the reference and test spheres over the same temperature range (from 15 to 14°C)? Why does the analysis require that the spheres have uniform temperatures at all times?

PROBLEM 1.48 KNOWN: Inner surface heating and new environmental conditions associated with a spherical shell of prescribed dimensions and material. FIND: (a) Governing equation for variation of wall temperature with time. Initial rate of temperature change, (b) Steady-state wall temperature, (c) Effect of convection coefficient on canister temperature. SCHEMATIC:

ASSUMPTIONS: (1) Negligible temperature gradients in wall, (2) Constant properties, (3) Uniform, time-independent heat flux at inner surface. PROPERTIES: Table A.1, Stainless Steel, AISI 302: ρ = 8055 kg/m3, c p = 510 J/kg⋅K.

 − E  ANALYSIS: (a) Performing an energy balance on the shell at an instant of time, E in out = Est . Identifying relevant processes and solving for dT/dt, 4 dT q′′i 4π ri2 − h 4π ro2 ( T − T∞ ) = ρ π ro3 − ri3 cp 3 dt

(

) (

)

(

)

dT 3 q′′ r 2 − hr 2 ( T − T ) . = i o ∞  3 3 dt ρ c r − r  i p o i

(

)

Substituting numerical values for the initial condition, find

dT  dt i

  W W 2 2 3 105 0.5m ) − 500 0.6m ) (500 − 300 ) K  ( ( m2 m2 ⋅ K  =  kg J  8055 510 (0.6 )3 − (0.5 )3  m3 3   kg K ⋅ m

dT  = −0.089 K/s . dt i

<

 = 0, it follows that (b) Under steady-state conditions with E st 2 2 q′′i 4π ri = h 4π ro ( T − T∞ )

(

) (

)

Continued …..

PROBLEM 1.48 (Cont.) 2

2 q′′i  ri  105 W/m2  0.5m  T = T∞ +   = 300K +   = 439K h  ro  500W/m 2 ⋅ K  0.6m 

<

(c) Parametric calculations were performed using the IHT First Law Model for an Isothermal Hollow Sphere. As shown below, there is a sharp increase in temperature with decreasing values of h < 1000 W/m2⋅K. For T > 380 K, boiling will occur at the canister surface, and for T > 410 K a condition known as film boiling (Chapter 10) will occur. The condition corresponds to a precipitous reduction in h and increase in T. 1000

Temperature, T(K)

900 800 700 600 500 400 300 100

400

800

2000

6000

10000

Convection coefficient, h(W/m^2.K)

Although the canister remains well below the melting point of stainless steel for h = 100 W/m2⋅K, boiling should be avoided, in which case the convection coefficient should be maintained at h > 1000 W/m2⋅K. COMMENTS: The governing equation of part (a) is a first order, nonhomogenous differential equation with constant coefficients. Its solution is θ = (S/R ) 1 − e−Rt + θ i e−Rt , where θ ≡ T − T∞ ,

) ( S ≡ 3qi′′ ri2 / ρ cp ( ro3 − ri3 ) , R = 3hro2 / ρ cp ( ro3 − ri3 ) . Note results for t → ∞ and for S = 0.

PROBLEM 1.49 KNOWN: Boiling point and latent heat of liquid oxygen. Diameter and emissivity of container. Free convection coefficient and temperature of surrounding air and walls. FIND: Mass evaporation rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Temperature of container outer surface equals boiling point of oxygen. ANALYSIS: (a) Applying an energy balance to a control surface about the container, it follows that, at any instant, E in − E out = 0 or qconv + q rad − q evap = 0 . The evaporative heat loss is equal to the product of the mass rate of vapor production and the heat of vaporization. Hence,  h ( T − T ) + εσ T 4 − T 4  A − m (1) s sur s  s  evap h fg = 0 ∞ 

(



(

)

)

 h ( T − T ) + εσ T 4 − T 4  π D2 ∞ s sur s     evap =  m h fg

(

)

10 W m 2 ⋅ K ( 298 − 263) K + 0.2 × 5.67 × 10−8 W m 2 ⋅ K 4 2984 − 2634 K 4  π (0.5 m )2    m evap = 214 kJ kg

 evap = m

(350 + 35.2 ) W / m2

(0.785 m2 ) = 1.41×10−3 kg s .

<

214 kJ kg

(b) Using the energy balance, Eq. (1), the mass rate of vapor production can be determined for the range of emissivity 0.2 to 0.94. The effect of increasing emissivity is to increase the heat rate into the container and, hence, increase the vapor production rate. Evaporation rate, mdot*1000 (kg/s)

1.9

1.8

1.7

1.6

1.5

1.4 0.2

0.4

0.6

0.8

1

Surface emissivity, eps

COMMENTS: To reduce the loss of oxygen due to vapor production, insulation should be applied to the outer surface of the container, in order to reduce qconv and qrad. Note from the calculations in part (a), that heat transfer by convection is greater than by radiation exchange.

PROBLEM 1.50 KNOWN: Frost formation of 2-mm thickness on a freezer compartment. Surface exposed to convection process with ambient air. FIND: Time required for the frost to melt, tm. SCHEMATIC:

ASSUMPTIONS: (1) Frost is isothermal at the fusion temperature, Tf, (2) The water melt falls away from the exposed surface, (3) Negligible radiation exchange at the exposed surface, and (4) Backside surface of frost formation is adiabatic. PROPERTIES: Frost, ρ f = 770 kg / m3 , h sf = 334 kJ / kg. ANALYSIS: The time tm required to melt a 2-mm thick frost layer may be determined by applying an energy balance, Eq 1.11b, over the differential time interval dt and to a differential control volume extending inward from the surface of the layer dx. From the schematic above, the energy in is the convection heat flux over the time period dt and the change in energy storage is the latent energy change within the control volume, As⋅dx.

Ein − E out = Est q′′conv As dt = dU"at

h As ( T∞ − Tf ) dt = − ρf As h sf dx

Integrating both sides of the equation and defining appropriate limits, find t 0 h ( T∞ − Tf ) m dt = − ρf hsf dx 0 xo

∫

tm = tm =

∫

ρf h sf x o h ( T∞ − Tf )

700 kg / m3 × 334 × 103 J / kg × 0.002m 2 W / m 2 ⋅ K ( 20 − 0 ) K

= 11, 690 s = 3.2 hour

<

COMMENTS: (1) The energy balance could be formulated intuitively by recognizing that the total heat in by convection during the time interval t m ( q′′cv ⋅ t m ) must be equal to the total latent energy for

melting the frost layer ( ρ x o h sf ). This equality is directly comparable to the derived expression above for tm. (2) Explain why the energy storage term in the analysis has a negative sign, and the limits of integration are as shown. Hint: Recall from the formulation of Eq. 1.11b, that the storage term represents the change between the final and initial states.

PROBLEM 1.51 KNOWN: Vertical slab of Woods metal initially at its fusion temperature, Tf, joined to a substrate.

(

)

Exposed surface is irradiated with laser source, G " W / m 2 . 2

 ′′ (kg/s⋅m ), and the material removed in a FIND: Instantaneous rate of melting per unit area, m m period of 2 s, (a) Neglecting heat transfer from the irradiated surface by convection and radiation exchange, and (b) Allowing for convection and radiation exchange.

SCHEMATIC:

ASSUMPTIONS: (1) Woods metal slab is isothermal at the fusion temperature, Tf, and (2) The melt runs off the irradiated surface. ANALYSIS: (a) The instantaneous rate of melting per unit area may be determined by applying an energy balance, Eq 1.11a, on the metal slab at an instant of time neglecting convection and radiation exchange from the irradiated surface.

E ′′in − E ′′out = E ′′st

α"G " =

d dM′′ ( −M′′h sf ) = −hsf dt dt

 ′′ is the time rate of change of mass in the control volume. Substituting values, where dM′′ / dt = m m  ′′m  ′′m = −60.6 × 10−3 kg / s ⋅ m 2 0.4 × 5000 W / m 2 = −33, 000 J / kg × m m The material removed in a 2s period per unit area is  ′′m ⋅ ∆t = 121 g / m 2 M′′2s = m (b) The energy balance considering convection and radiation exchange with the surroundings yields

< <

 ′′m α "G " − q′′cv − q′′rad = −h sf m q′′cv = h ( Tf − T∞ ) = 15 W / m 2 ⋅ K ( 72 − 20 ) K = 780 W / m 2

(

)

(

q′′rad = εσ Tf4 − T∞4 = 0.4 × 5.67 × 10−8 W / m 2 ⋅ K [72 + 273] − [20 + 273]

 ′′m = −32.3 × 10−3 kg / s ⋅ m 2 m

4

M 2s = 64 g / m 2

4

)K

4

= 154 W / m 2

<

COMMENTS: (1) The effects of heat transfer by convection and radiation reduce the estimate for the material removal rate by a factor of two. The heat transfer by convection is nearly 5 times larger than by radiation exchange. (2) Suppose the work piece were horizontal, rather than vertical, and the melt puddled on the surface rather than ran off. How would this affect the analysis? (3) Lasers are common heating sources for metals processing, including the present application of melting (heat transfer with phase change), as well as for heating work pieces during milling and turning (laser-assisted machining).

PROBLEM 1.52 KNOWN: Hot formed paper egg carton of prescribed mass, surface area and water content exposed to infrared heater providing known radiant flux. FIND: Whether water content can be reduced from 75% to 65% by weight during the 18s period carton is on conveyor. SCHEMATIC:

ASSUMPTIONS: (1) All the radiant flux from the heater bank is absorbed by the carton, (2) Negligible heat loss from carton by convection and radiation, (3) Negligible mass loss occurs from bottom side. PROPERTIES: Water (given): hfg = 2400 kJ/kg. ANALYSIS: Define a control surface about the carton, and write the conservation of energy requirement for an interval of time, ∆t, E in − E o ut = ∆ E st = 0

where Ein is due to the absorbed radiant flux, q ′′h , from the heater and Eout is the energy leaving due to evaporation of water from the carton. Hence. q ′′h ⋅ A s ⋅ ∆t = ∆M ⋅ h fg . For the prescribed radiant flux q ′′h , q ′′h A s∆t 5000 W / m2 × 0.0625 m2 × 18s ∆M = = = 0.00234 kg. h fg 2400 kJ / kg The chief engineer’s requirement was to remove 10% of the water content, or ∆M req = M × 0.10 = 0.220 kg × 0.10 = 0.022 kg which is nearly an order of magnitude larger than the evaporative loss. Considering heat losses by convection and radiation, the actual water removal from the carton will be less than ∆M. Hence, the purchase should not be recommended, since the desired water removal cannot be achieved.

<

PROBLEM 1.53 KNOWN: Average heat sink temperature when total dissipation is 20 W with prescribed air and surroundings temperature, sink surface area and emissivity. FIND: Sink temperature when dissipation is 30 W. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) All dissipated power in devices is transferred to the sink, (3) Sink is isothermal, (4) Surroundings and air temperature remain the same for both power levels, (5) Convection coefficient is the same for both power levels, (6) Heat sink is a small surface within a large enclosure, the surroundings. ANALYSIS: Define a control volume around the heat sink. Power dissipated within the devices is transferred into the sink, while the sink loses heat to the ambient air and surroundings by convection and radiation exchange, respectively. E in − E out = 0 (1) 4 = 0. Pe − hAs ( Ts − T∞ ) − Asεσ Ts4 − Tsur

)

(

Consider the situation when Pe = 20 W for which Ts = 42°C; find the value of h.

(

)

4  / (T − T ) h=  Pe / As − εσ Ts4 − Tsur   s ∞ h=  20 W/0.045 m 2 − 0.8 × 5.67 × 10−8 W/m 2 ⋅ K 4 3154 − 3004 K 4  / (315 − 300 ) K   h = 24.4 W / m2 ⋅ K.

)

(

For the situation when Pe = 30 W, using this value for h with Eq. (1), obtain 30 W - 24.4 W/m2 ⋅ K × 0.045 m 2 (Ts − 300 ) K

(

)

−0.045 m 2 × 0.8 × 5.67 × 10−8 W/m 2 ⋅ K 4 Ts4 − 3004 K 4 = 0

(

)

30 = 1.098 (Ts − 300 ) + 2.041× 10−9 Ts4 − 3004 . By trial-and-error, find Ts ≈ 322 K = 49 $ C.

<

COMMENTS: (1) It is good practice to express all temperatures in kelvin units when using energy balances involving radiation exchange. (2) Note that we have assumed As is the same for the convection and radiation processes. Since not all portions of the fins are completely exposed to the surroundings, As,rad is less than As,conv = As. (3) Is the assumption that the heat sink is isothermal reasonable?

PROBLEM 1.54 KNOWN: Number and power dissipation of PCBs in a computer console. Convection coefficient associated with heat transfer from individual components in a board. Inlet temperature of cooling air and fan power requirement. Maximum allowable temperature rise of air. Heat flux from component most susceptible to thermal failure. FIND: (a) Minimum allowable volumetric flow rate of air, (b) Preferred location and corresponding surface temperature of most thermally sensitive component.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Constant air properties, (3) Negligible potential and kinetic energy changes of air flow, (4) Negligible heat transfer from console to ambient air, (5) Uniform convection coefficient for all components. ANALYSIS: (a) For a control surface about the air space in the console, conservation of energy for an open system, Eq. (1.11e), reduces to

 =0  ( u + pv ) − m  ( u + pv ) + q − W m i o  = − P . Hence, with m  (ii − io ) = mc  p (Ti − To ) , where u + pv = i, q = 5Pb , and W f  p (To − Ti ) = 5 Pb + Pf mc For a maximum allowable temperature rise of 15°C, the required mass flow rate is

 = m

5 Pb + Pf 5 × 20 W + 25 W = = 8.28 ×10−3 kg/s cp ( To − Ti ) 1007 J/kg ⋅ K 15 $C

(

)

The corresponding volumetric flow rate is

∀=

 8.28 × 10−3 kg/s m = = 7.13 × 10−3 m3 / s 3 ρ 1.161 kg/m

<

(b) The component which is most susceptible to thermal failure should be mounted at the bottom of one of the PCBs, where the air is coolest. From the corresponding form of Newton’s law of cooling, q′′ = h ( Ts − Ti ) , the surface temperature is

q′′ 1× 104 W/m 2 $ Ts = Ti + = 20 C + = 70$ C h 200 W/m2 ⋅ K

<

COMMENTS: (1) Although the mass flow rate is invariant, the volumetric flow rate increases as the air is heated in its passage through the console, causing a reduction in the density. However, for the  , is small. (2) If the thermally prescribed temperature rise, the change in ρ, and hence the effect on ∀ sensitive component were located at the top of a PCB, it would be exposed to warmer air (To = 35°C) and the surface temperature would be Ts = 85°C.

PROBLEM 1.55 ′′ , and experiences for case (a): convection KNOWN: Top surface of car roof absorbs solar flux, qS,abs with air at T∞ and for case (b): the same convection process and radiation emission from the roof. FIND: Temperature of the plate, Ts , for the two cases. Effect of airflow on roof temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer to auto interior, (3) Negligible radiation from atmosphere. ANALYSIS: (a) Apply an energy balance to the control surfaces shown on the schematic. For an  − E instant of time, E in out = 0. Neglecting radiation emission, the relevant processes are convection

′′ between the plate and the air, q′′conv , and the absorbed solar flux, qS,abs . Considering the roof to have an area As , ′′ qS,abs ⋅ As − hAs ( Ts − T∞ ) = 0 ′′ Ts = T∞ + qS,abs /h Ts = 20$ C +

800W/m 2 12W/m 2 ⋅ K

= 20$ C + 66.7$ C = 86.7$ C

<

(b) With radiation emission from the surface, the energy balance has the form

′′ qS,abs ⋅ As − q conv − E ⋅ As = 0 ′′ qS,abs As − hAs ( Ts − T∞ ) − ε Asσ Ts4 = 0 . Substituting numerical values, with temperature in absolute units (K),

800

W m2

− 12

W m2 ⋅ K

(Ts − 293K ) − 0.8 × 5.67 ×10−8

W Ts4 = 0 2 4 m ⋅K

12Ts + 4.536 × 10−8 Ts4 = 4316 It follows that Ts = 320 K = 47°C.

< Continued.….

PROBLEM 1.55 (Cont.) (c) Parametric calculations were performed using the IHT First Law Model for an Isothermal Plane Wall. As shown below, the roof temperature depends strongly on the velocity of the auto relative to the ambient air. For a convection coefficient of h = 40 W/m2⋅K, which would be typical for a velocity of 55 mph, the roof temperature would exceed the ambient temperature by less than 10°C. 360

Temperature, Ts(K)

350 340 330 320 310 300 290 0

20

40

60

80

100

120

140

160

180

200

Convection coefficient, h(W/m^2.K)

COMMENTS: By considering radiation emission, Ts decreases, as expected. Note the manner in which q ′′conv is formulated using Newton’s law of cooling; since q ′′conv is shown leaving the control surface, the

rate equation must be h ( Ts − T∞ ) and not h ( T∞ − Ts ) .

PROBLEM 1.56 KNOWN: Detector and heater attached to cold finger immersed in liquid nitrogen. Detector surface of ε = 0.9 is exposed to large vacuum enclosure maintained at 300 K. FIND: (a) Temperature of detector when no power is supplied to heater, (b) Heater power (W) required to maintain detector at 195 K, (c) Effect of finger thermal conductivity on heater power. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through cold finger, (3) Detector and heater are very thin and isothermal at Ts , (4) Detector surface is small compared to enclosure surface. PROPERTIES: Cold finger (given): k = 10 W/m⋅K. ANALYSIS: Define a control volume about detector and heater and apply conservation of energy requirement on a rate basis, Eq. 1.11a,

E in − E out = 0 where

(1)

E in = q rad + q elec ;

E out = q cond

(2,3)

Combining Eqs. (2,3) with (1), and using the appropriate rate equations,

)

(

4 − T4 + q ε Asσ Tsur s elec = kAs (Ts − TL )/L .

(4)

(a) Where q elec = 0, substituting numerical values

(

)

0.9 × 5.67 ×10−8 W/m 2 ⋅ K 4 3004 − Ts4 K 4 = 10W/m ⋅ K (Ts − 77 ) K/0.050 m

(

)

5.103 × 10−8 3004 − Ts4 = 200 (Ts − 77 ) Ts = 79.1K

< Continued.….

PROBLEM 1.56 (Cont.) (b) When Ts = 195 K, Eq. (4) yields

)

(

0.9 × [π ( 0.005 m ) / 4] × 5.67 × 10−8 W/m 2 ⋅ K 4 3004 − 1954 K 4 + q elec 2

= 10W/m ⋅ K × [π (0.005 m ) /4] × (195 − 77 ) K / 0.050 m 2

<

qelec = 0.457 W = 457 mW

(c) Calculations were performed using the First Law Model for a Nonisothermal Plane Wall. With net radiative transfer to the detector fixed by the prescribed values of Ts and Tsur , Eq. (4) indicates that q elec increases linearly with increasing k.

Heater power, qelec(W)

19 17 15 13 11 9 7 5 3 1 -1 0

100

200

300

400

Thermal conductivity, k(W/m.K)

Heat transfer by conduction through the finger material increases with its thermal conductivity. Note that, for k = 0.1 W/m⋅K, q elec = -2 mW, where the minus sign implies the need for a heat sink, rather than a heat source, to maintain the detector at 195 K. In this case q rad exceeds q cond , and a heat sink would be needed to dispose of the difference. A conductivity of k = 0.114 W/m⋅K yields a precise balance between q rad and q cond . Hence to circumvent heaving to use a heat sink, while minimizing the heater power requirement, k should exceed, but remain as close as possible to the value of 0.114 W/m⋅K. Using a graphite fiber composite, with the fibers oriented normal to the direction of conduction, Table A.2 indicates a value of k ≈ 0.54 W/m⋅K at an average finger temperature of T = 136 K. For this value, q elec = 18 mW COMMENTS: The heater power requirement could be further reduced by decreasing ε.

PROBLEM 1.57 KNOWN: Conditions at opposite sides of a furnace wall of prescribed thickness, thermal conductivity and surface emissivity. FIND: Effect of wall thickness and outer convection coefficient on surface temperatures. Recommended values of L and h 2 . SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible radiation exchange at surface 1, (4) Surface 2 is exposed to large surroundings. ANALYSIS: The unknown temperatures may be obtained by simultaneously solving energy balance equations for the two surface. At surface 1,

q′′conv,1 = q′′cond

(

)

h1 T∞,1 − T1 = k (T1 − T2 )/L

(1)

At surface 2,

q′′cond = q′′conv + q′′rad

(

)

(

4 k ( T1 − T2 )/L = h 2 T2 − T∞,2 + εσ T24 − Tsur

)

(2)

Surface temperature, T(K)

Using the IHT First Law Model for a Nonisothermal Plane Wall, we obtain

1700 1500 1300 1100 900 700 500 300 0

0.1

0.2

0.3

0.4

0.5

Wall thickness, L(m) Inner surface temperature, T1(K) Outer surface temperature, T2(K)

Continued …..

PROBLEM 1.57 (Cont.) Both q′′cond and T2 decrease with increasing wall thickness, and for the prescribed value of h 2 = 10 W/m2⋅K, a value of L ≥ 0.275 m is needed to maintain T2 ≤ 373 K = 100 °C. Note that inner surface temperature T1 , and hence the temperature difference, T1 − T2 , increases with increasing L.

Surface temperature, T(K)

Performing the calculations for the prescribed range of h 2 , we obtain 1700 1500 1300 1100 900 700 500 300 0

10

20

30

40

50

Convection coefficient, h2(W/m^2.K) Inner surface temperature, T(K) Outer surface temperature, T(K)

For the prescribed value of L = 0.15 m, a value of h 2 ≥ 24 W/m2⋅K is needed to maintain T2 ≤ 373 K. The variation has a negligible effect on T1 , causing it to decrease slightly with increasing h 2 , but does have a strong influence on T2 . COMMENTS: If one wishes to avoid use of active (forced convection) cooling on side 2, reliance will have to be placed on free convection, for which h 2 ≈ 5 W/m2⋅K. The minimum wall thickness would then be L = 0.40 m.

PROBLEM 1.58 KNOWN: Furnace wall with inner surface temperature T1 = 352°C and prescribed thermal conductivity experiencing convection and radiation exchange on outer surface. See Example 1.5. FIND: (a) Outer surface temperature T2 resulting from decreasing the wall thermal conductivity k or increasing the convection coefficient h by a factor of two; benefit of applying a low emissivity coating (ε < 0.8); comment on the effectiveness of these strategies to reduce risk of burn injury when 2 T2 ≤ 65°C; and (b) Calculate and plot T2 as a function of h for the range 20 ≤ h ≤ 100 W/m ⋅K for three materials with k = 0.3, 0.6, and 1.2 W/m⋅K; what conditions will provide for safe outer surface temperatures. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Radiation exchange is between small surface and large enclosure, (4) Inner surface temperature remains constant for all conditions. ANALYSIS: (a) The surface (x = L) energy balance is

(

T −T 4 k 1 2 = h ( T2 − T∞ ) + εσ T24 − Tsur L

)

With T1 = 352°C, the effects of parameters h, k and ε on the outer surface temperature are calculated and tabulated below. Conditions Example 1.5 Decrease k by ½ Increase h by 2 Change k and h Decrease ε

k (W / m ⋅ K ) 1.2 0.6 1.2 0.6 1.2

(

h W / m2 ⋅ K 20 20 40 40 20

)

ε

T2 (°C )

0.8 0.8 0.8 0.8 0.1

100 69 73 51 115

(b) Using the energy balance relation in the Workspace of IHT, the outer surface temperature can be calculated and plotted as a function of the convection coefficient for selected values of the wall thermal conductivity. Continued …..

O u te r su rfa ce te m p e ra tu re , T2 (C )

PROBLEM 1.58 (Cont.) 100 80 60 40 20 20

40

60

80

100

C o n ve ctio n co e fficie n t, h (W /m ^2 .K ) k = 1 .2 W /m .K k = 0 .6 W /m .K k = 0 .3 W /m .K

COMMENTS: (1) From the parameter study of part (a), note that decreasing the thermal conductivity is more effective in reducing T2 than is increasing the convection coefficient. Only if both changes are made will T2 be in the safe range. (2) From part (a), note that applying a low emissivity coating is not beneficial. Did you suspect that before you did the analysis? Give a physical explanation for this result. (3) From the parameter study graph we conclude that safe wall conditions (T2 ≤ 65°C) can be 2 maintained for these conditions: with k = 1.2 W/m⋅K when h > 55 W/m ⋅K; with k = 0.6 W/m⋅K 2 when h > 25 W/m ⋅K; and with k = 0.3 W/m⋅K when h > 20 W/m⋅K.

PROBLEM 1.59 KNOWN: Inner surface temperature, thickness and thermal conductivity of insulation exposed at its outer surface to air of prescribed temperature and convection coefficient. FIND: Outer surface temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in the insulation, (3) Negligible radiation exchange between outer surface and surroundings. ANALYSIS: From an energy balance at the outer surface at an instant of time, q ′′cond = q conv ′′ . Using the appropriate rate equations, k

(T1 − T2 ) = h L

(T2 − T∞ ).

Solving for T2, find

(

)

( )

0.1 W/m ⋅ K W k 400$ C + 500 35$ C T1 + h T∞ 2 0.025m m ⋅K T2 = L = k W 0.1 W/m ⋅ K h+ 500 + L 0.025m m2 ⋅ K T2 = 37.9$ C.

<

COMMENTS: If the temperature of the surroundings is approximately that of the air, radiation exchange between the outer surface and the surroundings will be negligible, since T2 is small. In this case convection makes the dominant contribution to heat transfer from the outer surface, and assumption (3) is excellent.

PROBLEM 1.60 KNOWN: Thickness and thermal conductivity, k, of an oven wall. Temperature and emissivity, ε, of front surface. Temperature and convection coefficient, h, of air. Temperature of large surroundings. FIND: (a) Temperature of back surface, (b) Effect of variations in k, h and ε. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Radiation exchange with large surroundings. ANALYSIS: (a) Applying an energy balance, Eq. 1.13, at an instant of time to the front surface and substituting the appropriate rate equations, Eqs. 1.2, 1.3a and 1.7, find

)

(

T −T 4 . k 1 2 = h ( T2 − T∞ ) + εσ T24 − Tsur L Substituting numerical values, find T1 − T2 =

W W   4 4  −8 20 100 K + 0.8 × 5.67 × 10 400 K ) − (300 K )   = 200 K . (     0.7 W/m ⋅ K  m 2 ⋅ K m2 ⋅ K 4  0.05 m

<

Since T2 = 400 K, it follows that T1 = 600 K.

(b) Parametric effects may be evaluated by using the IHT First Law Model for a Nonisothermal Plane Wall. Changes in k strongly influence conditions for k < 20 W/m⋅K, but have a negligible effect for larger values, as T2 approaches T1 and the heat fluxes approach the corresponding limiting values 10000 Heat flux, q''(W/m^2)

Temperature, T2(K)

600

500

8000 6000 4000 2000

400

0 0

100

200

300

Thermal conductivity, k(W/m.K) 300 0

100

200

300

Thermal conductivity, k(W/m.K)

400

Conduction heat flux, q''cond(W/m^2) Convection heat flux, q''conv(W/m^2) Radiation heat flux, q''rad(W/m^2)

400

PROBLEM 1.60 (Cont.) The implication is that, for k > 20 W/m⋅K, heat transfer by conduction in the wall is extremely efficient relative to heat transfer by convection and radiation, which become the limiting heat transfer processes. Larger fluxes could be obtained by increasing ε and h and/or by decreasing T∞ and Tsur . With increasing h, the front surface is cooled more effectively ( T2 decreases), and although q′′rad decreases, the reduction is exceeded by the increase in q′′conv . With a reduction in T2 and fixed values

of k and L, q′′cond must also increase.

30000 Heat flux, q''(W/m^2)

Temperature, T2(K)

600

500

20000

10000

0 0

100

200

Convection coefficient, h(W/m^2.K) 400 0

100

Conduction heat flux, q''cond(W/m^2) Convection heat flux, q''conv(W/m^2) Radiation heat flux, q''rad(W/m^2)

200

Convection coefficient, h(W/m^2.K)

The surface temperature also decreases with increasing ε, and the increase in q′′rad exceeds the reduction in q′′conv , allowing q′′cond to increase with ε. 10000 Heat flux, q''(W/m^2)

575

Temperature, T2(K)

570

565

560

8000 6000 4000 2000 0

555

0

0.2

0.4

0.6

0.8

Emissivity 550 0

0.2

0.4

0.6

Emissivity

0.8

1

Conduction heat flux, q''cond(W/m^2) Convection heat flux, q''conv(W/m^2) Radiation heat flux, q''rad(W/m^2)

COMMENTS: Conservation of energy, of course, dictates that, irrespective of the prescribed conditions, q′′cond = q′′conv + q′′rad .

1

PROBLEM 1.61 KNOWN: Temperatures at 10 mm and 20 mm from the surface and in the adjoining airflow for a thick steel casting. FIND: Surface convection coefficient, h. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in the x-direction, (3) Constant properties, (4) Negligible generation. ANALYSIS: From a surface energy balance, it follows that

q ′′cond = q conv ′′ where the convection rate equation has the form

q′′conv = h ( T∞ − T0 ) , and q cond can be evaluated from the temperatures prescribed at surfaces 1 and 2. That is, from ′′ Fourier’s law,

T −T q′′cond = k 1 2 x 2 − x1

(50 − 40 ) C = 15, 000 W/m2 . W q′′cond = 15 m ⋅ K ( 20-10 ) ×10−3 m $

Since the temperature gradient in the solid must be linear for the prescribed conditions, it follows that

T0 = 60°C. Hence, the convection coefficient is

h= h=

q ′′cond T∞ − T0

15,000 W / m2 $

40 C

= 375 W / m2 ⋅ K.

<

COMMENTS: The accuracy of this procedure for measuring h depends strongly on the validity of the assumed conditions.

PROBLEM 1.62 KNOWN: Duct wall of prescribed thickness and thermal conductivity experiences prescribed heat flux q′′o at outer surface and convection at inner surface with known heat transfer coefficient. FIND: (a) Heat flux at outer surface required to maintain inner surface of duct at Ti = 85°C, (b) Temperature of outer surface, To , (c) Effect of h on To and q′′o . SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Constant properties, (4) Backside of heater perfectly insulated, (5) Negligible radiation. ANALYSIS: (a) By performing an energy balance on the wall, recognize that balance on the top surface, it follows that

q′′o = q′′cond . From an energy

q′′cond = q′′conv = q′′o . Hence, using the convection rate equation,

q′′o = q′′conv = h ( Ti − T∞ ) = 100 W / m 2 ⋅ K (85 − 30 ) C = 5500W /m2 . (b) Considering the duct wall and applying Fourier’s Law, $

q′′o = k

<

T −T ∆T =k o i L ∆X

q′′ L 5500 W/m 2 × 0.010 m $ = (85 + 2.8 ) C = 87.8$ C . To = Ti + o = 85$ C + k 20 W/m ⋅ K

<

(c) For Ti = 85°C, the desired results may be obtained by simultaneously solving the energy balance equations T − Ti T − Ti q′′o = k o k o = h ( Ti − T∞ ) and L L Using the IHT First Law Model for a Nonisothermal Plane Wall, the following results are obtained. 91 Surface temperature, To(C)

Heat flux, q''o(W/m^2)

12000 10000 8000 6000 4000 2000

90 89 88 87 86

0

85

0

40

80

120

160

Convection coefficient, h(W/m^2.K)

200

0

40

80

120

160

200

Convection coefficient, h(W/m^2.K)

Since q′′conv increases linearly with increasing h, the applied heat flux q′′o must be balanced by an increase in q′′cond , which, with fixed k, Ti and L, necessitates an increase in To . COMMENTS: The temperature difference across the wall is small, amounting to a maximum value of (To − Ti ) = 5.5°C for h = 200 W/m2⋅K. If the wall were thinner (L < 10 mm) or made from a material with higher conductivity (k > 20 W/m⋅K), this difference would be reduced.

PROBLEM 1.63 KNOWN: Dimensions, average surface temperature and emissivity of heating duct. Duct air inlet temperature and velocity. Temperature of ambient air and surroundings. Convection coefficient. FIND: (a) Heat loss from duct, (b) Air outlet temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Constant air properties, (3) Negligible potential and kinetic energy changes of air flow, (4) Radiation exchange between a small surface and a large enclosure. ANALYSIS: (a) Heat transfer from the surface of the duct to the ambient air and the surroundings is given by Eq. (1.10)

(

4 q = hAs ( Ts − T∞ ) + ε Asσ Ts4 − Tsur

)

2

where As = L (2W + 2H) = 15 m (0.7 m + 0.5 m) = 16.5 m . Hence,

( )

(

)

q = 4 W/m2 ⋅ K × 16.5 m2 45$ C + 0.5 ×16.5 m2 × 5.67 × 10−8 W/m2 ⋅ K 4 3234 − 2784 K 4 q = qconv + q rad = 2970 W + 2298 W = 5268 W

<

 = 0 and the third assumption, Eq. (1.11e) yields, (b) With i = u + pv, W  (ii − io ) = mc  p (Ti − To ) = q m where the sign on q has been reversed to reflect the fact that heat transfer is from the system.  = ρ VAc = 1.10 kg/m3 × 4 m/s (0.35m × 0.20m ) = 0.308 kg/s, the outlet temperature is With m q 5268 W To = Ti − = 58$ C − = 41$ C  p mc 0.308 kg/s ×1008 J/kg ⋅ K

<

COMMENTS: The temperature drop of the air is large and unacceptable, unless the intent is to use the duct to heat the basement. If not, the duct should be insulated to insure maximum delivery of thermal energy to the intended space(s).

PROBLEM 1.64 KNOWN: Uninsulated pipe of prescribed diameter, emissivity, and surface temperature in a room with fixed wall and air temperatures. See Example 1.2. FIND: (a) Which option to reduce heat loss to the room is more effective: reduce by a factor of two 2 the convection coefficient (from 15 to 7.5 W/m ⋅K) or the emissivity (from 0.8 to 0.4) and (b) Show 2 graphically the heat loss as a function of the convection coefficient for the range 5 ≤ h ≤ 20 W/m ⋅K for emissivities of 0.2, 0.4 and 0.8. Comment on the relative efficacy of reducing heat losses associated with the convection and radiation processes. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between pipe and the room is between a small surface in a much larger enclosure, (3) The surface emissivity and absorptivity are equal, and (4) Restriction of the air flow does not alter the radiation exchange process between the pipe and the room. ANALYSIS: (a) The heat rate from the pipe to the room per unit length is

(

4 q′ = q′ / L = q′conv + q′rad = h (π D )( Ts − T∞ ) + ε (π D )σ Ts4 − Tsur

)

Substituting numerical values for the two options, the resulting heat rates are calculated and compared with those for the conditions of Example 1.2. We conclude that both options are comparably effective.

(

h W / m2 ⋅ K

Conditions Base case, Example 1.2 Reducing h by factor of 2 Reducing ε by factor of 2

)

ε

15 7.5 15

0.8 0.8 0.4

q′ ( W / m ) 998 788 709

(b) Using IHT, the heat loss can be calculated as a function of the convection coefficient for selected values of the surface emissivity.

Heat loss, q' (/m)

1200

800

400

0 5

10

15

20

Convection coefficient, h (W/m^2.K) eps = 0.8, bare pipe eps = 0.4, coated pipe eps = 0.2, coated pipe

Continued …..

PROBLEM 1.64 (Cont.) COMMENTS: (1) In Example 1.2, Comment 3, we read that the heat rates by convection and radiation exchange were comparable for the base case conditions (577 vs. 421 W/m). It follows that reducing the key transport parameter (h or ε) by a factor of two yields comparable reductions in the heat loss. Coating the pipe to reduce the emissivity might to be the more practical option as it may be difficult to control air movement. (2) For this pipe size and thermal conditions (Ts and T∞), the minimum possible convection coefficient 2 is approximately 7.5 W/m ⋅K, corresponding to free convection heat transfer to quiescent ambient air. Larger values of h are a consequence of forced air flow conditions. (3) The Workspace for the IHT program to calculate the heat loss and generate the graph for the heat loss as a function of the convection coefficient for selected emissivities is shown below. It is good practice to provide commentary with the code making your solution logic clear, and to summarize the results. // Heat loss per unit pipe length; rate equation from Ex. 1.2 q' = q'cv + q'rad q'cv = pi*D*h*(Ts - Tinf) q'rad = pi*D*eps*sigma*(Ts^4 - Tsur^4) sigma = 5.67e-8 // Input parameters D = 0.07 Ts_C = 200 // Representing temperatures in Celsius units using _C subscripting Ts = Ts_C +273 Tinf_C = 25 Tinf = Tinf_C + 273 h = 15 // For graph, sweep over range from 5 to 20 Tsur_C = 25 Tsur = Tsur_C + 273 eps = 0.8 //eps = 0.4 // Values of emissivity for parameter study //eps = 0.2 /* Base case results Tinf Ts Tsur eps h 298 473 298 0.8 15

q' q'cv sigma 997.9 577.3 5.67E-8 */

q'rad

D

Tinf_C

Ts_C

Tsur_C

420.6

0.07

25

200

25

PROBLEM 1.65 KNOWN: Conditions associated with surface cooling of plate glass which is initially at 600°C. Maximum allowable temperature gradient in the glass. FIND: Lowest allowable air temperature, T∞ SCHEMATIC:

ASSUMPTIONS: (1) Surface of glass exchanges radiation with large surroundings at Tsur = T∞, (2) One-dimensional conduction in the x-direction. ANALYSIS: The maximum temperature gradient will exist at the surface of the glass and at the instant that cooling is initiated. From the surface energy balance, Eq. 1.12, and the rate equations, Eqs. 1.1, 1.3a and 1.7, it follows that

-k

)

(

dT 4 − h ( Ts − T∞ ) − εσ Ts4 − Tsur =0 dx

or, with (dT/dx)max = -15°C/mm = -15,000°C/m and Tsur = T∞,

−1.4

$C  W  W  −15, 000  = 5 (873 − T∞ ) K 2 K m⋅K  m m ⋅  

+0.8 × 5.67 × 10−8

W

8734 − T 4  K 4 . ∞ 

 m2 ⋅ K 4 

T∞ may be obtained from a trial-and-error solution, from which it follows that, for T∞ = 618K,

21,000

W

W W ≈ 1275 2 + 19,730 2 . m m m 2

Hence the lowest allowable air temperature is

T∞ ≈ 618K = 345$ C.

<

COMMENTS: (1) Initially, cooling is determined primarily by radiation effects. (2) For fixed T∞, the surface temperature gradient would decrease with increasing time into the cooling process. Accordingly, T∞ could be decreasing with increasing time and still keep within the maximum allowable temperature gradient.

PROBLEM 1.66 KNOWN: Hot-wall oven, in lieu of infrared lamps, with temperature Tsur = 200°C for heating a coated plate to the cure temperature. See Example 1.6. FIND: (a) The plate temperature Ts for prescribed convection conditions and coating emissivity, and (b) Calculate and plot Ts as a function of Tsur for the range 150 ≤ Tsur ≤ 250°C for ambient air temperatures of 20, 40 and 60°C; identify conditions for which acceptable curing temperatures between 100 and 110°C may be maintained. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss from back surface of plate, (3) Plate is small object in large isothermal surroundings (hot oven walls). ANALYSIS: (a) The temperature of the plate can be determined from an energy balance on the plate, considering radiation exchange with the hot oven walls and convection with the ambient air.

E ′′in − E ′′out = 0

(

or

)

q′′rad − q′′conv = 0

4 − T4 − h T − T = 0 εσ Tsur ( s ∞) s 0.5 × 5.67 × 10−8 W / m 2 ⋅ K 4

([200 + 273] − T ) K 4

4 s

4

− 15 W / m 2 ⋅ K (Ts − [20 + 273]) K = 0

<

Ts = 357 K = 84°C

(b) Using the energy balance relation in the Workspace of IHT, the plate temperature can be calculated and plotted as a function of oven wall temperature for selected ambient air temperatures. Plate temperature, Ts (C)

150

100

50 150

175

200

225

250

Oven wall temperature, Tsur (C) Tinf = 60 C Tinf = 40 C Tinf = 20 C

COMMENTS: From the graph, acceptable cure temperatures between 100 and 110°C can be maintained for these conditions: with T∞ = 20°C when 225 ≤ Tsur ≤ 240°C; with T∞ = 40°C when 205 ≤ Tsur ≤ 220°C; and with T∞ = 60°C when 175 ≤ Tsur ≤ 195°C.

PROBLEM 1.67 KNOWN: Operating conditions for an electrical-substitution radiometer having the same receiver temperature, Ts, in electrical and optical modes. FIND: Optical power of a laser beam and corresponding receiver temperature when the indicated electrical power is 20.64 mW. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Conduction losses from backside of receiver negligible in optical mode, (3) Chamber walls form large isothermal surroundings; negligible effects due to aperture, (4) Radiation exchange between the receiver surface and the chamber walls is between small surface and large enclosure, (5) Negligible convection effects. PROPERTIES: Receiver surface: ε = 0.95, αopt = 0.98. ANALYSIS: The schematic represents the operating conditions for the electrical mode with the optical beam blocked. The temperature of the receiver surface can be found from an energy balance on the receiver, considering the electrical power input, conduction loss from the backside of the receiver, and the radiation exchange between the receiver and the chamber.

E in − E out = 0

Pelec − q loss − q rad = 0

(

)

4 =0 Pelec − 0.05 Pelec − ε Asσ Ts4 − Tsur 20.64 × 10

−3

(

)

(

)

2 2 4 4 4 4 2 −8 W (1 − 0.05 ) − 0.95 π 0.015 / 4 m × 5.67 × 10 W / m ⋅ K Ts − 77 K = 0

Ts = 213.9 K

<

For the optical mode of operation, the optical beam is incident on the receiver surface, there is no electrical power input, and the receiver temperature is the same as for the electrical mode. The optical power of the beam can be found from an energy balance on the receiver considering the absorbed beam power and radiation exchange between the receiver and the chamber.

E in − E out = 0

α opt Popt − q rad = 0.98 Popt − 19.60 mW = 0 Popt = 19.99 mW

<

where qrad follows from the previous energy balance using Ts = 213.9K. COMMENTS: Recognizing that the receiver temperature, and hence the radiation exchange, is the same for both modes, an energy balance could be directly written in terms of the absorbed optical power and equivalent electrical power, αopt Popt = Pelec - qloss.

PROBLEM 1.68 KNOWN: Surface temperature, diameter and emissivity of a hot plate. Temperature of surroundings and ambient air. Expression for convection coefficient. FIND: (a) Operating power for prescribed surface temperature, (b) Effect of surface temperature on power requirement and on the relative contributions of radiation and convection to heat transfer from the surface.

SCHEMATIC:

ASSUMPTIONS: (1) Plate is of uniform surface temperature, (2) Walls of room are large relative to plate, (3) Negligible heat loss from bottom or sides of plate. ANALYSIS: (a) From an energy balance on the hot plate, Pelec = qconv + qrad = Ap ( q ′′conv + q ′′rad ). 1/3 Substituting for the area of the plate and from Eqs. (1.3a) and (1.7), with h = 0.70 (Ts - T∞) , it follows that

(

2 Pelec = π D / 4

) 

(

4 4 4/3 0.70 ( Ts − T∞ ) + εσ Ts − Tsur

) (

4 4 −8 2 4/3 Pelec = π ( 0.3m ) / 4  0.70 (175 ) 473 − 298 + 0.8 × 5.67 × 10  Pelec = 0.0707 m

)

W/m

2

2 2 685 W/m + 1913 W/m  = 48.4 W + 135.2 W = 190.6 W

2

<

(b) As shown graphically, both the radiation and convection heat rates, and hence the requisite electric power, increase with increasing surface temperature. E ffe c t o f s u rfa c e te m p e ra tu re o n e le c tric p o w e r a n d h e a t ra te s

H e a t ra te (W )

500 400 300 200 100 0 100

150

200

250

300

S u rfa c e te m p e ra tu re (C ) P e le c q ra d q co n v

However, because of its dependence on the fourth power of the surface temperature, the increase in radiation is more pronounced. The significant relative effect of radiation is due to the small 2 convection coefficients characteristic of natural convection, with 3.37 ≤ h ≤ 5.2 W/m ⋅K for 100 ≤ Ts < 300°C. COMMENTS: Radiation losses could be reduced by applying a low emissivity coating to the surface, which would have to maintain its integrity over the range of operating temperatures.

PROBLEM 1.69 KNOWN: Long bus bar of rectangular cross-section and ambient air and surroundings temperatures. Relation for the electrical resistivity as a function of temperature. FIND: (a) Temperature of the bar with a current of 60,000 A, and (b) Compute and plot the operating temperature of the bus bar as a function of the convection coefficient for the range 10 ≤ h ≤ 100 2 W/m ⋅K. Minimum convection coefficient required to maintain a safe-operating temperature below 120°C. Will increasing the emissivity significantly affect this result? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Bus bar is long, (3) Uniform bus-bar temperature, (3) Radiation exchange between the outer surface of the bus bar and its surroundings is between a small surface and a large enclosure. PROPERTIES: Bus-bar material, ρ e = ρ e,o [1 + α ( T − To )], ρ e,o = 0.0828 µΩ ⋅ m, To = 25°C, α = 0.0040 K

−1

.

ANALYSIS: (a) An energy balance on the bus-bar for a unit length as shown in the schematic above has the form

′ − E ′out + E ′gen = 0 E in

−q′rad − q′conv + I2 R ′e = 0

)

(

4 − h P T − T + I2 ρ / A = 0 −ε Pσ T 4 − Tsur ( e c ∞) where P = 2 ( H + W ) , R ′e = ρ e / A c and A c = H × W. Substituting numerical values, 4 −0.8 × 2 0.600 + 0.200 m × 5.67 × 10−8 W / m 2 ⋅ K 4 T 4 − 30 + 273 K 4

(

(

)

[

−10 W / m 2 ⋅ K × 2 ( 0.600 + 0.200 ) m ( T − [30 + 273]) K + ( 60, 000 A )

2

{0.0828 ×10

−6

]

)

}

Ω ⋅ m 1 + 0.0040 K −1 ( T − [25 + 273]) K  / ( 0.600 × 0.200 ) m 2 = 0



Solving for the bus-bar temperature, find



<

T = 426 K = 153°C.

(b) Using the energy balance relation in the Workspace of IHT, the bus-bar operating temperature is 2 calculated as a function of the convection coefficient for the range 10 ≤ h ≤ 100 W/m ⋅K. From this graph we can determine that to maintain a safe operating temperature below 120°C, the minimum convection coefficient required is

h min = 16 W / m 2 ⋅ K.

< Continued …..

PROBLEM 1.69 (Cont.) Using the same equations, we can calculate and plot the heat transfer rates by convection and radiation as a function of the bus-bar temperature. 3000 H e a t ra te s, q 'cv o r q 'ra d (W /m )

175

B a r te m p e ra tu re , T (C )

150 125 100 75

2000

1000

0

50

25

50

75

100

125

150

175

B u s b a r te m p e ra tu re , T (C )

25 0

20

40

60

80

100

C o n ve c tio n h e a t flu x, q 'cv R a d ia tio n e xc h a n g e , q 'ra d , e p s = 0 .8

C o n ve c tio n co e fficie n t, h (W /m ^2 .K )

Note that convection is the dominant mode for low bus-bar temperatures; that is, for low current flow. As the bus-bar temperature increases toward the safe-operating limit (120°C), convection and radiation exchange heat transfer rates become comparable. Notice that the relative importance of the radiation exchange rate increases with increasing bus-bar temperature. COMMENTS: (1) It follows from the second graph that increasing the surface emissivity will be only significant at higher temperatures, especially beyond the safe-operating limit. (2) The Workspace for the IHT program to perform the parametric analysis and generate the graphs is shown below. It is good practice to provide commentary with the code making your solution logic clear, and to summarize the results. /* Results for base case conditions: Ts_C q'cv q'rad rhoe H eps h 153.3 1973 1786 1.253E-7 0.6 0.8 10 */

I

Tinf_C

Tsur_C

W

alpha

6E4

30

30

0.2

0.004

// Surface energy balance on a per unit length basis -q'cv - q'rad + Edot'gen = 0 q'cv = h * P * (Ts - Tinf) P = 2 * (W + H) // perimeter of the bar experiencing surface heat transfer q'rad = eps * sigma * (Ts^4 - Tsur^4) * P sigma = 5.67e-8 Edot'gen = I^2 * Re' Re' = rhoe / Ac rhoe = rhoeo * ( 1 + alpha * (Ts - Teo)) Ac = W * H // Input parameters I = 60000 alpha = 0.0040 rhoeo = 0.0828e-6 Teo = 25 + 273 W = 0.200 H = 0.600 Tinf_C = 30 Tinf = Tinf_C + 273 h = 10 eps = 0.8 Tsur_C = 30 Tsur = Tsur_C + 273 Ts_C = Ts - 273

// temperature coefficient, K^-1; typical value for cast aluminum // electrical resistivity at the reference temperature, Teo; microohm-m // reference temperature, K

PROBLEM 1.70 KNOWN: Solar collector designed to heat water operating under prescribed solar irradiation and loss conditions. FIND: (a) Useful heat collected per unit area of the collector, q ′′u , (b) Temperature rise of the water flow, To − Ti , and (c) Collector efficiency. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) No heat losses out sides or back of collector, (3) Collector area is small compared to sky surroundings. PROPERTIES: Table A.6, Water (300K): cp = 4179 J/kg⋅K. ANALYSIS: (a) Defining the collector as the control volume and writing the conservation of energy requirement on a per unit area basis, find that

E in − E out + E gen = E st .

Identifying processes as per above right sketch,

q ′′solar − q ′′rad − q ′′conv − q ′′u = 0

= 0.9 q s′′; that is, 90% of the solar flux is absorbed in the collector (Eq. 1.6). Using the where q solar ′′ appropriate rate equations, the useful heat rate per unit area is

(

)

4 T4 q′′u = 0.9 q′′s − εσ Tcp − sky − h ( Ts − T∞ ) W W W q′′u = 0.9 × 700 3034 − 2634 K 4 − 10 − 0.94 × 5.67 × 10−8 (30 − 25)$ C 2 2 4 2 m m ⋅K m ⋅K

)

(

q ′′u = 630 W / m2 − 194 W / m2 − 50 W / m2 = 386 W / m2 .

<

(b) The total useful heat collected is q ′′u ⋅ A. Defining a control volume about the water tubing, the useful heat causes an enthalpy change of the flowing water. That is,

 p ( Ti − To ) q′′u ⋅ A=mc

or

(Ti − To ) = 386 W/m2 × 3m2 / 0.01kg/s × 4179J/kg ⋅ K=27.7$C.

(

)(

)

′′ = 386 W/m 2 / 700 W/m 2 = 0.55 or 55%. (c) The efficiency is η = q′′u / qS

< <

COMMENTS: Note how the sky has been treated as large surroundings at a uniform temperature Tsky.

PROBLEM 1.71 KNOWN: Surface-mount transistor with prescribed dissipation and convection cooling conditions. FIND: (a) Case temperature for mounting arrangement with air-gap and conductive paste between case  , subject to the constraint that T = 40°C. and circuit board, (b) Consider options for increasing E g c SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Transistor case is isothermal, (3) Upper surface experiences convection; negligible losses from edges, (4) Leads provide conduction path between case and board, (5) Negligible radiation, (6) Negligible energy generation in leads due to current flow, (7) Negligible convection from surface of leads. PROPERTIES: (Given): Air, k g,a = 0.0263 W/m⋅K; Paste, k g,p = 0.12 W/m⋅K; Metal leads, k " = 25 W/m⋅K. ANALYSIS: (a) Define the transistor as the system and identify modes of heat transfer.

E in − E out + E g = ∆E st = 0

−q conv − q cond,gap − 3q lead + E g = 0 T − Tb T − Tb  − hAs ( Tc − T∞ ) − k g As c − 3k " Ac c + Eg = 0 t L where As = L1 × L 2 = 4 × 8 mm2 = 32 × 10-6 m2 and A c = t × w = 0.25 × 1 mm2 = 25 × 10-8 m2. Rearranging and solving for Tc ,

{

}

Tc = hAs T∞ +  k g As /t + 3 ( k " Ac /L ) Tb + E g /  hAs + k g As /t + 3 ( k " Ac /L ) Substituting numerical values, with the air-gap condition ( k g,a = 0.0263 W/m⋅K)

{

(

)

Tc = 50W/m 2 ⋅ K × 32 × 10−6 m 2 × 20$ C +  0.0263W/m ⋅ K × 32 × 10−6 m 2 /0.2 × 10−3 m  $ −8 2 −3 −3 −3 −3 +3 25 W/m ⋅ K × 25 × 10 m /4 × 10 m  35 C / 1.600 × 10 + 4.208 × 10 + 4.688 × 10  W/K   

(

Tc = 47.0$ C .

)

}

< Continued.….

PROBLEM 1.71 (Cont.) With the paste condition ( k g,p = 0.12 W/m⋅K), Tc = 39.9°C. As expected, the effect of the conductive paste is to improve the coupling between the circuit board and the case. Hence, Tc decreases.

Power dissipation, Edotg(W)

(b) Using the keyboard to enter model equations into the workspace, IHT has been used to perform the desired calculations. For values of k " = 200 and 400 W/m⋅K and convection coefficients in the range from 50 to 250 W/m2⋅K, the energy balance equation may be used to compute the power dissipation for a maximum allowable case temperature of 40°C. 0.7

0.6

0.5

0.4

0.3 50

100

150

200

250

Convection coefficient, h(W/m^2.K) kl = 400 W/m.K kl = 200 W/m.K

As indicated by the energy balance, the power dissipation increases linearly with increasing h, as well as with increasing k " . For h = 250 W/m2⋅K (enhanced air cooling) and k " = 400 W/m⋅K (copper leads), the transistor may dissipate up to 0.63 W. COMMENTS: Additional benefits may be derived by increasing heat transfer across the gap separating the case from the board, perhaps by inserting a highly conductive material in the gap.

PROBLEM 1.72(a) KNOWN: Solar radiation is incident on an asphalt paving. FIND: Relevant heat transfer processes. SCHEMATIC:

The relevant processes shown on the schematic include: q S′′

Incident solar radiation, a large portion of which q S,abs ′′ , is absorbed by the asphalt surface,

q ′′rad

Radiation emitted by the surface to the air,

q conv Convection heat transfer from the surface to the air, and ′′ Conduction heat transfer from the surface into the asphalt. q cond ′′ Applying the surface energy balance, Eq. 1.12, q S,abs − q ′′rad − q conv = q cond ′′ ′′ ′′ . COMMENTS: (1) q cond and q conv could be evaluated from Eqs. 1.1 and 1.3, respectively. ′′ ′′ (2) It has been assumed that the pavement surface temperature is higher than that of the underlying pavement and the air, in which case heat transfer by conduction and convection are from the surface. (3) For simplicity, radiation incident on the pavement due to atmospheric emission has been ignored (see Section 12.8 for a discussion). Eq. 1.6 may then be used for the absorbed solar irradiation and Eq. 1.5 may be used to obtain the emitted radiation q ′′rad . (4) With the rate equations, the energy balance becomes ′′ qS,abs − ε σ Ts4 − h ( Ts − T∞ ) = − k

dT  . dx s

PROBLEM 1.72(b) KNOWN: Physical mechanism for microwave heating. FIND: Comparison of (a) cooking in a microwave oven with a conventional radiant or convection oven and (b) a microwave clothes dryer with a conventional dryer. (a) Microwave cooking occurs as a result of volumetric thermal energy generation throughout the food, without heating of the food container or the oven wall. Conventional cooking relies on radiant heat transfer from the oven walls and/or convection heat transfer from the air space to the surface of the food and subsequent heat transfer by conduction to the core of the food. Microwave cooking is more efficient and is achieved in less time. (b) In a microwave dryer, the microwave radiation would heat the water, but not the fabric, directly (the fabric would be heated indirectly by energy transfer from the water). By heating the water, energy would go directly into evaporation, unlike a conventional dryer where the walls and air are first heated electrically or by a gas heater, and thermal energy is subsequently transferred to the wet clothes. The microwave dryer would still require a rotating drum and air flow to remove the water vapor, but is able to operate more efficiently and at lower temperatures. For a more detailed description of microwave drying, see Mechanical Engineering, March 1993, page 120.

PROBLEM 1.72(c) KNOWN: Surface temperature of exposed arm exceeds that of the room air and walls. FIND: Relevant heat transfer processes. SCHEMATIC:

Neglecting evaporation from the surface of the skin, the only relevant heat transfer processes are: q conv

Convection heat transfer from the skin to the room air, and

q rad

Net radiation exchange between the surface of the skin and the surroundings (walls of the room).

You are not imagining things. Even though the room air is maintained at a fixed temperature (T∞ = 15°C), the inner surface temperature of the outside walls, Tsur, will decrease with decreasing outside air temperature. Upon exposure to these walls, body heat loss will be larger due to increased qrad. COMMENTS: The foregoing reasoning assumes that the thermostat measures the true room air temperature and is shielded from radiation exchange with the outside walls.

PROBLEM 1.72(d) KNOWN: Tungsten filament is heated to 2900 K in an air-filled glass bulb. FIND: Relevant heat transfer processes. SCHEMATIC:

The relevant processes associated with the filament and bulb include: q rad,f

Radiation emitted by the tungsten filament, a portion of which is transmitted through the glass,

q conv,f

Free convection from filament to air of temperature Ta,i < Tf ,

q rad,g,i

Radiation emitted by inner surface of glass, a small portion of which is intercepted by the filament,

q conv,g,i

Free convection from air to inner glass surface of temperature Tg,i < Ta,i ,

q cond,g

Conduction through glass wall,

q conv,g,o

Free convection from outer glass surface to room air of temperature Ta,o < Tg,o , and

q rad,g-sur

Net radiation heat transfer between outer glass surface and surroundings, such as the walls of a room, of temperature Tsur < Tg,o .

COMMENTS: If the glass bulb is evacuated, no convection is present within the bulb; that is, q conv,f = q conv,g,i = 0.

PROBLEM 1.72(e) KNOWN: Geometry of a composite insulation consisting of a honeycomb core. FIND: Relevant heat transfer processes. SCHEMATIC:

The above schematic represents the cross section of a single honeycomb cell and surface slabs. Assumed direction of gravity field is downward. Assuming that the bottom (inner) surface temperature exceeds the top (outer) surface temperature Ts,i > Ts,o , heat transfer is

(

)

in the direction shown. Heat may be transferred to the inner surface by convection and radiation, whereupon it is transferred through the composite by q cond,i

Conduction through the inner solid slab,

q conv,hc

Free convection through the cellular airspace,

q cond,hc

Conduction through the honeycomb wall,

q rad,hc

Radiation between the honeycomb surfaces, and

q cond,o

Conduction through the outer solid slab.

Heat may then be transferred from the outer surface by convection and radiation. Note that for a single cell under steady state conditions, q rad,i + q conv,i = q cond,i = q conv,hc + q cond,hc +q rad,hc = q cond,o = q rad,o + q conv,o . COMMENTS: Performance would be enhanced by using materials of low thermal conductivity, k, and emissivity, ε. Evacuating the airspace would enhance performance by eliminating heat transfer due to free convection.

PROBLEM 1.72(f) KNOWN: A thermocouple junction is used, with or without a radiation shield, to measure the temperature of a gas flowing through a channel. The wall of the channel is at a temperature much less than that of the gas. FIND: (a) Relevant heat transfer processes, (b) Temperature of junction relative to that of gas, (c) Effect of radiation shield. SCHEMATIC:

ASSUMPTIONS: (1) Junction is small relative to channel walls, (2) Steady-state conditions, (3) Negligible heat transfer by conduction through the thermocouple leads. ANALYSIS: (a) The relevant heat transfer processes are: q rad

Net radiation transfer from the junction to the walls, and

q conv

Convection transfer from the gas to the junction.

(b) From a surface energy balance on the junction, q conv = q rad or from Eqs. 1.3a and 1.7,

(

)

(

)

h A Tj − Tg = ε A σ Tj4 − Ts4 . To satisfy this equality, it follows that Ts < Tj < Tg . That is, the junction assumes a temperature between that of the channel wall and the gas, thereby sensing a temperature which is less than that of the gas.

(

)

(c) The measurement error Tg − Tj is reduced by using a radiation shield as shown in the schematic. The junction now exchanges radiation with the shield, whose temperature must exceed that of the channel wall. The radiation loss from the junction is therefore reduced, and its temperature more closely approaches that of the gas.

PROBLEM 1.72(g) KNOWN: Fireplace cavity is separated from room air by two glass plates, open at both ends. FIND: Relevant heat transfer processes. SCHEMATIC:

The relevant heat transfer processes associated with the double-glazed, glass fire screen are: q rad,1

Radiation from flames and cavity wall, portions of which are absorbed and transmitted by the two panes,

q rad,2

Emission from inner surface of inner pane to cavity,

q rad,3

Net radiation exchange between outer surface of inner pane and inner surface of outer pane,

q rad,4

Net radiation exchange between outer surface of outer pane and walls of room,

q conv,1

Convection between cavity gases and inner pane,

q conv2

Convection across air space between panes,

q conv,3

Convection from outer surface to room air,

q cond,1

Conduction across inner pane, and

q cond,2

Conduction across outer pane.

COMMENTS: (1) Much of the luminous portion of the flame radiation is transmitted to the room interior. (2) All convection processes are buoyancy driven (free convection).

PROBLEM 1.73(a) KNOWN: Room air is separated from ambient air by one or two glass panes. FIND: Relevant heat transfer processes. SCHEMATIC:

The relevant processes associated with single (above left schematic) and double (above right schematic) glass panes include. q conv,1

Convection from room air to inner surface of first pane,

q rad,1

Net radiation exchange between room walls and inner surface of first pane,

q cond,1

Conduction through first pane,

q conv,s

Convection across airspace between panes,

q rad,s

Net radiation exchange between outer surface of first pane and inner surface of second pane (across airspace),

q cond,2

Conduction through a second pane,

q conv,2

Convection from outer surface of single (or second) pane to ambient air,

q rad,2

Net radiation exchange between outer surface of single (or second) pane and surroundings such as the ground, and

qS

Incident solar radiation during day; fraction transmitted to room is smaller for double pane.

COMMENTS: Heat loss from the room is significantly reduced by the double pane construction.

PROBLEM 1.73(b) KNOWN: Configuration of a flat plate solar collector. FIND: Relevant heat transfer processes with and without a cover plate. SCHEMATIC:

The relevant processes without (above left schematic) and with (above right schematic) include: qS

Incident solar radiation, a large portion of which is absorbed by the absorber plate. Reduced with use of cover plate (primarily due to reflection off cover plate).

q rad,∞

Net radiation exchange between absorber plate or cover plate and surroundings,

q conv,∞

Convection from absorber plate or cover plate to ambient air,

q rad,a-c

Net radiation exchange between absorber and cover plates,

q conv,a-c

Convection heat transfer across airspace between absorber and cover plates,

q cond

Conduction through insulation, and

q conv

Convection to working fluid.

COMMENTS: The cover plate acts to significantly reduce heat losses by convection and radiation from the absorber plate to the surroundings.

PROBLEM 1.73(c) KNOWN: Configuration of a solar collector used to heat air for agricultural applications. FIND: Relevant heat transfer processes. SCHEMATIC:

Assume the temperature of the absorber plates exceeds the ambient air temperature. At the cover plates, the relevant processes are: q conv,a-i

Convection from inside air to inner surface,

q rad,p-i

Net radiation transfer from absorber plates to inner surface,

q conv,i-o

Convection across airspace between covers,

q rad,i-o

Net radiation transfer from inner to outer cover,

q conv,o-∞

Convection from outer cover to ambient air,

q rad,o

Net radiation transfer from outer cover to surroundings, and

qS

Incident solar radiation.

Additional processes relevant to the absorber plates and airspace are: q S,t

Solar radiation transmitted by cover plates,

q conv,p-a

Convection from absorber plates to inside air, and

q cond

Conduction through insulation.

PROBLEM 1.73(d) KNOWN: Features of an evacuated tube solar collector. FIND: Relevant heat transfer processes for one of the tubes. SCHEMATIC:

The relevant heat transfer processes for one of the evacuated tube solar collectors includes: qS

Incident solar radiation including contribution due to reflection off panel (most is transmitted),

q conv,o

Convection heat transfer from outer surface to ambient air,

q rad,o-sur

Net rate of radiation heat exchange between outer surface of outer tube and the surroundings, including the panel,

q S,t

Solar radiation transmitted through outer tube and incident on inner tube (most is absorbed),

q rad,i-o

Net rate of radiation heat exchange between outer surface of inner tube and inner surface of outer tube, and

q conv,i

Convection heat transfer to working fluid.

There is also conduction heat transfer through the inner and outer tube walls. If the walls are thin, the temperature drop across the walls will be small.

PROBLEM 2.1 KNOWN: Steady-state, one-dimensional heat conduction through an axisymmetric shape. FIND: Sketch temperature distribution and explain shape of curve. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Constant properties, (3) No internal heat generation.

 − E ANALYSIS: Performing an energy balance on the object according to Eq. 1.11a, E in out = 0, it follows that E in − E out = q x

$

and that q x ≠ q x x . That is, the heat rate within the object is everywhere constant. From Fourier’s law,

q x = − kA x

dT , dx

and since qx and k are both constants, it follows that

Ax

dT = Constant. dx

That is, the product of the cross-sectional area normal to the heat rate and temperature gradient remains a constant and independent of distance x. It follows that since Ax increases with x, then dT/dx must decrease with increasing x. Hence, the temperature distribution appears as shown above. COMMENTS: (1) Be sure to recognize that dT/dx is the slope of the temperature distribution. (2) What would the distribution be when T2 > T1? (3) How does the heat flux, q ′′x , vary with distance?

PROBLEM 2.2 KNOWN: Hot water pipe covered with thick layer of insulation. FIND: Sketch temperature distribution and give brief explanation to justify shape. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial) conduction, (3) No internal heat generation, (4) Insulation has uniform properties independent of temperature and position. ANALYSIS: Fourier’s law, Eq. 2.1, for this one-dimensional (cylindrical) radial system has the form

q r = − kA r

1 6

dT dT = − k 2πr dr dr

where A r = 2πr and  is the axial length of the pipe-insulation system. Recognize that for steadystate conditions with no internal heat generation, an energy balance on the system requires E in = E out since E g = E st = 0. Hence

qr = Constant. That is, qr is independent of radius (r). Since the thermal conductivity is also constant, it follows that

r

 dT "# = Constant. ! dr $

This relation requires that the product of the radial temperature gradient, dT/dr, and the radius, r, remains constant throughout the insulation. For our situation, the temperature distribution must appear as shown in the sketch. COMMENTS: (1) Note that, while qr is a constant and independent of r, q ′′r is not a constant. How

16

does q ′′r r vary with r? (2) Recognize that the radial temperature gradient, dT/dr, decreases with increasing radius.

PROBLEM 2.3 KNOWN: A spherical shell with prescribed geometry and surface temperatures. FIND: Sketch temperature distribution and explain shape of the curve. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in radial (spherical coordinates) direction, (3) No internal generation, (4) Constant properties. ANALYSIS: Fourier’s law, Eq. 2.1, for this one-dimensional, radial (spherical coordinate) system has the form

qr = −k Ar

(

)

dT dT = − k 4π r 2 dr dr

where Ar is the surface area of a sphere. For steady-state conditions, an energy balance on the system

 = E , since E = E = 0. Hence, yields E in out g st qin = q out = q r ≠ q r ( r ) .

That is, qr is a constant, independent of the radial coordinate. Since the thermal conductivity is constant, it follows that

 dT  r 2   = Constant.  dr  This relation requires that the product of the radial temperature gradient, dT/dr, and the radius 2 squared, r , remains constant throughout the shell. Hence, the temperature distribution appears as shown in the sketch. COMMENTS: Note that, for the above conditions, q r ≠ q r ( r ) ; that is, qr is everywhere constant. How does q ′′r vary as a function of radius?

PROBLEM 2.4 KNOWN: Symmetric shape with prescribed variation in cross-sectional area, temperature distribution and heat rate. FIND: Expression for the thermal conductivity, k. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction, (3) No internal heat generation. ANALYSIS: Applying the energy balance, Eq. 1.11a, to the system, it follows that, since

E in = E out ,

q x = Constant ≠ f ( x ). Using Fourier’s law, Eq. 2.1, with appropriate expressions for Ax and T, yields

dT dx d K 6000W=-k ⋅ (1-x ) m2 ⋅ 300 1 − 2x-x3  .  m dx  q x = −k A x

)

(

Solving for k and recognizing its units are W/m⋅K,

k=

-6000

(

)

(1-x ) 300 −2 − 3x 2 

=

20

(1 − x )

(

2 + 3x 2

)

.

<

COMMENTS: (1) At x = 0, k = 10W/m⋅K and k → ∞ as x → 1. (2) Recognize that the 1-D assumption is an approximation which becomes more inappropriate as the area change with x, and hence two-dimensional effects, become more pronounced.

PROBLEM 2.5 KNOWN: End-face temperatures and temperature dependence of k for a truncated cone. FIND: Variation with axial distance along the cone of q x , q ′′x , k, and dT / dx. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in x (negligible temperature gradients along y), (2) Steady-state conditions, (3) Adiabatic sides, (4) No internal heat generation. ANALYSIS: For the prescribed conditions, it follows from conservation of energy, Eq. 1.11a, that  = E for a differential control volume, E in out or q x = q x+dx . Hence

qx is independent of x.

$

Since A(x) increases with increasing x, it follows that q ′′x = q x / A x decreases with increasing x. Since T decreases with increasing x, k increases with increasing x. Hence, from Fourier’s law, Eq. 2.2,

q ′′x = − k

dT , dx

it follows that | dT/dx | decreases with increasing x.

PROBLEM 2.6 KNOWN: Temperature dependence of the thermal conductivity, k(T), for heat transfer through a plane wall. FIND: Effect of k(T) on temperature distribution, T(x). ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) No internal heat generation. ANALYSIS: From Fourier’s law and the form of k(T),

q ′′x = − k

1

6

dT dT = − k o + aT . dx dx

(1)

2 2 The shape of the temperature distribution may be inferred from knowledge of d T/dx = d(dT/dx)/dx. Since q ′′x is independent of x for the prescribed conditions,

1 !

6 "#$ 2 d 2 T  dT " −1 k o + aT6 2 − a ! dx #$ = 0. dx

dq ′′x d dT k o + aT =0 =dx dx dx

Hence,

d 2T

 "#2 ! $

-a dT = 2 k o + aT dx dx

%Kk o + aT = k > 0 where & dT " 2 K'! dx #$ > 0

from which it follows that for

a > 0: d 2 T / dx 2 < 0 a = 0: d 2 T / dx 2 = 0 a < 0: d 2 T / dx 2 > 0.

COMMENTS: The shape of the distribution could also be inferred from Eq. (1). Since T decreases with increasing x, a > 0: k decreases with increasing x = > | dT/dx | increases with increasing x a = 0: k = ko = > dT/dx is constant a < 0: k increases with increasing x = > | dT/dx | decreases with increasing x.

PROBLEM 2.7 KNOWN: Thermal conductivity and thickness of a one-dimensional system with no internal heat generation and steady-state conditions. FIND: Unknown surface temperatures, temperature gradient or heat flux. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat flow, (2) No internal heat generation, (3) Steady-state conditions, (4) Constant properties. ANALYSIS: The rate equation and temperature gradient for this system are dT dT T1 − T2 = q′′x = −k and . dx dx L Using Eqs. (1) and (2), the unknown quantities can be determined. dT ( 400 − 300 ) K (a) = = 200 K/m dx 0.5m q′′x = −25

(b)

W m⋅K

× 200

K m

<

= −5000 W/m 2 .

K  = 6250 W/m 2 m⋅K  m  K  dT   T2 = T1 − L   = 1000$ C-0.5m  -250  m  dx   q′′x = −25

W

×  −250

T2 = 225$ C.

(c) q′′x = −25

W m⋅K

× 200

< K m

 

= −5000 W/m 2

T2 = 80$ C-0.5m  200

(d)

K

= −20$ C.  m

<

q′′ 4000 W/m 2 K =− x =− = −160 dx k 25 W/m ⋅ K m

dT

T1 = L

( )

 dT  + T = 0.5m -160 K  + −5$ C  dx  2  m 

T1 = −85$ C.

(e)

(1,2)

dT

=−

q′′x

−3000 W/m 2 ) ( K =− = 120

25 W/m ⋅ K K  T2 = 30$ C-0.5m 120  = −30$ C. m  dx

k

m

2

a +b

2

<

PROBLEM 2.8 KNOWN: One-dimensional system with prescribed thermal conductivity and thickness. FIND: Unknowns for various temperature conditions and sketch distribution. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heat generation, (4) Constant properties. ANALYSIS: The rate equation and temperature gradient for this system are dT dT T2 − T1 q′′x = −k and . = dx dx L Using Eqs. (1) and (2), the unknown quantities for each case can be determined.

(a)

(b)

(c)

dT

( −20 − 50 ) K

= −280 K/m 0.25m W K  q′′x = −50 ×  −280  = 14.0 kW/m 2 . m⋅K  m dx

dT

=

q′′x = −50 T2 = L ⋅

 m⋅K  W

dT dx

× 160

K

m 

<

= −8.0 kW/m 2

 

+ T1 = 0.25m × 160

K

+ 70$ C.  m

<

T2 = 110$ C.

(d)

q′′x = −50

 m⋅K  W

T1 = T2 − L ⋅

×  −80

dT dx

K

m 

= 4.0 kW/m 2

 

= 40$ C − 0.25m  −80

K

m 

.∈

T1 = 60$ C. q′′x = −50

<

 m⋅K  W

×  200

K

2

= −10.0 kW/m m  dT K  = 30$ C − 0.25m  200  = −20$ C. T1 = T2 − L ⋅ dx m 

(e)

<

( −10 − ( −30 )) K

= 80 K/m 0.25m W  K × 80  = −4.0 kW/m 2 . q′′x = −50 m⋅K  m dx

=

(1,2)

<

PROBLEM 2.9 KNOWN: Plane wall with prescribed thermal conductivity, thickness, and surface temperatures. FIND: Heat flux, q ′′x , and temperature gradient, dT/dx, for the three different coordinate systems shown. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat flow, (2) Steady-state conditions, (3) No internal generation, (4) Constant properties. ANALYSIS: The rate equation for conduction heat transfer is

q ′′x = − k

dT , dx

(1)

where the temperature gradient is constant throughout the wall and of the form

 $ $

dT T L − T 0 = . dx L

(2)

Substituting numerical values, find the temperature gradients,



$

<

400 − 600 K dT T1 − T2 = = = −2000 K / m dx L 0.100m



$

<

600 − 400 K dT T2 − T1 = = = 2000 K / m. dx L 0.100m



$

<

(a)

600 − 400 K dT T2 − T1 = = = 2000 K / m dx L 0.100m

(b) (c)

The heat rates, using Eq. (1) with k = 100 W/m⋅K, are

(a)

q ′′x = −100

W × 2000 K / m = -200 kW / m2 m⋅ K

<

(b)

q ′′x = −100

W ( −2000 K / m) = +200 kW / m2 m⋅ K

<

(c)

q ′′x = −100

W × 2000 K / m = -200 kW / m2 m⋅ K

<

PROBLEM 2.10 KNOWN: Temperature distribution in solid cylinder and convection coefficient at cylinder surface. FIND: Expressions for heat rate at cylinder surface and fluid temperature. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: The heat rate from Fourier’s law for the radial (cylindrical) system has the form

q r = − kA r

dT . dr

2 Substituting for the temperature distribution, T(r) = a + br ,

1 6

q r = − k 2πrL 2br = - 4πkbLr 2 . At the outer surface ( r = ro), the conduction heat rate is

q r=ro = −4πkbLro2 .

<

From a surface energy balance at r = ro,

1

6 1 6

q r=ro = q conv = h 2πro L T ro − T∞ , Substituting for q r=ro and solving for T∞,

1 6

T∞ = T ro +

2 kbro h

T∞ = a + bro2 +

 !

2 kbro h

T∞ = a + bro ro +

"# $

2k . h

<

PROBLEM 2.11 KNOWN: Two-dimensional body with specified thermal conductivity and two isothermal surfaces of prescribed temperatures; one surface, A, has a prescribed temperature gradient. FIND: Temperature gradients, ∂T/∂x and ∂T/∂y, at the surface B. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) No heat generation, (4) Constant properties. ANALYSIS: At the surface A, the temperature gradient in the x-direction must be zero. That is, (∂T/∂x)A = 0. This follows from the requirement that the heat flux vector must be normal to an isothermal surface. The heat rate at the surface A is given by Fourier’s law written as

q ′y,A = − k ⋅ w A

 

W K ∂T = −10 × 2 m × 30 = −600W / m. m⋅ K m ∂y A

On the surface B, it follows that

∂T / ∂y$B = 0

<

in order to satisfy the requirement that the heat flux vector be normal to the isothermal surface B. Using the conservation of energy requirement, Eq. 1.11a, on the body, find

q ′y,A − q ′x,B = 0 Note that,

q ′x,B = − k ⋅ w B

or

q ′x,B = q ′y,A .

 

∂T ∂x B

and hence

W / m$ = 60 K / m. ∂T / ∂x$B = −kq⋅ w′y,AB = 10−W−600 / m ⋅ K × 1m

<

COMMENTS: Note that, in using the conservation requirement, q ′in = + q ′y,A and q ′out = + q ′x,B .

PROBLEM 2.12 KNOWN: Length and thermal conductivity of a shaft. Temperature distribution along shaft. FIND: Temperature and heat rates at ends of shaft. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x, (3) Constant properties. ANALYSIS: Temperatures at the top and bottom of the shaft are, respectively,

T(0) = 100°C

<

T(L) = -40°C.

Applying Fourier’s law, Eq. 2.1,

"

'

$

dT $ = −25 W / m ⋅ K 0.005 m2 −150 + 20x C / m dx q x = 0125 150 - 20x W. . q x = − kA



$

Hence,

qx(0) = 18.75 W

qx(L) = 16.25 W.

<

The difference in heat rates, qx(0) > qx(L), is due to heat losses q " from the side of the shaft. COMMENTS: Heat loss from the side requires the existence of temperature gradients over the shaft cross-section. Hence, specification of T as a function of only x is an approximation.

PROBLEM 2.13 KNOWN: A rod of constant thermal conductivity k and variable cross-sectional area Ax(x) = Aoeax where Ao and a are constants. FIND: (a) Expression for the conduction heat rate, qx(x); use this expression to determine the temperature distribution, T(x); and sketch of the temperature distribution, (b) Considering the presence of volumetric heat generation rate, q = q o exp ( −ax ) , obtain an expression for qx(x) when the left face, x = 0, is well insulated. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the rod, (2) Constant properties, (3) Steadystate conditions. ANALYSIS: Perform an energy balance on the control volume, A(x)⋅dx,

E in − E out + E g = 0 q x − q x + dx + q ⋅ A ( x ) ⋅ dx = 0 The conduction heat rate terms can be expressed as a Taylor series and substituting expressions for q and A(x),

−

d (q x ) + q o exp ( −ax ) ⋅ Ao exp (ax ) = 0 dx

q x = −k ⋅ A ( x )

(1)

dT dx

(2)

(a) With no internal generation, q o = 0, and from Eq. (1) find

−

d (q x ) = 0 dx

<

indicating that the heat rate is constant with x. By combining Eqs. (1) and (2)

−

d  dT   −k ⋅ A ( x )  = 0 dx  dx 

or

A (x )⋅

dT = C1 dx

(3)

<

Continued...

PROBLEM 2.13 (Cont.) That is, the product of the cross-sectional area and the temperature gradient is a constant, independent of x. Hence, with T(0) > T(L), the temperature distribution is exponential, and as shown in the sketch above. Separating variables and integrating Eq. (3), the general form for the temperature distribution can be determined,

Ao exp ( ax ) ⋅

dT = C1 dx

dT = C1Ao−1 exp ( − ax ) dx T ( x ) = −C1Ao a exp ( −ax ) + C2

<

We could use the two temperature boundary conditions, To = T(0) and TL = T(L), to evaluate C1 and C2 and, hence, obtain the temperature distribution in terms of To and TL. (b) With the internal generation, from Eq. (1),

−

d (q x ) + q o Ao = 0 dx

or

q x = q o Ao x

<

That is, the heat rate increases linearly with x. COMMENTS: In part (b), you could determine the temperature distribution using Fourier’s law and knowledge of the heat rate dependence upon the x-coordinate. Give it a try!

PROBLEM 2.14 KNOWN: Dimensions and end temperatures of a cylindrical rod which is insulated on its side. FIND: Rate of heat transfer associated with different rod materials. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction along cylinder axis, (2) Steady-state conditions, (3) Constant properties. PROPERTIES: The properties may be evaluated from Tables A-1 to A-3 at a mean temperature of 50°C = 323K and are summarized below. ANALYSIS: The heat transfer rate may be obtained from Fourier’s law. Since the axial temperature gradient is linear, this expression reduces to



$ 100 − 0$° C = 0.491m⋅° C$ ⋅ k

π 0.025m T −T q = kA 1 2 = k L 4

2

0.1m

Cu Al St.St. SiN Oak Magnesia Pyrex (pure) (2024) (302) (85%) _______________________________________________________________ k(W/m⋅K)

401

177

16.3

14.9

0.19

0.052

1.4

q(W)

197

87

8.0

7.3

0.093

0.026

0.69

<

COMMENTS: The k values of Cu and Al were obtained by linear interpolation; the k value of St.St. was obtained by linear extrapolation, as was the value for SiN; the value for magnesia was obtained by linear interpolation; and the values for oak and pyrex are for 300 K.

PROBLEM 2.15 KNOWN: One-dimensional system with prescribed surface temperatures and thickness. FIND: Heat flux through system constructed of these materials: (a) pure aluminum, (b) plain carbon steel, (c) AISI 316, stainless steel, (d) pyroceram, (e) teflon and (f) concrete. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No heat generation, (4) Constant thermal properties. PROPERTIES: The thermal conductivity is evaluated at the average temperature of the system, T = (T1+T2)/2 = (325+275)K/2 = 300K. Property values and table identification are shown below. ANALYSIS: For this system, Fourier’s law can be written as

q ′′x = − k

dT T −T = −k 2 1 . dx L

Substituting numerical values, the heat flux is

q ′′x = − k

275- 325$K = +2500 K ⋅ k 20 × 10-3 m

m

2 where q ′′x will have units W/m if k has units W/m⋅K. The heat fluxes for each system follow.

Thermal conductivity Material

(a) (b) (c) (d) (e) (f)

Pure Aluminum Plain carbon steel AISI 316, S.S. Pyroceram Teflon Concrete

Table A-1 A-1 A-1 A-2 A-3 A-3

k(W/m⋅K) 237 60.5 13.4 3.98 0.35 1.4

Heat flux

"

q ′′x kW / m2 593 151 33.5 9.95 0.88 3.5

' <

COMMENTS: Recognize that the thermal conductivity of these solid materials varies by more than two orders of magnitude.

PROBLEM 2.16 KNOWN: Different thicknesses of three materials: rock, 18 ft; wood, 15 in; and fiberglass insulation, 6 in. FIND: The insulating quality of the materials as measured by the R-value. PROPERTIES: Table A-3 (300K): Material

Thermal conductivity, W/m⋅K

Limestone Softwood Blanket (glass, fiber 10 kg/m3)

2.15 0.12 0.048

ANALYSIS: The R-value, a quantity commonly used in the construction industry and building technology, is defined as

R≡

 $

L in

"

'

k Btu ⋅ in / h ⋅ ft 2 ⋅$ F

.

2 The R-value can be interpreted as the thermal resistance of a 1 ft cross section of the material. Using the conversion factor for thermal conductivity between the SI and English systems, the R-values are:

Rock, Limestone, 18 ft:

in −1 ft R= = 14.5 Btu/h ⋅ ft 2 ⋅$ F W Btu/h ⋅ ft ⋅$ F in 2.15 × 0.5778 ×12 m⋅K W/m ⋅ K ft 18 ft × 12

(

)

"

'

Wood, Softwood, 15 in:

15 in

R= 0.12

W Btu / h ⋅ ft ⋅$ F in × 0.5778 × 12 m⋅ K W / m⋅ K ft

= 18 Btu / h ⋅ ft 2 ⋅$ F

−1

Insulation, Blanket, 6 in:

6 in

R= 0.048

W Btu / h ⋅ ft ⋅$ F in × 0.5778 × 12 m⋅ K W / m⋅ K ft

"

'

= 18 Btu / h ⋅ ft 2 ⋅$ F

COMMENTS: The R-value of 19 given in the advertisement is reasonable.

−1

PROBLEM 2.17 KNOWN: Electrical heater sandwiched between two identical cylindrical (30 mm dia. × 60 mm length) samples whose opposite ends contact plates maintained at To. FIND: (a) Thermal conductivity of SS316 samples for the prescribed conditions (A) and their average temperature, (b) Thermal conductivity of Armco iron sample for the prescribed conditions (B), (c) Comment on advantages of experimental arrangement, lateral heat losses, and conditions for which ∆T1 ≠ ∆T2. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat transfer in samples, (2) Steady-state conditions, (3) Negligible contact resistance between materials.

%

PROPERTIES: Table A.2, Stainless steel 316 T = 400 K : k ss = 15.2 W / m ⋅ K; Armco iron

%

T = 380 K : k iron = 716 . W / m ⋅ K.

ANALYSIS: (a) For Case A recognize that half the heater power will pass through each of the samples which are presumed identical. Apply Fourier’s law to a sample

q = kA c k=

∆T ∆x



$

0.5 100V × 0.353A × 0.015 m q∆x = = 15.0 W / m ⋅ K. 2 A c ∆T π 0.030 m / 4 × 25.0$ C



$

<

The total temperature drop across the length of the sample is ∆T1(L/∆x) = 25°C (60 mm/15 mm) = 100°C. Hence, the heater temperature is Th = 177°C. Thus the average temperature of the sample is



$

T = To + Th / 2 = 127° C = 400 K

<

.

We compare the calculated value of k with the tabulated value (see above) at 400 K and note the good agreement. (b) For Case B, we assume that the thermal conductivity of the SS316 sample is the same as that found in Part (a). The heat rate through the Armco iron sample is Continued …..

PROBLEM 2.17 (CONT.) q iron q iron



$

π 0.030 m 15.0° C = q heater − q ss = 100V × 0.601A − 15.0 W / m ⋅ K × × 4 0.015 m = 601 . − 10.6 W = 49.5 W



2

$

where

q ss = k ssA c ∆T2 / ∆x2 . Applying Fourier’s law to the iron sample,

k iron =

49.5 W × 0.015 m q iron ∆x2 = = 70.0 W / m ⋅ K. 2 A c ∆T2 π 0.030 m / 4 × 15.0° C



$

<

The total drop across the iron sample is 15°C(60/15) = 60°C; the heater temperature is (77 + 60)°C = 137°C. Hence the average temperature of the iron sample is



$

T = 137 + 77 ° C / 2 = 107° C = 380 K.

<

We compare the computed value of k with the tabulated value (see above) at 380 K and note the good agreement. (c) The principal advantage of having two identical samples is the assurance that all the electrical power dissipated in the heater will appear as equivalent heat flows through the samples. With only one sample, heat can flow from the backside of the heater even though insulated. Heat leakage out the lateral surfaces of the cylindrically shaped samples will become significant when the sample thermal conductivity is comparable to that of the insulating material. Hence, the method is suitable for metallics, but must be used with caution on nonmetallic materials. For any combination of materials in the upper and lower position, we expect ∆T1 = ∆T2. However, if the insulation were improperly applied along the lateral surfaces, it is possible that heat leakage will occur, causing ∆T1 ≠ ∆T2.

PROBLEM 2.18 KNOWN: Comparative method for measuring thermal conductivity involving two identical samples stacked with a reference material. FIND: (a) Thermal conductivity of test material and associated temperature, (b) Conditions for which ∆Tt,1 ≠ ∆Tt,2 SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer through samples and reference material, (3) Negligible thermal contact resistance between materials.

%

PROPERTIES: Table A.2, Armco iron T = 350 K : k r = 69.2 W / m ⋅ K. ANALYSIS: (a) Recognizing that the heat rate through the samples and reference material, all of the same diameter, is the same, it follows from Fourier’s law that

kt

∆Tt,1 ∆x

kt = kr

= kr

∆Tt,2 ∆Tr = kt ∆x ∆x

∆Tr 2.49° C = 69.2 W / m ⋅ K = 519 . W / m ⋅ K. ∆Tt 3.32° C

We should assign this value a temperature of 350 K.

< <

(b) If the test samples are identical in every respect, ∆Tt,1 ≠ ∆Tt,2 if the thermal conductivity is highly dependent upon temperature. Also, if there is heat leakage out the lateral surface, we can expect ∆Tt,2 < ∆Tt,1. Leakage could be influential, if the thermal conductivity of the test material were less than an order of magnitude larger than that of the insulating material.

PROBLEM 2.19 KNOWN: Identical samples of prescribed diameter, length and density initially at a uniform temperature Ti, sandwich an electric heater which provides a uniform heat flux q ′′o for a period of time ∆to. Conditions shortly after energizing and a long time after de-energizing heater are prescribed. FIND: Specific heat and thermal conductivity of the test sample material. From these properties, identify type of material using Table A.1 or A.2. SCHEMATIC:

ASSUMPTIONS: (1) One dimensional heat transfer in samples, (2) Constant properties, (3) Negligible heat loss through insulation, (4) Negligible heater mass. ANALYSIS: Consider a control volume about the samples and heater, and apply conservation of energy over the time interval from t = 0 to ∞

E in − E out = ∆E = E f − E i

16

P∆t o − 0 = Mc p T ∞ − Ti where energy inflow is prescribed by the Case A power condition and the final temperature Tf by Case B. Solving for cp,

cp =

P∆t o 15 W × 120 s = 3 M T ∞ − Ti 2 × 3965 kg / m π × 0.0602 / 4 m2 × 0.010 m 33.50 - 23.00 ° C

16

2

7

<

c p = 765 J / kg ⋅ K

2 where M = ρV = 2ρ(πD /4)L is the mass of both samples. For Case A, the transient thermal response of the heater is given by

Continued …..

PROBLEM 2.19 (Cont.) 1/ 2  t " To 1 t 6 − Ti = 2q ′′o ## πρ c k p ! $ 2 t  2q o′′ " k= # πρc p ! To 1 t 6 − Ti $

30 s k= π × 3965 kg / m3 × 765 J / kg ⋅ K

 2 × 2653 W / m "# !124.57 - 23.006° C $ 2

2

= 36.0 W / m ⋅ K

<

where

q o′′ =

P P 15 W = = = 2653 W / m2 . 2 2 2 2A s 2 πD / 4 2 π × 0.060 / 4 m

4

9 4

9

With the following properties now known,

ρ = 3965 kg/m

3

cp = 765 J/kg⋅K

k = 36 W/m⋅K

entries in Table A.1 are scanned to determine whether these values are typical of a metallic material. Consider the following, •

metallics with low ρ generally have higher thermal conductivities,

•

specific heats of both types of materials are of similar magnitude,

•

the low k value of the sample is typical of poor metallic conductors which generally have much higher specific heats,

•

more than likely, the material is nonmetallic.

From Table A.2, the second entry, polycrystalline aluminum oxide, has properties at 300 K corresponding to those found for the samples.

<

PROBLEM 2.20 KNOWN: Temperature distribution, T(x,y,z), within an infinite, homogeneous body at a given instant of time. FIND: Regions where the temperature changes with time. SCHEMATIC:

ASSUMPTIONS: (1) Constant properties of infinite medium and (2) No internal heat generation. ANALYSIS: The temperature distribution throughout the medium, at any instant of time, must satisfy the heat equation. For the three-dimensional cartesian coordinate system, with constant properties and no internal heat generation, the heat equation, Eq. 2.15, has the form

∂ 2T ∂x

2

+

∂ 2T ∂y

2

+

∂ 2T ∂z

2

=

1 ∂T . α ∂t

(1)

If T(x,y,z) satisfies this relation, conservation of energy is satisfied at every point in the medium. Substituting T(x,y,z) into the Eq. (1), first find the gradients, ∂T/∂x, ∂T/∂y, and ∂T/∂z.

1

6

1

6

1

6

1 ∂T ∂ ∂ ∂ 2x - y + 2 z + 2y = −4 y - x + 2z + . ∂x ∂y ∂z α ∂t Performing the differentiations,

2−4+2 =

1 ∂T . α ∂t

Hence,

∂T =0 ∂t which implies that, at the prescribed instant, the temperature is everywhere independent of time. COMMENTS: Since we do not know the initial and boundary conditions, we cannot determine the temperature distribution, T(x,y,z), at any future time. We can only determine that, for this special instant of time, the temperature will not change.

PROBLEM 2.21 KNOWN: Diameter D, thickness L and initial temperature Ti of pan. Heat rate from stove to bottom of pan. Convection coefficient h and variation of water temperature T∞(t) during Stage 1. Temperature TL of pan surface in contact with water during Stage 2. FIND: Form of heat equation and boundary conditions associated with the two stages. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in pan bottom, (2) Heat transfer from stove is uniformly distributed over surface of pan in contact with the stove, (3) Constant properties. ANALYSIS: Stage 1

∂ 2T

Heat Equation:

∂x 2

Boundary Conditions:

Initial Condition:

=

1 ∂T α ∂t

−k

qo ∂T = q′′o = ∂x x = 0 π D2 / 4

−k

∂T = h T ( L, t ) − T∞ ( t ) ∂x x = L

(

)

T ( x, 0 ) = Ti

Stage 2 Heat Equation:

Boundary Conditions:

d 2T dx 2 −k

=0

dT = q′′o dx x = 0

T ( L ) = TL COMMENTS: Stage 1 is a transient process for which T∞(t) must be determined separately. As a first approximation, it could be estimated by neglecting changes in thermal energy storage by the pan bottom and assuming that all of the heat transferred from the stove acted to increase thermal energy storage within the water. Hence, with q ≈ Mcp d T∞/dt, where M and cp are the mass and specific heat of the water in the pan, T∞(t) ≈ (q/Mcp) t.

PROBLEM 2.22 KNOWN: Steady-state temperature distribution in a cylindrical rod having uniform heat generation of q 1 = 5 × 107 W / m3 . FIND: (a) Steady-state centerline and surface heat transfer rates per unit length, q ′r . (b) Initial time rate of change of the centerline and surface temperatures in response to a change in the generation rate from q 1 to q 2 = 108 W / m3 . SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the r direction, (2) Uniform generation, and (3) Steady-state for q 1 = 5 × 107 W / m3 . ANALYSIS: (a) From the rate equations for cylindrical coordinates,

q ′′r = − k

∂T ∂r

q = -kA r

∂ T . ∂r

Hence,

1 6 ∂∂ Tr

q r = − k 2πrL or

q ′r = −2πkr

∂T ∂r

where ∂T/∂r may be evaluated from the prescribed temperature distribution, T(r). At r = 0, the gradient is (∂T/∂r) = 0. Hence, from Eq. (1) the heat rate is

16

<

q ′r 0 = 0. At r = ro, the temperature gradient is

"# $ "# $

 !

"#1 6 $

4

91

6

∂T K = −2 4.167 × 105 2 ro = −2 4.167 × 105 0.025m ∂ r r=r m o ∂T = −0.208 × 105 K / m. ∂ r r=r o

Continued …..

PROBLEM 2.22(Cont.) Hence, the heat rate at the outer surface (r = ro) per unit length is

1 6 1 q ′r 1 ro 6 = 0.980 × 105 W / m.

6

q ′r ro = −2π 30 W / m ⋅ K 0.025m −0.208 × 105 K / m

<

(b) Transient (time-dependent) conditions will exist when the generation is changed, and for the prescribed assumptions, the temperature is determined by the following form of the heat equation, Eq. 2.20

 !

"# $

∂T ∂T 1 ∂ kr + q 2 = ρc p ∂r ∂t r ∂r Hence

 !

 !

"# $

"# $

∂T ∂T 1 1 ∂ kr = + q 2 . ∂ t ρc p r ∂ r ∂r However, initially (at t = 0), the temperature distribution is given by the prescribed form, T(r) = 800 52 4.167×10 r , and

 !

"# $

4

∂T 1 ∂ k ∂ kr r -8.334 × 105 ⋅ r = ∂r r ∂r r ∂r =

4

k −16.668 × 105 ⋅ r r

9

9

= 30 W / m ⋅ K -16.668 × 105 K / m2

1

6

= −5 × 107 W / m3 the original q = q 1 . Hence, everywhere in the wall,

∂T 1 = −5 × 107 + 108 W / m3 3 ∂ t 1100 kg / m × 800 J / kg ⋅ K or

∂T = 56.82 K / s. ∂t

<

COMMENTS: (1) The value of (∂T/∂t) will decrease with increasing time, until a new steady-state condition is reached and once again (∂T/∂t) = 0. (2) By applying the energy conservation requirement, Eq. 1.11a, to a unit length of the rod for the

16 1 6

4 9

 ′in − E ′out + E ′gen = 0. Hence q ′r 0 − q ′r ro = − q 1 πro2 . steady-state condition, E

PROBLEM 2.23 KNOWN: Temperature distribution in a one-dimensional wall with prescribed thickness and thermal conductivity. FIND: (a) The heat generation rate, q , in the wall, (b) Heat fluxes at the wall faces and relation to

q .

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3) Constant properties. ANALYSIS: (a) The appropriate form of the heat equation for steady-state, one-dimensional conditions with constant properties is Eq. 2.15 re-written as

 "# ! $ Substituting the prescribed temperature distribution, d d " d 2bx = −2bk q = -k a + bx 2 9# = − k 4 dx ! dx $ dx q = -24-2000 C / m2 9 × 50 W / m ⋅ K = 2.0 × 105 W / m3 . (b) The heat fluxes at the wall faces can be evaluated from Fourier’s law, dT " q ′′x 1 x6 = − k . dx #$ x q = -k

d dT dx dx

$

<

Using the temperature distribution T(x) to evaluate the gradient, find

16

q ′′x x = − k

d a + bx 2 = −2 kbx. dx

The fluxes at x = 0 and x = L are then

16 q ′′x 1 L6 = −2 kbL = -2 × 50W / m ⋅ K4-2000 C / m2 9 × 0.050m q ′′x 1 L6 = 10,000 W / m2 . q ′′x 0 = 0

<

$

COMMENTS: From an overall energy balance on the wall, it follows that, for a unit area,

E in − E out + E g = 0

16 16

16 1 6

 =0 q ′′x 0 − q ′′x L + qL

q ′′ L − q ′′x 0 10,000 W / m2 − 0 q = x = = 2.0 × 105 W / m3. L 0.050m

<

PROBLEM 2.24 KNOWN: Wall thickness, thermal conductivity, temperature distribution, and fluid temperature. FIND: (a) Surface heat rates and rate of change of wall energy storage per unit area, and (b) Convection coefficient. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Constant k. ANALYSIS: (a) From Fourier’s law,

q ′′x = − k



$

∂T = 200 − 60x ⋅ k ∂x

q ′′in = q ′′x=0 = 200

°C W ×1 = 200 W / m2 m m⋅ K



$

q ′′out = q ′′x=L = 200 − 60 × 0.3 ° C / m × 1 W / m ⋅ K = 182 W / m2 .

< <

Applying an energy balance to a control volume about the wall, Eq. 1.11a,

E ′′in − E ′′out = E st ′′ E ′′st = q ′′in − q ′′out = 18 W / m2 .

<

(b) Applying a surface energy balance at x = L,

$

q ′′out = h T L − T∞ q ′′out 182 W / m2 h= = T L − T∞ 142.7 -100 ° C

$



$

<

h = 4.3 W / m2 ⋅ K. COMMENTS: (1) From the heat equation, 2

2

(∂T/∂t) = (k/ρcp) ∂ T/∂x = 60(k/ρcp), it follows that the temperature is increasing with time at every point in the wall. (2) The value of h is small and is typical of free convection in a gas.

PROBLEM 2.25 KNOWN: Analytical expression for the steady-state temperature distribution of a plane wall experiencing uniform volumetric heat generation q while convection occurs at both of its surfaces. FIND: (a) Sketch the temperature distribution, T(x), and identify significant physical features, (b) Determine q , (c) Determine the surface heat fluxes, q ′′x ( − L ) and q ′′x ( + L ) ; how are these fluxes related to the generation rate; (d) Calculate the convection coefficients at the surfaces x = L and x = +L, (e) Obtain an expression for the heat flux distribution, q ′′x ( x ) ; explain significant features of the distribution; (f) If the source of heat generation is suddenly deactivated ( q = 0), what is the rate of change of energy stored at this instant; (g) Determine the temperature that the wall will reach eventually with q = 0; determine the energy that must be removed by the fluid per unit area of the wall to reach this state. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform volumetric heat generation, (3) Constant properties. ANALYSIS: (a) Using the analytical expression in the Workspace of IHT, the temperature distribution appears as shown below. The significant features include (1) parabolic shape, (2) maximum does not occur at the mid-plane, T(-5.25 mm) = 83.3°C, (3) the gradient at the x = +L surface is greater than at x = -L. Find also that T(-L) = 78.2°C and T(+L) = 69.8°C for use in part (d). Temperature distribution 90

Temperature, T(x) (C)

85

80

75

70 -20

-10

0

10

20

x-coordinate, x (mm)

(b) Substituting the temperature distribution expression into the appropriate form of the heat diffusion equation, Eq. 2.15, the rate of volumetric heat generation can be determined.

d  dT  q  + =0 dx  dx  k

where

T ( x ) = a + bx + cx 2

d q q (0 + b + 2cx ) + = (0 + 2c ) + = 0 dx k k Continued …..

PROBLEM 2.25 (Cont.)

)

(

q = −2ck = −2 −2 ×104°C / m 2 5 W / m ⋅ K = 2 ×105 W / m3

<

(c) The heat fluxes at the two boundaries can be determined using Fourier’s law and the temperature distribution expression.

q′′x ( x ) = − k

dT dx

T ( x ) = a + bx + cx 2

where

q′′x ( − L ) = − k [0 + b + 2cx ]x

=− L

= − [b − 2cL] k

(

)

q′′x ( − L ) = −  −210°C / m − 2 −2 × 104°C / m 2 0.020m  × 5 W / m ⋅ K = −2950 W / m 2

<

q′′x ( + L ) = − ( b + 2cL ) k = +5050 W / m 2

<





From an overall energy balance on the wall as shown in the sketch below, E in − E out + E gen = 0, ?

 =0 + q′′x ( − L ) − q′′x ( + L ) + 2qL

or

− 2950 W / m 2 − 5050 W / m 2 + 8000 W / m 2 = 0

5 3 2  = 2 × 2 × 10 W / m × 0.020 m = 8000 W / m , so the equality is satisfied where 2qL

(d) The convection coefficients, hl and hr, for the left- and right-hand boundaries (x = -L and x= +L, respectively), can be determined from the convection heat fluxes that are equal to the conduction fluxes at the boundaries. See the surface energy balances in the sketch above. See also part (a) result for T(-L) and T(+L).

q′′cv," = q′′x ( − L )

h l T∞ − T ( − L ) = h l [20 − 78.2] K = −2950 W / m 2

h l = 51W / m 2 ⋅ K

<

q′′cv,r = q′′x ( + L )

h r T ( + L ) − T∞  = h r [69.8 − 20] K = +5050 W / m 2

h r = 101W / m 2 ⋅ K

<

(e) The expression for the heat flux distribution can be obtained from Fourier’s law with the temperature distribution

q′′x ( x ) = − k

dT = − k [0 + b + 2cx ] dx

(

)

q′′x ( x ) = −5 W / m ⋅ K  −210°C / m + 2 −2 × 104°C / m 2  x = 1050 + 2 × 105 x   Continued …..

<

PROBLEM 2.25 (Cont.) The distribution is linear with the x-coordinate. The maximum temperature will occur at the location where q′′x ( x max ) = 0,

x max = −

1050 W / m 2 2 ×105 W / m3

= −5.25 × 10−3 m = −5.25 mm

<

(f) If the source of the heat generation is suddenly deactivated so that q = 0, the appropriate form of the heat diffusion equation for the ensuing transient conduction is

k

∂  ∂T  ∂T   = ρ cp ∂x  ∂x  ∂t

2

At the instant this occurs, the temperature distribution is still T(x) = a + bx + cx . The right-hand term represents the rate of energy storage per unit volume, E ′′st = k

∂ ∂x

[0 + b + 2cx ] = k [0 + 2c] = 5 W / m ⋅ K × 2

(−2 ×104°C / m2 ) = −2 ×105 W / m3 <

(g) With no heat generation, the wall will eventually (t → ∞) come to equilibrium with the fluid, T(x,∞) = T∞ = 20°C. To determine the energy that must be removed from the wall to reach this state, apply the conservation of energy requirement over an interval basis, Eq. 1.11b. The “initial” state is that corresponding to the steady-state temperature distribution, Ti, and the “final” state has Tf = 20°C. We’ve used T∞ as the reference condition for the energy terms.

E′′in − E′′out = ∆E′′st = E′′f − E′′i − E′′out = ρ cp 2L ( Tf − T∞ ) − ρ cp ∫ E′′out = ρ cp ∫

with

E′′in = 0.

+L (T − T∞ ) dx −L i

+L +L  2 − T  dx = ρ c ax + bx 2 / 2 + cx 3 / 3 − T x  a bx cx + + p  ∞  ∞  − L  −L

E′′out = ρ cp  2aL + 0 + 2cx 3 / 3 − 2T∞ L   

(

E′′out = 2600 kg / m3 × 800 J / kg ⋅ K  2 × 82°C × 0.020m + 2 −2 ×104°C / m2 

)

(0.020m )3 / 3 − 2 (20°C ) 0.020m  

E′′out = 4.94 ×106 J / m 2

<

COMMENTS: (1) In part (a), note that the temperature gradient is larger at x = + L than at x = - L. This is consistent with the results of part (c) in which the conduction heat fluxes are evaluated.

Continued …..

PROBLEM 2.25 (Cont.) (2) In evaluating the conduction heat fluxes, q′′x ( x ) , it is important to recognize that this flux is in the positive x-direction. See how this convention is used in formulating the energy balance in part (c). (3) It is good practice to represent energy balances with a schematic, clearly defining the system or surface, showing the CV or CS with dashed lines, and labeling the processes. Review again the features in the schematics for the energy balances of parts (c & d). (4) Re-writing the heat diffusion equation introduced in part (b) as −

d  dT   −k  + q = 0 dx  dx 

recognize that the term in parenthesis is the heat flux. From the differential equation, note that if the differential of this term is a constant ( q / k ) , then the term must be a linear function of the x-coordinate. This agrees with the analysis of part (e). (5) In part (f), we evaluated E st , the rate of energy change stored in the wall at the instant the volumetric heat generation was deactivated. Did you notice that E st = −2 × 105 W / m3 is the same value of the deactivated q ? How do you explain this?

PROBLEM 2.26 KNOWN: Steady-state conduction with uniform internal energy generation in a plane wall; temperature distribution has quadratic form. Surface at x=0 is prescribed and boundary at x = L is insulated. FIND: (a) Calculate the internal energy generation rate, q , by applying an overall energy balance to the wall, (b) Determine the coefficients a, b, and c, by applying the boundary conditions to the prescribed form of the temperature distribution; plot the temperature distribution and label as Case 1, (c) Determine new values for a, b, and c for conditions when the convection coefficient is halved, and the generation rate remains unchanged; plot the temperature distribution and label as Case 2; (d) Determine new values for a, b, and c for conditions when the generation rate is doubled, and the 2 convection coefficient remains unchanged (h = 500 W/m ⋅K); plot the temperature distribution and label as Case 3. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with constant properties and uniform internal generation, and (3) Boundary at x = L is adiabatic. ANALYSIS: (a) The internal energy generation rate can be calculated from an overall energy balance on the wall as shown in the schematic below.

E ′′in − E ′′out + E ′′gen = 0

′′ = q′′conv E in

where

h ( T∞ − To ) + q L = 0

(1)

q = − h ( T∞ − To ) / L = −500 W / m 2 ⋅ K ( 20 − 120 ) °C / 0.050 m = 1.0 ×106 W / m3

<

2

(b) The coefficients of the temperature distribution, T(x) = a + bx + cx , can be evaluated by applying the boundary conditions at x = 0 and x = L. See Table 2.1 for representation of the boundary conditions, and the schematic above for the relevant surface energy balances. Boundary condition at x = 0, convection surface condition

E ′′in − E ′′out = q ′′conv − q′′x ( 0 ) = 0

where

q′′x (0 ) = − k

dT   dx  x = 0

h ( T∞ − To ) −  − k (0 + b + 2cx )x = 0  = 0   Continued …..

PROBLEM 2.26 (Cont.)

b = −h (T∞ − To ) / k = −500 W / m 2 ⋅ K ( 20 − 120 ) °C / 5 W / m ⋅ K = 1.0 × 104 K / m

<

Boundary condition at x = L, adiabatic or insulated surface

E in − E out = −q′′x ( L ) = 0

q′′x ( L ) = − k

where

dT   dx  x = L

k [0 + b + 2cx ]x = L = 0

(3)

c = − b / 2L = −1.0 × 104 K / m / ( 2 × 0.050m ) = −1.0 × 105 K / m 2

<

Since the surface temperature at x = 0 is known, T(0) = To = 120°C, find

T ( 0 ) = 120°C = a + b ⋅ 0 + c ⋅ 0

or

a = 120°C

(4)

<

Using the foregoing coefficients with the expression for T(x) in the Workspace of IHT, the temperature distribution can be determined and is plotted as Case 1 in the graph below. 2

(c) Consider Case 2 when the convection coefficient is halved, h2 = h/2 = 250 W/m ⋅K, q = 1 × 106 3

W/m and other parameters remain unchanged except that To ≠ 120°C. We can determine a, b, and c for the temperature distribution expression by repeating the analyses of parts (a) and (b). Overall energy balance on the wall, see Eqs. (1,4)

a = To = q L / h + T∞ = 1× 106 W / m3 × 0.050m / 250 W / m 2 ⋅ K + 20°C = 220°C

<

Surface energy balance at x = 0, see Eq. (2)

b = −h (T∞ − To ) / k = −250 W / m 2 ⋅ K ( 20 − 220 ) °C / 5 W / m ⋅ K = 1.0 ×104 K / m

<

Surface energy balance at x = L, see Eq. (3)

c = − b / 2L = −1.0 × 104 K / m / ( 2 × 0.050m ) = −1.0 × 105 K / m 2

<

The new temperature distribution, T2 (x), is plotted as Case 2 below. (d) Consider Case 3 when the internal energy volumetric generation rate is doubled, 6

3

2

q 3 = 2q = 2 × 10 W / m , h = 500 W/m ⋅K, and other parameters remain unchanged except that To ≠ 120°C. Following the same analysis as part (c), the coefficients for the new temperature

distribution, T (x), are

a = 220°C

b = 2 × 104 K / m

c = −2 × 105 K / m2

<

and the distribution is plotted as Case 3 below.

Continued …..

PROBLEM 2.26 (Cont.)

800 700

Te m p e ra tu re , T (C )

600 500 400 300 200 100 0

5

10

15

20

25

30

35

40

45

50

W a ll p o s itio n , x (m m ) 1 . h = 5 0 0 W /m ^2 .K , q d o t = 1 e 6 W /m ^3 2 . h = 2 5 0 W /m ^2 .K , q d o t = 1 e 6 W /m ^3 3 . h = 5 0 0 W /m ^2 .K , q d o t = 2 e 6 W /m ^3

COMMENTS: Note the following features in the family of temperature distributions plotted above. The temperature gradients at x = L are zero since the boundary is insulated (adiabatic) for all cases. The shapes of the distributions are all quadratic, with the maximum temperatures at the insulated boundary. By halving the convection coefficient for Case 2, we expect the surface temperature To to increase  ) but the relative to the Case 1 value, since the same heat flux is removed from the wall ( qL convection resistance has increased. By doubling the generation rate for Case 3, we expect the surface temperature To to increase relative  ). to the Case 1 value, since double the amount of heat flux is removed from the wall ( 2qL Can you explain why To is the same for Cases 2 and 3, yet the insulated boundary temperatures are quite different? Can you explain the relative magnitudes of T(L) for the three cases?

PROBLEM 2.27 KNOWN: Temperature distribution and distribution of heat generation in central layer of a solar pond. FIND: (a) Heat fluxes at lower and upper surfaces of the central layer, (b) Whether conditions are steady or transient, (c) Rate of thermal energy generation for the entire central layer. SCHEMATIC:

ASSUMPTIONS: (1) Central layer is stagnant, (2) One-dimensional conduction, (3) Constant properties ANALYSIS: (a) The desired fluxes correspond to conduction fluxes in the central layer at the lower and upper surfaces. A general form for the conduction flux is

 "# ! $ Hence,  A e-aL + B"# q ′′l = q cond ′′ 1 x=L6 = − k ! ka $ q ′′cond = − k

∂T A -ax e +B . = −k ∂x ka

1 6

(b) Conditions are steady if ∂T/∂t = 0. Applying the heat equation,

∂ 2T ∂x

2

+

q 1 ∂ T = k α ∂t

-

 !

"# $

A q ′′u = q ′′cond x=0 = − k +B . ka

<

A -ax A -ax 1 ∂ T e + e = α ∂t k k

Hence conditions are steady since

∂T/∂t = 0

<

(for all 0 ≤ × ≤ L).

(c) For the central layer, the energy generation is

I

I

L L E ′′g = q dx = A e-ax dx 0

A E g = − e -ax a

0

L 0

=−

4

9

4

9

A -aL A e −1 = 1 − e -aL . a a

Alternatively, from an overall energy balance,

4 1 69 4  A + B"# − k  A e-aL + B"# = A 41 − e-aL 9. E g = k ! ka $ ! ka $ a

q 2′′ − q1′′ + E g′′ = 0

<

1 69

E ′′g = q1′′ − q ′′2 = − q ′′cond x=0 − − q ′′cond x=L

COMMENTS: Conduction is in the negative x-direction, necessitating use of minus signs in the above energy balance.

PROBLEM 2.28 KNOWN: Temperature distribution in a semi-transparent medium subjected to radiative flux.

16

FIND: (a) Expressions for the heat flux at the front and rear surfaces, (b) Heat generation rate q x , (c) Expression for absorbed radiation per unit surface area in terms of A, a, B, C, L, and k.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in medium, (3) Constant properties, (4) All laser irradiation is absorbed and can be characterized by an internal volumetric heat generation term q x .

16

ANALYSIS: (a) Knowing the temperature distribution, the surface heat fluxes are found using Fourier’s law,

 dT "# = − k - A 1−a6e-ax + B"# ! dx $ ! ka 2 $  A "  A + kB"# Front Surface, x=0: q ′′x 106 = − k + ⋅ 1 + B# = − ! ka $ ! a $  A " A " Rear Surface, x=L: q ′′x 1 L6 = − k + e-aL + B# = − e-aL + kB#. ! ka $ !a $ q ′′x = − k

< <

(b) The heat diffusion equation for the medium is

d  dT    or q = -k     dx  dx  d  A -ax " q 1 x6 = − k + e + B# = Ae-ax . dx ! ka $ d dT q + =0 dx dx k

<

(c) Performing an energy balance on the medium, E in − E out + E g = 0

 g represents the absorbed irradiation. On a unit area basis recognize that E

16 1 6

4

9

A E ′′g = − E in ′′ + E out ′′ = − q ′′x 0 + q ′′x L = + 1 − e-aL . a  ′′g by integration over the volume of the medium, Alternatively, evaluate E

I

16

I

4

9

L A L L A E ′′g = q x dx = Ae -ax dx = - e-ax = 1 − e-aL . 0 0 0 a a

<

PROBLEM 2.29 KNOWN: Steady-state temperature distribution in a one-dimensional wall of thermal 3 2 conductivity, T(x) = Ax + Bx + Cx + D. FIND: Expressions for the heat generation rate in the wall and the heat fluxes at the two wall faces (x = 0,L). ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3) Homogeneous medium. ANALYSIS: The appropriate form of the heat diffusion equation for these conditions is d 2T

q + =0 dx 2 k

or

q = -k

d 2T dx 2

.

Hence, the generation rate is q = -k

 "# ! $

d dT d = −k 3Ax2 + 2 Bx + C + 0 dx dx dx

<

q = -k 6Ax + 2B

which is linear with the coordinate x. The heat fluxes at the wall faces can be evaluated from Fourier’s law, q ′′x = − k

dT = − k 3Ax 2 + 2Bx + C dx

using the expression for the temperature gradient derived above. Hence, the heat fluxes are: Surface x=0:

16

<

q ′′x 0 = − kC Surface x=L:

16

<

q ′′x L = − k 3AL2 + 2BL + C . COMMENTS: (1) From an overall energy balance on the wall, find E ′′in − E ′′out + E ′′g = 0

16 1 6

1 6 1 6

q ′′x 0 − q ′′x L + E g′′ = − kC − − k 3AL2 + 2 BL + C + E g′′ = 0 E g′′ = −3AkL2 − 2 BkL. From integration of the volumetric heat rate, we can also find E ′′g as

I

16

I

L L L E ′′g = q x dx = -k 6Ax + 2B dx = -k 3Ax 2 + 2 Bx 0

0

E ′′g = −3AkL − 2 BkL. 2

0

PROBLEM 2.30 KNOWN: Plane wall with no internal energy generation. FIND: Determine whether the prescribed temperature distribution is possible; explain your reasoning. With the temperatures T(0) = 0°C and T∞ = 20°C fixed, compute and plot the temperature T(L) as a function of the convection coefficient for the range 10 ≤ h ≤ 100 W/m2⋅K. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) No internal energy generation, (3) Constant properties, (4) No radiation exchange at the surface x = L, and (5) Steady-state conditions. ANALYSIS: (a) Is the prescribed temperature distribution possible? If so, the energy balance at the surface x = L as shown above in the Schematic, must be satisfied. E in − E out ? = ? 0 q′′x ( L ) − q′′cv ? = ? 0 (1,2) where the conduction and convection heat fluxes are, respectively, T ( L ) − T (0 ) dT  $ q′′x ( L ) = − k = −k = −4.5 W m ⋅ K × (120 − 0 ) C 0.18 m = −3000 W m 2  dx  x = L L q′′cv = h [T ( L ) − T∞ ] = 30 W m 2 ⋅ K × (120 − 20 ) C = 3000 W m2 $

Substituting the heat flux values into Eq. (2), find (-3000) - (3000) ≠ 0 and therefore, the temperature distribution is not possible. (b) With T(0) = 0°C and T∞ = 20°C, the temperature at the surface x = L, T(L), can be determined from an overall energy balance on the wall as shown above in the Schematic, T ( L ) − T (0 ) −k − h [T ( L ) − T∞ ] = 0 E in − E out = 0 q′′x (0) − q′′cv = 0 L −4.5 W m ⋅ K T ( L ) − 0$ C  0.18 m − 30 W m 2 ⋅ K  T ( L ) − 20$ C  = 0









<

T(L) = 10.9°C 20

Surface temperature, T(L) (C)

Using this same analysis, T(L) as a function of the convection coefficient can be determined and plotted. We don’t expect T(L) to be linearly dependent upon h. Note that as h increases to larger values, T(L) approaches T∞ . To what value will T(L) approach as h decreases?

16

12

8

4

0 0

20

40

60

Convection cofficient, h (W/m^2.K)

80

100

PROBLEM 2.31 KNOWN: Coal pile of prescribed depth experiencing uniform volumetric generation with convection, absorbed irradiation and emission on its upper surface. FIND: (a) The appropriate form of the heat diffusion equation (HDE) and whether the prescribed temperature distribution satisfies this HDE; conditions at the bottom of the pile, x = 0; sketch of the temperature distribution with labeling of key features; (b) Expression for the conduction heat rate at the location x = L; expression for the surface temperature Ts based upon a surface energy balance at x = L; evaluate Ts and T(0) for the prescribed conditions; (c) Based upon typical daily averages for GS and h, compute and plot Ts and T(0) for (1) h = 5 W/m2⋅K with 50 ≤ GS ≤ 500 W/m2, (2) GS = 400 W/m2 with 5 ≤ h ≤ 50 W/m2⋅K. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Uniform volumetric heat generation, (3) Constant properties, (4) Negligible irradiation from the surroundings, and (5) Steady-state conditions. PROPERTIES: Table A.3, Coal (300K): k = 0.26 W/m.K ANALYSIS: (a) For one-dimensional, steady-state conduction with uniform volumetric heat generation and constant properties the heat diffusion equation (HDE) follows from Eq. 2.16,

d  dT  q  + =0 dx  dx  k

(1)

Substituting the temperature distribution into the HDE, Eq. (1),  2  x2   2 qL d  qL 2x   q 1 −  T ( x ) = Ts + 0− 0 +  + ? = ?0  2 2 

2k



L 

dx  

2k 

L  

k

<

(2,3)

<

we find that it does indeed satisfy the HDE for all values of x. From Eq. (2), note that the temperature distribution must be quadratic, with maximum value at x = 0. At x = 0, the heat flux is

q′′x ( 0 ) = − k

 qL  2 dT  2x   k 0 = − + =0   0 − 2   dx  x = 0 2k   L   x = 0

so that the gradient at x = 0 is zero. Hence, the bottom is insulated. (b) From an overall energy balance on the pile, the conduction heat flux at the surface must be

 q′′x ( L ) = E ′′g = qL

< Continued...

PROBLEM 2.31 (Cont.) From a surface energy balance per unit area shown in the Schematic above, q′′x ( L ) − q′′cv + GS,abs − E = 0

E in − E out + E g = 0

 − h ( Ts − T∞ ) + 0.95GS − εσ Ts4 = 0 qL

(4)

20 W m ×1m − 5 W m ⋅K ( Ts − 298 K ) + 0.95 × 400 W m − 0.95 × 5.67 × 10 3

2

2

−8

2

4 4

W m ⋅K Ts = 0

<

Ts = 295.7 K =22.7°C From Eq. (2) with x = 0, find

30 W m 2 × (1m )  2 qL $ T ( 0 ) = Ts + = 22.7 C + = 61.1$ C 2k 2 × 0.26 W m ⋅ K 2

(5)

<

where the thermal conductivity for coal was obtained from Table A.3. (c) Two plots are generated using Eq. (4) and (5) for Ts and T(0), respectively; (1) with h = 5 W/m2⋅K for 50 ≤ GS ≤ 500 W/m2 and (2) with GS = 400 W/m2 for 5 ≤ h ≤ 50 W/m2⋅K. Solar irradiation, GS = 400 W/m^2 80

Convection coefficient, h = 5 W/m^2.K Temperature, Ts or T(0) (C)

Temperature, Ts or T(0) (C)

80

60

40

20

60

40

20

0

0

10

20

30

40

50

Convection coefficient, h (W/m^2.K)

-20 0

100

200

300

400

500

T0_C Ts_C

Solar irradiation, GS (W/m^2) T0_C Ts_C

From the T vs. h plot with GS = 400 W/m2, note that the convection coefficient does not have a major influence on the surface or bottom coal pile temperatures. From the T vs. GS plot with h = 5 W/m2⋅K, note that the solar irradiation has a very significant effect on the temperatures. The fact that Ts is less than the ambient air temperature, T∞ , and, in the case of very low values of GS, below freezing, is a consequence of the large magnitude of the emissive power E. COMMENTS: In our analysis we ignored irradiation from the sky, an environmental radiation effect 4 where T = you’ll consider in Chapter 12. Treated as large isothermal surroundings, Gsky = σ Tsky sky 30°C for very clear conditions and nearly air temperature for cloudy conditions. For low GS conditions we should consider Gsky, the effect of which will be to predict higher values for Ts and T(0).

PROBLEM 2.32 KNOWN: Cylindrical system with negligible temperature variation in the r,z directions. FIND: (a) Heat equation beginning with a properly defined control volume, (b) Temperature distribution T(φ) for steady-state conditions with no internal heat generation and constant properties, (c) Heat rate for Part (b) conditions.

SCHEMATIC:

ASSUMPTIONS: (1) T is independent of r,z, (2) ∆r = (ro - ri) α B , T x, t → Ts more rapidly for Material A. If α A < α B , T x, t → Ts more rapidly for Material B.

<

COMMENTS: Note that the prescribed function for T(x,t) does not reduce to Ti for t → 0. For times at or close to zero, the function is not a valid solution of the problem. At such times, the solution for T(x,t) must include additional terms. The solution is consideed in Section 5.5.1 of the text.

PROBLEM 2.53 2

KNOWN: Thin electrical heater dissipating 4000 W/m sandwiched between two 25-mm thick plates whose surfaces experience convection. FIND: (a) On T-x coordinates, sketch the steady-state temperature distribution for -L ≤ × ≤ +L; calculate values for the surfaces x = L and the mid-point, x = 0; label this distribution as Case 1 and explain key features; (b) Case 2: sudden loss of coolant causing existence of adiabatic condition on the x = +L surface; sketch temperature distribution on same T-x coordinates as part (a) and calculate values for x = 0, ± L; explain key features; (c) Case 3: further loss of coolant and existence of adiabatic condition on the x = - L surface; situation goes undetected for 15 minutes at which time power to the heater is deactivated; determine the eventual (t → ∞) uniform, steady-state temperature distribution; sketch temperature distribution on same T-x coordinates as parts (a,b); and (d) On T-t coordinates, sketch the temperature-time history at the plate locations x = 0, ± L during the transient period between the steady-state distributions for Case 2 and Case 3; at what location and when will the temperature in the system achieve a maximum value? SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal volumetric generation in plates, and (3) Negligible thermal resistance between the heater surfaces and the plates. ANALYSIS: (a) Since the system is symmetrical, the heater power results in equal conduction fluxes through the plates. By applying a surface energy balance on the surface x = +L as shown in the schematic, determine the temperatures at the mid-point, x = 0, and the exposed surface, x + L.

E in − E out = 0

q′′x ( + L ) − q′′conv = 0

q′′x ( + L ) = q′′o / 2

where

q′′o / 2 − h  T ( + L ) − T∞  = 0

(

)

T1 ( + L ) = q′′o / 2h + T∞ = 4000 W / m 2 / 2 × 400 W / m 2 ⋅ K + 20°C = 25°C

<

From Fourier’s law for the conduction flux through the plate, find T(0).

q′′x = q′′o / 2 = k T ( 0 ) − T ( + L ) / L

T1 ( 0 ) = T1 ( + L ) + q′′o L / 2k = 25°C + 4000 W / m 2 ⋅ K × 0.025m / ( 2 × 5 W / m ⋅ K ) = 35°C The temperature distribution is shown on the T-x coordinates below and labeled Case 1. The key features of the distribution are its symmetry about the heater plane and its linear dependence with distance. Continued …..

<

PROBLEM 2.53 (Cont.)

(b) Case 2: sudden loss of coolant with the existence of an adiabatic condition on surface x = +L. For this situation, all the heater power will be conducted to the coolant through the left-hand plate. From a surface energy balance and application of Fourier’s law as done for part (a), find T2 ( − L ) = q′′o / h + T∞ = 4000 W / m 2 / 400 W / m 2 ⋅ K + 20°C = 30°C T2 ( 0 ) = T2 ( − L ) + q′′o L / k = 30°C + 4000 W / m 2 × 0.025 m / 5 W / m ⋅ K = 50°C The temperature distribution is shown on the T-x coordinates above and labeled Case 2. The distribution is linear in the left-hand plate, with the maximum value at the mid-point. Since no heat flows through the right-hand plate, the gradient must zero and this plate is at the maximum temperature as well. The maximum temperature is higher than for Case 1 because the heat flux through the left-hand plate has increased two-fold.

< <

(c) Case 3: sudden loss of coolant occurs at the x = -L surface also. For this situation, there is no heat 2 transfer out of either plate, so that for a 15-minute period, ∆to, the heater dissipates 4000 W/m and then is deactivated. To determine the eventual, uniform steady-state temperature distribution, apply the conservation of energy requirement on a time-interval basis, Eq. 1.11b. The initial condition corresponds to the temperature distribution of Case 2, and the final condition will be a uniform, elevated temperature Tf = T3 representing Case 3. We have used T∞ as the reference condition for the energy terms. E′′in − E′′out + E′′gen = ∆E′′st = E′′f − E′′i (1) Note that E ′′in − E ′′out = 0 , and the dissipated electrical energy is E′′gen = q′′o ∆t o = 4000 W / m 2 (15 × 60 ) s = 3.600 × 106 J / m 2 For the final condition, E′′f = ρ c ( 2L )[Tf − T∞ ] = 2500 kg / m3 × 700 J / kg ⋅ K ( 2 × 0.025m ) [Tf − 20 ]°C E′′f = 8.75 × 104 [Tf − 20] J / m2 where Tf = T3, the final uniform temperature, Case 3. For the initial condition, E′′i = ρ c ∫

+L T2 ( x ) − T∞ dx = ρ c −L

[

]

{∫

0 +L T2 ( x ) − T∞ dx + T2 (0 ) − T∞ dx 0 −L

[

]

∫

[

where T2 ( x ) is linear for –L ≤ x ≤ 0 and constant at T2 ( 0 ) for 0 ≤ x ≤ +L.

T2 ( x ) = T2 ( 0 ) +  T2 ( 0 ) − T2 ( L ) x / L

]

}

(2)

(3)

(4)

−L ≤ x ≤ 0

T2 ( x ) = 50°C + [50 − 30]°Cx / 0.025m T2 ( x ) = 50°C + 800x

(5)

Substituting for T2 ( x ) , Eq. (5), into Eq. (4) Continued …..

PROBLEM 2.53 (Cont.)  0  E′′i = ρ c  ∫ [50 + 800x − T∞ ] dx + T2 ( 0 ) − T∞  L   −L  0   E′′i = ρ c  50x + 400x 2 − T∞ x  + T2 (0 ) − T∞  L   − L   

{

}

Ei′′ = ρ c −  −50L + 400L2 + T∞ L  + T2 ( 0 ) − T∞  L   E′′i = ρ cL {+50 − 400L − T∞ + T2 (0 ) − T∞ }

E′′i = 2500 kg / m3 × 700 J / kg ⋅ K × 0.025 m {+50 − 400 × 0.025 − 20 + 50 − 20}K E′′i = 2.188 × 106 J / m 2

(6)

Returning to the energy balance, Eq. (1), and substituting Eqs. (2), (3) and (6), find Tf = T3. 3.600 × 106 J / m 2 = 8.75 × 104 [T3 − 20] − 2.188 × 106 J / m 2

T3 = ( 66.1 + 20 ) °C = 86.1°C

<

The temperature distribution is shown on the T-x coordinates above and labeled Case 3. The distribution is uniform, and considerably higher than the maximum value for Case 2. (d) The temperature-time history at the plate locations x = 0, ± L during the transient period between the distributions for Case 2 and Case 3 are shown on the T-t coordinates below.

Note the temperatures for the locations at time t = 0 corresponding to the instant when the surface x = - L becomes adiabatic. These temperatures correspond to the distribution for Case 2. The heater remains energized for yet another 15 minutes and then is deactivated. The midpoint temperature, T(0,t), is always the hottest location and the maximum value slightly exceeds the final temperature T3.

PROBLEM 2.54 KNOWN: Radius and length of coiled wire in hair dryer. Electric power dissipation in the wire, and temperature and convection coefficient associated with air flow over the wire. FIND: (a) Form of heat equation and conditions governing transient, thermal behavior of wire during start-up, (b) Volumetric rate of thermal energy generation in the wire, (c) Sketch of temperature distribution at selected times during start-up, (d) Variation with time of heat flux at r = 0 and r = ro. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Constant properties, (3) Uniform volumetric heating, (4) Negligible radiation from surface of wire. ANALYSIS: (a) The general form of the heat equation for cylindrical coordinates is given by Eq. 2.20. For one-dimensional, radial conduction and constant properties, the equation reduces to

 r ∂T  + q = ρ c p ∂T = 1 ∂T   r ∂r  ∂r  k ∂t α ∂t

1 ∂

The initial condition is

T ( r, 0 ) = Ti

The boundary conditions are:

∂T / ∂r r = 0 = 0

−k

∂T ∂r r = r o

< < < <

= h [T ( ro , t ) − T∞ ]

(b) The volumetric rate of thermal energy generation is q =

E g

P 500 W 8 3 = elec = = 3.18 × 10 W / m 2 2 ∀ π ro L π ( 0.001m ) ( 0.5m )

<

Under steady-state conditions, all of the thermal energy generated within the wire is transferred to the air by convection. Performing an energy balance for a control surface about the wire, − E out + E g = 0, it follows that −2π ro L q ′′ ( ro , t → ∞ ) + Pelec = 0. Hence, q ′′ ( ro , t → ∞ ) =

Pelec 2π ro L

=

500 W 2π ( 0.001m ) 0.5m

5

= 1.59 × 10 W / m

2

<

COMMENTS: The symmetry condition at r = 0 imposes the requirement that ∂T / ∂r r = 0 = 0, and hence q ′′ ( 0, t ) = 0 throughout the process. The temperature at ro, and hence the convection heat flux, increases steadily during the start-up, and since conduction to the surface must be balanced by convection from the surface at all times, ∂T / ∂r r = r also increases during the start-up. o

PROBLEM 3.1 KNOWN: One-dimensional, plane wall separating hot and cold fluids at T∞,1 and T∞ ,2 , respectively. FIND: Temperature distribution, T(x), and heat flux, q ′′x , in terms of T∞,1 , T∞,2 , h1 , h 2 , k and L. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties, (4) Negligible radiation, (5) No generation. ANALYSIS: For the foregoing conditions, the general solution to the heat diffusion equation is of the form, Equation 3.2, T ( x ) = C1x + C2 . (1) The constants of integration, C1 and C2, are determined by using surface energy balance conditions at x = 0 and x = L, Equation 2.23, and as illustrated above, dT  dT  (2,3) −k  = h1 T∞,1 − T ( 0 ) −k  = h 2 T ( L ) − T∞,2  . dt  x=0 dx  x=L For the BC at x = 0, Equation (2), use Equation (1) to find − k ( C1 + 0 ) = h1  T∞,1 − (C1 ⋅ 0 + C2 )

(4)

and for the BC at x = L to find − k ( C1 + 0 ) = h 2 ( C1L + C2 ) − T∞,2  .

(5)

Multiply Eq. (4) by h2 and Eq. (5) by h1, and add the equations to obtain C1. Then substitute C1 into Eq. (4) to obtain C2. The results are T∞,1 − T∞,2 T∞,1 − T∞,2 C1 = − C2 = − + T∞,1 1 1 1 L 1 L k + h1  + +  +   h1 h 2 k   h1 h 2 k  T∞,1 − T∞,2  x 1  T (x ) = − <  +  + T∞,1. 1 1 L   k h1  h + h + k  2  1 

(

)

(

(

)

)

From Fourier’s law, the heat flux is a constant and of the form T∞,1 − T∞,2 dT q′′x = − k . = − k C1 = + dx 1 1 L h + h + k  2  1 

(

)

<

PROBLEM 3.2 KNOWN: Temperatures and convection coefficients associated with air at the inner and outer surfaces of a rear window. FIND: (a) Inner and outer window surface temperatures, Ts,i and Ts,o, and (b) Ts,i and Ts,o as a function of the outside air temperature T∞,o and for selected values of outer convection coefficient, ho. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible radiation effects, (4) Constant properties. PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m⋅K. ANALYSIS: (a) The heat flux may be obtained from Eqs. 3.11 and 3.12,

)

(

40$ C − −10$ C T∞,i − T∞,o q′′ = = 1 L 1 1 0.004 m 1 + + + + 2 ho k hi 65 W m ⋅ K 1.4 W m ⋅ K 30 W m2 ⋅ K q′′ =

50$ C

(0.0154 + 0.0029 + 0.0333) m

(

2

⋅K W

= 968 W m 2 .

)

Hence, with q′′ = h i T∞,i − T∞,o , the inner surface temperature is Ts,i = T∞ ,i −

q′′ hi

$

= 40 C −

968 W m 2 2

30 W m ⋅ K

<

= 7.7$ C

(

)

Similarly for the outer surface temperature with q′′ = h o Ts,o − T∞,o find Ts,o = T∞,o −

q′′

= −10$ C −

968 W m 2 2

= 4.9$ C

<

65 W m ⋅ K (b) Using the same analysis, Ts,i and Ts,o have been computed and plotted as a function of the outside air temperature, T∞,o, for outer convection coefficients of ho = 2, 65, and 100 W/m2⋅K. As expected, Ts,i and Ts,o are linear with changes in the outside air temperature. The difference between Ts,i and Ts,o increases with increasing convection coefficient, since the heat flux through the window likewise increases. This difference is larger at lower outside air temperatures for the same reason. Note that with ho = 2 W/m2⋅K, Ts,i - Ts,o, is too small to show on the plot. ho

Continued …..

Surface temperatures, Tsi or Tso (C)

PROBLEM 3.2 (Cont.) 40 30 20 10 0 -10 -20 -30 -30

-25

-20

-15

-10

-5

0

Outside air temperature, Tinfo (C) Tsi; ho = 100 W/m^2.K Tso; ho = 100 W/m^2.K Tsi; ho = 65 W/m^2.K Tso; ho = 65 W/m^2.K Tsi or Tso; ho = 2 W/m^.K

COMMENTS: (1) The largest resistance is that associated with convection at the inner surface. The values of Ts,i and Ts,o could be increased by increasing the value of hi. (2) The IHT Thermal Resistance Network Model was used to create a model of the window and generate the above plot. The Workspace is shown below. // Thermal Resistance Network Model: // The Network:

// Heat rates into node j,qij, through thermal resistance Rij q21 = (T2 - T1) / R21 q32 = (T3 - T2) / R32 q43 = (T4 - T3) / R43 // Nodal energy balances q1 + q21 = 0 q2 - q21 + q32 = 0 q3 - q32 + q43 = 0 q4 - q43 = 0 /* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal points at which there is no external source of heat. */ T1 = Tinfo // Outside air temperature, C //q1 = // Heat rate, W T2 = Tso // Outer surface temperature, C q2 = 0 // Heat rate, W; node 2, no external heat source T3 = Tsi // Inner surface temperature, C q3 = 0 // Heat rate, W; node 2, no external heat source T4 = Tinfi // Inside air temperature, C //q4 = // Heat rate, W // Thermal Resistances: R21 = 1 / ( ho * As ) R32 = L / ( k * As ) R43 = 1 / ( hi * As )

// Convection thermal resistance, K/W; outer surface // Conduction thermal resistance, K/W; glass // Convection thermal resistance, K/W; inner surface

// Other Assigned Variables: Tinfo = -10 // Outside air temperature, C ho = 65 // Convection coefficient, W/m^2.K; outer surface L = 0.004 // Thickness, m; glass k = 1.4 // Thermal conductivity, W/m.K; glass Tinfi = 40 // Inside air temperature, C hi = 30 // Convection coefficient, W/m^2.K; inner surface As = 1 // Cross-sectional area, m^2; unit area

PROBLEM 3.3 KNOWN: Desired inner surface temperature of rear window with prescribed inside and outside air conditions. FIND: (a) Heater power per unit area required to maintain the desired temperature, and (b) Compute and plot the electrical power requirement as a function of T∞,o for the range -30 ≤ T∞,o ≤ 0°C with ho of 2, 20, 65 and 100 W/m2⋅K. Comment on heater operation needs for low ho. If h ~ Vn, where V is the vehicle speed and n is a positive exponent, how does the vehicle speed affect the need for heater operation? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Uniform heater flux, q′′h , (4) Constant properties, (5) Negligible radiation effects, (6) Negligible film resistance. PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m⋅K. ANALYSIS: (a) From an energy balance at the inner surface and the thermal circuit, it follows that for a unit surface area, T∞ ,i − Ts,i 1 hi q ′′h =

Ts,i − T∞ ,o + q ′′h = L k + 1 ho

Ts,i − T∞ ,o L k + 1 ho

−

T∞ ,i − Ts,i 1 hi

(

$

=

0.004 m 1.4 W m ⋅ K

q ′′h = (1370 − 100 ) W m = 1270 W m 2

$

15 C − −10 C +

)

$

−

1 2

65 W m ⋅ K

$

25 C − 15 C 1 2

10 W m ⋅ K

<

2

(b) The heater electrical power requirement as a function of the exterior air temperature for different exterior convection coefficients is shown in the plot. When ho = 2 W/m2⋅K, the heater is unecessary, since the glass is maintained at 15°C by the interior air. If h ~ Vn, we conclude that, with higher vehicle speeds, the exterior convection will increase, requiring increased heat power to maintain the 15°C condition. Heater power (W/m^2)

3500 3000 2500 2000 1500 1000 500 0 -30

-20

-10

0

Exterior air temperature, Tinfo (C) h = 20 W/m^2.K h = 65 W/m^2.K h = 100 W/m^2.K

COMMENTS: With q′′h = 0, the inner surface temperature with T∞,o = -10°C would be given by

T∞ ,i − Ts,i T∞ ,i − T∞ ,o

=

1 hi 1 hi + L k + 1 ho

=

0.10 0.118

= 0.846,

or

$

( ) $

$

Ts,i = 25 C − 0.846 35 C = − 4.6 C .

PROBLEM 3.4 KNOWN: Curing of a transparent film by radiant heating with substrate and film surface subjected to known thermal conditions. FIND: (a) Thermal circuit for this situation, (b) Radiant heat flux, q′′o (W/m2), to maintain bond at curing temperature, To, (c) Compute and plot q′′o as a function of the film thickness for 0 ≤ Lf ≤ 1 mm, and (d) If the film is not transparent, determine q′′o required to achieve bonding; plot results as a function of Lf. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3) All the radiant heat flux q′′o is absorbed at the bond, (4) Negligible contact resistance. ANALYSIS: (a) The thermal circuit for this situation is shown at the right. Note that terms are written on a per unit area basis. (b) Using this circuit and performing an energy balance on the film-substrate interface, q′′o = q1′′ + q′′2

q′′o =

To − T∞ T −T + o 1 R ′′cv + R ′′f R s′′

where the thermal resistances are R ′′cv = 1 h = 1 50 W m 2 ⋅ K = 0.020 m 2 ⋅ K W R ′′f = L f k f = 0.00025 m 0.025 W m ⋅ K = 0.010 m 2 ⋅ K W R ′′s = Ls k s = 0.001m 0.05 W m ⋅ K = 0.020 m 2 ⋅ K W q′′o =

(60 − 20 )$ C [0.020 + 0.010] m2 ⋅ K

+ W

(60 − 30 )$ C 2

0.020 m ⋅ K W

= (133 + 1500 ) W m 2 = 2833 W m 2

<

(c) For the transparent film, the radiant flux required to achieve bonding as a function of film thickness Lf is shown in the plot below. (d) If the film is opaque (not transparent), the thermal circuit is shown below. In order to find q′′o , it is necessary to write two energy balances, one around the Ts node and the second about the To node.

. The results of the analyses are plotted below. Continued...

PROBLEM 3.4 (Cont.) Radiant heat flux, q''o (W/m^2)

7000

6000

5000

4000

3000

2000 0

0.2

0.4

0.6

0.8

1

Film thickness, Lf (mm) Opaque film Transparent film

COMMENTS: (1) When the film is transparent, the radiant flux is absorbed on the bond. The flux required decreases with increasing film thickness. Physically, how do you explain this? Why is the relationship not linear? (2) When the film is opaque, the radiant flux is absorbed on the surface, and the flux required increases with increasing thickness of the film. Physically, how do you explain this? Why is the relationship linear? (3) The IHT Thermal Resistance Network Model was used to create a model of the film-substrate system and generate the above plot. The Workspace is shown below. // Thermal Resistance Network Model: // The Network: // Heat rates into node j,qij, through thermal resistance Rij q21 = (T2 - T1) / R21 q32 = (T3 - T2) / R32 q43 = (T4 - T3) / R43 // Nodal energy balances q1 + q21 = 0 q2 - q21 + q32 = 0 q3 - q32 + q43 = 0 q4 - q43 = 0 /* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal points at which there is no external source of heat. */ T1 = Tinf // Ambient air temperature, C //q1 = // Heat rate, W; film side T2 = Ts // Film surface temperature, C q2 = 0 // Radiant flux, W/m^2; zero for part (a) T3 = To // Bond temperature, C q3 = qo // Radiant flux, W/m^2; part (a) T4 = Tsub // Substrate temperature, C //q4 = // Heat rate, W; substrate side // Thermal Resistances: R21 = 1 / ( h * As ) R32 = Lf / (kf * As) R43 = Ls / (ks * As)

// Convection resistance, K/W // Conduction resistance, K/W; film // Conduction resistance, K/W; substrate

// Other Assigned Variables: Tinf = 20 // Ambient air temperature, C h = 50 // Convection coefficient, W/m^2.K Lf = 0.00025 // Thickness, m; film kf = 0.025 // Thermal conductivity, W/m.K; film To = 60 // Cure temperature, C Ls = 0.001 // Thickness, m; substrate ks = 0.05 // Thermal conductivity, W/m.K; substrate Tsub = 30 // Substrate temperature, C As = 1 // Cross-sectional area, m^2; unit area

PROBLEM 3.5 KNOWN: Thicknesses and thermal conductivities of refrigerator wall materials. Inner and outer air temperatures and convection coefficients. FIND: Heat gain per surface area. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3) Negligible contact resistance, (4) Negligible radiation, (5) Constant properties. ANALYSIS: From the thermal circuit, the heat gain per unit surface area is

q′′ =

q′′ =

q′′ =

T∞,o − T∞,i

(1/ h i ) + (Lp / k p ) + ( Li / k i ) + (Lp / k p ) + (1/ h o )

(

)

( 25 − 4 ) °C

2 1/ 5 W / m 2 ⋅ K + 2 (0.003m / 60 W / m ⋅ K ) + (0.050m / 0.046 W / m ⋅ K ) 21°C

(0.4 + 0.0001 + 1.087 ) m2 ⋅ K / W

= 14.1 W / m 2

<

COMMENTS: Although the contribution of the panels to the total thermal resistance is negligible, that due to convection is not inconsequential and is comparable to the thermal resistance of the insulation.

PROBLEM 3.6 KNOWN: Design and operating conditions of a heat flux gage. FIND: (a) Convection coefficient for water flow (Ts = 27°C) and error associated with neglecting conduction in the insulation, (b) Convection coefficient for air flow (Ts = 125°C) and error associated with neglecting conduction and radiation, (c) Effect of convection coefficient on error associated with neglecting conduction for Ts = 27°C. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant k. ANALYSIS: (a) The electric power dissipation is balanced by convection to the water and conduction through the insulation. An energy balance applied to a control surface about the foil therefore yields ′′ = q ′′conv + q ′′cond = h ( Ts − T∞ ) + k (Ts − Tb ) L Pelec Hence, h= h=

′′ − k ( Ts − Tb ) L Pelec Ts − T∞

( 2000 − 8) W 2K

m2

=

2000 W m 2 − 0.04 W m ⋅ K ( 2 K ) 0.01m 2K

<

= 996 W m 2 ⋅ K

If conduction is neglected, a value of h = 1000 W/m2⋅K is obtained, with an attendant error of (1000 996)/996 = 0.40% (b) In air, energy may also be transferred from the foil surface by radiation, and the energy balance yields

(

)

4 ′′ = q ′′conv + q ′′rad + q′′cond = h (Ts − T∞ ) + εσ Ts4 − Tsur + k ( Ts − Tb ) L Pelec

Hence, h=

(

)

4 ′′ − εσ Ts4 − Tsur Pelec − k ( Ts − T∞ ) L

Ts − T∞ 2

=

=

2000 W m − 0.15 × 5.67 × 10

−8

2

W m ⋅K

4

(398

4

− 298

4

)K

4

− 0.04 W m ⋅ K (100 K) / 0.01m

100 K

( 2000 − 146 − 400 ) W 100 K

m2

= 14.5 W m 2 ⋅ K

< Continued...

PROBLEM 3.6 (Cont.) If conduction, radiation, or conduction and radiation are neglected, the corresponding values of h and the percentage errors are 18.5 W/m2⋅K (27.6%), 16 W/m2⋅K (10.3%), and 20 W/m2⋅K (37.9%). (c) For a fixed value of Ts = 27°C, the conduction loss remains at q′′cond = 8 W/m2, which is also the ′′ and q′′conv . Although this difference is not clearly shown in the plot for fixed difference between Pelec 10 ≤ h ≤ 1000 W/m2⋅K, it is revealed in the subplot for 10 ≤ 100 W/m2⋅K. 200

Power dissipation, P''elec(W/m^2)

Power dissipation, P''elec(W/m^2)

2000

1600

1200

800

400

0 0

200

400

600

800

Convection coefficient, h(W/m^2.K) No conduction With conduction

1000

160

120

80

40

0 0

20

40

60

80

100

Convection coefficient, h(W/m^2.K) No conduction With conduction

Errors associated with neglecting conduction decrease with increasing h from values which are significant for small h (h < 100 W/m2⋅K) to values which are negligible for large h. COMMENTS: In liquids (large h), it is an excellent approximation to neglect conduction and assume that all of the dissipated power is transferred to the fluid.

PROBLEM 3.7 KNOWN: A layer of fatty tissue with fixed inside temperature can experience different outside convection conditions. FIND: (a) Ratio of heat loss for different convection conditions, (b) Outer surface temperature for different convection conditions, and (c) Temperature of still air which achieves same cooling as moving air (wind chill effect). SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction through a plane wall, (2) Steady-state conditions, (3) Homogeneous medium with constant properties, (4) No internal heat generation (metabolic effects are negligible), (5) Negligible radiation effects. PROPERTIES: Table A-3, Tissue, fat layer: k = 0.2 W/m⋅K. ANALYSIS: The thermal circuit for this situation is

Hence, the heat rate is q=

Ts,1 − T∞ R tot

=

Ts,1 − T∞ L/kA + 1/ hA

.

Therefore, L 1  k + h  windy

q′′calm . = q′′windy L 1  k + h  calm Applying a surface energy balance to the outer surface, it also follows that q′′cond = q′′conv . Continued …..

PROBLEM 3.7 (Cont.) Hence,

(

) (

k Ts,1 − Ts,2 = h Ts,2 − T∞ L k T∞ + Ts,1 hL Ts,2 = . k 1+ hL

)

To determine the wind chill effect, we must determine the heat loss for the windy day and use it to evaluate the hypothetical ambient air temperature, T∞′ , which would provide the same heat loss on a calm day, Hence, q′′ =

′ Ts,1 − T∞ Ts,1 − T∞ = L 1 L 1  k + h   +  windy  k h  calm

From these relations, we can now find the results sought:

(a)

0.003 m 1 + q′′calm 0.2 W/m ⋅ K 65 W/m 2 ⋅ K 0.015 + 0.0154 = = 0.003 m 1 q′′windy 0.015 + 0.04 + 0.2 W/m ⋅ K 25 W/m 2 ⋅ K q′′calm = 0.553 q′′windy

<

−15$ C + (b)

Ts,2 

calm

=

1+

(

0.2 W/m ⋅ K

(c)

windy

=

36$ C

0.2 W/m ⋅ K

= 22.1$ C

<

= 10.8$ C

<

(25 W/m2 ⋅ K )(0.003 m)

−15$ C + Ts,2 

)

25 W/m 2 ⋅ K ( 0.003 m )

1+

0.2 W/m ⋅ K

(65 W/m2 ⋅ K )(0.003m ) (65 W/m2 ⋅ K )(0.003m )

′ = 36$ C − (36 + 15 ) C T∞ $

0.2 W/m ⋅ K

36$ C

(0.003/0.2 + 1/ 25 ) = −56.3$ C (0.003 / 0.2 + 1/ 65)

COMMENTS: The wind chill effect is equivalent to a decrease of Ts,2 by 11.3°C and -1

increase in the heat loss by a factor of (0.553) = 1.81.

<

PROBLEM 3.8 KNOWN: Dimensions of a thermopane window. Room and ambient air conditions. FIND: (a) Heat loss through window, (b) Effect of variation in outside convection coefficient for double and triple pane construction. SCHEMATIC (Double Pane):

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constant properties, (4) Negligible radiation effects, (5) Air between glass is stagnant. PROPERTIES: Table A-3, Glass (300 K): kg = 1.4 W/m⋅K; Table A-4, Air (T = 278 K): ka = 0.0245 W/m⋅K. ANALYSIS: (a) From the thermal circuit, the heat loss is T∞,i − T∞,o q= 1 1 L L L 1  + + +  +  A  h i k g k a k g h o  q=

(

20$ C − −10$ C

)

 1   1 0.007 m 0.007 m 0.007 m 1 + + + +      0.4 m 2   10 W m 2 ⋅ K 1.4 W m ⋅ K 0.0245 W m ⋅ K 1.4 W m ⋅ K 80 W m 2 ⋅ K     q=

30$ C

(0.25 + 0.0125 + 0.715 + 0.0125 + 0.03125 ) K

W

=

30$ C 1.021K W

= 29.4 W

<

(b) For the triple pane window, the additional pane and airspace increase the total resistance from 1.021 K/W to 1.749 K/W, thereby reducing the heat loss from 29.4 to 17.2 W. The effect of ho on the heat loss is plotted as follows. 30

Heat loss, q(W)

27

24

21

18

15 10

28

46

64

82

100

Outside convection coefficient, ho(W/m^2.K) Double pane Triple pane

Continued...

PROBLEM 3.8 (Cont.) Changes in ho influence the heat loss at small values of ho, for which the outside convection resistance is not negligible relative to the total resistance. However, the resistance becomes negligible with increasing ho, particularly for the triple pane window, and changes in ho have little effect on the heat loss. COMMENTS: The largest contribution to the thermal resistance is due to conduction across the enclosed air. Note that this air could be in motion due to free convection currents. If the corresponding convection coefficient exceeded 3.5 W/m2⋅K, the thermal resistance would be less than that predicted by assuming conduction across stagnant air.

PROBLEM 3.9 KNOWN: Thicknesses of three materials which form a composite wall and thermal conductivities of two of the materials. Inner and outer surface temperatures of the composite; also, temperature and convection coefficient associated with adjoining gas. FIND: Value of unknown thermal conductivity, kB. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible contact resistance, (5) Negligible radiation effects. ANALYSIS: Referring to the thermal circuit, the heat flux may be expressed as $ Ts,i − Ts,o 600 − 20 ) C ( q′′ = = 0.3 m 0.15 m 0.15 m L A L B LC + + + + kB 50 W/m ⋅ K k A k B k C 20 W/m ⋅ K

q′′=

580 W/m 2 . 0.018+0.15/k B

(1)

The heat flux may be obtained from

(

)

q′′=h T∞ − Ts,i = 25 W/m 2 ⋅ K (800-600 ) C $

(2)

q′′=5000 W/m 2 . Substituting for the heat flux from Eq. (2) into Eq. (1), find 0.15 580 580 = − 0.018 = − 0.018 = 0.098 kB q′′ 5000 k B = 1.53 W/m ⋅ K. COMMENTS: Radiation effects are likely to have a significant influence on the net heat flux at the inner surface of the oven.

<

PROBLEM 3.10 KNOWN: Properties and dimensions of a composite oven window providing an outer surface safe2 to-touch temperature Ts,o = 43°C with outer convection coefficient ho = 30 W/m ⋅K and ε = 0.9 when the oven wall air temperatures are Tw = Ta = 400°C. See Example 3.1. FIND: Values of the outer convection coefficient ho required to maintain the safe-to-touch condition when the oven wall-air temperature is raised to 500°C or 600°C. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in window with no contact resistance and constant properties, (3) Negligible absorption in window material, (4) Radiation exchange processes are between small surface and large isothermal surroundings. ANALYSIS: From the analysis in the Ex. 3.1 Comment 2, the surface energy balances at the inner and outer surfaces are used to determine the required value of ho when Ts,o = 43°C and Tw,i = Ta = 500 or 600°C.

)

(

4 − T4 + h T − T εσ Tw,i i( a s,i ) = s,i

Ts,i − Ts,o

(LA / k A ) + (LB / k B )

(

Ts,i − Ts,o

(LA / k A ) + (LB / k B )

)

(

4 − T4 = εσ Ts,o w,o + h o Ts,o − T∞

)

Using these relations in IHT, the following results were calculated: Tw,i, Ts(°C) 400 500 600

Ts,i(°C) 392 493 594

2

ho(W/m ⋅K) 30 40.4 50.7

COMMENTS: Note that the window inner surface temperature is closer to the oven air-wall temperature as the outer convection coefficient increases. Why is this so?

PROBLEM 3.11 KNOWN: Drying oven wall having material with known thermal conductivity sandwiched between thin metal sheets. Radiation and convection conditions prescribed on inner surface; convection conditions on outer surface. FIND: (a) Thermal circuit representing wall and processes and (b) Insulation thickness required to maintain outer wall surface at To = 40°C. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Thermal resistance of metal sheets negligible. ANALYSIS: (a) The thermal circuit is shown above. Note labels for the temperatures, thermal resistances and the relevant heat fluxes. (b) Perform energy balances on the i- and o- nodes finding

T∞,i − Ti R ′′cv,i

T −T + o i + q′′rad = 0 R ′′cd

(1)

Ti − To T∞,o − To + =0 R ′′cd R ′′cv,o

(2)

where the thermal resistances are

R ′′cv,i = 1/ h i = 0.0333 m 2 ⋅ K / W

(3)

R ′′cd = L / k = L / 0.05 m 2 ⋅ K / W

(4)

R ′′cv,o = 1/ h o = 0.0100 m 2 ⋅ K / W

(5)

Substituting numerical values, and solving Eqs. (1) and (2) simultaneously, find

<

L = 86 mm

COMMENTS: (1) The temperature at the inner surface can be found from an energy balance on the i-node using the value found for L.

T∞,i − Ti R ′′cv,o

+

T∞,o − Ti

R ′′cd + R ′′cv,i

+ q′′rad = 0

Ti = 298.3°C

It follows that Ti is close to T∞,i since the wall represents the dominant resistance of the system. (2) Verify that q′′i = 50 W / m 2 and q′′o = 150 W / m 2 . Is the overall energy balance on the system satisfied?

PROBLEM 3.12 KNOWN: Configurations of exterior wall. Inner and outer surface conditions. FIND: Heating load for each of the three cases. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible radiation effects. PROPERTIES: (T = 300 K): Table A.3: plaster board, kp = 0.17 W/m⋅K; urethane, kf = 0.026 W/m⋅K; wood, kw = 0.12 W/m⋅K; glass, kg = 1.4 W/m⋅K. Table A.4: air, ka = 0.0263 W/m⋅K. ANALYSIS: (a) The heat loss may be obtained by dividing the overall temperature difference by the total thermal resistance. For the composite wall of unit surface area, A = 1 m2, q=

T∞ ,i − T∞ ,o

(1 h i ) + ( L p k p ) + ( Lf k f ) + ( L w k w ) + (1 h o ) A

(

$

q=

$

20 C − −15 C

)

( 0.2 + 0.059 + 1.92 + 0.083 + 0.067 ) m 2 ⋅ K W  1m 2   $

q=

35 C 2.33 K W

<

= 15.0 W

(b) For the single pane of glass, q=

T∞ ,i − T∞ ,o

(1 h i ) + ( Lg k g ) + (1 h o ) A $

q=

$

35 C

( 0.2 + 0.002 + 0.067 ) m 2 ⋅ K W  1m 2  

=

35 C 0.269 K W

= 130.3 W

<

(c) For the double pane window, q=

T∞ ,i − T∞ ,o

(1 h i ) + 2 ( Lg k g ) + ( La k a ) + (1 h o ) A $

q=

35 C

( 0.2 + 0.004 + 0.190 + 0.067 ) m 2 ⋅ K W  1m 2  

$

=

35 C 0.461K W

= 75.9 W

<

COMMENTS: The composite wall is clearly superior from the standpoint of reducing heat loss, and the dominant contribution to its total thermal resistance (82%) is associated with the foam insulation. Even with double pane construction, heat loss through the window is significantly larger than that for the composite wall.

PROBLEM 3.13 KNOWN: Composite wall of a house with prescribed convection processes at inner and outer surfaces. FIND: (a) Expression for thermal resistance of house wall, Rtot; (b) Total heat loss, q(W); (c) Effect on heat loss due to increase in outside heat transfer convection coefficient, ho; and (d) Controlling resistance for heat loss from house. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Negligible contact resistance.

(

)

PROPERTIES: Table A-3, T = ( Ti + To ) / 2 = ( 20 − 15 ) C/2=2.5$C ≈ 300K : Fiberglass $

3

blanket, 28 kg/m , kb = 0.038 W/m⋅K; Plywood siding, ks = 0.12 W/m⋅K; Plasterboard, kp = 0.17 W/m⋅K. ANALYSIS: (a) The expression for the total thermal resistance of the house wall follows from Eq. 3.18. Lp L L 1 1 R tot = + + b + s + . < hiA k pA k bA ksA hoA (b) The total heat loss through the house wall is q = ∆T/R tot = ( Ti − To ) / R tot . Substituting numerical values, find 1

R tot =

0.01m

+

+

0.10m

30W/m 2 ⋅ K × 350m 2 0.17W/m ⋅ K × 350m 2 0.038W/m ⋅ K × 350m 2 0.02m 1 + + 2 2 0.12W/m ⋅ K × 350m 60W/m ⋅ K × 350m2 R tot = [9.52 + 16.8 + 752 + 47.6 + 4.76]× 10−5 $C/W = 831× 10−5 $C/W The heat loss is then, $

q=  20- (-15 ) C/831×10-5 $C/W=4.21 kW. 2

< -5

(c) If ho changes from 60 to 300 W/m ⋅K, Ro = 1/hoA changes from 4.76 × 10 °C/W to 0.95 -5

-5

× 10 °C/W. This reduces Rtot to 826 × 10 °C/W, which is a 0.5% decrease and hence a 0.5% increase in q. (d) From the expression for Rtot in part (b), note that the insulation resistance, Lb/kbA, is 752/830 ≈ 90% of the total resistance. Hence, this material layer controls the resistance of the wall. From part (c) note that a 5-fold decrease in the outer convection resistance due to an increase in the wind velocity has a negligible effect on the heat loss.

PROBLEM 3.14 KNOWN: Composite wall of a house with prescribed convection processes at inner and outer surfaces. FIND: Daily heat loss for prescribed diurnal variation in ambient air temperature. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction (negligible change in wall thermal energy storage over 24h period), (2) Negligible contact resistance. 3

PROPERTIES: Table A-3, T ≈ 300 K: Fiberglass blanket (28 kg/m ), kb = 0.038 W/m⋅K; Plywood, ks = 0.12 W/m⋅K; Plasterboard, kp = 0.17 W/m⋅K. ANALYSIS: The heat loss may be approximated as Q =

24h T − T ∞,i ∞,o

∫

0

R tot

dt where

L p L b Ls 1  + + +  +  A  h i k p k b k s h o   1  1 0.01m 0.1m 0.02m 1 R tot = + + + +  2 2 2 200m  30 W/m ⋅ K 0.17 W/m ⋅ K 0.038 W/m ⋅ K 0.12 W/m ⋅ K 60 W/m ⋅ K  R tot =

11

R tot = 0.01454 K/W.

Hence the heat rate is 12h 1  2π   Q=  ∫  293 −  273 + 5 sin R tot  24   0 

Q = 68.8

 t   dt + 

24h

∫

12

2π    293 −  273 + 11 sin 24  

   t   dt    

W   2π t  12  2π t  24   24   24  +  20t+11   cos   20t+5   cos K⋅h  K   24  0  24  12   2π   2π 

60 132     Q = 68.8   240 + ( −1 − 1) +  480 − 240 + (1 + 1)  W ⋅ h π π     Q = 68.8 {480-38.2+84.03} W ⋅ h Q=36.18 kW ⋅ h=1.302 × 108J. COMMENTS: From knowledge of the fuel cost, the total daily heating bill could be determined. For example, at a cost of 0.10$/kW⋅h, the heating bill would be $3.62/day.

<

PROBLEM 3.15 KNOWN: Dimensions and materials associated with a composite wall (2.5m × 6.5m, 10 studs each 2.5m high).

FIND: Wall thermal resistance. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Temperature of composite depends only on x (surfaces normal to x are isothermal), (3) Constant properties, (4) Negligible contact resistance.

PROPERTIES: Table A-3 (T ≈ 300K): Hardwood siding, kA = 0.094 W/m⋅K; Hardwood,

3

kB = 0.16 W/m⋅K; Gypsum, kC = 0.17 W/m⋅K; Insulation (glass fiber paper faced, 28 kg/m ), kD = 0.038 W/m⋅K. ANALYSIS: Using the isothermal surface assumption, the thermal circuit associated with a single unit (enclosed by dashed lines) of the wall is

( LA / k A A A ) =

0.008m = 0.0524 K/W 0.094 W/m ⋅ K (0.65m × 2.5m )

( LB / k BA B ) =

0.13m = 8.125 K/W 0.16 W/m ⋅ K ( 0.04m × 2.5m )

( LD /k D A D ) =

0.13m = 2.243 K/W 0.038 W/m ⋅ K ( 0.61m × 2.5m )

( LC / k C A C ) =

0.012m = 0.0434 K/W. 0.17 W/m ⋅ K ( 0.65m × 2.5m )

The equivalent resistance of the core is −1

R eq = (1/ R B + 1/ R D )

−1

= (1/ 8.125 + 1/ 2.243)

= 1.758 K/W

and the total unit resistance is R tot,1 = R A + R eq + R C = 1.854 K/W. With 10 such units in parallel, the total wall resistance is

(

R tot = 10 × 1/ R tot,1

)−1 = 0.1854 K/W.

COMMENTS: If surfaces parallel to the heat flow direction are assumed adiabatic, the thermal circuit and the value of Rtot will differ.

<

PROBLEM 3.16 KNOWN: Conditions associated with maintaining heated and cooled conditions within a refrigerator compartment. FIND: Coefficient of performance (COP). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state operating conditions, (2) Negligible radiation, (3) Compartment completely sealed from ambient air. ANALYSIS: The Case (a) experiment is performed to determine the overall thermal resistance to heat transfer between the interior of the refrigerator and the ambient air. Applying an energy balance to a control surface about the refrigerator, it follows from Eq. 1.11a that, at any instant, E g − E out = 0 Hence, q elec − q out = 0

(

where q out = T∞,i − T∞,o

)

R t . It follows that

T∞,i − T∞,o (90 − 25 )$ C = = 3.25$ C/W q elec 20 W For Case (b), heat transfer from the ambient air to the compartment (the heat load) is balanced by heat transfer to the refrigerant (qin = qout). Hence, the thermal energy transferred from the refrigerator over the 12 hour period is T∞,i − T∞,o Qout = q out ∆t = qin ∆t = ∆t Rt Rt =

( 25 − 5 )$ C

(12 h × 3600 s h ) = 266, 000 J 3.25$ C W The coefficient of performance (COP) is therefore Q 266, 000 COP = out = = 2.13 Win 125, 000 COMMENTS: The ideal (Carnot) COP is Tc 278 K COP )ideal = = = 13.9 Th − Tc ( 298 − 278 ) K Qout =

and the system is operating well below its peak possible performance.

<

PROBLEM 3.17 KNOWN: Total floor space and vertical distance between floors for a square, flat roof building. FIND: (a) Expression for width of building which minimizes heat loss, (b) Width and number of floors which minimize heat loss for a prescribed floor space and distance between floors. Corresponding heat loss, percent heat loss reduction from 2 floors. SCHEMATIC:

ASSUMPTIONS: Negligible heat loss to ground. ANALYSIS: (a) To minimize the heat loss q, the exterior surface area, As, must be minimized. From Fig. (a)

As = W 2 + 4WH = W 2 + 4WNf Hf where

Nf = Af W 2 Hence,

As = W 2 + 4WAf Hf W 2 = W 2 + 4A f Hf W The optimum value of W corresponds to

dAs 4Af Hf = 2W − =0 dW W2 or

Wop = ( 2Af Hf )

1/ 3

<

The competing effects of W on the areas of the roof and sidewalls, and hence the basis for an optimum, is shown schematically in Fig. (b). (b) For Af = 32,768 m2 and Hf = 4 m,

(

Wop = 2 × 32,768 m 2 × 4 m

)

1/ 3

= 64 m

< Continued …..

PROBLEM 3.17 (Cont.) Hence,

Nf =

Af W2

=

32, 768 m 2

(64 m )2

=8

<

and 2  2 4 × 32, 768 m × 4 m  $ q = UAs ∆T = 1W m 2 ⋅ K ( 64 m ) +  25 C = 307, 200 W 64 m  

<

For Nf = 2, W = (Af/Nf)1/2 = (32,768 m2/2)1/2 = 128 m 2  2 4 × 32, 768 m × 4 m  $ q = 1W m 2 ⋅ K (128 m ) +  25 C = 512,000 W 128 m   % reduction in q = (512,000 - 307,200)/512,000 = 40% COMMENTS: Even the minimum heat loss is excessive and could be reduced by reducing U.

<

PROBLEM 3.18 KNOWN: Concrete wall of 150 mm thickness experiences a flash-over fire with prescribed radiant flux and hot-gas convection on the fire-side of the wall. Exterior surface condition is 300°C, typical ignition temperature for most household and office materials. FIND: (a) Thermal circuit representing wall and processes and (b) Temperature at the fire-side of the wall; comment on whether wall is likely to experience structural collapse for these conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Constant properties. PROPERTIES: Table A-3, Concrete (stone mix, 300 K): k = 1.4 W/m⋅K. ANALYSIS: (a) The thermal cirucit is shown above. Note labels for the temperatures, thermal resistances and the relevant heat fluxes. (b) To determine the fire-side wall surface temperatures, perform an energy balance on the o-node.

T∞ − To T −T + q′′rad = L o R ′′cv R ′′cd where the thermal resistances are

R ′′cv = 1/ h i = 1/ 200 W / m 2 ⋅ K = 0.00500 m 2 ⋅ K / W R ′′cd = L / k = 0.150 m /1.4 W / m ⋅ K = 0.107 m 2 ⋅ K / W Substituting numerical values,

( 400 − To ) K 0.005 m 2 ⋅ K / W To = 515°C

+ 25, 000 W / m 2

(300 − To ) K 0.107 m 2 ⋅ K / W

=0

<

COMMENTS: (1) The fire-side wall surface temperature is within the 350 to 600°C range for which explosive spalling could occur. It is likely the wall will experience structural collapse for these conditions. (2) This steady-state condition is an extreme condition, as the wall may fail before near steady-state conditions can be met.

PROBLEM 3.19 KNOWN: Representative dimensions and thermal conductivities for the layers of fire-fighter’s protective clothing, a turnout coat. FIND: (a) Thermal circuit representing the turnout coat; tabulate thermal resistances of the layers and processes; and (b) For a prescribed radiant heat flux on the fire-side surface and temperature of Ti =.60°C at the inner surface, calculate the fire-side surface temperature, To. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through the layers, (3) Heat is transferred by conduction and radiation exchange across the stagnant air gaps, (3) Constant properties. PROPERTIES: Table A-4, Air (470 K, 1 atm): kab = kcd = 0.0387 W/m⋅K. ANALYSIS: (a) The thermal circuit is shown with labels for the temperatures and thermal resistances.

The conduction thermal resistances have the form R ′′cd = L / k while the radiation thermal resistances across the air gaps have the form

R ′′rad =

1 h rad

=

1 3 4σ Tavg

The linearized radiation coefficient follows from Eqs. 1.8 and 1.9 with ε = 1 where Tavg represents the average temperature of the surfaces comprising the gap

(

)

3 h rad = σ ( T1 + T2 ) T12 + T22 ≈ 4σ Tavg For the radiation thermal resistances tabulated below, we used Tavg = 470 K. Continued …..

PROBLEM 3.19 (Cont.) Shell (s)

Air gap (a-b)

Barrier (mb)

Air gap (c-d)

Liner (tl)

R ′′cd m ⋅ K / W

0.0259

0.04583

0.0259

0.00921

R ′′rad

2

0.04264

--

0.04264

--

--

R ′′gap

2

0.01611

--

0.01611

--

--

--

0.1043

( ) 0.01702 (m ⋅ K / W ) -(m ⋅ K / W ) -2

R ′′total

--

--

--

--

Total (tot) --

From the thermal circuit, the resistance across the gap for the conduction and radiation processes is

1 1 1 = + R ′′gap R ′′cd R ′′rad and the total thermal resistance of the turn coat is

R ′′tot = R ′′cd,s + R ′′gap,a − b + R ′′cd,mb + R ′′gap,c − d + R ′′cd,tl 2

(b) If the heat flux through the coat is 0.25 W/cm , the fire-side surface temperature To can be calculated from the rate equation written in terms of the overall thermal resistance.

q′′ = ( To − Ti ) / R ′′tot

(

To = 66°C + 0.25 W / cm 2 × 102 cm / m

) × 0.1043 m2 ⋅ K / W 2

To = 327°C COMMENTS: (1) From the tabulated results, note that the thermal resistance of the moisture barrier (mb) is nearly 3 times larger than that for the shell or air gap layers, and 4.5 times larger than the thermal liner layer. (2) The air gap conduction and radiation resistances were calculated based upon the average temperature of 470 K. This value was determined by setting Tavg = (To + Ti)/2 and solving the equation set using IHT with kair = kair (Tavg).

PROBLEM 3.20 KNOWN: Materials and dimensions of a composite wall separating a combustion gas from a liquid coolant. FIND: (a) Heat loss per unit area, and (b) Temperature distribution. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3) Constant properties, (4) Negligible radiation effects. PROPERTIES: Table A-1, St. St. (304) ( T ≈ 1000K ) : k = 25.4 W/m⋅K; Table A-2, Beryllium Oxide (T ≈ 1500K): k = 21.5 W/m⋅K. ANALYSIS: (a) The desired heat flux may be expressed as q′′=

$ T∞,1 − T∞,2 2600 − 100 ) C ( = 1 LA L 1 0.02 1  m 2 .K  1 0.01 + + R t,c + B + 0.05 + + + + h1 k A k B h 2  50 21.5 25.4 1000  W 

q′′=34,600 W/m 2 .

<

(b) The composite surface temperatures may be obtained by applying appropriate rate equations. From the fact that q′′=h1 T∞,1 − Ts,1 , it follows that

(

Ts,1 = T∞,1 −

(

)

q′′ 34, 600 W/m 2 1908$ C. = 2600$ C − 2 h1 50 W/m ⋅ K

)

With q′′= ( k A / LA ) Ts,1 − Tc,1 , it also follows that L q′′ 0.01m × 34,600 W/m 2 Tc,1 = Ts,1 − A = 1908$ C − = 1892$ C. kA 21.5 W/m ⋅ K

(

)

Similarly, with q′′= Tc,1 − Tc,2 / R t,c Tc,2 = Tc,1 − R t,c q′′=1892$C − 0.05

m2 ⋅ K W × 34, 600 = 162$ C 2 W m Continued …..

PROBLEM 3.20 (Cont.)

(

)

and with q′′= ( k B / LB ) Tc,2 − Ts,2 , L q′′ 0.02m × 34,600 W/m 2 Ts,2 = Tc,2 − B = 162$ C − = 134.6$ C. kB 25.4 W/m ⋅ K The temperature distribution is therefore of the following form:

< COMMENTS: (1) The calculations may be checked by recomputing q′′ from

(

)

q′′=h 2 Ts,2 − T∞,2 = 1000W/m2 ⋅ K (134.6-100 ) C=34,600W/m2 $

(2) The initial estimates of the mean material temperatures are in error, particularly for the stainless steel. For improved accuracy the calculations should be repeated using k values corresponding to T ≈ 1900°C for the oxide and T ≈ 115°C for the steel. (3) The major contributions to the total resistance are made by the combustion gas boundary layer and the contact, where the temperature drops are largest.

PROBLEM 3.21 KNOWN: Thickness, overall temperature difference, and pressure for two stainless steel plates. FIND: (a) Heat flux and (b) Contact plane temperature drop. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3) Constant properties. PROPERTIES: Table A-1, Stainless Steel (T ≈ 400K): k = 16.6 W/m⋅K. ANALYSIS: (a) With R ′′t,c ≈ 15 × 10−4 m 2 ⋅ K/W from Table 3.1 and L 0.01m = = 6.02 × 10−4 m 2 ⋅ K/W, k 16.6 W/m ⋅ K it follows that R ′′tot = 2 ( L/k ) + R ′′t,c ≈ 27 × 10−4 m 2 ⋅ K/W; hence q′′=

100$ C ∆T = = 3.70 × 104 W/m 2 . -4 2 R ′′tot 27 × 10 m ⋅ K/W

<

(b) From the thermal circuit, R ′′t,c 15 × 10−4 m 2 ⋅ K/W ∆Tc = = = 0.556. Ts,1 − Ts,2 R ′′tot 27 × 10-4 m 2 ⋅ K/W Hence,

(

)

(

)

∆Tc = 0.556 Ts,1 − Ts,2 = 0.556 100$ C = 55.6$ C.

<

COMMENTS: The contact resistance is significant relative to the conduction resistances. The value of R ′′t,c would diminish, however, with increasing pressure.

PROBLEM 3.22 KNOWN: Temperatures and convection coefficients associated with fluids at inner and outer surfaces of a composite wall. Contact resistance, dimensions, and thermal conductivities associated with wall materials. FIND: (a) Rate of heat transfer through the wall, (b) Temperature distribution. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Negligible radiation, (4) Constant properties. ANALYSIS: (a) Calculate the total resistance to find the heat rate, 1 L L 1 + A + R t,c + B + h1A k A A k BA h 2 A 0.01 0.3 0.02 1 K  1 R tot =  + + + + 10 × 5 0.1× 5 5 0.04 × 5 20 × 5  W K K R tot = [0.02 + 0.02 + 0.06 + 0.10 + 0.01] = 0.21 W W R tot =

q=

T∞,1 − T∞,2 R tot

$ 200 − 40 ) C ( = = 762 W.

0.21 K/W

(b) It follows that Ts,1 = T∞,1 − TA = Ts,1 −

q 762 W = 200$ C − = 184.8$ C h1A 50 W/K

qL A

$

= 184.8 C −

kAA

762W × 0.01m 0.1

W m⋅K

× 5m

$

= 169.6 C

2

K

TB = TA − qR t,c = 169.6$ C − 762W × 0.06 Ts,2 = TB −

qL B

$

k BA

T∞,2 = Ts,2 −

= 123.8 C −

q h 2A

W

762W × 0.02m 0.04

= 47.6$ C −

W m⋅K

× 5m

762W 100W/K

= 123.8$ C $

= 47.6 C

2

= 40$ C

<

PROBLEM 3.23 KNOWN: Outer and inner surface convection conditions associated with zirconia-coated, Inconel turbine blade. Thicknesses, thermal conductivities, and interfacial resistance of the blade materials. Maximum allowable temperature of Inconel. FIND: Whether blade operates below maximum temperature. Temperature distribution in blade, with and without the TBC. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction in a composite plane wall, (2) Constant properties, (3) Negligible radiation. ANALYSIS: For a unit area, the total thermal resistance with the TBC is R ′′tot,w = h o−1 + ( L k )Zr + R ′′t,c + ( L k )In + h i−1

(

)

R ′′tot,w = 10−3 + 3.85 × 10−4 + 10−4 + 2 × 10−4 + 2 × 10 −3 m 2 ⋅ K W = 3.69 × 10−3 m 2 ⋅ K W With a heat flux of T∞,o − T∞,i 1300 K q′′w = = = 3.52 × 105 W m 2 −3 2 R ′′tot,w 3.69 × 10 m ⋅ K W the inner and outer surface temperatures of the Inconel are

(

)

Ts,i(w) = T∞ ,i + ( q′′w h i ) = 400 K + 3.52 × 105 W m 2 500 W m 2 ⋅ K = 1104 K

(

Ts,o(w) = T∞ ,i + (1 h i ) + ( L k )In  q ′′w = 400 K + 2 × 10

−3

+ 2 × 10

−4

)m

2

(

5

⋅ K W 3.52 × 10 W m

2

) = 1174 K

Without the TBC, R ′′tot,wo = h o−1 + ( L k )In + h i−1 = 3.20 × 10 −3 m 2 ⋅ K W , and q ′′wo = ( T∞,o − T∞,i ) R ′′tot,wo = (1300 K)/3.20×10-3 m2⋅K/W = 4.06×105 W/m2. The inner and outer surface temperatures of the Inconel are then

(

)

Ts,i(wo) = T∞ ,i + ( q′′wo h i ) = 400 K + 4.06 × 105 W m2 500 W m 2 ⋅ K = 1212 K Ts,o(wo) = T∞ ,i +

[(1 h i ) + ( L k )In ] q′′wo = 400 K + ( 2 × 10−3 + 2 × 10−4 ) m 2 ⋅ K

(

5

W 4.06 × 10 W m

2

)

= 1293 K

Continued...

PROBLEM 3.23 (Cont.)

Temperature, T(K)

1300 1260 1220 1180 1140 1100 0

0.001

0.002

0.003

0.004

0.005

Inconel location, x(m) With TBC Without TBC

Use of the TBC facilitates operation of the Inconel below Tmax = 1250 K. COMMENTS: Since the durability of the TBC decreases with increasing temperature, which increases with increasing thickness, limits to the thickness are associated with reliability considerations.

PROBLEM 3.24 KNOWN: Size and surface temperatures of a cubical freezer. Materials, thicknesses and interface resistances of freezer wall. FIND: Cooling load. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant properties. PROPERTIES: Table A-1, Aluminum 2024 (~267K): kal = 173 W/m⋅K. Table A-1, Carbon steel AISI 1010 (~295K): kst = 64 W/m⋅K. Table A-3 (~300K): kins = 0.039 W/m⋅K. ANALYSIS: For a unit wall surface area, the total thermal resistance of the composite wall is

L L L R ′′tot = al + R ′′t,c + ins + R ′′t,c + st k al k ins kst R ′′tot =

0.00635m m2 ⋅ K 0.100m m2 ⋅ K 0.00635m + 2.5 × 10−4 + + 2.5 ×10−4 + 173 W / m ⋅ K W 0.039 W / m ⋅ K W 64 W / m ⋅ K

)

(

R ′′tot = 3.7 × 10−5 + 2.5 ×10−4 + 2.56 + 2.5 ×10−4 + 9.9 × 10−5 m 2 ⋅ K / W Hence, the heat flux is

q′′ =

Ts,o − Ts,i R ′′tot

 22 − ( −6 ) °C W =  = 10.9 2.56 m 2 ⋅ K / W m2

and the cooling load is

q = As q′′ = 6 W 2 q′′ = 54m 2 ×10.9 W / m 2 = 590 W COMMENTS: Thermal resistances associated with the cladding and the adhesive joints are negligible compared to that of the insulation.

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PROBLEM 3.25 KNOWN: Thicknesses and thermal conductivity of window glass and insulation. Contact resistance. Environmental temperatures and convection coefficients. Furnace efficiency and fuel cost. FIND: (a) Reduction in heat loss associated with the insulation, (b) Heat losses for prescribed conditions, (c) Savings in fuel costs for 12 hour period. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional heat transfer, (3) Constant properties. ANALYSIS: (a) The percentage reduction in heat loss is

 q′′ q′′ − q′′with R q = wo × 100% = 1 − with q′′wo q′′wo 

R ′′tot,wo     × 100% = 1 −  × 100%   R ′′tot, with 

where the total thermal resistances without and with the insulation, respectively, are

R ′′tot,wo = R ′′cnv,o + R ′′cnd,w + R ′′cnv,i =

1 Lw 1 + + ho k w hi

R ′′tot,wo = ( 0.050 + 0.004 + 0.200 ) m 2 ⋅ K / W = 0.254 m 2 ⋅ K / W R ′′tot,with = R ′′cnv,o + R ′′cnd,w + R ′′t,c + R ′′cnd,ins + R ′′cnv,i =

L 1 Lw 1 + + R ′′t,c + ins + ho k w kins h i

R ′′tot,with = (0.050 + 0.004 + 0.002 + 0.926 + 0.500 ) m 2 ⋅ K / W = 1.482 m 2 ⋅ K / W R q = (1 − 0.254 /1.482 ) × 100% = 82.9%

<

2

(b) With As = 12 m , the heat losses without and with the insulation are q wo = As T∞,i − T∞,o / R ′′tot,wo = 12 m 2 × 32°C / 0.254 m 2 ⋅ K / W = 1512 W

(

)

(

)

q with = As T∞,i − T∞,o / R ′′tot,with = 12 m 2 × 32°C /1.482 m 2 ⋅ K / W = 259 W

< <

(c) With the windows covered for 12 hours per day, the daily savings are S=

( q wo − q with ) ηf

∆t C g × 10

−6

MJ / J =

(1512 − 259 ) W 0.8

12h × 3600 s / h × $0.01 / MJ × 10

−6

MJ / J = $0.677

COMMENTS: (1) The savings may be insufficient to justify the cost of the insulation, as well as the daily tedium of applying and removing the insulation. However, the losses are significant and unacceptable. The owner of the building should install double pane windows. (2) The dominant contributions to the total thermal resistance are made by the insulation and convection at the inner surface.

PROBLEM 3.26 KNOWN: Surface area and maximum temperature of a chip. Thickness of aluminum cover and chip/cover contact resistance. Fluid convection conditions. FIND: Maximum chip power. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Negligible heat loss from sides and bottom, (4) Chip is isothermal. PROPERTIES: Table A.1, Aluminum (T ≈ 325 K): k = 238 W/m⋅K. ANALYSIS: For a control surface about the chip, conservation of energy yields E g − E out = 0 or Pc −

(Tc − T∞ ) A =0 ( L/k ) + R ′′t,c + (1/ h ) (85 − 25 )$ C

Pc,max = Pc,max =

(10-4m2 )

( 0.002 / 238 ) + 0.5 × 10−4 + (1/1000 ) m 2 ⋅ K/W   $ 4 2 − 60 × 10 C⋅m

(8.4 ×10-6 + 0.5 ×10−4 + 10−3 ) m2 ⋅ K/W

<

Pc,max = 5.7 W.

!

&

COMMENTS: The dominant resistance is that due to convection R conv > R t,c >> R cond .

PROBLEM 3.27 KNOWN: Operating conditions for a board mounted chip. FIND: (a) Equivalent thermal circuit, (b) Chip temperature, (c) Maximum allowable heat dissipation for dielectric liquid (ho = 1000 W/m2⋅K) and air (ho = 100 W/m2⋅K). Effect of changes in circuit board temperature and contact resistance. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible chip thermal resistance, (4) Negligible radiation, (5) Constant properties. PROPERTIES: Table A-3, Aluminum oxide (polycrystalline, 358 K): kb = 32.4 W/m⋅K. ANALYSIS: (a)

(b) Applying conservation of energy to a control surface about the chip ( E in − E out = 0 ) , q′′c − qi′′ − q′′o = 0 q′′c =

Tc − T∞,i

1 h i + ( L k )b + R ′′t,c

+

Tc − T∞ ,o 1 ho

With q „„c 3 – 10 4 W m 2 , ho = 1000 W/m2⋅K, kb = 1 W/m⋅K and R ′′t,c = 10 −4 m 2 ⋅ K W , 3 × 104 W m 2 =

Tc − 20$ C

(1 40 + 0.005 1 + 10−4 ) m2 ⋅ K W

+

Tc − 20$ C

(1 1000 ) m 2 ⋅ K

W

3 × 104 W m 2 = (33.2Tc − 664 + 1000Tc − 20, 000 ) W m 2 ⋅ K 1003Tc = 50,664

<

Tc = 49°C. (c) For Tc = 85°C and ho = 1000 W/m2⋅K, the foregoing energy balance yields

<

q′′c = 67,160 W m 2

with q′′o = 65,000 W/m and q′′i = 2160 W/m . Replacing the dielectric with air (ho = 100 W/m ⋅K), the following results are obtained for different combinations of kb and R ′′t,c . 2

2

2

Continued...

PROBLEM 3.27 (Cont.) R ′′t,c

kb (W/m⋅K)

q′′i (W/m )

q ′′o (W/m )

q ′′c (W/m )

2159 2574 2166 2583

6500 6500 6500 6500

8659 9074 8666 9083

2

2

2

(m2⋅K/W)

1 32.4 1 32.4

10-4 10-4 10-5 10-5

<

COMMENTS: 1. For the conditions of part (b), the total internal resistance is 0.0301 m2⋅K/W, while the outer resistance is 0.001 m2⋅K/W. Hence

( (

Tc − T∞,o q′′o = q′′i Tc − T∞,i

) R ′′o = 0.0301 = 30 . ) R ′′i 0.001

and only approximately 3% of the heat is dissipated through the board. 2. With ho = 100 W/m2⋅K, the outer resistance increases to 0.01 m2⋅K/W, in which case q ′′o q i′′ = R i′′ R ′′o = 0.0301/0.01 = 3.1 and now almost 25% of the heat is dissipated through the board. Hence, although measures to reduce R i′′ would have a negligible effect on q′′c for the liquid coolant, some improvement may be gained for air-cooled conditions. As shown in the table of part (b), use of an aluminum oxide board increase q′′i by 19% (from 2159 to 2574 W/m2) by reducing R i′′ from 0.0301 to 0.0253 m2⋅K/W. Because the initial contact resistance ( R ′′t,c = 10 −4 m 2 ⋅ K W ) is already much less than R i′′ , any reduction in its value would have a negligible effect on q′′i . The largest gain would be realized by increasing hi, since the inside convection resistance makes the dominant contribution to the total internal resistance.

PROBLEM 3.28 KNOWN: Dimensions, thermal conductivity and emissivity of base plate. Temperature and convection coefficient of adjoining air. Temperature of surroundings. Maximum allowable temperature of transistor case. Case-plate interface conditions. FIND: (a) Maximum allowable power dissipation for an air-filled interface, (b) Effect of convection coefficient on maximum allowable power dissipation. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Negligible heat transfer from the enclosure, to the surroundings. (3) One-dimensional conduction in the base plate, (4) Radiation exchange at surface of base plate is with large surroundings, (5) Constant thermal conductivity. 5

2

PROPERTIES: Aluminum-aluminum interface, air-filled, 10 µm roughness, 10 N/m contact pressure (Table 3.1): R ′′t,c = 2.75 × 10−4 m 2 ⋅ K / W. ANALYSIS: (a) With all of the heat dissipation transferred through the base plate,

Pelec = q =

Ts,c − T∞

(1)

R tot

where R tot = R t,c + R cnd + (1/ R cnv ) + (1/ R rad )

R tot =

and

R ′′t,c Ac

+

L kW 2

(

h r = εσ Ts,p + Tsur

+

−1

1  1    W2  h + hr 

(2)

2 + T2 ) (Ts,p sur )

(3)

To obtain Ts,p, the following energy balance must be performed on the plate surface,

q=

Ts,c − Ts,p R t,c + R cnd -4

2

(

)

(

= qcnv + q rad = hW 2 Ts,p − T∞ + h r W 2 Ts,p − Tsur -4

)

2

(4) -4

2

With Rt,c = 2.75 × 10 m ⋅K/W/2×10 m = 1.375 K/W, Rcnd = 0.006 m/(240 W/m⋅K × 4 × 10 m ) = 0.0625 K/W, and the prescribed values of h, W, T∞ = Tsur and ε, Eq. (4) yields a surface temperature of Ts,p = 357.6 K = 84.6°C and a power dissipation of Continued …..

PROBLEM 3.28 (Cont.)

<

Pelec = q = 0.268 W

The convection and radiation resistances are Rcnv = 625 m⋅K/W and Rrad = 345 m⋅K/W, where hr = 2 7.25 W/m ⋅K.

P o w e r d is s ip a tio n , P e le c (W )

(b) With the major contribution to the total resistance made by convection, significant benefit may be derived by increasing the value of h. 4 .5 4 3 .5 3 2 .5 2 1 .5 1 0 .5 0 0

20

40

60

80

100 120 140 160 180 200

C o n ve ctio n c o e ffic ie n t, h (W /m ^2 .K )

2

For h = 200 W/m ⋅K, Rcnv = 12.5 m⋅K/W and Ts,p = 351.6 K, yielding Rrad = 355 m⋅K/W. The effect of radiation is then negligible. 2

COMMENTS: (1) The plate conduction resistance is negligible, and even for h = 200 W/m ⋅K, the contact resistance is small relative to the convection resistance. However, Rt,c could be rendered negligible by using indium foil, instead of an air gap, at the interface. From Table 3.1, R ′′t,c = 0.07 × 10−4 m 2 ⋅ K / W, in which case Rt,c = 0.035 m⋅K/W. 2

(2) Because Ac < W , heat transfer by conduction in the plate is actually two-dimensional, rendering the conduction resistance even smaller.

PROBLEM 3.29 KNOWN: Conduction in a conical section with prescribed diameter, D, as a function of x in 1/2 the form D = ax . FIND: (a) Temperature distribution, T(x), (b) Heat transfer rate, qx. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in xdirection, (3) No internal heat generation, (4) Constant properties. PROPERTIES: Table A-2, Pure Aluminum (500K): k= 236 W/m⋅K. ANALYSIS: (a) Based upon the assumptions, and following the same methodology of Example 3.3, qx is a constant independent of x. Accordingly, q x = − kA 2

using A = πD /4 where D = ax 4q x

x dx

π a 2k

)

(

2  dT  dT = − k π ax1/2 / 4  dx   dx

∫x1

x

1/2

T T1

= −∫

(1)

. Separating variables and identifying limits, (2)

dT.

Integrating and solving for T(x) and then for T2, T ( x ) = T1 −

4q x

π a 2k

ln

x x1

T2 = T1 −

4q x

x ln 2 . π a 2k x1

(3,4)

Solving Eq. (4) for qx and then substituting into Eq. (3) gives the results, qx = −

π 2 a k ( T1 − T2 ) /1n ( x1 / x 2 ) 4

T ( x ) = T1 + ( T1 − T2 )

ln ( x/x1 )

ln ( x1 / x 2 )

.

(5)

<

From Eq. (1) note that (dT/dx)⋅x = Constant. It follows that T(x) has the distribution shown above. (b) The heat rate follows from Eq. (5), qx =

π W 25 × 0.52 m × 236 (600 − 400 ) K/ln = 5.76kW. 4 m⋅K 125

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PROBLEM 3.30 KNOWN: Geometry and surface conditions of a truncated solid cone. FIND: (a) Temperature distribution, (b) Rate of heat transfer across the cone. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x, (3) Constant properties. PROPERTIES: Table A-1, Aluminum (333K): k = 238 W/m⋅K.

(

)

ANALYSIS: (a) From Fourier’s law, Eq. (2.1), with A=π D 2 / 4 = π a 2 / 4 x 3 , it follows that 4q x dx

π a 2 x3

= − kdT.

Hence, since qx is independent of x, T 4q x x dx = − k ∫ dT ∫ 2 3 T1 π a x1 x or x 4q x  1  − = − k ( T − T1 ).   π a 2  2x 2  x1

Hence  1   − 1 . π a 2 k  x 2 x12  (b) From the foregoing expression, it also follows that T = T1 +

2q x

<

T2 − T1 π a 2k 2 1/x 2 − 1/ x 2  1   2 -1 π 1m 238 W/m ⋅ K ( 20 − 100 )$ C qx = × 2 ( 0.225 )−2 − ( 0.075)−2  m-2   qx =

( )

q x = 189 W.

<

COMMENTS: The foregoing results are approximate due to use of a one-dimensional model in treating what is inherently a two-dimensional problem.

PROBLEM 3.31 KNOWN: Temperature dependence of the thermal conductivity, k. FIND: Heat flux and form of temperature distribution for a plane wall. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction through a plane wall, (2) Steady-state conditions, (3) No internal heat generation. ANALYSIS: For the assumed conditions, qx and A(x) are constant and Eq. 3.21 gives L T q′′x ∫ dx = − ∫ 1 ( k o + aT )dT 0

q′′x =

To

(

)

1 a  k o ( To − T1 ) + To2 − T12  .  L 2 

From Fourier’s law, q′′x = − ( k o + aT ) dT/dx. Hence, since the product of (ko+aT) and dT/dx) is constant, decreasing T with increasing x implies, a > 0: decreasing (ko+aT) and increasing |dT/dx| with increasing x a = 0: k = ko => constant (dT/dx) a < 0: increasing (ko+aT) and decreasing |dT/dx| with increasing x. The temperature distributions appear as shown in the above sketch.

PROBLEM 3.32 KNOWN: Temperature dependence of tube wall thermal conductivity. FIND: Expressions for heat transfer per unit length and tube wall thermal (conduction) resistance. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) No internal heat generation. ANALYSIS: From Eq. 3.24, the appropriate form of Fourier’s law is dT dT = − k ( 2π rL ) dr dr dT q′r = −2π kr dr dT q′r = −2π rk o (1 + aT ) . dr q r = − kA r

Separating variables, q′ dr − r = k o (1 + aT ) dT 2π r and integrating across the wall, find q′ − r 2π q′ − r 2π q′ − r 2π

ro dr

∫ri

r ln o ri r ln o ri

r

T = k o ∫ o (1+aT ) dT Ti

 aT 2  To = ko T +  2  Ti   a   = k o ( To − Ti ) + To2 − Ti2  2  

(

)

 a  (T − T ) q′r = −2π k o 1 + ( To + Ti ) o i .  2  ln ( ro / ri ) It follows that the overall thermal resistance per unit length is ln ( ro / ri ) ∆T R ′t = . = q′r  a  2π k o 1 + ( To + Ti )  2  COMMENTS: Note the necessity of the stated assumptions to treating q′r as independent of r.

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PROBLEM 3.33 KNOWN: Steady-state temperature distribution of convex shape for material with k = ko(1 + αT) where α is a constant and the mid-point temperature is ∆To higher than expected for a linear temperature distribution. FIND: Relationship to evaluate α in terms of ∆To and T1, T2 (the temperatures at the boundaries). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heat generation, (4) α is positive and constant. ANALYSIS: At any location in the wall, Fourier’s law has the form dT q′′x = − k o (1 + α T ) . (1) dx Since q′′x is a constant, we can separate Eq. (1), identify appropriate integration limits, and integrate to obtain L

T2

∫0 q′′x dx = − ∫T1

k o (1 + α T )dT

 α T22   α T12   k o     . q′′x = − T2 + − T1 + L  2   2      We could perform the same integration, but with the upper limits at x = L/2, to obtain 2    α TL/2 α T12   2k o     q′′x = − TL/2 + − T1 + L  2   2      where T +T TL/2 = T ( L/2 ) = 1 2 + ∆To . 2 Setting Eq. (3) equal to Eq. (4), substituting from Eq. (5) for TL/2, and solving for α, it follows that 2∆To α= . 2 T22 + T12 / 2 − ( T1 + T2 ) / 2 + ∆To 

(

)

(2)

(3)

(4)

(5)

<

PROBLEM 3.34 KNOWN: Hollow cylinder of thermal conductivity k, inner and outer radii, ri and ro, respectively, and length L. FIND: Thermal resistance using the alternative conduction analysis method. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) No internal volumetric generation, (4) Constant properties. ANALYSIS: For the differential control volume, energy conservation requires that qr = qr+dr for steady-state, one-dimensional conditions with no heat generation. With Fourier’s law, q r = − kA

dT dT = − k ( 2π rL ) dr dr

(1)

where A = 2πrL is the area normal to the direction of heat transfer. Since qr is constant, Eq. (1) may be separated and expressed in integral form, T q r ro dr = − ∫ o k ( T ) dT. ∫ Ti 2π L ri r

Assuming k is constant, the heat rate is qr =

2π Lk ( Ti − To ) ln ( ro / ri )

.

Remembering that the thermal resistance is defined as R t ≡ ∆T/q it follows that for the hollow cylinder, Rt =

ln ( ro / ri ) 2π LK

.

COMMENTS: Compare the alternative method used in this analysis with the standard method employed in Section 3.3.1 to obtain the same result.

<

PROBLEM 3.35 KNOWN: Thickness and inner surface temperature of calcium silicate insulation on a steam pipe. Convection and radiation conditions at outer surface. FIND: (a) Heat loss per unit pipe length for prescribed insulation thickness and outer surface temperature. (b) Heat loss and radial temperature distribution as a function of insulation thickness. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties. PROPERTIES: Table A-3, Calcium Silicate (T = 645 K): k = 0.089 W/m⋅K. ANALYSIS: (a) From Eq. 3.27 with Ts,2 = 490 K, the heat rate per unit length is

(

2π k Ts,1 − Ts,2

q′ = q r L = q′ =

ln ( r2 r1 )

)

2π ( 0.089 W m ⋅ K )(800 − 490 ) K ln ( 0.08 m 0.06 m )

<

q′ = 603 W m .

(b) Performing an energy for a control surface around the outer surface of the insulation, it follows that q′cond = q′conv + q′rad Ts,1 − Ts,2

ln ( r2 r1 ) 2π k

(

=

Ts,2 − T∞

+

Ts,2 − Tsur

1 ( 2π r2 h ) 1 ( 2π r2 h r )

where h r = εσ Ts,2 + Tsur

2 2 + Tsur ) (Ts,2 ).

Solving this equation for Ts,2, the heat rate may be

determined from

(

)

(

)

q′ = 2π r2  h Ts,2 − T∞ + h r Ts,2 − Tsur  Continued...

PROBLEM 3.35 (Cont.) and from Eq. 3.26 the temperature distribution is T(r) =

Ts,1 − Ts,2 ln ( r1 r2 )

 r   + Ts,2  r2 

ln 

As shown below, the outer surface temperature of the insulation Ts,2 and the heat loss q′ decay precipitously with increasing insulation thickness from values of Ts,2 = Ts,1 = 800 K and q′ = 11,600 W/m, respectively, at r2 = r1 (no insulation). 800

10000

Heat loss, qprime(W/m)

Temperature, Ts2(K)

700

600

500

400

1000

300

100

0

0.04

0.08

0.12

0

0.04

Insulation thickness, (r2-r1) (m)

0.08

0.12

Insulation thickness, (r2-r1) (m)

Outer surface temperature

Heat loss, qprime

When plotted as a function of a dimensionless radius, (r - r1)/(r2 - r1), the temperature decay becomes more pronounced with increasing r2.

Temperature, T(r) (K)

800

700

600

500

400

300 0

0.2

0.4

0.6

0.8

1

Dimensionless radius, (r-r1)/(r2-r1) r2 = 0.20m r2 = 0.14m r2= 0.10m

Note that T(r2) = Ts,2 increases with decreasing r2 and a linear temperature distribution is approached as r2 approaches r1. COMMENTS: An insulation layer thickness of 20 mm is sufficient to maintain the outer surface temperature and heat rate below 350 K and 1000 W/m, respectively.

PROBLEM 3.36 KNOWN: Temperature and volume of hot water heater. Nature of heater insulating material. Ambient air temperature and convection coefficient. Unit cost of electric power. FIND: Heater dimensions and insulation thickness for which annual cost of heat loss is less than $50. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction through side and end walls, (2) Conduction resistance dominated by insulation, (3) Inner surface temperature is approximately that of the water (Ts,1 = 55°C), (4) Constant properties, (5) Negligible radiation. PROPERTIES: Table A.3, Urethane Foam (T = 300 K): k = 0.026 W/m⋅K. ANALYSIS: To minimize heat loss, tank dimensions which minimize the total surface area, As,t, should

(

)

be selected. With L = 4∀/πD2, As,t = π DL + 2 π D 2 4 = 4 ∀ D + π D 2 2 , and the tank diameter for which As,t is an extremum is determined from the requirement dAs,t dD = − 4∀ D2 + π D = 0 It follows that D = ( 4∀ π )

1/ 3

L = ( 4∀ π )

1/ 3

and

With d 2 As,t dD 2 = 8∀ D3 + π > 0 , the foregoing conditions yield the desired minimum in As,t. Hence, for ∀ = 100 gal × 0.00379 m3/gal = 0.379 m3,

<

Dop = L op = 0.784 m The total heat loss through the side and end walls is q=

Ts,1 − T∞

ln ( r2 r1 ) 2π kLop

+

1 h2π r2 Lop

(

2 Ts,1 − T∞

+

(

δ

2 k π Dop 4

+

) 1

) h (π Dop2 4)

We begin by estimating the heat loss associated with a 25 mm thick layer of insulation. With r1 = Dop/2 = 0.392 m and r2 = r1 + δ = 0.417 m, it follows that Continued...

PROBLEM 3.36 (Cont.) q=

ln ( 0.417 0.392 )

(55 − 20 )$ C

2π ( 0.026 W m ⋅ K ) 0.784 m

(2 W m ⋅ K ) 2π (0.417 m ) 0.784 m 2

2 (55 − 20 ) C $

+

0.025 m

(0.026 W q=

1

+

(2 W m2 ⋅ K )π 4 (0.784 m )2 2 (35 C ) = ( 48.2 + 23.1) W = 71.3 W

m ⋅ K )π 4 ( 0.784 m )

2

$

35$ C

(0.483 + 0.243) K

1

+

W

+

(1.992 + 1.036 ) K

W

The annual energy loss is therefore

(

)

Qannual = 71.3 W (365 days )( 24 h day ) 10−3 kW W = 625 kWh With a unit electric power cost of $0.08/kWh, the annual cost of the heat loss is C = ($0.08/kWh)625 kWh = $50.00 Hence, an insulation thickness of δ = 25 mm

<

will satisfy the prescribed cost requirement. COMMENTS: Cylindrical containers of aspect ratio L/D = 1 are seldom used because of floor space constraints. Choosing L/D = 2, ∀ = πD3/2 and D = (2∀/π)1/3 = 0.623 m. Hence, L = 1.245 m, r1 = 0.312m and r2 = 0.337 m. It follows that q = 76.1 W and C = $53.37. The 6.7% increase in the annual cost of the heat loss is small, providing little justification for using the optimal heater dimensions.

PROBLEM 3.37 KNOWN: Inner and outer radii of a tube wall which is heated electrically at its outer surface and is exposed to a fluid of prescribed h and T∞. Thermal contact resistance between heater and tube wall and wall inner surface temperature. FIND: Heater power per unit length required to maintain a heater temperature of 25°C. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible temperature drop across heater. ANALYSIS: The thermal circuit has the form

Applying an energy balance to a control surface about the heater, q′ = q′a + q′b To − Ti T −T q′ = + o ∞ ln ( ro / ri ) (1/hπ Do ) + R ′t,c 2π k $ $  25 − ( −10 ) C 25-5 ) C ( q′= + ln (75mm/25mm ) m ⋅ K 1/ 100 W/m 2 ⋅ K × π × 0.15m  + 0.01   2π × 10 W/m ⋅ K W

(

)

q′ = ( 728 + 1649 ) W/m q′=2377 W/m. COMMENTS: The conduction, contact and convection resistances are 0.0175, 0.01 and 0.021 m ⋅K/W, respectively,

<

PROBLEM 3.38 KNOWN: Inner and outer radii of a tube wall which is heated electrically at its outer surface. Inner and outer wall temperatures. Temperature of fluid adjoining outer wall. FIND: Effect of wall thermal conductivity, thermal contact resistance, and convection coefficient on total heater power and heat rates to outer fluid and inner surface. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible temperature drop across heater, (5) Negligible radiation. ANALYSIS: Applying an energy balance to a control surface about the heater, q′ = q′i + q′o q′ =

To − Ti T − T∞ + o ln ( ro ri ) (1 2π ro h ) + R ′t,c 2π k

Selecting nominal values of k = 10 W/m⋅K, R ′t,c = 0.01 m⋅K/W and h = 100 W/m2⋅K, the following parametric variations are obtained 3500

3000

3000

2500 Heat rate(W/m)

Heat rate (W/m)

2500 2000 1500 1000

2000 1500 1000 500

500 0

0

0

50

100

150

200

0

0.02

Thermal conductivity, k(W/m.K) qi q qo

0.04

0.06

0.08

0.1

Contact resistance, Rtc(m.K/W) qi q qo

Continued...

PROBLEM 3.38 (Cont.)

20000

Heat rate(W/m)

16000

12000

8000

4000

0 0

200

400

600

800

1000

Convection coefficient, h(W/m^2.K) qi q qo

For a prescribed value of h, q′o is fixed, while qi′ , and hence q′ , increase and decrease, respectively, with increasing k and R ′t,c . These trends are attributable to the effects of k and R ′t,c on the total (conduction plus contact) resistance separating the heater from the inner surface. For fixed k and R ′t,c , q′i is fixed, while q′o , and hence q′ , increase with increasing h due to a reduction in the convection resistance. COMMENTS: For the prescribed nominal values of k, R ′t,c and h, the electric power requirement is q′ = 2377 W/m. To maintain the prescribed heater temperature, q′ would increase with any changes which reduce the conduction, contact and/or convection resistances.

PROBLEM 3.39 KNOWN: Wall thickness and diameter of stainless steel tube. Inner and outer fluid temperatures and convection coefficients. FIND: (a) Heat gain per unit length of tube, (b) Effect of adding a 10 mm thick layer of insulation to outer surface of tube. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant properties, (4) Negligible contact resistance between tube and insulation, (5) Negligible effect of radiation. PROPERTIES: Table A-1, St. St. 304 (~280K): kst = 14.4 W/m⋅K. ANALYSIS: (a) Without the insulation, the total thermal resistance per unit length is

R ′tot = R ′conv,i + R ′cond,st + R ′conv,o = R ′tot =

1 2π (0.018m ) 400 W / m 2 ⋅ K

+

ln ( r2 / ri ) 1 1 + + 2π ri h i 2π k st 2π r2 h o ln ( 20 /18 )

2π (14.4 W / m ⋅ K )

+

1 2π (0.020m ) 6 W / m 2 ⋅ K

)

(

R ′tot = 0.0221 + 1.16 × 10−3 + 1.33 m ⋅ K / W = 1.35 m ⋅ K / W The heat gain per unit length is then

q′ =

T∞,o − T∞,i R ′tot

=

( 23 − 6 ) °C

1.35 m ⋅ K / W

<

= 12.6 W / m

(b) With the insulation, the total resistance per unit length is now R ′tot = R ′conv,i + R ′cond,st + R ′cond,ins + R ′conv,o , where R ′conv,i and R ′cond,st remain the same. The thermal resistance of the insulation is

R ′cond,ins =

ln ( r3 / r2 ) 2π k ins

=

ln (30 / 20 )

2π (0.05 W / m ⋅ K )

= 1.29 m ⋅ K / W

and the outer convection resistance is now

R ′conv,o =

1 1 = = 0.88 m ⋅ K / W 2π r3h o 2π ( 0.03m ) 6 W / m 2 ⋅ K

The total resistance is now

(

)

R ′tot = 0.0221 + 1.16 × 10−3 + 1.29 + 0.88 m ⋅ K / W = 2.20 m ⋅ K / W Continued …..

PROBLEM 3.39 (Cont.) and the heat gain per unit length is

q′ =

T∞,o − T∞,i R ′tot

=

17°C = 7.7 W / m 2.20 m ⋅ K / W

COMMENTS: (1) The validity of assuming negligible radiation may be assessed for the worst case condition corresponding to the bare tube. Assuming a tube outer surface temperature of Ts = T∞,i = 279K, large surroundings at Tsur = T∞,o = 296K, and an emissivity of ε = 0.7, the heat gain due to net

(

)

4 4 radiation exchange with the surroundings is q ′rad = εσ ( 2π r2 ) Tsur − Ts = 7.7 W / m. Hence, the net

rate of heat transfer by radiation to the tube surface is comparable to that by convection, and the assumption of negligible radiation is inappropriate. (2) If heat transfer from the air is by natural convection, the value of ho with the insulation would actually be less than the value for the bare tube, thereby further reducing the heat gain. Use of the insulation would also increase the outer surface temperature, thereby reducing net radiation transfer from the surroundings. (3) The critical radius is rcr = kins/h ≈ 8 mm < r2. Hence, as indicated by the calculations, heat transfer is reduced by the insulation.

PROBLEM 3.40 KNOWN: Diameter, wall thickness and thermal conductivity of steel tubes. Temperature of steam flowing through the tubes. Thermal conductivity of insulation and emissivity of aluminum sheath. Temperature of ambient air and surroundings. Convection coefficient at outer surface and maximum allowable surface temperature. FIND: (a) Minimum required insulation thickness (r3 – r2) and corresponding heat loss per unit length, (b) Effect of insulation thickness on outer surface temperature and heat loss. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction, (3) Negligible contact resistances at the material interfaces, (4) Negligible steam side convection resistance (T∞,i = Ts,i), (5) Negligible conduction resistance for aluminum sheath, (6) Constant properties, (7) Large surroundings. ANALYSIS: (a) To determine the insulation thickness, an energy balance must be performed at the outer surface, where q ′ = q ′conv,o + q ′rad . With q ′conv,o = 2π r3h o ( Ts,o − T∞,o ) , q ′rad = 2π r3 εσ

(T

)

(

)(

)

4 4 ′ ′ ′ ′ s,o − Tsur , q = Ts,i − Ts,o / R cond,st + R cond,ins , R cond,st = "n ( r2 / r1 ) / 2π k st ,

and R ′cond,ins

= "n ( r3 / r2 ) / 2π k ins , it follows that

(

2π Ts,i − Ts,o

)

(

)

(

)

4 − T4  = 2π r3  h o Ts,o − T∞,o + εσ Ts,o sur   n ( r2 / r1 ) n ( r3 / r2 )   + k st k ins 2π (848 − 323 ) K

(

"n 0.18 / 0.15

)

35 W / m ⋅ K

+

(

"n r3 / 0.18

)

= 2π r3

 6 W / m 2 ⋅ K ( 323 − 300 ) K + 0.20 × 5.67 × 10−8 W / m 2 ⋅ K 4 

(

4

323 − 300

4

)

K

4

 

0.10 W / m ⋅ K

A trial-and-error solution yields r3 = 0.394 m = 394 mm, in which case the insulation thickness is

t ins = r3 − r2 = 214 mm

<

The heat rate is then

q′ =

2π (848 − 323) K = 420 W / m n ( 0.18 / 0.15 ) n (0.394 / 0.18 ) + 35 W / m ⋅ K 0.10 W / m ⋅ K

<

(b) The effects of r3 on Ts,o and q ′ have been computed and are shown below. Conditioned …..

PROBLEM 3.40 (Cont.)

Ou te r s u rfa ce te m p e ra tu re , C

240 200 160 120 80 40 0 .2

0 .2 6

0 .3 2

0 .3 8

0 .4 4

0 .5

O u te r ra d iu s o f in s u la tio n , m Ts ,o

H e a t ra te s , W /m

2500 2000 1500 1000 500 0 0 .2

0 .2 6

0 .3 2

0 .3 8

0 .4 4

0 .5

O u te r ra d iu s o f in s u la tio n , m To ta l h e a t ra te C o n ve ctio n h e a t ra te R a d ia tio n h e a t ra te

Beyond r3 ≈ 0.40m, there are rapidly diminishing benefits associated with increasing the insulation thickness. COMMENTS: Note that the thermal resistance of the insulation is much larger than that for the tube wall. For the conditions of Part (a), the radiation coefficient is hr = 1.37 W/m, and the heat loss by radiation is less than 25% of that due to natural convection ( q ′rad = 78 W / m, q ′conv,o = 342 W / m ) .

PROBLEM 3.41 KNOWN: Thin electrical heater fitted between two concentric cylinders, the outer surface of which experiences convection. FIND: (a) Electrical power required to maintain outer surface at a specified temperature, (b) Temperature at the center

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Steady-state conditions, (3) Heater element has negligible thickness, (4) Negligible contact resistance between cylinders and heater, (5) Constant properties, (6) No generation.

ANALYSIS: (a) Perform an energy balance on the composite system to determine the power required to maintain T(r2) = Ts = 5°C. E ′in − E ′out + E gen = E st + q′elec − q′conv = 0. Using Newton’s law of cooling, q′elec = q′conv = h ⋅ 2π r2 ( Ts − T∞ ) q′elec = 50

W m2 ⋅ K

$

× 2π ( 0.040m ) 5 − ( −15 ) C=251 W/m.

<

(b) From a control volume about Cylinder A, we recognize that the cylinder must be isothermal, that is,

T(0) = T(r1). Represent Cylinder B by a thermal circuit: q′=

T ( r1 ) − Ts R ′B

For the cylinder, from Eq. 3.28, R ′B = ln r2 / r1 / 2π k B giving T ( r1 ) = Ts + q′R ′B = 5$ C+253.1

W ln 40/20 = 23.5$ C π m 2 × 1.5 W/m ⋅ K

Hence, T(0) = T(r1) = 23.5°C. Note that kA has no influence on the temperature T(0).

<

PROBLEM 3.42 KNOWN: Electric current and resistance of wire. Wire diameter and emissivity. Thickness, emissivity and thermal conductivity of coating. Temperature of ambient air and surroundings. Expression for heat transfer coefficient at surface of the wire or coating. FIND: (a) Heat generation per unit length and volume of wire, (b) Temperature of uninsulated wire, (c) Inner and outer surface temperatures of insulation. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction through insulation, (3) Constant properties, (4) Negligible contact resistance between insulation and wire, (5) Negligible radial temperature gradients in wire, (6) Large surroundings. ANALYSIS: (a) The rates of energy generation per unit length and volume are, respectively, 2 E ′g = I2 R ′elec = ( 20 A ) (0.01Ω / m ) = 4 W / m

<

2 q = E ′g / Ac = 4 E ′g / π D2 = 16 W / m / π ( 0.002m ) = 1.27 × 106 W / m3

<

(b) Without the insulation, an energy balance at the surface of the wire yields 4 E ′g = q′ = q′conv + q′rad = π D h ( T − T∞ ) + π D ε wσ T 4 − Tsur

)

(

where h = 1.25 [( T − T∞ ) / D ]1/ 4 . Substituting, 4 W / m = 1.25π ( 0.002m )

3/ 4

(T − 293 )5 / 4 + π (0.002m ) 0.3 × 5.67 × 10−8 W / m 2 ⋅ K 4

(T

4

)

− 2934 K 4

and a trial-and-error solution yields

<

T = 331K = 58°C (c) Performing an energy balance at the outer surface,

(

)

(

4 − T4 E ′g = q′ = q′conv + q′rad = π D h Ts,2 − T∞ + π D ε iσ Ts,2 sur 4 W / m = 1.25π ( 0.006m )

3/ 4

)

4 4 4 − 293 ) K (Ts,2 − 293)5 / 4 + π (0.006m ) 0.9 × 5.67 × 10−8 W / m 2 ⋅ K 4 (Ts,2

and an iterative solution yields the following value of the surface temperature

<

Ts,2 = 307.8 K = 34.8°C

The inner surface temperature may then be obtained from the following expression for heat transfer by conduction in the insulation. Continued …..

PROBLEM 3.42 (Cont.)

q′ =

Ts,i − T2 R ′cond

4W =

=

Ts,i − Ts,2

n ( r2 / r1 ) / 2π k i

(

2π ( 0.25 W / m ⋅ K ) Ts,i − 307.8 K

)

n 3

<

Ts,i = 310.6 K = 37.6°C As shown below, the effect of increasing the insulation thickness is to reduce, not increase, the surface temperatures.

S u rfa ce te m p e ra tu re s , C

50

45

40

35

30 0

1

2

3

4

In s u la tio n th ickn e s s , m m In n e r s u rfa ce te m p e ra tu re , C O u te r s u rfa ce te m p e ra tu re , C

This behavior is due to a reduction in the total resistance to heat transfer with increasing r2. Although

(

)

2 2 the convection, h, and radiation, h r = εσ (Ts,2 + Tsur ) Ts,2 + Tsur , coefficients decrease with

increasing r2, the corresponding increase in the surface area is more than sufficient to provide for a reduction in the total resistance. Even for an insulation thickness of t = 4 mm, h = h + hr = (7.1 + 5.4) 2 2 2 W/m ⋅K = 12.5 W/m ⋅K, and rcr = k/h = 0.25 W/m⋅K/12.5 W/m ⋅K = 0.020m = 20 mm > r2 = 5 mm. The outer radius of the insulation is therefore well below the critical radius.

PROBLEM 3.43 KNOWN: Diameter of electrical wire. Thickness and thermal conductivity of rubberized sheath. Contact resistance between sheath and wire. Convection coefficient and ambient air temperature. Maximum allowable sheath temperature. FIND: Maximum allowable power dissipation per unit length of wire. Critical radius of insulation. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction through insulation, (3) Constant properties, (4) Negligible radiation exchange with surroundings. ANALYSIS: The maximum insulation temperature corresponds to its inner surface and is independent of the contact resistance. From the thermal circuit, we may write

E ′g = q′ =

Tin,i − T∞

R ′cond + R ′conv

=

Tin,i − T∞

(

)

(

n rin,o / rin,i / 2π k  + 1/ 2π rin,o h  

)

where rin,i = D / 2 = 0.001m, rin,o = rin,i + t = 0.003m, and Tin,i = Tmax = 50°C yields the maximum allowable power dissipation. Hence,

(50 − 20 ) °C

E ′g,max =

"n 3

2π × 0.13 W / m ⋅ K

+

=

1

30°C

(1.35 + 5.31) m ⋅ K / W

= 4.51 W / m

<

2π ( 0.003m )10 W / m ⋅ K 2

The critical insulation radius is also unaffected by the contact resistance and is given by

rcr =

k 0.13 W / m ⋅ K = = 0.013m = 13mm h 10 W / m 2 ⋅ K

<

Hence, rin,o < rcr and E ′g,max could be increased by increasing rin,o up to a value of 13 mm (t = 12 mm). COMMENTS: The contact resistance affects the temperature of the wire, and for q ′ = E ′g,max = 4.51 W / m, the outer surface temperature of the wire is Tw,o = Tin,i + q ′ R ′t,c = 50°C + ( 4.51 W / m )

(3 × 10

−4

)

m ⋅ K / W / π ( 0.002m ) = 50.2°C. Hence, the temperature change across the contact 2

resistance is negligible.

PROBLEM 3.44  concentric KNOWN: Long rod experiencing uniform volumetric generation of thermal energy, q, with a hollow ceramic cylinder creating an enclosure filled with air. Thermal resistance per unit length due to radiation exchange between enclosure surfaces is R ′rad . The free convection 2

coefficient for the enclosure surfaces is h = 20 W/m ⋅K. FIND: (a) Thermal circuit of the system that can be used to calculate the surface temperature of the rod, Tr; label all temperatures, heat rates and thermal resistances; evaluate the thermal resistances; and (b) Calculate the surface temperature of the rod. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial conduction through the hollow cylinder, (3) The enclosure surfaces experience free convection and radiation exchange. ANALYSIS: (a) The thermal circuit is shown below. Note labels for the temperatures, thermal resistances and the relevant heat fluxes. Enclosure, radiation exchange (given):

R ′rad = 0.30 m ⋅ K / W

Enclosure, free convection:

1 1 = = 0.80 m ⋅ K / W 2 hπ Dr 20 W / m ⋅ K × π × 0.020m 1 1 R ′cv,cer = = = 0.40 m ⋅ K / W 2 hπ Di 20 W / m ⋅ K × π × 0.040m R ′cv,rod =

Ceramic cylinder, conduction:

R ′cd =

n ( Do / Di ) n (0.120 / 0.040 ) = = 0.10 m ⋅ K / W 2π k 2π ×1.75 W / m ⋅ K

The thermal resistance between the enclosure surfaces (r-i) due to convection and radiation exchange is

1 1 1 = + R ′enc R ′rad R ′cv,rod + R ′cv,cer 1  1  R ′enc =  +  0.30 0.80 + 0.40 

−1

m ⋅ K / W = 0.24 m ⋅ K / W

The total resistance between the rod surface (r) and the outer surface of the cylinder (o) is

R ′tot = R ′enc + R ′cd = ( 0.24 + 0.1) m ⋅ K / W = 0.34 m ⋅ K / W

Continued …..

PROBLEM 3.44 (Cont.)

(b) From an energy balance on the rod (see schematic) find Tr.

′ − E ′out + E ′gen = 0 E in −q + q ∀ = 0

(

)

− (Tr − Ti ) / R ′tot + q π D 2r / 4 = 0

(

)

− (Tr − 25 ) K / 0.34 m ⋅ K / W + 2 × 106 W / m3 π × 0.020m 2 / 4 = 0 Tr = 239°C

<

COMMENTS: In evaluating the convection resistance of the air space, it was necessary to define an average air temperature (T∞) and consider the convection coefficients for each of the space surfaces. As you’ll learn later in Chapter 9, correlations are available for directly estimating the convection coefficient (henc) for the enclosure so that qcv = henc (Tr – T1).

PROBLEM 3.45 KNOWN: Tube diameter and refrigerant temperature for evaporator of a refrigerant system. Convection coefficient and temperature of outside air. FIND: (a) Rate of heat extraction without frost formation, (b) Effect of frost formation on heat rate, (c) Time required for a 2 mm thick frost layer to melt in ambient air for which h = 2 W/m2⋅K and T‡ = 20°C. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible convection resistance for refrigerant flow T∞,i = Ts,1 , (3) Negligible tube wall conduction resistance, (4) Negligible radiation exchange at outer surface.

(

)

ANALYSIS: (a) The cooling capacity in the defrosted condition (δ = 0) corresponds to the rate of heat extraction from the airflow. Hence,

(

)

q′ = h2π r1 T∞,o − Ts,1 = 100 W m 2 ⋅ K ( 2π × 0.005 m )( −3 + 18 ) C $

<

q′ = 47.1W m (b) With the frost layer, there is an additional (conduction) resistance to heat transfer, and the extraction rate is T∞,o − Ts,1 T∞,o − Ts,1 q′ = = R ′conv + R ′cond 1 ( h2π r2 ) + ln ( r2 r1 ) 2π k For 5 ≤ r2 ≤ 9 mm and k = 0.4 W/m⋅K, this expression yields Thermal resistance, Rt(m.K/W)

Heat extraction, qprime(W/m)

50

45

40

0.4

0.3

0.2

0.1

0

35 0

0.001

0.002

0.003

Frost layer thickness, delta(m) Heat extraction, qprime(W/m)

0.004

0

0.001

0.002

0.003

0.004

Frost layer thickness, delta(m) Conduction resistance, Rtcond(m.K/W) Convection resistance, Rtconv(m.K/W)

Continued...

PROBLEM 3.45 (Cont.) The heat extraction, and hence the performance of the evaporator coil, decreases with increasing frost layer thickness due to an increase in the total resistance to heat transfer. Although the convection resistance decreases with increasing δ, the reduction is exceeded by the increase in the conduction resistance. (c) The time tm required to melt a 2 mm thick frost layer may be determined by applying an energy balance, Eq. 1.11b, over the differential time interval dt and to a differential control volume extending inward from the surface of the layer. E in dt = dEst = dU lat

(

)

h ( 2π rL ) T∞,o − Tf dt = −h sf ρ d∀ = −h sf ρ ( 2π rL ) dr

(

h T∞,o − Tf

tm =

) ∫0t m dt = − ρ hsf ∫rr21 dr

ρ h sf ( r2 − r1 )

(

h T∞,o − Tf

)

=

t m = 11, 690 s = 3.25 h

(

)

700 kg m3 3.34 × 105 J kg (0.002 m ) 2 W m 2 ⋅ K ( 20 − 0 ) C $

<

COMMENTS: The tube radius r1 exceeds the critical radius rcr = k/h = 0.4 W/m⋅K/100 W/m2⋅K = 0.004 m, in which case any frost formation will reduce the performance of the coil.

PROBLEM 3.46 KNOWN: Conditions associated with a composite wall and a thin electric heater. FIND: (a) Equivalent thermal circuit, (b) Expression for heater temperature, (c) Ratio of outer and inner heat flows and conditions for which ratio is minimized. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant properties, (3) Isothermal heater, (4) Negligible contact resistance(s). ANALYSIS: (a) On the basis of a unit axial length, the circuit, thermal resistances, and heat rates are as shown in the schematic.

 = E (b) Performing an energy balance for the heater, E in out , it follows that q′′h ( 2π r2 ) = q′i + q′o =

Th − T∞,i −1

( h i 2π r1 )

ln ( r2 r1 ) + 2π k B

+

Th − T∞,o −1

(h o 2π r3 )

ln ( r3 r2 ) + 2π k A

<

(c) From the circuit,

ln ( r2 r1 ) q′o 2π k B = × ln ( r r ) q′i Th − T∞,i ( h o 2π r3 )−1 + 3 2 2π k A

( (

Th − T∞,o

) )

( h i 2π r1 )−1 +

To reduce q′o q′i , one could increase kB, hi, and r3/r2, while reducing kA, ho and r2/r1. COMMENTS: Contact resistances between the heater and materials A and B could be important.

<

PROBLEM 3.47 KNOWN: Electric current flow, resistance, diameter and environmental conditions associated with a cable. FIND: (a) Surface temperature of bare cable, (b) Cable surface and insulation temperatures for a thin coating of insulation, (c) Insulation thickness which provides the lowest value of the maximum insulation temperature. Corresponding value of this temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r, (3) Constant properties. ANALYSIS: (a) The rate at which heat is transferred to the surroundings is fixed by the rate of heat generation in the cable. Performing an energy balance for a control surface about the cable, it follows that E = q or, for the bare cable, I2 R ′ L=h (π D L )(T − T ). With g

q′=I2 R ′e = ( 700A )

2

Ts = T∞ +

e

(6 ×10−4 Ω / m) = 294 W/m, it follows that

i

s

∞

q′ 294 W/m = 30$ C+ hπ D i 25 W/m 2 ⋅ K π (0.005m )

(

)

Ts = 778.7$ C.

<

(b) With a thin coating of insulation, there exist contact and convection resistances to heat transfer from the cable. The heat transfer rate is determined by heating within the cable, however, and therefore remains the same. Ts − T∞ Ts − T∞ q= = 1 R ′′t,c 1 R t,c + + hπ Di L π Di L hπ Di L π Di (Ts − T∞ ) q′= R ′′t,c + 1/ h and solving for the surface temperature, find q′ Ts = π Di

1 294 W/m  m2 ⋅ K m2 ⋅ K   ′′ $  R t,c + h  + T∞ = π ( 0.005m ) 0.02 W + 0.04 W  + 30 C  

Ts = 1153$ C.

< Continued …..

PROBLEM 3.47 (Cont.) The insulation temperature is then obtained from

T −T q= s i R t,c

or

Ti = Ts − qR t,c = 1153$ C − q

R ′′t,c

π Di L

= 1153$ C −

294

W m2 ⋅ K × 0.02 m W π (0.005m )

Ti = 778.7$ C.

<

(c) The maximum insulation temperature could be reduced by reducing the resistance to heat transfer from the outer surface of the insulation. Such a reduction is possible if Di < Dcr. From Example 3.4,

rcr =

k 0.5 W/m ⋅ K = = 0.02m. h 25 W/m 2 ⋅ K

Hence, Dcr = 0.04m > Di = 0.005m. To minimize the maximum temperature, which exists at the inner surface of the insulation, add insulation in the amount D − Di Dcr − Di (0.04 − 0.005 ) m t= o = = 2 2 2

<

t = 0.0175m. The cable surface temperature may then be obtained from q′=

R ′′t,c

π Di

+

Ts − T∞ ln ( Dcr / Di ) 2π k

+

1 hπ Dcr

=

Ts − 30$ C

ln ( 0.04/0.005 ) 0.02 m 2 ⋅ K/W + + 2π ( 0.5 W/m ⋅ K ) π ( 0.005m )

1 25

W 2

m ⋅K

π ( 0.04m )

Hence,

Ts − 30$ C Ts − 30$ C W 294 = = m (1.27+0.66+0.32 ) m ⋅ K/W 2.25 m ⋅ K/W Ts = 692.5$ C Recognizing that q = (Ts - Ti)/Rt,c, find Ti = Ts − qR t,c = Ts − q Ti = 318.2$ C.

R ′′t,c

π Di L

= 692.5$ C −

294

W m2 ⋅ K × 0.02 m W π (0.005m )

<

COMMENTS: Use of the critical insulation thickness in lieu of a thin coating has the effect of reducing the maximum insulation temperature from 778.7°C to 318.2°C. Use of the critical insulation thickness also reduces the cable surface temperature to 692.5°C from 778.7°C with no insulation or from 1153°C with a thin coating.

PROBLEM 3.48 KNOWN: Saturated steam conditions in a pipe with prescribed surroundings. FIND: (a) Heat loss per unit length from bare pipe and from insulated pipe, (b) Pay back period for insulation. SCHEMATIC: Steam Costs: 9 $4 for 10 J Insulation Cost: $100 per meter Operation time: 7500 h/yr ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constant properties, (4) Negligible pipe wall resistance, (5) Negligible steam side convection resistance (pipe inner surface temperature is equal to steam temperature), (6) Negligible contact resistance, (7) Tsur = T∞. PROPERTIES: Table A-6, Saturated water (p = 20 bar): Tsat = Ts = 486K; Table A-3, Magnesia, 85% (T ≈ 392K): k = 0.058 W/m⋅K. ANALYSIS: (a) Without the insulation, the heat loss may be expressed in terms of radiation and convection rates,

(

)

4 q′=επ Dσ Ts4 − Tsur + h (π D )( Ts − T∞ ) W q′=0.8π ( 0.2m ) 5.67 × 10−8 4864 − 2984 K 4 2 4 m ⋅K W +20 (π × 0.2m ) ( 486-298) K m2 ⋅ K

(

)

q′= (1365+2362 ) W/m=3727 W/m.

<

With the insulation, the thermal circuit is of the form

Continued …..

PROBLEM 3.48 (Cont.) From an energy balance at the outer surface of the insulation, q′cond = q′conv + q′rad Ts,i − Ts,o 4 − T4 = hπ Do Ts,o − T∞ + εσπ Do Ts,o sur ln ( Do / Di ) / 2π k 486 − Ts,o K W π (0.3m ) Ts,o − 298K = 20 2 ln ( 0.3m/0.2m ) m ⋅K 2π ( 0.058 W/m ⋅ K ) W 4 − 2984 K 4 . π (0.3m ) Ts,o +0.8 × 5.67 ×10-8 2 4 m ⋅K

(

(

)

)

(

)

(

)

(

)

By trial and error, we obtain Ts,o ≈ 305K in which case q′=

( 486-305) K = 163 W/m. ln ( 0.3m/0.2m ) 2π (0.055 W/m ⋅ K )

<

(b) The yearly energy savings per unit length of pipe due to use of the insulation is Savings Energy Savings Cost = × Yr ⋅ m Yr. Energy Savings J s h $4 = (3727 − 163) × 3600 × 7500 × Yr ⋅ m s⋅m h Yr 109 J Savings = $385 / Yr ⋅ m. Yr ⋅ m The pay back period is then Pay Back Period =

Insulation Costs $100 / m = Savings/Yr. ⋅ m $385/Yr ⋅ m

Pay Back Period = 0.26 Yr = 3.1 mo.

<

COMMENTS: Such a low pay back period is more than sufficient to justify investing in the insulation.

PROBLEM 3.49 KNOWN: Temperature and convection coefficient associated with steam flow through a pipe of prescribed inner and outer diameters. Outer surface emissivity and convection coefficient. Temperature of ambient air and surroundings. FIND: Heat loss per unit length. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constant properties, (4) Surroundings form a large enclosure about pipe. PROPERTIES: Table A-1, Steel, AISI 1010 (T ≈ 450 K): k = 56.5 W/m⋅K. ANALYSIS: Referring to the thermal circuit, it follows from an energy balance on the outer surface that T∞,i − Ts,o Ts,o − T∞,o Ts,o − Tsur = + R conv,i + R cond R conv,o R rad or from Eqs. 3.9, 3.28 and 1.7,

(1/π

T∞,i − Ts,o

Di hi ) + ln ( Do / Di ) / 2π k 523K − Ts,o

=

Ts,o − T∞,o

(1/π

Do h o )

(

4 4 + επ Doσ Ts,o − Tsur

=

)

Ts,o − 293K

−1 ln (75/60 ) π × 0.075m × 25 W/m 2 ⋅ K 2π × 56.5 W/m ⋅ K 4 +0.8π × ( 0.075m ) × 5.67 × 10−8 W/m 2 ⋅ K 4 Ts,o − 2934  K 4   523 − Ts,o Ts,o − 293 8 4 4 − = + 1.07 × 10 Ts,o − 293  .   0.0106+0.0006 0.170

(

π × 0.6m × 500 W/m 2 ⋅ K

)

−1

+

(

)

From a trial-and-error solution, Ts,o ≈ 502K. Hence the heat loss is

(

)

(

4 − T4 q′=π Do h o Ts,o − T∞,o + επ Doσ Ts,o sur q′=π ( 0.075m ) 25 W/m 2 ⋅ K (502-293) + 0.8 π ( 0.075m ) 5.67 × 10−8

)

5024 − 2434  K 4  m ⋅K  W

2

4

q′=1231 W/m+600 W/m=1831 W/m. COMMENTS: The thermal resistance between the outer surface and the surroundings is much larger than that between the outer surface and the steam.

<

PROBLEM 3.50 KNOWN: Temperature and convection coefficient associated with steam flow through a pipe of prescribed inner and outer radii. Emissivity of outer surface magnesia insulation, and convection coefficient. Temperature of ambient air and surroundings. FIND: Heat loss per unit length q ′ and outer surface temperature Ts,o as a function of insulation thickness. Recommended insulation thickness. Corresponding annual savings and temperature distribution. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constant properties, (4) Surroundings form a large enclosure about pipe. PROPERTIES: Table A-1, Steel, AISI 1010 (T ≈ 450 K): ks = 56.5 W/m⋅K. Table A-3, Magnesia, 85% (T ≈ 365 K): km = 0.055 W/m⋅K. ANALYSIS: Referring to the thermal circuit, it follows from an energy balance on the outer surface that T∞,i − Ts,o Ts,o − T∞,o Ts,o − Tsur = + R ′conv,i + R ′cond,s + R ′cond,m R ′conv,o R ′rad or from Eqs. 3.9, 3.28 and 1.7,

(1 2π r1h i ) + ln ( r2

T∞ ,i − Ts,o

r1 ) 2π k s + ln ( r3 r2 ) 2π k m

=

Ts,o − T∞ ,o

Ts,o − Tsur

+

(1 2π r3h o )  ( 2π r3 )εσ (Ts,o + Tsur ) 

(T

2 2 s,o + Tsur

)

−1

This expression may be solved for Ts,o as a function of r3, and the heat loss may then be determined by evaluating either the left-or right-hand side of the energy balance equation. The results are plotted as follows. Continued...

PROBLEM 3.50 (Cont.) 2000 Thermal resistance, Rprime(K/m.W)

2

Heat loss, qprime(W/m)

1600

1200

800

400

1.5

1

0.5

0 0.035

0 0.035

0.045

0.055

0.065

0.075

0.045

0.055

0.065

0.075

Outer radius of insulation, r3(m)

Outer radius of insulation, r3(m)

Insulation conduction resistance, Rcond,m Outer convection resistance, Rconv,o Radiation resistance, Rrad

q1

The rapid decay in q′ with increasing r3 is attributable to the dominant contribution which the insulation begins to make to the total thermal resistance. The inside convection and tube wall conduction resistances are fixed at 0.0106 m⋅K/W and 6.29×10-4 m⋅K/W, respectively, while the resistance of the insulation increases to approximately 2 m⋅K/W at r3 = 0.075 m. The heat loss may be reduced by almost 91% from a value of approximately 1830 W/m at r3 = r2 = 0.0375 m (no insulation) to 172 W/m at r3 = 0.0575 m and by only an additional 3% if the insulation thickness is increased to r3 = 0.0775 m. Hence, an insulation thickness of (r3 - r2) = 0.020 m is recommended, for which q′ = 172 W/m. The corresponding annual savings (AS) in energy costs is therefore $4 h s × 7000 × 3600 = $167 / m AS = [(1830 − 172 ) W m ] y h 109 J

<

The corresponding temperature distribution is

Local temperature, T(K)

500

460

420

380

340

300 0.038

0.042

0.046

0.05

0.054

0.058

Radial location in insulation, r(m) Tr

The temperature in the insulation decreases from T(r) = T2 = 521 K at r = r2 = 0.0375 m to T(r) = T3 = 309 K at r = r3 = 0.0575 m. Continued...

PROBLEM 3.50 (Cont.) COMMENTS: 1. The annual energy and costs savings associated with insulating the steam line are substantial, as is the reduction in the outer surface temperature (from Ts,o ≈ 502 K for r3 = r2, to 309 K for r3 = 0.0575 m). 2. The increase in R ′rad to a maximum value of 0.63 m⋅K/W at r3 = 0.0455 m and the subsequent decay is due to the competing effects of hrad and A′3 = (1 2π r3 ) . Because the initial decay in T3 = Ts,o with increasing r3, and hence, the reduction in hrad, is more pronounced than the increase in A′3 , R ′rad increases with r3. However, as the decay in Ts,o, and hence hrad, becomes less pronounced, the increase in A′3 becomes more pronounced and R ′rad decreases with increasing r3.

PROBLEM 3.51 KNOWN: Pipe wall temperature and convection conditions associated with water flow through the pipe and ice layer formation on the inner surface. FIND: Ice layer thickness δ. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Negligible pipe wall thermal resistance, (3) negligible ice/wall contact resistance, (4) Constant k. PROPERTIES: Table A.3, Ice (T = 265 K): k ≈ 1.94 W/m⋅K. ANALYSIS: Performing an energy balance for a control surface about the ice/water interface, it follows that, for a unit length of pipe,

q′conv = q′cond

(

)

h i ( 2π r1 ) T∞,i − Ts,i =

Ts,i − Ts,o

ln ( r2 r1 ) 2π k

Dividing both sides of the equation by r2,

ln ( r2 r1 )

( r2 r1 )

=

Ts,i − Ts,o k 1.94 W m ⋅ K 15$ C × = × = 0.097 $ 2 h i r2 T∞,i − Ts,i 3 C 2000 W m ⋅ K (0.05 m )

(

)

The equation is satisfied by r2/r1 = 1.114, in which case r1 = 0.050 m/1.114 = 0.045 m, and the ice layer thickness is

δ = r2 − r1 = 0.005 m = 5 mm

<

COMMENTS: With no flow, hi → 0, in which case r1 → 0 and complete blockage could occur. The pipe should be insulated.

PROBLEM 3.52 KNOWN: Inner surface temperature of insulation blanket comprised of two semi-cylindrical shells of different materials. Ambient air conditions. FIND: (a) Equivalent thermal circuit, (b) Total heat loss and material outer surface temperatures.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial conduction, (3) Infinite contact resistance between materials, (4) Constant properties. ANALYSIS: (a) The thermal circuit is, R ′conv,A = R ′conv,B = 1 / π r2 h R ′cond ( A ) = R ′cond ( B ) =

ln ( r2 / r1 )

<

π kA ln ( r2 / ri )

π kB

The conduction resistances follow from Section 3.3.1 and Eq. 3.28. Each resistance is larger by a factor of 2 than the result of Eq. 3.28 due to the reduced area.

(b) Evaluating the thermal resistances and the heat rate ( q′=q′A + q′B ) ,

(

R ′conv = π × 0.1m × 25 W/m 2 ⋅ K R ′cond ( A ) =

q′= q′=

ln ( 0.1m/0.05m )

π × 2 W/m ⋅ K

Ts,1 − T∞

R ′cond ( A ) + R ′conv

+

)

−1

= 0.1273 m ⋅ K/W

= 0.1103 m ⋅ K/W

R ′cond ( B ) = 8 R ′cond ( A ) = 0.8825 m ⋅ K/W

Ts,1 − T∞

R ′cond( B) + R ′conv

(500 − 300 ) K (500 − 300 ) K + = (842 + 198 ) W/m=1040 W/m. (0.1103+0.1273) m ⋅ K/W (0.8825+0.1273) m ⋅ K/W

<

Hence, the temperatures are

W m⋅K × 0.1103 = 407K m W W m⋅K Ts,2( B) = Ts,1 − q′BR ′cond ( B ) = 500K − 198 × 0.8825 = 325K. m W

Ts,2( A ) = Ts,1 − q′A R ′cond ( A ) = 500K − 842

(

< <

)

COMMENTS: The total heat loss can also be computed from q′= Ts,1 − T∞ / R equiv , −1 −1 −1   ′ ′ ′ ′ where R equiv =  R cond ( A ) + R conv,A + ( R cond(B) + R conv,B )  = 0.1923 m ⋅ K/W.   Hence q′= (500 − 300 ) K/0.1923 m ⋅ K/W=1040 W/m.

(

)

PROBLEM 3.53 KNOWN: Surface temperature of a circular rod coated with bakelite and adjoining fluid conditions. FIND: (a) Critical insulation radius, (b) Heat transfer per unit length for bare rod and for insulation at critical radius, (c) Insulation thickness needed for 25% heat rate reduction. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r, (3) Constant properties, (4) Negligible radiation and contact resistance. PROPERTIES: Table A-3, Bakelite (300K): k = 1.4 W/m⋅K. ANALYSIS: (a) From Example 3.4, the critical radius is k 1.4 W/m ⋅ K rcr = = = 0.01m. h 140 W/m 2 ⋅ K

<

(b) For the bare rod, q′=h (π Di ) ( Ti − T∞ ) q′=140

W m2 ⋅ K

(π × 0.01m ) ( 200 − 25)$ C=770 W/m

<

For the critical insulation thickness,

( 200 − 25) C Ti − T∞ q′= = ln ( rcr / ri ) ln (0.01m/0.005m ) 1 1 + + 2π rcr h 2π k 2π × 1.4 W/m ⋅ K 2π × (0.01m ) × 140 W/m 2 ⋅ K $

q′=

175$C = 909 W/m (0.1137+0.0788) m ⋅ K/W

<

(c) The insulation thickness needed to reduce the heat rate to 577 W/m is obtained from

( 200 − 25) C Ti − T∞ W q′= = = 577 ln ( r/ri ) ln ( r/0.005m ) m 1 1 + + 2 2π rh 2π k 2π ( r )140 W/m ⋅ K 2π × 1.4 W/m ⋅ K $

From a trial-and-error solution, find r ≈ 0.06 m. The desired insulation thickness is then

δ = ( r − ri ) ≈ (0.06 − 0.005 ) m=55 mm.

<

PROBLEM 3.54 KNOWN: Geometry of an oil storage tank. Temperature of stored oil and environmental conditions. FIND: Heater power required to maintain a prescribed inner surface temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in radial direction, (3) Constant properties, (4) Negligible radiation. PROPERTIES: Table A-3, Pyrex (300K): k = 1.4 W/m⋅K. ANALYSIS: The rate at which heat must be supplied is equal to the loss through the cylindrical and hemispherical sections. Hence, q=qcyl + 2q hemi = qcyl + qspher or, from Eqs. 3.28 and 3.36, q=

q=

Ts,i − T∞

ln ( Do / Di ) 1 + 2π Lk π Do Lh

+

Ts,i − T∞ 1  1 1  1 − +   2π k  Di Do  π Do2h

( 400 − 300 ) K ln 1.04 1 + 2π ( 2m )1.4 W/m ⋅ K π (1.04m ) 2m 10 W/m 2 ⋅ K

(

+

( 400 − 300 ) K

)

1 1 (1 − 0.962 ) m-1 + 2 2π (1.4 W/m ⋅ K ) π (1.04m ) 10 W/m 2 ⋅ K 100K 100K q= + -3 -3 -3 2.23 ×10 K/W + 15.30 × 10 K/W 4.32 ×10 K/W + 29.43 ×10-3 q = 5705W + 2963W = 8668W.

<

PROBLEM 3.55 KNOWN: Diameter of a spherical container used to store liquid oxygen and properties of insulating material. Environmental conditions. FIND: (a) Reduction in evaporative oxygen loss associated with a prescribed insulation thickness, (b) Effect of insulation thickness on evaporation rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Negligible conduction resistance of container wall and contact resistance between wall and insulation, (3) Container wall at boiling point of liquid oxygen. ANALYSIS: (a) Applying an energy balance to a control surface about the insulation, E in − E out = 0, it follows that q conv + q rad = q cond = q . Hence, T∞ − Ts,2 Tsur − Ts,2 Ts,2 − Ts,1 + = =q R t,conv R t,rad R t,cond

(

where R t,conv = 4π r22 h

)

−1

(

, R t,rad = 4π r22 h r

(

(1)

)

−1

1.9, the radiation coefficient is h r = εσ Ts,2 + Tsur

, R t,cond = (1 4π k )[(1 r1 ) − (1 r2 )] , and, from Eq. 2 2 + Tsur ) (Ts,2 ).

With t = 10 mm (r2 = 260 mm), ε =

0.2 and T∞ = Tsur = 298 K, an iterative solution of the energy balance equation yields Ts,2 ≈ 297.7 K, where Rt,conv = 0.118 K/W, Rt,rad = 0.982 K/W and Rt,cond = 76.5 K/W. With the insulation, it follows that the heat gain is qw ≈ 2.72 W Without the insulation, the heat gain is q wo =

T∞ − Ts,1 Tsur − Ts,1 + R t,conv R t,rad

where, with r2 = r1, Ts,1 = 90 K, Rt,conv = 0.127 K/W and Rt,rad = 3.14 K/W. Hence, qwo = 1702 W  = q/hfg, the percent reduction in evaporated oxygen is With the oxygen mass evaporation rate given by m

% Re duction = Hence, % Re duction =

  m wo − m w  m wo

q − qw × 100% = wo × 100% q wo

(1702 − 2.7 ) W 1702 W

× 100% = 99.8%

< Continued...

PROBLEM 3.55 (Cont.)  = (b) Using Equation (1) to compute Ts,2 and q as a function of r2, the corresponding evaporation rate, m  with r2 are plotted as follows. q/hfg, may be determined. Variations of q and m 10000

0.01

Evaporation rate, mdot(kg/s)

Heat gain, q(W)

1000

100

10

1

0.1

0.001

0.0001

1E-5

1E-6 0.25

0.26

0.27

0.28

0.29

Outer radius of insulation, r2(m)

0.3

0.25

0.26

0.27

0.28

0.29

0.3

Outer radius of insulation, r2(m)

Because of its extremely low thermal conductivity, significant benefits are associated with using even a  are achieved with r2 = 0.26 thin layer of insulation. Nearly three-order magnitude reductions in q and m -3  decrease from values of 1702 W and 8×10 kg/s at r2 = 0.25 m to 0.627 m. With increasing r2, q and m W and 2.9×10-6 kg/s at r2 = 0.30 m. COMMENTS: Laminated metallic-foil/glass-mat insulations are extremely effective and corresponding conduction resistances are typically much larger than those normally associated with surface convection and radiation.

PROBLEM 3.56 KNOWN: Sphere of radius ri, covered with insulation whose outer surface is exposed to a convection process. FIND: Critical insulation radius, rcr. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial (spherical) conduction, (3) Constant properties, (4) Negligible radiation at surface. ANALYSIS: The heat rate follows from the thermal circuit shown in the schematic, q= ( Ti − T∞ ) / R tot where R tot = R t,conv + R t,cond and R t,conv =

1 1 = hAs 4π hr 2

(3.9)

R t,cond =

1  1 1  −  4π k  ri r 

(3.36)

If q is a maximum or minimum, we need to find the condition for which d R tot = 0. dr It follows that d  1  1 1 1   1 1 1 1 − + = + − =0     dr  4π k  ri r  4π hr 2   4π k r 2 2π h r3  giving k h The second derivative, evaluated at r = rcr, is d  dR tot  1 1 3 1 = − + dr  dr  2π k r3 2π h r 4  r=r rcr = 2

cr

=

1



−

1

 ( 2k/h )3  2π k

+

3 1  1 1  3 =  −1 +  > 0 3 2π h 2k/h  ( 2k/h ) 2π k  2

Hence, it follows no optimum Rtot exists. We refer to this condition as the critical insulation radius. See Example 3.4 which considers this situation for a cylindrical system.

PROBLEM 3.57 KNOWN: Thickness of hollow aluminum sphere and insulation layer. Heat rate and inner surface temperature. Ambient air temperature and convection coefficient. FIND: Thermal conductivity of insulation. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant properties, (4) Negligible contact resistance, (5) Negligible radiation exchange at outer surface. PROPERTIES: Table A-1, Aluminum (523K): k ≈ 230 W/m⋅K. ANALYSIS: From the thermal circuit, T −T T1 − T∞ q= 1 ∞ = 1/ 1/r 1/ r 1 R tot 1 − 2 + r2 − 1/ r3 + 4π k A1 4π k I h4π r32 q=

( 250 − 20 )$ C

1/0.15 − 1/ 0.18 1/ 0.18 − 1/ 0.30 K 1   + + 2W 4π k I  4π ( 230 ) 30 4 0.3 π ( )( )  

= 80 W

or 3.84 × 10−4 +

0.177 230 + 0.029 = = 2.875. kI 80

Solving for the unknown thermal conductivity, find kI = 0.062 W/m⋅K. COMMENTS: The dominant contribution to the total thermal resistance is made by the insulation. Hence uncertainties in knowledge of h or kA1 have a negligible effect on the accuracy of the kI measurement.

<

PROBLEM 3.58 KNOWN: Dimensions of spherical, stainless steel liquid oxygen (LOX) storage container. Boiling point and latent heat of fusion of LOX. Environmental temperature. FIND: Thermal isolation system which maintains boil-off below 1 kg/day. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible thermal resistances associated with internal and external convection, conduction in the container wall, and contact between wall and insulation, (3) Negligible radiation at exterior surface, (4) Constant insulation thermal conductivity. PROPERTIES: Table A.1, 304 Stainless steel (T = 100 K): ks = 9.2 W/m⋅K; Table A.3, Reflective, aluminum foil-glass paper insulation (T = 150 K): ki = 0.000017 W/m⋅K. ANALYSIS: The heat gain associated with a loss of 1 kg/day is  q = mh fg =

1kg day 86, 400 s day

(2.13 ×105 J kg ) = 2.47 W (

)

With an overall temperature difference of T∞ − Tbp = 150 K, the corresponding total thermal resistance is ∆T 150 K R tot = = = 60.7 K W q 2.47 W Since the conduction resistance of the steel wall is R t,cond,s =

1 1  1 1 1  − = 2.4 × 10−3 K W  − =   4π k s  r1 r2  4π (9.2 W m ⋅ K )  0.35 m 0.40 m  1

it is clear that exclusive reliance must be placed on the insulation and that a special insulation of very low thermal conductivity should be selected. The best choice is a highly reflective foil/glass matted insulation which was developed for cryogenic applications. It follows that R t,cond,i = 60.7 K W =

1 1

 1 1 1 1 −   − =  4π k i  r2 r3  4π (0.000017 W m ⋅ K )  0.40 m r3 

which yields r3 = 0.4021 m. The minimum insulation thickness is therefore δ = (r3 - r2) = 2.1 mm. COMMENTS: The heat loss could be reduced well below the maximum allowable by adding more insulation. Also, in view of weight restrictions associated with launching space vehicles, consideration should be given to fabricating the LOX container from a lighter material.

PROBLEM 3.59 KNOWN: Diameter and surface temperature of a spherical cryoprobe. Temperature of surrounding tissue and effective convection coefficient at interface between frozen and normal tissue. FIND: Thickness of frozen tissue layer. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible contact resistance between probe and frozen tissue, (3) Constant properties. ANALYSIS: Performing an energy balance for a control surface about the phase front, it follows that q conv − q cond = 0 Hence,

( )(T∞ − Ts,2 ) = [(1 r1T)s,2− (1−rT2s,1)] 4π k

h 4π r22

r22 [(1 r1 ) − (1 r2 )] =

( ) h (T∞ − Ts,2 ) k Ts,2 − Ts,1

 r2   r2   k ( Ts,2 − Ts,1 ) 1.5 W m ⋅ K  30  =       − 1 =  r1   r1   hr1 ( T∞ − Ts,2 ) 50 W m 2 ⋅ K ( 0.0015 m )  37 

(

)

 r2   r2       − 1 = 16.2  r1   r1  

( r2 r1 ) = 4.56 It follows that r2 = 6.84 mm and the thickness of the frozen tissue is

δ = r2 − r1 = 5.34 mm

<

PROBLEM 3.60 KNOWN: Inner diameter, wall thickness and thermal conductivity of spherical vessel containing heat generating medium. Inner surface temperature without insulation. Thickness and thermal conductivity of insulation. Ambient air temperature and convection coefficient. FIND: (a) Thermal energy generated within vessel, (b) Inner surface temperature of vessel with insulation. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional, radial conduction, (3) Constant properties, (4) Negligible contact resistance, (5) Negligible radiation. ANALYSIS: (a) From an energy balance performed at an instant for a control surface about the pharmaceuticals, E g = q, in which case, without the insulation E g = q =

Ts,1 − T∞ 1 1 1 − +  4π k w  r1 r2  4π r 2 h 2 1

E g = q =

(

=

(50 − 25 ) °C 1  1 − 1 +   4π (17 W / m ⋅ K )  0.50m 0.51m  4π ( 0.51m )2 6 W / m 2 ⋅ K

25°C

1

)

1.84 × 10−4 + 5.10 × 10−2 K / W

= 489 W

<

(b) With the insulation,

 1 1 1  1 1 1 1  Ts,1 = T∞ + q  − + − +      4π k w  r1 r2  4π k i  r2 r3  4π r 2 h  3    K 1 1  1  1 4 −   Ts,1 = 25°C + 489 W 1.84 × 10 + − + 4π ( 0.04 )  0.51 0.53  4π ( 0.53)2 6  W    K Ts,1 = 25°C + 489 W 1.84 × 10−4 + 0.147 + 0.047  = 120°C  W

<

COMMENTS: The thermal resistance associated with the vessel wall is negligible, and without the insulation the dominant resistance is due to convection. The thermal resistance of the insulation is approximately three times that due to convection.

PROBLEM 3.61 KNOWN: Spherical tank of 1-m diameter containing an exothermic reaction and is at 200°C when 2 the ambient air is at 25°C. Convection coefficient on outer surface is 20 W/m ⋅K. FIND: Determine the thickness of urethane foam required to reduce the exterior temperature to 40°C. Determine the percentage reduction in the heat rate achieved using the insulation. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial (spherical) conduction through the insulation, (3) Convection coefficient is the same for bare and insulated exterior surface, and (3) Negligible radiation exchange between the insulation outer surface and the ambient surroundings. PROPERTIES: Table A-3, urethane, rigid foam (300 K): k = 0.026 W/m⋅K. ANALYSIS: (a) The heat transfer situation for the heat rate from the tank can be represented by the thermal circuit shown above. The heat rate from the tank is T − T∞ q= t

R cd + R cv where the thermal resistances associated with conduction within the insulation (Eq. 3.35) and convection for the exterior surface, respectively, are R cd =

(1/ rt − 1/ ro ) =

(1/ 0.5 − 1/ ro )

4π k

=

(1/ 0.5 − 1/ ro ) K / W

4π × 0.026 W / m ⋅ K 0.3267 1 1 1 R cv = = = = 3.979 × 10−3 ro−2 K / W 2 2 2 hAs 4π hro 4π × 20 W / m ⋅ K × ro

To determine the required insulation thickness so that To = 40°C, perform an energy balance on the onode.

Tt − To T∞ − To + =0 R cd R cv ( 200 − 40 ) K

ro = 0.5135 m

( 25 − 40 ) K

=0 3.979 ×10−3 ro2 K / W t = ro − ri = (0.5135 − 0.5000 ) m = 13.5 mm

(1/ 0.5 − 1/ ro ) / 0.3267 K / W

+

<

From the rate equation, for the bare and insulated surfaces, respectively,

qo =

( 200 − 25 ) K = 10.99 kW Tt − T∞ = 1/ 4π hrt2 0.01592 K / W

qins =

( 200 − 25 ) Tt − T∞ = = 0.994 kW R cd + R cv (0.161 + 0.01592 ) K / W

Hence, the percentage reduction in heat loss achieved with the insulation is,

qins − qo 0.994 − 10.99 ×100 = − ×100 = 91% qo 10.99

<

PROBLEM 3.62 KNOWN: Dimensions and materials used for composite spherical shell. Heat generation associated with stored material. FIND: Inner surface temperature, T1, of lead (proposal is flawed if this temperature exceeds the melting point). SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties at 300K, (4) Negligible contact resistance. PROPERTIES: Table A-1, Lead: k = 35.3 W/m⋅K, MP = 601K; St.St.: 15.1 W/m⋅K. ANALYSIS: From the thermal circuit, it follows that T −T 4  q= 1 ∞ = q  π r13  R tot 3  Evaluate the thermal resistances, 1   1 R Pb = 1/ ( 4π × 35.3 W/m ⋅ K )  − = 0.00150 K/W  0.25m 0.30m  1   1 R St.St. = 1/ ( 4π ×15.1 W/m ⋅ K )  − = 0.000567 K/W  0.30m 0.31m 

)

(

R conv = 1/ 4π × 0.312 m 2 × 500 W/m 2 ⋅ K  = 0.00166 K/W   R tot = 0.00372 K/W. The heat rate is q=5 × 105 W/m3 ( 4π / 3)( 0.25m ) = 32, 725 W. The inner surface 3

temperature is T1 = T∞ + R tot q=283K+0.00372K/W (32,725 W ) T1 = 405 K < MP = 601K.

<

Hence, from the thermal standpoint, the proposal is adequate. COMMENTS: In fabrication, attention should be given to maintaining a good thermal contact. A protective outer coating should be applied to prevent long term corrosion of the stainless steel.

PROBLEM 3.63 KNOWN: Dimensions and materials of composite (lead and stainless steel) spherical shell used to store radioactive wastes with constant heat generation. Range of convection coefficients h available for cooling. FIND: (a) Variation of maximum lead temperature with h. Minimum allowable value of h to maintain maximum lead temperature at or below 500 K. (b) Effect of outer radius of stainless steel shell on maximum lead temperature for h = 300, 500 and 1000 W/m2⋅K. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties at 300 K, (4) Negligible contact resistance. PROPERTIES: Table A-1, Lead: k = 35.3 W/m⋅K, St. St.: 15.1 W/m⋅K. ANALYSIS: (a) From the schematic, the maximum lead temperature T1 corresponds to r = r1, and from the thermal circuit, it may be expressed as T1 = T∞ + R tot q where q = q ( 4 3 )π r13 = 5 × 105 W m3 ( 4π 3)( 0.25 m ) = 32, 725 W . The total thermal resistance is 3

R tot = R cond,Pb + R cond,St.St + R conv

where expressions for the component resistances are provided in the schematic. Using the Resistance Network model and Thermal Resistance tool pad of IHT, the following result is obtained for the variation of T1 with h. Maximum Pb Temperature, T(r1) (K)

700

600

500

400

300 100

200

300

400

500

600

700

800

900

1000

Convection coefficient, h(W/m^2.K) T_1

Continued...

PROBLEM 3.63 (Cont.) To maintain T1 below 500 K, the convection coefficient must be maintained at

<

h ≥ 181 W/m2⋅K

Maximum Pb temperature, T(r1) (K)

(b) The effect of varying the outer shell radius over the range 0.3 ≤ r3 ≤ 0.5 m is shown below. 600

550

500

450

400

350 0.3

0.35

0.4

0.45

0.5

Outer radius of steel shell, r3(m) h = 300 W/m^2.K h = 500 W/m^2.K h = 1000 W/m^2.k

For h = 300, 500 and 1000 W/m2⋅K, the maximum allowable values of the outer radius are r3 = 0.365, 0.391 and 0.408 m, respectively. COMMENTS: For a maximum allowable value of T1 = 500 K, the maximum allowable value of the total thermal resistance is Rtot = (T1 - T∞)/q, or Rtot = (500 - 283)K/32,725 W = 0.00663 K/W. Hence, any increase in Rcond,St.St due to increasing r3 must be accompanied by an equivalent reduction in Rconv.

PROBLEM 3.64 KNOWN: Representation of the eye with a contact lens as a composite spherical system subjected to convection processes at the boundaries. FIND: (a) Thermal circuits with and without contact lens in place, (b) Heat loss from anterior chamber for both cases, and (c) Implications of the heat loss calculations.

SCHEMATIC: r1=10.2mm

k1=0.35 W/m⋅K

r2=12.7mm

k2=0.80 W/m⋅K

r3=16.5mm 2

T∞,i=37°C

hi=12 W/m ⋅K

T∞,o=21°C

ho=6 W/m ⋅K

2

ASSUMPTIONS: (1) Steady-state conditions, (2) Eye is represented as 1/3 sphere, (3) Convection coefficient, ho, unchanged with or without lens present, (4) Negligible contact resistance. ANALYSIS: (a) Using Eqs. 3.9 and 3.36 to express the resistance terms, the thermal circuits are:

Without lens:

<

With lens:

<

(b) The heat losses for both cases can be determined as q = (T∞,i - T∞,o)/Rt, where Rt is the thermal resistance from the above circuits. Without lens: R t,wo =

3

(

12W/m 2 ⋅ K4π 10.2 × 10-3m 3

+ 2

(

-3

6 W/m ⋅ K4π 12.7 × 10 m

)

2

R t,w = 191.2 K/W+13.2 K/W+

+

3

(

-3

6W/m ⋅ K4π 16.5 × 10 m

)

2

2

1  1  1 − m 4π × 0.35 W/m ⋅ K 10.2 12.7  10−3 3

= 191.2 K/W+13.2 K/W+246.7 K/W=451.1 K/W

With lens:

2

)

+

1  1  1 − m  4π × 0.80 W/m ⋅ K 12.7 16.5  10−3 3

= 191.2 K/W+13.2 K/W+5.41 K/W+146.2 K/W=356.0 K/W

Hence the heat loss rates from the anterior chamber are Without lens: q wo = (37 − 21)$ C/451.1 K/W=35.5mW With lens:

q w = (37 − 21) C/356.0 K/W=44.9mW $

< <

(c) The heat loss from the anterior chamber increases by approximately 20% when the contact lens is in place, implying that the outer radius, r3, is less than the critical radius.

PROBLEM 3.65 KNOWN: Thermal conductivity and inner and outer radii of a hollow sphere subjected to a uniform heat flux at its outer surface and maintained at a uniform temperature on the inner surface. FIND: (a) Expression for radial temperature distribution, (b) Heat flux required to maintain prescribed surface temperatures. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) No generation, (4) Constant properties. ANALYSIS: (a) For the assumptions, the temperature distribution may be obtained by integrating Fourier’s law, Eq. 3.33. That is, T q r r dr qr 1 r k dT or = − − = −k T − Ts,1 . ∫Ts,1 4π ∫r1 r 2 4π r r1 Hence, q 1 1  T ( r ) = Ts,1 + r  −  4π k  r r1 

(

)

or, with q′′2 ≡ q r / 4π r22 , q′′2 r22  1 1  T ( r ) = Ts,1 +  −  k  r r1 

<

(b) Applying the above result at r2, q′′2 =

(

k Ts,2 − Ts,1  1 1 r22  −   r2 r1 

) = 10 W/m ⋅ K (50 − 20 )$ C = −3000 W/m2 . 2 1

(0.1m )

1  1  0.1 − 0.05  m

<

COMMENTS: (1) The desired temperature distribution could also be obtained by solving the appropriate form of the heat equation, d  2 dT  r =0 dr  dr  and applying the boundary conditions T ( r1 ) = Ts,1 and − k

dT  = q′′2 . dr  r 2

(2) The negative sign on q ′′2 implies heat transfer in the negative r direction.

PROBLEM 3.66 KNOWN: Volumetric heat generation occurring within the cavity of a spherical shell of prescribed dimensions. Convection conditions at outer surface. FIND: Expression for steady-state temperature distribution in shell. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Steady-state conditions, (3) Constant properties, (4) Uniform generation within the shell cavity, (5) Negligible radiation. ANALYSIS: For the prescribed conditions, the appropriate form of the heat equation is d  2 dT  r =0 dr  dr  Integrate twice to obtain, dT r2 and = C1 dr

C T = − 1 + C2 . r

(1,2)

The boundary conditions may be obtained from energy balances at the inner and outer surfaces. At the inner surface (ri),  / 3k. E = q 4/3π r 3 = q dT/dr) = −qr = −k 4π r 2 dT/dr) g

(

i

)

cond,i

(

i

)

ri

ri

i

(3)

At the outer surface (ro),

q cond,o = − k4π ro2 dT/dr)ro = q conv = h4π ro2 T ( ro ) − T∞  dT/dr)ro = − ( h/k )  T ( ro ) − T∞  .

(4)

 3 / 3k. From Eqs. (1), (2) and (4) From Eqs. (1) and (3), C1 = −qr i   3   h   qr i +C −T  − = −  ∞ 2   k   3ro k 3kro2   3 3   qr qr C2 = i − i + T∞ . 3hro2 3ro k Hence, the temperature distribution is  3 qr i

 3  1 1  qr  3 qr i i +T . T=  − + ∞ 3k  r ro  3hro2 COMMENTS: Note that E g = q cond,i = q cond,o = q conv .

<

PROBLEM 3.67 KNOWN: Spherical tank of 3-m diameter containing LP gas at -60°C with 250 mm thickness of insulation having thermal conductivity of 0.06 W/m⋅K. Ambient air temperature and convection 2 coefficient on the outer surface are 20°C and 6 W/m ⋅K, respectively. FIND: (a) Determine the radial position in the insulation at which the temperature is 0°C and (b) If the insulation is pervious to moisture, what conclusions can be reached about ice formation? What effect will ice formation have on the heat gain? How can this situation be avoided? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial (spherical) conduction through the insulation, and (3) Negligible radiation exchange between the insulation outer surface and the ambient surroundings. ANALYSIS: (a) The heat transfer situation can be represented by the thermal circuit shown above. The heat gain to the tank is

q=

 20 − ( −60 ) K T∞ − Tt = = 612.4 W R ins + R cv 0.1263 + 4.33 × 10−3 K / W

(

)

where the thermal resistances for the insulation (see Table 3.3) and the convection process on the outer surface are, respectively, −1 1/ r − 1/ ro (1/1.50 − 1/1.75 ) m R ins = i = = 0.1263 K / W

4π k

4π × 0.06 W / m ⋅ K 1 1 1 R cv = = = = 4.33 ×10−3 K / W hAs h4π ro2 6 W / m 2 ⋅ K × 4π (1.75 m )2

To determine the location within the insulation where Too (roo) = 0°C, use the conduction rate equation, Eq. 3.35, −1  1 4π k (Too − Tt )  4π k (Too − Tt ) q= roo =  −  q (1/ ri − 1/ roo )  ri  and substituting numerical values, find −1

 1 4π × 0.06 W / m ⋅ K (0 − ( −60 )) K  roo =  −  612.4 W 1.5 m 

= 1.687 m

<

(b) With roo = 1.687 m, we’d expect the region of the insulation ri ≤ r ≤ roo to be filled with ice formations if the insulation is pervious to water vapor. The effect of the ice formation is to substantially increase the heat gain since kice is nearly twice that of kins, and the ice region is of thickness (1.687 – 1.50)m = 187 mm. To avoid ice formation, a vapor barrier should be installed at a radius larger than roo.

PROBLEM 3.68 KNOWN: Radius and heat dissipation of a hemispherical source embedded in a substrate of prescribed thermal conductivity. Source and substrate boundary conditions. FIND: Substrate temperature distribution and surface temperature of heat source. SCHEMATIC:

ASSUMPTIONS: (1) Top surface is adiabatic. Hence, hemispherical source in semi-infinite medium is equivalent to spherical source in infinite medium (with q = 8 W) and heat transfer is one-dimensional in the radial direction, (2) Steady-state conditions, (3) Constant properties, (4) No generation. ANALYSIS: Heat equation reduces to 1 d  2 dT  r 2dT/dr=C1 r =0 2 dr dr   r T ( r ) = −C1 / r+C2 . Boundary conditions: T ( ∞ ) = T∞

T ( ro ) = Ts

Hence, C2 = T∞ and Ts = −C1 / ro + T∞

and

C1 = ro ( T∞ − Ts ) .

The temperature distribution has the form T ( r ) = T∞ + ( Ts − T∞ ) ro / r

<

and the heat rate is q=-kAdT/dr = − k2π r 2  − ( Ts − T∞ ) ro / r 2  = k2π ro ( Ts − T∞ )   It follows that Ts − T∞ =

q 4W = = 50.9$ C -4 k2π ro 125 W/m ⋅ K 2π 10 m

(

)

Ts = 77.9$ C. COMMENTS: For the semi-infinite (or infinite) medium approximation to be valid, the substrate dimensions must be much larger than those of the transistor.

<

PROBLEM 3.69 KNOWN: Critical and normal tissue temperatures. Radius of spherical heat source and radius of tissue to be maintained above the critical temperature. Tissue thermal conductivity. FIND: General expression for radial temperature distribution in tissue. Heat rate required to maintain prescribed thermal conditions. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant k. ANALYSIS: The appropriate form of the heat equation is

1 d  dT  r =0 r 2 dr  dr  Integrating twice,

dT C1 = dr r 2 C T ( r ) = − 1 + C2 r

)

(

Since T → Tb as r → ∞, C2 = Tb. At r = ro, q = − k 4π ro2 dT dr = −4π kro2 C1 ro2 = -4πkC1. ro Hence, C1 = -q/4πk and the temperature distribution is

T (r ) =

q + Tb 4π kr

<

It follows that

q = 4π kr  T ( r ) − Tb  Applying this result at r = rc,

q = 4π ( 0.5 W m ⋅ K )( 0.005 m )( 42 − 37 ) C = 0.157 W $

<

COMMENTS: At ro = 0.0005 m, T(ro) = q ( 4π kro ) + Tb = 92°C. Proximity of this temperature to the boiling point of water suggests the need to operate at a lower power dissipation level.

PROBLEM 3.70 KNOWN: Cylindrical and spherical shells with uniform heat generation and surface temperatures. FIND: Radial distributions of temperature, heat flux and heat rate. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Uniform heat generation, (3) Constant k. ANALYSIS: (a) For the cylindrical shell, the appropriate form of the heat equation is 1 d  dT  q r + = 0 r dr  dr  k The general solution is T (r ) = −

q 2 r + C1 ln r + C2 4k

Applying the boundary conditions, it follows that T ( r1 ) = Ts,1 = −

q 2 r1 + C1 ln r1 + C2 4k

T ( r2 ) = Ts,2 = −

q 2 r2 + C1 ln r2 + C2 4k

which may be solved for

(

)

(

)

 C1 = ( q/4k ) r22 − r12 + Ts,2 − Ts,1  ln ( r2 /r1 ) 

C2 = Ts,2 + ( q 4k ) r22 − C1 ln r2 Hence,

ln ( r/r2 ) T ( r ) = Ts,2 + ( q 4k ) r22 − r 2 + ( q 4k ) r22 − r12 + Ts,2 − Ts,1    ln ( r2 /r1 )

)

(

)

(

(

)

<

With q′′ = − k dT/dr , the heat flux distribution is q′′ ( r ) =

q 2

(

)

(

)

k ( q 4k ) r22 − r12 + Ts,2 − Ts,1 

 r− 

r ln ( r2 /r1 )

< Continued...

PROBLEM 3.70 (Cont.) Similarly, with q = q′′ A(r) = q′′ (2πrL), the heat rate distribution is  q ( r ) = π Lqr −

2

)

(

(

)

2π Lk ( q 4k ) r22 − r12 + Ts,2 − Ts,1   

<

ln ( r2 /r1 )

(b) For the spherical shell, the heat equation and general solution are 1 d  2 dT  q r + = 0 r 2 dr  dr  k T(r) = − ( q 6k ) r 2 − C1/r + C2 Applying the boundary conditions, it follows that T ( r1 ) = Ts,1 = − ( q 6k ) r12 − C1/r1 + C2 T ( r2 ) = Ts,2 = − ( q 6k ) r22 − C1/r2 + C2 Hence,

(

)

(

) [(1 r1 ) − (1 r2 )]

C1 = ( q 6k ) r22 − r12 + Ts,2 − Ts,1    C2 = Ts,2 + ( q 6k ) r22 + C1/r2 and

(1 r ) − (1 r2 ) T ( r ) = Ts,2 + ( q 6k ) r22 − r 2 − ( q 6k ) r22 − r12 + Ts,2 − Ts,1    (1 r ) − (1 r ) 1 2

)

(

(

)

(

)

<

With q′′ (r) = - k dT/dr, the heat flux distribution is q′′ ( r ) =

q 3

(

(

)

( q 6 ) r 2 − r 2 + k ( T − T ) 2 1 s,2 s,1  1  r− 

(1 r1 ) − (1 r2 )

<

r2

)

and, with q = q′′ 4π r 2 , the heat rate distribution is 4π q 3 q (r ) = r − 3

(

)

(

)

4π (q 6 ) r22 − r12 + k Ts,2 − Ts,1 



(1 r1 ) − (1 r2 )

<

PROBLEM 3.71 KNOWN: Temperature distribution in a composite wall. FIND: (a) Relative magnitudes of interfacial heat fluxes, (b) Relative magnitudes of thermal conductivities, and (c) Heat flux as a function of distance x. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties. ANALYSIS: (a) For the prescribed conditions (one-dimensional, steady-state, constant k), the parabolic temperature distribution in C implies the existence of heat generation. Hence, since dT/dx increases with decreasing x, the heat flux in C increases with decreasing x. Hence, q′′3 > q′′4 However, the linear temperature distributions in A and B indicate no generation, in which case q′′2 = q3′′ (b) Since conservation of energy requires that q′′3,B = q′′3,C and dT/dx)B < dT/dx)C , it follows from Fourier’s law that k B > kC. Similarly, since q′′2,A = q′′2,B and dT/dx)A > dT/dx) B , it follows that k A < k B. (c) It follows that the flux distribution appears as shown below.

COMMENTS: Note that, with dT/dx)4,C = 0, the interface at 4 is adiabatic.

PROBLEM 3.72 KNOWN: Plane wall with internal heat generation which is insulated at the inner surface and subjected to a convection process at the outer surface. FIND: Maximum temperature in the wall. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with uniform volumetric heat generation, (3) Inner surface is adiabatic. ANALYSIS: From Eq. 3.42, the temperature at the inner surface is given by Eq. 3.43 and is the maximum temperature within the wall,  2 / 2k+Ts . To = qL The outer surface temperature follows from Eq. 3.46,  Ts = T∞ + qL/h W Ts = 92$ C+0.3 × 106 × 0.1m/500W/m2 ⋅ K=92$C+60$C=152$C. 3 m It follows that To = 0.3 × 106 W/m3 × ( 0.1m ) / 2 × 25W/m ⋅ K+152$C 2

To = 60$ C+152$C=212$C.

<

COMMENTS: The heat flux leaving the wall can be determined from knowledge of h, Ts and T∞ using Newton’s law of cooling. q′′conv = h ( Ts − T∞ ) = 500W/m 2 ⋅ K (152 − 92 ) C=30kW/m2 . $

This same result can be determined from an energy balance on the entire wall, which has the form E g − E out = 0 where

 E g = qAL

and

E out = q′′conv ⋅ A.

Hence,  q′′conv = qL=0.3 × 106 W/m3 × 0.1m=30kW/m2 .

PROBLEM 3.73 KNOWN: Composite wall with outer surfaces exposed to convection process. FIND: (a) Volumetric heat generation and thermal conductivity for material B required for special conditions, (b) Plot of temperature distribution, (c) T1 and T2, as well as temperature distributions corresponding to loss of coolant condition where h = 0 on surface A. SCHEMATIC: LA = 30 mm LB = 30 mm LC = 20 mm kA = 25 W/m⋅K kC = 50 W/m⋅K ASSUMPTIONS: (1) Steady-state, one-dimensional heat transfer, (2) Negligible contact resistance at interfaces, (3) Uniform generation in B; zero in A and C. ANALYSIS: (a) From an energy balance on wall B, E in − E out + E g = E st  − q1′′ − q′′2 + 2qL B =0

q B = ( q1′′ + q ′′2 ) 2L B .

To determine the heat fluxes, q1„„ and q „„2 , construct thermal circuits for A and C:

q1′′ = ( T1 − T∞ ) (1 h + L A k A )

q ′′2 = ( T2 − T∞ ) ( L C k C + 1 h )

 1 0.030 m  +   2  1000 W m ⋅ K 25 W m ⋅ K 

q1′′ = ( 261 − 25 ) C $

q1′′ = 236 C ( 0.001 + 0.0012 ) m ⋅ K W $

$

q ′′2 = 186 C ( 0.0004 + 0.001) m ⋅ K W $

2

q1′′ = 107, 273 W m

 0.020 m  1 +   2  50 W m ⋅ K 1000 W m ⋅ K 

q ′′2 = ( 211 − 25 ) C

2

2

q ′′2 = 132, 857 W m

2

Using the values for q′′1 and q′′2 in Eq. (1), find

(

q B = 106, 818 + 132,143 W m

2

) 2 × 0.030 m = 4.00 ×10

6

<

3

W m .

To determine kB, use the general form of the temperature and heat flux distributions in wall B, T(x) = −

q B 2k B

2

x + C1x + C 2



q ′′x (x) = − k B  −

q

 kB



x + C1 



(1,2)

there are 3 unknowns, C1, C2 and kB, which can be evaluated using three conditions, Continued...

PROBLEM 3.73 (Cont.) T ( − L B ) = T1 = −

q B 2k B

T ( + L B ) = T2 = −

q B 2k B

( − L B )2 − C1L B + C 2

where T1 = 261°C

(3)

( + L B )2 + C1L B + C2

where T2 = 211°C

(4)

where q1′′ = 107,273 W/m2

(5)

 q  q ′′x ( − L B ) = −q1′′ = − k B  − B ( − L B ) + C1   kB 

Using IHT to solve Eqs. (3), (4) and (5) simultaneously with q B = 4.00 × 106 W/m3, find

<

k B = 15.3 W m ⋅ K

(b) Following the method of analysis in the IHT Example 3.6, User-Defined Functions, the temperature distribution is shown in the plot below. The important features are (1) Distribution is quadratic in B, but non-symmetrical; linear in A and C; (2) Because thermal conductivities of the materials are different, discontinuities exist at each interface; (3) By comparison of gradients at x = -LB and +LB, find q′′2 > q1′′ . (c) Using the same method of analysis as for Part (c), the temperature distribution is shown in the plot below when h = 0 on the surface of A. Since the left boundary is adiabatic, material A will be isothermal at T1. Find T1 = 835°C

<

T2 = 360°C Loss of coolant on surface A

400

Temperature, T (C)

Temperature, T (C)

800

300

200

100

600

400

200

-60

-40

-20

0

20

Wall position, x-coordinate (mm) T_xA, kA = 25 W/m.K T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3 T_x, kC = 50 W/m.K

40

-60

-40

-20

0

20

Wall position, x-coordinate (mm) T_xA, kA = 25 W/m.K; adiabatic surface T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3 T_x, kC = 50 W/m.K

40

PROBLEM 3.74 KNOWN: Composite wall exposed to convection process; inside wall experiences a uniform heat generation. FIND: (a) Neglecting interfacial thermal resistances, determine T1 and T2, as well as the heat fluxes through walls A and C, and (b) Determine the same parameters, but consider the interfacial contact resistances. Plot temperature distributions. SCHEMATIC: k A = 25 W m ⋅ K

L A = 30 mm

k B = 15 W m ⋅ K

L B = 30 mm

k C = 50 W m ⋅ K

L C = 20 mm

q B

4 – 10 6 W m 3

ASSUMPTIONS: (1) One-dimensional, steady-state heat flow, (2) Negligible contact resistance between walls, part (a), (3) Uniform heat generation in B, zero in A and C, (4) Uniform properties, (5) Negligible radiation at outer surfaces. ANALYSIS: (a) The temperature distribution in wall B follows from Eq. 3.41, q L  x  T2 − T1 x T − T2 . + 1 T ( x ) = B B 1 − + 2  2k B  2 LB 2 LB  2

2

(1)

The heat fluxes to the neighboring walls are found using Fourier’s law, q′′x = − k

dT dx

.

 q T −T  At x = −L B : q′′x ( − L B ) − k B  + B ( L B ) + 2 1  = q1′′ (2) 2L B   kB  q T −T  At x = + L B : q′′x ( L B ) − k B  − B ( L B ) + 2 1  = q′′2 (3) 2L B   kB The heat fluxes, q′′1 and q′′2 , can be evaluated by thermal circuits.

Substituting numerical values, find

(

q1′′ = ( T∞ − T1 ) C (1 h + LA k A ) = ( 25 − T1 ) C 1 1000 W m 2 ⋅ K + 0.03 m 25 W m ⋅ K $

$

)

q1′′ = ( 25 − T1 ) C ( 0.001 + 0.0012 ) K W = 454.6 ( 25 − T1 ) $

(

(4)

q′′2 = ( T2 − T∞ ) C (1 h + LC k C ) = ( T2 − 25 ) C 1 1000 W m 2 ⋅ K + 0.02 m 50 W m ⋅ K $

q ′′2 = ( T2 − 25 ) C $

$

( 0.001 + 0.0004 ) K

W = 714.3 ( T2 − 25 ) .

)

(5) Continued...

PROBLEM 3.74 (Cont.) Substituting the expressions for the heat fluxes, Eqs. (4) and (5), into Eqs. (2) and (3), a system of two equations with two unknowns is obtained. T −T Eq. (2): −4 × 106 W m3 × 0.03 m + 15 W m ⋅ K 2 1 = q1′′ 2 × 0.03 m −1.2 × 105 W m 2 − 2.5 × 102 ( T2 − T1 ) W m 2 = 454.6 ( 25 − T1 ) 704.6 T1 − 250 T2 = 131, 365 Eq. (3):

(6)

+4 × 106 W m3 × 0.03 m − 15 W m ⋅ K

T2 − T1 2 × 0.03 m

= q′′2

+1.2 × 105 W m 2 − 2.5 × 102 (T2 − T1 ) W m 2 = 714.3 ( T2 − 25 ) 250 T1 − 964 T2 = −137,857 Solving Eqs. (6) and (7) simultaneously, find

(7)

T2 = 210.0°C T1 = 260.9°C From Eqs. (4) and (5), the heat fluxes at the interfaces and through walls A and C are, respectively, q1′′ = 454.6 ( 25 − 260.9 ) = −107, 240 W m 2

< < <

q′′2 = 714.3 ( 210 − 25 ) = +132,146 W m 2 . Note directions of the heat fluxes.

(b) Considering interfacial contact resistances, we will use a different approach. The general solution for the temperature and heat flux distributions in each of the materials is TA ( x ) = C1x + C2 TB ( x ) = −

q B 2k B

x 2 + C3 x + C4

TC ( x ) = C5 x + C6

q′′x = − k A C1

− (LA + LB ) ≤ x ≤ −LB

(1,2)

q q′′x = − B x + C3 kB

−LB ≤ x ≤ LB

(3,4)

+ L B ≤ x ≤ ( L B + LC )

(5,6)

q′′x = − k CC5

To determine C1 ... C6 and the distributions, we need to identify boundary conditions using surface energy balances. At x = -(LA + LB):

(7) −q′′x ( −L A − L B ) + q′′cv = 0 − ( − k A C1 ) + h [T∞ − TA ( − L A − L B )] (8) At x = -LB: The heat flux must be continuous, but the temperature will be discontinuous across the contact resistance. q′′x,A ( − L B ) = q′′x,B ( − L B )

(9)

q ′′x,A ( − L B ) = [T1A ( − L B ) − T1B ( − L B )] R ′′tc,AB

(10)

Continued...

PROBLEM 3.74 (Cont.)

At x = + LB: The same conditions apply as for x = -LB, q ′′x,B ( + L B ) = q ′′x,C ( + L B )

(11)

q ′′x,B ( + L B ) = [T2B ( + L B ) − T2C ( + L B )] R ′′tc,BC

(12)

At x = +(LB + LC): − q x,C ( L B + LC ) − q ′′cv = 0

(13)

− ( − k C C5 ) − h [TC ( L B + L C ) − T∞ ] = 0

(14)

Following the method of analysis in IHT Example 3.6, User-Defined Functions, we solve the system of equations above for the constants C1 ... C6 for conditions with negligible and prescribed values for the interfacial constant resistances. The results are tabulated and plotted below; q1′′ and q′′2 represent heat fluxes leaving surfaces A and C, respectively. T1B (°C) 260

T2B (°C) 210

T2C (°C) 210

q1′′ (kW/m )

q′′2 (kW/m )

R ′′tc = 0

T1A (°C) 260

106.8

132.0

R ′′tc ≠ 0

233

470

371

227

94.6

144.2

Conditions

2

500 Temperature, T (C)

Temperature, T (C)

500

2

300

100

300

100 -60

-40

-20

0

20

40

-60

-40

Wall position, x-coordinate (mm)

-20

0

20

40

Wall position, x-coordinate (mm)

T_xA, kA = 25 W/m.K T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3 T_x, kC = 50 W/m.K

T_xA, kA = 25 W/m.K T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3 T_x, kC = 50 W/m.K

COMMENTS: (1) The results for part (a) can be checked using an energy balance on wall B, E in − E out = − E g

q1′′ − q ′′2 = −q B × 2L B

where 2 q1′′ − q ′′2 = −107, 240 − 132,146 = 239, 386 W m

− q B L B = −4 × 10 W m × 2 ( 0.03 m ) = −240, 000 W m . 6

3

2

Hence, we have confirmed proper solution of Eqs. (6) and (7). (2) Note that the effect of the interfacial contact resistance is to increase the temperature at all locations. The total heat flux leaving the composite wall (q1 + q2) will of course be the same for both cases.

PROBLEM 3.75 KNOWN: Composite wall of materials A and B. Wall of material A has uniform generation, while wall B has no generation. The inner wall of material A is insulated, while the outer surface of material B experiences convection cooling. Thermal contact resistance between the materials is R ′′t,c = 10

−4

2

m ⋅ K / W . See Ex. 3.6 that considers the case without contact resistance.

FIND: Compute and plot the temperature distribution in the composite wall. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with constant properties, and (3) Inner surface of material A is adiabatic. ANALYSIS: From the analysis of Ex. 3.6, we know the temperature distribution in material A is parabolic with zero slope at the inner boundary, and that the distribution in material B is linear. At the interface between the two materials, x = LA, the temperature distribution will show a discontinuity. TA ( x ) =

q L2A 

x2 

1 − +T 2  1A  LA 

0 ≤ x ≤ LA

2 kA 

TB ( x ) = T1B − ( T1B − T2 )

x − LA

LA ≤ x ≤ LA + LB LB Considering the thermal circuit above (see also Ex. 3.6) including the thermal contact resistance, T − T∞ T1B − T∞ T − T∞ = = 2 q′′ = q L A = 1A R ′′tot R ′′cond,B + R ′′conv R ′′conv find TA(0) = 147.5°C, T1A = 122.5°C, T1B = 115°C, and T2 = 105°C. Using the foregoing equations in IHT, the temperature distributions for each of the materials can be calculated and are plotted on the Effect of therm al contact res is tance on tem perature distribution graph below. 150

140

T (C )

130

120

110

100 0

10

20

30

40

50

60

70

x (m m )

COMMENTS: (1) The effect of the thermal contact resistance between the materials is to increase the maximum temperature of the system. (2) Can you explain why the temperature distribution in the material B is not affected by the presence of the thermal contact resistance at the materials’ interface?

PROBLEM 3.76  KNOWN: Plane wall of thickness 2L, thermal conductivity k with uniform energy generation q. For case 1, boundary at x = -L is perfectly insulated, while boundary at x = +L is maintained at To = 50°C. For case 2, the boundary conditions are the same, but a thin dielectric strip with thermal resistance R ′′t = 0.0005 m 2 ⋅ K / W is inserted at the mid-plane. FIND: (a) Sketch the temperature distribution for case 1 on T-x coordinates and describe key features; identify and calculate the maximum temperature in the wall, (b) Sketch the temperature distribution for case 2 on the same T-x coordinates and describe the key features; (c) What is the temperature difference between the two walls at x = 0 for case 2? And (d) What is the location of the maximum temperature of the composite wall in case 2; calculate this temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in the plane and composite walls, and (3) Constant properties. ANALYSIS: (a) For case 1, the temperature distribution, T1(x) vs. x, is parabolic as shown in the schematic below and the gradient is zero at the insulated boundary, x = -L. From Eq. 3.43, 2 2 q 2L 5 × 106 W / m3 2 × 0.020 m

T1 ( − L ) − T1 ( + L ) =

( ) 2k

(

=

2 × 50 W / m ⋅ K

)

= 80°C

and since T1(+L) = To = 50°C, the maximum temperature occurs at x = -L,

T1 ( − L ) = T1 ( + L ) + 80°C = 130°C

(b) For case 2, the temperature distribution, T2(x) vs. x, is piece-wise parabolic, with zero gradient at x = -L and a drop across the dielectric strip, ∆TAB. The temperature gradients at either side of the dielectric strip are equal.

(c) For case 2, the temperature drop across the thin dielectric strip follows from the surface energy balance shown above.

q′′x ( 0 ) = ∆TAB / R ′′t

 q′′x ( 0 ) = qL

 = 0.0005 m 2 ⋅ K / W × 5 × 106 W / m3 × 0.020 m = 50°C. ∆TAB = R ′′t qL (d) For case 2, the maximum temperature in the composite wall occurs at x = -L, with the value,

T2 ( − L ) = T1 ( − L ) + ∆TAB = 130°C + 50°C = 180°C

<

PROBLEM 3.77 KNOWN: Geometry and boundary conditions of a nuclear fuel element. FIND: (a) Expression for the temperature distribution in the fuel, (b) Form of temperature distribution for the entire system. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3) Uniform generation, (4) Constant properties, (5) Negligible contact resistance between fuel and cladding. ANALYSIS: (a) The general solution to the heat equation, Eq. 3.39, d 2T dx 2 is

+

T=−

q =0 kf

( −L ≤ x ≤ +L )

q 2 x + C1x+C2 . 2k f

The insulated wall at x = - (L+b) dictates that the heat flux at x = - L is zero (for an energy balance applied to a control volume about the wall, E in = E out = 0). Hence dT  q = − ( −L ) + C1 = 0 dx  x =− L kf T=−

C1 = −

or

 qL kf

 q 2 qL x − x+C2 . 2k f kf

The value of Ts,1 may be determined from the energy conservation requirement that E g = qcond = qconv , or on a unit area basis.

(

) (

)

k q ( 2L ) = s Ts,1 − Ts,2 = h Ts,2 − T∞ . b Hence, Ts,1 = Ts,1 =

q ( 2 Lb ) ks q ( 2 Lb ) ks

+ Ts,2 +

q ( 2L ) h

where

Ts,2 =

q ( 2L ) h

+ T∞

+ T∞ . Continued …..

PROBLEM 3.77 (Cont.) Hence from Eq. (1), T ( L ) = Ts,1 =

q ( 2 Lb ) ks

+

q ( 2 L ) h

( )

 2 3 q L + T∞ = − + C2 2 kf

which yields  2b 2 3 L    + + C2 = T∞ + qL   ks h 2 k f  Hence, the temperature distribution for ( − L ≤ x ≤ +L ) is T=−

 q 2 qL  x − x+qL 2k f kf

 2b 2 3 L   + +  + T∞  ks h 2 k f 

(b) For the temperature distribution shown below,

( −L − b ) ≤ x ≤ −L: − L ≤ x ≤ +L: +L ≤ x ≤ L+b:

dT/dx=0, T=Tmax | dT/dx | ↑ with ↑ x (dT/dx ) is const.

<

PROBLEM 3.78 KNOWN: Thermal conductivity, heat generation and thickness of fuel element. Thickness and thermal conductivity of cladding. Surface convection conditions. FIND: (a) Temperature distribution in fuel element with one surface insulated and the other cooled by convection. Largest and smallest temperatures and corresponding locations. (b) Same as part (a) but with equivalent convection conditions at both surfaces, (c) Plot of temperature distributions. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state, (3) Uniform generation, (4) Constant properties, (5) Negligible contact resistance. ANALYSIS: (a) From Eq. C.1,

T (x ) =

q L2  x 2  Ts,2 − Ts,1 x Ts,1 + Ts,2 1 − + + 2k f  L2  2 L 2  

(1)

With an insulated surface at x = -L, Eq. C.10 yields

Ts,1 − Ts,2 =

2 q L2 kf

(2)

and with convection at x = L + b, Eq. C.13 yields

(

)

(

k U Ts,2 − T∞ = q L − f Ts,2 − Ts,1 2L Ts,1 − Ts,2 =

)

2 LU 2 q L2 Ts,2 − T∞ − kf kf

(

)

(3)

Substracting Eq. (2) from Eq. (3),

0=

2LU 4 q L2 Ts,2 − T∞ − kf kf

(

Ts,2 = T∞ +

)

 2 qL U

(4) Continued …..

PROBLEM 3.78 (Cont.) and substituting into Eq. (2)

 L 1   +  Ts,1 = T∞ + 2 qL  kf U 

(5)

Substituting Eqs. (4) and (5) into Eq. (1),

T (x ) = − -1

 2 3 L q 2 qL   + x − x + qL 2 kf kf  U 2 kf

  + T∞ 

-1

or, with U = h + b/ks,

T (x ) = −

 2b 2 3 L  q 2 qL   + + x − x + qL 2 kf kf  ks h 2 k f

  + T∞ 

(6)

<

The maximum temperature occurs at x = - L and is

 b 1 L   + + T ( − L ) = 2 qL  ks h k f

  + T∞ 

 0.003m 1 0.015 m  + +  + 200°C = 530°C  15 W / m ⋅ K 2 10, 000 W / m ⋅ K 60 W / m ⋅ K  

T ( − L ) = 2 × 2 × 10 W / m × 0.015 m  7

3

<

The lowest temperature is at x = + L and is

T (+L ) = −

 2b 2 3 L  2 3 qL   + + + qL 2 kf  ks h 2 k f

  + T∞ = 380°C 

<

(b) If a convection condition is maintained at x = - L, Eq. C.12 reduces to

(

)

(

k  − f Ts,2 − Ts,1 U T∞ − Ts,1 = −qL 2L

)

 2 2 LU 2 qL Ts,1 − Ts,2 = Ts,1 − T∞ − kf kf

(

)

(7)

Subtracting Eq. (7) from Eq. (3),

0=

(

2 LU Ts,2 − T∞ − Ts,1 + T∞ kf

)

or

Ts,1 = Ts,2

Hence, from Eq. (7) Continued …..

PROBLEM 3.78 (Cont.) Ts,1 = Ts,2 =

1 b   qL   +  + T∞ + T∞ = qL U  h ks 

(8)

Substituting into Eq. (1), the temperature distribution is

T (x ) =

1 b   2  x2  qL   +  + T∞ 1 −  + qL 2 k f  L2   h ks   

(9)

<

The maximum temperature is at x = 0 and is T (0 ) =

7

2 × 10 W / m

3

(0.015 m )2

2 × 60 W / m ⋅ K

7

3

 1 0.003 m  + + 200°C  10, 000 W / m 2 ⋅ K 15 W / m ⋅ K   

+ 2 × 10 W / m × 0.015 m 

T ( 0 ) = 37.5°C + 90°C + 200°C = 327.5°C

<

The minimum temperature at x = ± L is

 1 0.003m   + 200°C = 290°C Ts,1 = Ts,2 = 2 × 107 W / m3 ( 0.015 m )  +  10, 000 W / m 2 ⋅ K 15 W / m ⋅ K    (c) The temperature distributions are as shown. 550 Te m p e ra tu re , T(C )

500 450 400 350 300 250 200 -0 .0 1 5

-0 .0 0 9

-0 .0 0 3

0 .0 0 3

0 .0 0 9

0 .0 1 5

Fu e l e le m e n t lo c a tio n , x(m ) In s u la te d s u rfa c e S ym m e trica l co n ve c tio n c o n d itio n s

The amount of heat generation is the same for both cases, but the ability to transfer heat from both surfaces for case (b) results in lower temperatures throughout the fuel element. COMMENTS: Note that for case (a), the temperature in the insulated cladding is constant and equivalent to Ts,1 = 530°C.

<

PROBLEM 3.79 KNOWN: Wall of thermal conductivity k and thickness L with uniform generation q ; strip heater with uniform heat flux q′′o ; prescribed inside and outside air conditions (hi, T∞,i, ho, T∞,o). FIND: (a) Sketch temperature distribution in wall if none of the heat generated within the wall is lost to the outside air, (b) Temperatures at the wall boundaries T(0) and T(L) for the prescribed condition, (c) Value of q ′′o required to maintain this condition, (d) Temperature of the outer surface, T(L), if  q=0 but q′′o corresponds to the value calculated in (c). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Uniform volumetric generation, (4) Constant properties. ANALYSIS: (a) If none of the heat generated within the wall is lost to the outside of the chamber, the gradient at x = 0 must be zero. Since q is uniform, the temperature distribution is parabolic, with T(L) > T∞,i. (b) To find temperatures at the boundaries of wall, begin with the general solution to the appropriate form of the heat equation (Eq.3.40).

T (x ) = −

q 2 x + C1x+C2 2k

(1)

From the first boundary condition,

dT =0 dx x=o

→

C1 = 0.

(2)

Two approaches are possible using different forms for the second boundary condition. Approach No. 1: With boundary condition → T (0 ) = T1

T (x ) = −

q 2 x + T1 2k

(3)

To find T1, perform an overall energy balance on the wall

E in − E out + E g = 0

 − h T ( L ) − T∞,i  + qL=0

T ( L ) = T2 = T∞,i +

 qL h

(4) Continued …..

PROBLEM 3.79 (Cont.) and from Eq. (3) with x = L and T(L) = T2,

T (L) = −

q 2 L + T1 2k

or

T1 = T2 +

  2 q 2 qL qL L = T∞,i + + 2k h 2k

(5,6)

Substituting numerical values into Eqs. (4) and (6), find

T2 = 50$ C+1000 W/m3 × 0.200 m/20 W/m 2 ⋅ K=50$C+10$C=60$C

<

T1 = 60$ C+1000 W/m3 × ( 0.200 m ) / 2 × 4 W/m ⋅ K=65$C.

<

2

Approach No. 2: Using the boundary condition

−k

dT = h T ( L ) − T∞,i  dx x=L

yields the following temperature distribution which can be evaluated at x = 0,L for the required temperatures,

T (x ) = −

)

(

 q qL x 2 − L2 + + T∞,i . 2k h

(c) The value of q′′o when T(0) = T1 = 65°C follows from the circuit

q′′o =

T1 − T∞,o 1/ h o

q′′o = 5 W/m2 ⋅ K ( 65-25) C=200 W/m 2 . $

<

 (d) With q=0, the situation is represented by the thermal circuit shown. Hence, q′′o = q′′a + q′′b q′′o =

T1 − T∞ ,o 1/ h o

+

T1 − T∞,i L/k+1/h i

which yields

T1 = 55$ C.

<

PROBLEM 3.80 KNOWN: Wall of thermal conductivity k and thickness L with uniform generation and strip heater with uniform heat flux q′′o ; prescribed inside and outside air conditions ( T∞,i , hi, T∞,o , ho). Strip heater acts to guard against heat losses from the wall to the outside. FIND: Compute and plot q′′o and T(0) as a function of q for 200 ≤ q ≤ 2000 W/m3 and T∞,i = 30, 50 and 70°C. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Uniform volumetric generation, (4) Constant properties. ANALYSIS: If no heat generated within the wall will be lost to the outside of the chamber, the gradient at the position x = 0 must be zero. Since q is uniform, the temperature distribution must be parabolic as shown in the sketch. To determine the required heater flux q′′o as a function of the operation conditions q and T∞,i , the analysis begins by considering the temperature distribution in the wall and then surface energy balances at the two wall surfaces. The analysis is organized for easy treatment with equation-solving software. Temperature distribution in the wall, T(x): The general solution for the temperature distribution in the wall is, Eq. 3.40, T(x) = −

q 2k

x 2 + C1x + C2

and the guard condition at the outer wall, x = 0, requires that the conduction heat flux be zero. Using Fourier’s law, dT  (1) q′′x (0) = − k = − kC1 = 0 (C1 = 0 )  dx  x = 0 At the outer wall, x = 0, T(0) = C2 (2) Surface energy balance, x = 0: E in − E out = 0 q ′′o − q ′′cv,o − q ′′x ( 0 ) = 0

(

)

q ′′cv,o = h T(0) − T∞ ,o , q ′′x ( 0 ) = 0

(3) (4a,b) Continued...

PROBLEM 3.80 (Cont.) Surface energy balance, x = L: E in − E out = 0 q ′′x (L) − q ′′cv,i = 0 q ′′x (L) = − k

dT 



dx  x = L

(5)  = + qL

(6)

q ′′cv,i = h  T(L) − T∞ ,i 

 q L2 + T 0 − T  ( ) ∞,i   2k 

q ′′cv,i = h −

(7)

400 300 200 100 0 0

500

1000

1500

2000

Volumetric generation rate, qdot (W/m^3) Tinfi = 30 C Tinfi = 50 C Tinfi = 70 C

Wall temperature, T(0) (C)

Heater flux, q''o (W/m^2)

Solving Eqs. (1) through (7) simultaneously with appropriate numerical values and performing the parametric analysis, the results are plotted below.

120 100 80 60 40 20 0

500

1000

1500

2000

Volumetric generation rate, qdot (W/m^3) Tinfi = 30 C Tinfi = 50 C Tinfi = 70 C

From the first plot, the heater flux q′′o is a linear function of the volumetric generation rate q . As expected, the higher q and T∞,i , the higher the heat flux required to maintain the guard condition ( q′′x (0) = 0). Notice that for any q condition, equal changes in T∞,i result in equal changes in the required q′′o . The outer wall temperature T(0) is also linearly dependent upon q . From our knowledge of the temperature distribution, it follows that for any q condition, the outer wall temperature T(0) will track changes in T∞,i .

PROBLEM 3.81 KNOWN: Plane wall with prescribed nonuniform volumetric generation having one boundary insulated and the other isothermal. FIND: Temperature distribution, T(x), in terms of x, L, k, q o and To . SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in xdirection, (3) Constant properties. ANALYSIS: The appropriate form the heat diffusion equation is d  dT  q + = 0. dx  dx  k Noting that q = q ( x ) = q o (1 − x/L ) , substitute for q ( x ) into the above equation, separate variables and then integrate,  x2  x −  + C1. 2L   Separate variables and integrate again to obtain the general form of the temperature distribution in the wall, q  q  x 2 x 3  x2  dT = − o  x − T (x ) = − o  −  dx+C1dx  + C1x+C2 . k  2L  k  2 6L  q  dT  d  = − o k  dx 

 x 1 − L  dx

q dT =− o dx k

Identify the boundary conditions at x = 0 and x = L to evaluate C1 and C2. At x = 0, q T ( 0 ) = To = − o ( 0 − 0 ) + C1 ⋅ 0 + C2 hence, C2 = To k At x = L, q o  q L dT  L2  = 0 = − L − hence, C1 = o   + C1  dx  x=L k  2L  2k The temperature distribution is q o  x 2 x 3  q o L T (x ) = − x+To .  − + k  2 6L  2k

<

COMMENTS: It is good practice to test the final result for satisfying BCs. The heat flux at x = 0 can be found using Fourier’s law or from an overall energy balance L  E out = E g = ∫ qdV 0

to obtain

q′′out = q o L/2.

PROBLEM 3.82 KNOWN: Distribution of volumetric heating and surface conditions associated with a quartz window. FIND: Temperature distribution in the quartz. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible radiation emission and convection at inner surface (x = 0) and negligible emission from outer surface, (4) Constant properties. ANALYSIS: The appropriate form of the heat equation for the quartz is obtained by substituting the prescribed form of q into Eq. 3.39. d 2T α (1 − β ) q′′o -α x e + =0 k dx 2 Integrating, (1 − β ) q′′o e-α x + C dT =+ 1 dx k

T=−

(1 − β ) q′′ e-α x + C x+C o

kα

1

2

− k dT/dx) x=o = β q′′o − k dT/dx) x=L = h  T ( L ) − T∞ 

Boundary Conditions:

 (1-β )  −k  q′′o + C1  = β q′′o  k  C1 = −q′′o / k

Hence, at x = 0: At x = L:

 (1-β )   (1-β )  −k  q′′o e-α L + C1  = h q′′o e-α L + C1L+C2 − T∞   k   kα 

Substituting for C1 and solving for C2, q′′ C2 = o h Hence,

q′′ 1 − (1 − β ) e-α L  + q′′o + o(1-β ) e-α L + T . ∞   k kα

T (x ) =

(1 − β ) q′′o e-α L − e-α x  + q′′o kα





k

(L − x ) +

q′′o  1 − (1 − β ) e-α L  + T∞ . <    h

COMMENTS: The temperature distribution depends strongly on the radiative coefficients, α and β. For α → ∞ or β = 1, the heating occurs entirely at x = 0 (no volumetric heating).

PROBLEM 3.83 KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactive wastes. Surface convection conditions. FIND: Radial temperature distribution. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible temperature drop across container wall. ANALYSIS: The appropriate form of the heat equation is q o  r 2  1 d  dT  q 1 −  = − = − r   r dr  dr  k k  ro2    r

 4 q r 2 qr dT =− o + + C1 dr 2k 4kro2

q r 2 q r 4 T = − o + o + C1 ln r+C2 . 4k 16kr 2 o

From the boundary conditions, dT |r=0 = 0 → C1 = 0 dr

−k

dT |r=ro = h T ( ro ) − T∞ ) dr

 q r 2 q r 2  q o ro q o ro + − = h  − o o + o o + C2 − T∞  2 4 16k  4k  q r 3q r 2 C2 = o o + o o + T∞ . 4h 16k Hence  q r q r 2 3 1  r T ( r ) = T∞ + o o + o o  −  4h k 16 4  ro 

2 4  1 r    +   .  16  ro  

<

COMMENTS: Applying the above result at ro yields Ts = T ( ro ) = T∞ + ( q o ro ) / 4h The same result may be obtained by applying an energy balance to a control surface about the container, where E g = qconv . The maximum temperature exists at r = 0.

PROBLEM 3.84 KNOWN: Cylindrical shell with uniform volumetric generation is insulated at inner surface and exposed to convection on the outer surface.  h, T∞ and k, (b) FIND: (a) Temperature distribution in the shell in terms of ri , ro , q, Expression for the heat rate per unit length at the outer radius, q′ ( ro ). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial (cylindrical) conduction in shell, (3) Uniform generation, (4) Constant properties. ANALYSIS: (a) The general form of the temperature distribution and boundary conditions are T (r ) = −

q 2 r + C1 ln r+C2 4k

dT  q 1  = 0 = − ri + C1 + 0 dr r 2k ri

at r = ri:

C1 =

i

dT   = h T ( ro ) − T∞  dr r

−k

at r = ro:

q 2 r 2k i

surface energy balance

o

 q  q  q  1   q  k  − ro +  ri2 ⋅   = h  − ro2 +  ri2  ln ro + C2 − T∞  ro    2k   2k  4k   2k  qr C2 = − o 2h

  2   2 1 +  ri   + qro   ro   2k  

2      1 −  ri  ln ro  + T∞  2  ro    

Hence,

(

)

 i2  r  qr  q 2 2 qr T (r ) = ro − r + ln   − o 4k 2k  ro  2h

  2  1 +  ri   + T∞ .   ro    

<

(b) From an overall energy balance on the shell, q′ ( r ) = E ′ = q π r 2 − r 2 . r o

g

(o i )

<

Alternatively, the heat rate may be found using Fourier’s law and the temperature distribution, q′r ( r ) = − k ( 2π ro )

 q   2 1 qr dT  i  = −2π kro − ro + + 0 + 0  = q π ro2 − ri2  dr r 2k ro  2k  o  

(

)

PROBLEM 3.85 KNOWN: The solid tube of Example 3.7 with inner and outer radii, 50 and 100 mm, and a thermal conductivity of 5 W/m⋅K. The inner surface is cooled by a fluid at 30°C with a convection coefficient 2 of 1000 W/m ⋅K. 5

FIND: Calculate and plot the temperature distributions for volumetric generation rates of 1 × 10 , 5 5 6 3 × 10 , and 1 × 10 W/m . Use Eq. (7) with Eq. (10) of the Example 3.7 in the IHT Workspace. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant properties and (4) Uniform volumetric generation. ANALYSIS: From Example 3.7, the temperature distribution in the tube is given by Eq. (7),

T ( r ) = Ts,2 +

)

(

q 2 2 q 2  r2  r2 − r − r "n   4k 2k 2  r 

r1 ≤ r ≤ r2

The temperature at the inner boundary, Ts,1, follows from the surface energy balance, Eq. (10), π q r22 − r12 = h2π r1 Ts,1 − T∞

(

)

(

)

(1)

(2)

For the conditions prescribed in the schematic with q = 1× 105 W / m3 , Eqs. (1) and (2), with r = r1 and T(r) = Ts,1, are solved simultaneously to find Ts,2 = 69.3°C. Eq. (1), with Ts,2 now a known parameter, can be used to determine the temperature distribution, T(r). The results for different values of the generation rate are shown in the graph. Effect of generation rate on temperature distributions

Temperature, T(C)

500 400 300 200 100 0 50

60

70

80

90

100

Radial location, r (mm) qdot = 1e5 W/m^3 qdot = 5e5 W/m^3 qdot = 1e6 W/m^3

COMMENTS: (1) The temperature distributions are parabolic with a zero gradient at the insulated outer boundary, r = r2. The effect of increasing q is to increase the maximum temperature in the tube, which always occurs at the outer boundary. (2) The equations used to generate the graphical result in the IHT Workspace are shown below. // The temperature distribution, from Eq. 7, Example 3.7 T_r = Ts2 + qdot/(4*k) * (r2^2 – r^2) – qgot / (2*k) * r2^2*ln (r2/r) // The temperature at the inner surface, from Eq. 7 Ts1 = Ts2 + qdot / (4*k) * (r2^2 – r1^2) – qdot / (2*k) * r2^2 * ln (r2/r1) // The energy balance on the surface, from Eq. 10 pi * qdot * (r2^2 – r1^2) = h * 2 * pi * r1 * (Ts1 – Tinf)

PROBLEM 3.86 KNOWN: Diameter, resistivity, thermal conductivity, emissivity, voltage, and maximum temperature of heater wire. Convection coefficient and air exit temperature. Temperature of surroundings. FIND: Maximum operating current, heater length and power rating. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Uniform wire temperature, (3) Constant properties, (4) Radiation exchange with large surroundings. ANALYSIS: Assuming a uniform wire temperature, Tmax = T(r = 0) ≡ To ≈ Ts, the maximum volumetric heat generation may be obtained from Eq. (3.55), but with the total heat transfer coefficient, ht = h + hr, used in lieu of the convection coefficient h. With

(

)

h r = εσ ( Ts + Tsur ) Ts + Tsur = 0.20 × 5.67 × 10 2

2

−8

2

W/m ⋅K

4

(1473 + 323 ) K

(

2

1473 + 323

)

2

2

2

K = 46.3 W / m ⋅ K

h t = ( 250 + 46.3) W / m 2 ⋅ K = 296.3 W / m 2 ⋅ K

)

(

2 296.3 W / m 2 ⋅ K 2 ht q max = (Ts − T∞ ) = (1150°C ) = 1.36 × 109 W / m3 ro 0.0005m Hence, with

2 I 2 R e I ( ρe L / A c ) I 2 ρ e I2 ρe q = = = = 2 ∀ LAc A c2 π D2 / 4

)

(

1/ 2

 q  Imax =  max   ρe 

1/ 2

π D 2  1.36 × 109 W / m3   =  10−6 Ω ⋅ m  4  

π (0.001m ) = 29.0 A 4 2

<

Also, with ∆E = I Re = I (ρeL/Ac),

∆E ⋅ Ac L= = Imax ρe

2 110 V π ( 0.001m ) / 4   = 2.98m 29.0 A 10−6 Ω ⋅ m

(

)

<

and the power rating is

Pelec = ∆E ⋅ Imax = 110 V ( 29 A ) = 3190 W = 3.19 kW

<

COMMENTS: To assess the validity of assuming a uniform wire temperature, Eq. (3.53) may be used to compute the centerline temperature corresponding to q max and a surface temperature of 2

1200°C. It follows that To =

q ro

4k

( 0.0005m )2 4 ( 25 W / m ⋅ K ) 9

+ Ts =

1.36 × 10 W / m

3

+ 1200°C = 1203°C. With only a

3°C temperature difference between the centerline and surface of the wire, the assumption is excellent.

PROBLEM 3.87 KNOWN: Energy generation in an aluminum-clad, thorium fuel rod under specified operating conditions. FIND: (a) Whether prescribed operating conditions are acceptable, (b) Effect of q and h on acceptable operating conditions. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in r-direction, (2) Steady-state conditions, (3) Constant properties, (4) Negligible temperature gradients in aluminum and contact resistance between aluminum and thorium. PROPERTIES: Table A-1, Aluminum, pure: M.P. = 933 K; Table A-1, Thorium: M.P. = 2023 K, k ≈ 60 W/m⋅K. ANALYSIS: (a) System failure would occur if the melting point of either the thorium or the aluminum were exceeded. From Eq. 3.53, the maximum thorium temperature, which exists at r = 0, is T(0) =

2  qr o

4k

+ Ts = TTh,max

where, from the energy balance equation, Eq. 3.55, the surface temperature, which is also the aluminum temperature, is Ts = T∞ +

 qr o

2h

= TAl

Hence, TAl = Ts = 95$ C +

7 × 108 W m3 × 0.0125 m 2

14, 000 W m ⋅ K

7 × 108 W m3 ( 0.0125m )

= 720$ C = 993 K

2

TTh,max =

4 × 60 W m ⋅ K

+ 993 K = 1449 K

<

Although TTh,max < M.P.Th and the thorium would not melt, Tal > M.P.Al and the cladding would melt under the proposed operating conditions. The problem could be eliminated by decreasing q , increasing h or using a cladding material with a higher melting point. (b) Using the one-dimensional, steady-state conduction model (solid cylinder) of the IHT software, the following radial temperature distributions were obtained for parametric variations in q and h. Continued...

PROBLEM 3.87 (Cont.) 1600

1200 Temperature, T(K)

Temperature, T(K)

1500 1400 1300 1200 1100 1000

1000 800 600

900

400

800 0

0

0.002 0.004 0.006 0.008 0.01 0.012 0.014

0.002 0.004 0.006 0.008 0.01 0.012 0.014 Radius, r(m)

Radius, r(m)

qdot = 2E8, h = 2000 W/m^2.K qdot = 2E8, h = 3000 W/m^2.K qdot = 2E8, h = 5000 W/m^2.K qdot = 2E8, h = 10000 W/m^2.K

h = 10000 W/m^2.K, qdot = 7E8 W/m^3 h = 10000 W/m^2.K, qdot = 8E8 W/m^3 h = 10000 W/m^2.K, qdot = 9E9 W/m^3

For h = 10,000 W/m2⋅K, which represents a reasonable upper limit with water cooling, the temperature of the aluminum would be well below its melting point for q = 7 × 108 W/m3, but would be close to the melting point for q = 8 × 108 W/m3 and would exceed it for q = 9 × 108 W/m3. Hence, under the best of conditions, q ≈ 7 × 108 W/m3 corresponds to the maximum allowable energy generation. However, if coolant flow conditions are constrained to provide values of h < 10,000 W/m2⋅K, volumetric heating would have to be reduced. Even for q as low as 2 × 108 W/m3, operation could not be sustained for h = 2000 W/m2⋅K. The effects of q and h on the centerline and surface temperatures are shown below.

Surface temperature, Ts (K)

Centerline temperature, T(0) (K)

2000 2000 1600 1200 800 400 0

1600 1200 800 400 0 1E8

1E8

2.8E8

4.6E8

6.4E8

8.2E8

Energy generation, qdot (W/m^3) h = 2000 W/m^2.K h = 5000 W/m^2.K h = 10000 W/m^2.K

1E9

2.8E8

4.6E8

6.4E8

8.2E8

1E9

Energy generation, qdot (W/m^3) h = 2000 W/m^2.K h = 5000 W/m^2.K h = 10000 W/m^2.K

For h = 2000 and 5000 W/m2⋅K, the melting point of thorium would be approached for q ≈ 4.4 × 108 and 8.5 × 108 W/m3, respectively. For h = 2000, 5000 and 10,000 W/m2⋅K, the melting point of aluminum would be approached for q ≈ 1.6 × 108, 4.3 × 108 and 8.7 × 108 W/m3. Hence, the envelope of acceptable operating conditions must call for a reduction in q with decreasing h, from a maximum of q ≈ 7 × 108 W/m3 for h = 10,000 W/m2⋅K. COMMENTS: Note the problem which would arise in the event of a loss of coolant, for which case h would decrease drastically.

PROBLEM 3.88 KNOWN: Radii and thermal conductivities of reactor fuel element and cladding. Fuel heat generation rate. Temperature and convection coefficient of coolant. FIND: (a) Expressions for temperature distributions in fuel and cladding, (b) Maximum fuel element temperature for prescribed conditions, (c) Effect of h on temperature distribution. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible contact resistance, (4) Constant properties. ANALYSIS: (a) From Eqs. 3.49 and 3.23, the heat equations for the fuel (f) and cladding (c) are 1 d  dTf  q  =− kf r dr  dr 

1 d  dTc  r =0 r dr  dr 

(0 ≤ r ≤ r1 )

( r1 ≤ r ≤ r2 )

Hence, integrating both equations twice, dTf dr dTc dr

=− =

 qr

2k f

+

C1 kf r

C3 k cr

Tf = −

2  qr

C + 1 ln r + C 2 4k f k f

C Tc = 3 ln r + C4 kc

(1,2)

(3,4)

The corresponding boundary conditions are: dTf dr )r = 0 = 0 −k f

dTf dr

dT   = −k c c   dr r = r r = r1 1

Tf ( r1 ) = Tc ( r1 ) − kc

dTc  = h [Tc ( r2 ) − T∞ ]  dr r = r

(5,6) (7,8)

2

Note that Eqs. (7) and (8) are obtained from surface energy balances at r1 and r2, respectively. Applying Eq. (5) to Eq. (1), it follows that C1 = 0. Hence, Tf = −

2  qr

+ C2 4k f From Eq. (6), it follows that 2  C ln r qr − 1 + C2 = 3 1 + C4 4k f kc

(9)

(10) Continued...

PROBLEM 3.88 (Cont.) Also, from Eq. (7),  qr 1

C =− 3 2 r1

2  qr C3 = − 1 2

or

(11)

C  C Finally, from Eq. (8), − 3 = h  3 ln r2 + C4 − T∞  or, substituting for C3 and solving for C4 r2  kc  C4 =

2  qr 1

2  qr 1

ln r2 + T∞ + 2r2 h 2k c Substituting Eqs. (11) and (12) into (10), it follows that C2 =

2  qr 1

4k f

2k c

+

2  qr 1

2r2 h

+

2  qr 1

2k c

ln r2 + T∞

2  r qr ln 2 + 1 T∞ 4k f 2k c r1 2r2 h Substituting Eq. (13) into (9),

C2 =

2  qr 1

−

2  qr 1 ln r1

(12)

+

(

2  qr 1

)

(13)

2  r qr ln 2 + 1 + T∞ 4k f 2k c r1 2r2 h Substituting Eqs. (11) and (12) into (4),

Tf =

Tc =

q

r12 − r 2 +

2  qr 1

2  qr 1

2  r qr ln 2 + 1 + T∞ . 2k c r 2r2 h

(14)

<

(15)

<

(b) Applying Eq. (14) at r = 0, the maximum fuel temperature for h = 2000 W/m2⋅K is Tf ( 0 ) =

2 × 108 W m3 × ( 0.006 m )

2 × 108 W m3 × (0.006 m )

2

4× 2 W m⋅ K

2

+

2 × 25 W m ⋅ K

2 × 108 W m3 ( 0.006 m )

ln

0.009 m 0.006 m

2

+

2 × ( 0.09 m ) 2000 W m 2 ⋅ K

+ 300 K

Tf ( 0 ) = (900 + 58.4 + 200 + 300 ) K = 1458 K . (c) Temperature distributions for the prescribed values of h are as follows: 600

1300

Temperature, Tc(K)

Temperature, Tf(K)

1500

<

1100 900 700 500 300 0

0.001

0.002

0.004

0.005

Radius in fuel element, r(m) h = 2000 W/m^2.K h = 5000 W/m^2.K h = 10000 W/m^2.K

0.006

500

400

300 0.006

0.007

0.008

0.009

Radius in cladding, r(m) h = 2000 W/m^2.K h = 5000 W/m^2.K h = 10000 W/m^2.K

Continued...

PROBLEM 3.88 (Cont.) Clearly, the ability to control the maximum fuel temperature by increasing h is limited, and even for h → ∞, Tf(0) exceeds 1000 K. The overall temperature drop, Tf(0) - T∞, is influenced principally by the low thermal conductivity of the fuel material. 2 8  COMMENTS: For the prescribed conditions, Eq. (14) yields, Tf(0) - Tf(r1) = qr 1 4k f = (2×10 W/m3)(0.006 m)3/8 W/m⋅K = 900 K, in which case, with no cladding and h → ∞, Tf(0) = 1200 K. To reduce Tf(0) below 1000 K for the prescribed material, it is necessary to reduce q .

PROBLEM 3.89 KNOWN: Dimensions and properties of tubular heater and external insulation. Internal and external convection conditions. Maximum allowable tube temperature. FIND: (a) Maximum allowable heater current for adiabatic outer surface, (3) Effect of internal convection coefficient on heater temperature distribution, (c) Extent of heat loss at outer surface. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Constant properties, (3) Uniform heat generation, (4) Negligible radiation at outer surface, (5) Negligible contact resistance. ANALYSIS: (a) From Eqs. 7 and 10, respectively, of Example 3.7, we know that

(

q 2 r2 q 2 2 r2 ln − r2 − r1 2k r1 4k

Ts,2 − Ts,1 = and Ts,1 = T∞ ,1 +

(

q r22 − r12

)

(1)

)

(2)

2h1r1

Hence, eliminating Ts,1, we obtain Ts,2 − T∞ ,1 =

)

(

2  qr 2 

)

(

 r 1 k ln 2 − 1 − r12 r22 + 1 − r12 r22   2k  r1 2 h1r1 

Substituting the prescribed conditions (h1 = 100 W/m2⋅K),

(

)(

Ts,2 − T∞ ,1 = 1.237 × 10−4 m3 ⋅ K W q W m3

)

Hence, with Tmax corresponding to Ts,2, the maximum allowable value of q is 1400 − 400 q max = = 8.084 × 106 W m3 −4 1.237 × 10 with q =

I 2 Re ∀

=

I2 ρe L Ac ρe I2 = 2 LA c π r 2 − r  2 1



(

I max = π r22 − r12

)

1/ 2

 q     ρe 

)

(

(

2

= π 0.035 − 0.025

2

)

6 3 1/ 2  2 8.084 × 10 W m m  = 6406 A  −6  

 0.7 × 10 Ω ⋅ m 

<

Continued …..

PROBLEM 3.89 (Cont.) (b) Using the one-dimensional, steady-state conduction model of IHT (hollow cylinder; convection at inner surface and adiabatic outer surface), the following temperature distributions were obtained.

Temperature, T(K)

1500 1300 1100 900 700 500 300 0.025

0.027

0.029

0.031

0.033

0.035

Radius, r(m) h = 100 W/m^2.K h = 500 W/m^2.K h = 1000 W/m^2.K

The results are consistent with key implications of Eqs. (1) and (2), namely that the value of h1 has no effect on the temperature drop across the tube (Ts,2 - Ts,1 = 30 K, irrespective of h1), while Ts,1 decreases with increasing h1. For h1 = 100, 500 and 1000 W/m2⋅K, respectively, the ratio of the temperature drop between the inner surface and the air to the temperature drop across the tube, (Ts,1 - T∞,1)/(Ts,2 - Ts,1), decreases from 970/30 = 32.3 to 194/30 = 6.5 and 97/30 = 3.2. Because the outer surface is insulated, the heat rate to the airflow is fixed by the value of q and, irrespective of h1,

(

)

q′ ( r1 ) = π r22 − r12 q = −15, 240 W

<

(c) Heat loss from the outer surface of the tube to the surroundings depends on the total thermal resistance ln ( r3 r2 ) 1 R tot = + 2π Lk i 2π r3Lh 2 or, for a unit area on surface 2, r ln ( r3 r2 ) r R ′′tot,2 = ( 2π r2 L ) R tot = 2 + 2 ki r3h 2 Again using the capabilities of IHT (hollow cylinder; convection at inner surface and heat transfer from outer surface through R ′′tot,2 ), the following temperature distributions were determined for the tube and insulation. Insulation Temperature, T(K)

Tube temperature, T(K)

1200 1160 1120 1080 1040 1000 0.025

1200 1100 1000 900 800 700 600 500

0.027

0.029

0.031

Radius, r(m) delta =0.025 m delta = 0.050 m

0.033

0.035

0

0.2

0.4

0.6

0.8

1

Dimensionless radius, (r-r2)/(r3-r2) r3 = 0.060 m r3 = 0.085 m

Continued...

PROBLEM 3.89 (Cont.) Heat losses through the insulation, q′ ( r2 ) , are 4250 and 3890 W/m for δ = 25 and 50 mm, respectively,

with corresponding values of q′ ( r1 ) equal to -10,990 and -11,350 W/m. Comparing the tube temperature distributions with those predicted for an adiabatic outer surface, it is evident that the losses reduce tube wall temperatures predicted for the adiabatic surface and also shift the maximum temperature from r = 0.035 m to r ≈ 0.033 m. Although the tube outer and insulation inner surface temperatures, Ts,2 = T(r2), increase with increasing insulation thickness, Fig. (c), the insulation outer surface temperature decreases. COMMENTS: If the intent is to maximize heat transfer to the airflow, heat losses to the ambient should be reduced by selecting an insulation material with a significantly smaller thermal conductivity.

PROBLEM 3.90 KNOWN: Electric current I is passed through a pipe of resistance R ′e to melt ice under steady-state conditions. FIND: (a) Temperature distribution in the pipe wall, (b) Time to completely melt the ice. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant properties, (4) Uniform heat generation in the pipe wall, (5) Outer surface of the pipe is adiabatic, (6) Inner surface is at a constant temperature, Tm. 3

PROPERTIES: Table A-3, Ice (273K): ρ = 920 kg/m ; Handbook Chem. & Physics, Ice: 5

Latent heat of fusion, hsf = 3.34×10 J/kg. ANALYSIS: (a) The appropriate form of the heat equation is Eq. 3.49, and the general solution, Eq. 3.51 is T (r ) = −

q 2 r + C1lnr+C2 4k

where  q=

(

I2R ′e

π r22 − r12

)

.

Applying the boundary condition ( dT/dr )r = 0, it follows that 2 0=

 2 C1 qr + 2k r2  22 qr

Hence

C1 =

and

 qr q T ( r ) = − r 2 + 2 lnr+C2 . 4k 2k

2k 2

Continued …..

PROBLEM 3.90 (Cont.) Applying the second boundary condition, T ( r1 ) = Tm , it follows that 2

 qr q Tm = − r12 + 2 lnr1 + C2 . 4k 2k Solving for C2 and substituting into the expression for T(r), find T ( r ) = Tm +

 22 qr 2k

ln

)

(

r q 2 2 − r − r1 . r1 4k

<

(b) Conservation of energy dictates that the energy required to completely melt the ice, Em, must equal the energy which reaches the inner surface of the pipe by conduction through the wall during the melt period. Hence from Eq. 1.11b ∆Est = Ein − Eout + Egen ∆Est = E m = t m ⋅ qcond,r1 or, for a unit length of pipe,   dT   ρ π r12 h sf = t m  − k ( 2π r1 )     dr  r1  

( )

 qr  2 qr   ρ π r12 h sf = −2π r1kt m  2 − 1   2kr1 2k   

( ) ( )

)

(

ρ π r12 h sf = − t m q π r22 − r12 . Dropping the minus sign, which simply results from the fact that conduction is in the negative r direction, it follows that tm =

ρ h sf r12

(

q r22 − r12

)

=

ρ h sf π r12 I2R ′e

.

With r1 = 0.05m, I = 100 A and R ′e = 0.30 Ω/m, it follows that 920kg/m3 × 3.34 ×105 J/kg × π × (0.05m )

2

tm = or

(100A )2 × 0.30Ω / m

t m = 804s.

<

COMMENTS: The foregoing expression for tm could also be obtained by recognizing that all of the energy which is generated by electrical heating in the pipe wall must be transferred to the ice. Hence, I2 R ′e t m = ρ h sf π r12 .

PROBLEM 3.91 KNOWN: Materials, dimensions, properties and operating conditions of a gas-cooled nuclear reactor. FIND: (a) Inner and outer surface temperatures of fuel element, (b) Temperature distributions for different heat generation rates and maximum allowable generation rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible contact resistance, (5) Negligible radiation. PROPERTIES: Table A.1, Thoriun: Tmp ≈ 2000 K; Table A.2, Graphite: Tmp ≈ 2300 K. ANALYSIS: (a) The outer surface temperature of the fuel, T2, may be determined from the rate equation T − T∞ q′ = 2 R ′tot where R ′tot =

ln ( r3 r2 ) 2π k g

+

1 2π r3h

=

ln (14 11)

2π (3 W m ⋅ K )

+

(

1

2π ( 0.014 m ) 2000 W m ⋅ K 2

)

= 0.0185 m ⋅ K W

and the heat rate per unit length may be determined by applying an energy balance to a control surface about the fuel element. Since the interior surface of the element is essentially adiabatic, it follows that

)

(

(

)

q′ = q π r22 − r12 = 108 W m3 × π 0.0112 − 0.0082 m 2 = 17, 907 W m Hence, T2 = q′R ′tot + T∞ = 17, 907 W m ( 0.0185 m ⋅ K W ) + 600 K = 931K

<

With zero heat flux at the inner surface of the fuel element, Eq. C.14 yields T1 = T2 +

2  qr 2

2  r  r 2  qr  1 − 1  − 1 ln  2  4k t  r 2  2k t  r1   2 

T1 = 931K +

2 2 108 W m3 ( 0.011m )   0.008 2  108 W m3 ( 0.008 m )  0.011  ln  − 1 −     4 × 57 W m ⋅ K 2 × 57 W m ⋅ K  0.008    0.011  

Continued...

PROBLEM 3.91 (Cont.)

<

T1 = 931K + 25 K − 18 K = 938 K

(b) The temperature distributions may be obtained by using the IHT model for one-dimensional, steadystate conduction in a hollow tube. For the fuel element ( q > 0), an adiabatic surface condition is prescribed at r1, while heat transfer from the outer surface at r2 to the coolant is governed by the thermal resistance R ′′tot,2 = 2π r2 R ′tot = 2π(0.011 m)0.0185 m⋅K/W = 0.00128 m2⋅K/W. For the graphite ( q =

2500

2500

2100

2100

Temperature, T(K)

Temperature, T(K)

0), the value of T2 obtained from the foregoing solution is prescribed as an inner boundary condition at r2, while a convection condition is prescribed at the outer surface (r3). For 1 × 108 ≤ q ≤ 5 × 108 W/m3, the following distributions are obtained.

1700 1300 900 500 0.008

0.009

0.01

Radial location in fuel, r(m) qdot = 5E8 qdot = 3E8 qdot = 1E8

0.011

1700 1300 900 500 0.011

0.012

0.013

0.014

Radial location in graphite, r(m) qdot = 5E8 qdot = 3E8 qdot = 1E8

The comparatively large value of kt yields small temperature variations across the fuel element, while the small value of kg results in large temperature variations across the graphite. Operation at q = 5 × 108 W/m3 is clearly unacceptable, since the melting points of thorium and graphite are exceeded and approached, respectively. To prevent softening of the materials, which would occur below their melting points, the reactor should not be operated much above q = 3 × 108 W/m3. COMMENTS: A contact resistance at the thorium/graphite interface would increase temperatures in the fuel element, thereby reducing the maximum allowable value of q .

PROBLEM 3.92 KNOWN: Long rod experiencing uniform volumetric generation encapsulated by a circular sleeve exposed to convection. FIND: (a) Temperature at the interface between rod and sleeve and on the outer surface, (b) Temperature at center of rod. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction in rod and sleeve, (2) Steady-state conditions, (3) Uniform volumetric generation in rod, (4) Negligible contact resistance between rod and sleeve. ANALYSIS: (a) Construct a thermal circuit for the sleeve,

where 2 q′=E ′gen = q π D12 / 4 = 24, 000 W/m3 × π × ( 0.20 m ) / 4 = 754.0 W/m

R s′ =

ln ( r2 / r1 ) 2π k s

R conv =

=

ln ( 400/200 ) 2π × 4 W/m ⋅ K

= 2.758 × 10−2 m ⋅ K/W

1 1 = = 3.183 × 10−2 m ⋅ K/W hπ D 2 25 W/m 2 ⋅ K × π × 0.400 m

The rate equation can be written as q′=

T1 − T∞ T −T = 2 ∞ R ′s + R ′conv R ′conv

)

(

T1 = T∞ + q′ ( R ′s + R ′conv ) = 27$ C+754 W/m 2.758 × 10-2 + 3.183 × 10−2 K/W ⋅ m=71.8$C T2 = T∞ + q′R ′conv = 27$ C+754 W/m × 3.183 × 10-2m ⋅ K/W=51.0$C.

< <

(b) The temperature at the center of the rod is T ( 0 ) = To =

 12 qr 4k r

24, 000 W/m3 ( 0.100 m )

2

+ T1 =

4 × 0.5 W/m ⋅ K

+ 71.8$ C=192$C.

<

COMMENTS: The thermal resistances due to conduction in the sleeve and convection are comparable. Will increasing the sleeve outer diameter cause the surface temperature T2 to increase or decrease?

PROBLEM 3.93 KNOWN: Radius, thermal conductivity, heat generation and convection conditions associated with a solid sphere. FIND: Temperature distribution. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant properties, (4) Uniform heat generation. ANALYSIS: Integrating the appropriate form of the heat diffusion equation, d  2 dT    kr dr  + q=0 r 2 dr  1

r2

 3 dT qr =− + C1 dr 3k

T (r ) = −

or

 dT qr C =− + 1 dr 3k r 2

 2 C1 qr − + C2 . 6k r

The boundary conditions are: −k

 2 d  2 dT  qr r = − dr  dr  k

dT  =0 dr  r=0

hence

C1 = 0, and

dT  = h T ( ro ) − T∞  . dr  r o

Substituting into the second boundary condition (r = ro), find  qr  o  o2  o qr  o2 qr qr = h + C2 − T∞  + + T∞ . C2 = 3 3h 6k  6k  The temperature distribution has the form  q 2 2 qr T (r ) = ro − r + o + T∞ . 6k 3h COMMENTS: To verify the above result, obtain T(ro) = Ts,  qr Ts = o + T∞ 3h Applying energy balance to the control volume about the sphere,

(

)

4  q  π ro3  = h4π ro2 ( Ts − T∞ ) 3 

find

Ts =

o qr + T∞ . 3h

<

PROBLEM 3.94 KNOWN: Radial distribution of heat dissipation of a spherical container of radioactive wastes. Surface convection conditions. FIND: Radial temperature distribution. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible temperature drop across container wall. ANALYSIS: The appropriate form of the heat equation is 1 r2

q o d  2 dT  q r =− =− dr  dr  k k

  2  1 −  r   .   ro    

q  r 3 r 5  dT  + C1 =− o − dr k  3 5ro2    q  r 2 r 4  C1 − + C2 . T=− o − k  6 20ro2  r   From the boundary conditions, r2

Hence

dT/dr |r=0 = 0

and

− kdT/dr |r=ro = h  T ( ro ) − T∞ 

it follows that C1 = 0 and  q  r 2 r 2   r  r q o  o − o  = h  − o  o − o  + C2 − T∞   k  6 20   3 5 C2 =

2roq o 7q o ro2 + + T∞ . 15h 60k

2 4  o2  7 1  r  2ro q o qr 1  r   Hence T ( r ) = T∞ + + −   +   . 15h k  60 6  ro  20  ro     COMMENTS: Applying the above result at ro yields

<

Ts = T ( ro ) = T∞ + ( 2ro q o /15h ). The same result may be obtained by applying an energy balance to a control surface about the container, where E g = qconv . The maximum temperature exists at r = 0.

PROBLEM 3.95 KNOWN: Dimensions and thermal conductivity of a spherical container. Thermal conductivity and volumetric energy generation within the container. Outer convection conditions. FIND: (a) Outer surface temperature, (b) Container inner surface temperature, (c) Temperature distribution within and center temperature of the wastes, (d) Feasibility of operating at twice the energy generation rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional radial conduction. ANALYSIS: (a) For a control volume which includes the container, conservation of energy yields  −q E g − E out = 0 , or qV conv = 0 . Hence

( )

(

q ( 4 3 ) π ri3 = h4π ro2 Ts,o − T∞ 5

)

3

and with q = 10 W/m , Ts,o = T∞ +

3  qr i

3hro2

= 25$ C +

105 W m 2 ( 0.5 m )

3

3000 W m ⋅ K ( 0.6 m ) 2

2

<

= 36.6$ C .

(b) Performing a surface energy balance at the outer surface, E in − E out = 0 or q cond − q conv = 0 . Hence 4π k ss Ts,i − Ts,o = h4π ro2 Ts,o − T∞ (1 ri ) − (1 ro )

(

)

(

)

(

)

2 1000 W m ⋅ K  ro  $ Ts,i = Ts,o + − 1  ro ( Ts,o − T∞ ) = 36.6 C + ( 0.2 ) 0.6 m 11.6$ C = 129.4$ C .  k ss  ri 15 W m ⋅ K 

h

<

(c) The heat equation in spherical coordinates is d  dT  2  k rw  r 2 = 0.  + qr dr  dr  Solving, r2

dT

=−

3  qr

+ C1

dr 3k rw Applying the boundary conditions, dT and =0 dr r = 0 C1 = 0

and

and

T (r ) = −

2  qr

C − 1 + C2 6k rw r

T ( ri ) = Ts,i 2  C2 = Ts,i + qr i 6k rw .

Continued...

PROBLEM 3.95 (Cont.) Hence

ri2 − r 2 ) ( 6k rw q

T ( r ) = Ts,i +

<

At r = 0, T ( 0 ) = Ts,i +

2  qr i

6k rw

= 129.4$ C +

105 W m3 ( 0.5 m )

2

6 ( 20 W m ⋅ K )

<

= 337.7$ C

(d) The feasibility assessment may be performed by using the IHT model for one-dimensional, steadystate conduction in a solid sphere, with the surface boundary condition prescribed in terms of the total thermal resistance

( )

ri2 2 R ′′tot,i = 4π ri R tot = R ′′cnd,i + R ′′cnv,i =

[(1 ri ) − (1 ro )] + 1  ri 2 k ss





h  ro 

where, for ro = 0.6 m and h = 1000 W/m2⋅K, R ′′cnd,i = 5.56 × 10-3 m2⋅K/W, R ′′cnv,i = 6.94 × 10-4 m2⋅K/W,

Center temperature, T(0) (C)

and R ′′tot,i = 6.25 × 10-3 m2⋅K/W. Results for the center temperature are shown below. 675

625

575

525

475 0

2000

4000

6000

8000

10000

Convection coefficient, h(W/m^2.K) ro = 0.54 m ro = 0.60 m

Clearly, even with ro = 0.54 m = ro,min and h = 10,000 W/m2⋅K (a practical upper limit), T(0) > 475°C and the desired condition can not be met. The corresponding resistances are R ′′cnd,i = 2.47 × 10-3 m2⋅K/W, R ′′cnv,i = 8.57 × 10-5 m2⋅K/W, and R ′′tot,i = 2.56 × 10-3 m2⋅K/W. The conduction resistance remains dominant, and the effect of reducing R ′′cnv,i by increasing h is small. The proposed extension is not

feasible. COMMENTS: A value of q = 1.79 × 105 W/m3 would allow for operation at T(0) = 475°C with ro = 0.54 m and h = 10,000 W/m2⋅K.

PROBLEM 3.96 KNOWN: Carton of apples, modeled as 80-mm diameter spheres, ventilated with air at 5°C and experiencing internal volumetric heat generation at a rate of 4000 J/kg⋅day. FIND: (a) The apple center and surface temperatures when the convection coefficient is 7.5 W/m2⋅K, and (b) Compute and plot the apple temperatures as a function of air velocity, V, for the range 0.1 ≤ V ≤ 1 m/s, when the convection coefficient has the form h = C1V0.425, where C1 = 10.1 W/m2⋅K⋅(m/s)0.425. SCHEMATIC:

ASSUMPTIONS: (1) Apples can be modeled as spheres, (2) Each apple experiences flow of ventilation air at T‡ = 5°C, (3) One-dimensional radial conduction, (4) Constant properties and (5) Uniform heat generation. ANALYSIS: (a) From Eq. C.24, the temperature distribution in a solid sphere (apple) with uniform generation is T(r) =

2  qr o

r2   1 −  + Ts 6k  r 2   o 

(1)

To determine Ts, perform an energy balance on the apple as shown in the sketch above, with volume V = 4 3 Sro3 , − q cv + q ∀ = 0 E in − E out + E g = 0

( )(Ts − T∞ ) + q (4 3π ro3 ) = 0 −7.5 W m 2 ⋅ K ( 4π × 0.0402 m 2 )(Ts − 5 C ) + 38.9 W m3 ( 4 3 π × 0.0403 m3 ) = 0 − h 4π ro2



(2)

$

where the volumetric generation rate is q = 4000 J kg ⋅ day q = 4000 J kg ⋅ day × 840 kg m3 × (1day 24 hr ) × (1hr 3600 s ) q = 38.9 W m3 and solving for Ts, find

<

Ts = 5.14$ C From Eq. (1), at r = 0, with Ts, find T(0) =

38.9 W m3 × 0.0402 m 2 6 × 0.5 W m ⋅ K

+ 5.14$ C = 0.12$ C + 5.14$ C = 5.26$ C

< Continued...

PROBLEM 3.96 (Cont.) (b) With the convection coefficient depending upon velocity, h = C1V 0.425 with C1 = 10.1 W/m2⋅K⋅(m/s)0.425, and using the energy balance of Eq. (2), calculate and plot Ts as a function of ventilation air velocity V. With very low velocities, the center temperature is nearly 0.5°C higher than the air. From our earlier calculation we know that T(0) - Ts = 0.12°C and is independent of V.

Center temperature, T(0) (C)

5.4

5.3

5.2 0

0.2

0.4

0.6

0.8

1

Ventilation air velocity, V (m/s)

COMMENTS: (1) While the temperature within the apple is nearly isothermal, the center temperature will track the ventilation air temperature which will increase as it passes through stacks of cartons. (2) The IHT Workspace used to determine Ts for the base condition and generate the above plot is shown below. // The temperature distribution, Eq (1), T_r = qdot * ro^2 / (4 * k) * ( 1- r^2/ro^2 ) + Ts // Energy balance on the apple, Eq (2) - qcv + qdot * Vol = 0 Vol = 4 / 3 * pi * ro ^3 // Convection rate equation: qcv = h* As * ( Ts - Tinf ) As = 4 * pi * ro^2 // Generation rate: qdot = qdotm * (1/24) * (1/3600) * rho // Assigned variables: ro = 0.080 k = 0.5 qdotm = 4000 rho = 840 r=0 h = 7.5 //h = C1 * V^0.425 //C1 = 10.1 //V = 0.5 Tinf = 5

// Generation rate, W/m^3; Conversions: days/h and h/sec

// Radius of apple, m // Thermal conductivity, W/m.K // Generation rate, J/kg.K // Specific heat, J/kg.K // Center, m; location for T(0) // Convection coefficient, W/m^2.K; base case, V = 0.5 m/s // Correlation // Air velocity, m/s; range 0.1 to 1 m/s // Air temperature, C

PROBLEM 3.97 KNOWN: Plane wall, long cylinder and sphere, each with characteristic length a, thermal  conductivity k and uniform volumetric energy generation rate q. FIND: (a) On the same graph, plot the dimensionless temperature, [ T ( x or r ) − T ( a ) ]/[ q a /2k], vs. the dimensionless characteristic length, x/a or r/a, for each shape; (b) Which shape has the smallest temperature difference between the center and the surface? Explain this behavior by comparing the ratio of the volume-to-surface area; and (c) Which shape would be preferred for use as a nuclear fuel element? Explain why? 2

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties and (4) Uniform volumetric generation. ANALYSIS: (a) For each of the shapes, with T(a) = Ts, the dimensionless temperature distributions can be written by inspection from results in Appendix C.3. 2 T ( x ) − Ts x Plane wall, Eq. C.22 = 1−   a  2 / 2k qa Long cylinder, Eq. C.23

Sphere, Eq. C.24

2 1 r  1 −     2 / 2k 2   a   qa T ( r ) − Ts 1   r 2  = 1 −     2 / 2k 3   a   qa

T ( r ) − Ts

=

The dimensionless temperature distributions using the foregoing expressions are shown in the graph below. Dimensionless temperature distribution (T_x,r-Ts) / (qdot*a^2/2*k)

1 0.8 0.6 0.4 0.2 0 0

0.2

0.4

0.6

0.8

1

Dimensionless length, x/a or r/a Plane wall, 2a Long cylinder, a Sphere, a

Continued …..

PROBLEM 3.97 (Cont.) (b) The sphere shape has the smallest temperature difference between the center and surface, T(0) – T(a). The ratio of volume-to-surface-area, ∀/As, for each of the shapes is Plane wall

∀ a (1× 1) = =a As (1×1)

Long cylinder

∀ π a 2 ×1 a = = As 2π a ×1 2

Sphere

∀ 4π a 3 / 3 a = = As 3 4π a 2

The smaller the ∀/As ratio, the smaller the temperature difference, T(0) – T(a). (c) The sphere would be the preferred element shape since, for a given ∀/As ratio, which controls the generation and transfer rates, the sphere will operate at the lowest temperature.

PROBLEM 3.98 KNOWN: Radius, thickness, and incident flux for a radiation heat gauge. FIND: Expression relating incident flux to temperature difference between center and edge of gauge. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r (negligible temperature drop across foil thickness), (3) Constant properties, (4) Uniform incident flux, (5) Negligible heat loss from foil due to radiation exchange with enclosure wall, (6) Negligible contact resistance between foil and heat sink.

ANALYSIS: Applying energy conservation to a circular ring extending from r to r + dr, q r + q′′i ( 2π rdr ) = q r+dr ,

q r = − k ( 2π rt )

dT , dr

q r+dr = q r +

dq r dr. dr

Rearranging, find that q′′i ( 2π rdr ) =

d  dT ( −k2π rt )  dr  dr  dr 

d  dT  q′′ r = − i r.   dr  dr  kt Integrating, r

dT q′′r 2 = − i + C1 dr 2kt

and

q′′r 2 T ( r ) = − i + C1lnr+C2 . 4kt

With dT/dr|r=0 =0, C1 = 0 and with T(r = R) = T(R), q′′R 2 T (R ) = − i + C2 4kt

or

q′′R 2 C2 = T ( R ) + i . 4kt

Hence, the temperature distribution is T (r ) =

(

)

qi′′ R 2 − r 2 + T ( R ). 4kt

Applying this result at r = 0, it follows that q′′i =

4kt

4kt  T ( 0 ) − T ( R ) = ∆T. R2 R2

COMMENTS: This technique allows for determination of a radiation flux from measurement of a temperature difference. It becomes inaccurate if emission from the foil becomes significant.

<

PROBLEM 3.98 KNOWN: Radius, thickness, and incident flux for a radiation heat gauge. FIND: Expression relating incident flux to temperature difference between center and edge of gauge. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r (negligible temperature drop across foil thickness), (3) Constant properties, (4) Uniform incident flux, (5) Negligible heat loss from foil due to radiation exchange with enclosure wall, (6) Negligible contact resistance between foil and heat sink.

ANALYSIS: Applying energy conservation to a circular ring extending from r to r + dr, q r + q′′i ( 2π rdr ) = q r+dr ,

q r = − k ( 2π rt )

dT , dr

q r+dr = q r +

dq r dr. dr

Rearranging, find that q′′i ( 2π rdr ) =

d  dT ( −k2π rt )  dr  dr  dr 

d  dT  q′′ r = − i r.   dr  dr  kt Integrating, r

dT q′′r 2 = − i + C1 dr 2kt

and

q′′r 2 T ( r ) = − i + C1lnr+C2 . 4kt

With dT/dr|r=0 =0, C1 = 0 and with T(r = R) = T(R), q′′R 2 T (R ) = − i + C2 4kt

or

q′′R 2 C2 = T ( R ) + i . 4kt

Hence, the temperature distribution is T (r ) =

(

)

qi′′ R 2 − r 2 + T ( R ). 4kt

Applying this result at r = 0, it follows that q′′i =

4kt

4kt  T ( 0 ) − T ( R ) = ∆T. R2 R2

COMMENTS: This technique allows for determination of a radiation flux from measurement of a temperature difference. It becomes inaccurate if emission from the foil becomes significant.

<

PROBLEM 3.99 KNOWN: Net radiative flux to absorber plate. FIND: (a) Maximum absorber plate temperature, (b) Rate of energy collected per tube. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (x) conduction along absorber plate, (3) Uniform radiation absorption at plate surface, (4) Negligible losses by conduction through insulation, (5) Negligible losses by convection at absorber plate surface, (6) Temperature of absorber plate at x = 0 is approximately that of the water. PROPERTIES: Table A-1, Aluminum alloy (2024-T6): k ≈ 180 W/m⋅K. ANALYSIS: The absorber plate acts as an extended surface (a conduction-radiation system), and a differential equation which governs its temperature distribution may be obtained by applying Eq.1.11a to a differential control volume. For a unit length of tube q′x + q′′rad ( dx ) − q′x+dx = 0. With

q′x+dx = q′x +

and

q′x = − kt

dq′x dx dx

dT dx

it follows that, q′′rad − d 2T dx 2

d  dT  − kt =0  dx  dx 

q′′ + rad = 0 kt

Integrating twice it follows that, the general solution for the temperature distribution has the form, q′′ T ( x ) = − rad x 2 + C1x+C2 . 2kt Continued …..

PROBLEM 3.99 (Cont.) The boundary conditions are: T ( 0 ) = Tw dT  =0 dx  x=L/2

C2 = Tw q′′ L C1 = rad 2kt

Hence, q′′ T ( x ) = rad x ( L − x ) + Tw . 2kt The maximum absorber plate temperature, which is at x = L/2, is therefore q′′ L2 Tmax = T ( L/2 ) = rad + Tw . 8kt The rate of energy collection per tube may be obtained by applying Fourier’s law at x = 0. That is, energy is transferred to the tubes via conduction through the absorber plate. Hence,   dT  q′=2  − k t  dx  x=0   where the factor of two arises due to heat transfer from both sides of the tube. Hence, q′= − Lq′′rad . 800

W

2 0.2m ) ( 2

m + 60$ C W   8 180  ( 0.006m ) m ⋅ K 

Hence

Tmax =

or

Tmax = 63.7$ C

and

q′ = −0.2m × 800 W/m 2

or

q′ = −160 W/m.

< <

COMMENTS: Convection losses in the typical flat plate collector, which is not evacuated, would reduce the value of q ′.

PROBLEM 3.100 KNOWN: Surface conditions and thickness of a solar collector absorber plate. Temperature of working fluid. FIND: (a) Differential equation which governs plate temperature distribution, (b) Form of the temperature distribution.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Adiabatic bottom surface, (4) Uniform radiation flux and convection coefficient at top, (5) Temperature of absorber plate at x = 0 corresponds to that of working fluid. ANALYSIS: (a) Performing an energy balance on the differential control volume,

q′x + dq′rad = q′x+dx + dq′conv where

q′x+dx = q′x + ( dq′x / dx ) dx dq′rad = q′′rad ⋅ dx dq′conv = h ( T − T∞ ) ⋅ dx

Hence,

q′′rad dx= ( dq′x / dx ) dx+h ( T − T∞ ) dx.

From Fourier’s law, the conduction heat rate per unit width is q′′ d 2T h q′x = − k t dT/dx T − T∞ ) + rad = 0. − ( 2

dx

kT

<

kt

(b) Defining θ = T − T∞ , d 2T/dx 2 = d 2θ / dx 2 and the differential equation becomes, d 2θ dx 2

−

q′′ h θ + rad = 0. kt kt

It is a second-order, differential equation with constant coefficients and a source term, and its general solution is of the form θ = C1e+λ x + C2e-λ x + S/λ 2 1/ 2 λ = ( h/kt ) , S=q′′rad / kt. where Appropriate boundary conditions are:

θ (0 ) = To − T∞ ≡ θ 0 ,

dθ /dx) x=L = 0.

θ o = C1 + C2 + S/λ 2

Hence,

dθ /dx) x=L = C1 λ e+λ L − C2 λ e-λ L = 0

)(

(

C1 = θ 0 − S/λ 2 / 1 + e2λ L

Hence,

(

θ = θ 0 − S/λ 2

)

)

C2 = C1 e2λ L

(

)(

C2 = θ 0 − S/λ 2 / 1 + e-2λ L

 eλ x e -λ x  +   + S/λ 2 . 2 L -2 L λ λ 1+e  1+e

) <

PROBLEM 3.101 KNOWN: Dimensions of a plate insulated on its bottom and thermally joined to heat sinks at its ends. Net heat flux at top surface. FIND: (a) Differential equation which determines temperature distribution in plate, (b) Temperature distribution and heat loss to heat sinks.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in x (W,L>>t), (3) Constant properties, (4) Uniform surface heat flux, (5) Adiabatic bottom, (6) Negligible contact resistance.

ANALYSIS: (a) Applying conservation of energy to the differential control volume, qx + dq

= qx +dx, where qx+dx = qx + (dqx/dx) dx and dq=q′′o ( W ⋅ dx ). Hence, ( dq x / dx ) − q′′o W=0. From Fourier’s law, q x = − k ( t ⋅ W ) dT/dx. Hence, the differential equation for the temperature distribution is −

q′′ + o = 0. dx 2 kt

d 2T

d  dT  ktW − q′′o W=0  dx  dx 

<

(b) Integrating twice, the general solution is, q′′ T ( x ) = − o x 2 + C1 x +C2 2kt and appropriate boundary conditions are T(0) = To, and T(L) = To. Hence, To = C2, and q′′ To = − o L2 + C1L+C2 2kt

and

q′′ L C1 = o . 2kt

Hence, the temperature distribution is

(

)

q′′ L T ( x ) = − o x 2 − Lx + To . 2kt

<

Applying Fourier’s law at x = 0, and at x = L, q′′ WL L  q′′   q ( 0 ) = − k ( Wt ) dT/dx) x=0 = −kWt  − o   x −  =− o 2  x=0 2  kt   q′′ WL L  q′′   q ( L ) = − k ( Wt ) dT/dx) x=L = − kWt  − o   x −  =+ o 2  x=L 2  kt   Hence the heat loss from the plates is q=2 ( q′′o WL/2 ) = q′′o WL.

<

COMMENTS: (1) Note signs associated with q(0) and q(L). (2) Note symmetry about x = L/2. Alternative boundary conditions are T(0) = To and dT/dx)x=L/2=0.

PROBLEM 3.102 KNOWN: Dimensions and surface conditions of a plate thermally joined at its ends to heat sinks at different temperatures. FIND: (a) Differential equation which determines temperature distribution in plate, (b) Temperature distribution and an expression for the heat rate from the plate to the sinks, and (c) Compute and plot temperature distribution and heat rates corresponding to changes in different parameters. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x (W,L >> t), (3) Constant properties, (4) Uniform surface heat flux and convection coefficient, (5) Negligible contact resistance. ANALYSIS: (a) Applying conservation of energy to the differential control volume q x + dq o = q x + dx + dq conv

where

q x + dx = q x + ( dq x dx ) dx

dq conv = h ( T − T∞ )( W ⋅ dx )

Hence, dq x

q x + q ′′o ( W ⋅ dx ) = q x + ( dq x dx ) dx + h ( T − T∞ )( W ⋅ dx )

dx

Using Fourier’s law, q x = − k ( t ⋅ W ) dT dx , − ktW

d2T

+ hW ( T − T∞ ) = q′′o

d 2T

−

h

dx 2 dx 2 kt (b) Introducing θ ≡ T − T∞ , the differential equation becomes

(T − T∞ ) +

q′′o kt

+ hW ( T − T∞ ) = q ′′o W .

<

= 0.

q′′ θ + o =0. kt dx 2 kt This differential equation is of second order with constant coefficients and a source term. With d 2θ

−

h

λ 2 ≡ h kt and S ≡ q′′o kt , it follows that the general solution is of the form θ = C1e + λ x + C2 e−λ x + S λ 2 . Appropriate boundary conditions are: θ (0) = To − T∞ ≡ θ o

(1) θ (L) = TL − T∞ ≡ θ L

(2,3)

Substituting the boundary conditions, Eqs. (2,3) into the general solution, Eq. (1),

θ o = C1e0 + C2e0 + S λ 2 θ L = C1e + λ L + C2e − λ L + S λ 2 To solve for C2, multiply Eq. (4) by -e+λL and add the result to Eq. (5),

( ) ( ) C2 = (θ L − θ o e + λ L ) − S λ 2 ( −e + λ L + 1) ( −e+ λ L + e −λ L )  

(4,5)

−θ o e + λ L + θ L = C2 −e + λ L + e − λ L + S λ 2 −e + λ L + 1

(6) Continued...

PROBLEM 3.102 (Cont.) Substituting for C2 from Eq. (6) into Eq. (4), find

{(

)

) (−e+λL + e−λL )} − S λ 2

(

C1 = θ o −  θ L − θ o e+ λ L − S λ 2 −e + λ L + 1   

(7)

Using C1 and C2 from Eqs. (6,7) and Eq. (1), the temperature distribution can be expressed as



θ (x) =  e



+λ x

−

(

)

(

sinh ( λ x ) + λ L  sinh ( λ x )  + λ L sinh ( λ x ) +λ L e + 1− e θo + θ L + − 1 − e  sinh ( λ L ) sinh L sinh L λ λ ( ) ( )  

) λS (8) 2

< The heat rate from the plate is q p = −q x ( 0 ) + q x ( L ) and using Fourier’s law, the conduction heat rates, with Ac = W⋅t, are q x ( 0 ) = − kAc

  0  dθ  eλ L λ kA e λ λ θo + θL = − −     c dx  x = 0 sinh ( λ L )  sinh (λ L )  



 1 − e+ λ L

+ −

 sinh (λ L )

q x ( L ) = − kAc

 S    λ 2 

λ − λ

<

  λ L  λ cosh ( λ L ) dθ  eλ L kA e cosh L λ λ λ θL = − −  θo + ( )   c dx  x = L sinh ( λ L ) sinh ( λ L )   



 1 − e+ λ L

+ −

 sinh (λ L )

 S    λ 2 

λ cosh ( λ L ) − λ e + λ L 

<

(c) For the prescribed base-case conditions listed below, the temperature distribution (solid line) is shown in the accompanying plot. As expected, the maximum temperature does not occur at the midpoint, but slightly toward the x-origin. The sink heat rates are q′′x ( 0 ) = −17.22 W

q′′x ( L ) = 23.62 W

<

Temperature, T(x) (C)

300

200

100

0 0

20

40

60

80

100

Distance, x (mm) q''o = 20,000 W/m^2; h = 50 W/m^2.K q''o = 30,000 W/m^2; h = 50 W/m^2.K q''o = 20,000 W/m^2; h = 200 W/m^2.K q''o = 4927 W/m^2 with q''x(0) = 0; h = 200 W/m^2.K

The additional temperature distributions on the plot correspond to changes in the following parameters, with all the remaining parameters unchanged: (i) q′′o = 30,000 W/m2, (ii) h = 200 W/m2⋅K, (iii) the value of q′′o for which q′′x (0) = 0 with h = 200 W/m2⋅K. The condition for the last curve is q′′o = 4927 W/m2 for which the temperature gradient at x = 0 is zero. Base case conditions are: q′′o = 20,000 W/m2, To = 100°C, TL = 35°C, T∞ = 25°C, k = 25 W/m⋅K, h = 50 W/m2⋅K, L = 100 mm, t = 5 mm, W = 30 mm.

PROBLEM 3.103 KNOWN: Thin plastic film being bonded to a metal strip by laser heating method; strip dimensions and thermophysical properties are prescribed as are laser heating flux and convection conditions. FIND: (a) Expression for temperature distribution for the region with the plastic strip, -w1/2 ≤ x ≤ w1/2, (b) Temperature at the center (x = 0) and the edge of the plastic strip (x = ± w1/2) when the laser flux is 10,000 W/m2; (c) Plot the temperature distribution for the strip and point out special features. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction only, (3) Plastic film has negligible thermal resistance, (4) Upper and lower surfaces have uniform convection coefficients, (5) Edges of metal strip are at air temperature (T∞), that is, strip behaves as infinite fin so that w2 → ∞, (6) All the incident laser heating flux q′′o is absorbed by the film. PROPERTIES: Metal strip (given): ρ = 7850 kg/m3, cp = 435 J/kg⋅m3, k = 60 W/m⋅K. ANALYSIS: (a) The strip-plastic film arrangement can be modeled as an infinite fin of uniform cross section a portion of which is exposed to the laser heat flux on the upper surface. The general solutions for the two regions of the strip, in terms of θ ≡ T ( x ) − T∞ , are θ1 ( x ) = C1e

0 ≤ x ≤ w1 2

+ mx

+ C 2e

− mx

+M m

θ 2 ( x ) = C3 e

+ mx

+ C4 e

(1)

m = ( 2h kd )

M = q ′′o P 2kA c = q ′′o kd

w1 2 ≤ x ≤ ∞

2 1/ 2

− mx

. Four boundary conditions can be identified to evaluate the constants: dθ1

At x = 0:

(0 ) = 0 = C1me0 − C 2 me−0 + 0

dx θ ( w1 2 ) = θ 2 ( w1 2 )

At x = w1/2:

C1e

+ m w1 2

+ C 2e

− m w1 2

→

2

+ mw1 2

dθ1 ( w1 2 ) / dx = dθ 2 ( w1 2 ) / dx

At x = w1/2:

mC1e

+ m w1 2

− mC 2 e

− m w1 2

+ 0 = mC3e

(4)

C1 = C 2

+ M m = C 3e

+ m w1 2

+ C 4e

M m 2e

(5)

− m w1 2

− mC 4 e

∞ −∞ At x → ∞: θ 2 ( ∞ ) = 0 = C3 e + C 4 e → C3 = 0 With C3 = 0 and C1 = C2, combine Eqs. (6 and 7) to eliminate C4 to find

C1 = C 2 = −

(2,3)

− mw1 2

(6)

(7) (8)

2

m w1 2

.

(9)

and using Eq. (6) with Eq. (9) find C 4 = M m sinh ( mw1 2 ) e 2

− mx1 / 2

(10) Continued...

PROBLEM 3.103 (Cont.) Hence, the temperature distribution in the region (1) under the plastic film, 0 ≤ x ≤ w1/2, is 2

M m

θ1 ( x ) = −

2e

m w1 w

(e

+ mx

+e

− mx

) + mM = mM (1 − e 2

− m w1 2

2

cosh mx

)

(11)

<

and for the region (2), x ≥ w1/2, θ2 (x ) =

M m

2

sinh ( mw1 2 ) e

− mx

(12)

(b) Substituting numerical values into the temperature distribution expression above, θ1(0) and θ1(w1/2) can be determined. First evaluate the following parameters: M = 10, 000 W m

(

2

60 W m ⋅ K × 0.00125 m = 133, 333 K m

2

m = 2 × 10 W m ⋅ K 60 W m ⋅ K × 0.00125 m

)

1/ 2

2

= 16.33 m

−1

Hence, for the midpoint x = 0, θ1 ( 0 ) =

133, 333 K m

(

16.33 m

)

2

−1 2

(

)

1 − exp −16.33 m −1 × 0.020 m × cosh ( 0 ) = 139.3 K  

<

T1 ( 0 ) = θ1 ( 0 ) + T∞ = 139.3 K + 25 C = 164.3 C . $

For the position x = w1/2 = 0.020 m,

(

θ1 ( w1 2 ) = 500.0 1 − 0.721cosh 16.33 m



$

−1

× 0.020 m

) = 120.1K <

T1 ( w1 2 ) = 120.1K + 25 C = 145.1 C . $

$

(c) The temperature distributions, θ1(x) and θ2(x), are shown in the plot below. Using IHT, Eqs. (11) and (12) were entered into the workspace and a graph created. The special features are noted: (1) No gradient at midpoint, x = 0; symmetrical distribution.

(3) Temperature excess and gradient approach zero with increasing value of x.

Strip temperature, T (C)

(2) No discontinuity of gradient at w1/2 (20 mm).

180

140

100

60

20 0

50

100

150

200

250

300

x-coordinate, x (mm) Region 1 - constant heat flux, q''o Region 2 - x >= w1/2

COMMENTS: How wide must the strip be in order to satisfy the infinite fin approximation such that θ2 (x → ∞) = 0? For x = 200 mm, find θ2(200 mm) = 6.3°C; this would be a poor approximation. When x = 300 mm, θ2(300 mm) = 1.2°C; hence when w2/2 = 300 mm, the strip is a reasonable approximation to an infinite fin.

PROBLEM 3.104 KNOWN: Thermal conductivity, diameter and length of a wire which is annealed by passing an electrical current through the wire. FIND: Steady-state temperature distribution along wire.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction along the wire, (3) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient h. ANALYSIS: Applying conservation of energy to a differential control volume,

q x + E g − dq conv − q x+dx = 0 q x+dx = q x +

dq x dx dx

(

) E g = q (π D2 / 4 ) dx.

q x = − k π D2 / 4 dT/dx

dq conv = h (π D dx ) ( T − T∞ ) Hence,

(

k π D2 / 4

) dx 2 dx+q (π D2 / 4)dx − h (π Ddx ) (T − T∞ ) = 0 d 2T

d 2θ

or, with θ ≡ T − T∞ ,

dx 2

−

4h q θ + =0 kD k

The solution (general and particular) to this nonhomogeneous equation is of the form

q θ = C1 emx + C2 e-mx + km 2 2

where m = (4h/kD). The boundary conditions are: dθ  = 0 = m C1 e0 − mC2 e0 →  dx  x=0

(

)

q θ ( L ) = 0 = C1 emL + e-mL + → km 2

C1 = C2 C1 =

2  −q/km

emL + e-mL

= C2

The temperature distribution has the form

T = T∞ −

 emx + e-mx  q  cosh mx  − 1 = T∞ − − 1 .   km 2  emL +e-mL  km 2  cosh mL  q

<

COMMENTS: This process is commonly used to anneal wire and spring products. To check the result, note that T(L) = T(-L) = T∞.

PROBLEM 3.105 KNOWN: Electric power input and mechanical power output of a motor. Dimensions of housing, mounting pad and connecting shaft needed for heat transfer calculations. Temperature of ambient air, tip of shaft, and base of pad. FIND: Housing temperature.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in pad and shaft, (3) Constant properties, (4) Negligible radiation. ANALYSIS: Conservation of energy yields

Pelec − Pmech − q h − q p − q s = 0 q h = h h A h ( Th − T∞ ) ,

θ L = 0,

qs =

(

π 2 / 4  D3h k s s  

(

t

cosh mL − θ L / θ b sinh mL

1/ 2

1/ 2 mL = 4h s L2 / k s D ,

Hence

qs = M

π2 3  M=  D hsks   4   

)

(

(Th − T∞ ) ,

qp = k p W 2

)

1/ 2

(Th − T∞ ).

(Th − T∞ )

tanh 4h s L2 / k s D

)

1/ 2

Substituting, and solving for (Th - T∞), Th − T∞ =

((

) ( ) 1/ 2 (4hsL2 / ksD) = 3.87,

h h A h + k p W 2 / t+ π 2 / 4 D3h s k s

((π / 4) D h k ) 2

Pelec − Pmech

3

1/ 2

s s

= 6.08 W/K,

1/ 2

/ tanh 4h s L2 / k s D

)

1/ 2

tanhmL=0.999

25 − 15 ) × 103 W 104 W ( Th − T∞ = = 10 × 2+0.5 ( 0.7 )2 / 0.05 + 6.08 / 0.999  W/K ( 20+4.90+6.15 ) W/K   Th − T∞ = 322.1K

Th = 347.1$ C

<

COMMENTS: (1) Th is large enough to provide significant heat loss by radiation from the

(

)

4 = 4347 housing. Assuming an emissivity of 0.8 and surroundings at 25°C, q rad = ε A h Th4 − Tsur

W, which compares with q conv = hA h ( Th − T∞ ) = 5390 W. Radiation has the effect of decreasing Th. (2) The infinite fin approximation, qs = M, is excellent.

PROBLEM 3.106 KNOWN: Dimensions and thermal conductivity of pipe and flange. Inner surface temperature of pipe. Ambient temperature and convection coefficient. FIND: Heat loss through flange. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction in pipe and flange, (3) Constant thermal conductivity, (4) Negligible radiation exchange with surroundings. ANALYSIS: From the thermal circuit, the heat loss through the flanges is

q=

Ts,i − T∞ R t,w + R t,f

=

Ts,i − T∞

n ( Do / Di ) / 4π tk  + (1/ hAf ηf )

Since convection heat transfer only occurs from one surface of a flange, the connected flanges may be modeled as a single annular fin of thickness t ′ = 2t = 30 mm. Hence, r2c = ( Df / 2 ) + t ′ / 2 = 0.140 m,

(

) (

)

(

)

2 2 − r12 = 2π r2c − Do / 2 = 2π 0.1402 − 0.062 m 2 = 0.101m 2 , Lc = L + t ′ / 2 = A f = 2π r2c

( Df

(

)

1/ 2 2/2 2 h / kA p = 0.188. With r2c/r1 = − Do ) / 2 + t = 0.065 m, A p = Lc t ′ = 0.00195 m , Lc

r2c/(Do/2) = 1.87, Fig. 3.19 yields ηf = 0.94. Hence,

q=

q=

300°C − 20°C

(

n (1.25 ) / 4π × 0.03m × 40 W / m ⋅ K  + 1/10 W / m 2 ⋅ K × 0.101m 2 × 0.94 280°C = 262 W (0.0148 + 1.053) K / W

) <

COMMENTS: Without the flange, heat transfer from a section of pipe of width t ′ = 2t is −1 q = Ts,i − T∞ / R t,w + R t,cnv , where R t,cnv = ( h × π Do t ′ ) = 7.07 K / W. Hence, q = 39.5 W, and there is significant heat transfer enhancement associated with the extended surfaces afforded by the flanges.

(

)(

)

PROBLEM 3.107 KNOWN: TC wire leads attached to the upper and lower surfaces of a cylindrically shaped solder bead. Base of bead attached to cylinder head operating at 350°C. Constriction resistance at base and TC wire convection conditions specified. FIND: (a) Thermal circuit that can be used to determine the temperature difference between the two intermediate metal TC junctions, (T1 – T2); label temperatures, thermal resistances and heat rates; and (b) Evaluate (T1 – T2) for the prescribed conditions. Comment on assumptions made in building the model. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in solder bead; no losses from lateral and top surfaces; (3) TC wires behave as infinite fins, (4) Negligible thermal contact resistance between TC wire terminals and bead. ANALYSIS: (a) The thermal circuit is shown above. Note labels for the temperatures, thermal resistances and the relevant heat fluxes. The thermal resistances are as follows: Constriction (con) resistance, see Table 4.1, case 10

R con = 1/ ( 2k bead Dsol ) = 1/ ( 2 × 40 W / m ⋅ K × 0.006 m ) = 2.08 K / W TC (tc) wires, infinitely long fins; Eq. 3.80 −0.5

R tc,1 = R tc,2 = R fin = ( hPk w Ac )

(

P = π D w , A c = π D2w / 4

R tc = 100 W / m 2 ⋅ K × π 2 × ( 0.003 m ) × 70 W / m ⋅ K / 4 3

)

−0.5

= 46.31 K / W

Solder bead (sol), cylinder Dsol and Lsol

R sol = Lsol / ( k sol Asol )

(

2 /4 Asol = π Dsol

)

R sol = 0.010 m / 10 W / m ⋅ K × π (0.006 m ) / 4 = 35.37 K / W 2

(b) Perform energy balances on the 1- and 2-nodes, solve the equations simultaneously to find T1 and T2, from which (T1 – T2) can be determined. Continued …..

PROBLEM 3.107 (Cont.) Node 1

T2 − T1 Thead − T1 T∞ − T1 + + =0 R sol R con R tc,1

Node 2

T∞ − T2 T1 − T2 + =0 R tc,2 R sol

Substituting numerical values with the equations in the IHT Workspace, find

T1 = 359°C

T2 = 199.2°C

T1 − T2 = 160°C

COMMENTS: (1) With this arrangement, the TC indicates a systematically low reading of the cylinder head. The size of the solder bead (Lsol) needs to be reduced substantially. (2) The model neglects heat losses from the top and lateral sides of the solder bead, the effect of which would be to increase our estimate for (T1 – T2). Constriction resistance is important; note that Thead – T1 = 26°C.

PROBLEM 3.108 KNOWN: Rod (D, k, 2L) that is perfectly insulated over the portion of its length –L ≤ x ≤ 0 and experiences convection (T∞, h) over the portion 0 ≤ x ≤ + L. One end is maintained at T1 and the other is separated from a heat sink at T3 with an interfacial thermal contact resistance R ′′tc . FIND: (a) Sketch the temperature distribution T vs. x and identify key features; assume T1 > T3 > T2; (b) Derive an expression for the mid-point temperature T2 in terms of thermal and geometric parameters of the system, (c) Using, numerical values, calculate T2 and plot the temperature distribution. Describe key features and compare to your sketch of part (a). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in rod for –L ≤ x ≤ 0, (3) Rod behaves as one-dimensional extended surface for 0 ≤ x ≤ +L, (4) Constant properties. ANALYSIS: (a) The sketch for the temperature distribution is shown below. Over the insulated portion of the rod, the temperature distribution is linear. A temperature drop occurs across the thermal contact resistance at x = +L. The distribution over the exposed portion of the rod is nonlinear. The minimum temperature of the system could occur in this portion of the rod.

(b) To derive an expression for T2, begin with the general solution from the conduction analysis for a fin of uniform cross-sectional area, Eq. 3.66. θ ( x ) = C1emx + C2e− mx 0 ≤ x ≤ +L (1) where m = (hP/kAc) conditions.

1/2

and θ = T(x) - T∞. The arbitrary constants are determined from the boundary

At x = 0, thermal resistance of rod

q x ( 0 ) = − kAc

θ − θ (0 ) dθ  = kAc 1  dx  x = 0 L

m C1e0 − m C 2e0 =

(

θ1 = T1 − T∞

)

1 θ1 − C1e0 + C2e0    L

(2) Continued …..

PROBLEM 3.108 (Cont.)

At x=L, thermal contact resistance

q x ( + L ) = − kAc

θ (L ) − θ3 dθ  =  dx  x = L R ′′tc / A c

θ 3 = T3 − T∞

1  mL C1e − k  m C1emL − m C2e− mL  = + C 2e− mL − θ 3     R ′′tc  

(3)

Eqs. (2) and (3) cannot be rearranged easily to find explicit forms for C1 and C2. The constraints will be evaluated numerically in part (c). Knowing C1 and C2, Eq. (1) gives

θ 2 = θ (0 ) = T2 − T∞ = C1 e0 + C2e0

(4)

(c) With Eqs. (1-4) in the IHT Workspace using numerical values shown in the schematic, find T2 = 62.1°C. The temperature distribution is shown in the graph below. Temperature distribution in rod

Temperature, T(x) (C)

200

150

100

50

0 -50

-30

-10

10

30

50

x-coordinate, x (mm)

COMMENTS: (1) The purpose of asking you to sketch the temperature distribution in part (a) was to give you the opportunity to identify the relevant thermal processes and come to an understanding of the system behavior. (2) Sketch the temperature distributions for the following conditions and explain their key features: (a) R ′′tc = 0, (b) R ′′tc → ∞, and (c) the exposed portion of the rod behaves as an infinitely long fin; that is, k is very large.

PROBLEM 3.109 KNOWN: Long rod in oven with air temperature at 400°C has one end firmly pressed against surface of a billet; thermocouples imbedded in rod at locations 25 and 120 mm from the billet indicate 325 and 375°C, respectively. FIND: The temperature of the billet, Tb. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Rod is infinitely long with uniform crosssectional area, (3) Uniform convection coefficient along rod. ANALYSIS: For an infinitely long rod of uniform cross-sectional area, the temperature distribution is

θ ( x ) = θ be-mx

(1)

where

θ ( x ) = T ( x ) − T∞

θ b = T (0 ) − T∞ = Tb − T∞ .

Substituting values for T1 and T2 at their respective distances, x1 and x2, into Eq. (1), it is possible to evaluate m,

θ ( x1 ) θ be-mx1 -m x − x = = e ( 1 2) -mx θ (x2 ) θ e 2 b

(325 − 400 )$ C = e-m(0.025−0.120 )m (375-400 )$ C

m=11.56.

Using the value for m with Eq. (1) at location x1, it is now possible to determine the rod base or billet temperature,

θ ( x1 ) = T1 − T∞ = (Tb − T∞ ) e-mx

(325 − 400 )$ C= (Tb − 400 )$ C e−11.56×0.025 Tb = 300$ C. COMMENTS: Using the criteria mL ≥ 2.65 (see Example 3.8) for the infinite fin approximation, the insertion length should be ≥ 229 mm to justify the approximation,

<

PROBLEM 3.110 KNOWN: Temperature sensing probe of thermal conductivity k, length L and diameter D is mounted on a duct wall; portion of probe Li is exposed to water stream at T∞,i while other end is exposed to ambient air at T∞,o ; convection coefficients hi and ho are prescribed. FIND: (a) Expression for the measurement error, ∆Terr = Ttip − T∞,i , (b) For prescribed T∞,i and

T∞,o , calculate ∆Terr for immersion to total length ratios of 0.225, 0.425, and 0.625, (c) Compute and plot the effects of probe thermal conductivity and water velocity (hi) on ∆Terr . SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in probe, (3) Probe is thermally isolated from the duct, (4) Convection coefficients are uniform over their respective regions. PROPERTIES: Probe material (given): k = 177 W/m⋅K. ANALYSIS: (a) To derive an expression for ∆Terr = Ttip - T∞,i , we need to determine the temperature distribution in the immersed length of the probe Ti(x). Consider the probe to consist of two regions: 0 ≤ xi ≤ Li, the immersed portion, and 0 ≤ xo ≤ (L - Li), the ambient-air portion where the origin corresponds to the location of the duct wall. Use the results for the temperature distribution and fin heat rate of Case A, Table 3.4: Temperature distribution in region i:

Ti ( x i ) − T∞,i cosh ( mi ( Li − xi )) + ( hi mi k ) sinh ( Li − xi ) θi = = θ b,i To − T∞,i cosh ( mi Li ) + ( h i mi k ) sinh ( mi Li )

(1)

and the tip temperature, Ttip = Ti(Li) at xi = Li, is

Ttip − T∞,i To − T∞,i

=A=

cosh (0 ) + ( hi mi k ) sinh ( 0 )

(2)

cosh ( mi Li ) + ( h i mi k ) sinh ( mi Li )

and hence

(

∆Terr = Ttip − T∞,i = A To − T∞,i

)

(3)

<

where To is the temperature at xi = xo = 0 which at present is unknown, but can be found by setting the fin heat rates equal, that is, q f ,o

 q f ,i

(4) Continued...

PROBLEM 3.110 (Cont.)

( h o PkAc )1/ 2 θ b,o ⋅ B = − ( hi PkAc )1/ 2 θ b,i ⋅ C Solving for To, find

θ b,o θ b,i

=

To − T∞,o To − T∞,i

= − ( h i PkAc )

1/ 2

1/ 2     h C i To = T∞,o +  T  ∞,i  ho  B     

θ b,i ⋅ C

1/ 2     h C i 1 +   B   ho   

(5)

where the constants B and C are,

B=

sinh ( mo Lo ) + ( h o mo k ) cosh ( mo Lo ) cosh ( mo Lo ) + ( h o mo k ) sinh ( mo Lo )

(6)

C=

sinh ( mi Li ) + ( h i mi k ) cosh ( mi Li ) cosh ( mi Li ) + ( h i mi k ) sinh ( mi Li )

(7)

(b) To calculate the immersion error for prescribed immersion lengths, Li/L = 0.225, 0.425 and 0.625, we use Eq. (3) as well as Eqs. (2, 6, 7 and 5) for A, B, C, and To, respectively. Results of these calculations are summarized below. Li/L 0.225

Lo (mm) 155

Li (mm) 45

A 0.2328

B 0.5865

C 0.9731

To (°C) 76.7

∆Terr (°C) -0.76

0.425

115

85

0.0396

0.4639

0.992

77.5

-0.10

0.625

75

125

0.0067

0.3205

0.9999

78.2

-0.01

2.5 Temperature error, Tinfo - Ttip (C)

(c) The probe behaves as a fin having ends exposed to the cool ambient air and the hot ambient water whose temperature is to be measured. If the thermal conductivity is decreased, heat transfer along the probe length is likewise decreased, the tip temperature will be closer to the water temperature. If the velocity of the water decreases, the convection coefficient will decrease, and the difference between the tip and water temperatures will increase.

< < <

2

1.5

1

0.5

0 0.2

0.3

0.4

0.5

0.6

Immersion length ratio, Li/L Base case: k = 177 W/m.K; ho = 1100 W/m^2.K Low velocity flow: k = 177 W/m.K; ho = 500 W/m^2.K Low conductivity probe: k = 50 W/m.K; ho = 1100 W/m^2.K

0.7

PROBLEM 3.111 KNOWN: Rod protruding normally from a furnace wall covered with insulation of thickness Lins with the length Lo exposed to convection with ambient air. FIND: (a) An expression for the exposed surface temperature To as a function of the prescribed thermal and geometrical parameters. (b) Will a rod of Lo = 100 mm meet the specified operating limit, T0 ≤ 100°C? If not, what design parameters would you change? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in rod, (3) Negligible thermal contact resistance between the rod and hot furnace wall, (4) Insulated section of rod, Lins, experiences no lateral heat losses, (5) Convection coefficient uniform over the exposed portion of the rod, Lo, (6) Adiabatic tip condition for the rod and (7) Negligible radiation exchange between rod and its surroundings. ANALYSIS: (a) The rod can be modeled as a thermal network comprised of two resistances in series: the portion of the rod, Lins, covered by insulation, Rins, and the portion of the rod, Lo, experiencing convection, and behaving as a fin with an adiabatic tip condition, Rfin. For the insulated section: R ins = Lins kA c

(1)

For the fin, Table 3.4, Case B, Eq. 3.76, 1 R fin = θ b q f = ( hPkAc )1/ 2 tanh ( mLo )

(2)

m = ( hP kA c ) Ac = π D2 4 P = πD From the thermal network, by inspection, To − T∞ T − T∞ R fin To = T∞ + = w (Tw − T∞ ) R fin R ins + R fin R ins + R fin (b) Substituting numerical values into Eqs. (1) - (6) with Lo = 200 mm, 6.298 To = 25$ C + ( 200 − 25 )$ C = 109$ C 6.790 + 6.298 1/ 2

R ins =

0.200 m 60 W m ⋅ K × 4.909 × 10

R fin = 1

(

( hPkA c ) =

−4

m

2

0.0347 W 2 K 2

(15 W m

2

= 6.790 K W

)

1/ 2

A c = π ( 0.025 m )

2

(3,4,5)

(6)

< <

4 = 4.909 × 10

−4

m

2

tanh ( 6.324 × 0.200 ) = 6.298 K W

⋅ K × π ( 0.025 m ) × 60 W m ⋅ K × 4.909 × 10

−4

m

2

) = 0.0347 W

2

K

2

Continued...

PROBLEM 3.111 (Cont.) m = ( hP kA c )

1/ 2

(

= 15 W m ⋅ K × π ( 0.025 m ) 60 W m ⋅ K × 4.909 × 10 2

−4

m

2

)

1/ 2

= 6.324 m

−1

Consider the following design changes aimed at reducing To ≤ 100°C. (1) Increasing length of the fin portions: with Lo = 200 mm, the fin already behaves as an infinitely long fin. Hence, increasing Lo will not result in reducing To. (2) Decreasing the thermal conductivity: backsolving the above equation set with T0 = 100°C, find the required thermal conductivity is k = 14 W/m⋅K. Hence, we could select a stainless steel alloy; see Table A.1. (3) Increasing the insulation thickness: find that for To = 100°C, the required insulation thickness would be Lins = 211 mm. This design solution might be physically and economically unattractive. (4) A very practical solution would be to introduce thermal contact resistance between the rod base and the furnace wall by “tack welding” (rather than a continuous bead around the rod circumference) the rod in two or three places. (5) A less practical solution would be to increase the convection coefficient, since to do so, would require an air handling unit. COMMENTS: (1) Would replacing the rod by a thick-walled tube provide a practical solution? (2) The IHT Thermal Resistance Network Model and the Thermal Resistance Tool for a fin with an adiabatic tip were used to create a model of the rod. The Workspace is shown below. // Thermal Resistance Network Model: // The Network:

// Heat rates into node j,qij, through thermal resistance Rij q21 = (T2 - T1) / R21 q32 = (T3 - T2) / R32 // Nodal energy balances q1 + q21 = 0 q2 - q21 + q32 = 0 q3 - q32 = 0 /* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal points at which there is no external source of heat. */ T1 = Tw // Furnace wall temperature, C //q1 = // Heat rate, W T2 = To // To, beginning of rod exposed length q2 = 0 // Heat rate, W; node 2; no external heat source T3 = Tinf // Ambient air temperature, C //q3 = // Heat rate, W // Thermal Resistances: // Rod - conduction resistance R21 = Lins / (k * Ac) // Conduction resistance, K/W Ac = pi * D^2 / 4 // Cross sectional area of rod, m^2 // Thermal Resistance Tools - Fin with Adiabatic Tip: R32 = Rfin // Resistance of fin, K/W /* Thermal resistance of a fin of uniform cross sectional area Ac, perimeter P, length L, and thermal conductivity k with an adiabatic tip condition experiencing convection with a fluid at Tinf and coefficient h, */ Rfin = 1/ ( tanh (m*Lo) * (h * P * k * Ac ) ^ (1/2) ) // Case B, Table 3.4 m = sqrt(h*P / (k*Ac)) P = pi * D // Perimeter, m // Other Assigned Variables: Tw = 200 // Furnace wall temperature, C k = 60 // Rod thermal conductivity, W/m.K Lins = 0.200 // Insulated length, m D = 0.025 // Rod diameter, m h = 15 // Convection coefficient, W/m^2.K Tinf = 25 // Ambient air temperature,C Lo = 0.200 // Exposed length, m

PROBLEM 3.112 KNOWN: Rod (D, k, 2L) inserted into a perfectly insulating wall, exposing one-half of its length to an airstream (T∞, h). An electromagnetic field induces a uniform volumetric energy generation ( q ) in the imbedded portion. FIND: (a) Derive an expression for Tb at the base of the exposed half of the rod; the exposed region may be approximated as a very long fin; (b) Derive an expression for To at the end of the imbedded half of the rod, and (c) Using numerical values, plot the temperature distribution in the rod and describe its key features. Does the rod behave as a very long fin? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in imbedded portion of rod, (3) Imbedded portion of rod is perfectly insulated, (4) Exposed portion of rod behaves as an infinitely long fin, and (5) Constant properties. ANALYSIS: (a) Since the exposed portion of the rod (0 ≤ x ≤ + L) behaves as an infinite fin, the fin heat rate using Eq. 3.80 is q x ( 0 ) = q f = M = ( hPkAc )

1/ 2

( Tb − T∞ )

(1)

From an energy balance on the imbedded portion of the rod, q f = q Ac L

(2)

2

Combining Eqs. (1) and (2), with P = πD and Ac = πD /4, find −1/ 2

−1/ 2

1/ 2  Tb = T∞ + q f ( hPkAc ) (3) = T∞ + qA c L ( hPk ) (b) The imbedded portion of the rod (-L ≤ x ≤ 0) experiences one-dimensional heat transfer with uniform q . From Eq. 3.43,

To =

2  qL

+ Tb 2k (c) The temperature distribution T(x) for the rod is piecewise parabolic and exponential, T ( x ) − Tb = T ( x ) − T∞ Tb − T∞

2  qL x

2

 

2k  L 

= exp ( −mx )

−L ≤ x ≤ 0 0 ≤ x ≤ +L Continued …..

< <

PROBLEM 3.112 (Cont.) The gradient at x = 0 will be continuous since we used this condition in evaluating Tb. The distribution is shown below with To = 105.4°C and Tb = 55.4°C. T(x) over embedded and exposed portions of rod 120

Temperature, T(x)

100

80

60

40

20 -50 -40 -30 -20 -10

0

10

Radial position, x

20

30

40

50

PROBLEM 3.113 KNOWN: Very long rod (D, k) subjected to induction heating experiences uniform volumetric generation ( q ) over the center, 30-mm long portion. The unheated portions experience convection (T∞, h). FIND: Calculate the temperature of the rod at the mid-point of the heated portion within the coil, To, and at the edge of the heated portion, Tb. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with uniform q in portion of rod within the coil; no convection from lateral surface of rod, (3) Exposed portions of rod behave as infinitely long fins, and (4) Constant properties. ANALYSIS: The portion of the rod within the coil, 0 ≤ x ≤ + L, experiences one-dimensional conduction with uniform generation. From Eq. 3.43,

 2 qL To = + Tb 2k

(1)

The portion of the rod beyond the coil, L ≤ x ≤ ∞, behaves as an infinitely long fin for which the heat rate from Eq. 3.80 is

q f = q x ( L ) = ( hPkAc )

1/ 2

(Tb − T∞ )

(2)

2

where P = πD and Ac = πD /4. From an overall energy balance on the imbedded portion of the rod as illustrated in the schematic above, find the heat rate as

E in − E out + E gen = 0  cL = 0 −q f + qA  cL q f = qA

(3)

Combining Eqs. (1-3), −1/ 2

2  1/ Tb = T∞ + qA c L ( hPk )

(4)

 2 qL −1/ 2 2  1/ To = T∞ + + qA c L ( hPk ) 2k

(5)

and substituting numerical values find

To = 305°C

Tb = 272°C

<

PROBLEM 3.114 KNOWN: Dimensions, end temperatures and volumetric heating of wire leads. Convection coefficient and ambient temperature. FIND: (a) Equation governing temperature distribution in the leads, (b) Form of the temperature distribution. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in x, (3) Uniform volumetric heating, (4) Uniform h (both sides), (5) Negligible radiation. ANALYSIS: (a) Performing an energy balance for the differential control volume,

E in − E out + E g = 0

dT  dT d  dT    c dx = 0 −  − kAc −  kAc  dx − hPdx (T − T∞ ) + qA dx  dx dx  dx  

− kAc d 2T dx 2

 q x − q x + dx − dq conv + qdV =0

−

hP q (T − T∞ ) + = 0 kAc k

<

 c /hP ) and m 2 ≡ hP/kAc , the (b) With a reduced temperature defined as Θ ≡ T − T∞ − ( qA differential equation may be rendered homogeneous, with a general solution and boundary conditionsas shown d 2Θ Θ ( x ) = C1emx + C2e−mx − m 2Θ = 0 2

dx

Θb = C1 + C2

Θc = C1emL + C2e−mL

it follows that

C1 =

Θb e−mL − Θc e− mL − emL

C2 =

Θc − Θb emL e− mL − emL

Θbe− mL − Θc ) e mx + ( Θc − Θbe mL ) e− mx ( Θ (x ) = e− mL − emL

COMMENTS: If q is large and h is small, temperatures within the lead may readily exceed the prescribed boundary temperatures.

<

PROBLEM 3.115 KNOWN: Disk-shaped electronic device (D, Ld, kd) dissipates electrical power (Pe) at one of its surfaces. Device is bonded to a cooled base (To) using a thermal pad (Lp, kA). Long fin (D, kf) is bonded to the heat-generating surface using an identical thermal pad. Fin is cooled by convection (T∞, h). FIND: (a) Construct a thermal circuit of the system, (b) Derive an expression for the temperature of the heat-generating device, Td, in terms of circuit thermal resistance, To and T∞; write expressions for the thermal resistances; and (c) Calculate Td for the prescribed conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through thermal pads and device; no losses from lateral surfaces; (3) Fin is infinitely long, (4) Negligible contact resistance between components of the system, and (5) Constant properties. ANALYSIS: (a) The thermal circuit is shown below with thermal resistances associated with conduction (pads, Rp; device, Rd) and for the long fin, Rf.

(b) To obtain an expression for Td, perform an energy balance about the d-node

E in − E out = q a + q b + Pe = 0

(1)

Using the conduction rate equation with the circuit

qa =

To − Td Rf + Rd

T −T qb = ∞ d R p + Rf

Combine with Eq. (1), and solve for Td,

Td =

(

)

(

(2,3)

Pe + To / R p + R d + T∞ / R p + R f

(

)

(

1/ R p + R d + 1/ R p + R f

)

)

(4) 2

where the thermal resistances with P = πD and Ac = πD /4 are

−1/ 2 R f = ( hPk f Ac ) R d = Ld / k d A c (c) Substituting numerical values with the foregoing relations, find

R p = Lp / k p Ac

R p = 1.061 K / W

R d = 4.244 K / W

(5,6,7)

R f = 5.712 K / W

and the device temperature as

Td = 62.4°C

<

COMMENTS: What fraction of the power dissipated in the device is removed by the fin? Answer: qb/Pe = 47%.

PROBLEM 3.116 KNOWN: Dimensions and thermal conductivity of a gas turbine blade. Temperature and convection coefficient of gas stream. Temperature of blade base and maximum allowable blade temperature. FIND: (a) Whether blade operating conditions are acceptable, (b) Heat transfer to blade coolant. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction in blade, (2) Constant k, (3) Adiabatic blade tip, (4) Negligible radiation. ANALYSIS: Conditions in the blade are determined by Case B of Table 3.4. (a) With the maximum temperature existing at x = L, Eq. 3.75 yields

T ( L ) − T∞ Tb − T∞

=

1 cosh mL

m = ( hP/kA c )

1/ 2

(

= 250W/m 2 ⋅ K × 0.11m/20W/m ⋅ K × 6 ×10−4 m 2

m = 47.87 m-1 and

)

1/ 2

mL = 47.87 m-1 × 0.05 m = 2.39

From Table B.1, cosh mL = 5.51. Hence,

T ( L ) = 1200$ C + (300 − 1200)$ C/5.51 = 1037 $C

<

and the operating conditions are acceptable.

(

(b) With M = ( hPkA c )1/ 2 Θ b = 250W/m 2 ⋅ K × 0.11m × 20W/m ⋅ K × 6 × 10 −4 m 2

) (−900 C ) = −517W , 1/ 2

$

Eq. 3.76 and Table B.1 yield

qf = M tanh mL = −517W ( 0.983) = −508W Hence, q b = −q f = 508W

<

COMMENTS: Radiation losses from the blade surface and convection from the tip will contribute to reducing the blade temperatures.

PROBLEM 3.117 KNOWN: Dimensions of disc/shaft assembly. Applied angular velocity, force, and torque. Thermal conductivity and inner temperature of disc. FIND: (a) Expression for the friction coefficient µ, (b) Radial temperature distribution in disc, (c) Value of µ for prescribed conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant k, (4) Uniform disc contact pressure p, (5) All frictional heat dissipation is transferred to shaft from base of disc. ANALYSIS: (a) The normal force acting on a differential ring extending from r to r+dr on the contact surface of the disc may be expressed as dFn = p2π rdr . Hence, the tangential force is dFt = µ p2π rdr , in which case the torque may be expressed as dτ = 2πµ pr 2dr For the entire disc, it follows that r 2π µ pr23 τ = 2πµ p 2 r 2dr = o 3

∫

where p = F π r22 . Hence,

µ=

3 τ 2 Fr2

<

(b) Performing an energy balance on a differential control volume in the disc, it follows that

qcond,r + dq fric − q cond,r + dr = 0

(

)

(

)

With dq fric = ω dτ = 2 µ Fω r 2 r22 dr , q cond,r + dr = q cond,r + dq cond,r dr dr , and

qcond,r = − k ( 2π rt ) dT dr , it follows that

d ( rdT dr ) 2 µ Fω r 2 r22 dr + 2π kt dr = 0 dr

(

or

d ( rdT dr ) dr

Integrating twice,

)

=−

µ Fω 2 r π ktr22

Continued...

PROBLEM 3.117 (Cont.) dT µ Fω 2 C1 r + =− dr r 3π ktr22 T=−

µ Fω 9π ktr22

r3 + C1nr + C2

Since the disc is well insulated at r = r2 , dT dr

C1 =

r2

= 0 and

µ Fω r2 3π kt

With T ( r1 ) = T1 , it also follows that

C2 = T1 +

µ Fω

r13 − C1nr1 2 9π ktr2

Hence,

T ( r ) = T1 −

µ Fω 9π ktr22

(r3 − r13 ) + µ3Fπωktr2 n rr1

<

(c) For the prescribed conditions,

µ=

3 8N ⋅ m = 0.333 2 200N ( 0.18m )

<

Since the maximum temperature occurs at r = r2,

µ Fω r2 Tmax = T ( r2 ) = T1 − 9π kt

  3  1 −  r1   + µ Fω r2 n  r2    r2   3π kt  r1   

With ( µ Fω r2 3π kt ) = ( 0.333 × 200N × 40rad/s × 0.18m 3π × 15W/m ⋅ K × 0.012m ) = 282.7$ C , 3 282.7$ C   0.02    0.18  $ 1 −  Tmax = 80 C −   + 282.7 Cn   3   0.18    0.02  $

Tmax = 80$ C − 94.1$ C + 621.1$ C = 607$ C COMMENTS: The maximum temperature is excessive, and the disks should be actively cooled (by convection) at their outer surfaces.

<

PROBLEM 3.118 KNOWN: Extended surface of rectangular cross-section with heat flow in the longitudinal direction. FIND: Determine the conditions for which the transverse (y-direction) temperature gradient is negligible compared to the longitudinal gradient, such that the 1-D analysis of Section 3.6.1 is valid by finding: (a) An expression for the conduction heat flux at the surface, q′′y ( t ) , in terms of Ts and To, assuming the transverse temperature distribution is parabolic, (b) An expression for the convection heat flux at the surface for the x-location; equate the two expressions, and identify the parameter that determines the ratio (To – Ts)/(Ts - T∞); and (c) Developing a criterion for establishing the validity of the 1-D assumption used to model an extended surface. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform convection coefficient and (3) Constant properties. ANALYSIS: (a) Referring to the schematics above, the conduction heat flux at the surface y = t at any x-location follows from Fourier’s law using the parabolic transverse temperature distribution.

q′′y ( t ) = − k

 2y  2k ∂T  Ts ( x ) − To ( x ) = − k   Ts ( x ) − To ( x )  =−  2 t  ∂y  y = t t y=t 

(1)

(b) The convection heat flux at the surface of any x-location follows from the rate equation

q′′cv = h  Ts ( x ) − T∞ 

(2)

Performing a surface energy balance as represented schematically above, equating Eqs. (1) and (2) provides

q′′y ( t ) = q′′cv

2k Ts ( x ) − To ( x ) = h Ts ( x ) − T∞  t  Ts ( x ) − To ( x ) ht = −0.5 = −0.5 Bi Ts ( x ) − T∞ ( x ) k

−

(3)

where Bi = ht/k, the Biot number, represents the ratio of the convection to the conduction thermal resistances,

Bi =

R ′′cd t / k = R ′′cv 1/ h

(4)

(c) The transverse gradient (heat flow) will be negligible compared to the longitudinal gradient when Bi > t). ANALYSIS: (a) The fin heat transfer rate for Cases A, B and D are given by Eqs. (3.72), (3.76) and 2

1/2

2

1/2

(3.80), where M ≈ (2 hw tk) (Tb - T∞) = (2 × 100 W/m ⋅K × 0.001m × 180 W/m⋅K) (75°C) w = 1/2 2 1/2 -1 -1 450 w W, m≈ (2h/kt) = (200 W/m ⋅K/180 W/m⋅K ×0.001m) = 33.3m , mL ≈ 33.3m × 0.010m 2 -1 = 0.333, and (h/mk) ≈ (100 W/m ⋅K/33.3m × 180 W/m⋅K) = 0.0167. From Table B-1, it follows that sinh mL ≈ 0.340, cosh mL ≈ 1.057, and tanh mL ≈ 0.321. From knowledge of qf, Eqs. (3.86), (3.81) and (3.83) yield ηf ≈

q ′f

h ( 2L + t )θ b

, εf

≈

q ′f

θ , R ′t,f = b ht θ b q ′f

Case A: From Eq. (3.72), (3.86), (3.81), (3.83) and (3.70), q ′f =

ηf =

εf =

M sinh mL + ( h / mk ) cosh mL

w cosh mL + ( h / mk ) sinh mL 151 W / m

100 W / m ⋅ K ( 0.021m ) 75°C 2

151 W / m 100 W / m ⋅ K ( 0.001m ) 75°C 2

T ( L ) = T∞ +

= 450 W / m

0.340 + 0.0167 × 1.057 1.057 + 0.0167 × 0.340

= 151 W / m

<

= 0.96

= 20.1, R ′t,f =

θb

cosh mL + ( h / mk ) sinh mL

<

75°C 151 W / m

= 25°C +

<

= 0.50 m ⋅ K / W

75°C 1.057 + ( 0.0167 ) 0.340

= 95.6°C

<

Case B: From Eqs. (3.76), (3.86), (3.81), (3.83) and (3.75) q ′f =

M w

tanh mL = 450 W / m ( 0.321) = 144 W / m

< < <

ηf = 0.92, ε f = 19.2, R ′t,f = 0.52 m ⋅ K / W T ( L ) = T∞ +

θb cosh mL

= 25°C +

75°C 1.057

= 96.0°C

Continued …..

PROBLEM 3.121 (Cont.) Case D (L → ∞): From Eqs. (3.80), (3.86), (3.81), (3.83) and (3.79) q ′f =

M w

<

= 450 W / m

ηf = 0, ε f = 60.0, R ′t,f = 0.167 m ⋅ K / W, T ( L ) = T∞ = 25°C

<

(b) The effect of h on the heat rate is shown below for the aluminum and stainless steel fins. Va ria tio n o f q f' w ith h (k= 1 8 0 W /m .K )

H e a t ra te , q f'(W /m )

1500

1000

500

0 0

200

400

600

800

1000

C o n ve ctio n co e fficie n t, h (W /m ^2 .K ) q fA' q fB ' q fD '

Va ria tio n o f q f' w ith h (k = 1 5 W /m .K )

H e a t ra te , q f'(W /m )

400 300 200 100 0 0

200

400

600

800

1000

C o n ve c tio n c o e ffic ie n t, h (W /m ^2 .K ) q fA' q fB ' q fD '

For both materials, there is little difference between the Case A and B results over the entire range of h. The difference (percentage) increases with decreasing h and increasing k, but even for the worst 2 case condition (h = 10 W/m ⋅K, k = 180 W/m⋅K), the heat rate for Case A (15.7 W/m) is only slightly larger than that for Case B (14.9 W/m). For aluminum, the heat rate is significantly over-predicted by the infinite fin approximation over the entire range of h. For stainless steel, it is over-predicted for 2 small values of h, but results for all three cases are within 1% for h > 500 W/m ⋅K. COMMENTS: From the results of Part (a), we see there is a slight reduction in performance (smaller values of q ′f , ηf and ε f , as well as a larger value of R ′t,f ) associated with insulating the tip. Although ηf = 0 for the infinite fin, q′f and εf are substantially larger than results for L = 10 mm, indicating that performance may be significantly improved by increasing L.

PROBLEM 3.122 KNOWN: Thickness, length, thermal conductivity, and base temperature of a rectangular fin. Fluid temperature and convection coefficient. FIND: (a) Heat rate per unit width, efficiency, effectiveness, thermal resistance, and tip temperature for different tip conditions, (b) Effect of fin length and thermal conductivity on the heat rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction along fin, (3) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient, (6) Fin width is much longer than thickness (w >> t). ANALYSIS: (a) The fin heat transfer rate for Cases A, B and D are given by Eqs. (3.72), (3.76) and 2

1/2

2

1/2

(3.80), where M ≈ (2 hw tk) (Tb - T∞) = (2 × 100 W/m ⋅K × 0.001m × 180 W/m⋅K) (75°C) w = 1/2 2 1/2 -1 -1 450 w W, m≈ (2h/kt) = (200 W/m ⋅K/180 W/m⋅K ×0.001m) = 33.3m , mL ≈ 33.3m × 0.010m 2 -1 = 0.333, and (h/mk) ≈ (100 W/m ⋅K/33.3m × 180 W/m⋅K) = 0.0167. From Table B-1, it follows that sinh mL ≈ 0.340, cosh mL ≈ 1.057, and tanh mL ≈ 0.321. From knowledge of qf, Eqs. (3.86), (3.81) and (3.83) yield ηf ≈

q ′f

h ( 2L + t )θ b

, εf ≈

q ′f

θ , R ′t,f = b ht θ b q ′f

Case A: From Eq. (3.72), (3.86), (3.81), (3.83) and (3.70), q ′f =

ηf =

εf =

M sinh mL + ( h / mk ) cosh mL

w cosh mL + ( h / mk ) sinh mL 151 W / m

100 W / m ⋅ K ( 0.021m ) 75°C 2

151 W / m 100 W / m ⋅ K ( 0.001m ) 75°C 2

T ( L ) = T∞ +

= 450 W / m

0.340 + 0.0167 × 1.057 1.057 + 0.0167 × 0.340

= 151 W / m

<

= 0.96

= 20.1, R ′t,f =

θb

cosh mL + ( h / mk ) sinh mL

<

75°C 151 W / m

= 25°C +

<

= 0.50 m ⋅ K / W

75°C 1.057 + ( 0.0167 ) 0.340

= 95.6°C

<

Case B: From Eqs. (3.76), (3.86), (3.81), (3.83) and (3.75) q ′f =

M w

tanh mL = 450 W / m ( 0.321) = 144 W / m

< < <

ηf = 0.92, ε f = 19.2, R ′t,f = 0.52 m ⋅ K / W T ( L ) = T∞ +

θb cosh mL

= 25°C +

75°C 1.057

= 96.0°C

Continued …..

PROBLEM 3.122 (Cont.) Case D (L → ∞): From Eqs. (3.80), (3.86), (3.81), (3.83) and (3.79) q ′f =

M w

<

= 450 W / m

ηf = 0, ε f = 60.0, R ′t,f = 0.167 m ⋅ K / W, T ( L ) = T∞ = 25°C

<

(b) The effect of L on the heat rate is shown below for the aluminum and stainless steel fins. Va ria tio n o f q f' w ith L (k= 1 8 0 W /m .K )

H e a t ra te , q f'(W /m )

500 400 300 200 100 0 0

0 .0 1

0 .0 2

0 .0 3

0 .0 4

0 .0 5

Fin le n g th , L (m ) q fA' q fB ' q fD '

Va ria tio n o f q f' w ith L (k= 1 5 W /m .K )

H e a t ra te , q f'(W /m )

150 120 90 60 30 0 0

0 .0 1

0 .0 2

0 .0 3

0 .0 4

0 .0 5

Fin le n g th , L (m ) q fA' q fB ' q fD '

For both materials, differences between the Case A and B results diminish with increasing L and are within 1% of each other at L ≈ 27 mm and L ≈ 13 mm for the aluminum and steel, respectively. At L = 3 mm, results differ by 14% and 13% for the aluminum and steel, respectively. The Case A and B results approach those of the infinite fin approximation more quickly for stainless steel due to the larger temperature gradients, |dT/dx|, for the smaller value of k. COMMENTS: From the results of Part (a), we see there is a slight reduction in performance (smaller values of q ′f , ηf and ε f , as well as a larger value of R ′t,f ) associated with insulating the tip. Although ηf = 0 for the infinite fin, q′f and εf are substantially larger than results for L = 10 mm, indicating that performance may be significantly improved by increasing L.

PROBLEM 3.123 KNOWN: Length, thickness and temperature of straight fins of rectangular, triangular and parabolic profiles. Ambient air temperature and convection coefficient. FIND: Heat rate per unit width, efficiency and volume of each fin. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient. ANALYSIS: For each fin,

q′f = q′max = ηf hA′f θ b ,

V′ = A p 1/2

2

1/2

-1

where ηf depends on the value of m = (2h/kt) = (100 W/m ⋅K/185 W/m⋅K × 0.003m) = 13.4m -1 and the product mL = 13.4m × 0.015m = 0.201 or mLc = 0.222. Expressions for ηf, A′f and Ap are obtained from Table 3-5. Rectangular Fin:

ηf =

tanh mLc 0.218 = = 0.982, A′f = 2 Lc = 0.033m mLc 0.222

<

)

(

q′ = 0.982 50 W / m 2 ⋅ K 0.033m (80°C ) = 129.6 W / m, V′ = tL = 4.5 × 10−5 m 2

<

Triangular Fin: ηf =

1 I1 ( 2mL )

mL I0 ( 2 mL )

(

=

0.205

( 0.201)1.042

1/ 2 2

= 0.978, A ′f = 2  L + ( t / 2 )



2



)

q ′ = 0.978 50 W / m ⋅ K 0.030m (80°C ) = 117.3 W / m, V ′ = ( t / 2 ) L = 2.25 × 10 2

<

= 0.030m −5

m

2

<

Parabolic Fin:

ηf =

1/ 2

 4 ( mL )2 + 1  

(

( )

= 0.963, A′f =  C1L + L2 / t ln ( t / L + C1 ) = 0.030m

2 +1

)





q′f = 0.963 50 W / m 2 ⋅ K 0.030m (80°C ) = 115.6 W / m, V′ = ( t / 3) L = 1.5 × 10−5 m 2

< <

COMMENTS: Although the heat rate is slightly larger (~10%) for the rectangular fin than for the triangular or parabolic fins, the heat rate per unit volume (or mass) is larger and largest for the triangular and parabolic fins, respectively.

PROBLEM 3.124 KNOWN: Melting point of solder used to join two long copper rods. FIND: Minimum power needed to solder the rods. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction along the rods, (3) Constant properties, (4) No internal heat generation, (5) Negligible radiation exchange with surroundings, (6) Uniform h, and (7) Infinitely long rods. PROPERTIES: Table A-1: Copper T = ( 650 + 25 ) C ≈ 600K: k = 379 W/m ⋅ K. $

ANALYSIS: The junction must be maintained at 650°C while energy is transferred by conduction from the junction (along both rods). The minimum power is twice the fin heat rate for an infinitely long fin, q min = 2q f = 2 ( hPkAc )

1/ 2

(Tb − T∞ ).

Substituting numerical values,  W q min = 2 10 (π × 0.01m )  m2 ⋅ K

1/ 2

W  π  2 379 m ⋅ K  4 ( 0.01m )  

(650 − 25)$ C.

Therefore, q min = 120.9 W. COMMENTS: Radiation losses from the rods may be significant, particularly near the junction, thereby requiring a larger power input to maintain the junction at 650°C.

<

PROBLEM 3.125 KNOWN: Dimensions and end temperatures of pin fins. FIND: (a) Heat transfer by convection from a single fin and (b) Total heat transfer from a 1 2 m surface with fins mounted on 4mm centers. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction along rod, (3) Constant properties, (4) No internal heat generation, (5) Negligible radiation. PROPERTIES: Table A-1, Copper, pure (323K): k ≈ 400 W/m⋅K. ANALYSIS: (a) By applying conservation of energy to the fin, it follows that q conv = qcond,i − q cond,o where the conduction rates may be evaluated from knowledge of the temperature distribution. The general solution for the temperature distribution is

θ ( x ) = C1 emx + C2 e-mx

θ ≡ T − T∞ .

The boundary conditions are θ(0) ≡ θo = 100°C and θ(L) = 0. Hence

θ o = C1 + C2 0 = C1 emL + C2 e-mL C2 = C1 e2mL

Therefore, C1 =

θ o e2mL

θo

, C2 = − 1 − e 2mL 1 − e 2mL and the temperature distribution has the form

θ=

θo

emx − e2mL-mx  . 

 1 − e2mL 

The conduction heat rate can be evaluated by Fourier’s law, qcond = − kAc

kAcθ o dθ m  emx + e 2mL-mx  =−   dx 1 − e 2mL 

or, with m = ( hP/kA c )

1/ 2

,

θ o ( hPkAc )

1/ 2

q cond = −

1 − e2mL

 emx + e2mL-mx  .   Continued …..

PROBLEM 3.125 (Cont.) Hence at x = 0,

θ o ( hPkAc )

1/ 2

q cond,i = −

1 − e2mL

(1 + e2mL )

at x = L

θ ( hPkA c ) q cond,o = − o 1 − e2mL Evaluating the fin parameters:

1/ 2

1/ 2

 hP  m=   kA c 

1/ 2

 4h  =   kD  1/ 2

( hPkAc )

1/ 2

π 2 3  D hk  =  4 

(2emL ) 1/ 2

 4 × 100 W/m 2 ⋅ K  =   400 W/m ⋅ K × 0.001m 

1/ 2

π 2 W W  3 = × ( 0.001m ) × 100 × 400  m ⋅ K   4 m2 ⋅ K

mL = 31.62 m-1 × 0.025m = 0.791, The conduction heat rates are q cond,i = q cond,o =

(

−100K 9.93 × 10-3 W/K

(

= 31.62 m-1

−3.865

emL = 2.204,

W K

e2mL = 4.865

) × 5.865 = 1.507 W

−100K 9.93 × 10-3 W/K

-3.865 and from the conservation relation,

= 9.93 × 10−3

) × 4.408 = 1.133 W <

q conv = 1.507 W − 1.133 W = 0.374 W.

(b) The total heat transfer rate is the heat transfer from N = 250×250 = 62,500 rods and the 2 heat transfer from the remaining (bare) surface (A = 1m - NAc). Hence,

(

)

q = N q cond,i + hAθ o = 62,500 (1.507 W ) + 100W/m 2 ⋅ K 0.951 m 2 100K q = 9.42 × 104 W+0.95 × 104 W=1.037 ×105 W. COMMENTS: (1) The fins, which cover only 5% of the surface area, provide for more than 90% of the heat transfer from the surface. (2) The fin effectiveness, ε ≡ q cond,i / hA cθ o , is ε = 192, and the fin efficiency, η ≡ ( q conv / hπ DLθ o ) , is η = 0.48.

(3) The temperature distribution, θ(x)/θo, and the conduction term, qcond,i, could have been obtained directly from Eqs. 3.77and 3.78, respectively. (4) Heat transfer by convection from a single fin could also have been obtained from Eq. 3.73.

PROBLEM 3.126 KNOWN: Pin fin of thermal conductivity k, length L and diameter D connecting two devices (Lg,kg) experiencing volumetric generation of thermal energy (q ). Convection conditions are prescribed (T∞, h).

FIND: Expression for the device surface temperature Tb in terms of device, convection and fin parameters. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Pin fin is of uniform cross-section with constant h, (3) Exposed surface of device is at a uniform temperature Tb, (4) Backside of device is insulated, (5) Device experiences 1-D heat conduction with uniform volumetric generation, (6) Constant properties, and (7) No contact resistance between fin and devices. ANALYSIS: Recognizing symmetry, the pin fin is modeled as a fin of length L/2 with insulated tip. Perform a surface energy balance, E in − E out = 0 q d − qs − q f = 0

(1)

The heat rate qd can be found from an energy balance on the entire device to find E in − E out + E g = 0  −q d + qV =0  q d = qA g Lg

(2)

The fin heat rate, qf, follows from Case B, q f = M tanh mL/2 = ( hPkAc )

1/ 2

)

(

Table 3.4

(Tb − T∞ ) tanh ( mL/2 ) ,

P/A c = π D/ π D 2 / 4 = 4 / D

and

m = ( hP/kAc )

1/ 2

PA c = π 2D3 / 4.

(3,4) (5,6)

Hence, the heat rate expression can be written as

(

( (

)

 g Lg = h Ag − Ac (Tb − T∞ ) + hk π 2 D3 / 4 qA

))

1/ 2

  4h 1/ 2 L  tanh   ⋅  ( Tb − T∞ )   kD  2  

(7)

Solve now for Tb,    4h 1/ 2 L   1/ 2 2 3   g Lg / h Ag − Ac + hk π D / 4 Tb = T∞ + qA tanh   ⋅    kD  2     

(

)

( (

))

(8) <

PROBLEM 3.127 KNOWN: Positions of equal temperature on two long rods of the same diameter, but different thermal conductivity, which are exposed to the same base temperature and ambient air conditions. FIND: Thermal conductivity of rod B, kB. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Rods are infinitely long fins of uniform cross-sectional area, (3) Uniform heat transfer coefficient, (4) Constant properties. ANALYSIS: The temperature distribution for the infinite fin has the form T ( x ) − T∞ θ = = e-mx To − T∞ θb

1/ 2

 hP  m=   kAc 

.

(1,2)

For the two positions prescribed, xA and xB, it was observed that TA ( x A ) = TB ( x B )

or

θ A ( x A ) = θ B ( x B ).

(3)

Since θb is identical for both rods, Eq. (1) with the equality of Eq. (3) requires that m A x A = m Bx B Substituting for m from Eq. (2) gives 1/ 2

 hP     k A Ac 

1/ 2

 hP  xA =    k BAc 

x B.

Recognizing that h, P and Ac are identical for each rod and rearranging, 2

x  kB =  B  kA  xA  2

 0.075m  kB =  × 70 W/m ⋅ K = 17.5 W/m ⋅ K.  0.15m  COMMENTS: This approach has been used as a method for determining the thermal conductivity. It has the attractive feature of not requiring power or temperature measurements, assuming of course, a reference material of known thermal conductivity is available.

<

PROBLEM 3.128 KNOWN: Slender rod of length L with ends maintained at To while exposed to convection cooling (T∞ < To, h). FIND: Temperature distribution for three cases, when rod has thermal conductivity (a) kA, (b) kB < kA, and (c) kA for 0 ≤ x ≤ L/2 and kB for L/2 ≤ x ≤ L. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, and (4) Negligible thermal resistance between the two materials (A, B) at the midspan for case (c). ANALYSIS: (a, b) The effect of thermal conductivity on the temperature distribution when all other conditions (To, h, L) remain the same is to reduce the minimum temperature with decreasing thermal conductivity. Hence, as shown in the sketch, the mid-span temperatures are TB (0.5L) < TA (0.5L) for kB < kA. The temperature distribution is, of course, symmetrical about the mid-span. (c) For the composite rod, the temperature distribution can be reasoned by considering the boundary condition at the mid-span. q′′x,A ( 0.5L ) = q′′x,B ( 0.5L )

−k A

dT 

dT  = −k B   dx A,x=0.5L dx B,x=0.5L

Since kA > kB, it follows that  dT   dT  0.5L, and the temperature distribution is not symmetrical about the mid-span. COMMENTS: (1) Recognize that the area under the curve on the T-x coordinates is proportional to the fin heat rate. What conclusions can you draw regarding the relative magnitudes of qfin for cases (a), (b) and (c)? (2) If L is increased substantially, how would the temperature distribution be affected?

PROBLEM 3.129 KNOWN: Base temperature, ambient fluid conditions, and temperatures at a prescribed distance from the base for two long rods, with one of known thermal conductivity. FIND: Thermal conductivity of other rod. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction along rods, (3) Constant properties, (4) Negligible radiation, (5) Negligible contact resistance at base, (6) Infinitely long rods, (7) Rods are identical except for their thermal conductivity. ANALYSIS: With the assumption of infinitely long rods, the temperature distribution is T − T∞ θ = = e-mx θ b Tb − T∞ or 1/ 2

T − T∞  hP  ln = − mx =   Tb − T∞  kA 

x

Hence, for the two rods, T − T  ln  A ∞  1/ 2  Tb − T∞  =  k B     TB − T∞   k A  ln    Tb − T∞   T − T∞  ln  A ln Tb − T∞  1/ 2  1/2 1/2 kB = kA = ( 200 ) T − T  ln ln  B ∞  T T −  b ∞ k B = 56.6 W/m ⋅ K.

75 − 25 100 − 25 = 7.524 60 − 25 100 − 25

<

COMMENTS: Providing conditions for the two rods may be maintained nearly identical, the above method provides a convenient means of measuring the thermal conductivity of solids.

PROBLEM 3.130 KNOWN: Arrangement of fins between parallel plates. Temperature and convection coefficient of air flow in finned passages. Maximum allowable plate temperatures. FIND: (a) Expressions relating fin heat transfer rates to end temperatures, (b) Maximum power dissipation for each plate.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fins, (3) Constant properties, (4) Negligible radiation, (5) All of the heat is dissipated to the air, (6) Uniform h, (7) Negligible variation in T∞, (8) Negligible contact resistance. PROPERTIES: Table A.1, Aluminum (pure), 375 K: k = 240 W/m⋅K. ANALYSIS: (a) The general solution for the temperature distribution in fin is θ ( x ) ≡ T ( x ) − T∞ = C1emx + C 2e-mx

θ (0 ) = θ o = To − T∞ ,

Boundary conditions: Hence

θ o = C1 + C2

θ ( L ) = θ L = TL − T∞ .

θ L = C1emL + C2e-mL

θ L = C1e mL + (θ o − C1 ) e-mL C1 =

Hence

θ L − θ oe-mL emL − e-mL

θ (x ) =

C2 = θ o −

θ L − θ o e-mL emL − e-mL

=

θ o emL − θ L emL − e-mL

.

m x-L m L-x θ Lemx − θ oe ( ) + θ oe ( ) − θ Le-mx

e mL − e-mL

(

m L-x -m L-x θ o e ( ) − e ( )  + θ L emx − e-mx  θ (x ) =  mL -mL e −e

)

θ sinh m (L-x ) + θ Lsinh mx . θ (x ) = o sinh mL The fin heat transfer rate is then

q f = − kAc Hence

θ m dT  θ m  cosh m ( L − x ) + L cosh mx  . = − kDt  − o dx sinh mL  sinh mL  θ m   θo m − L   tanh mL sinh mL 

<

θ m   θom − L .  sinh mL tanh mL 

<

q f,o = kDt 

q f,L = kDt 

Continued …..

PROBLEM 3.130 (Cont.) 1/ 2

(b)

 hP  m=    kAc 

1/ 2

 50 W/m 2 ⋅ K ( 2 × 0.1 m+2 × 0.001 m )   =  240 W/m ⋅ K × 0.1 m × 0.001 m   

= 35.5 m-1

mL = 35.5 m-1 × 0.012 m = 0.43 sinh mL = 0.439

tanh mL = 0.401

θ o = 100 K

θ L = 50 K

 100 K × 35.5 m-1 50 K × 35.5 m-1   q f,o = 240 W/m ⋅ K × 0.1 m × 0.001 m  −   0.401 0.439   q f,o = 115.4 W

(from the top plate)

 100 K × 35.5 m-1 50 K × 35.5 m-1   = ⋅ × × − q f,L 240 W/m K 0.1 m 0.001 m    0.439 0.401   q f,L = 87.8 W.

(into the bottom plate)

Maximum power dissipations are therefore q o,max = Nf q f,o + ( W − Nf t ) Dhθ o q o,max = 50 × 115.4 W+ ( 0.200 − 50 × 0.001) m × 0.1 m ×150 W/m 2 ⋅ K ×100 K

<

q o,max = 5770 W+225 W = 5995 W q L,max = − N f q f,L + ( W − Nf t ) Dhθ o q L,max = −50 × 87.8W + ( 0.200 − 50 × 0.001) m × 0.1 m ×150 W/m 2 ⋅ K × 50 K

<

q L,max = −4390 W+112W = −4278 W.

COMMENTS: (1) It is of interest to determine the air velocity needed to prevent excessive heating of the air as it passes between the plates. If the air temperature change is restricted to

 air = m

∆T∞ = 5 K, its flowrate must be

q tot 1717 W = = 0.34 kg/s. cp ∆T∞ 1007 J/kg ⋅ K × 5 K

Its mean velocity is then  air m 0.34 kg/s Vair = = = 163 m/s. 3 ρair Ac 1.16 kg/m × 0.012 m (0.2 − 50 × 0.001) m Such a velocity would be impossible to maintain. To reduce it to a reasonable value, e.g. 10 m/s, Ac would have to be increased substantially by increasing W (and hence the space between fins) and by increasing L. The present configuration is impractical from the standpoint that 1717 W could not be transferred to air in such a small volume. (2) A negative value of qL,max implies that heat must be transferred from the bottom plate to the air to maintain the plate at 350 K.

PROBLEM 3.131 KNOWN: Conditions associated with an array of straight rectangular fins. FIND: Thermal resistance of the array. SCHEMATIC:

ASSUMPTIONS: (1) Constant properties, (2) Uniform convection coefficient, (3) Symmetry about midplane. ANALYSIS: (a) Considering a one-half section of the array, the corresponding resistance is −1

R t,o = (ηo hA t )

where A t = NA f + A b . With S = 4 mm and t = 1 mm, it follows that N = W1 /S = 250, 2(L/2)W2 = 0.008 m2, Ab = W2(W1 - Nt) = 0.75 m2, and At = 2.75 m2. The overall surface efficiency is ηo = 1 −

NA f At

Af =

(1 − ηf )

where the fin efficiency is

ηf =

tanh m ( L 2 ) m (L 2)

1/ 2

and

 hP    kA c 

m=

 h ( 2t + 2W2 )   ktW2  

1/ 2

=

1/ 2

 2h    kt 

≈

= 38.7m −1

With m(L/2) = 0.155, it follows that ηf = 0.992 and ηo = 0.994. Hence

(

R t,o = 0.994 × 150W/m 2 ⋅ K × 2.75m 2

)

−1

= 2.44 × 10 −3 K/W

<

(b) The requirements that t ≥ 0.5 m and (S - t) > 2 mm are based on manufacturing and flow passage restriction constraints. Repeating the foregoing calculations for representative values of t and (S - t), we obtain S (mm) 2.5 3 3 4 4 5 5

N 400 333 333 250 250 200 200

t (mm) 0.5 0.5 1 0.5 2 0.5 3

Rt,o (K/W) 0.00169 0.00193 0.00202 0.00234 0.00268 0.00264 0.00334

COMMENTS: Clearly, the thermal performance of the fin array improves (Rt,o decreases) with increasing N. Because ηf ≈ 1 for the entire range of conditions, there is a slight degradation in performance (Rt,o increases) with increasing t and fixed N. The reduced performance is associated with the reduction in surface area of the exposed base. Note that the overall thermal resistance for the -2 entire fin array (top and bottom) is Rt,o/2 = 1.22 × 10 K/W.

PROBLEM 3.132 KNOWN: Width and maximum allowable temperature of an electronic chip. Thermal contact resistance between chip and heat sink. Dimensions and thermal conductivity of heat sink. Temperature and convection coefficient associated with air flow through the heat sink. FIND: (a) Maximum allowable chip power for heat sink with prescribed number of fins, fin thickness, and fin pitch, and (b) Effect of fin thickness/number and convection coefficient on performance. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional heat transfer, (3) Isothermal chip, (4) Negligible heat transfer from top surface of chip, (5) Negligible temperature rise for air flow, (6) Uniform convection coefficient associated with air flow through channels and over outer surfaces of heat sink, (7) Negligible radiation. ANALYSIS: (a) From the thermal circuit,

T −T Tc − T∞ qc = c ∞ = R tot R t,c + R t,b + R t,o

( )

where R t,c = R ′′t,c / W 2 = 2 × 10 −6 m 2 ⋅ K / W / ( 0.02m )2 = 0.005 K / W and R t,b = L b / k W 2 = 0.003m /180 W / m ⋅ K ( 0.02m ) = 0.042 K / W. From Eqs. (3.103), (3.102), and (3.99) 2

R t,o =

1 , ηo h A t

ηo = 1 −

N Af (1 − ηf ) , At -4

2

A t = N Af + A b 2

2

where Af = 2WLf = 2 × 0.02m × 0.015m = 6 × 10 m and Ab = W – N(tW) = (0.02m) – 11(0.182 -3 -4 2 1/2 2 × 10 m × 0.02m) = 3.6 × 10 m . With mLf = (2h/kt) Lf = (200 W/m ⋅K/180 W/m⋅K × 0.182 × -3 1/2 10 m) (0.015m) = 1.17, tanh mLf = 0.824 and Eq. (3.87) yields

ηf =

tanh mLf 0.824 = = 0.704 mLf 1.17 -3

2

It follows that At = 6.96 × 10 m , ηo = 0.719, Rt,o = 2.00 K/W, and

qc =

(85 − 20 ) °C

(0.005 + 0.042 + 2.00 ) K / W

<

= 31.8 W

(b) The following results are obtained from parametric calculations performed to explore the effect of decreasing the number of fins and increasing the fin thickness. Continued …..

PROBLEM 3.132 (Cont.) N

t(mm)

ηf

6 7 8 9 10 11

1.833 1.314 0.925 0.622 0.380 0.182

0.957 0.941 0.919 0.885 0.826 0.704

Rt,o (K/W) 2.76 2.40 2.15 1.97 1.89 2.00

2

qc (W)

At (m )

23.2 26.6 29.7 32.2 33.5 31.8

0.00378 0.00442 0.00505 0.00569 0.00632 0.00696

Although ηf (and ηo) increases with decreasing N (increasing t), there is a reduction in At which yields a minimum in Rt,o, and hence a maximum value of qc, for N = 10. For N = 11, the effect of h on the performance of the heat sink is shown below. Heat rate as a function of convection coefficient (N=11)

Heat rate, qc(W)

150

100

50

0 100 200 300 400 500 600 700 800 900 1000 Convection coefficient, h(W/m2.K)

2

With increasing h from 100 to 1000 W/m ⋅K, Rt,o decreases from 2.00 to 0.47 K/W, despite a decrease in ηf (and ηo) from 0.704 (0.719) to 0.269 (0.309). The corresponding increase in qc is significant. COMMENTS: The heat sink significantly increases the allowable heat dissipation. If it were not used and heat was simply transferred by convection from the surface of the chip with h = 100 2 2 W/m ⋅K, Rtot = 2.05 K/W from Part (a) would be replaced by Rcnv = 1/hW = 25 K/W, yielding qc = 2.60 W.

PROBLEM 3.133 KNOWN: Number and maximum allowable temperature of power transistors. Contact resistance between transistors and heat sink. Dimensions and thermal conductivity of heat sink. Temperature and convection coefficient associated with air flow through and along the sides of the heat sink. FIND: (a) Maximum allowable power dissipation per transistor, (b) Effect of the convection coefficient and fin length on the transistor power. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional heat transfer, (3) Isothermal transistors, (4) Negligible heat transfer from top surface of heat sink (all heat transfer is through the heat sink), (5) Negligible temperature rise for the air flow, (6) Uniform convection coefficient, (7) Negligible radiation. ANALYSIS: (a) From the thermal circuit,

Nt qt =

Tt − T∞

(R t,c )equiv + R t,b + R t,o

For the array of transistors, the corresponding contact resistance is the equivalent resistance associated with the component resistances, in which case,

(R t,c )equiv =  N t (1/ R t,c )

−1

−1

= (9 / 0.045 K / W )

= 5 × 10−3 K / W

The thermal resistance associated with the base of the heat sink is

R t,b =

Lb k (W )

2

=

0.006m 180 W / m ⋅ K ( 0.150m )

2

= 1.48 ×10−3 K / W

From Eqs. (3.103), (3.102) and (3.99), the thermal resistance associated with the fin array and the corresponding overall efficiency and total surface area are

R t,o =

1 , ηo h A t

ηo = 1 −

Nf Af (1 − ηf ) , At

A t = Nf Af + Ab -3

2

Each fin has a surface area of Af ≈ 2 W Lf = 2 × 0.15m × 0.03m = 9 × 10 m , and the area of the 2 2 -2 2 exposed base is Ab = W – Nf (tW) = (0.15m) – 25 (0.003m × 0.15m) = 1.13 × 10 m . With mLf = 1/2 2 1/2 (2h/kt) Lf = (200 W/m ⋅K/180 W/m⋅K × 0.003m) (0.03m) = 0.577, tanh mLf = 0.520 and Eq. (3.87) yields

ηf =

tanh mLf 0.520 = = 0.902 mLf 0.577 -3

-2

2

2

Hence, with At = [25 (9 × 10 ) + 1.13 × 10 ]m = 0.236m , Continued …..

ηo = 1 −

PROBLEM 3.133 (Cont.)

(

25 0.009m 2 0.236m 2

) (1 − 0.901) = 0.907

(

R t,o = 0.907 ×100 W / m 2 ⋅ K × 0.236m 2

)

−1

= 0.0467 K / W

The heat rate per transistor is then

qt =

(100 − 27 ) °C 1 = 152 W 9 (0.0050 + 0.0015 + 0.0467 ) K / W

<

(b) As shown below, the transistor power dissipation may be enhanced by increasing h and/or Lf.

Tra n s is to r p o w e r (W )

600 500 400 300 200 100 0 100

200

300

400

500

600

700

800

900 1000

C o n ve c tio n c o e ffic ie n t, h (W /m ^2 .K )

250 240 H e a t ra te , q t(W )

230 220 210 200 190 180 170 160 150 0 .0 3

0 .0 4

0 .0 5

0 .0 6

0 .0 7

0 .0 8

0 .0 9

0 .1

Fin le n g th , L f(m )

However, in each case, the effect of the increase diminishes due to an attendant reduction in ηf. For 2 example, as h increases from 100 to 1000 W/m ⋅K for Lf = 30 mm, ηf decreases from 0.902 to 0.498. COMMENTS: The heat sink significantly increases the allowable transistor power. If it were not 2 2 2 used and heat was simply transferred from a surface of area W = 0.0225 m with h = 100 W/m ⋅K, 2 -1 the corresponding thermal resistance would be Rt,cnv = (hW ) K/W = 0.44 and the transistor power would be qt = (Tt - T∞)/Nt Rt,cnv = 18.4 W.

PROBLEM 3.134 KNOWN: Geometry and cooling arrangement for a chip-circuit board arrangement. Maximum chip temperature. FIND: (a) Equivalent thermal circuit, (b) Maximum chip heat rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in chipboard assembly, (3) Negligible pin-chip contact resistance, (4) Constant properties, (5) Negligible chip thermal resistance, (6) Uniform chip temperature. PROPERTIES: Table A.1, Copper (300 K): k ≈ 400 W/m⋅K. ANALYSIS: (a) The thermal circuit is

Rf =

cosh mL+ ( h o / mk ) sinh mL θb = 16q f 16 h PkA 1/ 2 sinh mL+ h / mk cosh mL  ( o ) ( o c,f )  

(b) The maximum chip heat rate is q c = 16q f + q b + qi . Evaluate these parameters  h P m= o  kAc,f 

(

1/ 2

  

 4h = o  kDp 

1/ 2

   

)

1/ 2

 4 × 1000 W/m 2 ⋅ K   =  400 W/m ⋅ K × 0.0015 m   

mL = 81.7 m-1 × 0.015 m = 1.23,

( h/mk ) =

1000 W/m 2 ⋅ K 81.7 m-1 × 400 W/m ⋅ K

(

M = h oπ D p kπ Dp2 / 4

(

)

1/ 2

sinh mL = 1.57,

= 81.7 m-1

cosh mL = 1.86

= 0.0306

θb

)

1/ 2

3 M = 1000 W/m 2 ⋅ K π 2 / 4 ( 0.0015 m ) 400 W/m ⋅ K   

(55 C) = 3.17 W. $

Continued …..

PROBLEM 3.134 (Cont.) The fin heat rate is qf = M

sinh mL+ ( h/mk ) cosh mL

cosh mL+ ( h/mk ) sinh mL

= 3.17 W

1.57+0.0306 × 1.86 1.86+0.0306 × 1.57

q f = 2.703 W. The heat rate from the board by convection is 2 2 q b = h o A bθ b = 1000 W/m 2 ⋅ K ( 0.0127 m ) − (16π / 4 )( 0.0015 m )  55$ C  

q b = 7.32 W. The convection heat rate is

(55 C) qi = = (1/hi + R ′′t,c + Lb / k b ) (1/ Ac ) (1/40+10-4 + 0.005 /1) m2 ⋅ K/W (0.0127 m )2

Tc − T∞,i

$

qi = 0.29 W. Hence, the maximum chip heat rate is q c = 16 ( 2.703) + 7.32 + 0.29  W = [43.25 + 7.32 + 0.29] W

<

q c = 50.9 W.

COMMENTS: (1) The fins are extremely effective in enhancing heat transfer from the chip (assuming negligible contact resistance). Their effectiveness is ε = q f / π Dp2 / 4 h oθ b = 2.703

(

)

W/0.097 W = 27.8 2

2

(2) Without the fins, qc = 1000 W/m ⋅K(0.0127 m) 55°C + 0.29 W = 9.16 W. Hence the fins provide for a (50.9 W/9.16 W) × 100% = 555% enhancement of heat transfer. 2

(3) With the fins, the chip heat flux is 50.9 W/(0.0127 m) or q′′c = 3.16 × 105 W/m 2 = 31.6 2 W/cm . (4) If the infinite fin approximation is made, qf = M = 3.17 W, and the actual fin heat transfer is overestimated by 17%.

PROBLEM 3.135 KNOWN: Geometry of pin fin array used as heat sink for a computer chip. Array convection and chip substrate conditions. FIND: Effect of pin diameter, spacing and length on maximum allowable chip power dissipation. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in chip-board assembly, (3) Negligible pin-chip contact resistance, (4) Constant properties, (5) Negligible chip thermal resistance, (6) Uniform chip temperature. ANALYSIS: The total power dissipation is q c = qi + q t , where Tc − T∞,i qi = = 0.3W 1 h i + R ′′t,c + L b k b A c

(

)

and qt =

Tc − T∞,o R t,o

The resistance of the pin array is −1

R t,o = (ηo h o A t ) where

ηo = 1 −

NAf At

(1 − ηf )

A t = NA f + A b

(

A f = π D p L c = π D p L p + D p /4

)

Subject to the constraint that N D p … 9 mm, the foregoing expressions may be used to compute qt as a function of D p for L p = 15 mm and values of N = 16, 25 and 36. Using the IHT Performance Calculation, Extended Surface Model for the Pin Fin Array, we obtain Continued... 1/ 2

PROBLEM 3.135 (CONT.) 35 Heat rate, qt(W)

30 25 20 15 10 5 0 0.5

0.9

1.3

1.7

2.1

2.5

Pin diameter, Dp(mm) N = 36 N = 25 N = 16

Clearly, it is desirable to maximize the number of pins and the pin diameter, so long as flow passages are not constricted to the point of requiring an excessive pressure drop to maintain the prescribed convection coefficient. The maximum heat rate for the fin array ( q t = 33.1 W) corresponds to N = 36 and D p = 1.5 mm. Further improvement could be obtained by using N = 49 pins of diameter D p = 1.286 mm, which yield q t = 37.7 W. Exploring the effect of L p for N = 36 and D p = 1.5 mm, we obtain

Heat rate, qt(W)

60

50

40

30 10

20

30

40

50

Pin length, Lp(mm) N = 36, Dp = 1.5 mm

Clearly, there are benefits to increasing L p , although the effect diminishes due to an attendant reduction in ηf (from ηf = 0.887 for L p = 15 mm to ηf = 0.471 for L p = 50 mm). Although a heat dissipation rate of q t = 56.7 W is obtained for L p = 50 mm, package volume constraints could preclude such a large fin length. COMMENTS: By increasing N, D p and/or L p , the total surface area of the array, A t , is increased, thereby reducing the array thermal resistance, R t,o . The effects of D p and N are shown for L p = 15 mm. Resistance, Rt,o(K/W)

8 6 4 2 0 0.5

1

1.5 Pin diameter, Dp(mm)

N = 16 N = 25 N = 36

2

2.5

PROBLEM 3.136 KNOWN: Copper heat sink dimensions and convection conditions. FIND: (a) Maximum allowable heat dissipation for a prescribed chip temperature and interfacial chip/heat-sink contact resistance, (b) Effect of fin length and width on heat dissipation. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in chip-heat sink assembly, (3) Constant k, (4) Negligible chip thermal resistance, (5) Negligible heat transfer from back of chip, (6) Uniform chip temperature. ANALYSIS: (a) For the prescribed system, the chip power dissipation may be expressed as Tc − T∞ qc = R t,c + R cond,b + R t,o where R t,c =

R ′′t,c Wc2

R cond,b =

=

5 × 10 −6 m 2 ⋅ K W

(0.016m )

2

Lb

=

kWc2

= 0.0195 K W

0.003m 400 W m ⋅ K ( 0.016m )

2

= 0.0293 K W

The thermal resistance of the fin array is −1

R t,o = (ηo hA t ) where ηo = 1 − and

NAf At

(1 − ηf )

(

A t = NA f + A b = N ( 4wL c ) + Wc2 − Nw 2

) Continued...

PROBLEM 3.136 (Cont.) With w = 0.25 mm, S = 0.50 mm, Lf = 6 mm, N = 1024, and Lc ≈ Lf + w 4 = 6.063 × 10−3 m, it follows that A f = 6.06 × 10−6 m 2 and A t = 6.40 ×10−3 m 2 . The fin efficiency is

ηf =

tanh mLc mLc

1/ 2 1/ 2 where m = ( hP kA c ) = 245 m-1 and mLc = 1.49. It follows that ηf = 0.608 and = ( 4h kw ) ηo = 0.619, in which case

)

(

R t,o = 0.619 × 1500 W m 2 ⋅ K × 6.40 × 10−3 m 2 = 0.168 K W and the maximum allowable heat dissipation is

qc =

(85 − 25)$ C

(0.0195 + 0.0293 + 0.168 ) K

W

<

= 276W

(b) The IHT Performance Calculation, Extended Surface Model for the Pin Fin Array has been used to determine q c as a function of Lf for four different cases, each of which is characterized by the closest allowable fin spacing of (S - w) = 0.25 mm.

Maximum heat rate, qc(W)

Case A B C D

w (mm) 0.25 0.35 0.45 0.55

S (mm) 0.50 0.60 0.70 0.80

N 1024 711 522 400

340 330 320 310 300 290 280 270 6

7

8

9

10

Fin length, Lf(mm) w = 0.25 mm, S = 0.50 mm, N =1024 w = 0.35 mm, S = 0.60 mm, N = 711 w = 0.45 mm, S = 0.70 mm, N = 522 w = 0.55 mm, S = 0.80 mm, N = 400

With increasing w and hence decreasing N, there is a reduction in the total area A t associated with heat transfer from the fin array. However, for Cases A through C, the reduction in A t is more than balanced by an increase in ηf (and ηo ), causing a reduction in R t,o and hence an increase in q c . As the fin efficiency approaches its limiting value of ηf = 1, reductions in A t due to increasing w are no longer balanced by increases in ηf , and q c begins to decrease. Hence there is an optimum value of w, which depends on Lf . For the conditions of this problem, Lf = 10 mm and w = 0.55 mm provide the largest heat dissipation.

Problem 3.137 KNOWN: Two finned heat sinks, Designs A and B, prescribed by the number of fins in the array, N, fin dimensions of square cross-section, w, and length, L, with different convection coefficients, h. FIND: Determine which fin arrangement is superior. Calculate the heat rate, qf, efficiency, ηf, and effectiveness, εf, of a single fin, as well as, the total heat rate, qt, and overall efficiency, ηo, of the array. Also, compare the total heat rates per unit volume. SCHEMATIC:

Fin dimensions Cross section Length w x w (mm) L (mm) 1x1 30 3x3 7

Design A B

Number of fins 6x9 14 x 17

Convection coefficient (W/m2⋅K) 125 375

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fins, (3) Convection coefficient is uniform over fin and prime surfaces, (4) Fin tips experience convection, and (5) Constant properties. ANALYSIS: Following the treatment of Section 3.6.5, the overall efficiency of the array, Eq. (3.98), is

ηo =

qt q max

=

qt hA tθ b

(1)

where At is the total surface area, the sum of the exposed portion of the base (prime area) plus the fin surfaces, Eq. 3.99,

A t = N ⋅ Af + A b

(2)

where the surface area of a single fin and the prime area are Af = 4 ( L × W ) + w 2

(3)

A b = b1× b2 − N ⋅ A c

(4)

Combining Eqs. (1) and (2), the total heat rate for the array is

q t = Nηf hAf θ b + hA bθ b

(5)

where ηf is the efficiency of a single fin. From Table 4.3, Case A, for the tip condition with convection, the single fin efficiency based upon Eq. 3.86,

ηf =

qf hAf θ b

(6) Continued...

PROBLEM 3.137 (Cont.) where

qf = M

sinh(mL) + ( h mk ) cosh(mL)

(7)

cosh(mL) + ( h mk ) sinh(mL)

M = ( hPkA c )

1/ 2

θb

m = ( hP kAc )

1/ 2

P = 4w

Ac = w 2

(8,9,10)

The single fin effectiveness, from Eq. 3.81,

εf =

qf hA cθ b

(11)

Additionally, we want to compare the performance of the designs with respect to the array volume, vol

q′′′f = qf ∀ = qf

( b1⋅ b2 ⋅ L )

(12)

The above analysis was organized for easy treatment with equation-solving software. Solving Eqs. (1) through (11) simultaneously with appropriate numerical values, the results are tabulated below. Design A B

qt (W) 113 165

qf (W) 1.80 0.475

ηo

ηf

εf

0.804 0.909

0.779 0.873

31.9 25.3

q „„„ f (W/m3) 1.25×106 7.81×106

COMMENTS: (1) Both designs have good efficiencies and effectiveness. Clearly, Design B is superior because the heat rate is nearly 50% larger than Design A for the same board footprint. Further, the space requirement for Design B is four times less (∀ = 2.12×10-5 vs. 9.06×10-5 m3) and the heat rate per unit volume is 6 times greater. (2) Design A features 54 fins compared to 238 fins for Design B. Also very significant to the performance comparison is the magnitude of the convection coefficient which is 3 times larger for Design B. Estimating convection coefficients for fin arrays (and tube banks) is discussed in Chapter 7.6. Of concern is how the fins alter the flow past the fins and whether the convection coefficient is uniform over the array. (3) The IHT Extended Surfaces Model, for a Rectangular Pin Fin Array could have been used to solve this problem.

PROBLEM 3.138 KNOWN: Geometrical characteristics of a plate with pin fin array on both surfaces. Inner and outer convection conditions. FIND: (a) Heat transfer rate with and without pin fin arrays, (b) Effect of using silver solder to join the pins and the plate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant k, (3) Negligible radiation. PROPERTIES: Table A-1: Copper, T ≈ 315 K, k = 400 W/m⋅K. ANALYSIS: (a) The heat rate may be expressed as T∞,i − T∞,o q= R t,o(c),i + R w + R t,o(c),o where

(

R t,o(c) = ηo(c) hA t

ηo(c) = 1 −

)−1 ,

NA f 

ηf  1 −  , A t  C1 

A t = NA f + A b , A f = π D p Lc ≈ π D p ( L + D 4 ) ,

(

)

A b = W 2 − NAc,b = W 2 − N π D 2p 4 ,

ηf =

tanh mLc mLc

,

(

m = 4h kD p

)1/ 2 , Continued...

PROBLEM 3.138 (Cont.)

(

)

C1 = 1 + ηf hA f R ′′t,c A c,b , and Rw =

Lw W2k

.

Calculations may be expedited by using the IHT Performance Calculation, Extended Surface Model for the Pin Fin Array. For R ′′t,c = 0, C1 = 1, and with W = 0.160 m, Rw = 0.005 m/(0.160 m)2 400 W/m⋅K = 4.88 × 10-4 K/W. For the prescribed array geometry, we also obtain A c,b = 1.26 × 10-5 m2, A f = 2.64 × 10-4 m2, A b = 2.06 × 10-2 m2, and At = 0.126 m2. On the outer surface, where h o = 100 W/m2⋅K, m = 15.8 m-1, ηf = 0.965, ηo = 0.970 and R t,o = 0.0817 K/W. On the inner surface, where h i = 5 W/m2⋅K, m = 3.54 m-1, ηf = 0.998, ηo = 0.999 and R t,o = 1.588 K/W. Hence, the heat rate is q=

(65 − 20 )$ C

(1.588 + 4.88 ×10

−4

)

+ 0.0817 K W

<

= 26.94W

Without the fins, q=

T∞,i − T∞,o (65 − 20 )$ C = = 5.49W (1 h i A w ) + R w + (1 h o A w ) 7.81 + 4.88 ×10−4 + 0.39

)

(

<

Hence, the fin arrays provide nearly a five-fold increase in heat rate. (b) With use of the silver solder, ηo(c),o = 0.962 and R t,o(c),o = 0.0824 K/W. Also, ηo(c),i = 0.998 and R t,o(c),i = 1.589 K/W. Hence

q=

(

(65 − 20 )$ C

)

1.589 + 4.88 × 10−4 + 0.0824 K W

= 26.92W

Hence, the effect of the contact resistance is negligible. COMMENTS: The dominant contribution to the total thermal resistance is associated with internal conditions. If the heat rate must be increased, it should be done by increasing hi.

<

PROBLEM 3.139 KNOWN: Long rod with internal volumetric generation covered by an electrically insulating sleeve and supported with a ribbed spider. FIND: Combination of convection coefficient, spider design, and sleeve thermal conductivity which enhances volumetric heating subject to a maximum centerline temperature of 100°C. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial heat transfer in rod, sleeve and hub, (3) Negligible interfacial contact resistances, (4) Constant properties, (5) Adiabatic outer surface. ANALYSIS: The system heat rate per unit length may be expressed as T1 − T∞ q′ = q π ro2 = R ′sleeve + R ′hub + R ′t,o where

( )

R ′sleeve =

ln ( r1 ro )

, R ′hub =

2π k s NA′f ηo = 1 − (1 − ηf ) , A′t

ηf =

tanh m ( r3 − r2 ) m ( r3 − r2 )

,

ln ( r2 r1 )

= 3.168 × 10−4 m ⋅ K W , R ′t,o =

2π k r

A′f = 2 ( r3 − r2 ) , m = ( 2h k r t )

1/ 2

1 , ηo hA′t

A′t = NA′f + ( 2π r3 − Nt ) , .

The rod centerline temperature is related to T1 through  2 qr To = T ( 0 ) = T1 + o

4k

Calculations may be expedited by using the IHT Performance Calculation, Extended Surface Model for the Straight Fin Array. For base case conditions of ks = 0.5 W/m⋅K, h = 20 W/m2⋅K, t = 4 mm and N = 12, R ′sleeve = 0.0580 m⋅K/W, R ′t,o = 0.0826 m⋅K/W, ηf = 0.990, q′ = 387 W/m, and q = 1.23 × 106 W/m3. As shown below, q may be increased by increasing h, where h = 250 W/m2⋅K represents a reasonable upper limit for airflow. However, a more than 10-fold increase in h yields only a 63% increase in q . Continued...

Heat generation, qdot(W/m^3)

PROBLEM 3.139 (Cont.) 2E6 1.8E6 1.6E6 1.4E6 1.2E6 1E6 0

50

100

150

200

250

Convection coefficient, h(W/m^2.K) t = 4 mm, N = 12, ks = 0.5 W/m.K

The difficulty is that, by significantly increasing h, the thermal resistance of the fin array is reduced to 0.00727 m⋅K/W, rendering the sleeve the dominant contributor to the total resistance.

Heat generation, qdotx1E-6(W/m

Similar results are obtained when N and t are varied. For values of t = 2, 3 and 4 mm, variations of N in the respective ranges 12 ≤ N ≤ 26, 12 ≤ N ≤ 21 and 12 ≤ N ≤ 17 were considered. The upper limit on N was fixed by requiring that (S - t) ≥ 2 mm to avoid an excessive resistance to airflow between the ribs. As shown below, the effect of increasing N is small, and there is little difference between results for the three values of t. 2.1 2.08 2.06 2.04 2.02 2 12

14

16

18

20

22

24

26

Number of ribs, N t = 2 mm, N: 12 - 26, h = 250 W/m^2.K t = 3 mm, N: 12 - 21, h = 250 W/m^2.K t = 4 mm, N: 12 -17, h = 250 W/m^2.K

Heat generation, qdot(W/m^3)

In contrast, significant improvement is associated with changing the sleeve material, and it is only necessary to have ks ≈ 25 W/m⋅K (e.g. a boron sleeve) to approach an upper limit to the influence of ks. 4E6 3.6E6 3.2E6 2.8E6 2.4E6 2E6 0

20

40

60

80

100

Sleeve conductivity, ks(W/m.K) t = 4 mm, N = 12, h = 250 W/m^2.K

For h = 250 W/m2⋅K and ks = 25 W/m⋅K, only a slight improvement is obtained by increasing N. Hence, the recommended conditions are: h = 250 W m 2 ⋅ K,

k s = 25 W m ⋅ K,

N = 12,

t = 4mm

<

COMMENTS: The upper limit to q is reached as the total thermal resistance approaches zero, in which case T1 → T∞. Hence q max = 4k ( To − T∞ ) ro2 = 4.5 × 106 W m3 .

PROBLEM 3.140 KNOWN: Geometrical and convection conditions of internally finned, concentric tube air heater. FIND: (a) Thermal circuit, (b) Heat rate per unit tube length, (c) Effect of changes in fin array. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in radial direction, (3) Constant k, (4) Adiabatic outer surface. ANALYSIS: (a) For the thermal circuit shown schematically, −1

R ′conv,i = ( h i 2π r1 ) where

ηo = 1 −

NA′f (1 − ηf ) , A′t

(b)

q′ =

,

R ′cond = ln ( r2 r1 ) 2π k , and

A′f = 2L = 2 ( r3 − r2 ) ,

A′t = NA′f + ( 2π r2 − Nt ) , and

(T∞,i − T∞,o )

R ′conv,i + R ′cond + R ′t,o Substituting the known conditions, it follows that

(

R ′conv,i = 5000 W m 2 ⋅ K × 2π × 0.013m

)

−1

−1

R ′t,o = (ηo h o A′t )

ηf =

,

tanh mL mL

.

= 2.45 × 10−3 m ⋅ K W

R ′cond = ln (0.016m 0.013m ) 2π ( 20 W m ⋅ K ) = 1.65 × 10−3 m ⋅ K W

(

R ′t,o = 0.575 × 200 W m 2 ⋅ K × 0.461m

)

−1

= 18.86 × 10 −3 m ⋅ K W

where ηf = 0.490. Hence, $ 90 − 25 ) C ( q′ = ( 2.45 + 1.65 + 18.86 ) ×10−3 m ⋅ K

= 2831W m

<

W

(c) The small value of ηf suggests that some benefit may be gained by increasing t, as well as by increasing N. With the requirement that Nt ≤ 50 mm, we use the IHT Performance Calculation, Extended Surface Model for the Straight Fin Array to consider the following range of conditions: t = 2 mm, 12 ≤ N ≤ 25; t = 3 mm, 8 ≤ N ≤ 16; t = 4 mm, 6 ≤ N ≤ 12; t = 5 mm, 5 ≤ N ≤ 10. Calculations based on the foregoing model are plotted as follows. Continued...

PROBLEM 3.140 (Cont.)

Heat rate, q'(w/m)

5000

4000

3000

2000 5

10

15

20

25

Number of fins, N t = 2 mm t = 3 mm t = 4 mm t = 5 mm

By increasing t from 2 to 5 mm, ηf increases from 0.410 to 0.598. Hence, for fixed N, q′ increases with increasing t. However, from the standpoint of maximizing q′t , it is clearly preferable to use the larger number of thinner fins. Hence, subject to the prescribed constraint, we would choose t = 2 mm and N = 25, for which q′ = 4880 W/m. COMMENTS: (1) The air side resistance makes the dominant contribution to the total resistance, and efforts to increase q′ by reducing R′t,o are well directed. (2) A fin thickness any smaller than 2 mm would be difficult to manufacture.

PROBLEM 3.141 KNOWN: Dimensions and number of rectangular aluminum fins. Convection coefficient with and without fins. FIND: Percentage increase in heat transfer resulting from use of fins.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible radiation, (5) Negligible fin contact resistance, (6) Uniform convection coefficient.

PROPERTIES: Table A-1, Aluminum, pure: k ≈ 240 W/m⋅K. ANALYSIS: Evaluate the fin parameters

Lc = L+t/2 = 0.05025m A p = Lc t = 0.05025m × 0.5 × 10-3m=25.13 × 10-6 m 2 1/ 2

(

)

= ( 0.05025m )

(

)

= 0.794

L3/2 c h w / kA p L3/2 c h w / kA p

1/ 2

1/ 2

3/ 2

  30 W/m 2 ⋅ K    240 W/m ⋅ K × 25.13 × 10-6m 2 

It follows from Fig. 3.18 that ηf ≈ 0.72. Hence, q f = ηf q max = 0.72 h w 2wL θ b qf = 0.72 × 30 W/m 2 ⋅ K × 2 × 0.05m × ( w θ b ) = 2.16 W/m ⋅ K ( w θ b ) With the fins, the heat transfer from the walls is

q w = N qf + (1 − Nt ) w h w θ b

)

(

W ( w θ b ) + 1m − 250 × 5 ×10−4 m × 30 W/m2 ⋅ K ( w θ b ) m⋅K W q w = (540 + 26.3) ( w θ b ) = 566 w θ b . m⋅K q w = 250 × 2.16

Without the fins, qwo = hwo 1m × w θb = 40 w θb. Hence the percentage increase in heat transfer is q w − q wo (566 − 40 ) w θ b < = = 13.15 = 1315% q wo 40 w θ b 1/2

COMMENTS: If the infinite fin approximation is made, it follows that qf = (hPkAc) 1/2

-4 1/2

=[hw2wkwt] θb = (30 × 2 × 240 × 5×10 ) overestimated.

w θb = 2.68 w θb. Hence, qf is

θb

PROBLEM 3.142 KNOWN: Dimensions, base temperature and environmental conditions associated with rectangular and triangular stainless steel fins. FIND: Efficiency, heat loss per unit width and effectiveness associated with each fin. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient. PROPERTIES: Table A-1, Stainless Steel 304 (T = 333 K): k = 15.3 W/m⋅K. ANALYSIS: For the rectangular fin, with Lc = L + t/2, evaluate the parameter 1/ 2

 2 h kA 1/ 2 = 0.023m 3/ 2 L3/  ( ) c p

(

 75 W m 2 ⋅ K  15.3 W m ⋅ K ( 0.023m )( 0.006 m ) 

)

= 0.66 .

Hence, from Fig. 3.18, the fin efficiency is

<

ηf ≈ 0.79

From Eq. 3.86, the fin heat rate is q f = ηf hAf θ b = ηf hPLcθ b = ηf h2wLcθ b or, per unit width,

(

)

q q′f = f = 0.79 75 W m 2 ⋅ K 2 ( 0.023m ) 80$ C = 218 W m . w

<

From Eq. 3.81, the fin effectiveness is

εf =

qf q′f × w 218 W m = = = 6.06 . 2 hA c,bθ b h ( t × w )θ b 75 W m ⋅ K ( 0.006 m ) 80$ C

<

For the triangular fin with 1/ 2

 2 h kA 1/ 2 = 0.02 m 3/ 2 L3/  ( ) c p

(

)

 75 W m 2 ⋅ K   (15.3 W m ⋅ K )( 0.020 m )( 0.003m ) 

= 0.81 ,

find from Figure 3.18,

<

ηf ≈ 0.78 , From Eq. 3.86 and Table 3.5 find 1/ 2

2 q′f = ηf hA′f θ b = ηf h2  L2 + ( t 2 )   

θb 1/ 2

2 2 q′f = 0.78 × 75 W m 2 ⋅ K × 2 ( 0.02 ) + ( 0.006 2 )   

(

)

m 80$ C = 187 W m .

<

and from Eq. 3.81, the fin effectiveness is

εf =

q′f × w 187 W m = = 5.19 h ( t × w )θ b 75 W m 2 ⋅ K ( 0.006 m ) 80$ C

COMMENTS: Although it is 14% less effective, the triangular fin offers a 50% weight savings.

<

PROBLEM 3.143 KNOWN: Dimensions, base temperature and environmental conditions associated with a triangular, aluminum fin. FIND: (a) Fin efficiency and effectiveness, (b) Heat dissipation per unit width. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible radiation and base contact resistance, (5) Uniform convection coefficient. PROPERTIES: Table A-1, Aluminum, pure (T ≈ 400 K): k = 240 W/m⋅K. ANALYSIS: (a) With Lc = L = 0.006 m, find A p = Lt 2 = ( 0.006 m )( 0.002 m ) 2 = 6 × 10−6 m 2 ,

 1/ 2 3/ 2 L3c/ 2 h kA p = ( 0.006 m )  

(

1/ 2

  −6 2  240 W m K 6 10 m ⋅ × ×  

)

40 W m2 ⋅ K

= 0.077

and from Fig. 3.18, the fin efficiency is

<

ηf ≈ 0.99 . From Eq. 3.86 and Table 3.5, the fin heat rate is 1/ 2

2 q f = ηf q max = ηf hAf (tri)θ b = 2ηf hw  L2 + ( t 2 ) 





θb .

From Eq. 3.81, the fin effectiveness is 1/ 2

εf =

qf hAc,bθ b

=

2 2ηf hw  L2 + ( t 2 ) 



g ( w ⋅ t )θ b



θb

1/ 2

=

2 2ηf  L2 + ( t 2 ) 





t

1/ 2

εf =

2 2 2 × 0.99 ( 0.006 ) + ( 0.002 2 ) 



m



0.002 m

<

= 6.02

(b) The heat dissipation per unit width is 1/ 2

2 q′f = ( q f w ) = 2ηf h  L2 + ( t 2 ) 





θb 1/ 2

2 2 q′f = 2 × 0.99 × 40 W m 2 ⋅ K ( 0.006 ) + ( 0.002 2 ) 





m × ( 250 − 20 ) C = 110.8 W m . $

<

COMMENTS: The triangular profile is known to provide the maximum heat dissipation per unit fin mass.

PROBLEM 3.144 KNOWN: Dimensions and base temperature of an annular, aluminum fin of rectangular profile. Ambient air conditions. FIND: (a) Fin heat loss, (b) Heat loss per unit length of tube with 200 fins spaced at 5 mm increments. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible radiation and contact resistance, (5) Uniform convection coefficient. PROPERTIES: Table A-1, Aluminum, pure (T ≈ 400 K): k = 240 W/m⋅K. ANALYSIS: (a) The fin parameters for use with Figure 3.19 are

r2c = r2 + t 2 = (12.5 mm + 10 mm ) + 0.5 mm = 23mm = 0.023m r2c r1 = 1.84

Lc = L + t 2 = 10.5 mm = 0.0105 m

A p = Lc t = 0.0105 m × 0.001m = 1.05 ×10−5 m 2  2 h kA 1/ 2 = 0.0105 m 3/ 2  L3/ ( )  c p

(

)

1/ 2

  −5 2   240 W m ⋅ K × 1.05 ×10 m  25 W m 2 ⋅ K

= 0.15 .

Hence, the fin effectiveness is ηf ≈ 0.97, and from Eq. 3.86 and Fig. 3.5, the fin heat rate is

(

)

2 − r2 θ q f = ηf q max = ηf hAf (ann)θ b = 2πηf h r2,c 1 b 2 2 q f = 2π × 0.97 × 25 W m 2 ⋅ K × ( 0.023m ) − ( 0.0125 m )  225$ C = 12.8 W .  

<

(b) Recognizing that there are N = 200 fins per meter length of the tube, the total heat rate considering contributions due to the fin and base (unfinned surfaces is

q′ = N′q f + h (1 − N′t ) 2π r1θ b

(

)

q′ = 200 m −1 × 12.8 W + 25 W m 2 ⋅ K 1 − 200 m −1 × 0.001m × 2π × ( 0.0125 m ) 225$ C q′ = ( 2560 W + 353 W ) m = 2.91kW m .

<

COMMENTS: Note that, while covering only 20% of the tube surface area, the tubes account for more than 85% of the total heat dissipation.

PROBLEM 3.145 KNOWN: Dimensions and base temperature of aluminum fins of rectangular profile. Ambient air conditions. FIND: (a) Fin efficiency and effectiveness, (b) Rate of heat transfer per unit length of tube. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction in fins, (3) Constant properties, (4) Negligible radiation, (5) Negligible base contact resistance, (6) Uniform convection coefficient. PROPERTIES: Table A-1, Aluminum, pure (T ≈ 400 K): k = 240 W/m⋅K. ANALYSIS: (a) The fin parameters for use with Figure 3.19 are r2c = r2 + t 2 = 40 mm + 2 mm = 0.042 m

Lc = L + t 2 = 15 mm + 2 mm = 0.017 m

r2c r1 = 0.042 m 0.025 m = 1.68

A p = Lc t = 0.017 m × 0.004 m = 6.8 × 10 −5 m 2

(

L3c/ 2 h kA p

)1/ 2 = (0.017 m )3 / 2 40 W

1/ 2

m 2 ⋅ K 240 W m ⋅ K × 6.8 × 10−5 m 2 



= 0.11

The fin efficiency is ηf ≈ 0.97. From Eq. 3.86 and Fig. 3.5, 2 q f = ηf q max = ηf hAf (ann)θ b = 2πηf h  r2c − r12  θ b   2 2 q f = 2π × 0.97 × 40 W m 2 ⋅ K ( 0.042 ) − ( 0.025 )  m 2 × 180$ C = 50 W





<

From Eq. 3.81, the fin effectiveness is

εf =

qf hAc,bθ b

=

50 W 40 W m 2 ⋅ K 2π ( 0.025 m )( 0.004 m )180$ C

= 11.05

<

(b) The rate of heat transfer per unit length is q′ = N′q f + h (1 − N′t )( 2π r1 )θ b q′ = 125 × 50 W m + 40 W m 2 ⋅ K (1 − 125 × 0.004 )( 2π × 0.025 m ) × 180$ C q′ = ( 6250 + 565 ) W m = 6.82 kW m COMMENTS: Note the dominant contribution made by the fins to the total heat transfer.

<

PROBLEM 3.146 KNOWN: Dimensions, base temperature, and contact resistance for an annular, aluminum fin. Ambient fluid conditions. FIND: Fin heat transfer with and without base contact resistance. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient. PROPERTIES: Table A-1, Aluminum, pure (T ≈ 350 K): k ≈ 240 W/m⋅K. ANALYSIS: With the contact resistance, the fin heat loss is q f =

Tw − T∞

R t,c + R f

where

R t,c = R ′′t,c A b = 2 × 10−4 m 2 ⋅ K W 2π ( 0.015 m )( 0.002 m ) = 1.06 K W . From Eqs. 3.83 and 3.86, the fin resistance is θ θb θb 1 . Rf = b = = = q f ηf q max ηf hA f θ b 2π hη r 2 − r 2 f 2,c 1

(

Evaluating parameters, r2,c = r2 + t 2 = 30 mm + 1mm = 0.031m

(

Lc = L + t 2 = 0.016 m A p = Lc t = 3.2 × 10−5 m 2

r2c r1 = 0.031 0.015 = 2.07 Z L3c/ 2 h kA p

)

)1/ 2 = (0.016 m )3 / 2 75 W

1/ 2

m 2 ⋅ K 240 W m ⋅ K × 3.2 × 10−5 m 2 



= 0.20

find the fin efficiency from Figure 3.19 as ηf = 0.94. Hence, Rf =

qf =

(

)

1

2 2 2π 75 W m ⋅ K 0.94 ( 0.031m ) − ( 0.015 m )    2

= 3.07 K W

(100 − 25 )$ C = 18.2 W . (1.06 + 3.07 ) K W

<

Without the contact resistance, Tw = Tb and qf =

θb Rf

=

75$ C 3.07 K W

= 24.4 W .

COMMENTS: To maximize fin performance, every effort should be made to minimize contact resistance.

<

PROBLEM 3.147 KNOWN: Dimensions and materials of a finned (annular) cylinder wall. Heat flux and ambient air conditions. Contact resistance. FIND: Surface and interface temperatures (a) without and (b) with an interface contact resistance. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Constant properties, (3) Uniform h over surfaces, (4) Negligible radiation. ANALYSIS: The analysis may be performed per unit length of cylinder or for a 4 mm long section. The following calculations are based on a unit length. The inner surface temperature may be obtained from T −T q′ = i ∞ = q′′i ( 2π ri ) = 105 W/m 2 × 2π × 0.06 m = 37, 700 W/m R ′tot where R ′tot = R ′c + R ′t,c + R ′w + R ′equiv ;

−1

R ′equiv = (1/ R ′f + 1/ R ′b )

.

R ′c , Conduction resistance of cylinder wall: R ′c =

ln ( r1 / ri )

ln ( 66/60 )

=

2π k 2π (50 W/m ⋅ K ) R ′t,c , Contact resistance:

= 3.034 ×10−4 m ⋅ K/W

R ′t,c = R ′′t,c / 2π r1 = 10−4 m 2 ⋅ K/W/2π × 0.066 m = 2.411×10−4 m ⋅ K/W R ′w , Conduction resistance of aluminum base: R ′w =

ln ( rb / r1 ) 2π k

=

ln ( 70/66 ) 2π × 240 W/m ⋅ K

= 3.902 × 10−5 m ⋅ K/W

R ′b , Resistance of prime or unfinned surface: R ′b =

1 1 = = 454.7 × 10−4 m ⋅ K/W 2 hA′b 100 W/m ⋅ K × 0.5 × 2π (0.07 m )

R ′f , Resistance of fins: The fin resistance may be determined from T −T 1 R ′f = b ∞ = q′f ηf hA′f The fin efficiency may be obtained from Fig. 3.19, r2c = ro + t/2 = 0.096 m

Lc = L+t/2 = 0.026 m Continued …..

PROBLEM 3.147 (Cont.) A p = Lc t = 5.2 × 10−5 m 2

r2c / r1 = 1.45

(

L3/2 c h/kA p

)

1/ 2

= 0.375

Fig. 3.19 → ηf ≈ 0.88. The total fin surface area per meter length A′f = 250 π ro2 − rb2 × 2  = 250 m -1  2π 0.0962 − 0.07 2  m 2 = 6.78 m.    

)

(

Hence

)

(

R ′f = 0.88 ×100 W/m 2 ⋅ K × 6.78 m   

−1

(

= 16.8 ×10−4 m ⋅ K/W

)

1/ R ′equiv = 1/16.8 ×10−4 + 1/ 454.7 ×10−4 W/m ⋅ K = 617.2 W/m ⋅ K R ′equiv = 16.2 × 10−4 m ⋅ K/W. Neglecting the contact resistance,

R ′tot = (3.034 + 0.390 + 16.2 )10−4 m ⋅ K/W = 19.6 × 10−4 m ⋅ K/W

Ti = q′R ′tot + T∞ = 37, 700 W/m × 19.6 × 10-4 m ⋅ K/W+320 K = 393.9 K

<

T1 = Ti − q′R ′w = 393.9 K − 37, 700 W/m × 3.034 × 10-4 m ⋅ K/W = 382.5 K

<

Tb = T1 − q′R ′b = 382.5 K − 37, 700 W/m × 3.902 × 10-5 m ⋅ K/W = 381.0 K. < Including the contact resistance,

(

)

R ′tot = 19.6 ×10−4 + 2.411× 10−4 m ⋅ K/W = 22.0 ×10−4 m ⋅ K/W Ti = 37, 700 W/m × 22.0 × 10-4 m ⋅ K/W+320 K = 402.9 K

<

T1,i = 402.9 K − 37, 700 W/m × 3.034 × 10-4 m ⋅ K/W = 391.5 K

<

T1,o = 391.5 K − 37, 700 W/m × 2.411× 10-4 m ⋅ K/W = 382.4 K

<

Tb = 382.4 K − 37, 700 W/m × 3.902 × 10-5 m ⋅ K/W = 380.9 K.

<

COMMENTS: (1) The effect of the contact resistance is small.

(2) The effect of including the aluminum fins may be determined by computing Ti without the fins. In this case R ′tot = R ′c + R ′conv , where 1 1 = = 241.1× 10−4 m ⋅ K/W. R ′conv = 2 h2π r1 100 W/m ⋅ K 2π (0.066 m ) Hence, R tot = 244.1×10−4 m ⋅ K/W, and Ti = q′R ′tot + T∞ = 37, 700 W/m × 244.1×10-4 m ⋅ K/W+320 K = 1240 K. Hence, the fins have a significant effect on reducing the cylinder temperature. (3) The overall surface efficiency is ηo = 1 − ( A′f / A′t )(1 − ηf ) = 1 − 6.78 m/7.00 m (1 − 0.88 ) = 0.884. It follows that q′=ηo h o A′tθ b = 37, 700 W/m, which agrees with the prescribed value.

PROBLEM 3.148 KNOWN: Dimensions and materials of a finned (annular) cylinder wall. Combustion gas and ambient air conditions. Contact resistance. FIND: (a) Heat rate per unit length and surface and interface temperatures, (b) Effect of increasing the fin thickness. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Constant properties, (3) Uniform h over surfaces, (4) Negligible radiation. ANALYSIS: (a) The heat rate per unit length is Tg − T∞ q′ = R ′tot where R ′tot = R ′g + R ′w + R ′t,c + R ′b + R ′t,o , and

(

R ′g = h g 2π ri

R ′w =

)−1 = (150 W

ln ( r1 ri )

ln (66 60 )

=

2π k w

m 2 ⋅ K × 2π × 0.06m

2π ( 50 W m ⋅ K )

(

)

)

−1

= 0.0177m ⋅ K W ,

= 3.03 × 10−4 m ⋅ K W ,

R ′t,c = R ′′t,c 2π r1 = 10−4 m 4 ⋅ K W 2π × 0.066m = 2.41 × 10−4 m ⋅ K W R ′b =

ln ( rb r1 )

=

2π k

ηo = 1 −

= 3.90 × 10−5 m ⋅ K W ,

2π × 240 W m ⋅ K

−1

R t,o = (ηo hA′t )

ln ( 70 66 )

,

N′A f (1 − ηf ) , A′t

(

2 − rb2 A f = 2π roc

)

A′t = N′A f + (1 − N′t ) 2π rb

ηf =

( 2rb

m ) K1 ( mrb ) I1 ( mroc ) − I1 ( mrb ) K1 ( mroc )

(roc2 − rb2 ) I0 (mr1 ) K1 (mroc ) + K0 (mrb ) I1 (mroc )

roc = ro + ( t 2 ) ,

m = ( 2h kt )

1/ 2

Continued...

PROBLEM 3.148 (Cont.) Once the heat rate is determined from the foregoing expressions, the desired interface temperatures may be obtained from Ti = Tg − q′R ′g

( ) T1,o = Tg − q′ ( R ′g + R ′w + R ′t,c ) Tb = Tg − q′ ( R ′g + R ′w + R ′t,c + R ′b ) T1,i = Tg − q′ R ′g + R ′w

For the specified conditions we obtain A′t = 7.00 m, ηf = 0.902, ηo = 0.906 and R ′t,o = 0.00158 m⋅K/W. It follows that q′ = 39, 300 W m Ti = 405K,

T1,i = 393K,

T1,o = 384K,

Tb = 382K

< <

(b) The Performance Calculation, Extended Surface Model for the Circular Fin Array may be used to assess the effects of fin thickness and spacing. Increasing the fin thickness to t = 3 mm, with δ = 2 mm, reduces the number of fins per unit length to 200. Hence, although the fin efficiency increases ( Kf = 0.930), the reduction in the total surface area ( A′t = 5.72 m) yields an increase in the resistance of the fin array ( R „t ,o = 0.00188 m⋅K/W), and hence a reduction in the heat rate ( q′ = 38,700 W/m) and an increase in the interface temperatures ( Ti = 415 K, T1,i = 404 K, T1,o = 394 K, and Tb = 393 K). COMMENTS: Because the gas convection resistance exceeds all other resistances by at least an order of magnitude, incremental changes in R t,o will not have a significant effect on q „ or the interface temperatures.

PROBLEM 3.149 KNOWN: Dimensions of finned aluminum sleeve inserted over transistor. Contact resistance and convection conditions. FIND: Measures for increasing heat dissipation. SCHEMATIC: See Example 3.10. ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from top and bottom of transistor, (3) One-dimensional radial heat transfer, (4) Constant properties, (5) Negligible radiation. ANALYSIS: With 2π r2 = 0.0188 m and Nt = 0.0084 m, the existing gap between fins is extremely small (0.87 mm). Hence, by increasing N and/or t, it would become even more difficult to maintain satisfactory airflow between the fins, and this option is not particularly attractive. Because the fin efficiency for the prescribed conditions is close to unity ( ηf = 0.998), there is little advantage to replacing the aluminum with a material of higher thermal conductivity (e.g. Cu with k ~ 400 W/m⋅K). However, the large value of ηf suggests that significant benefit could be gained by increasing the fin length, L = r3  r2 . It is also evident that the thermal contact resistance is large, and from Table 3.2, it’s clear that a significant reduction could be effected by using indium foil or a conducting grease in the contact zone. Specifically, a reduction of R ′′t,c from 10-3 to 10-4 or even 10-5 m2⋅K/W is certainly feasible. Table 1.1 suggests that, by increasing the velocity of air flowing over the fins, a larger convection coefficient may be achieved. A value of h = 100 W/m2⋅K would not be unreasonable. As options for enhancing heat transfer, we therefore use the IHT Performance Calculation, Extended Surface Model for the Straight Fin Array to explore the effect of parameter variations over the ranges 10 ≤ L ≤ 20 mm, 10-5 ≤ R ′′t,c ≤ 10-3 m2⋅K/W and 25 ≤ h ≤ 100 W/m2⋅K. As shown below, there is a significant enhancement in heat transfer associated with reducing R ′′t,c from 10-3 to 10-4 m2⋅K/W, for which R t,c decreases from 13.26 to 1.326 K/W. At this value of R ′′t,c , the reduction in R t,o from 23.45 to 12.57 K/W which accompanies an increase in L from 10 to 20 mm becomes significant, yielding a heat rate of q t = 4.30 W for R ′′t,c = 10-4 m2⋅K/W and L = 20 mm. However, since R t,o >> R t,c , little benefit is gained by further reducing R ′′t,c to 10-5 m2⋅K/W.

Heat rate, qt(W)

5 4 3 2 1 0 0.01

0.012

0.014

0.016

0.018

0.02

Fin length, L(m) h = 25 W/m^2.K, R''t,c = E-3 m^2.K/W h = 25 W/m^2.K, R''t,c = E-4 m^2.K/W h = 25 W/m^2.K, R''t,c = E-5m^2.K/W

Continued...

PROBLEM 3.149 (Cont.) To derive benefit from a reduction in R ′′t,c to 10-5 m2⋅K/W, an additional reduction in R t,o must be made. This can be achieved by increasing h, and for L = 20 mm and h = 100 W/m2⋅K, R t,o = 3.56 K/W. With R ′′t,c = 10-5 m2⋅K/W, a value of q t = 16.04 W may be achieved.

Heat rate, qt(W)

20 16 12 8 4 0 0.01

0.012

0.014

0.016

0.018

0.02

Fin length, L(m) h = 25 W/m^2.K, R''t,c = E-5 m^2.K/W h = 50 W/m^2.K, R''t,c = E-5 m^2.K/W h = 100 W/m^2.K, R''t,c = E-5 m^2.K/W

COMMENTS: In assessing options for enhancing heat transfer, the limiting (largest) resistance(s) should be identified and efforts directed at their reduction.

PROBLEM 3.150 KNOWN: Diameter and internal fin configuration of copper tubes submerged in water. Tube wall temperature and temperature and convection coefficient of gas flow through the tube. FIND: Rate of heat transfer per tube length. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional fin conduction, (3) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient, (6) Tube wall may be unfolded and represented as a plane wall with four straight, rectangular fins, each with an adiabatic tip. ANALYSIS: The rate of heat transfer per unit tube length is: q′t = ηo hA′t Tg − Ts

(

ηo = 1 −

NA′f (1 − ηf A′t

)

)

NA′f = 4 × 2L = 8 ( 0.025m ) = 0.20m

A′t = NA′f + A′b = 0.20m + (π D − 4t ) = 0.20m + (π × 0.05m − 4 × 0.005m ) = 0.337m For an adiabatic fin tip, q M tanh mL ηf = f = q max h ( 2L ⋅ 1) Tg − Ts

(

M = [h2 (1m + t ) k (1m × t )]

1/ 2

mL = {[h2 (1m + t )]

)

(Tg − Ts ) ≈ 30 W

(

m ⋅ K ( 2m ) 400 W m ⋅ K 0.005m 2

 30 W m 2 ⋅ K ( 2m ) 1/ 2 [k (1m × t )]} L ≈  400 W m ⋅ K 0.005m 2 

(

2

)

1/ 2

 

( 400K ) = 4382W

1/ 2

)

   

0.025m = 0.137

Hence, tanh mL = 0.136, and 4382W ( 0.136 ) 595W ηf = = = 0.992 2 2 600W 30 W m ⋅ K 0.05m ( 400K )

(

ηo = 1 −

0.20 0.337

(

)

(1 − 0.992 ) = 0.995

)

q′t = 0.995 30 W m 2 ⋅ K 0.337m ( 400K ) = 4025 W m

(

)

COMMENTS: Alternatively, q′t = 4q′f + h ( A′t − A′f ) Tg − Ts . Hence, q′ = 4(595 W/m) + 30 W/m2⋅K (0.137 m)(400 K) = (2380 + 1644) W/m = 4024 W/m.

PROBLEM 3.151 KNOWN: Internal and external convection conditions for an internally finned tube. Fin/tube dimensions and contact resistance. FIND: Heat rate per unit tube length and corresponding effects of the contact resistance, number of fins, and fin/tube material. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient on finned surfaces, (6) Tube wall may be unfolded and approximated as a plane surface with N straight rectangular fins. PROPERTIES: Copper: k = 400 W/m⋅K; St.St.: k = 20 W/m⋅K. ANALYSIS: The heat rate per unit length may be expressed as Tg − Tw q′ = R ′t,o(c) + R ′cond + R ′conv,o where

(

)

R t,o(c) = ηo(c) h g A′t , A′t = NA′f + ( 2π r1 − Nt ) , R ′cond =

ln ( r2 r1 ) 2π k

, and

ηo(c) = 1 −

NA′f A′t

A′f = 2r1 ,

 ηf  1 −  ,  C1 

(

(

ηf = tanh mr1 mr1 , m = 2h g kt −1

R ′conv,o = ( 2π r2 h w )

)

C1 = 1 + ηf h g A′f R ′′t,c A′c,b ,

)1/ 2

A′c,b = t ,

.

Using the IHT Performance Calculation, Extended Surface Model for the Straight Fin Array, the following results were obtained. For the base case, q′ = 3857 W/m, where R ′t,o(c) = 0.101 m⋅K/W, R ′cond = 7.25 × 10-5 m⋅K/W and R ′conv,o = 0.00265 m⋅K/W. If the contact resistance is eliminated ( R ′′t,c = 0), q „ = 3922 W/m, where R ′t,o = 0.0993 m⋅K/W. If the number of fins is increased to N = 8, q′ = 5799 W/m, with R ′t,o(c) = 0.063 m⋅K/W. If the material is changed to stainless steel, q′ = 3591

W/m, with R ′t,o(c) = 0.107 m⋅K/W and R ′cond = 0.00145 m⋅K/W. COMMENTS: The small reduction in q „ associated with use of stainless steel is perhaps surprising, in view of the large reduction in k. However, because h g is small, the reduction in k does not significantly reduce the fin efficiency ( ηf changes from 0.994 to 0.891). Hence, the heat rate remains large. The influence of k would become more pronounced with increasing h g .

PROBLEM 3.152 KNOWN: Design and operating conditions of a tubular, air/water heater. FIND: (a) Expressions for heat rate per unit length at inner and outer surfaces, (b) Expressions for inner and outer surface temperatures, (c) Surface heat rates and temperatures as a function of volumetric heating q for prescribed conditions. Upper limit to q . SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Constant properties, (3) One-dimensional heat transfer. PROPERTIES: Table A-1: Aluminum, T = 300 K, k a = 237 W/m⋅K. ANALYSIS: (a) Applying Equation C.8 to the inner and outer surfaces, it follows that 2  qr   ri2  o   1 −  + ( Ts,o − Ts,i ) ln ( ro ri )  4k s  r 2    o   2   2π k s  qr ri2  2 o ′  q ( ro ) = qπ ro −   1 −  + ( Ts,o − Ts,i ) ln ( ro ri )  4k s  r 2    o  

2π k s

q′ ( ri ) = q π ri2 −

< <

(b) From Equations C.16 and C.17, energy balances at the inner and outer surfaces are of the form

(

)

h i T∞,i − Ts,i =

(

 qr i

)

2

Uo Ts,o − T∞,o =

−

 qr o

2

2  qr   ri2  o ks   1 −  + ( Ts,o − Ts,i ) 2 4k   s  ro  

<

ri ln ( ro ri )

−

2  qr   r2  k s  o  1 − i  + Ts,o − Ts,i   4k s  r 2  





o



(

ro ln ( ro ri )

)



<

Accounting for the fin array and the contact resistance, Equation 3.104 may be used to cast the overall heat transfer coefficient U o in the form Uo =

(

q′ ( ro )

A′w Ts,o − T∞,o

)

=

1 A′ = t ηo(c) h o A′w R ′t,o(c) A′w

where ηo(c) is determined from Equations 3.105a,b and A′w = 2π ro . Continued...

PROBLEM 3.152 (Cont.)

Surface temperatures, Ts(K)

(c) For the prescribed conditions and a representative range of 107 ≤ q ≤ 108 W/m3, use of the relations of part (b) with the capabilities of the IHT Performance Calculation Extended Surface Model for a Circular Fin Array yields the following graphical results. 500 460 420 380 340 300 1E7

2E7

3E7

4E7

5E7

6E7

7E7

8E7

9E7

1E8

Heat generation, qdot(W/m^3) Inner surface temperature, Ts,i Outer surface temperature, Ts,o

It is in this range that the upper limit of Ts,i = 373 K is exceeded for q = 4.9 × 107 W/m3, while the corresponding value of Ts,o = 379 K is well below the prescribed upper limit. The expressions of part

Surface heat rates, q'(W/m)

(a) yield the following results for the surface heat rates, where heat transfer in the negative r direction corresponds to q′ ( ri ) < 0. 50000 30000 10000 -10000 -30000 -50000 1E7

2E7

3E7

4E7

5E7

6E7

7E7

8E7

9E7

1E8

Heat generation, qdot(W/m^3) q'(ri) q'(ro)

For q = 4.9 × 107 W/m3, q′ ( ri ) = -2.30 × 104 W/m and q′ ( ro ) = 1.93 × 104 W/m. COMMENTS: The foregoing design provides for comparable heat transfer to the air and water streams. This result is a consequence of the nearly equivalent thermal resistances associated with heat transfer −1 from the inner and outer surfaces. Specifically, R ′conv,i = ( h i 2π ri ) = 0.00318 m⋅K/W is slightly smaller than R′t,o(c) = 0.00411 m⋅K/W, in which case q′ ( ri ) is slightly larger than q′ ( ro ) , while Ts,i is slightly smaller than Ts,o . Note that the solution must satisfy the energy conservation requirement, π ro2 − ri2 q = q′ ( ri ) + q′ ( ro ) .

(

)

PROBLEM 4.1 KNOWN: Method of separation of variables (Section 4.2) for two-dimensional, steady-state conduction.

2

FIND: Show that negative or zero values of λ , the separation constant, result in solutions which cannot satisfy the boundary conditions. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties. 2

ANALYSIS: From Section 4.2, identification of the separation constant λ leads to the two ordinary differential equations, 4.6 and 4.7, having the forms d 2X d 2Y + λ 2X = 0 − λ 2Y = 0 (1,2) 2 2

dx dy and the temperature distribution is

θ ( x,y ) = X ( x ) ⋅ Y (y ).

(3)

2

Consider now the situation when λ = 0. From Eqs. (1), (2), and (3), find that X = C1 + C 2x, Y = C3 + C 4y and θ ( x,y ) = ( C1 + C 2x ) ( C3 + C 4y ) .

(4)

Evaluate the constants - C1, C2, C3 and C4 - by substitution of the boundary conditions: x = 0: θ ( 0,y) = ( C1 + C 2 ⋅ 0 )( C3 + C4 y ) = 0 C1 = 0 y =0: θ ( x,0) = ( 0 + C2 X)( C3 + C4 ⋅ 0 ) = 0 C3 = 0 x = L: θ ( L,0) = ( 0 + C2 L)( 0 + C4 y ) = 0 C2 = 0 y = W: θ ( x,W ) = ( 0 + 0 ⋅ x )( 0 + C4 W ) = 1 0≠1 2

The last boundary condition leads to an impossibility (0 ≠ 1). We therefore conclude that a λ value of zero will not result in a form of the temperature distribution which will satisfy the boundary 2 conditions. Consider now the situation when λ < 0. The solutions to Eqs. (1) and (2) will be

and

X = C5e-λ x + C6e +λ x , Y = C7cos λ y + C8sin λy θ ( x,y ) =  C5 e-λ x + C6 e +λ x  [ C7 cos λ y + C8 sin λ y ] .  

Evaluate the constants for the boundary conditions. y = 0 : θ ( x,0) =  C5 e-λ x + C6 e-λ x  [ C7 cos 0 + C8 sin 0 ] = 0   0 0   x = 0 : θ ( 0,y) =  C5 e + C6 e  [ 0 + C8sin λ y] = 0  

(5,6) (7) C7 = 0 C8 = 0

If C8 = 0, a trivial solution results or C5 = -C6. x = L: θ ( L,y ) = C5  e-xL − e+xL  C8 sin λ y = 0.   From the last boundary condition, we require C5 or C8 is zero; either case leads to a trivial solution with either no x or y dependence.

PROBLEM 4.2 KNOWN: Two-dimensional rectangular plate subjected to prescribed uniform temperature boundary conditions. FIND: Temperature at the mid-point using the exact solution considering the first five non-zero terms; assess error resulting from using only first three terms. Plot the temperature distributions T(x,0.5) and T(1,y). SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties. ANALYSIS: From Section 4.2, the temperature distribution is n +1 + 1  nπ x  sinh ( nπ y L ) T − T1 2 θ ( −1) θ ( x, y ) ≡ sin  . (1,4.19) = ⋅ T2 − T1 π n L  sinh ( nπ W L )  n =1 Considering now the point (x,y) = (1.0,0.5) and recognizing x/L = 1/2, y/L = 1/4 and W/L = 1/2, n +1 + 1  nπ  sinh ( nπ 4 ) T − T1 2 θ ( −1) θ (1, 0.5) ≡ sin  . = ⋅ T2 − T1 π n 2  sinh ( nπ 2 )  n =1 When n is even (2, 4, 6 ...), the corresponding term is zero; hence we need only consider n = 1, 3, 5, 7 and 9 as the first five non-zero terms.

∑

∑

θ (1, 0.5) =

2   π  sinh (π 4 ) 2  3π + sin  2sin   π   2  sinh (π 2 ) 3  2

2  5π sin  5  2

θ (1, 0.5) =

 sinh (5π 4 ) 2  7π + sin    sinh (5π 2 ) 7  2

 sinh (3π 4 ) +   sinh (3π 2 )

 sinh (7π 4 ) 2  9π + sin    sinh (7π 2 ) 9  2

 sinh (9π 4 )     sinh (9π 2 ) 

2 [0.755 − 0.063 + 0.008 − 0.001 + 0.000] = 0.445 π

(2)

T (1, 0.5) = θ (1, 0.5)( T2 − T1 ) + T1 = 0.445 (150 − 50 ) + 50 = 94.5$ C .

<

Using Eq. (1), and writing out the first five terms of the series, expressions for θ(x,0.5) or T(x,0.5) and θ(1,y) or T(1,y) were keyboarded into the IHT workspace and evaluated for sweeps over the x or y variable. Note that for T(1,y), that as y → 1, the upper boundary, T(1,1) is greater than 150°C. Upon examination of the magnitudes of terms, it becomes evident that more than 5 terms are required to provide an accurate solution.

T(x,0.5) or T(1,y), C

If only the first three terms of the series, Eq. (2), are considered, the result will be θ(1,0.5) = 0.46; that is, there is less than a 0.2% effect. 150 130 110 90 70 50 0

0.2

0.4

0.6

0.8

x or y coordinate (m) T(1,y) T(x,0.5)

1

PROBLEM 4.3 KNOWN: Temperature distribution in the two-dimensional rectangular plate of Problem 4.2. FIND: Expression for the heat rate per unit thickness from the lower surface (0 ≤ x ≤ 2, 0) and result based on first five non-zero terms of the infinite series. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties. ANALYSIS: The heat rate per unit thickness from the plate along the lower surface is x =2

x =2

x =2

∂T ∂θ q′out = − ∫ dq′y ( x, 0 ) = − ∫ − k dx = k ( T2 − T1 ) ∫ dx ∂ y y =0 ∂y x =0 x =0 x =0 y =0

(1)

where from the solution to Problem 4.2, n +1 + 1  nπ x  sinh (nπ y L ) T − T1 2 ∞ ( −1) θ≡ sin  . = ∑  T2 − T1 π n  L  sinh ( nπ W L )

(2)

n =1

Evaluate the gradient of θ from Eq. (2) and substitute into Eq. (1) to obtain

q′out = k ( T2 − T1 )

x=2

∫

x =0

n +1 + 1  nπ x  ( nπ L ) cosh ( nπ y L ) 2 ∞ ( −1) sin  ∑  π n sinh ( nπ W L )  L  n =1

dx y =0

n +1 2   +1 2 ∞ ( −1) 1  nπ x    q′out = k ( T2 − T1 ) ∑ − cos   π n sinh ( nπ W L )  L   x =0  n =1   n +1 +1 2 ∞ ( −1) 1 1 − cos ( nπ ) q′out = k ( T2 − T1 ) ∑ π n sinh ( nπ L ) 

<

n =1

To evaluate the first five, non-zero terms, recognize that since cos(nπ) = 1 for n = 2, 4, 6 ..., only the nodd terms will be non-zero. Hence, Continued …..

PROBLEM 4.3 (Cont.) $ q′out = 50 W m ⋅ K (150 − 50 ) C

+

( −1)6 + 1 5

⋅

1 sinh (5π 2 )

( −1) + 1 ⋅ 2  ( −1) + 1 1 1 ⋅ ⋅ (2) (2) +  π 1 sinh (π 2 ) 3 sinh (3π 2 ) 

(2 ) +

2

( −1)8 + 1 7

4

⋅

1 sinh ( 7π 2 )

(2 ) +

( −1)10 + 1 9

⋅

1 sinh ( 9π 2 )

q′out = 3.183kW m [1.738 + 0.024 + 0.00062 + (...)] = 5.611kW m



( 2 ) 

<

COMMENTS: If the foregoing procedure were used to evaluate the heat rate into the upper surface, ′ =− qin

x =2

∫

dq′y ( x, W ) , it would follow that

x =0

q′in = k ( T2 − T1 )

n +1 +1 2 ∞ ( −1) coth ( nπ 2 ) 1 − cos ( nπ ) ∑ π n n =1

However, with coth(nπ/2) ≥ 1, irrespective of the value of n, and with

∞

∑ (−1)n +1 + 1

n =1

n being a

divergent series, the complete series does not converge and q′in → ∞ . This physically untenable condition results from the temperature discontinuities imposed at the upper left and right corners.

PROBLEM 4.4 KNOWN: Rectangular plate subjected to prescribed boundary conditions. FIND: Steady-state temperature distribution. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties. ANALYSIS: The solution follows the method of Section 4.2. The product solution is

(

T ( x,y ) = X ( x ) ⋅ Y ( y ) = ( C1cosλ x + C2 sinλ x ) C3e -λ y + C4e +λ y

)

and the boundary conditions are: T(0,y) = 0, T(a,y) = 0, T(x,0) = 0, T(x.b) = Ax. Applying BC#1, T(0,y) = 0, find C1 = 0. Applying BC#2, T(a,y) = 0, find that λ = nπ/a with n = 1,2,…. Applying BC#3, T(x,0) = 0, find that C3 = -C4. Hence, the product solution is nπ T ( x,y ) = X ( x ) ⋅ Y ( y ) = C2 C4 sin  x  e + λ y − e- λ y .  a  Combining constants and using superposition, find

(

T ( x,y ) =

)

∞

 nπ x   nπ y  Cn sin  sinh    . a a    n =1

∑

To evaluate Cn, use orthogonal functions with Eq. 4.16 to find a nπ x  nπ b  a nπ x  Cn = ∫ Ax ⋅ sin  ⋅ dx/sinh  sin 2  dx, ∫   0  a   a  0  a  noting that y = b. The numerator, denominator and Cn, respectively, are: a

2 2   a 2  nπ x  − ax cos  nπ x   = Aa − cos n π = Aa −1 n+1 , A ∫ x ⋅ sin ⋅ dx = A  sin  ( ) ( ) [ ]  a  nπ  a   0 a nπ nπ   nπ  0 a

nπ x

a 1 a  nπ b  a 2 nπ x  nπ b   1  2nπ x    nπ b  sinh  sin ⋅ dx = sinh  x− sin  = ⋅ sinh  , ∫      0 a 4n π  a   a  2  a  0 2  a 

Aa 2 a nπ b nπ b  . ( −1) n+1 / sinh   = 2Aa ( −1) n+1 / nπ sinh  nπ 2  a   a  Hence, the temperature distribution is  nπ y  sinh  ∞ −1 n+1 2 Aa ( ) ⋅ sin  n π x   a  . T ( x,y ) = ∑  a  nπ b  π n =1 n sinh   a  Cn =

<

PROBLEM 4.5 KNOWN: Long furnace of refractory brick with prescribed surface temperatures and material thermal conductivity. FIND: Shape factor and heat transfer rate per unit length using the flux plot method SCHEMATIC:

ASSUMPTIONS: (1) Furnace length normal to page, l, >> cross-sectional dimensions, (2) Twodimensional, steady-state conduction, (3) Constant properties. ANALYSIS: Considering the cross-section, the cross-hatched area represents a symmetrical element. Hence, the heat rte for the entire furnace per unit length is

q′ =

q S = 4 k ( T1 − T2 ) l l

(1)

where S is the shape factor for the symmetrical section. Selecting three temperature increments ( N = 3), construct the flux plot shown below.

From Eq. 4.26, and from Eq. (1),

S=

Ml N

or

S M 8.5 = = = 2.83 l N 3

q′ = 4 × 2.83 ×1.2

W ( 600 − 60 )o C = 7.34 kW/m. m ⋅K

< <

COMMENTS: The shape factor can also be estimated from the relations of Table 4.1. The symmetrical section consists of two plane walls (horizontal and vertical) with an adjoining edge. Using the appropriate relations, the numerical values are, in the same order,

S=

0.75m 0.5m l + 0.54l + l = 3.04l 0.5m 0.5m

Note that this result compares favorably with the flux plot result of 2.83l.

PROBLEM 4.6 KNOWN: Hot pipe embedded eccentrically in a circular system having a prescribed thermal conductivity. FIND: The shape factor and heat transfer per unit length for the prescribed surface temperatures. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Length l >> diametrical dimensions. ANALYSIS: Considering the cross-sectional view of the pipe system, the symmetrical section shown above is readily identified. Selecting four temperature increments (N = 4), construct the flux plot shown below.

For the pipe system, the heat rate per unit length is q′ =

q W = kS ( T1 − T2 ) = 0.5 × 4.26 (150 − 35 )o C = 245 W/m. l m ⋅K

<

COMMENTS: Note that in the lower, right-hand quadrant of the flux plot, the curvilinear squares are irregular. Further work is required to obtain an improved plot and, hence, obtain a more accurate estimate of the shape factor.

PROBLEM 4.7 KNOWN: Structural member with known thermal conductivity subjected to a temperature difference. FIND: (a) Temperature at a prescribed point P, (b) Heat transfer per unit length of the strut, (c) Sketch the 25, 50 and 75°C isotherms, and (d) Same analysis on the shape but with adiabatic-isothermal boundary conditions reversed. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: (a) Using the methodology of Section 4.3.1, construct a flux plot. Note the line of symmetry which passes through the point P is an isotherm as shown above. It follows that

T ( P ) = ( T1 + T2 ) 2 = (100 + 0 ) C 2 = 50$ C . $

<

(b) The flux plot on the symmetrical section is now constructed to obtain the shape factor from which the heat rate is determined. That is, from Eq. 4.25 and 4.26,

q = kS ( T1 − T2 ) and S = M N .

(1,2)

From the plot of the symmetrical section,

So = 4.2 4 = 1.05 . For the full section of the strut,

M = M o = 4.2 but N = 2No = 8. Hence,

S = So 2 = 0.53 and with q′ = q  , giving

q′  = 75 W m ⋅ K × 0.53 (100 − 0 ) C = 3975 W m . $

<

(c) The isotherms for T = 50, 75 and 100°C are shown on the flux plot. The T = 25°C isotherm is symmetric with the T = 75°C isotherm. (d) By reversing the adiabatic and isothermal boundary conditions, the two-dimensional shape appears as shown in the sketch below. The symmetrical element to be flux plotted is the same as for the strut, except the symmetry line is now an adiabat. Continued...

PROBLEM 4.7 (Cont.)

From the flux plot, find Mo = 3.4 and No = 4, and from Eq. (2)

So = M o N o = 3.4 4 = 0.85

S = 2So = 1.70

and the heat rate per unit length from Eq. (1) is

q′ = 75 W m ⋅ K × 1.70 (100 − 0 ) C = 12, 750 W m $

<

From the flux plot, estimate that T(P) ≈ 40°C.

<

COMMENTS: (1) By inspection of the shapes for parts (a) and (b), it is obvious that the heat rate for the latter will be greater. The calculations show the heat rate is greater by more than a factor of three. (2) By comparing the flux plots for the two configurations, and corresponding roles of the adiabats and isotherms, would you expect the shape factor for parts (a) to be the reciprocal of part (b)?

PROBLEM 4.8 KNOWN: Relative dimensions and surface thermal conditions of a V-grooved channel. FIND: Flux plot and shape factor. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: With symmetry about the midplane, only one-half of the object need be considered as shown below. Choosing 6 temperature increments (N = 6), it follows from the plot that M ≈ 7. Hence from Eq. 4.26, the shape factor for the half section is M 7 S = l = l = 1.17l. N 6 For the complete system, the shape factor is then S = 2.34l.

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PROBLEM 4.9 KNOWN: Long conduit of inner circular cross section and outer surfaces of square cross section. FIND: Shape factor and heat rate for the two applications when outer surfaces are insulated or maintained at a uniform temperature. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties and (3) Conduit is very long. ANALYSIS: The adiabatic symmetry lines for each of the applications is shown above. Using the flux plot methodology and selecting four temperature increments (N = 4), the flux plots are as shown below.

For the symmetrical sections, S = 2So, where So = M  /N and the heat rate for each application is q = 2(So/  )k(T1 - T2).

Application A B

M 10.3 6.2

N 4 4

So/  2.58 1.55

q′ (W/m) 11,588 6,975

< <

COMMENTS: (1) For application A, most of the heat lanes leave the inner surface (T1) on the upper portion. (2) For application B, most of the heat flow lanes leave the inner surface on the upper portion (that is, lanes 1-4). Because the lower, right-hand corner is insulated, the entire section experiences small heat flows (lane 6 + 0.2). Note the shapes of the isotherms near the right-hand, insulated boundary and that they intersect the boundary normally.

PROBLEM 4.10 KNOWN: Shape and surface conditions of a support column. FIND: (a) Heat transfer rate per unit length. (b) Height of a rectangular bar of equivalent thermal resistance. SCHEMATIC:

ASSUMPTIONS: (1)Steady-state conditions, (2) Negligible three-dimensional conduction effects, (3) Constant properties, (4) Adiabatic sides. PROPERTIES: Table A-1, Steel, AISI 1010 (323K): k = 62.7 W/m⋅K. ANALYSIS: (a) From the flux plot for the half section, M ≈ 5 and N ≈ 8. Hence for the full section Ml ≈ 1.25l N q = Sk ( T1 − T2 ) S= 2

q′ ≈ 1.25 × 62.7

W (100 − 0)o C m ⋅K

<

q′ ≈ 7.8 kW/m. (b) The rectangular bar provides for one-dimensional heat transfer. Hence, q=kA

Hence,

( T1 − T2 ) H

(T − T ) = k ( 0.3l ) 1 2 H

(

)

o 0.3k ( T1 − T2 ) 0.3m ( 62.7 W/m ⋅ K ) 100 C H= = = 0.24m. q′ 7800 W/m

<

COMMENTS: The fact that H < 0.3m is consistent with the requirement that the thermal resistance of the trapezoidal column must be less than that of a rectangular bar of the same height and top width (because the width of the trapezoidal column increases with increasing distance, x, from the top). Hence, if the rectangular bar is to be of equivalent resistance, it must be of smaller height.

PROBLEM 4.11 KNOWN: Hollow prismatic bars fabricated from plain carbon steel, 1m in length with prescribed temperature difference. FIND: Shape factors and heat rate per unit length. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties. PROPERTIES: Table A-1, Steel, Plain Carbon (400K), k = 57 W/m⋅K. ANALYSIS: Construct a flux plot on the symmetrical sections (shaded-regions) of each of the bars.

The shape factors for the symmetrical sections are, So,A =

Ml 4 = l = 1l N 4

So,B =

Ml 3.5 = l = 0.88l. N 4

Since each of these sections is ¼ of the bar cross-section, it follows that SA = 4 ×1l = 4l

SB = 4 × 0.88l = 3.5l.

<

The heat rate per unit length is q′ = q/l = k ( S/l )(T1 − T2 ) , q′A = 57

W × 4 (500 − 300 ) K = 45.6 kW/m m⋅K

<

q′B = 57

W × 3.5 ( 500 − 300 ) K = 39.9 kW/m. m ⋅K

<

PROBLEM 4.12 KNOWN: Two-dimensional, square shapes, 1 m to a side, maintained at uniform temperatures as prescribed, perfectly insulated elsewhere. FIND: Using the flux plot method, estimate the heat rate per unit length normal to the page if the thermal conductivity is 50 W/m⋅K ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties. ANALYSIS: Use the methodology of Section 4.3.1 to construct the flux plots to obtain the shape factors from which the heat rates can be calculated. With Figure (a), begin at the lower-left side making the isotherms almost equally spaced, since the heat flow will only slightly spread toward the right. Start sketching the adiabats in the vicinity of the T2 surface. The dashed line represents the adiabat which separates the shape into two segments. Having recognized this feature, it was convenient to identify partial heat lanes. Figure (b) is less difficult to analyze since the isotherm intervals are nearly regular in the lower left-hand corner.

The shape factors are calculated from Eq. 4.26 and the heat rate from Eq. 4.25.

S′ =

M 0.5 + 3 + 0.5 + 0.5 + 0.2 = N 6

S′ =

M 4.5 = = 0.90 N 5

S′ = 0.70

q′ = kS′ ( T1 − T2 )

q′ = kS′ ( T1 − T2 )

q′ = 10 W m ⋅ K × 0.70 (100 − 0 ) K = 700 W m q′ = 10 W m ⋅ K × 0.90 (100 − 0 ) K = 900 W m

<

COMMENTS: Using a finite-element package with a fine mesh, we determined heat rates of 956 and 915 W/m, respectively, for Figures (a) and (b). The estimate for the less difficult Figure (b) is within 2% of the numerical method result. For Figure (a), our flux plot result was 27% low.

PROBLEM 4.13 KNOWN: Uniform media of prescribed geometry. FIND: (a) Shape factor expressions from thermal resistance relations for the plane wall, cylindrical shell and spherical shell, (b) Shape factor expression for the isothermal sphere of diameter D buried in an infinite medium. ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform properties. ANALYSIS: (a) The relationship between the shape factor and thermal resistance of a shape follows from their definitions in terms of heat rates and overall temperature differences.

q = kS∆T

( 4.25 ),

q=

∆T Rt

(3.19 ) ,

S = 1/kR t

(4.27)

Using the thermal resistance relations developed in Chapter 3, their corresponding shape factors are:

Rt =

Plane wall:

Rt =

Cylindrical shell:

ln ( r2 / r1 )

S=

2π Lk

L kA

2π L lnr2 / r1 .

S=

A . L

<

<

(L into the page)

Rt =

Spherical shell:

1 1 1  − 4π k  r1 r2 

S=

4π . l/r1 − l/r2

< (b) The shape factor for the sphere of diameter D in an infinite medium can be derived using the alternative conduction analysis of Section 3.1. For this situation, qr is a constant and Fourier’s law has the form dT q r = −k 4π r 2 .

(

) dr

Separate variables, identify limits and integrate. ∞ T2 q dr − r ∫D / 2 = ∫ T dT 1 4π k r2 D q r = 4π k   ( T1 − T 2 ) 2

qr  1 ∞ q  2 − −  = − r  0 −  = ( T2 − T1 )  4π k  r  D/2 4π k  D or

S = 2π D.

<

COMMENTS: Note that the result for the buried sphere, S = 2πD, can be obtained from the expression for the spherical shell with r2 = ∞. Also, the shape factor expression for the “isothermal sphere buried in a semi-infinite medium” presented in Table 4.1 provides the same result with z → ∞.

PROBLEM 4.14 KNOWN: Heat generation in a buried spherical container. FIND: (a) Outer surface temperature of the container, (b) Representative isotherms and heat flow lines. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Soil is a homogeneous medium with constant properties. PROPERTIES: Table A-3, Soil (300K): k = 0.52 W/m⋅K. ANALYSIS: (a) From an energy balance on the container, q = E& g and from the first entry in Table 4.1, q=

2π D k ( T1 −T 2 ). l − D/4z

Hence, T1 = T2 +

q 1− D/4z 500W 1− 2m/40m = 20o C+ = 92.7 o C W k 2π D 2π ( 2m ) 0.52 m ⋅K

(b) The isotherms may be viewed as spherical surfaces whose center moves downward with increasing radius. The surface of the soil is an isotherm for which the center is at z = ∞.

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PROBLEM 4.15 KNOWN: Temperature, diameter and burial depth of an insulated pipe. FIND: Heat loss per unit length of pipe. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through insulation, two-dimensional through soil, (3) Constant properties, (4) Negligible oil convection and pipe wall conduction resistances. PROPERTIES: Table A-3, Soil (300K): k = 0.52 W/m⋅K; Table A-3, Cellular glass (365K): k = 0.069 W/m⋅K. ANALYSIS: The heat rate can be expressed as T −T q= 1 2 R tot where the thermal resistance is Rtot = Rins + Rsoil. From Eq. 3.28, R ins =

ln ( D2 / D1 )

=

ln ( 0.7m/0.5m )

2π Lk ins 2π L × 0.069 W/m ⋅ K From Eq. 4.27 and Table 4.1, R soil =

=

0.776m ⋅ K/W . L

cosh -1 ( 2z/D2 ) cosh -1 ( 3/0.7 ) 1 0.653 = = = m ⋅ K/W. SKsoil 2π Lksoil 2π × ( 0.52 W/m ⋅ K ) L L

Hence, q=

(120 − 0)o C

1 m ⋅K ( 0.776 + 0.653) L W

q′ = q/L = 84 W/m.

= 84

W ×L m

<

COMMENTS: (1) Contributions of the soil and insulation to the total resistance are approximately the same. The heat loss may be reduced by burying the pipe deeper or adding more insulation. (2) The convection resistance associated with the oil flow through the pipe may be significant, in which case the foregoing result would overestimate the heat loss. A calculation of this resistance may be based on results presented in Chapter 8. (3) Since z > 3D/2, the shape factor for the soil can also be evaluated from S = 2πL/ ln (4z/D) of Table 4.1, and an equivalent result is obtained.

PROBLEM 4.16 KNOWN: Operating conditions of a buried superconducting cable. FIND: Required cooling load. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Two-dimensional conduction in soil, (4) One-dimensional conduction in insulation. ANALYSIS: The heat rate per unit length is q′ = q′ =

Tg − Tn R′g + R′I

Tg − Tn

 kg ( 2π /ln ( 4z/Do ) )   

−1

+ ln ( Do / Di ) / 2π ki

where Tables 3.3 and 4.1 have been used to evaluate the insulation and ground resistances, respectively. Hence, q′ =

( 300 − 77 ) K

(1.2 W/m ⋅ K ) ( 2π /ln ( 8/0.2 ) )    223 K q′ = ( 0.489+22.064 ) m ⋅ K/W q′ = 9.9 W/m.

−1

+ ln ( 2 ) / 2π × 0.005 W/m ⋅ K

<

COMMENTS: The heat gain is small and the dominant contribution to the thermal resistance is made by the insulation.

PROBLEM 4.17 KNOWN: Electrical heater of cylindrical shape inserted into a hole drilled normal to the surface of a large block of material with prescribed thermal conductivity. FIND: Temperature reached when heater dissipates 50 W with the block at 25°C. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Block approximates semi-infinite medium with constant properties, (3) Negligible heat loss to surroundings above block surface, (4) Heater can be approximated as isothermal at T1. ANALYSIS: The temperature of the heater surface follows from the rate equation written as T1 = T2 + q/kS where S can be estimated from the conduction shape factor given in Table 4.1 for a “vertical cylinder in a semi-infinite medium,” S = 2π L/l n ( 4L/D ) . Substituting numerical values, find 4 × 0.1m  S = 2π × 0.1m/l n  = 0.143m.  0.005m  The temperature of the heater is then T1 = 25°C + 50 W/(5 W/m⋅K × 0.143m) = 94.9°C.

<

COMMENTS: (1) Note that the heater has L >> D, which is a requirement of the shape factor expression. (2) Our calculation presumes there is negligible thermal contact resistance between the heater and the medium. In practice, this would not be the case unless a conducting paste were used. (3) Since L >> D, assumption (3) is reasonable. (4) This configuration has been used to determine the thermal conductivity of materials from measurement of q and T1.

PROBLEM 4.18 KNOWN: Surface temperatures of two parallel pipe lines buried in soil. FIND: Heat transfer per unit length between the pipe lines. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties, (4) Pipe lines are buried very deeply, approximating burial in an infinite medium, (5) Pipe length >> D1 or D2 and w > D1 or D2. ANALYSIS: The heat transfer rate per unit length from the hot pipe to the cool pipe is q′ =

q S = k ( T1 −T 2 ) . L L

The shape factor S for this configuration is given in Table 4.1 as S=

2π L  4w 2 − D 2 − D 2  -1 1 2 cosh  2D1D2    

.

Substituting numerical values,  4 × ( 0.5m ) 2 − ( 0.1m ) 2 − ( 0.075m ) 2  S − 1  = 2π /cosh -1(65.63) = 2π /cosh  L  2 × 0.1m × 0.075m    S = 2π /4.88 = 1.29. L Hence, the heat rate per unit length is o

q′ = 1.29 × 0.5W/m ⋅ K (175 − 5) C = 110 W/m.

<

COMMENTS: The heat gain to the cooler pipe line will be larger than 110 W/m if the soil temperature is greater than 5°C. How would you estimate the heat gain if the soil were at 25°C?

PROBLEM 4.19 KNOWN: Tube embedded in the center plane of a concrete slab. FIND: (a) The shape factor and heat transfer rate per unit length using the appropriate tabulated relation, (b) Shape factor using flux plot method. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Constant properties, (4) Concrete slab infinitely long in horizontal plane, L >> z. PROPERTIES: Table A-3, Concrete, stone mix (300K): k = 1.4 W/m⋅K. ANALYSIS: (a) If we relax the restriction that z >> D/2, the embedded tube-slab system corresponds to the fifth case of Table 4.1. Hence,

S=

2π L ln (8z/ π D )

where L is the length of the system normal to the page, z is the half-thickness of the slab and D is the diameter of the tube. Substituting numerical values, find

S = 2π L/l n( 8× 50mm/π 50mm ) = 6.72L. Hence, the heat rate per unit length is

q′ =

q S W = k ( T1 − T2 ) = 6.72 ×1.4 ( 85 − 20 )o C = 612 W. L L m ⋅K

(b) To find the shape factor using the flux plot method, first identify the symmetrical section bounded by the symmetry adiabats formed by the horizontal and vertical center lines. Selecting four temperature increments (N = 4), the flux plot can then be constructed.

From Eq. 4.26, the shape factor of the symmetrical section is

So = ML/N = 6L/4 = 1.5L. For the tube-slab system, it follows that S = 4So = 6.0L, which compares favorably with the result obtained from the shape factor relation.

PROBLEM 4.20 KNOWN: Dimensions and boundary temperatures of a steam pipe embedded in a concrete casing. FIND: Heat loss per unit length. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible steam side convection resistance, pipe wall resistance and contact resistance (T1 = 450K), (3) Constant properties. PROPERTIES: Table A-3, Concrete (300K): k = 1.4 W/m⋅K. ANALYSIS: The heat rate can be expressed as q = Sk∆T1-2 = Sk ( T1 − T2 ) From Table 4.1, the shape factor is S=

2π L . 1.08 w  ln   D 

Hence,  q  2π k ( T1 − T2 ) q′ =   =  L  ln 1.08 w   D  q′ =

2π × 1.4W/m ⋅ K × ( 450 − 300 ) K = 1122 W/m. 1.08 × 1.5m   ln   0.5m 

<

COMMENTS: Having neglected the steam side convection resistance, the pipe wall resistance, and the contact resistance, the foregoing result overestimates the actual heat loss.

PROBLEM 4.21 KNOWN: Thin-walled copper tube enclosed by an eccentric cylindrical shell; intervening space filled with insulation. FIND: Heat loss per unit length of tube; compare result with that of a concentric tube-shell arrangement. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Thermal resistances of copper tube wall and outer shell wall are negligible, (4) Two-dimensional conduction in insulation. ANALYSIS: The heat loss per unit length written in terms of the shape factor S is q′ = k ( S/ l )(T1 − T2 ) and from Table 4.1 for this geometry,

 D2 + d 2 − 4z 2  S = 2π /cosh-1  . l 2Dd   Substituting numerical values, all dimensions in mm,

120 2 + 30 2 − 4 ( 20 ) 2  S -1  = 2π /cosh-1 (1.903) = 4.991. = 2π /cosh  l  2 ×120 × 30    Hence, the heat loss is o

q′ = 0.05W/m ⋅ K × 4.991( 85 − 35 ) C = 12.5 W/m.

<

If the copper tube were concentric with the shell, but all other conditions were the same, the heat loss would be

q′c =

2π k ( T1 − T2 ) ln ( D2 / D1 )

using Eq. 3.27. Substituting numerical values,

(

W ( 85 − 35 )o C/ln 120 / 30 m⋅ K q′c = 11.3 W/m. q′c = 2 π × 0.05

)

COMMENTS: As expected, the heat loss with the eccentric arrangement is larger than that for the concentric arrangement. The effect of the eccentricity is to increase the heat loss by (12.5 - 11.3)/11.3 ≈ 11%.

PROBLEM 4.22 KNOWN: Cubical furnace, 350 mm external dimensions, with 50 mm thick walls. FIND: The heat loss, q(W). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties. PROPERTIES: Table A-3, Fireclay brick ( T = (T1 + T2 ) / 2 = 610K ) : k ≈ 1.1 W/m ⋅ K. ANALYSIS: Using relations for the shape factor from Table 4.1, A 0.25 × 0.25m 2 = = 1.25m L 0.05m

Plane Walls (6)

SW =

Edges (12)

SE = 0.54D = 0.54 × 0.25m = 0.14m

Corners (8)

SC = 0.15L = 0.15 × 0.05m = 0.008m.

The heat rate in terms of the shape factors is q = kS ( T1 − T2 ) = k ( 6S W + 12SE + 8SC ) ( T1 − T2 ) W q = 1.1 ( 6 ×1.25m+12 × 0.14m+0.15 × 0.008m ) ( 600 − 75 )o C m⋅K q = 5.30 kW. COMMENTS: Note that the restrictions for SE and SC have been met.

<

PROBLEM 4.23 KNOWN: Dimensions, thermal conductivity and inner surface temperature of furnace wall. Ambient conditions. FIND: Heat loss. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Uniform convection coefficient over entire outer surface of container. ANALYSIS: From the thermal circuit, the heat loss is

q=

Ts,i − T∞ R cond(2D) + R conv

where Rconv = 1/hAs,o = 1/6(hW2) = 1/6[5 W/m2⋅K(5 m)2] = 0.00133 K/W. From Eq. (4.27), the twodimensional conduction resistance is

R cond(2D) =

1 Sk

where the shape factor S must include the effects of conduction through the 8 corners, 12 edges and 6 plane walls. Hence, using the relations for Cases 8 and 9 of Table 4.1,

S = 8 (0.15 L ) + 12 × 0.54 ( W − 2L ) + 6 As,i L where As,i = (W - 2L)2. Hence,

S = 8 (0.15 × 0.35 ) + 12 × 0.54 ( 4.30 ) + 6 (52.83) m S = ( 0.42 + 27.86 + 316.98 ) m = 345.26m and Rcond(2D) = 1/(345.26 m × 1.4 W/m⋅K) = 0.00207 K/W. Hence $ 1100 − 25 ) C ( q= (0.00207 + 0.00133) K

W

= 316 kW

<

COMMENTS: The heat loss is extremely large and measures should be taken to insulate the furnace.

PROBLEM 4.24 KNOWN: Platen heated by passage of hot fluid in poor thermal contact with cover plates exposed to cooler ambient air. FIND: (a) Heat rate per unit thickness from each channel, q ′i , (b) Surface temperature of cover plate, Ts , (c) q′i and Ts if lower surface is perfectly insulated, (d) Effect of changing centerline spacing on q′i and Ts SCHEMATIC: D=15 mm LA=30 mm Ti=150°C

Lo=60 mm LB=7.5 mm 2 hi=1000 W/m ⋅K 2

T∞=25°C ho=200 W/m ⋅K kA=20 W/m⋅K kB=75 W/m⋅K R ′′t,c = 2.0 ×10

−4

2

m ⋅ K/W

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction in platen, but one-dimensional in coverplate, (3) Temperature of interfaces between A and B is uniform, (4) Constant properties. ANALYSIS: (a) The heat rate per unit thickness from each channel can be determined from the following thermal circuit representing the quarter section shown.

The value for the shape factor is S′ = 1.06 as determined from the flux plot shown on the next page. Hence, the heat rate is q′i = 4 ( Ti − T∞ ) / R′tot (1) R ′tot = [1/1000 W/m 2 ⋅ K ( π 0.015m/4 ) + 1/20 W/m ⋅ K ×1.06

+ 2.0 × 10−4 m2 ⋅ K/W ( 0.060m/2 ) + 0.0075m/75 W/m ⋅ K ( 0.060m/2) + 1/200 W/m2 ⋅ K ( 0.060m/2 )]

R ′tot = [0.085 + 0.047 + 0.0067 + 0.0033 + 0.1667 ] m ⋅ K/W R ′tot = 0.309 m ⋅ K/W q′i = 4 (150 − 25) K/0.309 m ⋅ K/W = 1.62 kW/m.

<

(b) The surface temperature of the cover plate also follows from the thermal circuit as Ts − T∞ q′i / 4 = 1/ho ( Lo / 2 ) Continued …..

(2)

PROBLEM 4.24 (Cont.) q′ 1 1.62 kW Ts = T∞ + i = 25o C + × 0.167 m ⋅ K/W 4 ho ( Lo / 2 ) 4 Ts = 25o C + 67.6o C ≈ 93oC.

<

(c,d) The effect of the centerline spacing on q′i and Ts can be understood by examining the relative magnitudes of the thermal resistances. The dominant resistance is that due to the ambient air convection process which is inversely related to the spacing Lo. Hence, from Eq. (1), the heat rate will increase nearly linearly with an increase in Lo, q′i ~

1 1 ≈ ~ L o. R ′tot 1 / h o ( Lo / 2 )

From Eq. (2), find q′ 1 -1 ∆T = Ts − T∞ = i ~ qi′ ⋅ L-1 o ~ Lo ⋅ Lo ≈ 1. 4 h o ( Lo / 2 ) Hence we conclude that ∆T will not increase with a change in Lo. Does this seem reasonable? What effect does Lo have on Assumptions (2) and (3)? If the lower surface were insulated, the heat rate would be decreased nearly by half. This follows again from the fact that the overall resistance is dominated by the surface convection process. The temperature difference, Ts - T∞, would only increase slightly.

PROBLEM 4.25 KNOWN: Long constantan wire butt-welded to a large copper block forming a thermocouple junction on the surface of the block. FIND: (a) The measurement error (Tj - To) for the thermocouple for prescribed conditions, and (b) Compute and plot (Tj - To) for h = 5, 10 and 25 W/m2⋅K for block thermal conductivity 15 ≤ k ≤ 400 W/m⋅K. When is it advantageous to use smaller diameter wire? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Thermocouple wire behaves as a fin with constant heat transfer coefficient, (3) Copper block has uniform temperature, except in the vicinity of the junction. PROPERTIES: Table A-1, Copper (pure, 400 K), kb = 393 W/m⋅K; Constantan (350 K), kt ≈ 25 W/m⋅K. ANALYSIS: The thermocouple wire behaves as a long fin permitting heat to flow from the surface thereby depressing the sensing junction temperature below that of the block To. In the block, heat flows into the circular region of the wire-block interface; the thermal resistance to heat flow within the block is approximated as a disk of diameter D on a semi-infinite medium (kb, To). The thermocouple-block combination can be represented by a thermal circuit as shown above. The thermal resistance of the fin follows from the heat rate expression for an infinite fin, Rfin = (hPktAc)-1/2. From Table 4.1, the shape factor for the disk-on-a-semi-infinite medium is given as S = 2D and hence Rblock = 1/kbS = 1/2kbD. From the thermal circuit, R block 1.27 To − Tj = (To − T∞ ) = (125 − 25 )$ C ≈ 0.001(125 − 25 )$ C = 0.1$ C . R fin + R block 1273 + 1.27

<

with P = πD and Ac = πD2/4 and the thermal resistances as

(

 R fin =  10 W m 2 ⋅ K (π 4 ) 25 W m ⋅ K × 1 × 10−3 m 

)  3

−1/ 2

= 1273 K W

R block = (1 2 ) × 393 W m ⋅ K × 10−3 m = 1.27 K W . (b) We keyed the above equations into the IHT workspace, performed a sweep on kb for selected values of h and created the plot shown. When the block thermal conductivity is low, the error (To - Tj) is larger, increasing with increasing convection coefficient. A smaller diameter wire will be advantageous for low values of kb and higher values of h. 5

Error, To-Tj (C)

4 3 2 1 0 0

100

200

300

Block thermal conductivity, kb (W/m.K) h = 25 W/m^2.K; D = 1 mm h = 10 W/m^2.K; D = 1mm h = 5 W/m^2.K; D = 1mm h = 25 W/m^2.K; D = 5 mm

400

PROBLEM 4.26 KNOWN: Dimensions, shape factor, and thermal conductivity of square rod with drilled interior hole. Interior and exterior convection conditions. FIND: Heat rate and surface temperatures. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, (3) Uniform convection coefficients at inner and outer surfaces. ANALYSIS: The heat loss can be expressed as

q=

T∞,1 − T∞,2 R conv,1 + R cond(2D) + R conv,2

where −1

R conv,1 = ( h1π D1L )

−1

R cond(2D) = (Sk )

(

= 50 W m 2 ⋅ K × π × 0.25 m × 2 m −1

= (8.59 m × 150 W m ⋅ K ) −1

R conv,2 = ( h 2 × 4wL )

(

)

−1

= 0.01273K W

= 0.00078 K W

= 4 W m 2 ⋅ K × 4 m × 1m

)

−1

= 0.0625 K W

Hence, $ 300 − 25 ) C ( q= = 3.62 kW

0.076 K W

T1 = T∞,1 − qR conv,1 = 300$ C − 46$ C = 254$ C T2 = T∞,2 + qR conv,2 = 25$ C + 226$ C = 251$ C

< < <

COMMENTS: The largest resistance is associated with convection at the outer surface, and the conduction resistance is much smaller than both convection resistances. Hence, (T2 - T∞,2) > (T∞,1 - T1) >> (T1 - T2).

PROBLEM 4.27 KNOWN: Long fin of aluminum alloy with prescribed convection coefficient attached to different base materials (aluminum alloy or stainless steel) with and without thermal contact resistance R′′t, j at the junction. FIND: (a) Heat rate qf and junction temperature Tj for base materials of aluminum and stainless steel, (b) Repeat calculations considering thermal contact resistance, R′′t, j , and (c) Plot as a function of h for the range 10 ≤ h ≤ 1000 W/m2⋅K for each base material. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Infinite fin. PROPERTIES: (Given) Aluminum alloy, k = 240 W/m⋅K, Stainless steel, k = 15 W/m⋅K. ANALYSIS: (a,b) From the thermal circuits, the heat rate and junction temperature are

T −T Tb − T∞ qf = b ∞ = R tot R b + R t, j + R f

(1)

Tj = T∞ + qf R f

(2)

and, with P = πD and Ac = πD2/4, from Tables 4.1 and 3.4 find −1

R b = 1 Sk b = 1 ( 2Dk b ) = ( 2 × 0.005 m × k b )

R t, j = R ′′t, j A c = 3 ×10−5 m 2 ⋅ K W π ( 0.005 m )

2

−1/ 2

R f = ( hPkA c )

Base Al alloy St. steel

4 = 1.528 K W

2 = 50 W m 2 ⋅ K π 2 (0.005 m ) 240 W m ⋅ K 4   

Rb (K/W) 0.417 6.667

Without R′′t, j qf (W) Tj (°C) 4.46 98.2 3.26 78.4

−1/ 2

= 16.4 K W

With R′′t, j qf (W) 4.09 3.05

Tj (°C) 92.1 75.1

(c) We used the IHT Model for Extended Surfaces, Performance Calculations, Rectangular Pin Fin to calculate qf for 10 ≤ h ≤ 100 W/m2⋅K by replacing R ′′tc (thermal resistance at fin base) by the sum of the contact and spreading resistances, R′′t, j + R ′′b .

Continued...

PROBLEM 4.27 (Cont.)

Fin heat rate, qf (W)

6

5

4

3

2

1 0

20

40

60

80

100

Convection coefficient, h (W/m^2.K) Base material - aluminum alloy Base material - stainless steel

COMMENTS: (1) From part (a), the aluminum alloy base material has negligible effect on the fin heat rate and depresses the base temperature by only 2°C. The effect of the stainless steel base material is substantial, reducing the heat rate by 27% and depressing the junction temperature by 25°C. (2) The contact resistance reduces the heat rate and increases the temperature depression relatively more with the aluminum alloy base. (3) From the plot of qf vs. h, note that at low values of h, the heat rates are nearly the same for both materials since the fin is the dominant resistance. As h increases, the effect of R ′′b becomes more important.

PROBLEM 4.28 KNOWN: Igloo constructed in hemispheric shape sits on ice cap; igloo wall thickness and inside/outside convection coefficients (hi, ho) are prescribed. FIND: (a) Inside air temperature T∞,i when outside air temperature is T∞,o = -40°C assuming occupants provide 320 W within igloo, (b) Perform parameter sensitivity analysis to determine which variables have significant effect on Ti. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Convection coefficient is the same on floor and ceiling of igloo, (3) Floor and ceiling are at uniform temperature, (4) Floor-ice cap resembles disk on semi-infinite medium, (5) One-dimensional conduction through igloo walls. PROPERTIES: Ice and compacted snow (given): k = 0.15 W/m⋅K. ANALYSIS: (a) The thermal circuit representing the heat loss from the igloo to the outside air and through the floor to the ice cap is shown above. The heat loss is T∞,i − T∞,o T∞,i − Tic q= . + R cv,c + R wall + R cv,o R cv,f + R cap 2

Convection, ceiling:

R cv,c =

Convection, outside:

R cv,o =

Convection, floor:

R cv,f =

Conduction, wall:

R wall = 2 

Conduction, ice cap:

R cap =

( )

h i 4π ri2

=

2

( )

h o 4π ro2 1

( )

h i π ri2

=

2

=

2

2 15 W m 2 ⋅ K × 4π ( 2.3 m )

2

1 6 W m 2 ⋅ K × π (1.8 m )

2

= 0.00201K W

= 0.01637 K W

1  1  1 1    1 − 1  m  = 0.1281K W −  = 2       4π × 0.15 W m ⋅ K  1.8 2.3    4π k  ri ro 

1

=

1

1

=

kS 4kri 4 × 0.15 W m ⋅ K × 1.8 m where S was determined from the shape factor of Table 4.1. Hence, T∞,i − ( −40 ) C $

q = 320 W =

= 0.00819 K W

6 W m ⋅ K × 4π (1.8 m ) 2

( 0.00818 + 0.1281 + 0.0020 ) K

= 0.9259 K W

T∞,i − ( −20 ) C $

W

320 W = 7.232( T∞,i + 40) + 1.06( T∞,i + 20)

+

( 0.01637 + 0.9259 ) K T∞,i = 1.1°C.

W

< Continued...

PROBLEM 4.28 (Cont.) (b) Begin the parameter sensitivity analysis to determine important variables which have a significant influence on the inside air temperature by examining the thermal resistances associated with the processes present in the system and represented by the network. Process Convection, outside Conduction, wall Convection, ceiling Convection, floor Conduction, ice cap

Symbols Rcv,o Rwall Rcv,c Rcv,f Rcap

Value (K/W) 0.0020 0.1281 0.0082 0.0164 0.9259

R21 R32 R43 R54 R65

It follows that the convection resistances are negligible relative to the conduction resistance across the igloo wall. As such, only changes to the wall thickness will have an appreciable effect on the inside air temperature relative to the outside ambient air conditions. We don’t want to make the igloo walls thinner and thereby allow the air temperature to dip below freezing for the prescribed environmental conditions. Using the IHT Thermal Resistance Network Model, we used the circuit builder to construct the network and perform the energy balances to obtain the inside air temperature as a function of the outside convection coefficient for selected increased thicknesses of the wall.

Air temperature, Tinfi (C)

25

20

15

10

5

0 0

20

40

60

80

100

Outside coefficient, ho (W/m^2.K) Wall thickness, (ro-ri) = 0.5 m (ro-ri) = 0.75 m (ro-ri) = 1.0 m

COMMENTS: (1) From the plot, we can see that the influence of the outside air velocity which controls the outside convection coefficient ho is negligible. (2) The thickness of the igloo wall is the dominant thermal resistance controlling the inside air temperature.

PROBLEM 4.29 KNOWN: Diameter and maximum allowable temperature of an electronic component. Contact resistance between component and large aluminum heat sink. Temperature of heat sink and convection conditions at exposed component surface. FIND: (a) Thermal circuit, (b) Maximum operating power of component. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible heat loss from sides of chip. ANALYSIS: (a) The thermal circuit is:

where R2D,cond is evaluated from the shape factor S = 2D of Table 4.1. (b) Performing an energy balance for a control surface about the component, Tc − Tb P = q air + q sink = h π D2 / 4 ( Tc − T∞ ) + R ′′t,c / π D2 / 4 + 1/2Dk

(

)

P = 25 W/m 2 ⋅ K (π /4 )( 0.01 m ) 2 75o C +

P = 0.15 W +

(

)

75o C  0.5 ×10 -4 / (π / 4 )( 0.01)2  + ( 0.02 × 237 )−1 K/W  

{

75o C = 0.15 W + 88.49 W=88.6 W. ( 0.64+0.21) K/W

}

<

COMMENTS: The convection resistance is much larger than the cumulative contact and conduction resistance. Hence, virtually all of the heat dissipated in the component is transferred through the block. The two-dimensional conduction resistance is significantly underestimated by use of the shape factor S = 2D. Hence, the maximum allowable power is less than 88.6 W.

PROBLEM 4.30 KNOWN: Disc-shaped electronic devices dissipating 100 W mounted to aluminum alloy block with prescribed contact resistance. FIND: (a) Temperature device will reach when block is at 27°C assuming all the power generated by the device is transferred by conduction to the block and (b) For the operating temperature found in part (a), the permissible operating power with a 30-pin fin heat sink. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Device is at uniform temperature, T1, (3) Block behaves as semi-infinite medium. PROPERTIES: Table A.1, Aluminum alloy 2024 (300 K): k = 177 W/m⋅K. ANALYSIS: (a) The thermal circuit for the conduction heat flow between the device and the block shown in the above Schematic where Re is the thermal contact resistance due to the epoxy-filled interface,

R e = R ′′t,c A c = R ′′t,c

(π D2 4)

(

R e = 5 × 10−5 K ⋅ m 2 W π ( 0.020 m )

2

) 4 = 0.159 K W

The thermal resistance between the device and the block is given in terms of the conduction shape factor, Table 4.1, as

R b = 1 Sk = 1 ( 2Dk ) R b = 1 ( 2 × 0.020 m × 177 W m ⋅ K ) = 0.141K W From the thermal circuit,

T1 = T2 + qd ( R b + R e ) T1 = 27$ C + 100 W ( 0.141 + 0.159 ) K W T1 = 27$ C + 30$ C = 57$ C

<

(b) The schematic below shows the device with the 30-pin fin heat sink with fins and base material of copper (k = 400 W/m⋅K). The airstream temperature is 27°C and the convection coefficient is 1000 W/m2⋅K. Continued...

PROBLEM 4.30 (Cont.)

The thermal circuit for this system has two paths for the device power: to the block by conduction, qcd, and to the ambient air by conduction to the fin array, qcv,

qd =

T1 − T2 T1 − T∞ + R b + R e R e + R c + R fin

(3)

where the thermal resistance of the fin base material is

Rc =

Lc 0.005 m = = 0.03979 K W k c Ac 400 W m ⋅ K π 0.022 4 m 2

)

(

(4)

and Rfin represents the thermal resistance of the fin array (see Section 3.6.5),

1 ηo hA t

(5, 3.103)

NAf (1 − ηf ) At

(6, 3.102)

R fin = R t,o =

ηo = 1 −

where the fin and prime surface area is

A t = NAf + A b

(3.99)

(

)

A t = N (π Df L ) + π Dd2 4 − N π Df2 4    where Af is the fin surface area, Dd is the device diameter and Df is the fin diameter. 2 2 A t = 30 (π × 0.0015 m × 0.015 m ) + π (0.020 m ) 4 − 30 π (0.0015 m ) 4   

(

)

At = 0.06362 m2 + 0.0002611 m2 = 0.06388 m2 Using the IHT Model, Extended Surfaces, Performance Calculations, Rectangular Pin Fin, find the fin efficiency as ηf = 0.8546 (7) Continued...

PROBLEM 4.30 (Cont.) Substituting numerical values into Eq. (6), find

ηo = 1 −

30 × π × 0.0015 m × 0.015 m 0.06388 m 2

(1 − 0.8546 )

ηo = 0.8552 and the fin array thermal resistance is

R fin =

1 0.8552 ×1000 W m 2 ⋅ K × 0.06388 m 2

= 0.01831K W

Returning to Eq. (3), with T1 = 57°C from part (a), the permissible heat rate is

qd =

(57 − 27 )$ C (0.141 + 0.159 ) K

W

+

(57 − 27 )$ C

(0.159 + 0.03979 + 0.01831) K

W

qd = 100 W + 138.2 W = 238 W

<

COMMENTS: (1) Recognize in the part (b) analysis, that thermal resistances of the fin base and array are much smaller than the resistance due to the epoxy contact interfaces. (2) In calculating the fin efficiency, ηf, using the IHT Model it is not necessary to know the base temperature as ηf depends only upon geometric parameters, thermal conductivity and the convection coefficient.

PROBLEM 4.31 KNOWN: Dimensions and surface temperatures of a square channel. Number of chips mounted on outer surface and chip thermal contact resistance. FIND: Heat dissipation per chip and chip temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady state, (2) Approximately uniform channel inner and outer surface temperatures, (3) Two-dimensional conduction through channel wall (negligible end-wall effects), (4) Constant thermal conductivity. ANALYSIS: The total heat rate is determined by the two-dimensional conduction resistance of the channel wall, q = (T2 – T1)/Rt,cond(2D), with the resistance determined by using Eq. 4.27 with Case 11 of Table 4.1. For W/w = 1.6 > 1.4 R t,cond(2D) =

0.930 ln ( W / w ) − 0.050 2π L k

=

0.387 2π ( 0.160m ) 240 W / m ⋅ K

= 0.00160 K / W

The heat rate per chip is then

qc =

(50 − 20 ) °C = 156.3 W T2 − T1 = N R t,cond ( 2D ) 120 ( 0.0016 K / W )

<

and, with qc = (Tc – T2)/Rt,c, the chip temperature is

Tc = T2 + R t,c qc = 50°C + ( 0.2 K / W )156.3 W = 81.3°C

<

COMMENTS: (1) By acting to spread heat flow lines away from a chip, the channel wall provides an excellent heat sink for dissipating heat generated by the chip. However, recognize that, in practice, there will be temperature variations on the inner and outer surfaces of the channel, and if the prescribed values of T1 and T2 represent minimum and maximum inner and outer surface temperatures, respectively, the rate is overestimated by the foregoing analysis. (2) The shape factor may also be determined by combining the expression for a plane wall with the result of Case 8 (Table 4.1). With S = [4(wL)/(W-w)/2] + 4(0.54 L) = 2.479 m, Rt,cond(2D) = 1/(Sk) = 0.00168 K/W.

PROBLEM 4.32 KNOWN: Dimensions and thermal conductivity of concrete duct. Convection conditions of ambient air. Inlet temperature of water flow through the duct. FIND: (a) Heat loss per duct length near inlet, (b) Minimum allowable flow rate corresponding to maximum allowable temperature rise of water. SCHEMATIC:

ASSUMPTIONS: (1) Steady state, (2) Negligible water-side convection resistance, pipe wall conduction resistance, and pipe/concrete contact resistance (temperature at inner surface of concrete corresponds to that of water), (3) Constant properties, (4) Negligible flow work and kinetic and potential energy changes. ANALYSIS: (a) From the thermal circuit, the heat loss per unit length near the entrance is

q′ =

Ti − T∞ Ti − T∞ = R ′cond( 2D ) + R ′conv ln (1.08 w / D ) 1 + 2π k h ( 4w )

where R ′cond ( 2D ) is obtained by using the shape factor of Case 6 from Table 4.1 with Eq. (4.27). Hence, q′ =

90°C (90 − 0 ) °C = ln (1.08 × 0.3m / 0.15m ) 1 (0.0876 + 0.0333 ) K ⋅ m / W + 2 2π (1.4 W / m ⋅ K ) 25 W / m ⋅ K (1.2m )

= 745 W / m

<

(b) From Eq. (1.11e), with q = q′L and ( Ti − To ) = 5°C,

 = m

745 W / m (100m ) q′L q′L = = = 3.54 kg / s u i − u o c (Ti − To ) 4207 J / kg ⋅ K (5°C )

<

COMMENTS: The small reduction in the temperature of the water as it flows from inlet to outlet induces a slight departure from two-dimensional conditions and a small reduction in the heat rate per  is therefore obtained in part (b). unit length. A slightly conservative value (upper estimate) of m

PROBLEM 4.33 KNOWN: Dimensions and thermal conductivities of a heater and a finned sleeve. Convection conditions on the sleeve surface. FIND: (a) Heat rate per unit length, (b) Generation rate and centerline temperature of heater, (c) Effect of fin parameters on heat rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady state, (2) Constant properties, (3) Negligible contact resistance between heater and sleeve, (4) Uniform convection coefficient at outer surfaces of sleeve, (5) Uniform heat generation, (6) Negligible radiation. ANALYSIS: (a) From the thermal circuit, the desired heat rate is

q′ =

Ts − T∞ T −T = s ∞ R ′cond( 2D ) + R ′t,o R ′tot

The two-dimensional conduction resistance, may be estimated from Eq. (4.27) and Case 6 of Table 4.2

R ′cond ( 2D ) =

ln (1.08w / D ) ln ( 2.16 ) 1 = = = 5.11× 10−4 m ⋅ K / W S′k s 2π k s 2π ( 240 W / m ⋅ K )

The thermal resistance of the fin array is given by Eq. (3.103), where ηo and At are given by Eqs. 1/2 (3.102) and (3.99) and ηf is given by Eq. (3.89). With Lc = L + t/2 = 0.022 m, m = (2h/kst) = 32.3 -1 m and mLc = 0.710,

ηf =

tanh mLc 0.61 = = 0.86 mLc 0.71

A′t = NA′f + A′b = N ( 2L + t ) + ( 4w − Nt ) = 0.704m + 0.096m = 0.800m

ηo = 1 −

NA′f 0.704m 1 − ηf ) = 1 − ( (0.14 ) = 0.88 A′t 0.800m −1

R ′t,o = (ηo h A′t ) q′ =

(

(

= 0.88 × 500 W / m 2 ⋅ K × 0.80m

(300 − 50 ) °C

)

5.11× 10−4 + 2.84 × 10−3 m ⋅ K / W

)

−1

= 2.84 ×10−3 m ⋅ K / W

= 74, 600 W / m

 if h is replaced by an overall coefficient based on the (b) Eq. (3.55) may be used to determine q, surface area of the heater. From Eq. (3.32),

<

PROBLEM 4.33 (Cont.) −1

Us A′s = Usπ D = ( R ′tot )

−1

= (3.35 m ⋅ K / W )

= 298 W / m ⋅ K

Us = 298 W / m ⋅ K / (π × 0.02m ) = 4750 W / m 2 ⋅ K

)

(

q = 4 Us ( Ts − T∞ ) / D = 4 4750 W / m 2 ⋅ K ( 250°C ) / 0.02m = 2.38 × 108 W / m3

<

From Eq. (3.53) the centerline temperature is

T (0 ) =

2 q ( D / 2 )

4 kh

2.38 × 108 W / m3 (0.01m )

2

+ Ts =

4 ( 400 W / m ⋅ K )

+ 300°C = 315°C

<

(c) Subject to the prescribed constraints, the following results have been obtained for parameter variations corresponding to 16 ≤ N ≤ 40, 2 ≤ t ≤ 8 mm and 20 ≤ L ≤ 40 mm. N 16 16 28 32 40 40

t(mm) 4 8 4 3 2 2

L(mm)

ηf

q′ ( W / m )

20 20 20 20 20 40

0.86 0.91 0.86 0.83 0.78 0.51

74,400 77,000 107,900 115,200 127,800 151,300

Clearly there is little benefit to simply increasing t, since there is no change in A′t and only a marginal increase in ηf . However, due to an attendant increase in A′t , there is significant benefit to increasing N for fixed t (no change in ηf ) and additional benefit in concurrently increasing N while

decreasing t. In this case the effect of increasing A′t exceeds that of decreasing ηf . The same is true for increasing L, although there is an upper limit at which diminishing returns would be reached. The upper limit to L could also be influenced by manufacturing constraints. COMMENTS: Without the sleeve, the heat rate would be q′ = π Dh ( Ts − T∞ ) = 7850 W / m,

which is well below that achieved by using the increased surface area afforded by the sleeve.

PROBLEM 4.34 KNOWN: Dimensions of chip array. Conductivity of substrate. Convection conditions. Contact resistance. Expression for resistance of spreader plate. Maximum chip temperature. FIND: Maximum chip heat rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Constant thermal conductivity, (3) Negligible radiation, (4) All heat transfer is by convection from the chip and the substrate surface (negligible heat transfer from bottom or sides of substrate). ANALYSIS: From the thermal circuit,

T − T∞ Th − T∞ q = q h + qsp = h + R h,cnv R t (sp ) + R t,c + R sp,cnv

(

R h,cnv = h As,n R t (sp ) =

−1

)

( )

= hL2h

−1

2 = 100 W / m 2 ⋅ K ( 0.005m )   

1 − 1.410 A r + 0.344 A3r + 0.043 A5r + 0.034 A7r 4 k sub Lh

R t,c =

R ′′t,c L2h

=

(

0.5 × 10−4 m 2 ⋅ K / W

(0.005m )2

)

R sp,cnv =  h Asub − As,h 

q=

−1

=

−1

= 400 K / W

1 − 0.353 + 0.005 + 0 + 0

4 (80 W / m ⋅ K )( 0.005m )

= 2.000 K / W

(

)

= 100 W / m 2 ⋅ K 0.010m 2 − 0.005m 2 



= 0.408 K / W

−1

70°C 70°C + = 0.18 W + 0.52 W = 0.70 W 400 K / W ( 0.408 + 2 + 133.3) K / W

= 133.3 K / W

<

COMMENTS: (1) The thermal resistances of the substrate and the chip/substrate interface are much less than the substrate convection resistance. Hence, the heat rate is increased almost in proportion to the additional surface area afforded by the substrate. An increase in the spacing between chips (Sh) would increase q correspondingly. (2) In the limit A r → 0, R t (sp ) reduces to 2π 1/ 2 k sub D h for a circular heat source and 4k sub L h for a square source.

PROBLEM 4.35 KNOWN: Internal corner of a two-dimensional system with prescribed convection boundary conditions. FIND: Finite-difference equations for these situations: (a) Horizontal boundary is perfectly insulated and vertical boundary is subjected to a convection process (T∞,h), (b) Both boundaries are perfectly insulated; compare result with Eq. 4.45. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties, (4) No internal generation. ANALYSIS: Consider the nodal network shown above and also as Case 2, Table 4.2. Having defined the control volume – the shaded area of unit thickness normal to the page – next identify the heat transfer processes. Finally, perform an energy balance wherein the processes are expressed using appropriate rate equations. (a) With the horizontal boundary insulated and the vertical boundary subjected to a convection process, the energy balance results in the following finite-difference equation:

E& in − E& out = 0

q1 + q2 + q3 + q4 + q5 + q6 = 0

T − Tm,n −T  ∆x  T  ∆y  k ( ∆y ⋅1) m-1,n + k  ⋅1 m,n-1 m,n + h  ⋅1 T∞ − Tm,n ∆x ∆y  2   2 

(

)

− Tm,n T −T  ∆y  T + 0 + k  ⋅1 m+1,n + k ( ∆x ⋅ 1) m,n+1 m,n = 0. ∆x ∆y  2  Letting ∆x = ∆y, and regrouping, find

(

) (

)

2 Tm-1,n + Tm,n+1 + Tm+1,n + Tm,n-1 +

h∆x h∆x  T∞ − 6 + Tm,n = 0. k k  

<

(b) With both boundaries insulated, the energy balance would have q3 = q4 = 0. The same result would be obtained by letting h = 0 in the previous result. Hence,

(

) (

)

2 Tm-1,n + Tm,n+1 + Tm+1,n + Tm,n-1 − 6 Tm,n = 0.

<

Note that this expression compares exactly with Eq. 4.45 when h = 0, which corresponds to insulated boundaries.

PROBLEM 4.36 KNOWN: Plane surface of two-dimensional system. FIND: The finite-difference equation for nodal point on this boundary when (a) insulated; compare result with Eq. 4.46, and when (b) subjected to a constant heat flux. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction with no generation, (2) Constant properties, (3) Boundary is adiabatic. ANALYSIS: (a) Performing an energy balance on the control volume, (∆x/2)⋅∆y, and using the conduction rate equation, it follows that

E& in − E& out = 0

q1 + q2 + q3 = 0

(1,2)

T − Tm,n −T −T  ∆x  T  ∆x  T k ( ∆y ⋅1) m-1,n + k  ⋅1 m,n-1 m,n + k  ⋅1 m,n+1 m,n = 0. (3) ∆x ∆y ∆y  2   2  Note that there is no heat rate across the control volume surface at the insulated boundary. Recognizing that ∆x =∆y, the above expression reduces to the form

2Tm-1,n + Tm,n-1 + Tm,n+1 − 4Tm,n = 0.

(4) <

The Eq. 4.46 of Table 4.3 considers the same configuration but with the boundary subjected to a convection process. That is,

( 2Tm-1,n + Tm,n-1 + Tm,n+1 ) + 2hk∆x T∞ − 2  h∆k x + 2  Tm,n = 0.

(5)

Note that, if the boundary is insulated, h = 0 and Eq. 4.46 reduces to Eq. (4). (b) If the surface is exposed to a constant heat flux, q ′′o , the energy balance has the form q1 + q2 + q3 + q ′′o ⋅∆y = 0 and the finite difference equation becomes

q ′′ ∆x 2Tm-1,n + Tm,n-1 + Tm,n+1 − 4Tm,n = − o . k

<

COMMENTS: Equation (4) can be obtained by using the “interior node” finite-difference equation, Eq. 4.33, where the insulated boundary is treated as a symmetry plane as shown below.

PROBLEM 4.37 KNOWN: External corner of a two-dimensional system whose boundaries are subjected to prescribed conditions. FIND: Finite-difference equations for these situations: (a) Upper boundary is perfectly insulated and side boundary is subjected to a convection process, (b) Both boundaries are perfectly insulated; compare result with Eq. 4.47. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties, (4) No internal generation. ANALYSIS: Consider the nodal point configuration shown in the schematic and also as Case 4, Table 4.2. The control volume about the node – shaded area above of unit thickness normal to the page – has dimensions, (∆x/2)(∆y/2)⋅1. The heat transfer processes at the surface of the CV are identified as q1, q2 ⋅⋅⋅. Perform an energy balance wherein the processes are expressed using the appropriate rate equations. (a) With the upper boundary insulated and the side boundary subjected to a convection process, the energy balance has the form

E& in − E& out = 0

q1 + q2 + q3 + q4 = 0

− Tm,n  ∆y  T  ∆x k  ⋅1 m-1,n +k  ∆x  2   2

−T T  ∆y ⋅1 m,n-1 m,n + h  ∆y   2

(1,2)  ⋅1 T∞ − Tm,n + 0 = 0. 

(

)

Letting ∆x = ∆y, and regrouping, find

Tm,n-1 + Tm-1,n +

h∆x 1 h∆x  T∞ − 2  + 1 Tm,n = 0. k 2 k 

(3) <

(b) With both boundaries insulated, the energy balance of Eq. (2) would have q3 = q4 = 0. The same result would be obtained by letting h = 0 in the finite-difference equation, Eq. (3). The result is

Tm,n-1 + Tm-1,n − 2Tm,n = 0.

<

Note that this expression is identical to Eq. 4.47 when h = 0, in which case both boundaries are insulated. COMMENTS: Note the convenience resulting from formulating the energy balance by assuming that all the heat flow is into the node.

PROBLEM 4.38 KNOWN: Conduction in a one-dimensional (radial) cylindrical coordinate system with volumetric generation. FIND: Finite-difference equation for (a) Interior node, m, and (b) Surface node, n, with convection. SCHEMATIC:

(a) Interior node, m (b) Surface node with convection, n ASSUMPTIONS: (1) Steady-state, one-dimensional (radial) conduction in cylindrical coordinates, (2) Constant properties. ANALYSIS: (a) The network has nodes spaced at equal ∆r increments with m = 0 at the center; hence, r = m∆r (or n∆r). The control volume is V = 2π r ⋅ ∆r ⋅ l = 2π ( m ∆r ) ∆r ⋅ l. The energy

& =0 balance is E& in +E& g = q a +qb +qV  k  2π 

  ∆r   Tm-1 − Tm + k 2π r − 2  l  ∆r  

 ∆r   Tm+1 − Tm & + q  2π ( m ∆r ) ∆rl  = 0. r+ 2  l  ∆r 

Recognizing that r = m∆r, canceling like terms, and regrouping find & ∆r 2 1 1 qm   m − T + m+ T − 2mT + = 0. m   m-1   m+1



2



2

k

<

(b) The control volume for the surface node is V = 2 π r ⋅ ( ∆r/2 ) ⋅ l. The energy balance is & E& in +E& g = q d + q conv + qV=0. Use Fourier’s law to express qd and Newton’s law of cooling for qconv to obtain

 k  2π 

∆r   ∆r   Tn-1 − Tn  + h [ 2π rl ] ( T∞ − Tn ) + q& 2π ( n∆r ) l  = 0. r − 2  l  ∆r 2   

Let r = n∆r, cancel like terms and regroup to find

& ∆r 2 hn ∆r   1  hn ∆r  qn  1  n − 2  Tn-1 −   n − 2  + k  Tn + 2k + k T∞ = 0.  

< COMMENTS: (1) Note that when m or n becomes very large compared to ½, the finite-difference equation becomes independent of m or n. Then the cylindrical system approximates a rectangular one. (2) The finite-difference equation for the center node (m = 0) needs to be treated as a special case. The control volume is V = π ( ∆r / 2 ) l and the energy balance is 2

 ∆r 2  ∆r T −T & = k  2π   l  1 0 + q& π   l  = 0. E& in +E& g = q a + qV   2   ∆ r   2   Regrouping, the finite-difference equation is

− To + T1 +

2 q& ∆r

4k

= 0.

PROBLEM 4.39 KNOWN: Two-dimensional cylindrical configuration with prescribed radial (∆r) and angular (∆φ) spacings of nodes. FIND: Finite-difference equations for nodes 2, 3 and 1. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction in cylindrical coordinates (r,φ), (3) Constant properties. ANALYSIS: The method of solution is to define the appropriate control volume for each node, to identify relevant processes and then to perform an energy balance. (a) Node 2. This is an interior node with control volume as shown above. The energy balance is E& = q ′ + q′ + q′ + q′ = 0. Using Fourier’s law for each process, find in

a

b

c

d

 3  (T − T ) (T − T ) k   ri + ∆r  ∆φ  5 2 + k ( ∆r ) 3 2 + 2   ∆r ( ri + ∆ r ) ∆φ  T − T ( ) (T − T )  1   + k   ri + ∆r  ∆φ  i 2 + k ( ∆r ) 1 2 = 0. 2   ∆r ( ri + ∆r ) ∆φ  Canceling terms and regrouping yields, 2 2   ( ∆r ) 1 3  ( ∆r ) 1     T2 +  ri + ∆r  T5 + T3 + T1 ) + ri + ∆r Ti = 0. −2 ( ri + ∆r ) + (  2  2     ( ∆φ ) 2 ( ri + ∆r )  ( ri + ∆ r ) ( ∆φ ) 2

(b) Node 3. The adiabatic surface behaves as a symmetry surface. We can utilize the result of Part (a) to write the finite-difference equation by inspection as 2  ( ∆r )2 1  3  2 ( ∆r ) 1     −2 ( ri + ∆r ) + T3 + ri + ∆r  T6 + ⋅ T2 + ri + ∆r  Ti = 0. 2 2 2    2   ( ∆φ ) ( ri + ∆r )  ( ri + ∆ r ) ( ∆φ )

(c) Node 1. The energy balance is q ′a + qb′ + q′c + q′d = 0. Substituting,

 3  ∆φ  ( T4 − T1 ) ( T2 − T1) + k  ri + ∆r  + k ( ∆r )  2  2  ∆r ( ri + ∆r ) ∆φ   1  ∆φ  ( Ti − T1 ) + k   ri + ∆r  + h ( ∆r )( T∞ − T1 ) = 0 2  2  ∆r  This expression could now be rearranged.

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PROBLEM 4.40 KNOWN: Heat generation and thermal boundary conditions of bus bar. Finite-difference grid. FIND: Finite-difference equations for selected nodes. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties. ANALYSIS: (a) Performing an energy balance on the control volume, (∆x/2)(∆y/2)⋅1, find the FDE for node 1,

k ( ∆y/2 ⋅1) To − T1  ∆x  + hu  ⋅ 1 ( T∞ − T1) + ( T2 − T1 ) R′′t,c / ( ∆y/2) 1 ∆x  2  k ( ∆x/2 ⋅1) + ( T6 − T1 ) + &q ( ∆x/2 )( ∆y/2) 1 = 0 ∆y ∆x/kR′′t,c To + ( h u ∆x/k ) T∞ +T 2 + T6

(

)

(

)

2 + q& ( ∆x ) / 2 k −  ∆x/kR′′t,c + ( hu ∆x/k ) + 2 T1 = 0.

<

(b) Performing an energy balance on the control volume, (∆x)(∆y/2)⋅1, find the FDE for node 13,

h l ( ∆x ⋅1)( T∞ − T13 ) + ( k/ ∆x )( ∆y/2 ⋅1)( T12 − T13 ) + ( k/ ∆y )( ∆x ⋅1)( T8 − T13 ) + ( k/ ∆x )( ∆y/2 ⋅1)( T14 − T13 ) + q& ( ∆x ⋅ ∆y/2 ⋅1) = 0

( h l ∆x/k ) T∞ +1 / 2 (T12 + 2T8 + T14 ) + q& ( ∆x )2 /2k − ( h l ∆x/k + 2 ) T13 = 0.

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COMMENTS: For fixed To and T∞, the relative amounts of heat transfer to the air and heat sink are determined by the values of h and R ′′t,c.

PROBLEM 4.41 KNOWN: Nodal point configurations corresponding to a diagonal surface boundary subjected to a convection process and to the tip of a machine tool subjected to constant heat flux and convection cooling. FIND: Finite-difference equations for the node m,n in the two situations shown. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties. ANALYSIS: (a) The control volume about node m,n has triangular shape with sides ∆x and ∆y while the diagonal (surface) length is

2 ∆x. The heat rates associated with the control volume are due to

conduction, q1 and q2, and to convection, qc . Performing an energy balance, find

E& in − E& out = 0 q 1 + q 2 + qc = 0 Tm,n-1 − Tm,n T − Tm,n k ( ∆x ⋅1) + k ( ∆y ⋅1) m+1,n +h ∆y ∆x

(

)(

)

2 ∆x ⋅ 1 T∞ − Tm,n = 0.

Note that we have considered the tool to have unit depth normal to the page. Recognizing that ∆x = ∆y, dividing each term by k and regrouping, find

Tm,n-1 + Tm+1,n + 2 ⋅

h∆x h∆x  T∞ −  2 + 2 ⋅ Tm,n = 0. k k  

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(b) The control volume about node m,n has triangular shape with sides ∆x/2 and ∆y/2 while the lower diagonal surface length is

2 ( ∆x/2 ). The heat rates associated with the control volume are due to

the constant heat flux, qa, to conduction, qb, and to the convection process, qc . Perform an energy balance,

E& in − E& out = 0 q a + q b + qc = 0 ∆x  ∆y  Tm+1,n − Tm,n ∆x     q′′o ⋅  ⋅1 + k ⋅  ⋅1 +h ⋅ 2 ⋅ T∞ − Tm,n = 0. ∆x 2   2  2  

(

)

Recognizing that ∆x = ∆y, dividing each term by k/2 and regrouping, find

Tm+1,n + 2 ⋅

h∆x ∆x  h∆x  ⋅ T∞ + q′′o ⋅ − 1+ 2 ⋅ Tm,n = 0. k k  k 

COMMENTS: Note the appearance of the term h∆x/k in both results, which is a dimensionless parameter (the Biot number) characterizing the relative effects of convection and conduction.

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PROBLEM 4.42 KNOWN: Nodal point on boundary between two materials. FIND: Finite-difference equation for steady-state conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties, (4) No internal heat generation, (5) Negligible thermal contact resistance at interface. ANALYSIS: The control volume is defined about nodal point 0 as shown above. The conservation of energy requirement has the form 6

∑ qi = q1 + q2 + q3 + q4 + q5 + q6 = 0

i =1

since all heat rates are shown as into the CV. Each heat rate can be written using Fourier’s law,

∆y T1 − T0 T −T ∆y T3 − T0 ⋅ + k A ⋅ ∆x ⋅ 2 0 + k A ⋅ ⋅ 2 ∆x ∆y 2 ∆x ∆y T3 − T0 T4 − T0 ∆y T1 − T0 + kB ⋅ ⋅ + kB ⋅ ∆x ⋅ + kB ⋅ ⋅ = 0. 2 ∆x ∆y 2 ∆x

kA ⋅

Recognizing that ∆x = ∆y and regrouping gives the relation,

1 kA 1 kB −T0 + T1 + T2 + T3 + T = 0. 4 2 (k A + k B ) 4 2 (k A + kB ) 4

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COMMENTS: Note that when kA = kB, the result agrees with Eq. 4.33 which is appropriate for an interior node in a medium of fixed thermal conductivity.

PROBLEM 4.43 KNOWN: Two-dimensional grid for a system with no internal volumetric generation. FIND: Expression for heat rate per unit length normal to page crossing the isothermal boundary. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional heat transfer, (3) Constant properties. ANALYSIS: Identify the surface nodes (Ts ) and draw control volumes about these nodes. Since there is no heat transfer in the direction parallel to the isothermal surfaces, the heat rate out of the constant temperature surface boundary is q′ = qa′ + qb′ + qc′ + qd′ + qe′ + qf′ For each q ′i , use Fourier’s law and pay particular attention to the manner in which the crosssectional area and gradients are specified. T −T T −T T −T q′ = k ( ∆y/2 ) 1 s + k ( ∆y ) 2 s + k ( ∆y ) 3 s ∆x ∆x ∆x T5 − Ts T6 − Ts T7 − Ts + k ( ∆x ) + k ( ∆x ) + k ( ∆x/2) ∆y ∆y ∆y Regrouping with ∆x = ∆y, find q′ = k [ 0.5T1 + T2 + T3 + T5 + T6 + 0.5T7 − 5Ts ].

<

COMMENTS: Looking at the corner node, it is important to recognize the areas associated with q′c and q′d (∆y and ∆x, respectively).

PROBLEM 4.44 KNOWN: One-dimensional fin of uniform cross section insulated at one end with prescribed base temperature, convection process on surface, and thermal conductivity. FIND: Finite-difference equation for these nodes: (a) Interior node, m and (b) Node at end of fin, n, where x = L. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction. ANALYSIS: (a) The control volume about node m is shown in the schematic; the node spacing and control volume length in the x direction are both ∆x. The uniform cross-sectional area and fin perimeter are Ac and P, respectively. The heat transfer process on the control surfaces, q1 and q2, represent conduction while qc is the convection heat transfer rate between the fin and ambient fluid. Performing an energy balance, find

E& in − E& out = 0 q 1 + q 2 + qc = 0 Tm-1 − Tm T −T kAc + kAc m+1 m + hP∆x ( T∞ − Tm ) = 0. ∆x ∆x Multiply the expression by ∆x/kAc and regroup to obtain

Tm-1 + Tm+1 +

  hP hP ⋅ ∆x 2T∞ −  2 + ∆x 2  Tm = 0 kAc kAc  

1 R cond ( 2D ) . COMMENTS: To check our finite-difference solution, we could assess its consistency with conservation of energy requirements. For example, an energy balance performed at the inner surface requires a balance between convection from the surface and conduction to the surface, which may be expressed as

q′ = k ( ∆x ⋅1)

(T5 − T1 ) + k ∆y

( ∆x ⋅1)

T6 − T2 T −T + k ( ∆x / 2 ⋅ 1) 7 3 ∆y ∆y

Substituting the temperatures corresponding to h = 5000 W / m 2 ⋅ K, the expression yields q ′ = 10, 340 W / m, and, as it must be, conservation of energy is precisely satisfied. Results of the

(

)

analysis may also be checked by using the expression q ′ = ( Ts − T∞ ) / R ′cond ( 2D ) + R ′conv , where, for 2 h = 5000 W / m ⋅ K, R ′ = 1/ 4hw = 2.5 × 10−3 m ⋅ K / W, and from Eq. (4.27) and Case 11 of conv

(

)

Table 4.1, R ′cond = [0.930 ln ( W / w ) − 0.05] / 2π k = 3.94 × 10 −4 m ⋅ K / W. Hence,

(

q ′ = (50 − 20 ) °C / 2.5 × 10

−3

+ 3.94 × 10

−4

) m ⋅ K / W = 10, 370 W / m, and the agreement with the

finite-difference solution is excellent. Note that, even for h = 5000 W / m 2 ⋅ K, R ′conv >> R ′cond ( 2D ).

PROBLEM 4.50 KNOWN: Steady-state temperatures (°C) associated with selected nodal points in a two-dimensional system. FIND: (a) Temperatures at nodes 1, 2 and 3, (b) Heat transfer rate per unit thickness from the system surface to the fluid. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties. ANALYSIS: (a) Using the finite-difference equations for Nodes 1, 2 and 3: Node 1, Interior node, Eq. 4.33: T1 =

T1 =

1 ⋅ ∑ Tneighbors 4

1 (172.9 + 137.0 + 132.8 + 200.0 )o C = 160.7oC 4

<

Node 2, Insulated boundary, Eq. 4.46 with h = 0, Tm,n = T2

(

)

1 Tm-1,n + Tm+1,n + 2Tm,n-1 4 1 T2 = (129.4 + 45.8 + 2 ×103.5 )o C = 95.6oC 4 T2 =

<

Node 3, Plane surface with convection, Eq. 4.46, Tm,n = T3 2h ∆x  h ∆x + 2  T = 2T 2 + T + T + T∞ 3 m-1,n m,n+1 m,n-1  k  k

(

)

h ∆x/k = 50W/m 2 ⋅ K × 0.1m/1.5W/m ⋅ K = 3.33

2 ( 3.33 + 2 ) T3 = ( 2 × 103.5 + 45.8 + 67.0 ) °C + 2 × 3.33 × 30 o C 1

( 319.80 + 199.80) °C=48.7°C 10.66 (b) The heat rate per unit thickness from the surface to the fluid is determined from the sum of the convection rates from each control volume surface. q ′conv = qa′ + qb′ + qc′ + q ′d T3 =

q i = h∆y i ( Ti − T∞ ) q ′conv = 50

 0.1 m 45.8 − 30.0 °C + ( )  m ⋅K  2 W 2

0.1m ( 48.7 − 30.0 ) °C + 0.1m ( 67.0 − 30.0 ) °C + 0.1m + ( 200.0 − 30.0) °C 2 

q ′conv = ( 39.5 + 93.5 + 185.0 + 425 ) W/m = 743 W/m.

<

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PROBLEM 4.51 KNOWN: Nodal temperatures from a steady-state finite-difference analysis for a cylindrical fin of prescribed diameter, thermal conductivity and convection conditions ( T∞ , h). FIND: (a) The fin heat rate, qf, and (b) Temperature at node 3, T3. SCHEMATIC:

T0 = 100.0°C T1 = 93.4°C T2 = 89.5°C

ASSUMPTIONS: (a) The fin heat rate, qf, is that of conduction at the base plane, x = 0, and can be  − E found from an energy balance on the control volume about node 0, E in out = 0 ,

qf + q1 + qconv = 0

or

qf = −q1 − qconv .

Writing the appropriate rate equation for q1 and qconv, with Ac = πD2/4 and P = πD, T1 − T0 π kD2 q f = − kAc − hP ( ∆x 2 )( T∞ − T0 ) = − (T1 − T0 ) − (π 2 ) Dh∆x (T∞ − T0 )

∆x

4∆x

Substituting numerical values, with ∆x = 0.010 m, find

π × 15 W m ⋅ K (0.012 m )2 qf = − (93.4 − 100 )$ C 4 × 0.010 m π $ − × 0.012 m × 25 W m 2 ⋅ K × 0.010 m ( 25 − 100 ) C 2 qf = (1.120 + 0.353) W = 1.473 W .

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(b) To determine T3, derive the finite-difference equation for node 2, perform an energy balance on the  − E control volume shown above, E in out = 0 ,

qcv + q3 + q1 = 0

T −T T −T hP∆x ( T∞ − T2 ) + kA c 3 2 + kAc 1 2 = 0 ∆x ∆x

T3 = −T1 + 2T2 −

hP∆x 2 2 ∆x [T∞ − T2 ] kAc

Substituting numerical values, find

T2 = 89.2$ C COMMENTS: Note that in part (a), the convection heat rate from the outer surface of the control volume is significant (25%). It would have been poor approximation to ignore this term.

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PROBLEM 4.52 KNOWN: Long rectangular bar having one boundary exposed to a convection process (T∞, h) while the other boundaries are maintained at a constant temperature (Ts). FIND: (a) Using a grid spacing of 30 mm and the Gauss-Seidel method, determine the nodal temperatures and the heat rate per unit length into the bar from the fluid, (b) Effect of grid spacing and convection coefficient on the temperature field. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties. ANALYSIS: (a) With the grid spacing ∆x = ∆y = 30 mm, three nodes are created. Using the finitedifference equations as shown in Table 4.2, but written in the form required of the Gauss-Seidel method (see Section 4.5.2), and with Bi = h∆x/k = 100 W/m2⋅K × 0.030 m/1 W/m⋅K = 3, we obtain: Node 1:

T1 =

1 1 1 (T2 + Ts + BiT∞ ) = (T2 + 50 + 3 ×100 ) = (T2 + 350) 5 5 ( Bi + 2 )

(1)

Node 2:

T2 =

1 1 1 (T1 + 2Ts + T3 ) = (T1 + T3 + 2 × 50 ) = (T1 + T3 + 100 ) 4 4 4

(2)

Node 3:

T3 =

1 1 1 (T2 + 3Ts ) = (T2 + 3 × 50 ) = ( T2 + 150 ) 4 4 4

(3)

Denoting each nodal temperature with a superscript to indicate iteration step, e.g. T1k , calculate values as shown below. k 0

T1 85

T2 60

T3 (°C) 55

1 2 3 4

82.00 81.85 81.71 81.69

59.25 58.54 58.46 58.45

52.31 52.14 52.12 52.11

← initial guess

By the 4th iteration, changes are of order 0.02°C, suggesting that further calculations may not be necessary. Continued...

PROBLEM 4.52 (Cont.) In finite-difference form, the heat rate from the fluid to the bar is

q′conv = h ( ∆x 2 )( T∞ − Ts ) + h∆x ( T∞ − T1 ) + h ( ∆x 2 )( T∞ − Ts ) q′conv = h∆x ( T∞ − Ts ) + h∆x ( T∞ − T1 ) = h∆x ( T∞ − Ts ) + ( T∞ − T1 ) $

q′conv = 100 W m 2 ⋅ K × 0.030 m (100 − 50 ) + (100 − 81.7 ) C = 205 W m .

<

(b) Using the Finite-Difference Equations option from the Tools portion of the IHT menu, the following two-dimensional temperature field was computed for the grid shown in schematic (b), where x and y are in mm and the temperatures are in °C. y\x 0 15 30 45 60 75 90

0 50 50 50 50 50 50 50

15 80.33 63.58 56.27 52.91 51.32 50.51 50

30 85.16 67.73 58.58 54.07 51.86 50.72 50

45 80.33 63.58 56.27 52.91 51.32 50.51 50

60 50 50 50 50 50 50 50

The improved prediction of the temperature field has a significant influence on the heat rate, where, accounting for the symmetrical conditions,

q′ = 2h ( ∆x 2 )( T∞ − Ts ) + 2h ( ∆x )( T∞ − T1 ) + h ( ∆x )( T∞ − T2 ) q′ = h ( ∆x ) ( T∞ − Ts ) + 2 ( T∞ − T1 ) + ( T∞ − T2 ) $

q′ = 100 W m 2 ⋅ K ( 0.015 m ) 50 + 2 (19.67 ) + 14.84 C = 156.3 W m

<

Additional improvements in accuracy could be obtained by reducing the grid spacing to 5 mm, although the requisite number of finite-difference equations would increase from 12 to 108, significantly increasing problem set-up time. An increase in h would increase temperatures everywhere within the bar, particularly at the heated surface, as well as the rate of heat transfer by convection to the surface. COMMENTS: (1) Using the matrix-inversion method, the exact solution to the system of equations (1, 2, 3) of part (a) is T1 = 81.70°C, T2 = 58.44°C, and T3 = 52.12°C. The fact that only 4 iterations were required to obtain agreement within 0.01°C is due to the close initial guesses. (2) Note that the rate of heat transfer by convection to the top surface of the rod must balance the rate of heat transfer by conduction to the sides and bottom of the rod. NOTE TO INSTRUCTOR: Although the problem statement calls for calculations with ∆x = ∆y = 5 mm and for plotting associated isotherms, the instructional value and benefit-to-effort ratio are small. Hence, it is recommended that this portion of the problem not be assigned.

PROBLEM 4.53 KNOWN: Square shape subjected to uniform surface temperature conditions. FIND: (a) Temperature at the four specified nodes; estimate the midpoint temperature To, (b) Reducing the mesh size by a factor of 2, determine the corresponding nodal temperatures and compare results, and (c) For the finer grid, plot the 75, 150, and 250°C isotherms. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties. ANALYSIS: (a) The finite-difference equation for each node follows from Eq. 4.33 for an interior point written in the form, Ti = 1/4∑Tneighbors. Using the Gauss-Seidel iteration method, Section 4.5.2, the finitedifference equations for the four nodes are:

) ( T2k = 0.25 (100 + 200 + T4k −1 + T1k −1 ) = 0.25T1k −1 + 0.25T4k −1 + 75.0 T3k = 0.25 ( T1k −1 + T4k −1 + 300 + 50 ) = 0.25T1k −1 + 0.25T4k −1 + 87.5 T4k = 0.25 ( T2k −1 + 200 + 300 + T3k −1 ) = 0.25T2k −1 + 0.25T3k −1 + 125.0 T1k = 0.25 100 + T2k −1 + T3k −1 + 50 = 0.25T2k −1 + 0.25T3k −1 + 37.5

The iteration procedure using a hand calculator is implemented in the table below. Initial estimates are entered on the k = 0 row. k 0 1 2 3 4 5 6 7

T1 (°C) 100 112.50 123.44 119.93 119.05 118.83 118.77 118.76

T2 (°C) 150 165.63 158.60 156.40 156.40 156.29 156.26 156.25

T3 (°C) 150 178.13 171.10 169.34 168.90 168.79 168.76 168.76

T4 (°C) 250 210.94 207.43 206.55 206.33 206.27 206.26 206.25

< Continued...

PROBLEM 4.53 (Cont.) By the seventh iteration, the convergence is approximately 0.01°C. The midpoint temperature can be estimated as

To = ( T1 + T2 + T3 + T4 ) 2 = (118.76 + 156.25 + 168.76 + 206.25) C 4 = 162.5$ C $

(b) Because all the nodes are interior ones, the nodal equations can be written by inspection directly into the IHT workspace and the set of equations solved for the nodal temperatures (°C). Mesh Coarse Fine

T1 118.76 117.4

To 162.5 162.5

T2 156.25 156.1

T3 168.76 168.9

T4 206.25 207.6

The maximum difference for the interior points is 1.5°C (node 4), but the estimate at the center, To, is the same, independently of the mesh size. In terms of the boundary surface temperatures,

To = (50 + 100 + 200 + 300 ) C 4 = 162.5$ C $

Why must this be so? (c) To generate the isotherms, it would be necessary to employ a contour-drawing routine using the tabulated temperature distribution (°C) obtained from the finite-difference solution. Using these values as a guide, try sketching a few isotherms. 50 50 50 50 50 -

100 86.0 88.2 99.6 123.0 173.4 300

100 105.6 117.4 137.1 168.9 220.7 300

100 119 138.7 162.5 194.9 240.6 300

100 131.7 156.1 179.2 207.6 246.8 300

100 151.6 174.6 190.8 209.4 239.0 300

200 200 200 200 200 -

COMMENTS: Recognize that this finite-difference solution is only an approximation to the temperature distribution, since the heat conduction equation has been solved for only four (or 25) discrete points rather than for all points if an analytical solution had been obtained.

PROBLEM 4.54 KNOWN: Long bar of square cross section, three sides of which are maintained at a constant temperature while the fourth side is subjected to a convection process. FIND: (a) The mid-point temperature and heat transfer rate between the bar and fluid; a numerical technique with grid spacing of 0.2 m is suggested, and (b) Reducing the grid spacing by a factor of 2, find the midpoint temperature and the heat transfer rate. Also, plot temperature distribution across the surface exposed to the fluid. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties. ANALYSIS: (a) Considering symmetry, the nodal network is shown above. The matrix inversion method of solution will be employed. The finite-difference equations are: Nodes 1, 3, 5 Nodes 2, 4, 6 Nodes 7, 8 -

Interior nodes, Eq. 4.33; written by inspection. Also can be treated as interior points, considering symmetry. On a plane with convection, Eq. 4.46; noting that h∆x/k = 10 W/m2⋅K × 0.2 m/2W/m⋅K = 1, find Node 7: (2T5 + 300 + T8) + 2×1⋅100 - 2(1+2)T7 = 0 Node 8: (2T6 + T7 + T7) + 2×1⋅100 - 2(1+2)T8 = 0

The solution matrix [T] can be found using a stock matrix program using the [A] and [C] matrices shown below to obtain the solution matrix [T] (Eq. 4.52). Alternatively, the set of equations could be entered into the IHT workspace and solved for the nodal temperatures.

 −4 1 1 0 0 0 0 0   2 −4 0 1 0 0 0 0   1 0 −4 1 1 0 0 0   2 −4 0 1 0 0  A=0 1 0 0 1 0 −4 1 1 0  0 0 0 1 2 −4 0 1   0 0 0 0 2 0 −6 1   0 0 0 0 0 2 2 −6 

 −600   −300   −300    C= 0  −300  0   −500   −200 

 292.2   289.2   279.7    T =  272.2  254.5  240.1 198.1  179.4 

From the solution matrix, [T], find the mid-point temperature as T4 = 272.2°C

< Continued...

PROBLEM 4.54 (Cont.) The heat rate by convection between the bar and fluid is given as,

q′conv = 2 ( q′a + q′b + q′c ) q′conv = 2x  h ( ∆x 2 )( T8 − T∞ ) + h ( ∆x )( T7 − T∞ ) + h ( ∆x 2 )(300 − T∞ ) q′conv = 2x 10 W m 2 ⋅ K × ( 0.2 m 2 ) (179.4 − 100 ) + 2 (198.1 − 100 ) + (300 − 100 ) K   

<

q′conv = 952 W m .

(b) Reducing the grid spacing by a factor of 2, the nodal arrangement will appear as shown. The finitedifference equation for the interior and centerline nodes were written by inspection and entered into the IHT workspace. The IHT Finite-Difference Equations Tool for 2-D, SS conditions, was used to obtain the FDE for the nodes on the exposed surface.

The midpoint temperature T13 and heat rate for the finer mesh are T13 = 271.0°C

q′ = 834 W/m

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COMMENTS: The midpoint temperatures for the coarse and finer meshes agree closely, T4 = 272°C vs. T13 = 271.0°C, respectively. However, the estimate for the heat rate is substantially influenced by the mesh size; q′ = 952 vs. 834 W/m for the coarse and finer meshes, respectively.

PROBLEM 4.55 KNOWN: Volumetric heat generation in a rectangular rod of uniform surface temperature. FIND: (a) Temperature distribution in the rod, and (b) With boundary conditions unchanged, heat generation rate causing the midpoint temperature to reach 600 K. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, (3) Uniform volumetric heat generation. ANALYSIS: (a) From symmetry it follows that six unknown temperatures must be determined. Since all nodes are interior ones, the finite-difference equations may be obtained from Eq. 4.39 written in the form

Ti = 1 2 ∑ Tneighbors + 1 4 ( q ( ∆x∆y1) k ) . With q ( ∆x∆y ) 4k = 62.5 K, the system of finite-difference equations is

T1 = 0.25 ( Ts + T2 + T4 + Ts ) + 15.625

(1)

T2 = 0.25 ( Ts + T3 + T5 + T1 ) + 15.625

(2)

T3 = 0.25 ( Ts + T2 + T6 + T2 ) + 15.625

(3)

T4 = 0.25 ( T1 + T5 + T1 + Ts ) + 15.625

(4)

T5 = 0.25 ( T2 + T6 + T2 + T4 ) + 15.625

(5)

T6 = 0.25 ( T3 + T5 + T3 + T5 ) + 15.625

(6)

With Ts = 300 K, the set of equations was written directly into the IHT workspace and solved for the nodal temperatures, T1

T2

T3

T4

T5

T6 (K)

348.6

368.9

374.6

362.4

390.2

398.0

<

(b) With the boundary conditions unchanged, the q required for T6 = 600 K can be found using the same set of equations in the IHT workspace, but with these changes: (1) replace the last term on the RHS (15.625) of Eqs. (1-6) by q (∆x∆y)/4k = (0.005 m)2 q /4×20 W/m⋅K = 3.125 × 10-7 q and (2) set T6 = 600 K. The set of equations has 6 unknown, five nodal temperatures plus q . Solving find

q = 1.53 × 108 W m3

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PROBLEM 4.56 KNOWN: Flue of square cross section with prescribed geometry, thermal conductivity and inner and outer surface temperatures. FIND: Heat loss per unit length from the flue, q′. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, (3) No internal generation. ANALYSIS: Taking advantage of symmetry, the nodal network using the suggested 75mm grid spacing is shown above. To obtain the heat rate, we first need to determine the unknown temperatures T1, T2, T3 and T4. Recognizing that these nodes may be treated as interior nodes, the nodal equations from Eq. 4.33 are (T2 + 25 + T2 + 350) - 4T1 = 0 (T1 + 25 + T3 + 350) - 4T2 = 0 (T2 + 25 + T4 + 350) - 4T3 = 0 (T3 + 25 + 25 + T3) - 4T4 = 0. The Gauss-Seidel iteration method is convenient for this system of equations and following the procedures of Section 4.5.2, they are rewritten as, T1k = 0.50 T2k-1 + 93.75

T2k = 0.25 T1k + 0.25 T3k-1 + 93.75 T3k = 0.25 T2k + 0.25 T4k-1 + 93.75 T4k = 0.50 T3k + 12.5.

The iteration procedure is implemented in the table on the following page, one row for each iteration k. The initial estimates, for k = 0, are all chosen as (350 + 25)/2 ≈ 185°C. Iteration is continued until the maximum temperature difference is less than 0.2°C, i.e., ε < 0.2°C. Note that if the system of equations were organized in matrix form, Eq. 4.52, diagonal dominance would exist. Hence there is no need to reorder the equations since the magnitude of the diagonal element is greater than that of other elements in the same row. Continued …..

PROBLEM 4.56 (Cont.) k

T1(°C)

T2(°C)

T3(°C)

T4(°C)

0 1 2 3 4 5 6 7

185 186.3 187.1 187.4 184.9 184.2 184.0 183.9

185 186.6 187.2 182.3 180.8 180.4 180.3 180.3

185 186.6 167.0 163.3 162.5 162.3 162.3 162.2

185 105.8 96.0 94.2 93.8 93.7 93.6 93.6

← initial estimate

← ε 0.1, the internal conduction resistance is not negligible. Therefore significant transverse temperature gradients exist, and the one-dimensional conduction assumption in the fin is a poor one. Continued …..

PROBLEM 4.66 (Cont.) (2) From the table, with k = 5 W/m⋅K (Bi = 1), the 2-D fin heat rate obtained from the FEA analysis is 20% lower than that for the 1-D analytical analysis. This is as expected since the 2-D model accounts for transverse thermal resistance to heat flow. Note, however, that analyses predict the same tip temperature, a consequence of the fin approximating an infinitely long fin (mL = 20.2 >> 2.56; see Ex. 3.8 Comments). (3) For the k = 5 W/m⋅K case, the FEHT isotherms show considerable curvature in the region near the fin base. For example, at x = 10 and 20 mm, the difference between the centerline and surface temperatures are 15 and 7°C. (4) From the table, with increasing thermal conductivity, note that Bi decreases, and the onedimensional heat transfer assumption becomes more appropriate. The difference for the case when k = 500 W/m⋅K is mostly due to the approximate manner in which the heat rate is calculated in the FEA software.

PROBLEM 4.67 KNOWN: Long rectangular bar having one boundary exposed to a convection process (T∞, h) while the other boundaries are maintained at constant temperature Ts. FIND: Using the finite-element method of FEHT, (a) Determine the temperature distribution, plot the isotherms, and identify significant features of the distribution, (b) Calculate the heat rate per unit length (W/m) into the bar from the air stream, and (c) Explore the effect on the heat rate of increasing the convection coefficient by factors of two and three; explain why the change in the heat rate is not proportional to the change in the convection coefficient. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, two dimensional conduction, (2) Constant properties. ANALYSIS: (a) The symmetrical section shown in the schematic is drawn in FEHT with the specified boundary conditions and material property. The View | Temperature Contours command is used to represent ten isotherms (isopotentials) that have minimum and maximum values of 53.9°C and 85.9°C, respectively.

Because of the symmetry boundary condition, the isotherms are normal to the center-plane indicating an adiabatic surface. Note that the temperature change along the upper surface of the bar is substantial (≈ 40°C), whereas the lower half of the bar has less than a 3°C change. That is, the lower half of the bar is largely unaffected by the heat transfer conditions at the upper surface. (b, c) Using the View | Heat Flows command considering the upper surface boundary with selected convection coefficients, the heat rates into the bar from the air stream were calculated. h W / m2 ⋅ K 100 200 300

(

q′ ( W / m )

)

128

175

206

Increasing the convection coefficient by factors of 2 and 3, increases the heat rate by 37% and 61%, respectively. The heat rate from the bar to the air stream is controlled by the thermal resistances of the bar (conduction) and the convection process. Since the conduction resistance is significant, we should not expect the heat rate to change proportionally to the change in convection resistance.

PROBLEM 4.68 KNOWN: Log rod of rectangular cross-section of Problem 4.55 that experiences uniform heat generation while its surfaces are maintained at a fixed temperature. Use the finite-element software FEHT as your analysis tool. FIND: (a) Represent the temperature distribution with representative isotherms; identify significant features; and (b) Determine what heat generation rate will cause the midpoint to reach 600 K with unchanged boundary conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, and (2) Two-dimensional conduction with constant properties. ANALYSIS: (a) Using FEHT, do the following: in Setup, enter an appropriate scale; Draw the outline of the symmetrical section shown in the above schematic; Specify the Boundary Conditions (zero heat flux or adiabatic along the symmetrical lines, and isothermal on the edges). Also Specify the Material Properties and Generation rate. Draw three Element Lines as shown on the annotated version of the FEHT screen below. To reduce the mesh, hit Draw/Reduce Mesh until the desired fineness is achieved (256 elements is a good choice).

Continued …

PROBLEM 4.68 (Cont.) After hitting Run, Check and then Calculate, use the View/Temperature Contours and select the 10isopotential option to display the isotherms as shown in an annotated copy of the FEHT screen below.

The isotherms are normal to the symmetrical lines as expected since those surfaces are adiabatic. The isotherms, especially near the center, have an elliptical shape. Along the x = 0 axis and the y = 10 mm axis, the temperature gradient is nearly linear. The hottest point is of course the center for which the temperature is

(T(0, 10 mm) = 401.3 K.

<

The temperature of this point can be read using the View/Temperatures or View|Tabular Output command. (b) To determine the required generation rate so that T(0, 10 mm) = 600 K, it is necessary to re-run the model with several guessed values of q . After a few trials, find

q = 1.48 × 108 W / m3

<

PROBLEM 4.69 KNOWN: Symmetrical section of a flow channel with prescribed values of q and k, as well as the surface convection conditions. See Problem 4.5(S). FIND: Using the finite-element method of FEHT, (a) Determine the temperature distribution and plot the isotherms; identify the coolest and hottest regions, and the region with steepest gradients; describe the heat flow field, (b) Calculate the heat rate per unit length (W/m) from the outer surface A to the adjacent fluid, (c) Calculate the heat rate per unit length (W/m) to surface B from the inner fluid, and (d) Verify that the results are consistent with an overall energy balance on the section. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties. ANALYSIS: (a) The symmetrical section shown in the schematic is drawn in FEHT with the specified boundary conditions, material property and generation. The View | Temperature Contours command is used to represent ten isotherms (isopotentials) that have minimum and maximum values of 82.1°C and 125.2°C.

The hottest region of the section is the upper vertical leg (left-hand corner). The coolest region is in the lower horizontal leg at the far right-hand boundary. The maximum and minimum section temperatures (125°C and 77°C), respectively, are higher than either adjoining fluid. Remembering that heat flow lines are normal to the isotherms, heat flows from the hottest corner directly to the inner fluid and downward into the lower leg and then flows out surface A and the lower portion of surface B. Continued …..

PROBLEM 4.69 (Cont.) (b, c) Using the View | Heat Flows command considering the boundaries for surfaces A and B, the heat rates are:

q′s = 1135 W / m

q′B = −1365 W / m.

<

From an energy balance on the section, we note that the results are consistent since conservation of energy is satisfied.

E ′in − E ′out + E g = 0 −q′A + q′B + q ∀′ = 0 −1135 W / m + ( −1365 W / m ) + 2500 W / m = 0

<

where q ∀′ = 1× 106 W / m3 × [25 × 50 + 25 × 50]× 10−6 m 2 = 2500 W / m. COMMENTS: (1) For background on setting up this problem in FEHT, see the tutorial example of the User’s Manual. While the boundary conditions are different, and the internal generation term is to be included, the procedure for performing the analysis is the same. (2) The heat flow distribution can be visualized using the View | Temperature Gradients command. The direction and magnitude of the heat flow is represented by the directions and lengths of arrows. Compare the heat flow distribution to the isotherms shown above.

PROBLEM 4.70 KNOWN: Hot-film flux gage for determining the convection coefficient of an adjoining fluid stream by measuring the dissipated electric power, Pe , and the average surface temperature, Ts,f. FIND: Using the finite-element method of FEHT, determine the fraction of the power dissipation that is conducted into the quartz substrate considering three cases corresponding to convection coefficients 2 of 500, 1000 and 2000 W/m ⋅K. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant substrate properties, (3) Uniform convection coefficient over the hot-film and substrate surfaces, (4) Uniform power dissipation over hot film. ANALYSIS: The symmetrical section shown in the schematic above (right) is drawn into FEHT specifying the substrate material property. On the upper surface, a convection boundary condition

(

(T∞,h) is specified over the full width W/2. Additionally, an applied uniform flux Pe′′, W / m 2

)

boundary condition is specified for the hot-film region (w/2). The remaining surfaces of the twodimensional system are specified as adiabatic. In the schematic below, the electrical power dissipation Pe′ (W/m) in the hot film is transferred by convection from the film surface, q′cv,f , and from the adjacent substrate surface, q′cv,s .

The analysis evaluates the fraction, F, of the dissipated electrical power that is conducted into the substrate and convected to the fluid stream,

F = q′cv,s / Pe′ = 1 − q′cv,f / Pe′ where Pe′ = Pe′′ ( w / 2 ) = 5000 W / m 2 × ( 0.002 m ) = 10 W / m. After solving for the temperature distribution, the View|Heat Flow command is used to evaluate q′cv,f for the three values of the convection coefficient. Continued …..

PROBLEM 4.70 (Cont.) 2

h(W/m ⋅K)

Case 1 2 3

q′cv,f ( W / m )

500 1000 2000

5.64 6.74 7.70

F(%)

Ts,f (°C)

43.6 32.6 23.3

30.9 28.6 27.0

COMMENTS: (1) For the ideal hot-film flux gage, there is negligible heat transfer to the substrate, and the convection coefficient of the air stream is calculated from the measured electrical power, Pe′′, the average film temperature (by a thin-film thermocouple), Ts,f, and the fluid stream temperature, T∞, as h = Pe′′ / Ts,f − T∞ . The purpose in performing the present analysis is to estimate a correction

(

)

factor to account for heat transfer to the substrate. (2) As anticipated, the fraction of the dissipated electrical power conducted into the substrate, F, decreases with increasing convection coefficient. For the case of the largest convection coefficient, F amounts to 25%, making it necessary to develop a reliable, accurate heat transfer model to estimate the applied correction. Further, this condition limits the usefulness of this gage design to flows with high convection coefficients. (3) A reduction in F, and hence the effect of an applied correction, could be achieved with a substrate material having a lower thermal conductivity than quartz. However, quartz is a common substrate material for fabrication of thin-film heat-flux gages and thermocouples. By what other means could you reduce F? (4) In addition to the tutorial example in the FEHT User’s Manual, the solved models for Examples 4.3 and 4.4 are useful for developing skills helpful in solving this problem.

PROBLEM 4.71 KNOWN: Hot-plate tool for micro-lithography processing of 300-mm silicon wafer consisting of an aluminum alloy equalizing block (EB) heated by ring-shaped main and trim electrical heaters (MH and TH) providing two-zone control. FIND: The assignment is to size the heaters, MH and TH, by specifying their applied heat fluxes, q′′mh and q′′th , and their radial extents, ∆rmh and ∆rth , to maintain an operating temperature of 140°C with a uniformity of 0.1°C. Consider these steps in the analysis: (a) Perform an energy balance on the EB to obtain an initial estimate for the heater fluxes with q′′mh = q′′th extending over the full radial limits; using FEHT, determine the upper surface temperature distribution and comment on whether the desired uniformity has been achieved; (b) Re-run your FEHT code with different values of the heater fluxes to obtain the best uniformity possible and plot the surface temperature distribution; (c) Re-run your FEHT code for the best arrangement found in part (b) using the representative distribution of the convection coefficient (see schematic for h(r) for downward flowing gas across the upper surface of the EB; adjust the heat flux of TH to obtain improved uniformity; and (d) Suggest changes to the design for improving temperature uniformity. SCHEMATIC:

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ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction with uniform and constant properties in EB, (3) Lower surface of EB perfectly insulated, (4) Uniform convection coefficient over upper EB surface, unless otherwise specified and (5) negligible radiation exchange between the EB surfaces and the surroundings. ANALYSIS: (a) To obtain initial estimates for the MH and TH fluxes, perform an overall energy balance on the EB as illustrated in the schematic below.

E in − E out = 0

(

)

(

)

q′′mhπ r22 − r12 + q′′thπ r42 − r32 − h π ro2 + 2π ro w  ( Ts − T∞ ) = 0   Continued …..

PROBLEM 4.71 (Cont.) Substituting numerical values and letting q′′mh = q′′th , find

q′′mh = q′′th = 2939 W / m 2

<

Using FEHT, the analysis is performed on an axisymmetric section of the EB with the nodal arrangement as shown below.

The Temperature Contour view command is used to create the temperature distribution shown below. The temperatures at the center (T1) and the outer edge of the wafer (r = 150 mm, T14) are read from the Tabular Output page. The Temperature Gradients view command is used to obtain the heat flow distribution when the line length is proportional to the magnitude of the heat rate.

From the analysis results, for this base case design ( q′′mh = q′′th ) , the temperature difference across the radius of the wafer is 1.7°C, much larger than the design goal of 0.1°C. The upper surface temperature distribution is shown in the graph below.

Continued …..

PROBLEM 4.71 (Cont.)

EB surface temperature distribution 141.5

141

T(r,z), (C)

140.5

140

139.5

139

138.5 0

20

40

60

80

100

120

140

160

Radial position, r (mm)

(b) From examination of the results above, we conclude that if q′′mh is reduced and q′′th increased, the EB surface temperature uniformity could improve. The results of three trials compared to the base case are tabulated below. Trial

(

q′′mh W/m

2

)

(

q′′th W / m2

)

T1

T14

(°C )

(°C )

T1 − T14 ( °C )

Base

2939

2939

141.1

139.3

1.8

1

2880 (-2%)

2997 (+2%)

141.1

139.4

1.7

2

2880 (-2%)

3027 (+3%)

141.7

140.0

1.7

3

2910 (-1%)

2997 (+2%)

141.7

139.9

1.8

2939

2939

141.7

139.1

2.6

Part (d) 2939 k=150 W/m⋅K

2939

140.4

139.5

0.9

Part (d) 2939 k=300 W/m⋅K

2939

140.0

139.6

0.4

Part (c)

The strategy of changing the heater fluxes (trials 1-3) has not resulted in significant improvements in the EB surface temperature uniformity. Continued …..

PROBLEM 4.71 (Cont.) (c) Using the same FEHT code as with part (b), base case, the boundary conditions on the upper surface of the EB were specified by the function h(r) shown in the schematic. The value of h(r) 2 ranged from 5.4 to 13.5 W/m ⋅K between the centerline and EB edge. The result of the analysis is tabulated above, labeled as part (c). Note that the temperature uniformity has become significantly poorer. (d) There are at least two options that should be considered in the re-design to improve temperature uniformity. Higher thermal conductivity material for the EB. Aluminum alloy is the material most widely used in practice for reasons of low cost, ease of machining, and durability of the heated surface. The results of analyses for thermal conductivity values of 150 and 300 W/m⋅K are tabulated above, labeled as part (d). Using pure or oxygen-free copper could improve the temperature uniformity to better than 0.5°C. Distributed heater elements. The initial option might be to determine whether temperature uniformity could be improved using two elements, but located differently. Another option is a single element heater spirally embedded in the lower portion of the EB. By appropriately positioning the element as a function of the EB radius, improved uniformity can be achieved. This practice is widely used where precise and uniform temperature control is needed.

PROBLEM 4.72 KNOWN: Straight fin of uniform cross section with insulated end. FIND: (a) Temperature distribution using finite-difference method and validity of assuming onedimensional heat transfer, (b) Fin heat transfer rate and comparison with analytical solution, Eq. 3.76, (c) Effect of convection coefficient on fin temperature distribution and heat rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fin, (3) Constant properties, (4) Uniform film coefficient. ANALYSIS: (a) From the analysis of Problem 4.45, the finite-difference equations for the nodal arrangement can be directly written. For the nodal spacing ∆x = 4 mm, there will be 12 nodes. With " >> w representing the distance normal to the page,

)

(

hP h ⋅ 2 h ⋅2 2 500 W m 2 ⋅ K × 2 4 × 10−3 mm = 0.0533 ⋅ ∆x 2 ≈ ∆x 2 = ∆x = 3 − kA c k ⋅⋅w kw 50 W m ⋅ K × 6 ×10 m Node 1: Node n: Node 12:

100 + T2 + 0.0533 × 30 − ( 2 + 0.0533) T1 = 0 or

-2.053T1 + T2 = -101.6

Tn +1 + Tn −1 + 1.60 − 2.0533Tn = 0 or T11 + ( 0.0533 2 ) 30 − ( 0.0533 2 + 1) T12 = 0 or

Tn −1 − 2.053Tn + Tn −1 = −1.60 T11 − 1.0267T12 = −0.800

Using matrix notation, Eq. 4.52, where [A] [T] = [C], the A-matrix is tridiagonal and only the non-zero terms are shown below. A matrix inversion routine was used to obtain [T]. Tridiagonal Matrix A Nonzero Terms a1,1 a1,2 a2,1 a2,2 a2,3 a3,2 a3,3 a3,4 a4,3 a4,4 a4,5 a5,4 a5,5 a5,6 a6,5 a6,6 a6,7 a7,6 a7,7 a7,8 a8,7 a8,8 a8,9 a9,8 a9,9 a9,10 a10,9 a10,10 a10,11 a11,10 a11,11 a11,12 a12,11 a12,12 a12,13

1 1 1 1 1 1 1 1 1 1 1

Column Matrices Values -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -1.027

1 1 1 1 1 1 1 1 1 1 1 1

Node 1 2 3 4 5 6 7 8 9 10 11 12

C -101.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -0.8

T 85.8 74.5 65.6 58.6 53.1 48.8 45.5 43.0 41.2 39.9 39.2 38.9

The assumption of one-dimensional heat conduction is justified when Bi ≡ h(w/2)/k < 0.1. Hence, with Bi = 500 W/m2⋅K(3 × 10-3 m)/50 W/m⋅K = 0.03, the assumption is reasonable. Continued...

PROBLEM 4.72 (Cont.) (b) The fin heat rate can be most easily found from an energy balance on the control volume about Node 0, T −T  ∆x  q′f = q1′ + q′conv = k ⋅ w 0 1 + h  2  ( T0 − T∞ ) ∆x  2 

(

q′f = 50 W m ⋅ K 6 × 10−3 m

)

(100 − 85.8 )$ C 4 × 10

−3



+ 500 W m2 ⋅ K  2 ⋅

 

m

4 × 10−3 m  2

$  (100 − 30 ) C  

q′f = (1065 + 140 ) W m = 1205 W m . From Eq. 3.76, the fin heat rate is q = ( hPkA c )

1/ 2

<

⋅ θ b ⋅ tanh mL .

Substituting numerical values with P = 2(w +  ) ≈ 2  and Ac = w⋅  , m = (hP/kAc)1/2 = 57.74 m-1 and M = (hPkAc)1/2 = 17.32  W/K. Hence, with θb = 70°C, q′ = 17.32 W K × 70 K × tanh (57.44 × 0.048 ) = 1203 W m and the finite-difference result agrees very well with the exact (analytical) solution. (c) Using the IHT Finite-Difference Equations Tool Pad for 1D, SS conditions, the fin temperature distribution and heat rate were computed for h = 10, 100, 500 and 1000 W/m2⋅K. Results are plotted as follows. 100

1800 1500

80 70

Heat rate, q'(W/m)

Temperature, T(C)

90

60 50 40 30 0

8

16

24

32

40

48

1200 900 600 300

Fin location, x(mm) h = 10 W/m^2.K h = 100 W/m^2.K h = 500 W/m^2.K h = 1000 W/m^2.K

0 0

200

400

600

800

1000

Convection coefficient, h(W/m^2.K)

The temperature distributions were obtained by first creating a Lookup Table consisting of 4 rows of nodal temperatures corresponding to the 4 values of h and then using the LOOKUPVAL2 interpolating function with the Explore feature of the IHT menu. Specifically, the function T_EVAL = LOOKUPVAL2(t0467, h, x) was entered into the workspace, where t0467 is the file name given to the Lookup Table. For each value of h, Explore was used to compute T(x), thereby generating 4 data sets which were placed in the Browser and used to generate the plots. The variation of q′ with h was simply generated by using the Explore feature to solve the finite-difference model equations for values of h incremented by 10 from 10 to 1000 W/m2⋅K. Although q′f increases with increasing h, the effect of changes in h becomes less pronounced. This trend is a consequence of the reduction in fin temperatures, and hence the fin efficiency, with increasing h. For 10 ≤ h ≤ 1000 W/m2⋅K, 0.95 ≥ ηf ≥ 0.24. Note the nearly isothermal fin for h = 10 W/m2⋅K and the pronounced temperature decay for h = 1000 W/m2⋅K.

PROBLEM 4.73 KNOWN: Pin fin of 10 mm diameter and length 250 mm with base temperature of 100°C experiencing radiation exchange with the surroundings and free convection with ambient air. FIND: Temperature distribution using finite-difference method with five nodes. Fin heat rate and relative contributions by convection and radiation. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fin, (3) Constant properties, (4) Fin approximates small object in large enclosure, (5) Fin tip experiences convection and radiation, (6) hfc = 2.89[0.6 + 0.624(T - T∞)1/6]2. ANALYSIS: To apply the finite-difference method, define the 5-node system shown above where ∆x = L/5. Perform energy balances on the nodes to obtain the finite-difference equations for the nodal temperatures. Interior node, n = 1, 2, 3 or 4

E in − E out = 0

qa + q b + qc + qd = 0

(1)

T T −T −T h r,n P∆x (Tsur − Tn ) + kA c n +1 n + h fc,n P∆x (T∞ − Tn ) + kA c n −1 n = 0 ∆x ∆x

(2)

where the free convection coefficient is 1/ 6  2

h fc,n = 2.89 0.6 + 0.624 ( Tn − T∞ )  and the linearized radiation coefficient is 2 h r,n = εσ (Tn + Tsur ) Tn2 + Tsur with P = πD and Ac = πD2/4.

(



)

(3)

(4) (5,6)

Tip node, n = 5

E in − E out = 0 qa + q b + qc + qd + qe = 0 h r,5 ( P∆x 2 )(Tsur − T5 ) + h r,5A c (Tsur − T5 ) + h fc,5Ac (T∞ − T5 ) T −T + h fc,5 ( P∆x 2 )( T∞ − T5 ) + kAc 4 5 = 0 ∆x

(7) Continued...

PROBLEM 4.73 (Cont.) Knowing the nodal temperatures, the heat rates are evaluated as: Fin Heat Rate: Perform an energy balance around Node b.

E in − E out = 0 qa + q b + q c + q fin = 0 h r,b ( P∆x 2 )(Tsur − Tb ) + h fc,b ( P∆x 2 )(T∞ − Tb ) + kA c

(T1 − Tb ) + q ∆x

fin = 0

(8)

where hr,b and hfc,b are evaluated at Tb. Convection Heat Rate: To determine the portion of the heat rate by convection from the fin surface, we need to sum contributions from each node. Using the convection heat rate terms from the foregoing energy balances, for, respectively, node b, nodes 1, 2, 3, 4 and node 5.

qcv = − q b )b − ∑ qc )1− 4 − ( qc + q d )5

(9)

Radiation Heat Rate: In the same manner,

q rad = − qa )b − ∑ q b )1− 4 − ( qa + q b )5 The above equations were entered into the IHT workspace and the set of equations solved for the nodal temperatures and the heat rates. Summary of key results including the temperature distribution and heat rates is shown below. Node

b

1

2

3

4

5

Fin

Tj (°C) qcv (W) qfin (W) qrad (W) hcv (W/m2⋅K) hrod (W/m2⋅K)

100 0.603 10.1 1.5

58.5 0.451 8.6 1.4

40.9 0.183 7.3 1.3

33.1 0.081 6.4 1.3

29.8 0.043 5.7 1.2

28.8 0.015 5.5 1.2

1.375 1.604 0.229 -

<

COMMENTS: From the tabulated results, it is evident that free convection is the dominant node. Note that the free convection coefficient varies almost by a factor of two over the length of the fin.

PROBLEM 4.74 KNOWN: Thin metallic foil of thickness, t, whose edges are thermally coupled to a sink at temperature Tsink is exposed on the top surface to an ion beam heat flux, q „„s , and experiences radiation exchange with the vacuum enclosure walls at Tsur. FIND: Temperature distribution across the foil. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction in the foil, (2) Constant properties, (3) Upper and lower surfaces of foil experience radiation exchange, (4) Foil is of unit length normal to the page. ANALYSIS: The 10-node network representing the foil is shown below.

From an energy balance on node n, E in − E out = 0 , for a unit depth, q′a + q′b + q′c + q′d + q′e = 0

q′′s ∆x + h r,n ∆x ( Tsur − Tn ) + k ( t )( Tn +1 − Tn ) ∆x + h r,n ∆x ( Tsur − Tn ) + k ( t )( Tn −1 − Tn ) ∆x = 0 (1)

where the linearized radiation coefficient for node n is

(

2 h r,n = εσ ( Tsur + Tn ) Tsur + Tn2

Solving Eq. (1) for Tn find,

)

(

(2)

)

(

)

(

)

Tn = ( Tn +1 + Tn −1 ) + 2h r,n ∆x 2 kt Tsur + ∆x 2 kt qs′′   h r,n ∆x 2 kt + 2  (3)     which, considering symmetry, applies also to node 1. Using IHT for Eqs. (3) and (2), the set of finitedifference equations was solved for the temperature distribution (K): T1 374.1

T2 374.0

T3 373.5

T4 372.5

T5 370.9

T6 368.2

T7 363.7

T8 356.6

T9 345.3

T10 327.4 Continued...

PROBLEM 4.74 (Cont.) COMMENTS: (1) If the temperature gradients were excessive across the foil, it would wrinkle; most likely since its edges are constrained, the foil will bow. (2) The IHT workspace for the finite-difference analysis follows: // The nodal equations: T1 = ( (T2 + T2) + A1 * Tsur + B *q''s ) / ( A1 + 2) A1= 2 * hr1 * deltax^2 / (k * t) hr1 = eps * sigma * (Tsur + T1) * (Tsur^2 + T1^2) sigma = 5.67e-8 B = deltax^2 / (k * t) T2 = ( (T1 + T3) + A2 * Tsur + B *q''s ) / ( A2 + 2) A2= 2 * hr2 * deltax^2 / (k * t) hr2 = eps * sigma * (Tsur + T2) * (Tsur^2 + T2^2) T3 = ( (T2 + T4) + A3 * Tsur + B *q''s ) / ( A3 + 2) A3= 2 * hr3 * deltax^2 / (k * t) hr3 = eps * sigma * (Tsur + T3) * (Tsur^2 + T3^2) T4 = ( (T3 + T5) + A4 * Tsur + B *q''s ) / ( A4 + 2) A4= 2 * hr4 * deltax^2 / (k * t) hr4 = eps * sigma * (Tsur + T4) * (Tsur^2 + T4^2) T5 = ( (T4 + T6) + A5 * Tsur + B *q''s ) / ( A5 + 2) A5= 2 * hr5 * deltax^2 / (k * t) hr5 = eps * sigma * (Tsur + T5) * (Tsur^2 + T5^2) T6 = ( (T5 + T7) + A6 * Tsur + B *q''s ) / ( A6 + 2) A6= 2 * hr6 * deltax^2 / (k * t) hr6 = eps * sigma * (Tsur + T6) * (Tsur^2 + T6^2) T7 = ( (T6 + T8) + A7 * Tsur + B *q''s ) / ( A7 + 2) A7= 2 * hr7 * deltax^2 / (k * t) hr7 = eps * sigma * (Tsur + T7) * (Tsur^2 + T7^2) T8 = ( (T7 + T9) + A8 * Tsur + B *q''s ) / ( A8 + 2) A8= 2 * hr8 * deltax^2 / (k * t) hr8 = eps * sigma * (Tsur + T8) * (Tsur^2 + T8^2) T9 = ( (T8 + T10) + A9 * Tsur + B *q''s ) / ( A9 + 2) A9= 2 * hr9 * deltax^2 / (k * t) hr9 = eps * sigma * (Tsur + T9) * (Tsur^2 + T9^2) T10 = ( (T9 + Tsink) + A10 * Tsur + B *q''s ) / ( A10 + 2) A10= 2 * hr10 * deltax^2 / (k * t) hr10 = eps * sigma * (Tsur + T10) * (Tsur^2 + T10^2) // Assigned variables deltax = L / 10 L = 0.150 t = 0.00025 eps = 0.45 Tsur = 300 k = 40 Tsink = 300 q''s = 600

// Spatial increment, m // Foil length, m // Foil thickness, m // Emissivity // Surroundings temperature, K // Foil thermal conductivity, W/m.K // Sink temperature, K // Ion beam heat flux, W/m^2

/* Data Browser results: Temperature distribution (K) and linearized radiation cofficients (W/m^2.K): T1 T2 374.1 374

T3 373.5

T4 372.5

T5 370.9

T6 368.2

T7 363.7

T8 356.6

T9 345.3

T10 327.4

hr1 3.956

hr3 3.943

hr4 3.926

hr5 3.895

hr6 3.845

hr7 3.765

hr8 3.639

hr9 3.444

hr10 3.157 */

hr2 3.953

PROBLEM 4.75 KNOWN: Electrical heating elements with known dissipation rate embedded in a ceramic plate of known thermal conductivity; lower surface is insulated, while upper surface is exposed to a convection process. FIND: (a) Temperature distribution within the plate using prescribed grid spacing, (b) Sketch isotherms to illustrate temperature distribution, (c) Heat loss by convection from exposed surface (compare with element dissipation rate), (d) Advantage, if any, in not setting ∆x = ∆y, (e) Effect of grid size and convection coefficient on the temperature field. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, two-dimensional conduction in ceramic plate, (2) Constant properties, (3) No internal generation, except for Node 7 (or Node 15 for part (e)), (4) Heating element approximates a line source of negligible wire diameter. ANALYSIS: (a) The prescribed grid for the symmetry element shown above consists of 12 nodal points. Nodes 1-3 are points on a surface experiencing convection; nodes 4-6 and 8-12 are interior nodes. Node 7 is a special case of the interior node having a generation term; because of symmetry, q′ht = 25 W/m. The finite-difference equations are derived as follows: Continued...

PROBLEM 4.75 (Cont.) Surface Node 2. From an energy balance on the prescribed control volume with ∆x/∆y = 3, E in − E out = q′a + q′b + q′c + q′d = 0;

k

T −T ∆y T1 − T2 ∆y T3 − T2 + h∆x (T∞ − T2 ) + k + k∆x 5 2 = 0 . 2 ∆x 2 ∆x ∆y

Regrouping, find 2 2   ∆x    ∆x  ∆x ∆x  T2 1 + 2N T5 + 2N T∞ +1+ 2 = T1 + T3 + 2     ∆y ∆y   ∆y  ∆y    

where N = h∆x/k = 100 W/m2⋅K × 0.006 m/2 W/m⋅K = 0.30 K. Hence, with T∞ = 30°C,

T2 = 0.04587T1 + 0.04587T3 + 0.82569T5 + 2.4771

(1)

From this FDE, the forms for nodes 1 and 3 can also be deduced. Interior Node 7. From an energy balance on the prescribed control volume, with ∆x/∆y = 3,  ′ = 2 q ′ and E′ represents the conduction terms. Hence, E ′in − E ′g = 0 , where E g in ht q′a + q′b + q′c + q′d + 2q′ht = 0 , or T − T7 T − T7 T − T7 T − T7 k∆y 8 + k∆x 4 + k∆y 8 + k∆x 10 + 2q′ht = 0 ∆x ∆y ∆x ∆y Regrouping, 2 2 2   ∆x 2 2q′  ∆x   ∆x    ∆x   ∆x    T7 1 +  T4 + T8 +  T10 + ht  +1+  = T8 +       k  ∆y    ∆y   ∆y    ∆y   ∆y 

Recognizing that ∆x/∆y = 3, q′ht = 25 W/m and k = 2 W/m⋅K, the FDE is

T7 = 0.0500T8 + 0.4500T4 + 0.0500T8 + 0.4500T10 + 3.7500 (2) The FDEs for the remaining nodes may be deduced from this form. Following the procedure described in Section 4.5.2 for the Gauss-Seidel method, the system of FDEs has the form: k

k −1

k

k

k

k

T1 = 0.09174T2

k −1

+ 0.8257T4

k −1

T2 = 0.04587T1 + 0.04587T3

k −1

T3 = 0.09174T2 + 0.8257T6

+ 2.4771 k −1

+ 0.8257T5

+ 2.4771

k

k

k −1

k

k

k

k −1

k

k

k

k −1

k

k

k −1

k

k

k

k −1

k

k

k

k −1

k

k

k −1

k

k

k −1

k

k

k

T4 = 0.4500T1 + 0.1000T5

+ 2.4771

k −1

+ 0.4500T7

T5 = 0.4500T2 + 0.0500T4 + 0.0500T6

k −1

+ 0.4500T8

T6 = 0.4500T3 + 0.1000T5 + 0.4500T9 T7 = 0.4500T4 + 0.1000T8

k −1

+ 0.4500T10

T8 = 0.4500T5 + 0.0500T7 + 0.0500T9

+ 3.7500 k −1

+ 0.4500T11

T9 = 0.4500T6 + 0.1000T8 + 0.4500T12 T10 = 0.9000T7 + 0.1000T11 T11 = 0.9000T8 + 0.0500T10

k −1

+ 0.0500T12

T12 = 0.9000T9 + 0.1000T11

Continued …..

PROBLEM 4.75 (Cont.) Note the use of the superscript k to denote the level of iteration. Begin the iteration procedure with rational estimates for Ti (k = 0) and prescribe the convergence criterion as ε ≤ 0.1. k/Ti

1

2

3

4

5

6

7

8

9

10

11

12

0 1 2

55.0 57.4 57.1

50.0 51.7 51.6

45.0 46.0 46.9

61.0 60.4 59.7

54.0 53.8 53.2

47.0 48.1 48.7

65.0 63.5 64.3

56.0 54.6 54.3

49.0 49.6 49.9

60.0 62.7 63.4

55.0 54.8 54.5

50.0 50.1 50.4

∞

55.80

49.93

47.67

59.03

51.72

49.19

63.89

52.98

50.14

62.84

53.35

50.46

The last row with k = ∞ corresponds to the solution obtained by matrix inversion. It appears that at least 20 iterations would be required to satisfy the convergence criterion using the Gauss-Seidel method. (b) Selected isotherms are shown in the sketch of the nodal network.

Note that the isotherms are normal to the adiabatic surfaces. (c) The heat loss by convection can be expressed as 1 1  q′conv = h  ∆x ( T1 − T∞ ) + ∆x ( T2 − T∞ ) + ∆x ( T3 − T∞ ) 2 2   2

q ′conv = 100 W m ⋅ K × 0.006 m

 1 (55.80 − 30 ) + ( 49.93 − 30 ) + 1 ( 47.67 − 30 ) = 25.00 W m .  2  2

<

As expected, the heat loss by convection is equal to the heater element dissipation. This follows from the conservation of energy requirement. (d) For this situation, choosing ∆x = 3∆y was advantageous from the standpoint of precision and effort. If we had chosen ∆x = ∆y = 2 mm, there would have been 28 nodes, doubling the amount of work, but with improved precision. (e) Examining the effect of grid size by using the Finite-Difference Equations option from the Tools portion of the IHT Menu, the following temperature field was computed for ∆x = ∆y = 2 mm, where x and y are in mm and the temperatures are in °C. y\x 0 2 4 6

0 55.04 58.71 66.56 63.14

2 53.88 56.61 59.70 59.71

4 52.03 54.17 55.90 56.33

6 50.32 52.14 53.39 53.80

8 49.02 50.67 51.73 52.09

10 48.24 49.80 50.77 51.11

12 47.97 49.51 50.46 50.78 Continued …..

PROBLEM 4.75 (Cont.) Agreement with the results of part (a) is excellent, except in proximity to the heating element, where T15 = 66.6°C for the fine grid exceeds T7 = 63.9°C for the coarse grid by 2.7°C. For h = 10 W/m2⋅K, the maximum temperature in the ceramic corresponds to T15 = 254°C, and the heater could still be operated at the prescribed power. With h = 10 W/m2⋅K, the critical temperature of T15 = 400°C would be reached with a heater power of approximately 82 W/m. COMMENTS: (1) The method used to obtain the rational estimates for Ti (k = 0) in part (a) is as follows. Assume 25 W/m is transferred by convection uniformly over the surface; find Tsurf ≈ 50°C. Set T2 = 50°C and recognize that T1 and T3 will be higher and lower, respectively. Assume 25 W/m is conducted uniformly to the outer nodes; find T5 - T2 ≈ 4°C. For the remaining nodes, use intuition to guess reasonable values. (2) In selecting grid size (and whether ∆x = ∆y), one should consider the region of largest temperature gradients. NOTE TO INSTRUCTOR: Although the problem statement calls for calculations with ∆x = ∆y = 1 mm, the instructional value and benefit-to-effort ratio are small. Hence, consideration of this grid size is not recommended.

PROBLEM 4.76 KNOWN: Silicon chip mounted in a dielectric substrate. One surface of system is convectively cooled while the remaining surfaces are well insulated. FIND: Whether maximum temperature in chip will exceed 85°C. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Negligible contact resistance between chip and substrate, (4) Upper surface experiences uniform convection coefficient, (5) Other surfaces are perfectly insulated. ANALYSIS: Performing an energy balance on the chip assuming it is perfectly insulated from the substrate, the maximum temperature occurring at the interface with the dielectric substrate will be, according to Eqs. 3.43 and 3.46, Tmax =

2 q& ( H/4 )

2k c

+

q& ( H/4 ) h

7

+ T∞ =

10 W/m

3

( 0.003 m ) 2

2 × 50 W/m ⋅ K

7

+

10 W/m

3

( 0.003 m) 2

500 W/m ⋅ K

o

o

+ 20 C = 80.9 C.

Since Tmax < 85°C for the assumed situation, for the actual two-dimensional situation with the conducting dielectric substrate, the maximum temperature should be less than 80°C. Using the suggested grid spacing of 3 mm, construct the nodal network and write the finite-difference equation for each of the nodes taking advantage of symmetry of the system. Note that we have chosen to not locate nodes on the system surfaces for two reasons: (1) fewer total number of nodes, 20 vs. 25, and (2) Node 5 corresponds to center of chip which is likely the point of maximum temperature. Using these numerical values,

h∆x 500 W/m2 ⋅K × 0.003 m = = 0.30 ks 5 W/m ⋅ K h∆x 500 W/m2 ⋅K × 0.003 m = = 0.030 kc 5 W/m ⋅ K q& ∆x∆y = 1.800 kc

α=

2 2 = = 1.818 ( k s / k c ) +1 5/50 +1

2 2 = = 0.182 ( k c / k s ) +1 50/5 +1 1 γ = = 0.0910 kc / ks + 1 β=

find the nodal equations: Node 1

T −T T −T k s∆x 6 1 + k s∆y 2 1 + h∆x ( T∞ − T1 ) = 0 ∆y ∆x Continued …..

PROBLEM 4.76 (Cont.) 

− 2 +



 h ∆x T1 + T2 + T6 = − T∞  ks  ks

h ∆x

− 2.30T1 + T2 + T6 = − 6.00

(1)

T1 − 3.3T2 + T3 + T7 = −6.00

Node 2 (2) Node 3 T − T3 k s ∆y 2 + ∆x

T4 − T3 T − T3 + k s∆ x 8 + h∆ x ( T∞ − T3 ) = 0 ( ∆x/2 ) / kc ∆y + ( ∆x/2 ) / ks ∆y ∆y

T2 − ( 2 + α + ( h∆ x/k s ) T3 ) + α T4 + T8 = − ( h ∆x/k ) T∞ T2 − 4.12T3 + 1.82T4 + T8 = −6.00

(3)

Node 4 T3 − T4 T − T4 T9 − T4 + k c ∆y 5 + ( ∆x/2 ) / ks ∆y + ( ∆x/2 ) / kc ∆y ∆x ( ∆y/2 ) / ks ∆x + ( ∆y/2 ) k c∆x + q& ( ∆ x∆y ) + h ∆ x ( T∞ − T4 ) = 0

β T3 − (1 + 2 β + [ h∆ x/k c ]) T4 + T5 + β T9 = − ( h∆ x/kc ) T∞ − q&∆ x∆ y/kc 0.182T3 − 1.39T4 + T5 + 0.182T9 = −2.40

(4)

Node 5 T − T5 T10 − T5 k c ∆y 4 + + h ( ∆x/2 )( T∞ − T5 ) + q& ∆y ( ∆x/2 ) = 0 ∆x ( ∆y/2) / ks ( ∆x/2 ) + ( ∆y/2 ) / k c ( ∆x/2 ) 2T4 − 2.21T5 + 0.182T10 = −2.40

(5)

Nodes 6 and 11

k s∆x ( T1 − T 6 ) / ∆ y + ks∆y ( T7 − T 6 ) / ∆x + k s∆x ( T11 − T6 ) / ∆y = 0 T1 − 3T6 + T7 + T11 = 0

T6 − 3T11 + T12 + T16 = 0

(6,11)

Nodes 7, 8, 12, 13, 14 Treat as interior points, T2 + T6 − 4T7 + T8 + T12 = 0

T3 + T7 − 4T8 + T9 + T13 = 0

T7 + T11 − 4T12 + T13 + T17 = 0

(7,8) (12,13) (14)

T8 + T12 − 4T13 + T14 + T18 = 0

T9 + T13 − 4T14 + T15 + T19 = 0

Node 9 T − T9 T4 − T9 T − T9 T − T9 k s ∆y 8 + + k s∆y 10 + k s∆x 14 =0 ∆x ( ∆y/2 ) / k c ∆x + ( ∆y/2 ) / k s∆x ∆x ∆y 1.82T4 + T8 − 4.82T9 + T10 + T14 = 0

(9)

Node 10 Using the result of Node 9 and considering symmetry, 1.82T5 + 2T9 − 4.82T10 + T15 = 0

Node 15 Interior point considering symmetry T10 + 2T14 − 4T15 + T20 = 0 Node 16 By inspection,

(10) (15)

T11 − 2T16 + T17 = 0

(16) Continued …..

PROBLEM 4.76 (Cont.) Nodes 17, 18, 19, 20 T12 + T16 − 3T17 + T18 = 0 T14 + T18 − 3T19 + T20 = 0

T13 + T17 − 3T18 + T19 = 0 T15 + 2T19 − 3T20 = 0

(17,18) (19,20)

Using the matrix inversion method, the above system of finite-difference equations is written in matrix notation, Eq. 4.52, [A][T] = [C] where -2.3 1 0 0 0 1 -3.3 1 0 0 0 1 -4.12 1.82 0 0 0 .182 -1.39 1 0 0 0 2 -2.21 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1.82 0 0 0 0 0 1.82 [A] = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 0 0 0 0 -3 1 0 0 0 1 0 0 0 0 0 0 0 0 0

0 1 0 0 0 1 -4 1 0 0 0 1 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 .182 0 0 0 0 0 0 0 0 0 0 0 0 0 .182 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 -4 1 0 0 0 1 0 0 0 0 0 0 0 1 -4.82 1 0 0 0 1 0 0 0 0 0 0 0 2 -4.82 0 0 0 0 1 0 0 0 0 0 0 0 0 -3 1 0 0 0 1 0 0 0 0 0 0 0 1 -4 1 0 0 0 1 0 0 0 1 0 0 0 1 -4 1 0 0 0 1 0 0 0 1 0 0 0 1 -4 1 0 0 0 1 0 0 0 1 0 0 0 2 -4 0 0 0 0 1 0 0 0 1 0 0 0 0 -2 1 0 0 0 0 0 0 0 1 0 0 0 1 -3 1 0 0 0 0 0 0 0 1 0 0 0 1 -3 1 0 0 0 0 0 0 0 1 0 0 0 1 -3 1 0 0 0 0 0 0 0 1 0 0 0 2 -3

[C] =

-6 -6 -6 -2.4 -2.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

and the temperature distribution (°C), in geometrical representation, is

34.46 37.13 38.56 39.16

36.13 38.37 39.38 39.77

40.41 40.85 40.81 40.76

45.88 43.80 42.72 41.70

46.23 44.51 42.78 42.06

The maximum temperature is T5 = 46.23°C which is indeed less than 85°C.

<

COMMENTS: (1) The convection process for the energy balances of Nodes 1 through 5 were simplified by assuming the node temperature is also that of the surface. Considering Node 2, the energy balance processes for qa, qb and qc are identical (see Eq. (2)); however,

q conv =

T∞ − T2 ≈ h ( T∞ − T2 ) 1/h + ∆y/2k 2

-4

where h∆y/2k = 5 W/m ⋅K×0.003 m/2×50 W/m⋅K = 1.5×10 simplification is justified.

0 will be a constant since the flux, q′′x ( 0 ), is a constant. Noting that To = T(0,∞), the steady-state temperature distribution will be linear such that T − T ( L,∞ ) q′′o = k o = h T ( L,∞ ) − T∞  . L (b) The heat flux at the front surface, x = L, is given by q′′x ( L,t ) = −k ( dT/dx ) x=L . From the temperature distribution, we can construct the heat flux-time plot.

COMMENTS: At early times, the temperature and heat flux at x = L will not change from their initial values. Hence, we show a zero slope for q′′x ( L,t ) at early times. Eventually, the value of q′′x ( L,t ) will reach the steady-state value which is q′′o .

PROBLEM 5.2 KNOWN: Plane wall whose inner surface is insulated and outer surface is exposed to an airstream at T∞. Initially, the wall is at a uniform temperature equal to that of the airstream. Suddenly, a radiant source is switched on applying a uniform flux, q′′o , to the outer surface. FIND: (a) Sketch temperature distribution on T-x coordinates for initial, steady-state, and two intermediate times, (b) Sketch heat flux at the outer surface, q′′x ( L,t ) , as a function of time. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal generation, E g = 0, (4) Surface at x = 0 is perfectly insulated, (5) All incident radiant power is absorbed and negligible radiation exchange with surroundings. ANALYSIS: (a) The temperature distributions are shown on the T-x coordinates and labeled accordingly. Note these special features: (1) Gradient at x = 0 is always zero, (2) gradient is more steep at early times and (3) for steady-state conditions, the radiant flux is equal to the convective heat flux (this follows from an energy balance on the CS at x = L), q ′′o = q ′′conv = h [T ( L,∞ ) − T∞ ].

(b) The heat flux at the outer surface, q′′x ( L,t ) , as a function of time appears as shown above. COMMENTS: The sketches must reflect the initial and boundary conditions: T(x,0) = T∞ ∂ T −k x=0 = 0 ∂ x ∂ T −k x=L = h  T ( L,t ) − T∞  − q′′o ∂ x

uniform initial temperature. insulated at x = 0. surface energy balance at x = L.

PROBLEM 5.3 KNOWN: Microwave and radiant heating conditions for a slab of beef. FIND: Sketch temperature distributions at specific times during heating and cooling. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Uniform internal heat generation for microwave, (3) Uniform surface heating for radiant oven, (4) Heat loss from surface of meat to surroundings is negligible during the heating process, (5) Symmetry about midplane. ANALYSIS:

COMMENTS: (1) With uniform generation and negligible surface heat loss, the temperature distribution remains nearly uniform during microwave heating. During the subsequent surface cooling, the maximum temperature is at the midplane. (2) The interior of the meat is heated by conduction from the hotter surfaces during radiant heating, and the lowest temperature is at the midplane. The situation is reversed shortly after cooling begins, and the maximum temperature is at the midplane.

PROBLEM 5.4 KNOWN: Plate initially at a uniform temperature Ti is suddenly subjected to convection process (T∞,h) on both surfaces. After elapsed time to, plate is insulated on both surfaces. FIND: (a) Assuming Bi >> 1, sketch on T - x coordinates: initial and steady-state (t → ∞) temperature distributions, T(x,to) and distributions for two intermediate times to < t < ∞, (b) Sketch on T - t coordinates midplane and surface temperature histories, (c) Repeat parts (a) and (b) assuming Bi to, (5) T(0, t < to) < T∞. ANALYSIS: (a,b) With Bi >> 1, appreciable temperature gradients exist in the plate following exposure to the heating process.

On T-x coordinates: (1) initial, uniform temperature, (2) steady-state conditions when t → ∞, (3) distribution at to just before plate is covered with insulation, (4) gradients are always zero (symmetry), and (5) when t > to (dashed lines) gradients approach zero everywhere. (c) If Bi (ρc)m and kAl > km. Hence, the aluminum can store more energy and can be charged (or discharged) more quickly.

PROBLEM 5.34 KNOWN: Thickness, properties and initial temperature of steel slab. Convection conditions. FIND: Heating time required to achieve a minimum temperature of 550°C in the slab. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible radiation effects, (3) Constant properties. 2

ANALYSIS: With a Biot number of hL/k = (250 W/m ⋅K × 0.05m)/48 W/m⋅K = 0.260, a lumped capacitance analysis should not be performed. At any time during heating, the lowest temperature in the slab is at the midplane, and from the one-term approximation to the transient thermal response of a plane wall, Eq. (5.41), we obtain

(550 − 800 ) °C 0.417 C exp ζ 2 Fo T −T θ o∗ = o ∞ = = = 1 − 1 Ti − T∞ ( 200 − 800 ) °C

(

)

With ζ1 ≈ 0.488 rad and C1 ≈ 1.0396 from Table 5.1 and α = k / ρ c = 1.115 × 10 −5 m 2 / s,

)

(

−ζ12 α t / L2 = ln (0.401) = −0.914 t=

0.841(0.05m )

2

0.914 L2

=

ζ12α

(0.488 ) 1.115 ×10−5 m 2 / s 2

<

= 861s

COMMENTS: The surface temperature at t = 861s may be obtained from Eq. (5.40b), where ∗

∗

( ) = 0.417 cos (0.488 rad ) = 0.368. Hence, T (L, 792s ) ≡ T = T

θ = θ o cos ζ 1x

∗

s

∞ + 0.368 ( Ti − T∞ )

= 800°C − 221°C = 579°C. Assuming a surface emissivity of ε = 1 and surroundings that are at Tsur = T∞ = 800°C, the radiation heat transfer coefficient corresponding to this surface temperature is

(

)

h r = εσ ( Ts + Tsur ) Ts + Tsur = 205 W / m ⋅ K. Since this value is comparable to the convection 2

2

2

coefficient, radiation is not negligible and the desired heating will occur well before t = 861s.

PROBLEM 5.35 KNOWN: Pipe wall subjected to sudden change in convective surface condition. See Example 5.4. FIND: (a) Temperature of the inner and outer surface of the pipe, heat flux at the inner surface, and energy transferred to the wall after 8 min; compare results to the hand calculations performed for the Text Example; (b) Time at which the outer surface temperature of the pipe, T(0,t), will reach 25°C; (c) Calculate and plot on a single graph the temperature distributions, T(x,t) vs. x, for the initial condition, the final condition and the intermediate times of 4 and 8 min; explain key features; (d) Calculate and plot the temperature-time history, T(x,t) vs. t, for the locations at the inner and outer pipe surfaces, x = 0 and L, and for the range 0 ≤ t ≤ 16 min. Use the IHT | Models | Transient Conduction | Plane Wall model as the solution tool. SCHEMATIC:

ASSUMPTIONS: (1) Pipe wall can be approximated as a plane wall, (2) Constant properties, (3) Outer surface of pipe is adiabatic. ANALYSIS: The IHT model represents the series solution for the plane wall providing temperatures and heat fluxes evaluated at (x,t) and the total energy transferred at the inner wall at (t). Selected portions of the IHT code used to obtain the results tabulated below are shown in the Comments. (a) The code is used to evaluate the tabulated parameters at t = 8 min for locations x = 0 and L. The agreement is very good between the one-term approximation of the Example and the multipleterm series solution provided by the IHT model. T(L, 8min), °C T(0, 8 min), °C

Q′ (8 min ) × 10−7 , J / m q′′x ( L, 8 min ) , W / m 2

Text Ex 5.4

IHT Model

45.2 42.9

45.4 43.1

-2.73

-2.72

-7400

-7305

(b) To determine the time to for which T(0,t) = 25°C, the IHT model is solved for to after setting x = 0 and T_xt = 25°C. Find, to = 4.4 min.

<

T e m p e ra tu re , T(x ,t) (C )

(c) The temperature distributions, T(x,t) vs x, for the initial condition (t = 0), final condition ( t → ∞) and intermediate times of 4 and 8 min. are shown on the graph below. T e m p e ra tu re d is trib u tio n s , T(x,t) vs . x

60 40 20 0 -2 0 0

10

20

30

40

W a ll lo c a tio n , x (m m ) In itia l co n d itio n , t = 0 t = 4 m in t = 8 m in S te a d y-s ta te c o n d itio n , t > 3 0 m in

Continued …..

PROBLEM 5.35 (Cont.) The final condition corresponds to the steady-state temperature, T (x,∞) = T∞. For the intermediate times, the gradient is zero at the insulated boundary (x = 0, the pipe exterior). As expected, the temperature at x = 0 will be less than at the boundary experiencing the convection process with the hot oil, x = L. Note, however, that the difference is not very significant. The gradient at the inner wall, x = L, decreases with increasing time.

(d) The temperature history T(x,t) for the locations at the inner and outer pipe surfaces are shown in the graph below. Note that the temperature difference between the two locations is greatest at the start of the transient process and decreases with increasing time. After a 16 min. duration, the pipe temperature is almost uniform, but yet 3 or 4°C from the steady-state condition. T e m p e ra tu re -tim e h is to ry, T (x,t) vs . t

T e m p e ra tu re , T (x ,t) (C )

60

40

20

0

-2 0 0

2

4

6

8

10

12

14

16

T im e , t (m in ) O u te r s u rfa c e , x = 0 In n e r s u rfa c e , x = L

COMMENTS: (1) Selected portions of the IHT code for the plane wall model are shown below. Note the relation for the pipe volume, vol, used in calculating the total heat transferred per unit length over the time interval t. // Models | Transient Conduction | Plane Wall // The temperature distribution is T_xt = T_xt_trans("Plane Wall",xstar,Fo,Bi,Ti,Tinf) // Eq 5.39 //T_xt = 25 // Part (b) surface temperature, x = 0 // The heat flux in the x direction is q''_xt = qdprime_xt_trans("Plane Wall",x,L,Fo,Bi,k,Ti,Tinf) // Eq 2.6 // The total heat transfer from the wall over the time interval t is QoverQo = Q_over_Qo_trans("Plane Wall",Fo,Bi) // Eq 5.45 Qo = rho * cp * vol * (Ti - Tinf) // Eq 5.44 //vol = 2 * As * L // Appropriate for wall of 2L thickness vol = pi * D * L // Pipe wall of diameter D, thickness L and unit length Q = QoverQo * Qo // Total energy transfered per unit length

(2) Can you give an explanation for why the inner and outer surface temperatures are not very different? What parameter provides a measure of the temperature non-uniformity in a system during a transient conduction process?

PROBLEM 5.36 KNOWN: Thickness, initial temperature and properties of furnace wall. Convection conditions at inner surface. FIND: Time required for outer surface to reach a prescribed temperature. Corresponding temperature distribution in wall and at intermediate times. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in a plane wall, (2) Constant properties, (3) Adiabatic outer surface, (4) Fo > 0.2, (5) Negligible radiation from combustion gases. ANALYSIS: The wall is equivalent to one-half of a wall of thickness 2L with symmetric convection 2 conditions at its two surfaces. With Bi = hL/k = 100 W/m ⋅K × 0.15m/1.5 W/m⋅K = 10 and Fo > 0.2, the one-term approximation, Eq. 5.41 may be used to compute the desired time, where θ o∗ = (To − T∞ ) / (Ti − T∞ ) = 0.215. From Table 5.1, C1 = 1.262 and ζ1 = 1.4289. Hence, ln θ o∗ / C1 ln ( 0.215 /1.262 )

Fo = −

(

ζ12

)=−

= 0.867

(1.4289 )2

0.867 ( 0.15m ) Fo L2 = = 33,800s 3 α 1.5 W / m ⋅ K / 2600 kg / m × 1000 J / kg ⋅ K 2

t=

)

(

<

The corresponding temperature distribution, as well as distributions at t = 0, 10,000, and 20,000 s are plotted below 1000

Te m p e ra tu re , C

800 600 400 200 0 0

0 .2

0 .4

0 .6

0 .8

1

D im e n s io n le s s lo ca tio n , x/L t= 0 s t= 1 0 ,0 0 0 s t= 2 0 ,0 0 0 s t= 3 3 ,8 0 0 s

COMMENTS: Because Bi >>1, the temperature at the inner surface of the wall increases much more rapidly than at locations within the wall, where temperature gradients are large. The temperature gradients decrease as the wall approaches a steady-state for which there is a uniform temperature of 950°C.

PROBLEM 5.37 KNOWN: Thickness, initial temperature and properties of steel plate. Convection conditions at both surfaces. FIND: Time required to achieve a minimum temperature. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in plate, (2) Symmetric heating on both sides, (3) Constant properties, (4) Negligible radiation from gases, (5) Fo > 0.2. 2

ANALYSIS: The smallest temperature exists at the midplane and, with Bi = hL/k = 500 W/m ⋅K × 0.050m/45 W/m⋅K = 0.556 and Fo > 0.2, may be determined from the one-term approximation of Eq. 5.41. From Table 5.1, C1 = 1.076 and ζ1 = 0.682. Hence, with θ o∗ = (To - T∞)/(Ti - T∞) = 0.375,

Fo = −

(

ln θ o∗ / C1

ζ12

) = − ln (0.375 /1.076) = 2.266 (0.682 )2

2.266 ( 0.05m ) Fo L2 = = 491s 3 α 45 W / m ⋅ K / 7800 kg / m × 500 J / kg ⋅ K 2

t=

(

)

COMMENTS: From Eq. 5.40b, the corresponding surface temperature is

Ts = T∞ + ( Ti − T∞ )θ o∗ cos (ζ1 ) = 700°C − 400°C × 0.375 × 0.776 = 584°C Because Bi is not much larger than 0.1, temperature gradients in the steel are moderate.

<

PROBLEM 5.38 KNOWN: Plate of thickness 2L = 25 mm at a uniform temperature of 600°C is removed from a hot pressing operation. Case 1, cooled on both sides; case 2, cooled on one side only. FIND: (a) Calculate and plot on one graph the temperature histories for cases 1 and 2 for a 500second cooling period; use the IHT software; Compare times required for the maximum temperature in the plate to reach 100°C; and (b) For both cases, calculate and plot on one graph, the variation with time of the maximum temperature difference in the plate; Comment on the relative magnitudes of the temperature gradients within the plate as a function of time. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the plate, (2) Constant properties, and (3) For case 2, with cooling on one side only, the other side is adiabatic. 3

PROPERTIES: Plate (given): ρ = 3000 kg/m , c = 750 J/kg⋅K, k = 15 W/m⋅K. ANALYSIS: (a) From IHT, call up Plane Wall, Transient Conduction from the Models menu. For case 1, the plate thickness is 25 mm; for case 2, the plate thickness is 50 mm. The plate center (x = 0) temperature histories are shown in the graph below. The times required for the center temperatures to reach 100°C are t1 = 164 s

<

t2 = 367 s

(b) The plot of T(0, t) – T(1, t), which represents the maximum temperature difference in the plate during the cooling process, is shown below. Plate center temperature histories

Temperature difference history

600

150

T(0,t) - T(L,t) (C)

T(0,t) (C)

500 400 300 200

100

50

100 0

0 0

100

200

300

Time, t (s) Cooling - both sides Cooling - one side only

400

500

0

100

200

300

400

500

Time (s) Cooling - both sides Cooling - one side only

COMMENTS: (1) From the plate center-temperature history graph, note that it takes more than twice as long for the maximum temperature to reach 100°C with cooling on only one side. (2) From the maximum temperature-difference graph, as expected, cooling from one side creates a larger maximum temperature difference during the cooling process. The effect could cause microstructure differences, which could adversely affect the mechanical properties within the plate.

PROBLEM 5.39 KNOWN: Properties and thickness L of ceramic coating on rocket nozzle wall. Convection conditions. Initial temperature and maximum allowable wall temperature. FIND: (a) Maximum allowable engine operating time, tmax, for L = 10 mm, (b) Coating inner and outer surface temperature histories for L = 10 and 40 mm. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in a plane wall, (2) Constant properties, (3) Negligible thermal capacitance of metal wall and heat loss through back surface, (4) Negligible contact resistance at wall/ceramic interface, (5) Negligible radiation. ANALYSIS: (a) Subject to assumptions (3) and (4), the maximum wall temperature corresponds to the ceramic temperature at x = 0. Hence, for the ceramic, we wish to determine the time tmax at which T(0,t) = To(t) = 1500 K. With Bi = hL/k = 5000 W/m2⋅K(0.01 m)/10 W/m⋅K = 5, the lumped capacitance method cannot be used. Assuming Fo > 0.2, obtaining ζ1 = 1.3138 and C1 = 1.2402 from Table 5.1, and evaluating θ o* = ( To − T∞ ) ( Ti − T∞ ) = 0.4, Equation 5.41 yields

Fo = −

(

ln θ o* C1

ζ12

) = − ln (0.4 1.2402) = 0.656 (1.3138 )2

confirming the assumption of Fo > 0.2. Hence,

t max =

( ) = 0.656 (0.01m )2 = 10.9s

Fo L2

<

6 × 10−6 m 2 s

α

(b) Using the IHT Lumped Capacitance Model for a Plane Wall, the inner and outer surface temperature histories were computed and are as follows:

Temperature, T(K)

2300 1900 1500 1100 700 300 0

30

60

90

120

150

Time, t(s) L = 0.01, x = L L = 0.01, x = 0 L = 0.04, x = L L = 0.04, x = 0

Continued...

PROBLEM 5.39 (Cont.) The increase in the inner (x = 0) surface temperature lags that of the outer surface, but within t ≈ 45s both temperatures are within a few degrees of the gas temperature for L = 0.01 m. For L = 0.04 m, the increased thermal capacitance of the ceramic slows the approach to steady-state conditions. The thermal response of the inner surface significantly lags that of the outer surface, and it is not until t ≈ 137s that the inner surface reaches 1500 K. At this time there is still a significant temperature difference across the ceramic, with T(L,tmax) = 2240 K. COMMENTS: The allowable engine operating time increases with increasing thermal capacitance of the ceramic and hence with increasing L.

PROBLEM 5.40 KNOWN: Initial temperature, thickness and thermal diffusivity of glass plate. Prescribed surface temperature. FIND: (a) Time to achieve 50% reduction in midplane temperature, (b) Maximum temperature gradient at that time. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties. ANALYSIS: Prescribed surface temperature is analogous to h → ∞ and T∞ = Ts . Hence, Bi = ∞. Assume validity of one-term approximation to series solution for T (x,t). (a) At the midplane,

(

T −T θ o∗ = o s = 0.50 = C1exp −ζ12Fo Ti − Ts

)

ζ1tan ζ1 = Bi = ∞ → ζ1 = π /2. Hence C1 =

4sinζ1 4 = = 1.273 2ζ1 + sin ( 2ζ1 ) π

Fo = −

(

ln θ o∗ / C1 ζ12

) = 0.379

2 FoL2 0.379 ( 0.01 m ) t= = = 63 s. α 6 × 10−7 m 2 / s

(

<

)

(b) With θ ∗ = C1exp −ζ12 Fo cos ζ1x∗

(T − T ) ∂ T ( Ti − Ts ) ∂θ ∗ = = − i s ζ 1C1exp −ζ 12 Fo sinζ 1x ∗ ∂ x L L ∂ x∗

(

)

300o C π ∂ T/ ∂ x max = ∂ T/ ∂ x ∗ = − 0.5 = −2.36 ×10 4 oC/m. x =1 0.01 m 2 COMMENTS: Validity of one-term approximation is confirmed by Fo > 0.2.

<

PROBLEM 5.41 KNOWN: Thickness and properties of rubber tire. Convection heating conditions. Initial and final midplane temperature. FIND: (a) Time to reach final midplane temperature. (b) Effect of accelerated heating. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in a plane wall, (2) Constant properties, (3) Negligible radiation. ANALYSIS: (a) With Bi = hL/k = 200 W/m2⋅K(0.01 m)/0.14 W/m⋅K = 14.3, the lumped capacitance method is clearly inappropriate. Assuming Fo > 0.2, Eq. (5.41) may be used with C1 = 1.265 and ζ1 ≈ 1.458 rad from Table 5.1 to obtain

)

(

T −T θ o* = o ∞ = C1 exp −ζ12Fo = 1.265exp ( −2.126 Fo ) Ti − T∞ With θ o* = ( To − T∞ ) ( Ti − T∞ ) = (-50)/(-175) = 0.286, Fo = − ln ( 0.286 1.265 ) 2.126 = 0.70 = α t f L2

0.7 ( 0.01m )

2

tf =

6.35 × 10−8 m 2 s

<

= 1100s

(b) The desired temperature histories were generated using the IHT Transient Conduction Model for a Plane Wall, with h = 5 × 104 W/m2⋅K used to approximate imposition of a surface temperature of 200°C.

Temperature, T(C)

200

150

100

50

0 0

200

400

600

800

1000

1200

Time, t(s) x = 0, h = 200 W/m^2.K x = L, h = 200 W/m^2.K x = 0, h = 5E4 W/m^2.K x = L, h = 5E4W/m^2.K

The fact that imposition of a constant surface temperature (h → ∞) does not significantly accelerate the heating process should not be surprising. For h = 200 W/m2⋅K, the Biot number is already quite large (Bi = 14.3), and limits to the heating rate are principally due to conduction in the rubber and not to convection at the surface. Any increase in h only serves to reduce what is already a small component of the total thermal resistance. COMMENTS: The heating rate could be accelerated by increasing the steam temperature, but an upper limit would be associated with avoiding thermal damage to the rubber.

PROBLEM 5.42 KNOWN: Stack or book comprised of 11 metal plates (p) and 10 boards (b) each of 2.36 mm thickness and prescribed thermophysical properties. FIND: Effective thermal conductivity, k, and effective thermal capacitance, (ρcp). SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible contact resistance between plates and boards. 3

PROPERTIES: Metal plate (p, given): ρp = 8000 kg/m , cp,p = 480 J/kg⋅K, kp = 12 3 W/m⋅K; Circuit boards (b, given): ρb = 1000 kg/m , cp,b = 1500 J/kg⋅K, kb = 0.30 W/m⋅K. ANALYSIS: The thermal resistance of the book is determined as the sum of the resistance of the boards and plates, R ′′tot = NR ′′b + MR ′′p where M,N are the number of plates and boards in the book, respectively, and R ′′i = Li / ki where Li and ki are the thickness and thermal conductivities, respectively.

(

)

R ′′tot = M L p / k p + N ( L b / k b ) R ′′tot = 11 ( 0.00236 m/12 W/m ⋅ K ) + 10 ( 0.00236 m/0.30 W/m ⋅ K ) R ′′tot = 2.163 × 10

−3

+ 7.867 × 10

−2

= 8.083 × 10

−2

K/W.

The effective thermal conductivity of the book of thickness (10 + 11) 2.36 mm is 0.04956 m

k = L/R ′′tot =

8.083 × 10-2 K/W The thermal capacitance of the stack is

(

)

C′′tot = M ρ p L p c p + N ( ρ b L b c b )

(

<

= 0.613 W/m ⋅ K.

) (

C′′tot = 11 8000 kg/m × 0.00236 m × 480 J/kg ⋅ K + 10 1000 kg/m × 0.00236 m × 1500 J/kg ⋅ K 3

4

4

5

3

)

2

C′′tot = 9.969 × 10 + 3.540 × 10 = 1.35 × 10 J/m ⋅ K.

The effective thermal capacitance of the book is

( ρ cp ) = C′′tot / L = 1.351×105 J/m2 ⋅ K/0.04956 m = 2.726 ×106 J/m3 ⋅ K. < COMMENTS: The results of the analysis allow for representing the stack as a homogeneous -7 2 medium with effective properties: k = 0.613 W/m⋅K and α = (k/ρcp) = 2.249×10 m /s. See for example, Problem 5.38.

PROBLEM 5.43 KNOWN: Stack of circuit board-pressing plates, initially at a uniform temperature, is subjected by upper/lower platens to a higher temperature. FIND: (a) Elapsed time, te, required for the mid-plane to reach cure temperature when platens are suddenly changed to Ts = 190°C, (b) Energy removal from the stack needed to return its temperature to Ti. SCHEMATIC:

6 3 -7 PROPERTIES: Stack (given): k = 0.613 W/m⋅K, ρcp = 2.73×10 J/m ⋅K; α = k/ρcp = 2.245×10 2 m /s.

ANALYSIS: (a) Recognize that sudden application of surface temperature corresponds to h → ∞, or -1 Bi = 0 (Heisler chart) or Bi → ∞ (100, Table 5.1). With Ts = T∞,

T (0,t ) − Ts (170 − 190 ) C θ o∗ = = = 0.114. Ti − Ts (15 − 190 )$ C $

Using Eq. 5.41 with values of ζ1 = 1.552 and C1 = 1.2731 at Bi = 100 (Table 5.1), find Fo θ o∗ = C1exp −ζ12 Fo

(

Fo = −

ζ1

2 where Fo = αt/L ,

t=

α

(

)

ln θ o∗ / C1 = − 2

1

FoL2

)

=

(

1

(1.552 )2

1.002 25 ×10−3 m

)

ln ( 0.114/1.2731) = 1.002

2

2.245 × 10−7 m 2 / s

= 2.789 × 103 s = 46.5 min.

<

The Heisler chart, Figure D.1, could also be used to find Fo from values of θ o∗ and Bi -1 = 0. (b) The energy removal is equivalent to the energy gained by the stack per unit area for the time interval 0 → te. With Q′′o corresponding to the maximum amount of energy that could be transferred,

(

Q′′o = ρ c ( 2L )( Ti − T∞ ) = 2.73 × 10 J/m ⋅ K 2 × 25 × 10 6

3

-3

)

m (15 − 190 ) K = −2.389 × 10 J/m 7

2.

Q′′ may be determined from Eq. 5.46, sin (1.552rad ) Q′′ sinζ1 ∗ θo = 1 − = 1− × 0.114 = 0.795 Q′′o 1.552rad ζ1 We conclude that the energy to be removed from the stack per unit area to return it to Ti is Q′′ = 0.795Q′′o = 0.795 × 2.389 × 107 J/m2 = 1.90 × 107 J/m 2 .

<

PROBLEM 5.44 KNOWN: Car windshield, initially at a uniform temperature of -20°C, is suddenly exposed on its interior surface to the defrost system airstream at 30°C. The ice layer on the exterior surface acts as an insulating layer. FIND: What airstream convection coefficient would allow the exterior surface to reach 0°C in 60 s? SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, transient conduction in the windshield, (2) Constant properties, (3) Exterior surface is perfectly insulated. PROPERTIES: Windshield (Given): ρ = 2200 kg/m3, cp = 830 J/kg⋅K and k = 1.2 W/m⋅K. ANALYSIS: For the prescribed conditions, from Equations 5.31 and 5.33,

θ (0, 60s ) θ o T (0, 60s ) − T∞ (0 − 30 ) C = 0.6 = = = θi θi Ti − T∞ ( −20 − 30 )$ C $

Fo =

kt

ρ cL2

=

1.2 W m ⋅ K × 60 2200 kg m3 × 830 J kg ⋅ K × (0.005 m )

2

= 1.58

The single-term series approximation, Eq. 5.41, along with Table 5.1, requires an iterative solution to find an appropriate Biot number. Alternatively, the Heisler charts, Appendix D, Figure D.1, for the midplane temperature could be used to find

Bi −1 = k hL = 2.5 h = 1.2 W m ⋅ K 2.5 × 0.005 m = 96 W m 2 ⋅ K

<

COMMENTS: Using the IHT, Transient Conduction, Plane Wall Model, the convection coefficient can be determined by solving the model with an assumed h and then sweeping over a range of h until the T(0,60s) condition is satisfied. Since the model is based upon multiple terms of the series, the result of h = 99 W/m2⋅K is more precise than that found using the chart.

PROBLEM 5.45 KNOWN: Thickness, initial temperature and properties of plastic coating. Safe-to-touch temperature. Convection coefficient and air temperature. FIND: Time for surface to reach safe-to-touch temperature. Corresponding temperature at plastic/wood interface. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in coating, (2) Negligible radiation, (3) Constant properties, (4) Negligible heat of reaction, (5) Negligible heat transfer across plastic/wood interface. 2

ANALYSIS: With Bi = hL/k = 200 W/m ⋅K × 0.002m/0.25 W/m⋅K = 1.6 > 0.1, the lumped capacitance method may not be used. Applying the approximate solution of Eq. 5.40a, with C1 = 1.155 and ζ1 = 0.990 from Table 5.1,

(

) ( )

( 42 − 25 ) °C ∗ T − T∞ ∗ 2 = = 0.0971 = C1 exp −ζ 1 Fo cos ζ 1x = 1.155exp ( −0.980 Fo ) cos ( 0.99 ) θs = s Ti − T∞ ( 200 − 25 ) °C Hence, for x∗ = 1,

  0.0971 2 Fo = − ln  / 0.99 ) = 1.914   1.155cos ( 0.99 )  (   Fo L2 1.914 ( 0.002m ) = = 63.8s 7 2 − α 1.20 × 10 m / s 2

t=

<

From Eq. 5.41, the corresponding interface temperature is

(

)

To = T∞ + ( Ti − T∞ ) exp −ζ12 Fo = 25°C + 175°C exp ( −0.98 × 1.914 ) = 51.8°C

<

COMMENTS: By neglecting conduction into the wood and radiation from the surface, the cooling time is overpredicted and is therefore a conservative estimate. However, if energy generation due to solidification of polymer were significant, the cooling time would be longer.

PROBLEM 5.46 KNOWN: Inlet and outlet temperatures of steel rods heat treated by passage through an oven. FIND: Rod speed, V. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction (axial conduction is negligible), (2) Constant properties, (3) Negligible radiation. PROPERTIES: Table A-1, AISI 1010 Steel ( T ≈ 600K ) : k = 48.8 W/m⋅K, ρ = 7832 kg/m , 3

-5 2

cp = 559 J/kg⋅K, α = (k/ρcp) = 1.11×10 m /s. ANALYSIS: The time needed to traverse the rod through the oven may be found from Fig. D.4. T −T 600 − 750 θ o∗ = o ∞ = = 0.214 Ti − T∞ 50 − 750 k 48.8 W/m ⋅ K Bi-1 ≡ = = 15.6. hro 125 W/m 2 ⋅ K (0.025m ) Hence, Fo = α t/ro2 ≈ 12.2 2 t = 12.2 ( 0.025m ) /1.11× 10−5 m 2 / s = 687 s. The rod velocity is V=

L 5m = = 0.0073 m/s. t 687s

COMMENTS: (1) Since (h ro/2)/k = 0.032, the lumped capacitance method could have been used. From Eq. 5.5 it follows that t = 675 s. (2) Radiation effects decrease t and hence increase V, assuming there is net radiant transfer from the oven walls to the rod. (3) Since Fo > 0.2, the approximate analytical solution may be used. With Bi = hro/k =0.0641, Table 5.1 yields ζ1 = 0.3549 rad and C1 = 1.0158. Hence from Eq. 5.49c θ ∗  ln  o  = 12.4,  C1  which is in good agreement with the graphical result.

( )

Fo = − ζ12

−1

PROBLEM 5.47 KNOWN: Hot dog with prescribed thermophysical properties, initially at 6°C, is immersed in boiling water. FIND: Time required to bring centerline temperature to 80°C. SCHEMATIC:

ASSUMPTIONS: (1) Hot dog can be treated as infinite cylinder, (2) Constant properties. ANALYSIS: The Biot number, based upon Eq. 5.10, is 2 -3 h Lc h ro / 2 100 W/m ⋅ K 10 × 10 m/2 Bi ≡ = = = 0.96

(

k

0.52 W/m ⋅ K

k

)

Since Bi > 0.1, a lumped capacitance analysis is not appropriate. Using the Heisler chart, Figure D.4 with hr 100W/m 2 ⋅ K × 10 × 10-3m Bi ≡ o = = 1.92 or Bi-1 = 0.52

0.52 W/m ⋅ K

k

T ( 0,t ) − T∞ (80 − 100 ) C θ θ o∗ = o = = = 0.21 Ti − T∞ θi (6-100 )$ C $

and

(1)

(10 ×10-3m )

2

find

Fo = t∗ =

αt ro2

ro2 t = ⋅ Fo =

= 0.8

α

1.764 × 10−7 m 2 / s

<

× 0.8 = 453.5s = 7.6 min

α = k/ρ c = 0.52 W/m ⋅ K/880 kg/m3 × 3350 J/kg ⋅ K = 1.764 × 10−7 m 2 / s.

where

COMMENTS: (1) Note that Lc = ro/2 when evaluating the Biot number for the lumped capacitance analysis; however, in the Heisler charts, Bi ≡ hro/k. (2) The surface temperature of the hot dog follows from use of Figure D.5 with r/ro = 1 and Bi

-1

=

0.52; find θ(1,t)/θo ≈ 0.45. From Eq. (1), note that θo = 0.21 θi giving

θ (1, t ) = T ( ro , t ) − T∞ = 0.45θ o = 0.45 (0.21[Ti − T∞ ]) = 0.45 × 0.21[6 − 100] C = −8.9$C $

T ( ro , t ) = T∞ − 8.9$ C = (100 − 8.9 ) C = 91.1$C $

(3) Since Fo ≥ 0.2, the approximate solution for θ*, Eq. 5.49, is valid. From Table 5.1 with Bi = 1.92, find that ζ1 = 1.3245 rad and C1 = 1.2334. Rearranging Eq. 5.49 and substituting values,

Fo = −

(

)

ln θ o∗ / C1 = 2

1

ζ1

 0.213  ln   = 1.00 2 1.2334  (1.3245 rad ) 1

This result leads to a value of t = 9.5 min or 20% higher than that of the graphical method.

PROBLEM 5.48 KNOWN: Long rod with prescribed diameter and properties, initially at a uniform temperature, is heated in a forced convection furnace maintained at 750 K with a convection coefficient of h = 1000 W/m2⋅K. FIND: (a) The corresponding center temperature of the rod, T(0, to), when the surface temperature T(ro, to) is measured as 550 K, (b) Effect of h on centerline temperature history. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, radial conduction in rod, (2) Constant properties, (3) Rod, when initially placed in furnace, had a uniform (but unknown) temperature, (4) Fo ≥ 0.2. ANALYSIS: (a) Since the rod was initially at a uniform temperature and Fo ≥ 0.2, the approximate solution for the infinite cylinder is appropriate. From Eq. 5.49b,

(

)

( )

θ * r* , Fo = θ o* (Fo ) J 0 ζ1r*

(1)

where, for r* = 1, the dimensionless temperatures are, from Eq. 5.31,

θ * (1, Fo ) =

T ( ro , t o ) − T∞

θ o* ( Fo ) =

Ti − T∞

T (0, t o ) − T∞ Ti − T∞

(2,3)

Combining Eqs. (2) and (3) with Eq. (1) and rearranging,

T ( ro , t o ) − T∞ Ti − T∞

=

T ( 0, t o ) = T∞ +

T (0, t o ) − T∞ Ti − T∞

J 0 (ζ1 ⋅1)

1  T ( ro , t o ) − T∞  J 0 (ζ1 ) 

(4)

The eigenvalue, ζ1 = 1.0185 rad, follows from Table 5.1 for the Biot number

Bi =

2 hro 1000 W m ⋅ K (0.060 m 2 ) = = 0.60 . k 50 W m ⋅ K

From Table B-4, with ζ1 = 1.0185 rad, J0(1.0185) = 0.7568. Hence, from Eq. (4)

T ( 0, t o ) = 750 K +

1 [550 − 750] K = 486 K 0.7568

<

(b) Using the IHT Transient Conduction Model for a Cylinder, the following temperature histories were generated. Continued...

PROBLEM 5.48 (Cont.) Centerline temperature, To(K)

500

400

300 0

100

200

300

400

Time, t(s) h = 100 W/m^2.K h = 500 W/m^2.K h = 1000 W/m^2.K

The times required to reach a centerline temperature of 500 K are 367, 85 and 51s, respectively, for h = 100, 500 and 1000 W/m2⋅K. The corresponding values of the Biot number are 0.06, 0.30 and 0.60. Hence, even for h = 1000 W/m2⋅K, the convection resistance is not negligible relative to the conduction resistance and significant reductions in the heating time could still be effected by increasing h to values considerably in excess of 1000 W/m2⋅K. COMMENTS: For Part (a), recognize why it is not necessary to know Ti or the time to. We require that Fo ≥ 0.2, which for this sphere corresponds to t ≥ 14s. For this situation, the time dependence of the surface and center are the same.

PROBLEM 5.49 KNOWN: A long cylinder, initially at a uniform temperature, is suddenly quenched in a large oil bath. FIND: (a) Time required for the surface to reach 500 K, (b) Effect of convection coefficient on surface temperature history. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Fo > 0.2. ANALYSIS: (a) Check first whether lumped capacitance method is applicable. For h = 50 W/m2⋅K,

Bic =

2 hLc h ( ro 2 ) 50 W m ⋅ K ( 0.015 m / 2 ) = = = 0.221 . k k 1.7 W m ⋅ K

Since Bic > 0.1, method is not suited. Using the approximate series solution for the infinite cylinder,

)

(

(

) ( )

θ * r* , Fo = C1 exp −ζ12 Fo × J 0 ζ1r*

(1)

Solving for Fo and setting r* = 1, find

 θ*  ln   ζ12  C1J 0 (ζ1 )  T ( ro , t o ) − T∞ (500 − 350 ) K where θ * = (1, Fo ) = = = 0.231 . Ti − T∞ (1000 − 350 ) K Fo = −

1

From Table 5.1, with Bi = 0.441, find ζ1 = 0.8882 rad and C1 = 1.1019. From Table B.4, find J0(ζ1) = 0.8121. Substituting numerical values into Eq. (2),

Fo = −

1

(0.8882 )2

ln [0.231 1.1019 × 0.8121] = 1.72 .

From the definition of the Fourier number, Fo = α t ro2 , and α = k/ρc,

r2 ρc t = Fo o = Fo ⋅ ro2 α k t = 1.72 ( 0.015 m ) × 400 kg m3 × 1600 J kg ⋅ K 1.7 W m ⋅ K = 145s . 2

<

(b) Using the IHT Transient Conduction Model for a Cylinder, the following surface temperature histories were obtained. Continued...

PROBLEM 5.49 (Cont.)

Surface temperature, T(K)

1000 900 800 700 600 500 400 300 0

50

100

150

200

250

300

Time, t(s) h = 250 W/m^2.K h = 50 W/m^2.K

Increasing the convection coefficient by a factor of 5 has a significant effect on the surface temperature, greatly accelerating its approach to the oil temperature. However, even with h = 250 W/m2⋅K, Bi = 1.1 and the convection resistance remains significant. Hence, in the interest of accelerated cooling, additional benefit could be achieved by further increasing the value of h. COMMENTS: For Part (a), note that, since Fo = 1.72 > 0.2, the approximate series solution is appropriate.

PROBLEM 5.50 KNOWN: Long pyroceram rod, initially at a uniform temperature of 900 K, and clad with a thin metallic tube giving rise to a thermal contact resistance, is suddenly cooled by convection. FIND: (a) Time required for rod centerline to reach 600 K, (b) Effect of convection coefficient on cooling rate. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Thermal resistance and capacitance of metal tube are negligible, (3) Constant properties, (4) Fo ≥ 0.2. PROPERTIES: Table A-2, Pyroceram ( T = (600 + 900)K/2 = 750 K): ρ = 2600 kg/m3, c = 1100 J/kg⋅K, k = 3.13 W/m⋅K. ANALYSIS: (a) The thermal contact and convection resistances can be combined to give an overall heat transfer coefficient. Note that R ′t,c [m⋅K/W] is expressed per unit length for the outer surface. Hence, for h = 100 W/m2⋅K,

U=

1 1 = = 57.0 W m 2 ⋅ K . 1 h + R ′t,c (π D ) 1 100 W m 2 ⋅ K + 0.12 m ⋅ K W (π × 0.020 m )

Using the approximate series solution, Eq. 5.50c, the Fourier number can be expressed as Fo = − 1 ζ12 ln θ o* C1 .

( ) (

)

From Table 5.1, find ζ1 = 0.5884 rad and C1 = 1.0441 for Bi = Uro k = 57.0 W m 2 ⋅ K ( 0.020 m 2 ) 3.13 W m ⋅ K = 0.182 . The dimensionless temperature is

θ o* (0, Fo ) =

T ( 0, t ) − T∞ Ti − T∞

=

(600 − 300 ) K = 0.5. (900 − 300 ) K

Substituting numerical values to find Fo and then the time t,

Fo =

−1

(0.5884 )

2

ln

0.5 = 2.127 1.0441

r2 ρc t = Fo o = Fo ⋅ ro2 α k t = 2.127 ( 0.020 m 2 ) 2600 kg m3 × 1100 J kg ⋅ K 3.13 W m ⋅ K = 194s . 2

<

(b) The following temperature histories were generated using the IHT Transient conduction Model for a Cylinder. Continued...

900

900

800

800

Centerline temperature, (K)

Surface temperature, (K)

PROBLEM 5.50 (Cont.)

700 600 500 400 300

700 600 500 400 300

0

50

100

150 Time, t(s)

r = ro, h = 100 W/m^2.K r = ro, h = 500 W/m^2.K r = ro, h = 1000 W/m^2.K

200

250

300

0

50

100

150

200

250

300

Time, t(s) r = 0, h = 100 W/m^2.K r = 0, h = 500 W/m^2.K r = 0, h = 1000 W/m^2.K

While enhanced cooling is achieved by increasing h from 100 to 500 W/m2⋅K, there is little benefit associated with increasing h from 500 to 1000 W/m2⋅K. The reason is that for h much above 500 W/m2⋅K, the contact resistance becomes the dominant contribution to the total resistance between the fluid and the rod, rendering the effect of further reductions in the convection resistance negligible. Note that, for h = 100, 500 and 1000 W/m2⋅K, the corresponding values of U are 57.0, 104.8 and 117.1 W/m2⋅K, respectively. COMMENTS: For Part (a), note that, since Fo = 2.127 > 0.2, Assumption (4) is satisfied.

PROBLEM 5.51 KNOWN: Sapphire rod, initially at a uniform temperature of 800K is suddenly cooled by a convection process; after 35s, the rod is wrapped in insulation. FIND: Temperature rod reaches after a long time following the insulation wrap. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) No heat losses from the rod when insulation is applied. 3

PROPERTIES: Table A-2, Aluminum oxide, sapphire (550K): ρ = 3970 kg/m , c = 1068 J/kg⋅K, k = -5

22.3 W/m⋅K, α = 5.259×10

2

m /s.

ANALYSIS: First calculate the Biot number with Lc = ro/2, 2 h Lc h ( ro / 2 ) 1600 W/m ⋅ K ( 0.020 m/2)

Bi =

k

=

=

k

22.3 W/m ⋅ K

= 0.72.

Since Bi > 0.1, the rod cannot be approximated as a lumped capacitance system. The temperature distribution during the cooling process, 0 ≤ t ≤ 35s, and for the time following the application of insulation, t > 35s, will appear as

Eventually (t → ∞), the temperature of the rod will be uniform at T ( ∞ ) . To find T (∞ ) , write the conservation of energy requirement for the rod on a time interval basis,

E in − E out = ∆ E ≡ E final − E initial .

Using the nomenclature of Section 5.5.3 and basing energy relative to T∞, the energy balance becomes

−Q = ρ cV ( T ( ∞ ) − T∞ ) − Qo

where Qo = ρcV(Ti - T∞). Dividing through by Qo and solving for T (∞ ) , find

T ( ∞ ) = T∞ + ( Ti − T∞ )(1 − Q/Qo ) .

From the Groeber chart, Figure D.6, with hr 1600 W/m 2 ⋅ K × 0.020m Bi = o = = 1.43

k

(

22.3 W/m ⋅ K

)

(

)

Bi 2Fo = Bi 2 α t/ro2 = (1.43 ) 2 5.259 × 10-6 m 2 /s × 35s/ ( 0.020m ) 2 = 0.95. find Q/Qo ≈ 0.57. Hence,

T ( ∞ ) = 300K + ( 800 − 300 ) K (1-0.57 ) = 515 K.

<

COMMENTS: From use of Figures D.4 and D.5, find T(0,35s) = 525K and T(ro,35s) = 423K.

PROBLEM 5.52 KNOWN: Long bar of 70 mm diameter, initially at 90°C, is suddenly immersed in a water bath 2

(T∞ = 40°C, h = 20 W/m ⋅K). FIND: (a) Time, tf, that bar should remain in bath in order that, when removed and allowed to equilibrate while isolated from surroundings, it will have a uniform temperature T(r, ∞) = 55°C. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties. 3

PROPERTIES: Bar (given): ρ = 2600 kg/m , c = 1030 J/kg⋅K, k = 3.50 W/m⋅K, α = k/ρc = -6 2 1.31×10 m /s. ANALYSIS: Determine first whether conditions are space-wise isothermal 2 hLc h ( ro / 2) 20 W/m ⋅ K ( 0.035 m/2) Bi = = = = 0.10 k k 3.50 W/m ⋅ K

and since Bi ≥ 0.1, a Heisler solution is appropriate. (a) Consider an overall energy balance on the bar during the time interval ∆t = tf (the time the bar is in the bath). Ein − Eout = ∆E 0 − Q = E final − Einitial = Mc ( Tf − T∞ ) − Mc ( Ti − T∞ ) −Q = Mc ( Tf − T∞ ) − Q o Q T − T∞ ( 55 − 40 )o C = 0.70 = 1− f = 1− Qo Ti − T∞ ( 90 − 40 )o C

where Qo is the initial energy in the bar (relative to T∞; Eq. 5.44). With Bi = hro/k = 0.20 and 2 2 Q/Qo = 0.70, use Figure D.6 to find Bi Fo = 0.15; hence Fo = 0.15/Bi = 3.75 and 2 t f = Fo ⋅ ro2 / α = 3.75 ( 0.035 m ) /1.31×10− 6 m 2 / s = 3507 s.

< -1

(b) To determine T(ro, tf), use Figures D.4 and D.5 for θ(ro,t)/θi (Fo = 3.75, Bi = 5.0) and θo/θi -1 (Bi = 5.0, r/ro = 1, respectively, to find T (ro , t f ) = T∞ +

θ ( ro , t ) θ o ⋅ ⋅ θ i = 40o C + 0.25 × 0.90 ( 90 − 50 )o C = 49oC. θo θi

<

PROBLEM 5.53 KNOWN: Long plastic rod of diameter D heated uniformly in an oven to Ti and then allowed to convectively cool in ambient air (T∞, h) for a 3 minute period. Minimum temperature of rod should not be less than 200°C and the maximum-minimum temperature within the rod should not exceed 10°C. FIND: Initial uniform temperature Ti to which rod should be heated. Whether the 10°C internal temperature difference is exceeded. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Uniform and constant convection coefficients. 3

PROPERTIES: Plastic rod (given): k = 0.3 W/m⋅K, ρcp = 1040 kJ/m ⋅K. ANALYSIS: For the worst case condition, the rod cools for 3 minutes and its outer surface is at least 200°C in order that the subsequent pressing operation will be satisfactory. Hence, hro 8 W/m 2 ⋅K × 0.015 m Bi = = = 0.40 k 0.3 W/m ⋅ K αt k t 0.3 W/m ⋅ K 3 × 60s Fo = = ⋅ = × = 0.2308. ro2 ρ cp ro2 1040 × 103 J/m3 ⋅ K ( 0.015 m )2 Using Eq. 5.49a and ζ1 = 0.8516 rad and C1 = 1.0932 from Table 5.1, T (ro , t ) − T∞ θ∗ = = C1J 0 ζ1 ro∗ exp −ζ12 Fo . Ti − T∞

( ) (

)

With ro∗ = 1, from Table B.4, J 0 (ζ1 ×1) = J o ( 0.8516 ) = 0.8263, giving 200 − 25 = 1.0932 × 0.8263exp −0.85162 × 0.2308 Ti = 254o C. < Ti − 25 At this time (3 minutes) what is the difference between the center and surface temperatures of the rod? From Eq. 5.49b, θ ∗ T (ro , t ) − T∞ 200 − 25 = = = J 0 ζ1ro∗ = 0.8263 θo T ( 0,t ) − T∞ T ( 0,t ) − 25

)

(

( )

which gives T(0,t) = 237°C. Hence, o

∆T = T ( 0,180s ) − T ( ro ,180s ) = ( 237 − 200 ) C = 37oC.

<

Hence, the desired max-min temperature difference sought (10°C) is not achieved. COMMENTS: ∆T could be reduced by decreasing the cooling rate; however, h can not be made much smaller. Two solutions are (a) increase ambient air temperature and (b) non-uniformly heat rod in oven by controlling its residence time.

PROBLEM 5.54 KNOWN: Diameter and initial temperature of roller bearings. Temperature of oil bath and convection coefficient. Final centerline temperature. Number of bearings processed per hour. FIND: Time required to reach centerline temperature. Cooling load. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, radial conduction in rod, (2) Constant properties.

(

)

3

PROPERTIES: Table A.1, St. St. 304 T = 548 K : ρ=7900 kg/m , k = 19.0 W/m⋅K, cp = 546 -6

2

J/kg⋅K, α = 4.40 × 10 m /s. ANALYSIS: With Bi = h (ro/2)/k = 0.658, the lumped capacitance method can not be used. From the one-term approximation of Eq. 5.49 c for the centerline temperature,

(

)

T −T 50 − 30 2 θ o∗ = o ∞ = = 0.0426 = C1 exp −ζ12 Fo = 1.1382 exp  − ( 0.9287 ) Fo     Ti − T∞ 500 − 30 where, for Bi = hro/k = 1.316, C1 = 1.1382 and ζ1 = 0.9287 from Table 5.1.

Fo = −n ( 0.0374 ) / 0.863 = 3.81 t f = Fo ro2 / α = 3.81( 0.05 m ) / 4.40 × 10−6 = 2162s = 36 min 2

<

From Eqs. 5.44 and 5.51, the energy extracted from a single rod is

 2θ ∗  Q = ρ cV ( Ti − T∞ ) 1 − o J1 (ζ1 ) ζ1   With J1 (0.9287) = 0.416 from Table B.4,

 0.0852 × 0.416  2 Q = 7900 kg / m3 × 546 J / kg ⋅ K π ( 0.05m ) 1m  470 K 1 − = 1.53 × 107 J    0.9287   The nominal cooling load is

q=

N Q 10 × 1.53 × 107 J = = 70,800 W = 7.08 kW tf 2162s

COMMENTS: For a centerline temperature of 50°C, Eq. 5.49b yields a surface temperature of

T ( ro , t ) = T∞ + ( Ti − T∞ )θ o∗ Jo (ζ1 ) = 30°C + 470°C × 0.0426 × 0.795 = 45.9°C

<

PROBLEM 5.55 KNOWN: Long rods of 40 mm- and 80-mm diameter at a uniform temperature of 400°C in a curing oven, are removed and cooled by forced convection with air at 25°C. The 40-mm diameter rod takes 280 s to reach a safe-to-handle temperature of 60°C. FIND: Time it takes for a 80-mm diameter rod to cool to the same safe-to-handle temperature. Comment on the result? Did you anticipate this outcome? SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial (cylindrical) conduction in the rods, (2) Constant properties, and (3) Convection coefficient same value for both rods. 3

PROPERTIES: Rod (given): ρ = 2500 kg/m , c = 900 J/kg⋅K, k = 15 W/m⋅K. ANALYSIS: Not knowing the convection coefficient, the Biot number cannot be calculated to determine whether the rods behave as spacewise isothermal objects. Using the relations from Section 5.6, Radial Systems with Convection, for the infinite cylinder, Eq. 5.50, evaluate Fo = α t / ro2 , and knowing T(ro, to), a trial-and-error solution is required to find Bi = h ro/k and hence, h. Using the IHT Transient Conduction model for the Cylinder, the following results are readily calculated for the 40-mm rod. With to = 280 s,

Fo = 4.667

Bi = 0.264

h = 197.7 W / m 2 ⋅ K

For the 80-mm rod, with the foregoing value for h, with T(ro, to) = 60°C, find

Bi = 0.528

Fo = 2.413

t o = 579 s

<

COMMENTS: (1) The time-to-cool, to, for the 80-mm rod is slightly more than twice that for the 40-mm rod. Did you anticipate this result? Did you believe the times would be proportional to the diameter squared? (2) The simplest approach to explaining the relationship between to and the diameter follows from the lumped capacitance analysis, Eq. 5.13, where for the same θ/θi, we expect Bi⋅Foo to be a constant. That is,

h ⋅ ro α t o × =C k ro2 yielding to ~ ro (not ro2 ).

PROBLEM 5.56 KNOWN: Initial temperature, density, specific heat and diameter of cylindrical rod. Convection coefficient and temperature of air flow. Time for centerline to reach a prescribed temperature. Dependence of convection coefficient on flow velocity. FIND: (a) Thermal conductivity of material, (b) Effect of velocity and centerline temperature and temperature histories for selected velocities. SCHEMATIC:

ASSUMPTIONS: (1) Lumped capacitance analysis can not be used but one-term approximation for an infinite cylinder is appropriate, (2) One-dimensional conduction in r, (3) Constant properties, (4) Negligible radiation, (5) Negligible effect of thermocouple hole on conduction. ANALYSIS: (a) With θ o∗ =[To(0,1136s) - T∞]/(Ti - T∞) = (40 – 25)/(100 – 25) = 0.20, Eq. 5.49c yields Fo =

αt ro2

=

k t

ρ c p ro2

=

k (1136 s ) 1200 kg / m × 1250 J / kg ⋅ K × ( 0.02 m )

2

3

= − ln ( 0.2 / C1 ) / ζ12

(1)

Because C1 and ζ1 depend on Bi = hro/k, a trial-and-error procedure must be used. For example, a value of k may be assumed and used to calculate Bi, which may then be used to obtain C1 and ζ1 from Table 5.1. Substituting C1 and ζ1 into Eq. (1), k may be computed and compared with the assumed value. Iteration continues until satisfactory convergence is obtained, with

<

k ≈ 0.30 W / m ⋅ K and, hence, Bi = 3.67, C1 = 1.45, ζ1 = 1.87 and Fo = 0.568. For the above value of k, − ln (0.2 / C1 ) / ζ12 = 0.567, which equals the Fourier number, as prescribed by Eq. (1). 2

0.618

0.618

2.618

yields a value of C = 16.8 W⋅s /m (b) With h = 55 W/m ⋅K for V = 6.8 m/s, h = CV ⋅K. The desired variations of the centerline temperature with velocity (for t = 1136 s) and time (for V = 3, 10 and 20 m/s) are as follows: Continued …..

PROBLEM 5.56 (Cont.)

100 C e n te rline te m p e ra tu re , To (C )

C e n te rlin e te m p e ra tu re , To (C )

50

45

40

35

30 0

5

10

15

75

50

25 0

20

500

1000 Tim e , t(s )

Air ve lo city, V(m /s )

V=3 m /s V=1 0 m /s V=2 0 m /s

2

With increasing V from 3 to 20 m/s, h increases from 33 to 107 W/m ⋅K, and the enhanced cooling reduces the centerline temperature at the prescribed time. The accelerated cooling associated with increasing V is also revealed by the temperature histories, and the time required to achieve thermal equilibrium between the air and the cylinder decreases with increasing V. 2

COMMENTS: (1) For the smallest value of h = 33 W/m ⋅K, Bi ≡ h (ro/2)/k = 1.1 >> 0.1, and use of the lumped capacitance method is clearly inappropriate. (2) The IHT Transient Conduction Model for a cylinder was used to perform the calculations of Part (b). Because the model is based on the exact solution, Eq. 5.47a, it is accurate for values of Fo < 0.2, as well as Fo > 0.2. Although in principle, the model may be used to calculate the thermal conductivity for the conditions of Part (a), convergence is elusive and may only be achieved if the initial guesses are close to the correct results.

1500

PROBLEM 5.57 KNOWN: Diameter, initial temperature and properties of stainless steel rod. Temperature and convection coefficient of coolant. FIND: Temperature distributions for prescribed convection coefficients and times. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties. ANALYSIS: The IHT model is based on the exact solution to the heat equation, Eq. 5.47. The results are plotted as follows h =1 0 0 0 W /m ^2 -K

h =1 0 0 W /m ^2 -K

325

325

275 Te m p e re a tu re , C

Te m p e ra tu re , C

275 225 175 125

225 175 125

75

75

25

25 0

0 .2

0 .4

0 .6

0 .8

0

1

0 .2

0 .8

1

0 .8

1

t=0 s t=1 0 s t=5 0 s

t= 0 t= 1 0 0 s t= 5 0 0 s

h =5 0 0 0 W /m ^2 -K

2

larger than [T (ro,t) - T∞].

0 .6

D im e n s io n le s s ra d iu s , r*

D im e n s io n le s s ra d iu s , r*

325 275 Te m p e ra tu re , C

For h = 100 W/m ⋅K, Bi = hro/k = 0.1, and as expected, the temperature distribution is nearly uniform throughout the rod. For h = 1000 2 W/m ⋅K (Bi = 1), temperature variations within the rod are not negligible. In this case the centerline-to-surface temperature difference is comparable to the surface-to-fluid 2 temperature difference. For h = 5000 W/m ⋅K (Bi = 5), temperature variations within the rod are large and [T (0,t) – T (ro,t)] is substantially

0 .4

225 175 125 75 25 0

0 .2

0 .4

0 .6

D im e n s io n le s s ra d iu s , r* t=0 s t=1 s t=5 s t=2 5 s

COMMENTS: With increasing Bi, conduction within the rod, and not convection from the surface, becomes the limiting process for heat loss.

PROBLEM 5.58 KNOWN: A ball bearing is suddenly immersed in a molten salt bath; heat treatment to harden occurs at locations with T > 1000K. FIND: Time required to harden outer layer of 1mm. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Fo ≥ 0.2. ANALYSIS: Since any location within the ball whose temperature exceeds 1000K will be hardened, the problem is to find the time when the location r = 9mm reaches 1000K. Then a 1mm outer layer will be hardened. Begin by finding the Biot number. 5000 W/m2 ⋅ K ( 0.020m/2 ) hr Bi = o = = 1.00.

50 W/m ⋅ K

k

Using the one-term approximate solution for a sphere, find

Fo = −

( )

 1 ln θ ∗ / C1 sin ζ1r∗ 2 ∗  ζ1 ζ1r 1

 . 

From Table 5.1 with Bi = 1.00, for the sphere find ζ 1 = 15708 . rad and C1 = 1.2732. With r* = r/ro = (9mm/10mm) = 0.9, substitute numerical values.  (1000 − 1300 ) K  1 −1 Fo = ln  /1.2732 sin (1.5708 × 0.9 rad ) = 0.441. 1.5708 × 0.9  (1.5708)2  (300 − 1300 ) K From the definition of the Fourier number with α = k/ρc, 2 ro2 ρc kg  0.020m  2 t = Fo = Fo ⋅ ro = 0.441×   7800 3 × 500

α

k



2



m

J / 50 W/m ⋅ K = 3.4s. kg ⋅ K

<

COMMENTS: (1) Note the very short time required to harden the ball. At this time it can be easily shown the center temperature is T(0,3.4s) = 871 K. (2) The Heisler charts can also be used. From Fig. D.8, with Bi 0.69(±0.03). Since

θ = T − T∞ = 1000 − 1300 = −300K θ θ θo = ⋅ , θi θo θi θ o / θ i = 0.30 / 0.69 = 0.43 ( ±0.02 ).

and

Since

= 1.0 and r/ro = 0.9, read θ/θo =

θ i = Ti − T∞ = −1000K

it follows that

θ = 0.30. θi

-1

then

θ θ = 0.69 o θi θi

-1 From Fig. D.7 at θo/θi=0.43, Bi =1.0, read Fo = 0.45 (±0.03) and t = 3.5 (±0.2)s. Note the use of tolerances associated with reading the charts to ±5%.

PROBLEM 5.59 KNOWN: An 80mm sphere, initially at a uniform elevated temperature, is quenched in an oil bath with prescribed T∞, h. FIND: The center temperature of the sphere, T(0,t) at a certain time when the surface temperature is T(ro,t) = 150°C. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Initial uniform temperature within sphere, (3) Constant properties, (4) Fo ≥ 0.2. ANALYSIS: Check first to see if the sphere is spacewise isothermal. h ( ro / 3 ) 1000 W/m 2 ⋅ K × 0.040m/3 hL Bi c = c = = = 0.26. k k 50 W/m ⋅ K Since Bic > 0.1, lumped capacitance method is not appropriate. Recognize that when Fo ≥ 0.2, the time dependence of the temperature at any point within the sphere will be the same as the center. Using the Heisler chart method, Fig. D.8 provides the relation between T(ro,t) and T(0,t). Find first the Biot number, hro 1000 W/m 2 ⋅ K × 0.040m = = 0.80. k 50 W/m ⋅ K -1 With Bi = 1/0.80 = 1.25 and r/ro =1, read from Fig. D.8, Bi =

T ( ro , t ) − T∞ θ = = 0.67. θ o T ( 0,t ) − T∞ It follows that T ( 0,t ) = T∞ +

1 1  T (ro , t ) − T∞  = 50o C + 150 − 50 ]o C = 199 oC. [ 0.67 0.67

<

COMMENTS: (1) There is sufficient information to evaluate Fo; hence, we require that the time be sufficiently long after the start of quenching for this solution to be appropriate. (2)The approximate series solution could also be used to obtain T(0,t). For Bi = 0.80 from Table 5.1, ζ1 = 1.5044 rad. Substituting numerical values, r* = 1, θ∗ θ o∗

=

T (ro , t ) − T∞ T ( 0,t ) − T∞

=

It follows that T(0,t) = 201°C.

( )

1 sin ζ1r ∗ = sin (1.5044 rad ) = 0.663. 1.5044 ζ1r∗ 1

PROBLEM 5.60 KNOWN: Steel ball bearings at an initial, uniform temperature are to be cooled by convection while passing through a refrigerated chamber; bearings are to be cooled to a temperature such that 70% of the thermal energy is removed. FIND: Residence time of the balls in the 5m-long chamber and recommended drive velocity for the conveyor. SCHEMATIC:

ASSUMPTIONS: (1) Negligible conduction between ball and conveyor surface, (2) Negligible radiation exchange with surroundings, (3) Constant properties, (4) Uniform convection coefficient over ball’s surface. ANALYSIS: The Biot number for the lumped capacitance analysis is 2 hLc h ( ro / 3) 1000 W/m ⋅ K (0.1m/3) Bi ≡ = = = 0.67. k k 50 W/m ⋅ K

Since Bi > 0.1, lumped capacitance analysis is not appropriate. In Figure D.9, the internal energy change is shown as a function of Bi and Fo. For Q = 0.70 Qo

hro 1000 W/m 2 ⋅ K × 0.1m Bi = = = 2.0, k 50 W/m ⋅ K

and

2

find Bi Fo ≈ 1.2. The Fourier number is Fo =

αt ro2

=

2 × 10−5 m 2 / s × t

(0.1 m )

2

= 2.0 × 10−3 t

giving t=

Fo 2.0 × 10-3

=

1.2 / Bi 2 2.0 × 10-3

1.2 / ( 2.0 )

2

=

2.0 × 10−3

= 150s.

The velocity of the conveyor is expressed in terms of the length L and residence time t. Hence V=

L 5m = = 0.033m/s = 33mm/s. t 150s

COMMENTS: Referring to Eq. 5.10, note that for a sphere, the characteristic length is r 4 Lc = V/As = π ro3 / 4π ro2 = o . 3 3 However, when using the Heisler charts, note that Bi ≡ h ro/k.

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PROBLEM 5.61 KNOWN: Diameter and initial temperature of ball bearings to be quenched in an oil bath. FIND: (a) Time required for surface to cool to 100°C and the corresponding center temperature, (b) Oil bath cooling requirements. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction in ball bearings, (2) Constant properties. PROPERTIES: Table A-1, St. St., AISI 304, (T ≈ 500°C): k = 22.2 W/m⋅K, cp = 579 J/kg⋅K, 3 -6 2 ρ = 7900 kg/m , α = 4.85×10 m /s. ANALYSIS: (a) To determine whether use of the lumped capacitance method is suitable, first compute h ( ro / 3) 1000 W/m 2 ⋅ K ( 0.010m/3) Bi = = = 0.15. k 22.2 W/m ⋅ K We conclude that, although the lumped capacitance method could be used as a first approximation, the Heisler charts should be used in the interest of improving accuracy. Hence, with Bi -1 =

k 22.2 W/m ⋅ K = = 2.22 hro 1000 W/m 2 ⋅ K ( 0.01m )

and

r = 1, ro

Fig. D.8 gives θ ( ro , t ) ≈ 0.80. θo (t ) Hence, with θ ( ro , t ) θi

=

T (ro , t ) − T∞ Ti − T∞

=

100 − 40 = 0.074, 850 − 40 Continued …..

PROBLEM 5.61 (Cont.) it follows that θ o θ ( ro , t ) / θ i 0.074 = = = 0.093. θ i θ ( ro , t ) / θ o 0.80 From Fig. D.7, with θ o / θ i = 0.093 and Bi -1 = k/hro = 2.22, find t ∗ = Fo ≈ 2.0

( 0.01m )2 ( 2.0 ) ro2Fo t= = = 41s. α 4.85 × 10−6 m 2 / s

<

Also, θ o = To − T∞ = 0.093( Ti − T∞ ) = 0.093 ( 850 − 40 ) = 75o C To = 115o C 2

< 2

(b) With Bi Fo = (1/2.2) × 2.0 = 0.41, where Bi ≡ (hro/k) = 0.45, it follows from Fig. D.9 that for a single ball Q ≈ 0.93. Qo Hence, from Eq. 5.44, Q = 0.93 ρc pV ( Ti − T∞ ) Q = 0.93× 7900 kg/m 3 × 579 J/kg ⋅ K × Q = 1.44 × 104J

π ( 0.02m )3 × 810o C 6

is the amount of energy transferred from a single ball during the cooling process. Hence, the oil bath cooling rate must be q = 104 Q/3600s q = 4 ×10 4 W = 40 kW.

<

COMMENTS: If the lumped capacitance method is used, the cooling time, obtained from Eq. 5.5, would be t = 39.7s, where the ball is assumed to be uniformly cooled to 100°C. This result, and the fact that To - T(ro) = 15°C at the conclusion, suggests that use of the lumped capacitance method would have been reasonable. Note that, when using the Heisler charts, accuracy to better than 5% is seldom possible.

PROBLEM 5.62 KNOWN: Diameter and initial temperature of hailstone falling through warm air. FIND: (a) Time, tm, required for outer surface to reach melting point, T(ro,tm) = Tm = 0°C, (b) Centerpoint temperature at that time, (c) Energy transferred to the stone. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties. 3

PROPERTIES: Table A-3, Ice (253K): ρ = 920 kg/m , k = 2.03 W/m⋅K, cp = 1945 J/kg⋅K; α -6 2 = k/ρcp = 1.13 × 10 m /s. ANALYSIS: (a) Calculate the lumped capacitance Biot number, h ( ro / 3) 250 W/m 2 ⋅ K ( 0.0025m/3) = = 0.103. k 2.03 W/m ⋅ K Since Bi > 0.1, use the Heisler charts for which θ ( ro , t m ) T ( ro , tm ) − T∞ 0− 5 = = = 0.143 θi Ti − T∞ −30 − 5 k 2.03 W/m ⋅ K Bi -1 = = = 3.25. hro 250 W/m 2 ⋅K × 0.0025m Bi =

From Fig. D.8, find

θ ( ro , t m ) ≈ 0.86. θo ( t m )

It follows that

θo ( tm ) θ ( ro , t m ) / θ i 0.143 = ≈ ≈ 0.17. θi θ ( ro , t m ) / θ o ( t m ) 0.86

From Fig. D.7 find Fo ≈ 2.1. Hence, 2.1 ( 0.0025) 2 Fo ro2 tm ≈ = = 12s. α 1.13 ×10−6 m2 / s

<

(b) Since (θo/θi) ≈ 0.17, find To − T∞ ≈ 0.17 ( Ti − T∞ ) ≈ 0.17 ( −30 − 5 ) ≈ −6.0o C To ( tm ) ≈ −1.0o C. 2

<

2

(c) With Bi Fo = (1/3.25) ×2.1 = 0.2, from Fig. D.9, find Q/Qo ≈ 0.82. From Eq. 5.44,

(

Qo = ρ Vc pθ i = 920 kg/m 3

) (π /6)(0.005m )31945 (J/kg ⋅K )( −35K ) = −4.10 J

Q = 0.82 Qo = 0.82 ( −4.10 J ) = −3.4 J.

<

PROBLEM 5.63 KNOWN: Sphere quenching in a constant temperature bath. FIND: (a) Plot T(0,t) and T(ro,t) as function of time, (b) Time required for surface to reach 415 K, t ′ , (c) Heat flux when T(ro, t ′ ) = 415 K, (d) Energy lost by sphere in cooling to T(ro, t ′ ) = 415 K, (e) Steady-state temperature reached after sphere is insulated at t = t ′ , (f) Effect of h on center and surface temperature histories. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Uniform initial temperature. ANALYSIS: (a) Calculate Biot number to determine if sphere behaves as spatially isothermal object, 2 hLc h ( ro 3) 75 W m ⋅ K ( 0.015m 3) Bi = = = = 0.22 . k k 1.7 W m ⋅ K Hence, temperature gradients exist in the sphere and T(r,t) vs. t appears as shown above. (b) The Heisler charts may be used to find t ′ when T(ro, t ′ ) = 415 K. Using Fig. D.8 with r/ro = 1 and Bi-1 = k/hro = 1.7 W/m⋅K/(75 W/m2⋅K × 0.015 m) = 1.51, θ (1, t ′ ) θ o ≈ 0.72 . In order to enter Fig. D.7, we need to determine θo/θi, which is

θ o θ (1, t ′ ) θ (1, t ′ ) ( 415 − 320 ) K 0.72 = 0.275 = ≈ θi θo θi (800 − 320 ) K Hence, for Bi-1 = 1.51, Fo ≡ α t ′ ro2 ≈ 0.87 and

ρ cp 2 ro2 400 kg m3 × 1600 J kg ⋅ K 2 t ′ = Fo = Fo ⋅ ⋅ ro ≈ 0.87 × ( 0.015m ) = 74s α k 1.7 W m ⋅ K (c) The heat flux at the outer surface at time t „ is given by Newton’s law of cooling q′′ = h T ( ro , t′ ) − T∞  = 75 W m 2 ⋅ K [415 − 320] K = 7125 W / m2 . .

< <

The manner in which q′′ is calculated indicates that energy is leaving the sphere. (d) The energy lost by the sphere during the cooling process from t = 0 to t ′ can be determined from the Groeber chart, Fig. D.9. With Bi = 1/1.51 = 0.67 and Bi2Fo = (1/1.51)2 × 0.87 ≈ 0.4, the chart yields Q Qo ≈ 0.75 . The energy loss by the sphere with V = (πD3)/6 is therefore

(

)

Q ≈ 0.85Qo = 0.85 ρ π D3 6 cp ( Ti − T∞ )

(

)

Q ≈ 0.85 × 400 kg m3 π [0.030 m ] 6 1600 J kg ⋅ K (800 − 320 ) K = 3691J 3

< Continued...

PROBLEM 5.63 (Cont.) (e) If at time t ′ the surface of the sphere is perfectly insulated, eventually the temperature of the sphere will be uniform at T(∞). Applying conservation of energy to the sphere over a time interval, Ein - Eout = ∆E ≡ Efinal - Einitial. Hence, -Q = ρcV[T(∞) - T∞] - Qo, where Qo ≡ ρcV[Ti - T∞]. Dividing by Qo and regrouping, we obtain

T ( ∞ ) = T∞ + (1 − Q Qo )( Ti − T∞ ) ≈ 320 K + (1 − 0.75)(800 − 320 ) K = 440 K

<

(f) Using the IHT Transient Conduction Model for a Sphere, the following graphical results were generated. 800 90000

Heat flux, q''(ro,t) (W/m^2.K)

Temperature, T(K)

700 600 500 400 300 0

50

100

60000

30000

150 0

Time, t (s) h = 75 W/m^2.K, r = ro h = 75 W/m^2.K, r = 0 h = 200 W/m^2.K, r = ro h = 200 W/m^2.K, r = 0

0

50

100

150

Time, t(s) h = 75 W/m^2.K h = 200 W/m^2.K

The quenching process is clearly accelerated by increasing h from 75 to 200 W/m2⋅K and is virtually completed by t ≈ 100s for the larger value of h. Note that, for both values of h, the temperature difference [T(0,t) - T(ro,t)] decreases with increasing t. Although the surface heat flux for h = 200 W/m2⋅K is initially larger than that for h = 75 W/m2⋅K, the more rapid decline in T(ro,t) causes it to become smaller at t ≈ 30s. COMMENTS: 1. There is considerable uncertainty associated with reading Q/Qo from the Groeber chart, Fig. D.9, and it would be better to use the one-term approximation solutions of Section 5.6.2. With Bi = 0.662, from Table 5.1, find ζ1 = 1.319 rad and C1 = 1.188. Using Eq. 5.50, find Fo = 0.852 and t′ = 72.2 s. Using Eq. 5.52, find Q/Qo = 0.775 and T(∞) = 428 K. 2. Using the Transient Conduction/Sphere model in IHT based upon multiple-term series solution, the following results were obtained: t′ = 72.1 s; Q/Qo = 0.7745, and T(∞) = 428 K.

PROBLEM 5.64 KNOWN: Two spheres, A and B, initially at uniform temperatures of 800K and simultaneously quenched in large, constant temperature baths each maintained at 320K; properties of the spheres and convection coefficients. FIND: (a) Show in a qualitative manner, on T-t coordinates, temperatures at the center and the outer surface for each sphere; explain features of the curves; (b) Time required for the outer surface of each sphere to reach 415K, (c) Energy gained by each bath during process of cooling spheres to a surface temperature of 415K.

SCHEMATIC: Sphere A 150

ro (mm)

3

ρ (kg/m ) c (J/kg⋅K) k (W/m⋅K) 2 h (W/m ⋅K)

1600 400 170 5

Sphere B 15 400 1600 1.7 50

ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Uniform properties, (3) Constant convection coefficient. ANALYSIS: (a) From knowledge of the Biot number and the thermal time constant, it is possible to qualitatively represent the temperature distributions. From Eq. 5.10, with Lc = ro/3, find 5 W/m ⋅ K ( 0.150m/3 ) 2

Bi A = Bi =

h ( ro / 3 ) k

170 W/m ⋅ K

50 W/m ⋅ K ( 0.015m/3 )

= 1.47 × 10

−3

(1)

2

Bi B =

1.7 W/m ⋅ K

= 0.147

(2)

The thermal time constant for a lumped capacitance system from Eq. 5.7 is

 1  τ =  ( ρ Vc )  hAs  τ=

ρ ro c 3h

τB =

τA =

1600 kg/m3 × ( 0.150m ) 400 J/kg ⋅ K 3 × 5 W/m 2 ⋅ K

400 kg/m3 × ( 0.015m )1600 J/kg ⋅ K 3 × 50 W/m 2 ⋅ K

= 64s

= 6400s

(3) (4)

When Bi 0.1, hence gradients will be important. Note that the thermal time constant of A is much larger than for B; hence, A will cool much slower. See sketch for these features. (b) Recognizing that BiA < 0.1, Sphere A can be treated as spacewise isothermal and analyzed using the lumped capacitance method. From Eq. 5.6 and 5.7, with T = 415 K θ T − T∞ = = exp ( − t/τ ) (5) θi Ti − T∞ Continued …..

PROBLEM 5.64 (Cont.)  T − T∞   415 − 320  t A = −τ A  ln = 10,367s = 2.88h.  = −6400s  ln  800 − 320   Ti − T∞ 

<

Note that since the sphere is nearly isothermal, the surface and inner temperatures are approximately the same. Since BiB > 0.1, Sphere B must be treated by the Heisler chart method of solution beginning with Figure D.8. Using 2 hro 50 W/m ⋅ K × (0.015m ) Bi B ≡ = = 0.44 k 1.7 W/m ⋅ K

Bi-1 B = 2.27,

or

find that for r/ro = 1,

θ (1, t ) T ( ro , t ) − T∞ ( 415 − 320 ) = = = 0.8. θo θo θo

(6)

Using Eq. (6) and Figure D.7, find the Fourier number,

θ o (T ( ro , t ) − T∞ ) / 0.8 ( 415 − 320 ) K/0.8 = = = 0.25 Ti − T∞ θi (800 − 320 ) K

Fo =

αt ro2

= 1.3.

1.3 ( 0.015m ) Fo ro2 tB = = = 110s = 1.8 min α 2.656 ×10−6 m 2 / s 3 -6 2 where α = k/ρc = 1.7 W/m⋅K/400 kg/m × 1600 J/kg⋅K = 2.656×10 m /s. 2

<

(c) To determine the energy change by the spheres during the cooling process, apply the conservation of energy requirement on a time interval basis. Sphere A: Ein − Eout = ∆E

− Q A = ∆E = E ( t ) − E (0 ).

QA = ρ cV T ( t ) − Ti  = 1600kg/m3 × 400J/kg ⋅ K × ( 4/3)π (0.150m ) [415 − 800] K 3

QA = 3.483 × 106 J. Note that this simple expression is a consequence of the spacewise isothermal behavior. Sphere B:

<

− QB = E ( t ) − E (0 ).

Ein − E out = ∆E

For the nonisothermal sphere, the Groeber chart, Figure D.9, can be used to evaluate QB. 2

2

With Bi = 0.44 and Bi Fo = (0.44) ×1.3 = 2.52, find Q/Qo = 0.74. The energy transfer from the sphere during the cooling process, using Eq. 5.44, is QB = 0.74 Qo = 0.74  ρ cV ( Ti − T∞ ) QB = 0.75 × 400kg/m3 × 1600J/kg ⋅ K ( 4/3)π (0.015m ) (800 − 320 ) K = 3257 J. 3

COMMENTS: (1) In summary: Sphere A B

Bi = hro/k -3

4.41×10 0.44

τ s

$

t(s)

Q(J)

6400 64

10,370 110

3.48×10 3257

6

<

PROBLEM 5.65 KNOWN: Spheres of 40-mm diameter heated to a uniform temperature of 400°C are suddenly removed from an oven and placed in a forced-air bath operating at 25°C with a convection coefficient 2 of 300 W/m ⋅K. FIND: (a) Time the spheres must remain in the bath for 80% of the thermal energy to be removed, and (b) Uniform temperature the spheres will reach when removed from the bath at this condition and placed in a carton that prevents further heat loss. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction in the spheres, (2) Constant properties, and (3) No heat loss from sphere after removed from the bath and placed into the packing carton. 3

PROPERTIES: Sphere (given): ρ = 3000 kg/m , c = 850 J/kg⋅K, k = 15 W/m⋅K. ANALYSIS: (a) From Eq. 5.52, the fraction of thermal energy removed during the time interval ∆t = to is

Q = 1 − 3θ o∗ / ζ13 sin (ζ1 ) − ζ1 cos (ζ1 ) Qo

(1)

where Q/Qo = 0.8. The Biot number is Bi = hro / k = 300 W / m 2 ⋅ K × 0.020 m /15 W / m ⋅ K = 0.40 and for the one-term series approximation, from Table 5.1,

ζ1 = 1.0528 rad

C1 = 1.1164

(2)

The dimensionless temperature θ o∗ , Eq. 5.31, follows from Eq. 5.50.

(

θ o∗ = C1 exp −ζ12 Fo

)

(3)

where Fo = α t o / ro2 . Substituting Eq. (3) into Eq. (1), solve for Fo and to.

)

(

Q = 1 − 3 C1 exp −ζ12Fo / ζ13 sin (ζ1 ) − ζ1 cos (ζ1 ) Qo

(4)

Fo = 1.45

<

t o = 98.6 s

(b) Performing an overall energy balance on the sphere during the interval of time to ≤ t ≤ ∞,

Ein − E out = ∆E = E f − Ei = 0

(5)

where Ei represents the thermal energy in the sphere at to,

Ei = (1 − 0.8 ) Qo = (1 − 0.8 ) ρ cV (Ti − T∞ )

(6)

and Ef represents the thermal energy in the sphere at t = ∞,

(

Ef = ρ cV Tavg − T∞

)

(7)

Combining the relations, find the average temperature

(

)

ρ cV  Tavg − T∞ − (1 − 0.8 )(Ti − T∞ ) = 0   Tavg = 100°C

<

PROBLEM 5.66 KNOWN: Diameter, density, specific heat and thermal conductivity of Pyrex spheres in packed bed thermal energy storage system. Convection coefficient and inlet gas temperature. FIND: Time required for sphere to acquire 90% of maximum possible thermal energy and the corresponding center temperature. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial conduction in sphere, (2) Negligible heat transfer to or from a sphere by radiation or conduction due to contact with adjoining spheres, (3) Constant properties. 2

ANALYSIS: With Bi ≡ h(ro/3)/k = 75 W/m ⋅K (0.0125m)/1.4 W/m⋅K = 0.67, the approximate solution for one-dimensional transient conduction in a sphere is used to obtain the desired results. We first use Eq. (5.52) to obtain θ o∗ .

θ o∗ =

ζ13

 Q  1 −  3 sin (ζ1 ) − ζ1 cos (ζ1 )  Qo 

With Bi ≡ hro/k = 2.01, ζ1 ≈ 2.03 and C1 ≈ 1.48 from Table 5.1. Hence, 3 0.1( 2.03) 0.837 ∗ θo = = = 0.155

3  0.896 − 2.03 ( −0.443)

5.386

The center temperature is therefore

(

)

To = Tg,i + 0.155 Ti − Tg,i = 300°C − 42.7°C = 257.3°C From Eq. (5.50c), the corresponding time is θ∗  r2 t = − o ln  o  αζ12  C1 

(

<

)

where α = k / ρ c = 1.4 W / m ⋅ K / 2225 kg / m 3 × 835 J / kg ⋅ K = 7.54 × 10 −7 m 2 / s.

t=−

(0.0375m )2 ln (0.155 /1.48 ) 1, 020s = 2 7 2 − 7.54 × 10 m / s ( 2.03)

<

COMMENTS: The surface temperature at the time of interest may be obtained from Eq. (5.50b). With r ∗ = 1,

θ o∗ sin (ζ1 )  0.155 × 0.896  Ts = Tg,i + Ti − Tg,i = 300°C − 275°C   = 280.9°C ζ1 2.03  

(

)

<

PROBLEM 5.67 KNOWN: Initial temperature and properties of a solid sphere. Surface temperature after immersion in a fluid of prescribed temperature and convection coefficient. FIND: (a) Time to reach surface temperature, (b) Effect of thermal diffusivity and conductivity on thermal response. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Constant properties. ANALYSIS: (a) For k = 15 W/m⋅K, the Biot number is

Bi =

h ( ro 3) k

=

300 W m 2 ⋅ K ( 0.05 m 3) 15 W m ⋅ K

= 0.333 .

Hence, the lumped capacitance method cannot be used. From Equation 5.50a,

) ) (

sin ζ1r* T − T∞ 2 . = C1 exp −ζ1 Fo Ti − T∞ ζ1r*

(

At the surface, r* = 1. From Table 5.1, for Bi = 1.0, ζ1 = 1.5708 rad and C1 = 1.2732. Hence,

)

(

60 − 75 sin 90$ = 0.30 = 1.2732 exp −1.57082 Fo 25 − 75 1.5708 exp(-2.467Fo) = 0.370

Fo =

αt ro2

= 0.403

ro2 (0.05 m ) = 100s t = 0.403 = 0.403 α 10−5 m 2 s 2

<

(b) Using the IHT Transient Conduction Model for a Sphere to perform the parametric calculations, the effect of α is plotted for k = 15 W/m⋅K.

Continued...

75

75

65

65

Center temperature, T(C)

Surface temperature, T(C)

PROBLEM 5.67 (Cont.)

55

45

35

25

55

45

35

25 0

50

100

150

200

250

300

0

50

100

Time, t(s)

150

200

250

300

Time, t(s)

k = 15 W/m.K, alpha = 1E-4 m^2/s k = 15 W/m.K, alpha = 1E-5 m^2/s k = 15 W/m.K, alpha = 1E-6m^2/s

k = 15 W/m.K, alpha = 1E-4 m^2/s k = 15 W/m.K, alpha = 1E-5 m^2/s k = 15 W/m.K, alpha = 1E-6 m^2/s

For fixed k and increasing α, there is a reduction in the thermal capacity (ρcp) of the material, and hence the amount of thermal energy which must be added to increase the temperature. With increasing α, the material therefore responds more quickly to a change in the thermal environment, with the response at the center lagging that of the surface.

75

75

65

65

Center temperature, T(C)

Surface temperature, T(C)

The effect of k is plotted for α = 10-5 m2/s.

55

45

35

25

55

45

35

25 0

50

100

150

200

Time, t(s) k = 1.5 W/m.K, alpha = 1E-5 m^2/s k = 15 W/m.K, alpha = 1E-5 m^2/s k = 150W/m.K, alpha = 1E-5 m^2/s

250

300

0

50

100

150

200

250

300

Time, t(s) k = 1.5 W/m.K, alpha = 1E-5 m^2/s k = 15 W/m.K, alpha = 1E-5 m^2/s k =150 W/m.K, alpha = 1E-5m^2/s

With increasing k for fixed alpha, there is a corresponding increase in ρcp, and the material therefore responds more slowly to a thermal change in its surroundings. The thermal response of the center lags that of the surface, with temperature differences, T(ro,t) - T(0,t), during early stages of solidification increasing with decreasing k. COMMENTS: Use of this technique to determine h from measurement of T(ro) at a prescribed t requires an interative solution of the governing equations.

PROBLEM 5.68 KNOWN: Properties, initial temperature, and convection conditions associated with cooling of glass beads. FIND: (a) Time required to achieve a prescribed center temperature, (b) Effect of convection coefficient on center and surface temperature histories. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in r, (2) Constant properties, (3) Negligible radiation, (4) Fo ≥ 0.2. ANALYSIS: (a) With h = 400 W/m2⋅K, Bi ≡ h(ro/3)/k = 400 W/m2⋅K(0.0005 m)/1.4 W/m⋅K = 0.143 and the lumped capacitance method should not be used. From the one-term approximation for the center temperature, Eq. 5.50c, T −T 80 − 15 θ o* ≡ o ∞ = = 0.141 = C1 exp −ζ12 Fo Ti − T∞ 477 − 15 For Bi ≡ hro/k = 0.429, Table 5.1 yields ζ1 = 1.101 rad and C1 = 1.128. Hence,

)

(

Fo = −

θ*  1  0.141  ln  o  = − ln  = 1.715 2  1.128  ζ12  C1  1.101 ( ) 1

t = 1.715ro2

ρ cp k

3 2 2200 kg m × 800 J kg ⋅ K

= 1.715 ( 0.0015 m )

1.4 W m ⋅ K

<

= 4.85s

From Eq. 5.50b, the corresponding surface (r* = 1) temperature is sin ζ1 0.892 T ( ro , t ) = T∞ + ( Ti − T∞ )θ o* = 15$ C + 462$ C 0.141 = 67.8$ C ζ1 1.101 (b) The effect of h on the surface and center temperatures was determined using the IHT Transient Conduction Model for a Sphere.

)

500

500

400

400

Surface temperature, T(C)

Center temperature, T(C)

(

300

200

100

0

<

300

200

100

0 0

4

8

12 Time, t(s)

h = 100 W/m^2.K, r = 0 h = 400 W/m^2.K, r = 0 h = 1000 W/m^2.K, r = 0

16

20

0

4

8

12

16

20

Time, t(s) h = 100 W/m^2.K, r = ro h = 400 W/m^2.K, r = ro h = 1000 W/m^2.K, r = ro

Continued...

PROBLEM 5.68 (Cont.) The cooling rate increases with increasing h, particularly from 100 to 400 W/m2⋅K. The temperature difference between the center and surface decreases with increasing t and, during the early stages of solidification, with decreasing h. COMMENTS: Temperature gradients in the glass are largest during the early stages of solidification and increase with increasing h. Since thermal stresses increase with increasing temperature gradients, the propensity to induce defects due to crack formation in the glass increases with increasing h. Hence, there is a value of h above which product quality would suffer and the process should not be operated.

PROBLEM 5.69 KNOWN: Temperature requirements for cooling the spherical material of Ex. 5.4 in air and in a water bath. FIND: (a) For step 1, the time required for the center temperature to reach T(0,t) = 335°C while 2 cooling in air at 20°C with h = 10 W/m ⋅K; find the Biot number; do you expect radial gradients to be appreciable?; compare results with hand calculations in Ex. 5.4; (b) For step 2, time required for the 2 center temperature to reach T(0,t) = 50°C while cooling in water bath at 20°C with h = 6000 W/m ⋅K; and (c) For step 2, calculate and plot the temperature history, T(x,t) vs. t, for the center and surface of the sphere; explain features; when do you expect the temperature gradients in the sphere to the largest? Use the IHT Models | Transient Conduction | Sphere model as your solution tool. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the radial direction, (2) Constant properties. ANALYSIS: The IHT model represents the series solution for the sphere providing the temperatures evaluated at (r,t). A selected portion of the IHT code used to obtain results is shown in the Comments. (a) Using the IHT model with step 1 conditions, the time required for T(0,ta) = T_xt = 335°C with r = 0 and the Biot number are:

t a = 94.2 s

<

Bi = 0.0025

Radial temperature gradients will not be appreciable since Bi = 0.0025 > 0.1. The sphere does not behave as a space-wise isothermal object for the water-cooling process. (c) For the step 2 cooling process, the temperature histories for the center and surface of the sphere are calculated using the IHT model. Continued …..

PROBLEM 5.69 (Cont.) Te m p e ra tu re -tim e h is to ry, S te p 2

Te m p e ra tu re , T(r,t) (C )

400

300

200

100

0 0

1

2

3

4

5

6

Tim e , t (s ) S u rfa c e , r = ro C e n te r, r = 0

At early times, the difference between the center and surface temperature is appreciable. It is in this time region that thermal stresses will be a maximum, and if large enough, can cause fracture. Within 6 seconds, the sphere has a uniform temperature equal to that of the water bath. COMMENTS: Selected portions of the IHT sphere model codes for steps 1 and 2 are shown below. /* Results, for part (a), step 1, air cooling; clearly negligible gradient Bi Fo t T_xt Ti r ro 0.0025 25.13 94.22 335 400 0 0.005 */ // Models | Transient Conduction | Sphere - Step 1, Air cooling // The temperature distribution T(r,t) is T_xt = T_xt_trans("Sphere",rstar,Fo,Bi,Ti,Tinf) // Eq 5.47 T_xt = 335 // Surface temperature

/* Results, for part (b), step 2, water cooling; Ti = 335 C Bi Fo t T_xt Ti r ro 1.5 0.7936 2.976 50 335 0 0.005 */ // Models | Transient Conduction | Sphere - Step 2, Water cooling // The temperature distribution T(r,t) is T_xt = T_xt_trans("Sphere",rstar,Fo,Bi,Ti,Tinf) // Eq 5.47 //T_xt = 335 // Surface temperature from Step 1; initial temperature for Step 2 T_xt = 50 // Center temperature, end of Step 2

PROBLEM 5.70 KNOWN: Two large blocks of different materials – like copper and concrete – at room temperature, 23°C. FIND: Which block will feel cooler to the touch? SCHEMATIC:

ASSUMPTIONS: (1) Blocks can be treated as semi-infinite solid, (2) Hand or finger temperature is 37°C. 3

PROPERTIES: Table A-1, Copper (300K): ρ = 8933 kg/m , c = 385 J/kg⋅K, k = 401 3

W/m⋅K; Table A-3, Concrete, stone mix (300K): ρ = 2300 kg/m , c = 880 J/kg⋅K, k = 1.4 W/m⋅K. ANALYSIS: Considering the block as a semi-infinite solid, the heat transfer situation corresponds to a sudden change in surface temperature, Case 1, Figure 5.7. The sensation of coolness is related to the heat flow from the hand or finger to the block. From Eq. 5.58, the surface heat flux is q′′s ( t ) = k ( Ts − Ti ) / (πα t )

(1)

q′′s ( t ) ~ ( kρ c )

(2)

1/ 2

or 1/ 2

since α = k/ρ c.

Hence for the same temperature difference, Ts − Ti , and elapsed time, it follows that the heat fluxes for the two materials are related as 1/ 2  W kg J  1/ 2 ( kρ c )copper  401 m ⋅ K × 8933 m3 × 385 kg ⋅ K  q′′s,copper = = = 22.1 1/ 2 q′′s,concrete ( kρ c )1/2  W kg J  concrete 1.4 m ⋅ K × 2300 3 × 880 kg ⋅ K  m





Hence, the heat flux to the copper block is more than 20 times larger than to the concrete block. The copper block will therefore feel noticeably cooler than the concrete one.

PROBLEM 5.71 KNOWN: Asphalt pavement, initially at 50°C, is suddenly exposed to a rainstorm reducing the surface temperature to 20°C. 2

FIND: Total amount of energy removed (J/m ) from the pavement for a 30 minute period. SCHEMATIC:

ASSUMPTIONS: (1) Asphalt pavement can be treated as a semi-infinite solid, (2) Effect of rainstorm is to suddenly reduce the surface temperature to 20°C and is maintained at that level for the period of interest. 3

PROPERTIES: Table A-3, Asphalt (300K): ρ = 2115 kg/m , c = 920 J/kg⋅K, k = 0.062 W/m⋅K. ANALYSIS: This solution corresponds to Case 1, Figure 5.7, and the surface heat flux is given by Eq. 5.58 as q′′s ( t ) = k ( Ts − Ti ) / ( πα t )

1/2

(1)

The energy into the pavement over a period of time is the integral of the surface heat flux expressed as Q′′ = ∫ q ′′s ( t ) dt. 0 t

(2)

Note that q′′s ( t ) is into the solid and, hence, Q represents energy into the solid. Substituting Eq. (1) for q′′s ( t ) into Eq. (2) and integrating find t

Q′′ = k ( Ts − Ti ) / ( πα )1/2 ∫ t -1/2dt = 0

k ( Ts − Ti )

(πα )

1/2

× 2 t1/2 .

(3)

Substituting numerical values into Eq. (3) with k 0.062 W/m ⋅ K α= = = 3.18 ×10 −8 m 2 / s ρ c 2115 kg/m 3 × 920 J/kg ⋅ K find that for the 30 minute period, 0.062 W/m ⋅ K ( 20 − 50 ) K Q′′ = × 2 ( 30 × 60s )1/2 = −4.99 ×105 J/m 2 . 1/2 π × 3.18 ×10-8m2 / s

(

)

COMMENTS: Note that the sign for Q′′ is negative implying that energy is removed from the solid.

<

PROBLEM 5.72 KNOWN: Thermophysical properties and initial temperature of thick steel plate. Temperature of water jets used for convection cooling at one surface. FIND: Time required to cool prescribed interior location to a prescribed temperature. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in slab, (2) Validity of semi-infinite medium approximation, (3) Negligible thermal resistance between water jets and slab surface (Ts = T∞), (4) Constant properties. ANALYSIS: The desired cooling time may be obtained from Eq. (5.57). With T(0.025m, t) = 50°C,

T ( x, t ) − Ts Ti − Ts

=

(50 − 25) °C = 0.0909 = erf  x    (300 − 25) °C  2 αt 

x = 0.0807 2 αt t=

x2

(0.0807 ) 4α 2

=

(0.025m )2

(

0.0261 1.34 × 10−5 m 2 / s 3

)

<

= 1793s

-5

2

where α = k/ρc = 50 W/m⋅K/(7800 kg/m × 480 J/kg⋅K) = 1.34 × 10 m /s. 4

2

COMMENTS: (1) Large values of the convection coefficient (h ~ 10 W/m ⋅K) are associated with water jet impingement, and it is reasonable to assume that the surface is immediately quenched to the temperature of the water. (2) The surface heat flux may be determined from Eq. (5.58). In principle, 1/2 the flux is infinite at t = 0 and decays as t .

PROBLEM 5.73 KNOWN: Temperature imposed at the surface of soil initially at 20°C. See Example 5.5. FIND: (a) Calculate and plot the temperature history at the burial depth of 0.68 m for selected soil 7 2 thermal diffusivity values, α × 10 = 1.0, 1.38, and 3.0 m /s, (b) Plot the temperature distribution over -7 2 the depth 0 ≤ x ≤ 1.0 m for times of 1, 5, 10, 30, and 60 days with α = 1.38 × 10 m /s, (c) Plot the surface heat flux, q′′x ( 0, t ) , and the heat flux at the depth of the buried main, q′′x ( 0.68m, t ) , as a -7

2

function of time for a 60 day period with α = 1.38 × 10 m /s. Compare your results with those in the Comments section of the example. Use the IHT Models | Transient Conduction | Semi-infinite Medium model as the solution tool. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Soil is a semi-infinite medium, and (3) Constant properties. ANALYSIS: The IHT model corresponds to the case 1, constant surface temperature sudden boundary condition, Eqs. 5.57 and 5.58. Selected portions of the IHT code used to obtain the graphical results below are shown in the Comments. (a) The temperature history T(x,t) for x = 0.68 m with selected soil thermal diffusivities is shown below. The results are directly comparable to the graph shown in the Ex. 5.5 comments. x = 0.68 m, T(0,t) = Ts = -15C, T(x,0) = 20C

T(0.68 m, t) (C)

20

10

0

-10 0

15

30

45

60

Time, t (days) alpha = 1.00e-7 m^2/s alpha = 1.38e-7 m^2/s alpha = 3.00e-7 m^2/s

Continued …..

PROBLEM 5.73 (Cont.) (b) The temperature distribution T(x,t) for selected times is shown below. The results are directly comparable to the graph shown in the Ex. 5.5 comments. alpha = 1.38e-7 m^2/s , T(0,t) = -15C, T(x,0) = 20 C 20 15

T(x,t) (C)

10 5 0 -5 -10 -15 0

0.25

0.5

0.75

1

Depth, x (m) 1 day 5 days 10 days 30 days 60 days

(c) The heat flux from the soil, q′′x ( 0, t ) , and the heat flux at the depth of the buried main,

q′′x ( 0.68m, t ) , are calculated and plotted for the time period 0 ≤ t ≤ 60 days.

H e a t flu x , q ''(x ,t) (W /m ^ 2 )

0

a lp h a = 1 .3 8 e -7 m ^2 /s , k = 0 .5 2 W /m -K , T(0 ,t) = -1 5 C

-5 0

-1 0 0

-1 5 0

-2 0 0 0

15

30

45

60

Tim e , t (d a ys ) S u rfa c e h e a t flu x, x = 0 B u rie d -m a in d e p th , x = 0 .6 8 m

Both the surface and buried-main heat fluxes have a negative sign since heat is flowing in the negative x-direction. The surface heat flux is initially very large and, in the limit, approaches that of the buriedmain heat flux. The latter is initially zero, and since the effect of the sudden change in surface temperature is delayed for a time period, the heat flux begins to slowly increase. Continued …..

PROBLEM 5.73 (Cont.) COMMENTS: (1) Can you explain why the surface and buried-main heat fluxes are nearly the same at t = 60 days? Are these results consistent with the temperature distributions? What happens to the heat flux values for times much greater than 60 days? Use your IHT model to confirm your explanation. (2) Selected portions of the IHT code for the semi-infinite medium model are shown below.

// Models | Transient Conduction | Semi-infinite Solid | Constant temperature Ts /* Model: Semi-infinite solid, initially with a uniform temperature T(x,0) = Ti, suddenly subjected to prescribed surface boundary conditions. */ // The temperature distribution (Tx,t) is T_xt = T_xt_semi_CST(x,alpha,t,Ts,Ti) // Eq 5.55 // The heat flux in the x direction is q''_xt = qdprime_xt_semi_CST(x,alpha,t,Ts,Ti,k) //Eq 5.56 // Input parameters /* The independent variables for this system and their assigned numerical values are */ Ti = 20 // initial temperature, C k = 0.52 // thermal conductivity, W/m.K; base case condition alpha = 1.38e-7 // thermal diffusivity, m^2/s; base case //alpha = 1.0e-7 //alpha = 3.0e-7 // Calculating at x-location and time t, x=0 // m, surface // x = 0.68 // m, burial depth t = t_day * 24 * 3600 // seconds to days time covnersion //t_day = 60 //t_day = 1 //t_day = 5 //t_day = 10 //t_day = 30 t_day = 20 // Surface condition: constant surface temperature Ts = -15 // surface temperature, K

PROBLEM 5.74 KNOWN: Tile-iron, 254 mm to a side, at 150°C is suddenly brought into contact with tile over a subflooring material initially at Ti = 25°C with prescribed thermophysical properties. Tile adhesive softens in 2 minutes at 50°C, but deteriorates above 120°C. FIND: (a) Time required to lift a tile after being heated by the tile-iron and whether adhesive temperature exceeds 120°C, (2) How much energy has been removed from the tile-iron during the time it has taken to lift the tile. SCHEMATIC:

ASSUMPTIONS: (1) Tile and subflooring have same thermophysical properties, (2) Thickness of adhesive is negligible compared to that of tile, (3) Tile-subflooring behaves as semi-infinite solid experiencing one-dimensional transient conduction. PROPERTIES: Tile-subflooring (given): k = 0.15 W/m⋅K, ρcp = 1.5 × 106 J/m3⋅K, α = k/ρcp = 1.00 × 10-7 m2/s. ANALYSIS: (a) The tile-subflooring can be approximated as a semi-infinite solid, initially at a uniform temperature Ti = 25°C, experiencing a sudden change in surface temperature Ts = T(0,t) = 150°C. This corresponds to Case 1, Figure 5.7. The time required to heat the adhesive (xo = 4 mm) to 50°C follows from Eq. 5.57

T ( x o , t o ) − Ts Ti − Ts

  xo   = erf  2 (α t )1/ 2  o  

    50 − 150 0.004 m = erf   25 − 150  2 1.00 × 10−7 m 2 s × t 1/ 2    o  

(

(

0.80 = erf 6.325t o−1/ 2

)

)

to = 48.7s = 0.81 min using error function values from Table B.2. Since the softening time, ∆ts, for the adhesive is 2 minutes, the time to lift the tile is

t " = t o + ∆t s = ( 0.81 + 2.0 ) min = 2.81min .

<

To determine whether the adhesive temperature has exceeded 120°C, calculate its temperature at t " = 2.81 min; that is, find T(xo, t " )

    T ( x o , t " ) − 150 0.004 m = erf   25 − 150  2 1.0 × 10−7 m 2 s × 2.81× 60s 1/ 2     

(

)

Continued...

PROBLEM 5.74 (Cont.) T ( x o , t " ) − 150 = −125erf ( 0.4880 ) = 125 × 0.5098 T ( x o , t " ) = 86$ C

<

Since T(xo, t " ) < 120°C, the adhesive will not deteriorate. (b) The energy required to heat a tile to the lift-off condition is t" q′′ ( 0, t ) ⋅ As dt . 0 x

Q=∫

Using Eq. 5.58 for the surface heat flux q ′′s (t) = q ′′x (0,t), find

2k ( Ts − Ti ) t " k ( Ts − Ti ) dt 2 As As t1/ = " 1/ 2 1/ 2 1/ 2 0 t

Q=∫

Q=

(πα )

(πα ) $

2 × 0.15 W m ⋅ K (150 − 25 ) C

(

π ×1.00 ×10−7 m 2 s

)

1/ 2

× ( 0.254 m ) × ( 2.81× 60s ) 2

1/ 2

= 56 kJ

<

COMMENTS: (1) Increasing the tile-iron temperature would decrease the time required to soften the adhesive, but the risk of burning the adhesive increases. (2) From the energy calculation of part (b) we can estimate the size of an electrical heater, if operating continuously during the 2.81 min period, to maintain the tile-iron at a near constant temperature. The power required is

P = Q t " = 56 kJ 2.81× 60s = 330 W . Of course a much larger electrical heater would be required to initially heat the tile-iron up to the operating temperature in a reasonable period of time.

PROBLEM 5.75 KNOWN: Heat flux gage of prescribed thickness and thermophysical properties (ρ, cp, k) initially at a uniform temperature, Ti, is exposed to a sudden change in surface temperature T(0,t) = Ts. FIND: Relationships for time constant of gage when (a) backside of gage is insulated and (b) gage is imbedded in semi-infinite solid having the same thermophysical properties. Compare

(

)

with equation given by manufacturer, τ = 4d 2 ρ cp / π 2k. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties. ANALYSIS: The time constant τ is defined as the time required for the gage to indicate, following a sudden step change, a signal which is 63.2% that of the steady-state value. The manufacturer’s relationship for the time constant

)

(

τ = 4d 2 ρ cp / π 2k can be written in terms of the Fourier number as k 4 ατ τ Fo = = ⋅ = = 0.4053. d 2 ρ cp d 2 π 2 The Fourier number can be determined for the two different installations. (a) For the gage having its backside insulated, the surface and backside temperatures are Ts and T(0,t), respectively. From the sketch it follows that

θ o∗ =

T (0,τ ) − Ts Ti − Ts

From Eq. 5.41,

= 0.368.

(

)

θ o∗ = 0.368 = C1exp −ζ12Fo Using Table 5.1 with Bi = 100 (as the best approximation for Bi = hd/k → ∞, corresponding to sudden surface temperature change with h → ∞), ζ1 = 1.5552 rad and C1 = 1.2731. Hence, 0.368 = 1.2731exp(−1.55522 × Foa )

<

Foa = 0.513. Continued …..

PROBLEM 5.75 (Cont.) (b) For the gage imbedded in a semi-infinite medium having the same thermophysical properties, Table 5.7 (case 1) and Eq. 5.57 yield T ( x,τ ) − Ts 1/ 2 = 0.368 = erf d/2 (ατ )     Ti − Ts 1/ 2 d/2 (ατ ) = 0.3972 Fob =

ατ d2

=

1

( 2 × 0.3972 )2

= 1.585

<

COMMENTS: Both models predict higher values of Fo than that suggested by the manufacturer. It is understandable why Fob > Foa since for (b) the gage is thermally connected to an infinite medium, while for (a) it is isolated. From this analysis we conclude that the gage’s transient response will depend upon the manner in which it is applied to the surface or object.

PROBLEM 5.76 KNOWN: Procedure for measuring convection heat transfer coefficient, which involves melting of a surface coating. FIND: Melting point of coating for prescribed conditions. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in solid rod (negligible losses to insulation), (2) Rod approximated as semi-infinite medium, (3) Negligible surface radiation, (4) Constant properties, (5) Negligible thermal resistance of coating. -4

2

PROPERTIES: Copper rod (Given): k = 400 W/m⋅K, α = 10 m /s. ANALYSIS: Problem corresponds to transient conduction in a semi-infinite solid. Themal response is given by T ( x,t ) − Ti T∞ − Ti

1/ 2     h (α t ) t   x   .   erfc +    2 (α t )1/ 2  k    

    hx h 2α x   = erfc − exp  +  k  2 (α t )1/ 2   k2    

For x = 0, erfc(0) = 1 and T(x,t) = T(0,t) = Ts. Hence  h 2α Ts − Ti = 1 − exp   k2 T∞ − Ti 

 h (α t )1/ 2  t   erfc     k   

with h (α t m ) k

1/ 2

=

(

200 W/m 2 ⋅ K 10-4 m 2 / s × 400 s 400 W/m ⋅ K

)

1/ 2

= 0.1

Ts = Tm = Ti + ( T∞ − Ti ) 1 − exp ( 0.01) erfc ( 0.1) Ts = 25$ C + 275$ C [1-1.01× 0.888] = 53.5$ C. COMMENTS: Use of the procedure to evaluate h from measurement of tm necessitates iterative calculations.

<

PROBLEM 5.77 KNOWN: Irreversible thermal injury (cell damage) occurs in living tissue maintained at T ≥ 48°C for a duration ∆t ≥ 10s. FIND: (a) Extent of damage for 10 seconds of contact with machinery in the temperature range 50 to 100°C, (b) Temperature histories at selected locations in tissue (x = 0.5, 1, 5 mm) for a machinery temperature of 100°C. SCHEMATIC:

ASSUMPTIONS: (1) Portion of worker’s body modeled as semi-infinite medium, initially at a uniform temperature, 37°C, (2) Tissue properties are constant and equivalent to those of water at 37°C, (3) Negligible contact resistance. PROPERTIES: Table A-6, Water, liquid (T = 37°C = 310 K): ρ = 1/vf = 993.1 kg/m3, c = 4178 J/kg⋅K, k = 0.628 W/m⋅K, α = k/ρc = 1.513 × 10-7 m2/s. ANALYSIS: (a) For a given surface temperature -- suddenly applied -- the analysis is directed toward finding the skin depth xb for which the tissue will be at Tb ≥ 48°C for more than 10s? From Eq. 5.57,

T ( x b , t ) − Ts Ti − Ts

= erf  x b 2 (α t ) 

1/ 2 



= erf [ w ] .

For the two values of Ts, the left-hand side of the equation is $

Ts = 100 C :

( 48 − 100 )$ C 0.825 = (37 − 100 )$ C

$

Ts = 50 C :

( 48 − 50 )$ C 0.154 = (37 − 50 )$ C

The burn depth is

x b = [ w ] 2 (α t )

1/ 2

(

= [ w ] 2 1.513 × 10−7 m 2 s × t

)

1/ 2

= 7.779 × 10−4 [ w ] t1/ 2 . Continued...

PROBLEM 5.77 (Cont.) Using Table B.2 to evaluate the error function and letting t = 10s, find xb as Ts = 100°C:

xb = 7.779 × 10-4 [0.96](10s)1/2 = 2.362 × 103 m = 2.36 mm

Ts = 50°C:

xb = 7.779 × 10-4 [0.137](10s)1/2 = 3.37 × 103 m = 0.34 mm

< <

Recognize that tissue at this depth, xb, has not been damaged, but will become so if Ts is maintained for the next 10s. We conclude that, for Ts = 50°C, only superficial damage will occur for a contact period of 20s. (b) Temperature histories at the prescribed locations are as follows. 97

Temperature, T(C)

87 77 67 57 47 37 0

15

30

Time, t(s) x = 0.5 mm x = 1.0 mm x = 2.0 mm

The critical temperature of 48°C is reached within approximately 1s at x = 0.5 mm and within 7s at x = 2 mm. COMMENTS: Note that the burn depth xb increases as t1/2.

PROBLEM 5.78 KNOWN: Thermocouple location in thick slab. Initial temperature. Thermocouple measurement two minutes after one surface is brought to temperature of boiling water. FIND: Thermal conductivity of slab material. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Slab is semi-infinite medium, (3) Constant properties. 3

PROPERTIES: Slab material (given): ρ = 2200 kg/m , c = 700 J/kg⋅K. ANALYSIS: For the semi-infinite medium from Eq. 5.57,   x   = erf Ti − Ts  2 (α t )1/2      65 −100 0.01m  = erf  30 −100  2 (α ×120s )1/2      0.01m  = 0.5. erf   2 (α × 120s )1/2    T ( x,t ) − Ts

From Appendix B, find for erf w = 0.5 that w = 0.477; hence, 0.01m 2 (α ×120s )

1/2

= 0.477

(α × 120)1/2 = 0.0105

α = 9.156 ×10−7 m2 /s. It follows that since α = k/ρc, k = αρ c k = 9.156 ×10-7 m 2 / s × 2200 kg/m 3 × 700 J/kg ⋅ K k = 1.41 W/m⋅K.

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PROBLEM 5.79 KNOWN: Initial temperature, density and specific heat of a material. Convection coefficient and temperature of air flow. Time for embedded thermocouple to reach a prescribed temperature. FIND: Thermal conductivity of material. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Sample behaves as a semi-infinite modium, (3) Constant properties. ANALYSIS: The thermal response of the sample is given by Case 3, Eq. 5.60,

T ( x, t ) − Ti T∞ − Ti

 hx h 2α t     x h α t   x       exp erfc = erfc  − + +      2 αt  k k    2 α t   k 2     

where, for x = 0.01m at t = 300 s, [T(x,t) – Ti]/(T∞ - Ti) = 0.533. The foregoing equation must be solved iteratively for k, with α = k/ρcp. The result is

k = 0.45 W / m ⋅ K -7

<

2

with α = 4.30 × 10 m /s. COMMENTS: The solution may be effected by inserting the Transient Conduction/Semi-infinite Solid/Surface Conduction Model of IHT into the work space and applying the IHT Solver. However, the ability to obtain a converged solution depends strongly on the initial guesses for k and α.

PROBLEM 5.80 KNOWN: Very thick plate, initially at a uniform temperature, Ti, is suddenly exposed to a surface convection cooling process (T∞,h). FIND: (a) Temperatures at the surface and 45 mm depth after 3 minutes, (b) Effect of thermal diffusivity and conductivity on temperature histories at x = 0, 0.045 m. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Plate approximates semi-infinite medium, (3) Constant properties, (4) Negligible radiation. ANALYSIS: (a) The temperature distribution for a semi-infinite solid with surface convection is given by Eq. 5.60. T ( x, t ) − Ti T∞ − Ti

1/ 2       hx h 2α t     h (α t ) x x  − exp  +  . = erfc  +  erfc  2    2 (α t )1/ 2    k  2 (α t )1/ 2  k k        

At the surface, x = 0, and for t = 3 min = 180s, T ( 0,180s ) − 325$ C

(15 − 325 )$ C

  1002 W 2 m 4 K 2 × 5.6 × 10−6 m 2 s × 180s    = erfc ( 0 ) − exp  0 + 2    ( 20 W m ⋅ K )  1/ 2     2 −6 2   100 W m ⋅ K 5.6 × 10 m s × 180s  × erfc  0 +  20 W m ⋅ K      

(

)

= 1 − [exp ( 0.02520 )] × [erfc ( 0.159 )] = 1 − 1.02552 × (1 − 0.178 ) T ( 0,180s ) = 325$ C − (15 − 325 ) C ⋅ (1 − 1.0255 × 0.822 ) $

T ( 0,180s ) = 325$ C − 49.3$ C = 276$ C . At the depth x = 0.045 m, with t = 180s, T ( 0.045m,180s ) − 325 C $

(15 − 325 )$ C

<

       100 W m 2 ⋅ K × 0.045 m 0.045 m = erfc  −  exp  + 0.02520      1/ 2 20 W m ⋅ K  2 5.6 × 10−6 m 2 s × 180s             0.045 m × erfc  + 0.159  1/ 2   2 5.6 × 10−6 m 2 s × 180s     

(

)

)

(

= erfc ( 0.7087 ) + [exp ( 0.225 + 0.0252 )]× [erfc ( 0.7087 + 0.159 )] . T ( 0.045m,180s ) = 325$ C + (15 − 325 ) C [(1 − 0.684 ) − 1.284 (1 − 0.780 )] = 315$ C $

< Continued...

PROBLEM 5.80 (Cont.)

325

325

300

275 Temperature, T(C)

Temperature, T(C)

(b) The IHT Transient Conduction Model for a Semi-Infinite Solid was used to generate temperature histories, and for the two locations the effects of varying α and k are as follows.

275 250 225

225

175

125

200

75

175 0

50

100

150

200

250

0

300

50

100

200

250

300

250

300

k = 2 W/m.K, alpha = 5.6E-6m^2/s, x = 0 k = 20 W/m.K, alpha = 5.6E-6m^2/s, x = 0 k = 200 W/m.K, alpha = 5.6E-6m^2/s, x = 0

k = 20 W/m.K, alpha = 5.6E-5 m^2/s, x = 0 k = 20 W/m.K, alpha = 5.6E-6m^2/s, x = 0 k = 20 W/m.K, alpha = 5.6E-7m^2/s, x = 0

325

325

300

305 Temperature, T(C)

Temperature, T(C)

150 Time, t(s)

Time, t(s)

275

250

225

285

265

245

200

225

0

50

100

150

200

250

Time, t(s) k = 20 W/m.K, alpha = 5.6E-5 m^2.K, x = 45 mm k = 20 W/m.K, alpha = 5.6E-6m^2.K, x = 45 mm k = 20 W/m.K, alpha = 5.6E-7m^2.K, x = 45mm

300

0

50

100

150

200

Time, t(s) k = 2 W/m.K, alpha = 5.6E-6m^2/s, x = 45 mm k = 20 W/m.K, alpha = 5.6E-6m^2/s, x = 45 mm k = 200 W/m.K, alpha = 5.6E-6m^2/s, x = 45 mm

m

For fixed k, increasing alpha corresponds to a reduction in the thermal capacitance per unit volume (ρcp) of the material and hence to a more pronounced reduction in temperature at both surface and interior locations. Similarly, for fixed α, decreasing k corresponds to a reduction in ρcp and hence to a more pronounced decay in temperature. COMMENTS: In part (a) recognize that Fig. 5.8 could also be used to determine the required temperatures.

PROBLEM 5.81 KNOWN: Thick oak wall, initially at a uniform temperature of 25°C, is suddenly exposed to combustion products at 800°C with a convection coefficient of 20 W/m2⋅K. FIND: (a) Time of exposure required for the surface to reach an ignition temperature of 400°C, (b) Temperature distribution at time t = 325s. SCHEMATIC:

ASSUMPTIONS: (1) Oak wall can be treated as semi-infinite solid, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible radiation. PROPERTIES: Table A-3, Oak, cross grain (300 K): ρ = 545 kg/m3, c = 2385 J/kg⋅K, k = 0.17 W/m⋅K, α = k/ρc = 0.17 W/m⋅K/545 kg/m3 × 2385 J/kg⋅K = 1.31 × 10-7 m2/s. ANALYSIS: (a) This situation corresponds to Case 3 of Figure 5.7. The temperature distribution is given by Eq. 5.60 or by Figure 5.8. Using the figure with

T ( 0, t ) − Ti T∞ − Ti

=

400 − 25 = 0.48 800 − 25

x

and

2 (α t )

1/ 2

=0

we obtain h(αt)1/2/k ≈ 0.75, in which case t ≈ (0.75k/hα1/2)2. Hence, 1/ 2 2  2 7 2 − t ≈  0.75 × 0.17 W m ⋅ K 20 W m ⋅ K 1.31× 10 m s  = 310s  

)

(

<

(b) Using the IHT Transient Conduction Model for a Semi-infinite Solid, the following temperature distribution was generated for t = 325s. 400

Temperature, T(C)

325

250

175

100

25 0

0.005

0.01

0.015

0.02

0.025

0.03

Distance from the surface, x(m)

The temperature decay would become more pronounced with decreasing α (decreasing k, increasing ρcp) and in this case the penetration depth of the heating process corresponds to x ≈ 0.025 m at 325s. COMMENTS: The result of part (a) indicates that, after approximately 5 minutes, the surface of the wall will ignite and combustion will ensue. Once combustion has started, the present model is no longer appropriate.

PROBLEM 5.82 KNOWN: Thickness, initial temperature and thermophysical properties of concrete firewall. Incident radiant flux and duration of radiant heating. Maximum allowable surface temperatures at the end of heating. FIND: If maximum allowable temperatures are exceeded. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in wall, (2) Validity of semi-infinite medium approximation, (3) Negligible convection and radiative exchange with the surroundings at the irradiated surface, (4) Negligible heat transfer from the back surface, (5) Constant properties. ANALYSIS: The thermal response of the wall is described by Eq. (5.60)

T ( x, t ) = Ti +

2 q′′o (α t / π )

1/ 2

k

 − x 2  q′′ x  x   − o erfc  exp    4α t  k  2 αt   

where, α = k / ρ c p = 6.92 × 10 −7 m 2 / s and for t = 30 min = 1800s, 2q′′o (α t / π )1/ 2 / k = 284.5 K. Hence, at x = 0,

T ( 0,30 min ) = 25°C + 284.5°C = 309.5°C < 325°C

(

<

)

At x = 0.25m, − x 2 / 4α t = −12.54, q ′′o x / k = 1, 786K, and x / 2 (α t )1/ 2 = 3.54. Hence,

(

)

T ( 0.25m, 30 min ) = 25°C + 284.5°C 3.58 × 10−6 − 1786°C × ( ~ 0 ) ≈ 25°C

<

Both requirements are met. COMMENTS: The foregoing analysis is conservative since heat transfer at the irradiated surface due to convection and net radiation exchange with the environment have been neglected. If the emissivity of the surface and the temperature of the surroundings are assumed to be ε = 1 and Tsur =

(

)

4 298K, radiation exchange at Ts = 309.5°C would be q′′rad = εσ Ts4 − Tsur = 6, 080 W / m 2 ⋅ K,

which is significant (~ 60% of the prescribed radiation).

PROBLEM 5.83 KNOWN: Initial temperature of copper and glass plates. Initial temperature and properties of finger. FIND: Whether copper or glass feels cooler to touch. SCHEMATIC:

ASSUMPTIONS: (1) The finger and the plate behave as semi-infinite solids, (2) Constant properties, (3) Negligible contact resistance. 3

PROPERTIES: Skin (given): ρ = 1000 kg/m , c = 4180 J/kg⋅K, k = 0.625 W/m⋅K; Table A-1 3 (T = 300K), Copper: ρ = 8933 kg/m , c = 385 J/kg⋅K, k = 401 W/m⋅K; Table A-3 (T = 300K), 3 Glass: ρ = 2500 kg/m , c = 750 J/kg⋅K, k = 1.4 W/m⋅K. ANALYSIS: Which material feels cooler depends upon the contact temperature Ts given by Equation 5.63. For the three materials of interest, = 0.625 ×1000 × 4180 )1 / 2 = 1,616 J/m 2 ⋅ K ⋅s1/2 ( kρ c )1/2 skin ( 1/2 = 37,137 J/m 2 ⋅ K ⋅ s1/2 ( kρ c )1/2 cu = ( 401 × 8933 × 385 ) = 1.4 × 2500 × 750 )1/2 = 1,620 J/m 2 ⋅ K ⋅ s1/2. ( kρ c )1/2 glass ( Since ( kρ c )1/2 >> ( kρ c )1/2 , the copper will feel much cooler to the touch. From Equation cu glass 5.63, 1/2 TA,i + ( kρ c ) B TB,i ( kρ c )1/2 A Ts = + kρ c )1/2 ( kρ c )1/2 A ( B

Ts( cu ) =

1,616 ( 310 ) + 37,137 ( 300 )

Ts( glass ) =

1,616 + 37,137

= 300.4 K

1,616 ( 310 ) + 1,620 ( 300 ) 1,616 + 1,620

= 305.0 K.

< <

COMMENTS: The extent to which a material’s temperature is affected by a change in its thermal 1/2 environment is inversely proportional to (kρc) . Large k implies an ability to spread the effect by conduction; large ρc implies a large capacity for thermal energy storage.

PROBLEM 5.84 KNOWN: Initial temperatures, properties, and thickness of two plates, each insulated on one surface. FIND: Temperature on insulated surface of one plate at a prescribed time after they are pressed together. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible contact resistance. 3

PROPERTIES: Stainless steel (given): ρ = 8000 kg/m , c = 500 J/kg⋅K, k = 15 W/m⋅K. ANALYSIS: At the instant that contact is made, the plates behave as semi-infinite slabs and, since the (ρkc) product is the same for the two plates, Equation 5.63 yields a surface temperature of Ts = 350K. The interface will remain at this temperature, even after thermal effects penetrate to the insulated surfaces. The transient response of the hot wall may therefore be calculated from Equations 5.40 and 5.41. At the insulated surface (x* = 0), Equation 5.41 yields

(

To − Ts = C1 exp −ζ12 Fo Ti − Ts

)

where, in principle, h → ∞ and T∞ → Ts . From Equation 5.39c, Bi → ∞ yields ζ 1 = 1.5707, and from Equation 5.39b C1 =

4sinζ 1 = 1.273 2ζ1 + sin ( 2ζ1 ) αt

Also,

Fo =

Hence,

To − 350 = 1.273exp −1.5707 2 × 0.563 = 0.318 400 − 350

L2

=

3.75 ×10 −6 m 2 / s ( 60s )

( 0.02 m )2

(

= 0.563.

)

To = 365.9 K. COMMENTS: Since Fo > 0.2, the one-term approximation is appropriate.

<

PROBLEM 5.85 KNOWN: Thickness and properties of liquid coating deposited on a metal substrate. Initial temperature and properties of substrate. FIND: (a) Expression for time required to completely solidify the liquid, (b) Time required to solidify an alumina coating. SCHEMATIC:

ASSUMPTIONS: (1) Substrate may be approximated as a semi-infinite medium in which there is onedimensional conduction, (2) Solid and liquid alumina layers remain at fusion temperature throughout solidification (negligible resistance to heat transfer by conduction through solid), (3) Negligible contact resistance at the coating/substrate interface, (4) Negligible solidification contraction, (5) Constant properties. ANALYSIS: (a) Performing an energy balance on the solid layer, whose thickness S increases with t, the latent heat released at the solid/liquid interface must be balanced by the rate of heat conduction into the solid. Hence, per unit surface area,

ρ h sf

dS = q′′cond dt

1/ 2 where, from Eq. 5.58, q′′cond = k ( Tf − Ti ) (πα t ) . It follows that

dS ks (Tf − Ti ) = dt (παs t )1/ 2 k s ( Tf − Ti ) t dt δ ∫o dS = ρ h πα 1/ 2 ∫o t1/ 2 sf ( s )

ρ h sf

δ=

 Tf − Ti  1/ 2  t (παs )1/ 2  ρ hsf  2k s

πα  δρ hsf  t= s  4k s2  Tf − Ti 

2

<

(b) For the prescribed conditions, π 4 10−5 m 2 s

t=

(×

)  0.002 m × 3970 kg m3 × 3.577 ×106 J kg 2 = 0.43s

2 4 (120 W m ⋅ K ) 

2018 K

 

<

COMMENTS: Such solidification processes occur over short time spans and are typically termed rapid solidification.

PROBLEM 5.86 KNOWN: Properties of mold wall and a solidifying metal. FIND: (a) Temperature distribution in mold wall at selected times, (b) Expression for variation of solid layer thickness. SCHEMATIC:

ASSUMPTIONS: (1) Mold wall may be approximated as a semi-infinite medium in which there is onedimensional conduction, (2) Solid and liquid metal layers remain at fusion temperature throughout solidification (negligible resistance to heat transfer by conduction through solid), (3) Negligible contact resistance at mold/metal interface, (4) Constant properties. ANALYSIS: (a) As shown in schematic (b), the temperature remains nearly uniform in the metal (at Tf) throughout the process, while both the temperature and temperature penetration increase with time in the mold wall. (b) Performing an energy balance for a control surface about the solid layer, the latent energy released due to solidification at the solid/liquid interface is balanced by heat conduction into the solid, q′′lat =

q′′cond , where q′′lat = ρ h sf dS dt and q′′cond is given by Eq. 5.58. Hence,

ρ h sf

S

dS k w ( Tf − Ti ) = dt (πα w t )1/ 2

∫o dS = ρ S=

k w (Tf − Ti )

t dt

1/ 2 ∫o t1/ 2

h sf (πα w )

2k w ( Tf − Ti ) 1/ 2 t 1/ 2 ρ h sf (πα w )

<

COMMENTS: The analysis of part (b) would only apply until the temperature field penetrates to the exterior surface of the mold wall, at which point, it may no longer be approximated as a semi-infinite medium.

PROBLEM 5.87 KNOWN: Steel (plain carbon) billet of square cross-section initially at a uniform temperature of 30°C is placed in a soaking oven and subjected to a convection heating process with prescribed temperature and convection coefficient. FIND: Time required for billet center temperature to reach 600°C. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction in x1 and x2 directions, (2) Constant properties, (3) Heat transfer to billet is by convection only. PROPERTIES: Table A-1, Steel, plain carbon (T = (30+600)°C/2 = 588K = ≈ 600K): ρ = 3 -5 2 7854 kg/m , cp = 559 J/kg⋅K, k = 48.0 W/m⋅K, α =k/ρcp = 1.093 × 10 m /s. ANALYSIS: The billet corresponds to Case (e), Figure 5.11 (infinite rectangular bar). Hence, the temperature distribution is of the form

θ ∗ ( x1, x 2 , t ) = P ( x1, t ) × P ( x 2 , t )

where P(x,t) denotes the distribution corresponding to the plane wall. Because of symmetry in the x1 and x2 directions, the P functions are identical. Hence,

θ (0, 0, t ) θ o (0, t )  =  θi  θi 

2 Plane wall

θ = T − T∞  where θi = Ti − T∞ θ o = T (0,t ) − T∞

and L = 0.15m.

Substituting numerical values, find

θ o (0, t )  T (0,0,t ) − T∞  =  θi  Ti − T∞ 

1/ 2

1/ 2

 ( 600 − 750 )$ C   =  (30 − 750 )$ C   

= 0.46.

Consider now the Heisler chart for the plane wall, Figure D.1. For the values

θ θ o∗ = o ≈ 0.46 θi find t∗ = Fo =

αt L2

Bi-1 =

k 48.0 W/m ⋅ K = = 3.2 hL 100 W/m 2 ⋅ K × 0.15m

≈ 3.2.

Hence, 3.2 (0.15m ) 3.2 L2 = = 6587s = 1.83h. α 1.093 × 10−5 m 2 / s 2

t=

<

PROBLEM 5.88 KNOWN: Initial temperature of fire clay brick which is cooled by convection. FIND: Center and corner temperatures after 50 minutes of cooling. SCHEMATIC:

ASSUMPTIONS: (1) Homogeneous medium with constant properties, (2) Negligible radiation effects. 3

PROPERTIES: Table A-3, Fire clay brick (900K): ρ = 2050 kg/m , k = 1.0 W/m⋅K, cp = -6 2

960 J/kg⋅K. α = 0.51 × 10 m /s. ANALYSIS: From Fig. 5.11(h), the center temperature is given by T ( 0,0,0,t ) − T∞ Ti − T∞

= P1 (0, t ) × P2 ( 0, t ) × P3 ( 0, t )

where P1, P2 and P3 must be obtained from Fig. D.1. L1 = 0.03m:

Bi1 =

h L1 = 1.50 k

Fo1 =

L2 = 0.045m:

Bi 2 =

h L2 = 2.25 k

Fo 2 =

L3 = 0.10m:

Bi3 =

h L3 = 5.0 k

Fo3 =

αt L21

αt L22

αt L23

= 1.70 = 0.756 = 0.153

Hence from Fig. D.1, P1 (0, t ) ≈ 0.22 Hence,

P2 ( 0, t ) ≈ 0.50

T ( 0,0,0,t ) − T∞ Ti − T∞

P3 (0, t ) ≈ 0.85.

≈ 0.22 × 0.50 × 0.85 = 0.094

and the center temperature is T ( 0,0,0,t ) ≈ 0.094 (1600 − 313) K + 313K = 434K.

< Continued …..

PROBLEM 5.88 (Cont.) The corner temperature is given by T ( L1, L 2 , L3 , t ) − T∞ Ti − T∞

= P ( L1, t ) × P ( L2 , t ) × P ( L3 , t )

where P ( L1, t ) =

θ ( L1, t ) ⋅ P1 (0, t ) , etc. θo

and similar forms can be written for L2 and L3. From Fig. D.2,

θ ( L1, t ) ≈ 0.55 θo

θ ( L2 , t ) ≈ 0.43 θo

θ ( L3 , t ) ≈ 0.25. θo

Hence, P ( L1, t ) ≈ 0.55 × 0.22 = 0.12 P ( L 2 , t ) ≈ 0.43 × 0.50 = 0.22 P ( L3 , t ) ≈ 0.85 × 0.25 = 0.21 and T ( L1, L 2 , L3 , t ) − T∞ Ti − T∞

≈ 0.12 × 0.22 × 0.21 = 0.0056

or T ( L1, L2 , L3 , t ) ≈ 0.0056 (1600 − 313) K + 313K. The corner temperature is then T ( L1, L2 , L3 , t ) ≈ 320K.

<

COMMENTS: (1) The foregoing temperatures are overpredicted by ignoring radiation, which is significant during the early portion of the transient. (2) Note that, if the time required to reach a certain temperature were to be determined, an iterative approach would have to be used. The foregoing procedure would be used to compute the temperature for an assumed value of the time, and the calculation would be repeated until the specified temperature were obtained.

PROBLEM 5.89 KNOWN: Cylindrical copper pin, 100mm long × 50mm diameter, initially at 20°C; end faces are subjected to intense heating, suddenly raising them to 500°C; at the same time, the cylindrical surface is subjected to a convective heating process (T∞,h). FIND: (a) Temperature at center point of cylinder after a time of 8 seconds from sudden application of heat, (b) Consider parameters governing transient diffusion and justify simplifying assumptions that could be applied to this problem. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction, (2) Constant properties and convection heat transfer coefficient. 3 $ PROPERTIES: Table A-1, Copper, pure T ≈ (500 + 20 ) C/2 ≈ 500K : ρ = 8933 kg/m , c = 407

)

(

3 -4 2 J/kg⋅K, k = 386 W/m⋅K, α = k/ρc = 386 W/m⋅K/8933 kg/m × 407 J/kg⋅K = 1.064 × 10 m /s.

ANALYSIS: (1) The pin can be treated as a two-dimensional system comprised of an infinite cylinder whose surface is exposed to a convection process (T∞,h) and of a plane wall whose surfaces are maintained at a constant temperature (Te). This configuration corresponds to the short cylinder, Case (i) of Fig. 5.11,

θ ( r,x,t ) = C ( r,t ) × P ( x,t ) . θi

(1)

For the infinite cylinder, using Fig. D.4, with Bi =

hro k

=

(

100 W/m 2 ⋅ K 25 × 10-3m 385 W/m ⋅ K

C ( 0,8s ) =

find

) = 6.47 ×10−3

Fo =

and

αt ro2

1.064 × 10−

4m

=

2

s

× 8s

(25 ×10 m ) -3

θ ( 0,8s ) 

2

= 1.36,

 ≈ 1.  cyl

θi

(2)

For the infinite plane wall, using Fig. D.1, with Bi =

find

hL k

→∞

or

P ( 0,8s ) =

Bi

-1

→0

and

Fo =

αt L2

=

1.064 × 10−4 m 2 / s × 8s

(50 ×10 m ) -3

θ ( 0,8s )  θi

≈ 0.5.   wall

Combining Eqs. (2) and (3) with Eq. (1), find

2

= 0.34,

(3) θ ( 0, 0,8s ) θi

=

T ( 0,0,8s ) − T∞ Ti − T∞

≈ 1 × 0.5 = 0.5

T ( 0,0,8s ) = T∞ + 0.5 ( Ti − T∞ ) = 500 + 0.5 ( 20 − 500 ) = 260$ C. (b) The parameters controlling transient conduction with convective boundary conditions are the Biot and Fourier numbers. Since Bi 0.1, rod does not behave as spacewise isothermal object. Hence, treat rod as a semi-infinite cylinder, the multi-dimensional system Case (f), Fig. 5.11.

The product solution can be written as

θ ∗ ( r,x,t ) =

θ ( r,x,t ) θ ( r,t ) θ ( x,t ) = × = C r∗ , t ∗ × S x ∗ , t∗ θi θi θi

) (

(

)

Infinite cylinder, C(r*,t*). Using the Heisler charts with r* = r = 0 and  h ro  Bi-1 =

 k   

−1

 500 W/m 2 ⋅ K × 0.01m  =  12.4 W/m ⋅ K  

−1

= 2.48.

Evaluate α = k/ρc = 2.71 × 10 m /s, find Fo = α t/ro2 = 2.71× 10−6 m 2 / s × 30s/(0.01m) = -6 2

2

-1

0.812. From the Heisler chart, Fig. D.4, with Bi = 2.48 and Fo = 0.812, read C(0,t*) = θ(0,t)/θi = 0.61. Continued …..

PROBLEM 5.91 (Cont.) Semi-infinite medium, S(x*,t*). Recognize this as Case (3), Fig. 5.7. From Eq. 5.60, note that the LHS needs to be transformed as follows, T − Ti T − T∞ T − T∞ = 1− S ( x,t ) = . T∞ − Ti Ti − T∞ Ti − T∞ Thus,  1/ 2      2    h (α t )    x x   − exp  hx + h α t   erfc   . S ( x,t ) = 1 − erfc  + 1/ 2   2 1/ 2  k k       k     2 (α t )    2 (α t )   Evaluating this expression at the surface (x = 0) and 6mm from the exposed end, find    2 ⋅ K 2 2.71× 10−6 m 2 / s × 30s   500 W/m     S (0,30s ) = 1 − erfc (0 ) − exp 0 +      (12.4 W/m ⋅ K )2       

)

(

)

   2 ⋅ K 2.71× 10-6 m 2 / s × 30s 1/ 2   500 W/m      erfc 0 +  12.4 W/m ⋅ K        

(

{

}

S (0,30s ) = 1 − 1 − exp (0.1322 ) erfc ( 0.3636 ) = 0.693. Note that Table B.2 was used to evaluate the complementary error function, erfc(w).       0.006m S (6mm,30s ) = 1 − erfc  −  2 2.71×10-6m 2 / s × 30s 1/ 2     

)

(

   500 W/m 2 ⋅ K × 0.006m   exp     + 0.1322 erfc ( 0.3327 + 0.3636 )  = 0.835. 12.4 W/m ⋅ K      The product solution can now be evaluated for each location. At (0,0), T ( 0,0,30s ) − T∞ θ ∗ (0, 0, t ) = = C 0,t ∗ × S 0,t ∗ = 0.61× 0.693 = 0.423. Ti − T∞

( ) ( )

T ( 0,0,30s ) = T∞ + 0.423 (Ti − T∞ ) = 350K + 0.423 (850 − 350 ) K = 561K.

Hence, At (0,6mm),

( ) (

<

)

θ ∗ (0, 6mm,t ) = C 0,t∗ × S 6mm,t ∗ = 0.61× 0.835 = 0.509 T ( 0,6mm,30s ) = 604K.

<

COMMENTS: Note that the temperature at which the properties were evaluated was a good estimate.

PROBLEM 5.92 KNOWN: Stainless steel cylinder of Ex. 5.7, 80-mm diameter by 60-mm length, initially at 600 K, 2 suddenly quenched in an oil bath at 300 K with h = 500 W/m ⋅K. Use the Transient Conduction, Plane Wall and Cylinder models of IHT to obtain the following solutions. FIND: (a) Calculate the temperatures T(r,x,t) after 3 min: at the cylinder center, T(0, 0, 3 min), at the center of a circular face, T(0, L, 3 min), and at the midheight of the side, T(ro, 0, 3 min); compare your results with those in the example; (b) Calculate and plot temperature histories at the cylinder center, T(0, 0, t), the mid-height of the side, T(ro, 0, t), for 0 ≤ t ≤ 10 min; comment on the gradients and what effect they might have on phase transformations and thermal stresses; and (c) For 0 ≤ t ≤ 10 min, calculate and plot the temperature histories at the cylinder center, T(0, 0, t), for convection coefficients 2 of 500 and 1000 W/m ⋅K. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction in r- and x-coordinates, (2) Constant properties. 3

PROPERTIES: Stainless steel (Example 5.7): ρ = 7900 kg/m , c = 526 J/kg⋅K, k = 17.4 W/m⋅K. ANALYSIS: The following results were obtained using the Transient Conduction models for the Plane Wall and Cylinder of IHT. Salient portions of the code are provided in the Comments. (a) Following the methodology for a product solution outlined in Example 5.7, the following results were obtained at t = to = 3 min (r, x, t)

P(x, t)

C(r, t)

0, 0, to 0, L, to ro, 0, to

0.6357 0.4365 0.6357

0.5388 0.5388 0.3273

T(r, x, t)-IHT (K) 402.7 370.5 362.4

T(r, x, t)-Ex (K) 405 372 365 Continued …..

PROBLEM 5.92 (Cont.) The temperatures from the one-term series calculations of the Example 5.7 are systematically higher than those resulting from the IHT multiple-term series model, which is the more accurate method. (b) The temperature histories for the center and mid-height of the side locations are shown in the graph below. Note that at early times, the temperature difference between these locations, and hence the gradient, is large. Large differences could cause variations in microstructure and hence, mechanical properties, as well as induce residual thermal stresses. (c) Effect of doubling the convection coefficient is to increase the quenching rate, but much less than by a factor of two as can be seen in the graph below. Effect of increased conv. coeff. on quenching rate

600

600

500

500

T(0, 0, t) (C)

T(x, r, t) (C)

Quenching with h = 500 W/m^2.K

400

400

300

300

0

0

2

4

6

8

10

2

4

6

8

10

Time, t (min)

Time, t (min)

h = 500 W/m^2.K h = 1000 W/m^2.K

Mid-height of side (0,ro) Center (0, 0)

COMMENTS: From IHT menu for Transient Conduction | Plane Wall and Cylinder, the models were combined to solve the product solution. Key portions of the code, less the input variables, are copied below. // Plane wall temperature distribution // The temperature distribution is T_xtP = T_xt_trans("Plane Wall",xstar,FoP,BiP,Ti,Tinf) // The dimensionless parameters are xstar = x / L BiP = h * L / k // Eq 5.9 FoP= alpha * t / L^2 // Eq 5.33 alpha = k/ (rho * cp) // Dimensionless representation, P(x,t) P_xt = (T_xtP - Tinf ) / (Ti - Tinf) // Cylinder temperature distribution // The temperature distribution T(r,t) is T_rtC = T_xt_trans("Cylinder",rstar,FoC,BiC,Ti,Tinf) // The dimensionless parameters are rstar = r / ro BiC = h * ro / k FoC= alpha * t / ro^2 // Dimensionless representation, C(r,t) C_rt= (T_rtC - Tinf ) / (Ti - Tinf) // Product solution temperature distribution (T_xrt - Tinf) / (Ti - Tinf) = P_xt * C_rt

// Eq 5.39

// Eq 5.47

PROBLEM 5.93 p KNOWN: Stability criterion for the explicit method requires that the coefficient of the Tm term of the one-dimensional, finite-difference equation be zero or positive. p+1 FIND: For Fo > 1/2, the finite-difference equation will predict values of Tm which violate the Second law of thermodynamics. Consider the prescribed numerical values.

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Constant properties, (3) No internal heat generation. ANALYSIS: The explicit form of the finite-difference equation, Eq. 5.73, for an interior node is

(

)

p+1 p p p Tm . = Fo Tm+1 + Tm-1 + (1 − 2 Fo ) Tm p The stability criterion requires that the coefficient of Tm be zero or greater. That is, 1 or Fo ≤ . (1 − 2 Fo ) ≥ 0 2 For the prescribed temperatures, consider situations for which Fo = 1, ½ and ¼ and calculate p+1 Tm .

Fo = 1 Fo = 1/2 Fo = 1/4

p+1 Tm = 1(100 + 100 ) C + (1 − 2 ×1) 50$ C = 250$C $

p+1 Tm = 1/ 2 (100 + 100 ) C + (1 − 2 ×1/ 2 ) 50$ C = 100$C $

p+1 Tm = 1/ 4 (100 + 100 ) C + (1 − 2 ×1/ 4 ) 50$ C = 75$C. $

p+1 Plotting these distributions above, note that when Fo = 1, Tm is greater than 100°C, while p+1 for Fo = ½ and ¼ , Tm ≤ 100°C. The distribution for Fo = 1 is thermodynamically impossible: heat is flowing into the node during the time period ∆t, causing its temperature to rise; yet heat is flowing in the direction of increasing temperature. This is a violation of the Second law. When Fo = ½ or ¼, the node temperature increases during ∆t, but the temperature gradients for heat flow are proper. This will be the case when Fo ≤ ½, verifying the stability criterion.

PROBLEM 5.94 KNOWN: Thin rod of diameter D, initially in equilibrium with its surroundings, Tsur, suddenly passes a current I; rod is in vacuum enclosure and has prescribed electrical resistivity, ρe, and other thermophysical properties. FIND: Transient, finite-difference equation for node m. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Surroundings are much larger than rod, (3) Properties are constant and evaluated at an average temperature, (4) No convection within vacuum enclosure. ANALYSIS: The finite-difference equation is derived from the energy conservation requirement on the control volume, Ac∆x, where Ac = π D 2 / 4 The energy balance has the form

P = π D.

and

p T p+1 − Tm q a + q b − q rad + I2 R e = ρ cV m . ∆t

E in − E out + E g = E st

where E g = I 2R e and R e = ρe ∆x/A c . Using Fourier’s law to express the conduction terms, qa and qb, and Eq. 1.7 for the radiation exchange term, qrad, find

)

(

p p p+1 p T p − Tm T p − Tm 4,p 4 + I 2 ρ e ∆x = ρ cA ∆x Tm − Tm . + kA c m+1 − ε P∆xσ Tm − Tsur kAc m-1 c ∆x ∆x ∆t Ac p+1 Divide each term by ρcAc ∆x/∆t, solve for Tm and regroup to obtain p+1 = Tm

) 2 ε Pσ ∆t 4,p 4 + I ρ e ⋅ ∆t . − ⋅ T − T m sur ) Ac ρ c ( Ac2 ρ c (

  p ∆t ∆t k k p p ⋅ + Tm+1 − 2 ⋅ ⋅ − 1 Tm Tm-1 ρ c ∆x 2  ρ c ∆x 2 

2

Recognizing that Fo = α ∆t/∆x , regroup to obtain

(

)

p+1 p p p Tm = Fo Tm-1 + Tm+1 + (1 − 2 Fo ) Tm −

(

)

2 2 ε Pσ∆x 2 4,p 4 + I ρe ∆x ⋅ Fo. ⋅ Fo Tm − Tsur kAc kAc2

p The stability criterion is based upon the coefficient of the Tm term written as

Fo ≤ ½.

<

COMMENTS: Note that we have used the forward-difference representation for the time derivative; see Section 5.9.1. This permits convenient treatment of the non-linear radiation exchange term.

PROBLEM 5.95 KNOWN: One-dimensional wall suddenly subjected to uniform volumetric heating and convective surface conditions. FIND: Finite-difference equation for node at the surface, x = -L. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3) Uniform q . ANALYSIS: There are two types of finite-difference equations for the explicit and implicit methods of solution. Using the energy balance approach, both types will be derived. Explicit Method. Perform an energy balance on the surface node shown above, T p+1 − Top  = ρ cV o q conv + q cond + qV ∆t

E in − E out + E g = E st

(1)

)

(

p+1 p T1p − Top ∆x  ∆x  To − To   p h (1 ⋅1) T∞ − To + k (1 ⋅1) . + q 1 ⋅1 ⋅  = ρ c 1 ⋅1 ⋅  2 2 ∆x ∆t

(2)     For the explicit method, the temperatures on the LHS are evaluated at the previous time (p). The RHS provides a forward-difference approximation to the time derivative. Divide Eq. (2) by ρc∆x/2∆t and solve for Top+1. h∆t k∆t ∆t Top+1 = 2 T∞ − Top + 2 T1p − Top + q + Top . ρ c∆x ρc ρ c∆x 2 Introducing the Fourier and Biot numbers,

(

Fo ≡ ( k/ρ c ) ∆t/∆x 2

)

(

)

Bi ≡ h∆x/k

 p q ∆x 2  p+1 p To = 2 Fo  T1 + Bi ⋅ T∞ +  + (1 − 2 Fo − 2 Fo ⋅ Bi ) To . 2k 

(3)



The stability criterion requires that the coefficient of Top be positive. That is,

(1 − 2 Fo − 2 Fo ⋅ Bi ) ≥ 0

or

Fo ≤ 1/2 (1 + Bi ) .

(4) <

Implicit Method. Begin as above with an energy balance. In Eq. (2), however, the temperatures on the LHS are evaluated at the new (p+1) time. The RHS provides a backwarddifference approximation to the time derivative.

(

)

p+1 p T1p+1 − Top+1 − To  ∆x   ∆x  T p+1 h T∞ − To +k + q   = ρ c   o ∆x ∆t  2   2 



(1 + 2 Fo (Bi + 1)) Top+1 − 2 Fo ⋅ T1p+1 = Top + 2Bi ⋅ Fo ⋅ T∞ + Fo q∆kx

(5) 2

.

(6) <

COMMENTS: Compare these results (Eqs. 3, 4 and 6) with the appropriate expression in Table 5.2.

PROBLEM 5.96 KNOWN: Plane wall, initially having a linear, steady-state temperature distribution with boundaries maintained at T(0,t) = T1 and T(L,t) = T2, suddenly experiences a uniform volumetric heat generation due to the electrical current. Boundary conditions T1 and T2 remain fixed with time. FIND: (a) On T-x coordinates, sketch the temperature distributions for the following cases: initial conditions (t ≤ 0), steady-state conditions (t → ∞) assuming the maximum temperature exceeds T2, and two intermediate times; label important features; (b) For the three-nodal network shown, derive the finite-difference equation using either the implicit or explicit method; (c) With a time increment of ∆t = 5 s, obtain values of Tm for the first 45s of elapsed time; determine the corresponding heat fluxes at the boundaries; and (d) Determine the effect of mesh size by repeating the foregoing analysis using grids of 5 and 11 nodal points. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, transient conduction, (2) Uniform volumetric heat generation for t ≥ 0, (3) Constant properties. PROPERTIES: Wall (Given): ρ = 4000 kg/m3, c = 500 J/kg⋅K, k = 10 W/m⋅K. ANALYSIS: (a) The temperature distribution on T-x coordinates for the requested cases are shown below. Note the following key features: (1) linear initial temperature distribution, (2) non-symmetrical parabolic steady-state temperature distribution, (3) gradient at x = L is first positive, then zero and becomes negative, and (4) gradient at x = 0 is always positive. (b) Performing an energy balance on the control volume about node m above, for unit area, find

E in − E out + E g = E st

p +1 p T2 − Tm T1 − Tm Tm − Tm  k (1) + k (1) + q (1) ∆x = ρ (1) c∆x ∆x ∆x ∆t q∆t p +1 p Fo [T1 + T2 − 2Tm ] + = Tm − Tm ρ cp For the Tm term in brackets, use “p” for explicit and “p+1” for implicit form, p +1 p p p Tm Explicit: = Fo T1 + T2 + (1 − 2Fo ) Tm + q ∆t ρ cp

(

Implicit:

(

)

)

p +1  p Tm 1 + 2Fo ) = Fo T1p +1 + T2p +1 + q ∆t ρ cp + Tm   (

(1)

<

(2)

<

Continued...

PROBLEM 5.96 (Cont.) (c) With a time increment ∆t = 5s, the FDEs, Eqs. (1) and (2) become Explicit:

p +1 p Tm = 0.5Tm + 75

(3)

Implicit:

p +1 p Tm = Tm + 75 1.5

(4)

)

(

where Fo =

q ∆t

k∆t

ρ c∆x 2

=

10 W m ⋅ K × 5s 4000 kg m3 × 500 J kg ⋅ K ( 0.010 m )

2

= 0.25

2 × 107 W m3 × 5s

= 50 K 4000 kg m3 × 500 J kg ⋅ K Performing the calculations, the results are tabulated as a function of time,

ρc

=

p

t (s)

T1 (°C)

0 1 2 3 4 5 6 7 8 9

0 5 10 15 20 25 30 35 40 45

0 0 0 0 0 0 0 0 0 0

Tm (°C) Explicit 50 100.00 125.00 137.50 143.75 146.88 148.44 149.22 149.61 149.80

T2 (°C) Implicit 50 83.33 105.55 120.37 130.25 136.83 141.22 144.15 146.10 147.40

100 100 100 100 100 100 100 100 100 100

<

The heat flux at the boundaries at t = 45s follows from the energy balances on control volumes about the p boundary nodes, using the explicit results for Tm , Node 1:

E in − E out + E g = E st Tp − T q′′x ( 0, t ) + k m 1 + q ( ∆x 2 ) = 0 ∆x

(

p − T1 q′′x ( 0, t ) = − k Tm

) ∆x − q∆x 2 

(5)

q′′x ( 0, 45s ) = −10 W m ⋅ K (149.8 − 0 ) K 0.010 m − 2 × 107 W m3 × 0.010 m 2 q′′x ( 0, 45s ) = −149,800 W m 2 − 100, 000 W m2 = −249,800 W m2

Node 2:

<

T p − T2 k m − q′′x ( L, t ) + q ( ∆x 2 ) = 0 ∆x

(

p q′′x ( L, t ) = k Tm − T2

) ∆x + q∆x 2 = 0 

(6) Continued...

PROBLEM 5.96 (Cont.) q′′x ( L, t ) = 10 W m ⋅ K (149.80 − 100 ) C 0.010 m + 2 × 107 W m3 × 0.010 m 2 q′′x ( L, t ) = 49,800 W m 2 + 100, 000 W m 2 = +149,800 W m2

<

(d) To determine the effect of mesh size, the above analysis was repeated using grids of 5 and 11 nodal points, ∆x = 5 and 2 mm, respectively. Using the IHT Finite-Difference Equation Tool, the finitedifference equations were obtained and solved for the temperature-time history. Eqs. (5) and (6) were p used for the heat flux calculations. The results are tabulated below for t = 45s, where Tm (45s) is the center node, Mesh Size ∆x (mm) 10 5 2

p Tm (45s) (°C) 149.8 149.3 149.4

q′′x (0,45s) 2

kW/m -249.8 -249.0 -249.1

q′′x (L,45s) kW/m2 +149.8 +149.0 +149.0

COMMENTS: (1) The center temperature and boundary heat fluxes are quite insensitive to mesh size for the condition. (2) The copy of the IHT workspace for the 5 node grid is shown below. // Mesh size - 5 nodes, deltax = 5 mm // Nodes a, b(m), and c are interior nodes // Finite-Difference Equations Tool - nodal equations /* Node a: interior node; e and w labeled b and 1. */ rho*cp*der(Ta,t) = fd_1d_int(Ta,Tb,T1,k,qdot,deltax) /* Node b: interior node; e and w labeled c and a. */ rho*cp*der(Tb,t) = fd_1d_int(Tb,Tc,Ta,k,qdot,deltax) /* Node c: interior node; e and w labeled 2 and b. */ rho*cp*der(Tc,t) = fd_1d_int(Tc,T2,Tb,k,qdot,deltax) // Assigned Variables: deltax = 0.005 k = 10 rho = 4000 cp = 500 qdot = 2e7 T1 = 0 T2 = 100

/* Initial Conditions: Tai = 25 Tbi = 50 Tci = 75 */ /* Data Browser Results - Nodal temperatures at 45s Ta Tb Tc t 99.5 149.3 149.5 45 */ // Boundary Heat Fluxes - at t = 45s q''x0 = - k * (Taa - T1 ) / deltax - qdot * deltax / 2 q''xL = k * (Tcc - T2 ) / deltax + qdot * deltax /2 //where Taa = Ta (45s), Tcc = Tc(45s) Taa = 99.5 Tcc = 149.5 /* Data Browser results q''x0 q''xL -2.49E5 1.49E5 */

PROBLEM 5.97 KNOWN: Solid cylinder of plastic material (α = 6 × 10-7 m2/s), initially at uniform temperature of Ti = 20°C, insulated at one end (T4), while other end experiences heating causing its temperature T0 to increase linearly with time at a rate of a = 1°C/s. FIND: (a) Finite-difference equations for the 4 nodes using the explicit method with Fo = 1/2 and (b) Surface temperature T0 when T4 = 35°C. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, transient conduction in cylinder, (2) Constant properties, and (3) Lateral and end surfaces perfectly insulated. ANALYSIS: (a) The finite-difference equations using the explicit method for the interior nodes (m = 1, 2, 3) follow from Eq. 5.73 with Fo = 1/2,

)

(

(

p +1 p p p p p Tm = Fo Tm +1 + Tm −1 + (1 − 2Fo ) Tm = 0.5 Tm +1 + Tm −1

)

(1)

From an energy balance on the control volume node 4 as shown above yields with Fo = 1/2

(

qa + q b + 0 = ρ cV T4p +1 − T4p

E in − E out + E g = E st

(

0 + k T3p − T4p

) ∆t

) ∆x = ρ c (∆x 2)(T4p+1 − T4p ) ∆t

T4p +1 = 2FoT3p + (1 − 2Fo ) T4p = T3p

(2)

(b) Performing the calculations, the temperature-time history is tabulated below, where T0 = Ti +a⋅t where a = 1°C/s and t = p⋅∆t with,

Fo = α∆t ∆x 2 = 0.5 p 0 1 2 3 4 5 6 7

t (s) 0 30 60 90 120 150 180 210

∆t = 0.5 (0.006 m )

2

T0 (°C) 20 50 80 110 140 170 200 230

T1 (°C) 20 20 35 50 68.75 87.5 108.1 -

When T4(210s, p = 7) = 35°C, find T0(210s) = 230°C.

6 × 10−7 m 2 s = 30s T2 (°C) 20 20 20 27.5 35 46.25 57.5 -

T3 (°C) 20 20 20 20 23.75 27.5 35 -

T4 (°C) 20 20 20 20 20 23.75 27.5 35

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PROBLEM 5.98 KNOWN: A 0.12 m thick wall, with thermal diffusivity 1.5 × 10-6 m2/s, initially at a uniform temperature of 85°C, has one face suddenly lowered to 20°C while the other face is perfectly insulated. FIND: (a) Using the explicit finite-difference method with space and time increments of ∆x = 30 mm and ∆t = 300s, determine the temperature distribution within the wall 45 min after the change in surface temperature; (b) Effect of ∆t on temperature histories of the surfaces and midplane. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties. ANALYSIS: (a) The finite-difference equations for the interior points, nodes 0, 1, 2, and 3, can be determined from Eq. 5.73,

)

(

p +1 p p p Tm = Fo Tm −1 + Tm +1 + (1 − 2 Fo ) Tm

(1)

with

Fo = α∆t ∆x 2 = 1.5 × 10−6 m 2 s × 300s (0.03m ) = 1/ 2 . 2

(2)

Note that the stability criterion, Eq. 5.74, Fo ≤ 1/2, is satisfied. Hence, combining Eqs. (1) and (2), p +1 p p Tm = 1/ 2 Tm −1 + Tm +1 for m = 0, 1, 2, 3. Since the adiabatic plane at x = 0 can be treated as a

)

(

symmetry plane, Tm-1 = Tm+1 for node 0 (m = 0). The finite-difference solution is generated in the table below using t = p⋅∆t = 300 p (s) = 5 p (min). p 0 1 2 3 4 5 6 7 8 9

t(min) 0 10 20 30 40 45

T0 85 85 85 85 76.9 76.9 68.8 68.8 61.7 61.7

T1 85 85 85 76.9 76.9 68.8 68.8 61.7 61.7 55.6

T2 85 85 68.8 68.8 60.7 60.7 54.6 54.6 49.5 49.5

T3 85 52.5 52.5 44.4 44.4 40.4 40.4 37.3 37.3 34.8

TL(°C) 20 20 20 20 20 20 20 20 20 20

<

The temperature distribution can also be determined from the Heisler charts. For the wall,

Fo =

αt L2

=

1.5 × 10−6 m 2 s × ( 45 × 60 ) s

(0.12 m )2

= 0.28

and

Bi −1 =

k = 0. hL Continued...

PROBLEM 5.98 (Cont.) From Figure D.1, for Bi-1 = 0 and Fo = 0.28, find θ o θ i ≈ 0.55. Hence, for x = 0

θ = o Ti − T∞ θ i

To − T∞

θ $ To = T ( 0, t ) = T∞ + o ( Ti − T∞ ) = 20$ C + 0.55 (85 − 20 ) C = 55.8$ C . θi

or

This value is to be compared with 61.7°C for the finite-difference method. (b) Using the IHT Finite-Difference Equation Tool Pad for One-Dimensional Transient Conduction, temperature histories were computed and results are shown for the insulated surface (T0) and the midplane, as well as for the chilled surface (TL).

Temperature, T(C)

85 75 65 55 45 35 25 15 0

2000

4000

6000

8000

10000

12000

14000

16000

18000

Time, t(s) T0, deltat = 75 s T2, deltat = 75 s TL T0, deltat = 300 s T2, deltat = 300 s

Apart from small differences during early stages of the transient, there is excellent agreement between results obtained for the two time steps. The temperature decay at the insulated surface must, of course, lag that of the midplane.

PROBLEM 5.99 KNOWN: Thickness, initial temperature and thermophysical properties of molded plastic part. Convection conditions at one surface. Other surface insulated. FIND: Surface temperatures after one hour of cooling. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in product, (2) Negligible radiation, at cooled surface, (3) Negligible heat transfer at insulated surface, (4) Constant properties. ANALYSIS: Adopting the implicit scheme, the finite-difference equation for the cooled surface node is given by Eq. (5.88), from which it follows that p +1 p − 2Fo T9p +1 = 2 FoBi T∞ + T10 (1 + 2 Fo + 2 FoBi ) T10 The general form of the finite-difference equation for any interior node (1 to 9) is given by Eq. (5.89), (1 + 2 Fo ) Tmp +1 − Fo Tmp +−11 + Tmp ++11 = Tmp

)

(

The finite-difference equation for the insulated surface node may be obtained by applying the p

p

symmetry requirement to Eq. (5.89); that is, Tm +1 = Tm −1. Hence,

(1 + 2 Fo ) Top +1 − 2 Fo T1p +1 = Top 2

For the prescribed conditions, Bi = h∆x/k = 100 W/m ⋅K (0.006m)/0.30 W/m⋅K = 2. If the explicit method were used, the most restrictive stability requirement would be given by Eq. (5.79). Hence, for 2 -7 2 Fo (1+Bi) ≤ 0.5, Fo ≤ 0.167. With Fo = α∆t/∆x and α = k/ρc = 1.67 ×10 m /s, the corresponding restriction on the time increment would be ∆t ≤ 36s. Although no such restriction applies for the implicit method, a value of ∆t = 30s is chosen, and the set of 11 finite-difference equations is solved using the Tools option designated as Finite-Difference Equations, One-Dimensional, and Transient from the IHT Toolpad. At t = 3600s, the solution yields:

T10 (3600s ) = 24.1°C

T0 (3600s ) = 71.5°C

<

COMMENTS: (1) More accurate results may be obtained from the one-term approximation to the exact solution for one-dimensional, transient conduction in a plane wall. With Bi = hL/k = 20, Table 2

5.1 yields ζ1 = 1.496 rad and C1 = 1.2699. With Fo = αt/L = 0.167, Eq. (5.41) then yields To = T∞ +

(

)

(Ti - T∞) C1 exp −ζ 12 Fo = 72.4°C, and from Eq. (5.40b), Ts = T∞ + (Ti - T∞) cos (ζ1 ) = 24.5°C. Since the finite-difference results do not change with a reduction in the time step (∆t < 30s), the difference between the numerical and analytical results is attributed to the use of a coarse grid. To improve the accuracy of the numerical results, a smaller value of ∆x should be used. Continued …..

PROBLEM 5.99 (Cont.) (2) Temperature histories for the front and back surface nodes are as shown.

80

Te m p e ra tu re (C )

70 60 50 40 30 20 0

600

1200

1800

2400

3000

3600

Tim e (s ) In s u la te d s u rfa ce C o o le d s u rfa ce

Although the surface temperatures rapidly approaches that of the coolant, there is a significant lag in the thermal response of the back surface. The different responses are attributable to the small value of α and the large value of Bi.

PROBLEM 5.100 KNOWN: Plane wall, initially at a uniform temperature Ti = 25°C, is suddenly exposed to convection with a fluid at T∞ = 50°C with a convection coefficient h = 75 W/m2⋅K at one surface, while the other is exposed to a constant heat flux q′′o = 2000 W/m2. See also Problem 2.43. FIND: (a) Using spatial and time increments of ∆x = 5 mm and ∆t = 20s, compute and plot the temperature distributions in the wall for the initial condition, the steady-state condition, and two intermediate times, (b) On q′′x -x coordinates, plot the heat flux distributions corresponding to the four temperature distributions represented in part (a), and (c) On q′′x -t coordinates, plot the heat flux at x = 0 and x = L. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, transient conduction and (2) Constant properties. ANALYSIS: (a) Using the IHT Finite-Difference Equations, One-Dimensional, Transient Tool, the equations for determining the temperature distribution were obtained and solved with a spatial increment of ∆x = 5 mm. Using the Lookup Table functions, the temperature distributions were plotted as shown below. (b) The heat flux, q′′x (x,t), at each node can be evaluated considering the control volume shown with the schematic above

(

q′′x ( m, p ) = q′′x,a + q′′x,b

)

p p p p  Tm −1 − Tm Tm − Tm +1  p p   2 = k Tm 2 = k (1) − Tm +1 + k (1) − 1 ∆x ∆x  

(

) 2∆x

From knowledge of the temperature distribution, the heat flux at each node for the selected times is computed and plotted below. Heat flux, q''x(x,t) (W/m^2)

Temperature, T(x,t) (C)

160 140 120 100 80 60 40 20 0

10

20

30

40

2000 1500 1000 500 0

50

Wall coordinate, x (mm) Initial condition, t1200s

0

10

20

30

40

50

Wall coordinate, x (mm) Initial condition, t1200s

(c) The heat fluxes for the locations x = 0 and x = L, are plotted as a function of time. At the x = 0 surface, the heat flux is constant, q „„o = 2000 W/m2. At the x = L surface, the heat flux is given by Newton’s law of cooling, q′′x (L,t) = h[T(L,t) - T∞ ]; at t = 0, q′′x (L,0) = -1875 W/m2. For steady-state

conditions, the heat flux q′′x (x,∞) is everywhere constant at q „„o .

Continued...

PROBLEM 5.100 (Cont.) 2000

Heat flux (W/m^2)

1000

0

-1000

-2000 0

200

400

600

800

1000

1200

Elapsed time, t (s) q''x(0,t) - Heater flux q''x(L,t) - Convective flux

Comments: The IHT workspace using the Finite-Difference Equations Tool to determine the temperature distributions and heat fluxes is shown below. Some lines of code were omitted to save space on the page. // Finite-Difference Equations, One-Dimensional, Transient Tool: // Node 0 - Applied heater flux /* Node 0: surface node (w-orientation); transient conditions; e labeled 1. */ rho * cp * der(T0,t) = fd_1d_sur_w(T0,T1,k,qdot,deltax,Tinf0,h0,q''a0) q''a0 = 2000 // Applied heat flux, W/m^2; Tinf0 = 25 // Fluid temperature, C; arbitrary value since h0 is zero; no convection process h0 = 1e-20 // Convection coefficient, W/m^2.K; made zero since no convection process // Interior Nodes 1 - 9: /* Node 1: interior node; e and w labeled 2 and 0. */ rho*cp*der(T1,t) = fd_1d_int(T1,T2,T0,k,qdot,deltax) /* Node 2: interior node; e and w labeled 3 and 1. */ rho*cp*der(T2,t) = fd_1d_int(T2,T3,T1,k,qdot,deltax) ...... ...... /* Node 9: interior node; e and w labeled 10 and 8. */ rho*cp*der(T9,t) = fd_1d_int(T9,T10,T8,k,qdot,deltax) // Node 10 - Convection process: /* Node 10: surface node (e-orientation); transient conditions; w labeled 9. */ rho * cp * der(T10,t) = fd_1d_sur_e(T10,T9,k,qdot,deltax,Tinf,h,q''a) q''a = 0 // Applied heat flux, W/m^2; zero flux shown // Heat Flux Distribution at Interior Nodes, q''m: q''1 = k / deltax * (T0 - T2) / 2 q''2 = k / deltax * (T1 - T3) / 2 ...... ...... q''9 = k / deltax * (T8 - T10) / 2 // Heat flux at boundary x= L, q''10 q''xL = h * (T10 - Tinf) // Assigned Variables: deltax = 0.005 k = 1.5 alpha = 7.5e-6 cp = 1000 alpha = k / (rho * cp) qdot = 0 Ti = 25 Tinf = 50 h = 75

// Spatial increment, m // thermal conductivity, W/m.K // Thermal diffusivity, m^2/s // Specific heat, J/kg.K; arbitrary value // Defintion from which rho is calculated // Volumetric heat generation rate, W/m^3 // Initial temperature, C; used also for plotting initial distribution // Fluid temperature, K // Convection coefficient, W/m^2.K

// Solver Conditions: integrated t from 0 to 1200 with 1 s step, log every 2nd value

PROBLEM 5.101 KNOWN: Plane wall, initially at a uniform temperature To = 25°C, has one surface (x = L) suddenly exposed to a convection process with T∞ = 50°C and h = 1000 W/m2⋅K, while the other surface (x = 0) is maintained at To. Also, the wall suddenly experiences uniform volumetric heating with q = 1 × 107 W/m3. See also Problem 2.44. FIND: (a) Using spatial and time increments of ∆x = 4 mm and ∆t = 1s, compute and plot the temperature distributions in the wall for the initial condition, the steady-state condition, and two intermediate times, and (b) On q′′x -t coordinates, plot the heat flux at x = 0 and x = L. At what elapsed time is there zero heat flux at x = L? SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, transient conduction and (2) Constant properties. ANALYSIS: (a) Using the IHT Finite-Difference Equations, One-Dimensional, Transient Tool, the temperature distributions were obtained and plotted below. (b) The heat flux, q „„x (L,t), can be expressed in terms of Newton’s law of cooling,

(

)

q ′′x ( L, t ) = h T10 − T∞ . p

From the energy balance on the control volume about node 0 shown above,

(

q′′x ( 0, t ) + E g + q′′a = 0

q′′x ( 0, t ) = −q ( ∆x 2 ) − k T1 − To p

) ∆x

From knowledge of the temperature distribution, the heat fluxes are computed and plotted. 120 100 0

80

Heat flux (W/m^2)

Temperature, T(x,t) (C)

100000

60 40

-1E5

-2E5

20 0

10

20 Wall coordinate, x (mm)

Initial condition, t600s

30

40

-3E5 0

100

200

300

400

500

600

Elapsed time, t(s) q''x(0,t) q''x(L,t)

COMMENTS: The steady-state analytical solution has the form of Eq. 3.40 where C1 = 6500 m-1/°C and C2 = 25°C. Find q ′′x ( 0, ∞ ) = −3.25 × 105 W / m 2 and q′′x ( L ) = +7.5 × 104 W / m 2 . Comparing with the graphical results above, we conclude that steady-state conditions are not reached in 600 x.

PROBLEM 5.102 KNOWN: Fuel element of Example 5.8 is initially at a uniform temperature of 250°C with no internal generation; suddenly a uniform generation, q = 108 W/m3 , occurs when the element is inserted into the core while the surfaces experience convection (T∞,h). FIND: Temperature distribution 1.5s after element is inserted into the core. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3) q = 0, initially; at t > 0, q is uniform. ANALYSIS: As suggested, the explicit method with a space increment of 2mm will be used. Using the nodal network of Example 5.8, the same finite-difference equations may be used. Interior nodes, m = 1, 2, 3, 4 2  q ( ∆x )  p+1 p p p  + (1 − 2 Fo ) Tm Tm = Fo Tm-1 + Tm+1 + . 2     Midplane node, m = 0

(1)

p p . Same as Eq. (1), but with Tm-1 = Tm+1

Surface node, m = 5 2  q ( ∆x )   + (1 − 2Fo − 2Bi ⋅ Fo ) T p . (2) T5p+1 = 2 Fo T4p + Bi ⋅ T∞ + 5 2k     The most restrictive stability criterion is associated with Eq. (2), Fo(1+Bi) ≤ 1/2. Consider the following parameters: 2 h∆x 1100W/m ⋅ K × ( 0.002m ) Bi = = = 0.0733 k 30W/m ⋅ K 1/2 Fo ≤ = 0.466 (1 + Bi )

Fo ( ∆x )

2

∆t ≤

α

= 0.466

(0.002m )2

5 × 10−6 m 2 / s

= 0.373s. Continued …..

PROBLEM 5.102 (Cont.) To be well within the stability limit, select ∆t = 0.3s, which corresponds to Fo =

α∆t

=

5 × 10−6 m 2 / s × 0.3s

∆x 2 (0.002m )2 t = p∆t = 0.3p (s ).

= 0.375

Substituting numerical values with q = 108 W/m3 , the nodal equations become 2 T0p+1 = 0.375  2T1p + 108 W/m3 ( 0.002m ) / 30W/m ⋅ K  + (1 − 2 × 0.375) T0p   T0p+1 = 0.375  2T1p + 13.33 + 0.25 T0p  

T1p+1 = 0.375  T0p + T2p + 13.33 + 0.25 T1p   T2p+1 = 0.375  T1p + T3p + 13.33 + 0.25 T2p  

(3) (4) (5)

T3p+1 = 0.375  T2p + T4p + 13.33 + 0.25 T3p   T4p+1 = 0.375  T3p + T5p + 13.33 + 0.25 T4p  

(6) (7)

13.33   T5p+1 = 2 × 0.375  T4p + 0.0733 × 250 + + (1 − 2 × 0.375 − 2 × 0.0733 × 0.375 ) T5p  2   T5p+1 = 0.750  T4p + 24.99  + 0.195 T5p . (8)   The initial temperature distribution is Ti = 250°C at all nodes. The marching solution, following the procedure of Example 5.8, is represented in the table below. p 0 1 2 3 4

t(s) 0 0.3 0.6 0.9 1.2

T0 250 255.00 260.00 265.00 270.00

T1 250 255.00 260.00 265.00 270.00

T2 250 255.00 260.00 265.00 270.00

T3 250 255.00 260.00 265.00 269.96

T4 250 255.00 260.00 264.89 269.74

T5(°C) 250 254.99 259.72 264.39 268.97

5

1.5

275.00

275.00

274.98

274.89

274.53

273.50

The desired temperature distribution T(x, 1.5s), corresponds to p = 5. COMMENTS: Note that the nodes near the midplane (0,1) do not feel any effect of the coolant during the first 1.5s time period.

<

PROBLEM 5.103 KNOWN: Conditions associated with heat generation in a rectangular fuel element with surface cooling. See Example 5.8. FIND: (a) The temperature distribution 1.5 s after the change in operating power; compare your results with those tabulated in the example, (b) Calculate and plot temperature histories at the midplane (00) and surface (05) nodes for 0≤ t ≤ 400 s; determine the new steady-state temperatures, and approximately how long it will take to reach the new steady-state condition after the step change in operating power. Use the IHT Tools | Finite-Difference Equations | One-Dimensional | Transient conduction model builder as your solution tool. SCHEMATIC:

ASSUMPTIONS: (1) One dimensional conduction in the x-direction, (2) Uniform generation, and (3) Constant properties. ANALYIS: The IHT model builder provides the transient finite-difference equations for the implicit method of solution. Selected portions of the IHT code used to obtain the results tabulated below are shown in the Comments. (a) Using the IHT code, the temperature distribution (°C) as a function of time (s) up to 1.5 s after the step power change is obtained from the summarized results copied into the workspace 1 2 3 4 5 6

t 0 0.3 0.6 0.9 1.2 1.5

T00 357.6 358.1 358.6 359.1 359.6 360.1

T01 356.9 357.4 357.9 358.4 358.9 359.4

T02 354.9 355.4 355.9 356.4 356.9 357.4

T03 351.6 352.1 352.6 353.1 353.6 354.1

T04 346.9 347.4 347.9 348.4 348.9 349.3

T05 340.9 341.4 341.9 342.3 342.8 343.2

(b) Using the code, the mid-plane (00) and surface (05) node temperatures are plotted as a function of time. Te m p e ra tu re h is to ry a fte r s te p ch a n g e in p o w e r

Te m p e ra tu re , T(x,t) (C )

480

440

400

360

320 0

100

200

300

400

Tim e , t (s ) T0 0 , Mid -p la n e , x = 0 T0 5 , S u rfa c e , x = L

Continued …..

PROBLEM 5.103 (Cont.) Note that at t ≈ 240 s, the wall has nearly reached the new steady-state condition for which the nodal temperatures (°C) were found as: T00 465

T01 463.7

T02 459.7

T03 453

T04 443.7

T05 431.7

COMMENTS: (1) Can you validate the new steady-state nodal temperatures from part (b) by comparison against an analytical solution? (2) Will using a smaller time increment improve the accuracy of the results? Use your code with ∆t = 0.15 s to justify your explanation. (3) Selected portions of the IHT code to obtain the nodal temperature distribution using spatial and time increments of ∆x = 2 mm and ∆t = 0.3 s, respectively, are shown below. For the solveintegration step, the initial condition for each of the nodes corresponds to the steady-state temperature distribution with q1. // Tools | Finite-Difference Equations | One-Dimensional | Transient /* Node 00: surface node (w-orientation); transient conditions; e labeled 01. */ rho * cp * der(T00,t) = fd_1d_sur_w(T00,T01,k,qdot,deltax,Tinf01,h01,q''a00) q''a00 = 0 // Applied heat flux, W/m^2; zero flux shown Tinf01 = 20 // Arbitrary value h01 = 1e-8 // Causes boundary to behave as adiabatic /* Node 01: interior node; e and w labeled 02 and 00. */ rho*cp*der(T01,t) = fd_1d_int(T01,T02,T00,k,qdot,deltax) /* Node 02: interior node; e and w labeled 03 and 01. */ rho*cp*der(T02,t) = fd_1d_int(T02,T03,T01,k,qdot,deltax) /* Node 03: interior node; e and w labeled 04 and 02. */ rho*cp*der(T03,t) = fd_1d_int(T03,T04,T02,k,qdot,deltax) /* Node 04: interior node; e and w labeled 05 and 03. */ rho*cp*der(T04,t) = fd_1d_int(T04,T05,T03,k,qdot,deltax) /* Node 05: surface node (e-orientation); transient conditions; w labeled 04. */ rho * cp * der(T05,t) = fd_1d_sur_e(T05,T04,k,qdot,deltax,Tinf05,h05,q''a05) q''a05 = 0 // Applied heat flux, W/m^2; zero flux shown Tinf05 = 250 // Coolant temperature, C h05 = 1100 // Convection coefficient, W/m^2.K // Input parameters qdot = 2e7 // Volumetric rate, W/m^3, step change deltax = 0.002 // Space increment k = 30 // Thermophysical properties alpha = 5e-6 rho = 1000 alpha = k / (rho * cp) /* Steady-state conditions, with qdot1 = 1e7 W/m^3; initial conditions for step change T_x = 16.67 * (1 - x^2/L^2) + 340.91 // See text Seek T_x for x = 0, 2, 4, 6, 8, 10 mm; results used for Ti are Node T_x 00 357.6 01 356.9 02 354.9 03 351.6 04 346.9 05 340.9 */

PROBLEM 5.104 KNOWN: Conditions associated with heat generation in a rectangular fuel element with surface cooling. See Example 5.8. FIND: (a) The temperature distribution 1.5 s after the change in the operating power; compare results with those tabulated in the Example, and (b) Plot the temperature histories at the midplane, x = 0, and the surface, x = L, for 0 ≤ t ≤ 400 s; determine the new steady-state temperatures, and approximately how long it takes to reach this condition. Use the finite-element software FEHT as your solution tool. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Uniform generation, (3) Constant properties. ANALYSIS: Using FEHT, an outline of the fuel element is drawn of thickness 10 mm in the xdirection and arbitrary length in the y-direction. The boundary conditions are specified as follows: on the y-planes and the x = 0 plane, treat as adiabatic; on the x = 10 mm plane, specify the convection option. Specify the material properties and the internal generation with q1 . In the Setup menu, click on Steady-state, and then Run to obtain the temperature distribution corresponding to the initial temperature distribution, Ti ( x, 0 ) = T ( x, q1 ) , before the change in operating power to q 2 . Next, in the Setup menu, click on Transient; in the Specify | Internal Generation box, change the value to q 2 ; and in the Run command, click on Continue (not Calculate). (a) The temperature distribution 1.5 s after the change in operating power from the FEHT analysis and from the FDE analysis in the Example are tabulated below. x/L T(x/L, 1.5 s) FEHT (°C) FDE (°C)

0

0.2

0.4

360.1 360.08

359.4 359.41

357.4 357.41

0.6

0.8

1.0

354.1 349.3 354.07 349.37

343.2 343.27

The mesh spacing for the FEHT analysis was 0.5 mm and the time increment was 0.005 s. For the FDE analyses, the spatial and time increments were 2 mm and 0.3 s. The agreement between the results from the two numerical methods is within 0.1°C. (b) Using the FEHT code, the temperature histories at the mid-plane (x = 0) and the surface (x = L) are plotted as a function of time. Continued …..

PROBLEM 5.104 (Cont.)

From the distribution, the steady-state condition (based upon 98% change) is approached in 215 s. The steady-state temperature distributions after the step change in power from the FEHT and FDE analysis in the Example are tabulated below. The agreement between the results from the two numerical methods is within 0.1°C x/L T(x/L, ∞) FEHT (°C) FDE (°C)

0

0.2

0.4

0.6

0.8

465.0 465.15

463.7 463.82

459.6 459.82

453.0 453.15

443.6 443.82

1.0

431.7 431.82

COMMENTS: (1) For background information on the Continue option, see the Run menu in the FEHT Help section. Using the Run/Calculate command, the steady-state temperature distribution was determined for the q1 operating power. Using the Run|Continue command (after re-setting the generation to q 2 and clicking on Setup | Transient), this steady-state distribution automatically becomes the initial temperature distribution for the q 2 operating power. This feature allows for conveniently prescribing a non-uniform initial temperature distribution for a transient analysis (rather than specifying values on a node-by-node basis).

(2) Use the View | Tabular Output command to obtain nodal temperatures to the maximum number of significant figures resulting from the analysis. (3) Can you validate the new steady-state nodal temperatures from part (b) (with q 2 , t → ∞) by comparison against an analytical solution?

PROBLEM 5.105 KNOWN: Thickness, initial temperature, speed and thermophysical properties of steel in a thin-slab continuous casting process. Surface convection conditions. FIND: Time required to cool the outer surface to a prescribed temperature. Corresponding value of the midplane temperature and length of cooling section. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible radiation at quenched surfaces, (3) Symmetry about the midplane, (4) Constant properties. ANALYSIS: Adopting the implicit scheme, the finite-difference equaiton for the cooled surface node is given by Eq. (5.88), from which it follows that p +1 p 1 p − 2 Fo T9 + = 2 FoBi T∞ + T10 (1 + 2 Fo + 2 FoBi ) T10 The general form of the finite-difference equation for any interior node (1 to 9) is given by Eq. (5.89),

(1 + 2 Fo ) Tmp +1 − Fo

(Tmp+−11 + Tmp++11 ) = Tmp

The finite-difference equation for the midplane node may be obtained by applying the symmetry p

p

requirement to Eq. (5.89); that is, Tm +1 = Tm −1. Hence,

(1 + 2 Fo ) T0p +1 − 2 Fo T1p +1 = T0p 2

For the prescribed conditions, Bi = h∆x/k = 5000 W/m ⋅K (0.010m)/30 W/m⋅K = 1.67. If the explicit method were used, the stability requirement would be given by Eq. (5.79). Hence, for Fo(1 + Bi) ≤ 2 -6 2 0.5, Fo ≤ 0.187. With Fo = α∆t/∆x and α = k/ρc = 5.49 × 10 m /s, the corresponding restriction on the time increment would be ∆t ≤ 3.40s. Although no such restriction applies for the implicit method, a value of ∆t = 1s is chosen, and the set of 11 finite-difference equations is solved using the Tools option designated as Finite-Difference Equations, One-Dimensional and Transient from the IHT Toolpad. For T10 (t) = 300°C, the solution yields

<

t = 161s Continued …..

PROBLEM 5.105 (Cont.)

T0 ( t ) = 1364°C

<

With a casting speed of V = 15 mm/s, the length of the cooling section is

Lcs = Vt = 0.015 m / s (161s ) = 2.42m

<

2

COMMENTS: (1) With Fo = αt/L = 0.088 < 0.2, the one-term approximation to the exact solution for one-dimensional conduction in a plane wall cannot be used to confirm the foregoing results. However, using the exact solution from the Models, Transient Conduction, Plane Wall Option of IHT, values of T0 = 1366°C and Ts = 200.7°C are obtained and are in good agreement with the finitedifference predictions. The accuracy of these predictions could still be improved by reducing the value of ∆x. (2) Temperature histories for the surface and midplane nodes are plotted for 0 < t < 600s.

1500 T e m p e ra tu re (C )

1300 1100 900 700 500 300 100 0

100

200

300

400

500

600

T im e (s ) Mid p la n e C o o le d s u rfa c e

While T10 (600s) = 124°C, To (600s) has only dropped to 879°C. The much slower thermal response at the midplane is attributable to the small value of α and the large value of Bi = 16.67.

PROBLEM 5.106 KNOWN: Very thick plate, initially at a uniform temperature, Ti, is suddenly exposed to a convection cooling process (T∞,h). FIND: Temperatures at the surface and a 45mm depth after 3 minutes using finite-difference method with space and time increments of 15mm and 18s. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Plate approximates semi-infinite medium, (3) Constant properties. ANALYSIS: The grid network representing the plate is shown above. The finite-difference equation for node 0 is given by Eq. 5.82 for one-dimensional conditions or Eq. 5.77,

(

)

T0p+1 = 2 Fo T1p + Bi ⋅ T∞ + (1 − 2 Fo − 2 Bi ⋅ Fo ) T0p .

(1)

The numerical values of Fo and Bi are Fo =

α∆t ∆x 2

=

5.6 ×10−6 m2 /s × 18s

( 0.015m )2

= 0.448

(

)

2 -3 h∆x 100 W/m ⋅ K × 15 ×10 m Bi = = = 0.075. k 20 W/m ⋅ K

Recognizing that T∞ = 15°C, Eq. (1) has the form T0p+1 = 0.0359 T0p + 0.897 T1p + 1.01.

(2)

It is important to satisfy the stability criterion, Fo (1+Bi) ≤ 1/2. Substituting values, 0.448 (1+0.075) = 0.482 ≤ 1/2, and the criterion is satisfied. The finite-difference equation for the interior nodes, m = 1, 2…, follows from Eq. 5.73,

(

)

p+1 p p p Tm = Fo Tm+1 + Tm-1 + (1 − 2Fo ) Tm .

(3)

Recognizing that the stability criterion, Fo ≤ 1/2, is satisfied with Fo = 0.448,

(

)

p+1 p p p Tm = 0.448 Tm+1 + Tm-1 + 0.104Tm .

(4) Continued …..

PROBLEM 5.106 (Cont.) The time scale is related to p, the number of steps in the calculation procedure, and ∆t, the time increment, t = p∆t.

(5)

The finite-difference calculations can now be performed using Eqs. (2) and (4). The results are tabulated below. p 0 1 2 3 4 5 6 7 8 9 10

t(s) 0 18 36 54 72 90 108 126 144 162 180

T0 325 304.2 303.2 294.7 293.0 287.6 285.6 281.8 279.8 276.7 274.8

T1 325 324.7 315.3 313.7 307.8 305.8 301.6 299.5 296.2 294.1 291.3

T2 325 325 324.5 320.3 318.9 315.2 313.5 310.5 308.6 306.0 304.1

T3 325 325 325 324.5 322.5 321.5 319.3 317.9 315.8 314.3 312.4

T4 325 325 325 325 324.5 323.5 322.7 321.4 320.4 319.0

T5 325 325 325 325 325 324.5 324.0 323.3 322.5

T6 T7(K) 325 325 325 325 325 325 325 325 325 325 325 325 324.5 325 324.2

Hence, find T ( 0, 180s ) = T010 = 275o C

T ( 45mm, 180s ) = T310 = 312o C.

<

COMMENTS: (1) The above results can be readily checked against the analytical solution represented in Fig. 5.8 (see also Eq. 5.60). For x = 0 and t = 180s, find x =0 1/2 2 (α t )

(

1/2 100 W/m 2 ⋅ K 5.60× 10-6m 2 / s × 180s h (α t ) = k 20 W/m ⋅ K for which the figure gives T − Ti = 0.15 T∞ − Ti so that,

)

1/2

= 0.16

T ( 0, 180s ) = 0.15 ( T∞ − T i ) +T i = 0.15 (15 − 325)o C + 325oC T ( 0, 180s ) = 278o C. For x = 45mm, the procedure yields T(45mm, 180s) = 316°C. The agreement with the numerical solution is nearly within 1%.

PROBLEM 5.107 KNOWN: Sudden exposure of the surface of a thick slab, initially at a uniform temperature, to convection and to surroundings at a high temperature. FIND: (a) Explicit, finite-difference equation for the surface node in terms of Fo, Bi, Bir, (b) Stability criterion; whether it is more restrictive than that for an interior node and does it change with time, and (c) Temperature at the surface and at 30mm depth for prescribed conditions after 1 minute exposure. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Thick slab may be approximated as semi-infinite medium, (3) Constant properties, (4) Radiation exchange is between small surface and large surroundings. ANALYSIS: (a) The explicit form of the FDE for the surface node may be obtained by applying an energy balance to a control volume about the node. E ′′in − E ′′out = q′′conv + q′′rad + q′′cond = E ′′st

) (

(

)

T1p − Top p p h T∞ − To + h r Tsur − To + k ⋅1⋅ ∆x

p+1 p − To  ∆x  T (1) = ρ c  ⋅1 o ∆t  2  where the radiation process has been linearized, Eq. 1.8. (See also Comment 4, Example 5.9), 2  2 . h r = h pr Top , Tsur = εσ Top + Tsur  T0p  + Tsur (2)       Divide Eq. (1) by ρc∆x/2∆t and regroup using these definitions to obtain the FDE:

) (

(

Fo ≡ ( k/ρ c ) ∆t/∆x 2

(

)

Bi ≡ h∆x/k

)

Bi r ≡ h r ∆x/k

(3,4,5)

Top+1 = 2Fo Bi ⋅ T∞ + Bir ⋅ Tsur + T1p + (1 − 2 Bi ⋅ Fo − 2Bir ⋅ Fo − 2Fo ) Top .

(6) <

(b) The stability criterion for Eq. (6) requires that the coefficient of Top be positive. 1 − 2Fo ( Bi + Bi r + 1) ≥ 0

or

Fo ≤ 1/2 ( Bi + Bi r + 1).

(7) <

The stability criterion for an interior node, Eq. 5.74, is Fo ≤ 1/2. Since Bi + Bir > 0, the stability criterion of the surface node is more restrictive. Note that Bir is not constant but depends upon hr which increases with increasing Top (time). Hence, the restriction on Fo increases with increasing Top (time). Continued …..

PROBLEM 5.107 (Cont.) (c) Consider the prescribed conditions with negligible convection (Bi = 0). The FDEs for the thick slab are: Surface (0)

(

)

Top+1 = 2Fo Bi ⋅ Fo + Bi r ⋅ Tsur + T1 + (1 − 2Bi ⋅ Fo − 2Bi r ⋅ Fo − 2Fo ) Top p

)

(

(8)

p+1 p p p Tm = Fo Tm+1 + Tm-1 + (1 − 2Fo ) Tm

Interior (m≥1)

(9,5,7,3)

The stability criterion from Eq. (7) with Bi = 0 is, Fo ≤ 1/2 (1 + Bi r )

(10)

To proceed with the explicit, marching solution, we need to select a value of ∆t (Fo) that will satisfy the stability criterion. A few trial calculations are helpful. A value of ∆t = 15s 2 provides Fo = 0.105, and using Eqs. (2) and (5), hr(300K, 1000K) = 72.3 W/m ⋅K and Bir = 0.482. From the stability criterion, Eq. (10), find Fo ≤ 0.337. With increasing Top , hr and Bir 2

increase: hr(800K, 1000K) = 150.6 W/m ⋅K, Bir = 1.004 and Fo ≤ 0.249. Hence, if Top < 800K, ∆t = 15s or Fo = 0.105 satisfies the stability criterion. Using ∆t = 15s or Fo = 0.105 with the FDEs, Eqs. (8) and (9), the results of the solution are tabulated below. Note how h pr and Bi pr are evaluated at each time increment. Note that t = p⋅∆t, where ∆t = 15s. p

t(s)

To / h pr / Bi r

0

0

300 72.3 0.482

300

1

15

370.867 79.577 0.5305

2

30

3

4

T3

T4

300

300

300

300

300

300

300

426.079 85.984 0.5733

307.441

300

300

300

45

470.256 91.619 0.6108

319.117

300.781

300

300

60

502.289

333.061

302.624

300.082

300

After 60s(p = 4),

T1(K)

T2

To(0, 1 min) = 502.3K and T3(30mm, 1 min) = 300.1K.

….

<

COMMENTS: (1) The form of the FDE representing the surface node agrees with Eq. 5.82 if this equation is reduced to one-dimension. (2) We should recognize that the ∆t = 15s time increment represents a coarse step. To improve the accuracy of the solution, a smaller ∆t should be chosen.

PROBLEM 5.108 KNOWN: Thick slab of copper, initially at a uniform temperature, is suddenly exposed to a constant net radiant flux at one surface. See Example 5.9. FIND: (a) The nodal temperatures at nodes 00 and 04 at t = 120 s; that is, T00(0, 120 s) and T04(0.15 m, 120 s); compare results with those given by the exact solution in Comment 1; will a time increment of 0.12 s provide more accurate results?; and, (b) Plot the temperature histories for x = 0, 150 and 600 mm, and explain key features of your results. Use the IHT Tools | Finite-Difference Equations | OneDimensional | Transient conduction model builder to obtain the implicit form of the FDEs for the interior nodes. Use space and time increments of 37.5 mm and 1.2 s, respectively, for a 17-node network. For the surface node 00, use the FDE derived in Section 2 of the Example. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Slab of thickness 600 mm approximates a semi-infinite medium, and (3) Constant properties. ANALYSIS: The IHT model builder provides the implicit-method FDEs for the interior nodes, 01 – 15. The +x boundary condition for the node-16 control volume is assumed adiabatic. The FDE for the surface node 00 exposed to the net radiant flux was derived in the Example analysis. Selected portions of the IHT code used to obtain the following results are shown in the Comments. (a) The 00 and 04 nodal temperatures for t = 120 s are tabulated below using a time increment of ∆t = 1.2 s and 0.12 s, and compared with the results given from the exact analytical solution, Eq. 5.59. Node 00 04

FDE results (°C) ∆t = 1.2 s 119.3 45.09

∆t = 0.12 s 119.4 45.10

Analytical result (°C) Eq. 5.59 120.0 45.4

The numerical FDE-based results with the different time increments agree quite closely with one another. At the surface, the numerical results are nearly 1 °C less than the result from the exact analytical solution. This difference represents an error of -1% ( -1 °C / (120 – 20 ) °C x 100). At the x = 150 mm location, the difference is about -0.4 °C, representing an error of –1.5%. For this situation, the smaller time increment (0.12 s) did not provide improved accuracy. To improve the accuracy of the numerical model, it would be necessary to reduce the space increment, in addition to using the smaller time increment. (b) The temperature histories for x = 0, 150 and 600 mm (nodes 00, 04, and 16) for the range 0 ≤ t ≤ 150 s are as follows.

Continued …..

PROBLEM 5.108 (Cont.) Te m p e ra tu re h is to rie s fo r N o d e s 0 0 , 0 4 , a n d 1 6

Te m p e ra tu re , T(x,t)

120

80

40

0 0

50

100

150

Tim e , t (s ) T0 0 = T(0 , t) T0 4 = T(1 5 0 m m , t) T0 0 = T(6 0 0 m m , t)

As expected, the surface temperature, T00 = T(0,t), increases markedly at early times. As thermal penetration increases with increasing time, the temperature at the location x = 150 mm, T04 = T(150 mm, t), begins to increase after about 20 s. Note, however, the temperature at the location x = 600 mm, T16 = T(600 mm, t), does not change significantly within the 150 s duration of the applied surface heat flux. Our assumption of treating the +x boundary of the node 16 control volume as adiabatic is justified. A copper plate of 600-mm thickness is a good approximation to a semi-infinite medium at times less than 150 s. COMMENTS: Selected portions of the IHT code with the nodal equations to obtain the temperature distribution are shown below. Note how the FDE for node 00 is written in terms of an energy balance using the der (T,t) function. The FDE for node 16 assumes that the “east” boundary is adiabatic.

// Finite-difference equation, node 00; from Examples solution derivation; implicit method q''o + k * (T01 - T00) / deltax = rho * (deltax / 2) *cp * der (T00,t) // Finite-difference equations, interior nodes 01-15; from Tools /* Node 01: interior node; e and w labeled 02 and 00. */ rho*cp*der(T01,t) = fd_1d_int(T01,T02,T00,k,qdot,deltax) rho*cp*der(T02,t) = fd_1d_int(T02,T03,T01,k,qdot,deltax) ….. ….. rho*cp*der(T14,t) = fd_1d_int(T14,T15,T13,k,qdot,deltax) rho*cp*der(T15,t) = fd_1d_int(T15,T16,T14,k,qdot,deltax) // Finite-difference equation node 16; from Tools, adiabatic surface /* Node 16: surface node (e-orientation); transient conditions; w labeled 15. */ rho * cp * der(T16,t) = fd_1d_sur_e(T16,T15,k,qdot,deltax,Tinf16,h16,q''a16) q''a16 = 0 // Applied heat flux, W/m^2; zero flux shown Tinf16 = 20 // Arbitrary value h16 = 1e-8 // Causes boundary to behave as adiabatic

PROBLEM 5.109 KNOWN: Thick slab of copper as treated in Example 5.9, initially at a uniform temperature, is suddenly exposed to large surroundings at 1000°C (instead of a net radiant flux). FIND: (a) The temperatures T(0, 120 s) and T(0.15 m, 120s) using the finite-element software FEHT for a surface emissivity of 0.94 and (b) Plot the temperature histories for x = 0, 150 and 600 mm, and explain key features of your results. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Slab of thickness 600 mm approximates a semi-infinite medium, (3) Slab is small object in large, isothermal surroundings. ANALYSIS: (a) Using FEHT, an outline of the slab is drawn of thickness 600 mm in the x-direction and arbitrary length in the y-direction. Click on Setup | Temperatures in K, to enter all temperatures in kelvins. The boundary conditions are specified as follows: on the y-planes and the x = 600 mm plane, treat as adiabatic; on the surface (0,y), select the convection coefficient option, enter the linearized radiation coefficient after Eq. 1.9 written as 0.94 * 5.67e-8 * (T + 1273) * (T^2 + 1273^2) and enter the surroundings temperature, 1273 K, in the fluid temperature box. See the Comments for a view of the input screen. From View|Temperatures, find the results: T(0, 120 s) = 339 K = 66°C

<

T(150 mm, 120 s) = 305K = 32°C

(b) Using the View | Temperatures command, the temperature histories for x = 0, 150 and 600 mm (10 mm mesh, Nodes 18, 23 and 15, respectively) are plotted. As expected, the surface temperature increases markedly at early times. As thermal penetration increases with increasing time, the temperature at the location x = 150 mm begins to increase after about 30 s. Note, however, that the temperature at the location x = 600 mm does not change significantly within the 150 s exposure to the hot surroundings. Our assumption of treating the boundary at the x = 600 mm plane as adiabatic is justified. A copper plate of 600 mm is a good approximation to a semi-infinite medium at times less than 150 s.

Continued …..

PROBLEM 5.109 (Cont.)

COMMENTS: The annotated Input screen shows the outline of the slab, the boundary conditions, and the triangular mesh before using the Reduce-mesh option.

PPROBLEM 5.110 KNOWN: Electric heater sandwiched between two thick plates whose surfaces experience convection. Case 2 corresponds to steady-state operation with a loss of coolant on the x = -L surface. Suddenly, a second loss of coolant condition occurs on the x = +L surface, but the heater remains energized for the next 15 minutes. Case 3 corresponds to the eventual steady-state condition following the second loss of coolant event. See Problem 2.53. FIND: Calculate and plot the temperature time histories at the plate locations x = 0, ±L during the transient period between steady-state distributions for Case 2 and Case 3 using the finite-element approach with FEHT and the finite-difference method of solution with IHT (∆x = 5 mm and ∆t = 1 s). SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Heater has negligible thickness, and (4) Negligible thermal resistance between the heater surfaces and the plates. PROPERTIES: Plate material (given); ρ = 2500 kg/m3, c = 700 J/kg⋅K, k = 5 W/m⋅K. ANALYSIS: The temperature distribution for Case 2 shown in the above graph represents the initial condition for the period of time following the second loss of coolant event. The boundary conditions 2 at x = ±L are adiabatic, and the heater flux is maintained at q′′o = 4000 W/m for 0 ≤ t ≤ 15 min. Using FEHT, the heater is represented as a plate of thickness Lh = 0.5 mm with very low thermal capacitance (ρ = 1 kg/m and c = 1 J/kg⋅K), very high thermal conductivity (k= 10,000 W/m⋅K), and a 3

uniform volumetric generation rate of q = q′′o / L h = 4000 W / m 2 / 0.0005 m = 8.0 × 106 W/m for 0 ≤ t ≤ 900 s. In the Specify | Generation box, the generation was prescribed by the lookup file (see FEHT Help): ‘hfvst’,1,2,Time. This Notepad file is comprised of four lines, with the values on each line separated by a single tab space: 0 900 901 5000

8e6 8e6 0 0

The temperature-time histories are shown in the graph below for the surfaces x = - L (lowest curve, 13) and x = +L (19) and the center point x = 0 (highest curve, 14). The center point experiences the maximum temperature of 89°C at the time the heater is deactivated, t = 900 s. Continued …..

PROBLEM 5.110 For the finite-difference method of solution, the nodal arrangement for the system is shown below. The IHT model builder Tools | Finite-Difference Equations | One Dimensional can be used to obtain the FDEs for the internal nodes (02-04, 07-10) and the adiabatic boundary nodes (01, 11).

For the heater-plate interface node 06, the FDE for the implicit method is derived from an energy balance on the control volume shown in the schematic above.

E ′′in − E ′′out + E ′′gen = E ′′st q′′ + q′′ + q′′ = E ′′ a

b o st p +1 p +1 p +1 p +1 p +1 p T05 − T06 T07 T06 − T06 − T06 k +k + q′′o = ρ c∆x

∆x

∆x

∆t

The IHT code representing selected nodes is shown below for the adiabatic boundary node 01, interior node 02, and the heater-plates interface node 06. Note how the foregoing derived finite-difference equation in implicit form is written in the IHT Workspace. Note also the use of a Lookup Table for representing the heater flux vs. time. Continued …..

PROBLEM 5.110 (Cont.) // Finite-difference equations from Tools, Nodes 01, 02 /* Node 01: surface node (w-orientation); transient conditions; e labeled 02. */ rho * cp * der(T01,t) = fd_1d_sur_w(T01,T02,k,qdot,deltax,Tinf01,h01,q''a01) q''a01 = 0 // Applied heat flux, W/m^2; zero flux shown qdot = 0 // No internal generation Tinf01 = 20 // Arbitrary value h01 = 1e-6 // Causes boundary to behave as adiabatic /* Node 02: interior node; e and w labeled 03 and 01. */ rho*cp*der(T02,t) = fd_1d_int(T02,T03,T01,k,qdot,deltax) // Finite-difference equation from energy balance on CV, Node 06 k * (T05 - T06) / deltax + k * (T07 - T06)/ deltax + q''h = rho * cp * deltax * der(T06,t) q''h = LOOKUPVAL(qhvst,1,t,2) // Heater flux, W/m^2; specified by Lookup Table /* See HELP (Solver, Lookup Tables). The Look-up table file name "qhvst" contains 0 4000 900 4000 900.5 0 5000 0 */

The temperature-time histories using the IHT code for the plate locations x = 0, ±L are shown in the graphs below. We chose to show expanded presentations of the histories at early times, just after the second loss of coolant event, t = 0, and around the time the heater is deactivated, t = 900 s.

60 90

Temperature, T (C)

Temperature, T (C)

85

50

40

80

75

30

70

0

50

100

150

200

800

900

Time, t (s) Surface x = -L Center point, x = 0 Surface x = +L

1000

1100

1200

Time, t (s) Surface x = -L Center point, x = 0 Surface x = +L

COMMENTS: (1) The maximum temperature during the transient period is at the center point and occurs at the instant the heater is deactivated, T(0, 900s) = 89°C. After 300 s, note that the two surface temperatures are nearly the same, and never rise above the final steady-state temperature. (2) Both the FEHT and IHT methods of solution give identical results. Their steady-state solutions agree with the result of an energy balance on a time interval basis yielding Tss = 86.1°C.

PROBLEM 5.111 KNOWN: Plane wall of thickness 2L, initially at a uniform temperature, is suddenly subjected to convection heat transfer. FIND: The mid-plane, T(0,t), and surface, T(L,t), temperatures at t = 50, 100, 200 and 500 s, using the following methods: (a) the one-term series solution; determine also the Biot number; (b) the lumped capacitance solution; and (c) the two- and 5-node finite-difference numerical solutions. Prepare a table summarizing the results and comment on the relative differences of the predicted temperatures. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, and (2) Constant properties. ANALYSIS: (a) The results are tabulated below for the mid-plane and surface temperatures using the one-term approximation to the series solution, Eq. 5.40 and 5.41. The Biot number for the heat transfer process is

Bi = h L / k = 500 W / m 2 ⋅ K × 0.020 m / 15 W / m ⋅ K = 0.67 Since Bi >> 0.1, we expect an appreciable temperature difference between the mid-plane and surface as the tabulated results indicate (Eq. 5.10). (b) The results are tabulated below for the wall temperatures using the lumped capacitance method (LCM) of solution, Eq. 5.6. The LCM neglects the internal conduction resistance and since Bi = 0.67 >> 0.1, we expect this method to predict systematically lower temperatures (faster cooling) at the midplane compared to the one-term approximation. Solution method/Time(s)

50

100

200

500

Mid-plane, T(0,t) (°C) One-term, Eqs. 5.40, 5.41 Lumped capacitance 2-node FDE 5-node FDE

207.1 181.7 210.6 207.5

160.5 133.9 163.5 160.9

99.97 77.69 100.5 100.2

37.70 30.97 37.17 37.77

160.1 181.7 163.7 160.2

125.4 133.9 125.2 125.6

80.56 77.69 79.40 80.67

34.41 30.97 33.77 34.45

Surface, T(L,t) (°C) One-term, Eqs. 5.40, 5.41 Lumped capacitance 2-node FDE 5-node FDE

(c) The 2- and 5-node nodal networks representing the wall are shown in the schematic above. The implicit form of the finite-difference equations for the mid-plane, interior (if present) and surface nodes can be derived from energy balances on the nodal control volumes. The time-rate of change of the temperature is expressed in terms of the IHT integral intrinsic function, der(T,t). Continued …..

PROBLEM 5.111 (Cont.) Mid-plane node

k (T 2 − T 1) / ∆x = ρ c ( ∆x / 2 ) ⋅ der (T 1, t )

Interior node (5-node network)

k (T 1 − T 2 ) / ∆x + k (T 3 − T 2 ) / ∆x = ρ c ∆x ⋅ der (T 2, t )

Surface node (shown for 5-node network)

k (T 4 − T 5 ) / ∆x + h (T inf − T 5) = ρ c ( ∆x / 2 ) ⋅ der (T 5, t )

With appropriate values for ∆x, the foregoing FDEs were entered into the IHT workspace and solved for the temperature distributions as a function of time over the range 0 ≤ t ≤ 500 s using an integration time step of 1 s. Selected portions of the IHT codes for each of the models are shown in the Comments. The results of the analysis are summarized in the foregoing table. COMMENTS: (1) Referring to the table above, we can make the following observations about the relative differences and similarities of the estimated temperatures: (a) The one-term series model estimates are the most reliable, and can serve as the benchmark for the other model results; (b) The LCM model over estimates the rate of cooling, and poorly predicts temperatures since the model neglects the effect of internal resistance and Bi = 0.67 >> 0.1; (c) The 5-node model results are in excellent agreement with those from the one-term series solution; we can infer that the chosen space and time increments are sufficiently small to provide accurate results; and (d) The 2-node model under estimates the rate of cooling for early times when the time-rate of change is high; but for late times, the agreement is improved. (2) See the Solver | Intrinsic Functions section of IHT|Help or the IHT Examples menu (Example 5.3) for guidance on using the der(T,t) function. (3) Selected portions of the IHT code for the 2-node network model are shown below. // Writing the finite-difference equations – 2-node model // Node 1 k * (T2 - T1)/ deltax = rho * cp * (deltax / 2) * der(T1,t) // Node 2 k * (T1 - T2)/ deltax + h * (Tinf - T2) = rho * cp * (deltax / 2) * der(T2,t) // Input parameters L = 0.020 deltax = L rho = 7800 // density, kg/m^3 cp = 440 // specific heat, J/kg·K k = 15 // thermal conductivity, W/m.K h = 500 // convection coefficient, W/m^2·K Tinf = 25 // fluid temperature, K

(4) Selected portions of the IHT code for the 5-node network model are shown below. // Writing the finite-difference equations – 5-node model // Node 1 - midplane k * (T2 - T1)/ deltax = rho * cp * (deltax / 2) * der(T1,t) // Interior nodes k * (T1 - T2)/ deltax + k * (T3 - T2 )/ deltax = rho * cp * deltax * der(T2,t) k * (T2 - T3)/ deltax + k * (T4 - T3 )/ deltax = rho * cp * deltax * der(T3,t) k * (T3 - T4)/ deltax + k * (T5 - T4 )/ deltax = rho * cp * deltax * der(T4,t) // Node5 - surface k * (T4 - T5)/ deltax + h * (Tinf - T5) = rho * cp * (deltax / 2) * der(T5,t) // Input parameters L = 0.020 deltax = L / 4 …….. ……..

PROBLEM 5.112 2

KNOWN: Plastic film on metal strip initially at 25°C is heated by a laser (85,000 W/m for ∆ton = 10 s), to cure adhesive; convection conditions for ambient air at 25°C with coefficient 2

of 100 W/m ⋅K. FIND: Temperature histories at center and film edge, T(0,t) and T(x1,t), for 0 ≤ t ≤ 30 s, using an implicit, finite-difference method with ∆x = 4mm and ∆t = 1 s; determine whether adhesive is cured (Tc ≥ 90°C for ∆tc = 10s) and whether the degradation temperature of 200°C is exceeded. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Uniform convection coefficient on upper and lower surfaces, (4) Thermal resistance and mass of plastic film are negligible, (5) All incident laser flux is absorbed. 3

PROPERTIES: Metal strip (given): ρ = 7850 kg/m , cp = 435 J/kg⋅K, k = 60 W/m⋅K, α = -5 2 k/ρcp = 1.757 × 10 m /s. ANALYSIS: (a) Using a space increment of ∆x = 4mm, set up the nodal network shown below. Note that the film half-length is 22mm (rather than 20mm as in Problem 3.97) to simplify the finite-difference equation derivation.

Consider the general control volume and use the conservation of energy requirement to obtain the finite-difference equation. E in − E out = E st p T p+1 − Tm qa + q b + q laser + qconv = Mcp m ∆t Continued …..

PROBLEM 5.112 (Cont.) k ( d ⋅1)

p+1 p+1 Tm-1 − Tm

∆x

+ k (d ⋅1)

p+1 p+1 Tm+1 − Tm

(

∆x

)

p+1 p Tm − Tm p+1 ′′ + q o ( ∆x ⋅1) + 2h ( ∆x ⋅1) T∞ − Tm = ρ ( ∆x ⋅ d ⋅1) cp ∆t p p+1 Tm = (1 + 2Fo + 2Fo ⋅ Bi ) Tm

(1)

)

(

p+1 p+1 − Fo Tm+1 + Tm+1 − 2Fo ⋅ Bi ⋅ T∞ − Fo ⋅ Q

where Fo =

Bi =

Q=

α∆t ∆x 2

=

(

1.757 ×10−5 m 2 / s × 1s

h ∆x 2 / d k

(

= 1.098

(2)

) = 100 W/m2 ⋅ K (0.0042 / 0.00125) m = 0.0213 60 W/m ⋅ K

(3)

) = 85, 000 W/m2 (0.0042 / 0.0015) m = 18.133. 60 W/m ⋅ K

(4)

(0.004 m )2

q′′o ∆x 2 / d k

The results of the matrix inversion numerical method of solution (∆x = 4mm, ∆t = 1s) are shown below. The temperature histories for the center (m = 1) and film edge (m = 5) nodes, T(0,t) and T(x1,t), respectively, permit determining whether the adhesive has cured (T ≥ 90°C for 10 s).

Certainly the center region, T(0,t), is fully cured and furthermore, the degradation temperature (200°C) has not been exceeded. From the T(x1,t) distribution, note that ∆tc ≈ 8 sec, which is 20% less than the 10 s interval sought. Hence, the laser exposure (now 10 s) should be slightly increased and quite likely, the maximum temperature will not exceed 200°C.

PROBLEM 5.113 KNOWN: Insulated rod of prescribed length and diameter, with one end in a fixture at 200°C, reaches a uniform temperature. Suddenly the insulating sleeve is removed and the rod is subjected to a convection process. FIND: (a) Time required for the mid-length of the rod to reach 100°C, (b) Temperature history T(x,t ≤ t1), where t1 is time at which the midlength reaches 50°C. Temperature distribution at 0, 200s, 400s and t1. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional transient conduction in rod, (2) Uniform h along rod and at end, (3) Negligible radiation exchange between rod and surroundings, (4) Constant properties. ANALYSIS: (a) Choosing ∆x = 0.016 m, the finite-difference equations for the interior and end nodes are obtained. p +1 p Tm − Tm q a + q b + q c = ρ ⋅ A c ∆x ⋅ c p ⋅ Interior Point, m:

∆t

k ⋅ Ac

p p Tm −1 − Tm

∆x

+ kA c

p p Tm +1 − Tm

∆x

p +1 p Tm − Tm p + hP∆x T∞ − Tm = ρ A c ∆xc p

)

(

∆t

Regrouping, p +1 p p p Tm = Tm (1 − 2Fo − Bi ⋅ Fo ) + Fo Tm −1 + Tm +1 + Bi ⋅ FoT∞

(

where

Fo =

α∆t ∆x 2

(2)

)

Bi = h  ∆x 2 ( Ac P ) k .  

(1)

(3)

From Eq. (1), recognize that the stability of the numerical solution will be assured when the first term on the RHS is positive; that is Continued...

PROBLEM 5.113 (Cont.)

(1 − 2Fo − Bi ⋅ Fo ) ≥ 0

Fo ≤ 1 ( 2 + Bi ) .

or

(4)

p Nodal Point 1: Consider Eq. (1) for the special case that Tm −1 = To, which is independent of time. Hence, T1p +1 = T1p (1 − 2Fo − Bi ⋅ Fo ) + Fo To + T2p + Bi ⋅ FoT∞ . (5)

)

(

End Nodal Point 10:

p +1 p T10 − T10 ∆x qa + q b + qc = ρ ⋅ Ac ⋅ cp 2 ∆t

p +1 p ∆x ∆x T10 − T10 p p k ⋅ Ac + hA c T∞ − T10 + hP T∞ − T10 = ρ A c cp ∆x 2 2 ∆t p +1 p p Regrouping, T10 = T10 (1 − 2Fo − 2N ⋅ Fo − Bi ⋅ Fo ) + 2FoT9 + T∞ ( 2N ⋅ Fo + Bi ⋅ Fo ) where N = h∆x/k.

)

(

p T9p − T10

)

(

(6) (7)

The stability criterion is Fo ≤ 1/2(1 + N + Bi/2).

(8)

With the finite-difference equations established, we can now proceed with the numerical solution. Having already specified ∆x = 0.016 m, Bi can now be evaluated. Noting that Ac = πD2/4 and P = πD, giving Ac/P = D/4, Eq. (3) yields  2 0.010 m  Bi = 30 W m 2 ⋅ K ( 0.016 m ) (9)  14.8 W m ⋅ K = 0.208





4

From the stability criteria, Eqs. (4) and (8), for the finite-difference equations, it is recognized that Eq. (8) requires the greater value of Fo. Hence

1 0.208  Fo = 1 + 0.0324 +  = 0.440 2 2  where from Eq. (7), N =

(10)

30 W m 2 ⋅ K × 0.016 m = 0.0324 . 14.8 W m ⋅ K

(11)

From the definition of Fo, Eq. (2), we obtain the time increment 2 Fo ( ∆x ) 2 ∆t = = 0.440 0.016 m 3.63 × 10−6 m 2 s = 31.1s

(12)

and the time relation is t = p∆t = 31.1t.

(13)

α

(

)

Using the numerical values for Fo, Bi and N, the finite-difference equations can now be written (°C). Nodal Point m (2 ≤ m ≤ 9): p +1 p Tm = Tm (1 − 2 × 0.440 − 0.208 × 0.440 ) + 0.440 Tmp −1 + Tmp +1 + 0.208 × 0.440 × 25

)

(

(

)

p +1 p p p Tm = 0.029Tm + 0.440 Tm −1 + Tm +1 + 2.3 Nodal Point 1: + T1p 1 = 0.029T1p + 0.440 200 + T2p + 2.3 = 0.029T1p + 0.440T2p + 90.3

(

)

(14)

(15)

Nodal Point 10: p +1 p T10 = 0 × T10 + 2 × 0.440T9p + 25 ( 2 × 0.0324 × 0.440 + 0.208 × 0.440 ) = 0.880T9p + 3.0 (16) Continued...

PROBLEM 5.113 (Cont.) Using finite-difference equations (14-16) with Eq. (13), the calculations may be performed to obtain p 0 1 2 3 4 5 6 7 8

t(s) 0 31.1 62.2 93.3 124.4 155.5 186.6 217.7 248.8

T1 200 184.1 175.6 168.6 163.3 158.8 155.2 152.1 145.1

T2 200 181.8 166.3 154.8 145.0 137.1 130.2 124.5 119.5

T3 200 181.8 165.3 150.7 138.8 128.1 119.2 111.3 104.5

T4 200 181.8 165.3 150.7 137.0 125.3 114.8 105.7 97.6

T5 200 181.8 165.3 150.7 137.0 124.5 113.4 103.5 94.8

T6 200 181.8 165.3 150.7 137.0 124.3 113.0 102.9

T7 200 181.8 165.3 150.7 136.5 124.2 112.6 102.4

T8 200 181.8 165.3 149.7 136.3 123.4 112.3

T9 200 181.8 164.0 149.2 135.0 123.0 111.5

T10(°C) 200 179.0 163.0 147.3 134.3 121.8 111.2

<

Using linear interpolation between rows 7 and 8, we obtain T(L/2, 230s) = T5 ≈ 100°C.

(b) Using the option concerning Finite-Difference Equations for One-Dimensional Transient Conduction in Extended Surfaces from the IHT Toolpad, the desired temperature histories were computed for 0 ≤ t ≤ t1 = 930s. A Lookup Table involving data for T(x) at t = 0, 200, 400 and 930s was created. t(s)/x(mm) 0 200 400 930

0 200 200 200 200

16 200 157.8 146.2 138.1

32 200 136.7 114.9 99.23

48 200 127.0 97.32 74.98

64 200 122.7 87.7 59.94

80 200 121.0 82.57 50.67

96 200 120.2 79.8 44.99

112 200 119.6 78.14 41.53

128 200 118.6 76.87 39.44

144 200 117.1 75.6 38.2

160 200 114.7 74.13 37.55

and the LOOKUPVAL2 interpolating function was used with the Explore and Graph feature of IHT to create the desired plot. Temperature, T(C)

225 200 175 150 125 100 75 50 25 0

20

40

60

80

100

120

140

160

Fin location, x(mm) t= 0 t = 200 s t = 400 s t = 930 s

Temperatures decrease with increasing x and t, and except for early times (t < 200s) and locations in proximity to the fin tip, the magnitude of the temperature gradient, |dT/dx|, decreases with increasing x. The slight increase in |dT/dx| observed for t = 200s and x → 160 mm is attributable to significant heat loss from the fin tip. COMMENTS: The steady-state condition may be obtained by extending the finite-difference calculations in time to t ≈ 2650s or from Eq. 3.70.

PROBLEM 5.114 KNOWN: Tantalum rod initially at a uniform temperature, 300K, is suddenly subjected to a current flow of 80A; surroundings (vacuum enclosure) and electrodes maintained at 300K. FIND: (a) Estimate time required for mid-length to reach 1000K, (b) Determine the steadystate temperature distribution and estimate how long it will take to reach steady-state. Use a finite-difference method with a space increment of 10mm. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Surroundings are much larger than rod, (3) Properties are constant and evaluated at an average temperature. PROPERTIES: Table A-1, Tantalum ( T = (300+1000 ) K/2 = 650K ) : ρ = 16,600 kg/m , c 3

3

= 147 J/kg⋅K, k = 58.8 W/m⋅K, and α = k/ρc = 58.8 W/m⋅K/16,600 kg/m × 147 J/kg⋅K = -5 2

2.410 × 10 m /s. ANALYSIS: From the derivation of the previous problem, the finite-difference equation was found to be

)

(

p+1 p p p Tm = Fo Tm-1 + Tm+1 + (1 − 2Fo ) Tm −

Fo = α∆t/∆x 2

where

(

)

2 2 ε Pσ∆x 2 4,p 4 + I ρe ∆x ⋅ Fo Fo Tm − Tsur kAc kA c2

A c = π D2 / 4

P = π D.

(1) (2,3,4)

From the stability criterion, let Fo = 1/2 and numerically evaluate terms of Eq. (1). 2 0.1× 5.67 × 10−8 W/m 2 ⋅ K 4 × ( 0.01m ) 4 1 4,p 4 p+1 1 p p Tm = Tm-1 + Tm+1 − ⋅ Tm − [300K ] + 2 58.8 W/m ⋅ K × ( 0.003m ) 2

)

(

2 2 80A ) × 95 × 10−8 Ω ⋅ m (0.01m ) 1 ( + ⋅

(

58.8 W/m ⋅ K π [0.003m ] / 4

p+1 Tm =

(

2

)

)

2

(

)

2

1 p p 4,p Tm-1 + Tm+1 − 6.4285 × 10−12 Tm + 103.53. 2

(5)

Note that this form applies to nodes 0 through 5. For node 0, Tm-1 = Tm+1 = T1. Since Fo = 1/2, using Eq. (2), find that ∆t = ∆x 2 Fo/α = (0.01m ) × 1/ 2 / 2.410 × 10−5 m 2 / s = 2.07s. 2

(6)

Hence, t = p∆t = 2.07p.

(7) Continued …..

PROBLEM 5.114 (Cont.) (a) To estimate the time required for the mid-length to reach 1000K, that is To = 1000K, perform the forward-marching solution beginning with Ti = 300K at p = 0. The solution, as tabulated below, utilizes Eq. (5) for successive values of p. Elapsed time is determined by Eq. (7). P

t(s)

T0

T1

T2

T3

T4

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0

300 403.5 506.9 610.0 712.6 814.5 915.2 1010.9 1104.7 1190.9 1274.1 1348.2 1419.7 1479.8 1542.6 1605.3

300 403.5 506.9 610.0 712.6 814.5 911.9 1007.9 1096.8 1183.5 1261.6 1336.7 1402.4 1465.5 1538.2 1569.3

300 403.5 506.9 610.0 712.6 808.0 902.4 988.9 1073.8 1150.4 1224.9 1290.6 1353.9 1408.4 1460.9 1514.0

300 403.5 506.9 610.0 699.7 788.8 867.4 945.0 1014.0 1081.7 1141.5 1199.8 1250.5 1299.8 1341.2 1381.6

300 403.5 506.9 584.1 661.1 724.7 787.9 842.3 896.1 943.2 989.4 1029.9 1069.4 1103.6 1136.9 1164.8

10.4

20.7

31.1

T5

T6(°C)

300 403.5 455.1 506.7 545.2 583.5 615.1 646.6 673.6 700.3 723.6 746.5 766.5 786.0 802.9 819.3

Note that, at p ≈ 6.9 or t = 6.9 × 2.07 = 14.3s, the mid-point temperature is To ≈ 1000K.

300 300 300 300 300 300 300 300 300 300 300 300 300 300 300 300

<

(b) The steady-state temperature distribution can be obtained by continuing the marching solution until only small changes in Tm are noted. From the table above, note that at p = 15 or t = 31s, the temperature distribution is still changing with time. It is likely that at least 15 more calculation sets are required to see whether steady-state is being approached. COMMENTS: (1) This problem should be solved with a computer rather than a handcalculator. For such a situation, it would be appropriate to decrease the spatial increment in order to obtain better estimates of the temperature distribution. (2) If the rod were very long, the steady-state temperature distribution would be very flat at the mid-length x = 0. Performing an energy balance on the small control volume shown to the right, find E g − E out = 0 ρ ∆x 4 = 0. I2 e − εσ P∆x To4 − Tsur Ac

(

)

Substituting numerical values, find To = 2003K. It is unlikely that the present rod would ever reach this steady-state, maximum temperature. That is, the effect of conduction along the rod will cause the center temperature to be less than this value.

PROBLEM 5.115 KNOWN: Support rod spanning a channel whose walls are maintained at Tb = 300 K. Suddenly the rod is exposed to cross flow of hot gases with T∞ = 600 K and h = 75 W/m2⋅K. After the rod reaches steady-state conditions, the hot gas flow is terminated and the rod cools by free convection and radiation exchange with surroundings. FIND: (a) Compute and plot the midspan temperature as a function of elapsed heating time; compare the steady-state temperature distribution with results from an analytical model of the rod and (b) Compute the midspan temperature as a function of elapsed cooling time and determine the time required for the rod to reach the safe-to-touch temperature of 315 K. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Constant properties, (3) During heating process, uniform convection coefficient over rod, (4) During cooling process, free convection coefficient is of the form h = C∆Tn where C = 4.4 W/m2⋅K1.188 and n = 0.188, and (5) During cooling process, surroundings are large with respect to the rod. ANALYSIS: (a) The finite-difference equations for the 10-node mesh shown above can be obtained using the IHT Finite-Difference Equation, One-Dimensional, Transient Extended Surfaces Tool. The temperature-time history for the midspan position T10 is shown in the plot below. The steady-state temperature distribution for the rod can be determined from Eq. 3.75, Case B, Table 3.4. This case is treated in the IHT Extended Surfaces Model, Temperature Distribution and Heat Rate, Rectangular Pin Fin, for the adiabatic tip condition. The following table compares the steady-state temperature distributions for the numerical and analytical methods. Method Analytical Numerical

Temperatures (K) vs. Position x (mm) 10 20 30 40 386.1 443.4 479.5 499.4 386.0 443.2 479.3 499.2

0 300 300

50 505.8 505.6

The comparison is excellent indicating that the nodal mesh is sufficiently fine to obtain precise results.

Midspan temperature, T10 (K)

600

500

400

300 0

100

200

300

400

500

600

Elapsed heating time, t (s)

Continued...

PROBLEM 5.115 (Cont.) (b) The same finite-difference approach can be used to model the cooling process. In using the IHT tool, the following procedure was used: (1) Set up the FDEs with the convection coefficient expressed as hm = hfc,m + hr,m, the sum of the free convection and linearized radiation coefficients based upon nodal temperature Tm.

(

p h fc,m = C Tm − T∞

) )( )

(

 p 2 p 2  h r,m = εσ Tm + Tsur  Tm + Tsur     (2) For the initial solve, set hfc,m = hr,m = 5 W/m2⋅K and solve, (3) Using the solved results as the Initial Guesses for the next solve, allow hfc,m and hr,m to be unknowns. The temperature-time history for the midspan during the cooling process is shown in the plot below. The time to reach the safe-to-touch temperature, T10p = 315 K , is

<

t = 550 s

Midspan temperature, T10 (K)

600

500

400

300 0

200

400

600

Elapsed cooling time, t (s)

800

1000

PROBLEM 5.116 KNOWN: Thin metallic foil of thickness, w, whose edges are thermally coupled to a sink at temperature, Tsink, initially at a uniform temperature Ti = Tsink, is suddenly exposed on the top surface to an ion beam heat flux, qs′′ , and experiences radiation exchange with the vacuum enclosure walls at Tsur. Consider also the situation when the foil is operating under steady-state conditions when suddenly the ion beam is deactivated. FIND: (a) Compute and plot the midspan temperature-time history during the heating process; determine the elapsed time that this point on the foil reaches a temperature within 1 K of the steady-state value, and (b) Compute and plot the midspan temperature-time history during the cooling process from steady-state operation; determine the elapsed time that this point on the foil reaches the safe-to-touch temperature of 315 K. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, transient conduction in the foil, (2) Constant properties, (3) Upper and lower surfaces of foil experience radiation exchange with the large surroundings, (4) Ion beam incident on upper surface only, (4) Foil is of unit width normal to the page. ANALYSIS: (a) The finite-difference equations for the 10-node mesh shown above can be obtained using the IHT Finite-Difference Equation, One-Dimensional, Transient, Extended Surfaces Tool. In formulating the energy-balance functions, the following steps were taken: (1) the FDE function coefficient h must be identified for each node, e.g., h1 and (2) coefficient can be represented by the

(

)

2 linearized radiation coefficient, e.g., h1 = εσ (T1 + Tsur ) T12 + Tsur , (3) set q′′a = q′′o 2 since the ion

beam is incident on only the top surface of the foil, and (4) when solving, the initial condition corresponds to Ti = 300 K for each node. The temperature-time history of the midspan position is shown below. The time to reach within 1 K of the steady-state temperature (374.1 K) is

T10 ( t h ) = 373K

<

t h = 136s

(b) The same IHT workspace may be used to obtain the temperature-time history for the cooling process by taking these steps: (1) set q′′s = 0, (2) specify the initial conditions as the steady-state temperature (K) distribution tabulated below, T1 374.1

T2 374.0

T3 373.5

T4 372.5

T5 370.9

T6 368.2

T7 363.7

T8 356.6

T9 345.3

T10 327.4

(3) when performing the integration of the independent time variable, set the start value as 200 s and (4) save the results for the heating process in Data Set A. The temperature-time history for the heating and cooling processes can be made using Data Browser results from the Working and A Data Sets. The time required for the midspan to reach the safe-to-touch temperature is

T10 ( t c ) = 315 K

t c = 73s

< Continued...

PROBLEM 5.116 (Cont.) Midpsan temperature, T1 (K)

400

380

360

340

320

300 0

100

200

300

400

500

Heating or cooling time, t (s) Heating process Cooling process

COMMENTS: The IHT workspace using the Finite-Difference Equations Tool to determine the temperature-time distributions is shown below. Some of the lines of code were omitted to save space on the page. // Finite Difference Equations Tool: One-Dimensional, Transient, Extended Surface /* Node 1: extended surface interior node; transient conditions; e and w labeled 2 and 2. */ rho * cp * der(T1,t) = fd_1d_xsur_i(T1,T2,T2,k,qdot,Ac,P,deltax,Tinf, h1,q''a) q''a1 = q''s / 2 // Applied heat flux, W/m^2; on the upper surface only h1 = eps * sigma * (T1 + Tsur) * (T1^2 + Tsur^2) sigma = 5.67e-8 // Boltzmann constant, W/m^2.K^4 /* Node 2: extended surface interior node; transient conditions; e and w labeled 3 and 1. */ rho * cp * der(T2,t) = fd_1d_xsur_i(T2,T3,T1,k,qdot,Ac,P,deltax,Tinf, h2,q''a2) q''a2 = 0 // Applied heat flux, W/m^2; zero flux shown h2 = eps * sigma * (T2+ Tsur) * (T2^2 + Tsur^2) ....... ....... /* Node 10: extended surface interior node; transient conditions; e and w labeled sk and 9. */ rho * cp * der(T10,t) = fd_1d_xsur_i(T10,Tsk,T9,k,qdot,Ac,P,deltax,Tinf, h10,q''a) q''a10 = 0 // Applied heat flux, W/m^2; zero flux shown h10 = eps * sigma * (T10 + Tsur) * (T10^2 + Tsur^2) // Assigned variables deltax = L / 10 Ac = w * 1 P=2*1 L = 0.150 w = 0.00025 eps = 0.45 Tinf = Tsur Tsur = 300 k = 40 Tsk = 300 q''s = 600 q''s = 0 qdot = 0 alpha = 3e-5 rho = 1000 alpha = k / (rho * cp)

// Spatial increment, m // Cross-sectional area, m^2 // Perimeter, m // Overall length, m // Foil thickness, m // Foil emissivity // Fluid temperature, K // Surroundings temperature, K // Foil thermal conductivity // Sink temperature, K // Ion beam heat flux, W/m^2; for heating process // Ion beam heat flux, W/m^2; for cooling process // Foil volumetric generation rate, W/m^3 // Thermal diffusivity, m^2/s // Density, kg.m^3; arbitrary value // Definition

PROBLEM 5.117 KNOWN: Stack or book of steel plates (sp) and circuit boards (b) subjected to a prescribed platen heating schedule Tp(t). See Problem 5.42 for other details of the book. FIND: (a) Using the implicit numerical method with ∆x = 2.36mm and ∆t = 60s, find the mid-plane temperature T(0,t) of the book and determine whether curing will occur (> 170°C for 5 minutes), (b) Determine how long it will take T(0,t) to reach 37°C following reduction of the platen temperature to 15°C (at t = 50 minutes), (c) Validate code by using a sudden change of platen temperature from 15 to 190°C and compare with the solution of Problem 5.38. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible contact resistance between plates, boards and platens. 3

PROPERTIES: Steel plates (sp, given): ρsp = 8000 kg/m , cp,sp = 480 J/kg⋅K, ksp = 12 3 W/m⋅K; Circuit boards (b, given): ρb = 1000 kg/m , cp,b = 1500 J/kg⋅K, kb = 0.30 W/m⋅K. ANALYSIS: (a) Using the suggested space increment ∆x = 2.36mm, the model grid spacing treating the steel plates (sp) and circuit boards (b) as discrete elements, we need to derive the nodal equations for the interior nodes (2-11) and the node next to the platen (1). Begin by defining appropriate control volumes and apply the conservation of energy requirement. Effective thermal conductivity, ke: Consider an adjacent steel plate-board arrangement. The thermal resistance between the nodes i and j is ∆x ∆x/2 ∆x/2 = + ke kb ksp 2 2 = = W/m ⋅ K 1 / k b+ + 1 / k sp 1/0.3 + 1/12 = 0.585 W/m ⋅ K.

R ′′ij = ke ke

Odd-numbered nodes, 3 ≤ m ≤ 11 - steel plates (sp): Treat as interior nodes using Eq. 5.89 with ke 0.585 W/m ⋅ K = = 1.523 ×10 −7 m 2 / s ρsp csp 8000 kg/m3 × 480 J/kg ⋅ K αsp∆t 1.523 ×10 −7 m 2 / s × 60s Fo m = = = 1.641 2 ∆x 2 ( 0.00236 m )

αsp ≡

Continued …..

PROBLEM 5.117 (Cont.) to obtain, with m as odd-numbered,

(1 + 2Fom ) Tmp+1 − Fom

p+1 p+1 + Tm+1 ( Tm-1 ) = Tmp

(1)

Even-numbered nodes, 2 ≤ n ≤ 10 - circuit boards (b): Using Eq. 5.89 and evaluating α b and Fon k α b = e = 3.900 ×10−7 m 2 /s Fon = 4.201 ρ bc b

(1 + 2Fon ) Tnp+1 − Fon

p+1 p+1 + Tn+1 (Tn-1 ) = Tnp

(2)

Plate next to platen, n = 1 - steel plate (sp): The finite-difference equation for the plate node (n = 1) next to the platen follows from a control volume analysis. E& in − E& out = E& st q′′a + q′′b = ρsp ∆xc sp

T1p+1 − T1p ∆t

where q′′a = k sp

Tp ( t ) − T1p+1

T2p+1 − T1p+1

q′′b = k e ∆x/2 ∆x and Tp(t) = Tp(p) is the platen temperature which is changed with time according to the heating schedule. Regrouping find,   2k sp   p+1 p+1 2k sp FomTp ( p) = T1p  1 + Fom  1 +   T1 − Fom T2 − ke  ke   where 2ksp/ke = 2 × 12 W/m⋅K/0.585 W/m⋅K = 41.03.

(3)

Using the nodal Eqs. (1) -(3), an inversion method of solution was effected and the temperature distributions are shown on the following page. Temperature distributions - discussion: As expected, the temperatures of the nodes near the center of the book considerably lag those nearer the platen. The criterion for cure is T ≥ 170°C = 443 K for ∆tc = 5 min = 300 sec. From the temperature distributions, note that node 10 just reaches 443 K after 50 minutes and will not be cured. It appears that the region about node 5 will be cured. (b) The time required for the book to reach 37°C = 310 K can likewise be seen from the temperature distribution results. The plates/boards nearest the platen will cool to the safe handling temperature with 1000 s = 16 min, but those near the center of the stack will require in excess of 2000 s = 32 min. Continued …..

PROBLEM 5.117 (Cont.) (c) It is important when validating computer codes to have the program work a “problem” which has an exact analytical solution. You should select the problem such that all features of the code are tested.

PROBLEM 5.118 KNOWN: Reaction and composite clutch plates, initially at a uniform temperature, Ti = 40°C, are subjected to the frictional-heat flux shown in the engagement energy curve, q′′f vs. t . FIND: (a) On T-t coordinates, sketch the temperature histories at the mid-plane of the reaction plate, at the interface between the clutch pair, and at the mid-plane of the composite plate; identify key features; (b) Perform an energy balance on the clutch pair over a time interval basis and calculate the steady-state temperature resulting from a clutch engagement; (c) Obtain the temperature histories using the finite-element approach with FEHT and the finite-difference method of solution with IHT (∆x = 0.1 mm and ∆t = 1 ms). Calculate and plot the frictional heat fluxes to the reaction and composite plates, q′′rp and q′′cp , respectively, as a function of time. Comment on the features of the temperature and frictional-heat flux histories. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible heat transfer to the surroundings. 3

PROPERTIES: Steel, ρs = 7800 kg/m , cs = 500 J/kg⋅K, ks = 40 W/m⋅K; Friction material, ρfm = 3 1150 kg/m , cfm = 1650 J/kg⋅K, and kfm = 4 W/m⋅K. ANALYSIS: (a) The temperature histories for specified locations in the system are sketched on T-t coordinates below.

Initially, the temperature at all locations is uniform at Ti. Since there is negligible heat transfer to the surroundings, eventually the system will reach a uniform, steady-state temperature T(∞). During the engagement period, the interface temperature increases much more rapidly than at the mid-planes of the reaction (rp) and composite (cp) plates. The interface temperature should be the maximum within the system and could occur before lock-up, t = tlu. Continued …..

PROBLEM 5.118 (Cont.) (b) To determine the steady-state temperature following the engagement period, apply the conservation of energy requirement on the clutch pair on a time-interval basis, Eq. 1.11b.

The final and initial states correspond to uniform temperatures of T(∞) and Ti, respectively. The energy input is determined from the engagement energy curve, q′′f vs. t .

′′ E′′n − E′′out + E′′gen = ∆Est t lu

∫0

′′ = E′′out = 0 Ein

(

)

q′′f ( t ) dt = E′′f − E′′i =  ρscs L rp / 2 + Lcp / 2 + ρfm cfm Lfm  (Tf − Ti )  

Substituting numerical values, with Ti = 40°C and Tf = T(∞).

(

)

0.5 q′′o t lu =  ρs cs Lrp / 2 + Lcp / 2 + ρfm cfm Lfm  ( T (∞ ) − Ti )   0.5 × 1.6 × 107 W / m 2 × 0.100 s =  7800 kg / m3 × 500 J / kg ⋅ K ( 0.001 + 0.0005 ) m  +1150 kg / m3 × 1650 J / kg ⋅ K × 0.0005 m  ( T (∞ ) − 40 ) °C  T ( ∞ ) = 158°C

<

(c) Finite-element method of solution, FEHT. The clutch pair is comprised of the reaction plate (1 mm), an interface region (0.1 mm), and the composite plate (cp) as shown below.

Continued (2)...

PROBLEM 5.118 (Cont.) The external boundaries of the system are made adiabatic. The interface region provides the means to represent the frictional heat flux, specified with negligible thermal resistance and capacitance. The generation rate is prescribed as q = 1.6 ×1011 (1 − Time / 0.1) W / m3 0 ≤ Time ≤ t lu where the first coefficient is evaluated as q′′o / 0.1 × 10−3 m and the 0.1 mm parameter is the thickness of the region. Using the Run command, the integration is performed from 0 to 0.1 s with a time step of -6 1×10 s. Then, using the Specify|Generation command, the generation rate is set to zero and the Run|Continue command is executed. The temperature history is shown below.

(c) Finite-difference method of solution, IHT. The nodal arrangement for the clutch pair is shown below with ∆x = 0.1 mm and ∆t = 1 ms. Nodes 02-10, 13-16 and 18-21 are interior nodes, and their finite-difference equations (FDE) can be called into the Workspace using Tools|Finite Difference Equations|One-Dimenisonal|Transient. Nodes 01 and 22 represent the mid-planes for the reaction and composite plates, respectively, with adiabatic boundaries. The FDE for node 17 is derived from an energy balance on its control volume (CV) considering different properties in each half of the CV. The FDE for node 11 and 12 are likewise derived using energy balances on their CVs. At the interface, the following conditions must be satisfied

T11 = T12

q′′f = q′′rp + q′′cp

The frictional heat flux is represented by a Lookup Table, which along with the FDEs, are shown in the IHT code listed in Comment 2.

Continued (3)...

PROBLEM 5.118 (Cont.) The temperature and heat flux histories are plotted below. The steady-state temperature was found as 156.5° C, which is in reasonable agreement with the energy balance result from part (a).

Tem perature his tory for clutch pair, 100 m s lock-up tim e

Heat flux his tories for clutch pair during engagement 2 Heat flux, q'' (W/m^2 * 10^7)

250

Tem perature, T (C)

200

150

100

50

0

1.5

1

0.5

0

-0.5 0

200

400

600

800

1000

0

20

Engagem ent tim e, t (ms ) Midplane, reaction plate, T01 Interface, T11 or T12 Midplane, com pos ite plate, T22

40

60

80

Engagem ent tim e, t (ms ) Frictional heat flux, q''f Reaction plate, q''rp com pos ite plate, q''cp

COMMENTS: (1) The temperature histories resulting from the FEHT and IHT based solutions are in agreement. The interface temperature peaks near 225°C after 75 ms, and begins dropping toward the steady-state condition. The mid-plane of the reaction plate peaks around 100 ms, nearly reaching 200°C. The temperature of the mid-plane of the composite plate increases slowly toward the steadystate condition. (2) The calculated temperature-time histories for the clutch pair display similar features as expected from our initial sketches on T vs. t coordinates, part a. The maximum temperature for the composite is very high, subjecting the bonded frictional material to high thermal stresses as well as accelerating deterioration. For the reaction steel plate, the temperatures are moderate, but there is a significant gradient that could give rise to thermal stresses and hence, warping. Note that for the composite plate, the steel section is nearly isothermal and is less likely to experience warping. (2) The IHT code representing the FDE for the 22 nodes and the frictional heat flux relation is shown below. Note use of the Lookup Table for representing the frictional heat flux vs. time boundary condition for nodes 11 and 12. // Nodal equations, reaction plate (steel) /* Node 01: surface node (w-orientation); transient conditions; e labeled 02. */ rhos * cps * der(T01,t) = fd_1d_sur_w(T01,T02,ks,qdot,deltax,Tinf01,h01,q''a01) q''a01 = 0 // Applied heat flux, W/m^2; zero flux shown Tinf01 = 40 // Arbitrary value h01 = 1e-5 // Causes boundary to behave as adiabatic qdot = 0 /* Node 02: interior node; e and w labeled 03 and 01. */ rhos*cps*der(T02,t) = fd_1d_int(T02,T03,T01,ks,qdot,deltax) ……………………………. /* Node 10: interior node; e and w labeled 11 and 09. */ rhos*cps*der(T10,t) = fd_1d_int(T10,T11,T09,ks,qdot,deltax) /* Node 11: From an energy on the CV about node 11 */ ks * (T10 - T11) / deltax + q''rp = rhos * cps * deltax / 2 * der(T11,t)

Continued (4)...

100

PROBLEM 5.118 (Cont.) // Friction-surface interface conditions T11 = T12 q''f = LOOKUPVAL(HFVST16,1,t,2) // Applied heat flux, W/m^2; specified by Lookup Table /* See HELP (Solver, Lookup Tables). The look-up table, file name "HFVST16' contains 0 16e6 0.1 0 100 0 */ q''rp + q''cp = q''f // Frictional heat flux // Nodal equations - composite plate // Frictional material, nodes 12-16 /* Node 12: From an energy on the CV about node 12 */ kfm * (T13 - T12) / deltax + q''cp = rhofm * cpfm * deltax / 2 * der(T12,t) /* Node 13: interior node; e and w labeled 08 and 06. */ rhofm*cpfm*der(T13,t) = fd_1d_int(T13,T14,T12,kfm,qdot,deltax) ……………………………… /* Node 16: interior node; e and w labeled 11 and 09. */ rhofm*cpfm*der(T16,t) = fd_1d_int(T16,T17,T15,kfm,qdot,deltax) // Interface between friction material and steel, node 17 /* Node 17: From an energy on the CV about node 17 */ kfm * (T16 - T17) / deltax + ks * (T18 - T17) / deltax = RHS RHS = ( (rhofm * cpfm * deltax /2) + (rhos * cps * deltax /2) ) * der(T17,t) // Steel, nodes 18-22 /* Node 18: interior node; e and w labeled 03 and 01. */ rhos*cps*der(T18,t) = fd_1d_int(T18,T19,T17,ks,qdot,deltax) ………………………………. /* Node 22: interior node; e and w labeled 21 and 21. Symmetry condition. */ rhos*cps*der(T22,t) = fd_1d_int(T22,T21,T21,ks,qdot,deltax) // qdot = 0 // Input variables // Ti = 40 deltax = 0.0001 rhos = 7800 cps = 500 ks = 40 rhofm = 1150 cpfm = 1650 kfm = 4

// Initial temperature; entered during Solve // Steel properties

//Friction material properties

// Conversions, to facilitate graphing t_ms = t * 1000 qf_7 = q''f / 1e7 qrp_7 = q''rp / 1e7 qcp_7 = q''cp / 1e7

PROBLEM 5.119 KNOWN: Hamburger patties of thickness 2L = 10, 20 and 30 mm, initially at a uniform temperature Ti = 20°C, are grilled on both sides by a convection process characterized by T∞ = 100°C and h = 5000 W/m2⋅K. FIND: (a) Determine the relationship between time-to-doneness, td, and patty thickness. Doneness criteria is 60°C at the center. Use FEHT and the IHT Models|Transient Conduction|Plane Wall. (b) Using the results from part (a), estimate the time-to-doneness if the initial temperature is 5 °C rather than 20°C. Calculate values using the IHT model, and determine the relationship between time-todoneness and initial temperature. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, and (2) Constant properties are approximated as those of water at 300 K. 3

PROPERTIES: Table A-6, Water (300K), ρ = 1000 kg/m , c = 4179 J/kg⋅K, k = 0.613 W/m⋅K. ANALYSIS: (a) To determine T(0, td), the center point temperature at the time-to-doneness time, td, a one-dimensional shape as shown in the FEHT screen below is drawn, and the material properties, boundary conditions, and initial temperature are specified. With the Run|Calculate command, the early integration steps are made very fine to accommodate the large temperature-time changes occurring near x = L. Use the RunContinue command (see FEHT HELP) for the second and subsequent steps of the integration. This sequence of Start-(Step)-Stop values was used: 0 (0.001) 0.1 (0.01 ) 1 (0.1) 120 (1.0) 840 s.

Continued …..

PROBLEM 5.119 (Cont.) Using the View|Temperature vs. Time command, the temperature-time histories for the x/L = 0 (center), 0.5, and 1.0 (grill side) are plotted and shown below for the 2L = 10 mm thick patty.

Using the View|Temperatures command, the time slider can be adjusted to read td, when the center point, x = 0, reaches 60°C. See the summary table below. The IHT ready-to-solve model in Models|Transient Conduction|Plane Wall is based upon Eq. 5.40 and permits direct calculation of td when T(0,td) = 60°C for patty thickness 2L = 10, 20 and 30 mm and initial temperatures of 20 and 5°C. The IHT code is provided in Comment 3, and the results are tabulated below. Solution method

Time-to-doneness, t (s)

Ti ( C)

Patty thickness, 2L (mm)

FEHT IHT Eq. 5.40 (see Comment 4)

10

20

30

66.2 67.7 80.2 x

264.5 264.5 312.2 x x

591 590.4 699.1 x

20 20 5 5 20

Considering the IHT results for Ti = 20°C, note that when the thickness is doubled from 10 to 20 mm, td is (264.5/67.7=) 3.9 times larger. When the thickness is trebled, from 10 to 30 mm, td is 2 (590.4/67.7=) 8.7 times larger. We conclude that, td is nearly proportional to L , rather than linearly proportional to thickness. Continued …..

PROBLEM 5.119 (Cont.) (b) The temperature span for the cooking process ranges from T∞ = 100 to Ti = 20 or 5°C. The differences are (100-20 =) 80 or (100-5 =) 95°C. If td is proportional to the overall temperature span, then we expect td for the cases with Ti = 5°C to be a factor of (95/80 =) 1.19 higher (approximately 20% ) than with Ti = 20°C. From the tabulated results above, for the thickness 2L = 10, 20 and 30 mm, the td with Ti = 5°C are (80.2/67.7 = ) 1.18, (312 / 264.5 =) 1.18, and (699.1/590.4 =) 1.18, respectively, higher than with Ti = 20°C. We conclude that td is nearly proportional to the temperature span (T∞ - Ti). COMMENTS: (1) The results from the FEHT and IHT calculations are in excellent agreement. For this analysis, the FEHT model is more convenient to use as it provides direct calculations of the timeto-doneness. The FEHT tool allows the user to watch the cooking process. Use the ViewTemperature Contours command, click on the from start-to-stop button, and observe how color band changes represent the temperature distribution as a function of time. (2) It is good practice to check software tool analyses against hand calculations. Besides providing experience with the basic equations, you can check whether the tool was used or functioned properly. Using the one-term series solution, Eq. 5.40: T ( 0, t d ) − T∞ = C1 exp −ζ 2 Fo θ o∗ = Ti − T∞

)

(

C1, ζ = ( Bi ) , Table 5.1

Fo = α t d / L2 Ti (°C)

2L (mm)

θ o∗

Bi

C1

ζ1

Fo

td (s)

20 5

10 30

0.5000 0.4211

24.47 73.41

1.2707 1.2729

1.5068 1.5471

0.4108 0.4622

70.0 709

The results are slightly higher than those from the IHT model, which is based upon a multiple- rather than single-term series solution. (3) The IHT code used to obtain the tabulated results is shown below. Note that T_xt_trans is an intrinsic heat transfer function dropped into the Workspace from the Models window (see IHT Help|Solver|Intrinsic Functions|Heat Transfer Functions). // Models | Transient Conduction | Plane Wall /* Model: Plane wall of thickness 2L, initially with a uniform temperature T(x,0) = Ti, suddenly subjected to convection conditions (Tinf,h). */ // The temperature distribution is T_xt = T_xt_trans("Plane Wall",xstar,Fo,Bi,Ti,Tinf) // Eq 5.39 // The dimensionless parameters are xstar = x / L Bi = h * L / k // Eq 5.9 Fo= alpha * t / L^2 // Eq 5.33 alpha = k/ (rho * cp) // Input parameters x=0 // Center point of meat L = 0.005 // Meat half-thickness, m //L = 0.010 //L = 0.015 T_xt = 60 // Doneness temperature requirement at center, x = 0; C Ti = 20 // Initial uniform temperature //Ti = 5 rho = 1000 // Water properties at 300 K cp = 4179 k = 0.613 h = 5000 // Convection boundary conditions Tinf = 100

PROBLEM 5.120 KNOWN: A process mixture at 200°C flows at a rate of 207 kg/min onto a 1-m wide conveyor belt traveling with a velocity of 36 m/min. The underside of the belt is cooled by a water spray. FIND: The surface temperature of the mixture at the end of the conveyor belt, Te,s, using (a) IHT for writing and solving the FDEs, and (b) FEHT. Validate your numerical codes against an appropriate analytical method of solution. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction at any z-location, (2) Negligible heat transfer from mixture upper surface to ambient air, and (3) Constant properties. 3

PROPERTIES: Process mixture (m), ρm = 960 kg/m , cm = 1700 J/kg⋅K, and km = 1.5 W/m⋅K; 3 Conveyor belt (b), ρb = 8000 kg/ m , cb = 460 J/kg⋅K, and kb = 15 W/m⋅K. ANALYSIS: From the conservation of mass requirement, the thickness of the mixture on the conveyor belt can be determined.

 = ρm Ac V m

where

Ac = W Lm

207 kg / min× 1 min/ 60s = 960 kg / m3 × 1m × L m × 36 m / min×1min/ 60s Lm = 0.0060 m = 6 mm The time that the mixture is in contact with the steel conveyor belt, referred to as the residence time, is

t res = Lc / V = 30 m / (36 m / min× 1 min/ 60 s ) = 50 s The composite system comprised of the belt, Lb = 3 mm, and mixture, Lm = 6 mm, as represented in the schematic above, is initially at a uniform temperature T(x,0) = Ti = 200°C while at location z = 0, and suddenly is exposed to convection cooling (T∞, h). We will calculate the mixture upper surface temperature after 50 s, T(0, tres) = Te,s . (a) The nodal arrangement for the composite system is shown in the schematic below. The IHT model builder Tools|Finite-Difference Equations|Transient can be used to obtain the FDEs for nodes 01-12 and 14-19. Continued …..

PROBLEM 5.120 (Cont.)

For the mixture-belt interface node 13, the FDE for the implicit method is derived from an energy balance on the control volume about the node as shown above.

E ′′in − E ′′out = E ′′st q′′a + q′′b = E ′′st,m + E ′′st,b km

p +1 p +1 T12 − T13

∆x

+ kb

p +1 p +1 T14 − T13

∆x

= ( ρ mc m + ρbc b )( ∆x / 2 )

p +1 p T13 − T13

∆t

IHT code representing selected FDEs, nodes 01, 02, 13 and 19, is shown in Comment 4 below (∆x = 0.5 mm, ∆t = 0.1 s). Note how the FDE for node 13 derived above is written in the Workspace. From the analysis, find

<

Te,s = T(0, 50s) = 54.8°C

(b) Using FEHT, the composite system is drawn and the material properties, boundary conditions, and initial temperature are specified. The screen representing the system is shown below in Comment 5 with annotations on key features. From the analysis, find

<

Te,s = T(0, 50s) = 54.7°C

COMMENTS: (1) Both numerical methods, IHT and FEHT, yielded the same result, 55°C. For the safety of plant personnel working in the area of the conveyor exit, the mixture exit temperature should be lower, like 43°C. (2) By giving both regions of the composite the same properties, the analytical solution for the plane wall with convection, Section 5.5, Eq. 5.40, can be used to validate the IHT and FEHT codes. Using the IHT Models|Transient Conduction|Plane Wall for a 9-mm thickness wall with mixture thermophysical properties, we calculated the temperatures after 50 s for three locations: T(0, 50s) = 91.4°C; T(6 mm, 50s) = 63.6°C; and T(3 mm, 50s) = 91.4°C. The results from the IHT and FEHT codes agreed exactly. (3) In view of the high heat removal rate on the belt lower surface, it is reasonable to assume that negligible heat loss is occurring by convection on the top surface of the mixture. Continued …..

PROBLEM 5.120 (Cont.) (4) The IHT code representing selected FDEs, nodes 01, 02, 13 and 19, is shown below. The FDE for node 13 was derived from an energy balance, while the others are written from the Tools pad. // Finite difference equations from Tools, Nodes 01 -12 (mixture) and 14-19 (belt) /* Node 01: surface node (w-orientation); transient conditions; e labeled 02. */ rhom * cm * der(T01,t) = fd_1d_sur_w(T01,T02,km,qdot,deltax,Tinf01,h01,q''a01) q''a01 = 0 // Applied heat flux, W/m^2; zero flux shown qdot = 0 Tinf01 = 20 // Arbitrary value h01 = 1e-6 // Causes boundary to behave as adiabatic /* Node 02: interior node; e and w labeled 03 and 01. */ rhom*cm*der(T02,t) = fd_1d_int(T02,T03,T01,km,qdot,deltax) /* Node 19: surface node (e-orientation); transient conditions; w labeled 18. */ rhob * cb * der(T19,t) = fd_1d_sur_e(T19,T18,kb,qdot,deltax,Tinf19,h19,q''a19) q''a19 = 0 // Applied heat flux, W/m^2; zero flux shown Tinf19 = 30 h19 = 3000 // Finite-difference equation from energy balance on CV, Node 13 km*(T12 - T13)/deltax + kb*(T14 - T13)/deltax = (rhom*cm + rhob*cb) *(deltax/2)*der(T13,t)

(5) The screen from the FEHT analysis is shown below. It is important to use small time steps in the integration at early times. Use the View|Temperatures command to find the temperature of the mixture surface at tres = 50 s.

PROBLEM 5.121 KNOWN: Thin, circular-disc subjected to induction heating causing a uniform heat generation in a prescribed region; upper surface exposed to convection process. FIND: (a) Transient finite-difference equation for a node in the region subjected to induction heating, (b) Sketch the steady-state temperature distribution on T-r coordinates; identify important features. SCHEMATIC:

ASSUMPTIONS: (1) Thickness w > 1/2 and (m∆φ) >>1, the FDE takes the form of a 1-D cartesian system. Center Node (0,0). For the control volume, V = π ( ∆r/2 ) ⋅1. The energy balance is 2

E ′in − E ′out + E ′g = E ′st where E ′in = Σq′n . p

p

T1,n − To ∆r ∆r ∑ k ⋅  2 ∆φ  ⋅ ∆r + q π  2  n =0 N

= ρ c ⋅π

2

p 2 p+1  ∆r  To − To  2  ∆t

(5)

where N = (2π/∆φ) - 1, the total number of qn. Using the definition of Fo, find  q  1 N p p+1 2 p ∆r  + (1 − 4Fo ) To . < To = 4Fo  T1,n + ∑ N 1 4k +   n =0 By inspection, the stability criterion is Fo ≤ 1/4. (7) Surface Nodes (M,n). The control volume for the surface node is V = (M - ¼)∆r∆φ⋅∆r/2.1. From the energy balance,  = E ′st E ′in − E ′out + E ′g = (q1′ + q′2 )r + ( q′3 + q′4 )φ + qV

k ⋅ ( M − 1/2 ) ∆r ⋅ ∆φ p

p

p

TM-1,n − TM,n ∆r

+ h ( M∆r ⋅ ∆φ )

(

)

p p ∆r TM,n+1 − TM,n p ⋅ T∞ − TM,n +k ⋅

2

p

( M∆r ) ∆φ p+1

p

∆r TM,n-1 − TM,n ∆r  ∆r  TM,n − TM,n   . +k ⋅ ⋅ + q ( M − 1/4 ) ∆r ⋅ ∆φ ⋅ = ρ c ( M − 1/4 ) ∆r ⋅ ∆φ ⋅    2 2  2     ( M∆r ) ∆φ ∆t

2

Regrouping and using the definitions for Fo = α∆t/∆r and Bi = h∆r/k,  M − 1/2  1 q p p p p+1 Tm,n TM-1,n + TM,n+1 − TM,n-1 + 2Bi ⋅ T∞ + ∆r 2  = Fo 2 k ( M-1/4 ) M ( ∆φ )2  M − 1/4    M-1/2   M 1 p   TM,n + 1 − 2Fo  + Bi ⋅ + (8) < . M − 1/4 ( M − 1/4 ) M ( ∆φ )2    M-1/4    1  M − 1/2 M 1  . (9) The stability criterion is + Bi + Fo ≤  2  M − 1/4 M − 1/4 ( M − 1/4 ) M ( ∆φ )2   

(

)

To determine which stability criterion is most restrictive, compare Eqs. (4), (7) and (9). The most restrictive (lowest Fo) has the largest denominator. For small values of m, it is not evident whether Eq. (7) is more restrictive than Eq. (4); Eq. (4) depends upon magnitude of ∆φ. Likewise, it is not clear whether Eq. (9) will be more or less restrictive than Eq. (7). Numerical values must be substituted.

PROBLEM 5.123 KNOWN: Initial temperature distribution in two bars that are to be soldered together; interface contact resistance. FIND: (a) Explicit FDE for T4,2 in terms of Fo and Bi = ∆x/k R ′′t,c ; stability criterion, (b) T4,2 one time step after contact is made if Fo = 0.01 and value of ∆t; whether the stability criterion is satisfied. SCHEMATIC:

PROPERTIES: Table A-1, Steel, AISI 1010 (1000K): k = 31.3 W/m⋅K, c = 1168 J/kg⋅K, ρ = 7832 3 kh/m . ASSUMPTIONS: (1) Two-dimensional transient conduction, (2) Constant properties, (3) Interfacial solder layer has negligible thickness. ANALYSIS: (a) From an energy balance on the control volume V = (∆x/2)⋅∆y⋅1. E in − E out + E g = E st p+1

p

T4,2 − T4,2 q a + q b + q c + q d = ρ cV . ∆t Note that qa = ∆T/R ′′t,c Ac while the remaining qi are conduction terms,

(

1 R ′′t,c

(

)

)

(

p p T3,2 − T4,2 ∆y + k ( ∆x/2 )

( + k ( ∆x/2 )

p

p

T4,3 − T4,2

∆y p p T4,1 − T4,2

∆y

)

+ k ( ∆y

( )

p

p

T5,2 − T4,2

)

∆x

) = ρ c [(∆x / 2 )⋅ ∆y] T

p+1 p 4,2 − T4,2

∆t

.

Defining Fo ≡ ( k/ρ c ) ∆t/∆x 2 and Bic ≡ ∆y/R ′′t,c k, regroup to obtain

(

)

T4,2 = Fo T4,3 + 2T5,2 + T4,1 + 2Bi T3,2 + (1 − 4Fo − 2FoBi ) T4,2 . p+1

p

p

p

p

p

<

p

The stability criterion requires the coefficient of the T4,2 term be zero or positive,

(1 − 4Fo − 2FoBi ) ≥ 0

Fo ≤ 1/ ( 4 + 2Bi )

or

(

)

<

(b) For Fo = 0.01 and Bi = 0.020m/ 2 × 10-5m 2 ⋅ K/W × 31.3W/m ⋅ K = 31.95, T4,2 = 0.01 (1000 + 2 × 900 + 1000 + 2 × 31.95 × 700 ) K + (1 − 4 × 0.01 − 2 × 0.01 × 31.95 )1000K p+1

p+1

T4,2 = 485.30K + 321.00K = 806.3K. With Fo = 0.01, the time step is ∆t = Fo ∆x 2 ( ρ c/k ) = 0.01 ( 0.020m )

2

(7832kg/m3 ×1168J/kg ⋅ K/31.3W/m ⋅ K ) = 1.17s.

With Bi = 31.95 and Fo = 0.01, the stability criterion, Fo ≤ 0.015, is satisfied.

< < <

PROBLEM 5.124 KNOWN: Stainless steel cylinder of Ex. 5.7, 80-mm diameter by 60-mm length, initially at 600 K, 2 suddenly quenched in an oil bath at 300 K with h = 500 W/m ⋅K. Use the ready-to-solve model in the Examples menu of FEHT to obtain the following solutions. FIND: (a) Calculate the temperatures T(r, x ,t) after 3 min: at the cylinder center, T(0, 0, 3 mm), at the center of a circular face, T(0, L, 3 min), and at the midheight of the side, T(ro, 0, 3 min); compare your results with those in the example; (b) Calculate and plot temperature histories at the cylinder center, T(0, 0, t), the mid-height of the side, T(ro, 0, t), for 0 ≤ t ≤ 10 min; use the View/Temperature vs. Time command; comment on the gradients and what effect they might have on phase transformations and thermal stresses; (c) Using the results for the total integration time of 10 min, use the View/Temperature Contours command; describe the major features of the cooling process shown in this display; create and display a 10-isotherm temperature distribution for t = 3 min; and (d) For the locations of part (a), calculate the temperatures after 3 min if the convection coefficient is doubled (h = 2 1000 W/m ⋅K); for these two conditions, determine how long the cylinder needs to remain in the oil bath to achieve a safe-to touch surface temperature of 316 K. Tabulate and comment on the results of your analysis. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction in r- and x-coordinates, (2) Constant properties. 3

PROPERTIES: Stainless steel (Example 5.7): ρ = 7900 kg/m , c = 256 J/kg⋅K, k = 17.4 W/m⋅K. ANALYSIS: (a) The FEHT ready-to-solve model for Example 5.7 is accessed through the Examples menu and the annotated Input page is shown below. The following steps were used to obtain the solution: (1) Use the DrawReduce Mesh command three times to create the 512-element mesh; (2) In Run, click on Check, (3) In Run, press Calculate and hit OK to initiate the solver; and (4) Go to the View menu, select Tabular Output and read the nodal temperatures 4, 1, and 3 at t = to = 180 s. The tabulated results below include those from the n-term series solution used in the IHT software. Continued …..

PROBLEM 5.124 (Cont.) (r, x, to)

0, 0, to 0, L, to ro, 0, to

FEHT node

4 1 3

T(r, x, to) (K) FEHT 402.7 368.7 362.5

T(r, x, to) (K) 1-term series 405 372 365

T(r, x, to) (K) n-term series 402.7 370.5 362.4

Note that the one-term series solution results of Example 5.7 are systematically lower than those from the 512-element, finite-difference FEHT analyses. The FEHT results are in excellent agreement with the IHT n-term series solutions for the x = 0 plane nodes (4,3), except for the x = L plane node (1).

(b) Using the View Temperature vs. Time command, the temperature histories for nodes 4, 1, and 3 are plotted in the graph shown below. There is very small temperature difference between the locations on the surface, (node 1; 0, L) and (node 3; ro, 0). But, the temperature difference between these surface locations and the cylinder center (node 4; 0, 0) is large at early times. Such differences wherein locations cool at considerably different rates could cause variations in microstructure and hence, mechanical properties, as well as induce thermal stresses.

Continued …..

PROBLEM 5.124 (Cont.) (c) Use the View|Temperature Contours command with the shaded band option for the isotherm contours. Selecting the From Start to Stop time option, see the display of the contours as the cylinder cools during the quench process. The “movie” shows that cooling initiates at the corner (ro,L,t) and the isotherms quickly become circular and travel toward the center (0,0,t). The 10-isotherm distribution for t = 3 min is shown below.

2

(d) Using the FEHT model with convection coefficients of 500 and 1000 W/m ⋅K, the temperatures at t = to = 180 s for the three locations of part (a) are tabulated below. 2

2

h = 500 W/m ⋅K T(0, 0, to), K T(0, L, to), K T(ro, 0, to), K

h = 1000 W/m ⋅K

402.7 368.7 362.5

352.8 325.8 322.1

Note that the effect of doubling the convection coefficient is to reduce the temperature at these locations by about 40°C. The time the cylinder needs to remain in the oil bath to achieve the safe-totouch surface temperature of 316 K can be determined by examining the temperature history of the location (node1; 0, L). For the two convection conditions, the results are tabulated below. Doubling the coefficient reduces the cooling process time by 40 %. T(0, L, to) 316 316

2

h (W/m ⋅K) 500 1000

to (s) 370 219

PROBLEM 5.125 KNOWN: Flue of square cross-section, initially at a uniform temperature is suddenly exposed to hot flue gases. See Problem 4.57. FIND: Temperature distribution in the wall 5, 10, 50 and 100 hours after introduction of gases using the implicit finite-difference method.

SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional transient conduction, (2) Constant properties. -7

2

PROPERTIES: Flue (given): k = 0.85 W/m⋅K, α = 5.5 × 10 m /s. ANALYSIS: The network representing the flue cross-sectional area is shown with ∆x = ∆y = 50mm. Initially all nodes are at Ti = 25°C when suddenly the interior and exterior surfaces are exposed to convection processes, (T∞,i, hi) and (T∞,o, ho), respectively Referring to the network above, note that there are four types of nodes: interior (02, 03, 06, 07, 10, 11, 14, 15, 17, 18, 20); plane surfaces with convection (interior – 01, 05, 09); interior corner with convection (13), plane surfaces with convection (exterior – 04, 08, 12, 16, 19, 21); and, exterior corner with convection. The system of finitedifference equations representing the network is obtained using IHT|Tools|Finite-difference equations|Two-dimensional|Transient. The IHT code is shown in Comment 2 and the results for t = 5, 10, 50 and 100 hour are tabulated below.

(T18p+1 + T14p+1 + T18p+1 + T14p+1 ) = T17p

Node 17

p+1 − Fo (1 + 4Fo ) T17

Node 13

1 + 4Fo 1 + 1 Bi  T p+1 − 2 Fo 2T p+1 + T + 2T p+1 + T p+1 = T p + 4 Bi ⋅ Fo ⋅ T 9 i ∞ ,i 13 14 14 9 13   3 i  3 3 

(

)

(1 + 2Fo (2 + Bio )) T12p+1 − Fo (2T11p+1 + T16p+1 + T8p+1 ) = T12p + 2Bio ⋅ Fo ⋅ T∞,o

Node 12

p+1 p+1 p+1 p − 2Fo (T21 + T21 ) = T22 + 4Bio ⋅ Fo ⋅ T∞,o (1 + 4Fo (1 + Bio )) T22

Node 22

Numerical values for the relevant parameters are: α∆t 5.5 × 10−6 m2 / s × 3600s

Fo =

∆x 2

=

(0.050m )

2

= 7.92000

h ∆x 5 W/m 2 ⋅ K × 0.050m Bio = o = = 0.29412 k 0.85 W/m ⋅ K h ∆x 100 W/m 2 ⋅ K × 0.050m Bii = i = = 5.88235 k 0.85 W/m ⋅ K The system of FDEs can be represented in matrix notation, [A][T] = [C]. The coefficient matrix [A] and terms for the right-hand side matrix [C] are given on the following page.

Continued …..

PROBLEM 5.125 (Cont.)

For this problem a stock computer program was used to obtain the solution matrix [T]. The 0 = 298K. The results are tabulated below. initial temperature distribution was Tm T(m,n) (C) Node/time (h) T01

0

5

10

50

100

25

335.00

338.90

340.20

340.20

T02

25

248.00

274.30

282.90

282.90

T03

25

179.50

217.40

229.80

229.80

T04

25

135.80

170.30

181.60

181.60

T05

25

334.50

338.50

339.90

339.90

T06

25

245.30

271.90

280.80

280.80

T07

25

176.50

214.60

227.30

227.30

T08

25

133.40

168.00

179.50

179.50

T09

25

332.20

336.60

338.20

338.20

T10

25

235.40

263.40

273.20

273.20

T11

25

166.40

205.40

219.00

219.00

T12

25

125.40

160.40

172.70

172.70

T13

25

316.40

324.30

327.30

327.30

T14

25

211.00

243.00

254.90

254.90

T15

25

146.90

187.60

202.90

202.90

T16

25

110.90

146.70

160.20

160.20

T17

25

159.80

200.50

216.20

216.20

T18

25

117.40

160.50

177.50

177.50

T19

25

90.97

127.40

141.80

141.80

T20

25

90.62

132.20

149.00

149.00

T21

25

72.43

106.70

120.60

120.60

T22

25

59.47

87.37

98.89

98.89

COMMENTS: (1) Note that the steady-state condition is reached by t = 5 hours; this can be seen by comparing the distributions for t = 50 and 100 hours. Within 10 hours, the flue is within a few degrees of the steady-state condition. Continued …..

PROBLEM 5.125 (Cont.) (2) The IHT code for performing the numerical solution is shown in its entirety below. Use has been made of symmetry in writing the FDEs. The tabulated results above were obtained by copying from the IHT Browser and pasting the desired columns into EXCEL. // From Tools|Finite-difference equations|Two-dimensional|Transient // Interior surface nodes, 01, 05, 09, 13 /* Node 01: plane surface node, s-orientation; e, w, n labeled 05, 05, 02 . */ rho * cp * der(T01,t) = fd_2d_psur_s(T01,T05,T05,T02,k,qdot,deltax,deltay,Tinfi,hi,q''a) q''a = 0 // Applied heat flux, W/m^2; zero flux shown qdot = 0 rho * cp * der(T05,t) = fd_2d_psur_s(T05,T09,T01,T06,k,qdot,deltax,deltay,Tinfi,hi,q''a) rho * cp * der(T09,t) = fd_2d_psur_s(T09,T13,T05,T10,k,qdot,deltax,deltay,Tinfi,hi,q''a) /* Node 13: internal corner node, w-s orientation; e, w, n, s labeled 14, 09, 14, 09. */ rho * cp * der(T13,t) = fd_2d_ic_ws(T13,T14,T09,T14,T09,k,qdot,deltax,deltay,Tinfi,hi,q''a) // Interior nodes, 02, 03, 06, 07, 10, 11, 14, 15, 18, 20 /* Node 02: interior node; e, w, n, s labeled 06, 06, 03, 01. */ rho * cp * der(T02,t) = fd_2d_int(T02,T06,T06,T03,T01,k,qdot,deltax,deltay) rho * cp * der(T03,t) = fd_2d_int(T03,T07,T07,T04,T02,k,qdot,deltax,deltay) rho * cp * der(T06,t) = fd_2d_int(T06,T10,T02,T07,T05,k,qdot,deltax,deltay) rho * cp * der(T07,t) = fd_2d_int(T07,T11,T03,T08,T06,k,qdot,deltax,deltay) rho * cp * der(T10,t) = fd_2d_int(T10,T14,T06,T11,T09,k,qdot,deltax,deltay) rho * cp * der(T11,t) = fd_2d_int(T11,T15,T07,T12,T10,k,qdot,deltax,deltay) rho * cp * der(T14,t) = fd_2d_int(T14,T17,T10,T15,T13,k,qdot,deltax,deltay) rho * cp * der(T15,t) = fd_2d_int(T15,T18,T11,T16,T14,k,qdot,deltax,deltay) rho * cp * der(T17,t) = fd_2d_int(T17,T18,T14,T18,T14,k,qdot,deltax,deltay) rho * cp * der(T18,t) = fd_2d_int(T18,T20,T15,T19,T17,k,qdot,deltax,deltay) rho * cp * der(T20,t) = fd_2d_int(T20,T21,T18,T21,T18,k,qdot,deltax,deltay) // Exterior surface nodes, 04, 08, 12, 16, 19, 21, 22 /* Node 04: plane surface node, n-orientation; e, w, s labeled 08, 08, 03. */ rho * cp * der(T04,t) = fd_2d_psur_n(T04,T08,T08,T03,k,qdot,deltax,deltay,Tinfo,ho,q''a) rho * cp * der(T08,t) = fd_2d_psur_n(T08,T12,T04,T07,k,qdot,deltax,deltay,Tinfo,ho,q''a) rho * cp * der(T12,t) = fd_2d_psur_n(T12,T16,T08,T11,k,qdot,deltax,deltay,Tinfo,ho,q''a) rho * cp * der(T16,t) = fd_2d_psur_n(T16,T19,T12,T15,k,qdot,deltax,deltay,Tinfo,ho,q''a) rho * cp * der(T19,t) = fd_2d_psur_n(T19,T21,T16,T18,k,qdot,deltax,deltay,Tinfo,ho,q''a) rho * cp * der(T21,t) = fd_2d_psur_n(T21,T22,T19,T20,k,qdot,deltax,deltay,Tinfo,ho,q''a) /* Node 22: external corner node, e-n orientation; w, s labeled 21, 21. */ rho * cp * der(T22,t) = fd_2d_ec_en(T22,T21,T21,k,qdot,deltax,deltay,Tinfo,ho,q''a) // Input variables deltax = 0.050 deltay = 0.050 Tinfi = 350 hi = 100 Tinfo = 25 ho = 5 k = 0.85 alpha = 5.55e-7 alpha = k / (rho * cp) rho = 1000

// arbitrary value

(3) The results for t = 50 hour, representing the steady-state condition, are shown below, arranged according to the coordinate system. Tmn (C) x/y (mm)

0

50

100

150

200

250

300

0

181.60

179.50

172.70

160.20

141.80

120.60

98.89

149.00

50

229.80

227.30

219.00

202.90

177.50

100

282.90

280.80

273.20

172.70

216.20

150

340.20

339.90

338.20

327.30

In Problem 4.57, the temperature distribution was determined using the FDEs written for steady-state conditions, but with a finer network, ∆x = ∆y = 25 mm. By comparison, the results for the coarser network are slightly higher, within a fraction of 1°C, along the mid-section of the flue, but notably higher in the vicinity of inner corner. (For example, node 13 is 2.6°C higher with the coarser mesh.)

PROBLEM 5.126 KNOWN: Electrical heating elements embedded in a ceramic plate as described in Problem 4.75; initially plate is at a uniform temperature and suddenly heaters are energized. FIND: Time required for the difference between the surface and initial temperatures to reach 95% of the difference for steady-state conditions using the implicit, finite-difference method. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction, (2) Constant properties, (3) No internal generation except for Node 7, (4) Heating element approximates a line source; wire diameter is negligible. ANALYSIS: The grid for the symmetry element above consists of 12 nodes. Nodes 1-3 are points on a surface experiencing convection; nodes 4-12 are interior nodes; node 7 is a special case with internal generation and because of symmetry, q′ht = 25 W/m. Their finitedifference equations are derived as follows Surface Node 2. From an energy balance on the prescribed control volume with ∆x/∆y = 3, T p+1 − T2p E = E = q′ + q′ + q′ + q′ = ρ cV 2 in st a b c d ∆t p+1

∆y T1 k 2

p+1

− T2

∆x

p+1

∆y T3 +k 2

(

p+1 + h∆x T∞ − T2 p+1

− T2

∆x

+ k∆x

)

T5p+1 − T2p+1 ∆y

p+1

p

 ∆y  T2 − T2 . = ρ c  ∆x 2  ∆t  Continued …..

PROBLEM 5.126 (Cont.) Divide by k, use the following definitions, and regroup to obtain the finite-difference equations. N ≡ h∆x/k = 100 W/m 2 ⋅ K × 0.006m/2 W/m ⋅ K = 0.3000 Fo ≡ ( k/ρ c ) ∆t/∆x ⋅ ∆y = α∆t/∆x ⋅ ∆y =

(1)

1.5 × 10−6 m 2 / s × 1s/ (0.006 × 0.002 ) m 2 = 0.1250

(

) (

)

(2)

(

1  ∆y  p+1 p+1 p+1 1  ∆y  T T N T T − + − +   T3p+1 − T2p+1 ∞   1 2 2 2  ∆x  2  ∆x   ∆x  1 T2p+1 − T2p +   T5p+1 − T2p+1 = ∆ y 2Fo  

(

)

(

)

)

1  ∆y  p+1   ∆x  1  p+1 1  ∆x  p+1  ∆y  T − + N + +     ∆x  2Fo  T2 + 2  ∆y  T3 2  ∆x  1 ∆ y        ∆x  1 p +   T5p+1 = − NT∞ − T . 2Fo 2  ∆y 

(3)

Substituting numerical values for Fo and N, and using T∞ = 30°C and ∆x/∆y = 3, find 0.16667T1p+1 − 7.63333T2p+1 + 0.16667T3p+1 + 3.00000T5p+1 = 9.0000 − 4.0000T2p .

(4)

By inspection and use of Eq. (3), the FDEs for Nodes 1 and 3 can be inferred. Interior Node 7. From an energy balance on the prescribed control volume with ∆x/∆y = 3, ′ E ′in + E ′g = E st where E ′g = 2q′ht and E ′in represents the conduction terms −q′a + q′b + q′c + q′d , k∆y

T8p+1 − T7p+1 ∆x

+ k∆x

T4p+1 − T7p+1 ∆y

p+1 T10 − T7p+1

+ k∆y

T8p+1 − T7p+1 ∆x

+ 2q′ht = ρ c ( ∆x ⋅ ∆y ) ∆y Using the definition of Fo, Eq. (2), and regrouping, find 1  ∆x  p+1   ∆x   ∆y  1  p+1 T4 −    +   +  T7   2  ∆y  ∆ y ∆ x 2Fo       q′ 1  ∆x  p+1 1 p  ∆y  +   T8p+1 +   T10 = − ht − T 2  ∆y  k 2Fo 7  ∆x  + k∆x

T7p+1 − T7p ∆t

.

(5)

p+1 1.50000T4p+1 − 7.33333T7p+1 + 0.33333T8p+1 + 1.50000T10 = −12.5000 − 4.0000T7p .

Continued …..

(6)

PROBLEM 5.126 (Cont.) Recognizing the form of Eq. (5), it is a simple matter to infer the FDE for the remaining interior points for which q ht = 0. In matrix notation [A][T] = [C], the coefficient matrix [A] and RHS matrix [C] are:

Recall that the problem asks for the time required to reach 95% of the difference for steady-state conditions. This provides information on approximately how long it takes for the plate to come to a steady operating condition. If you worked Problem 4.71, you know the steady-state temperature distribution. Then you can proceed to find the p Tm values with increasing time until the first node reaches the required limit. We should not expect the nodes to reach their limit at the same time.

Not knowing the steady-state temperature distribution, use the implicit FDE in matrix form above to step through time → ∞ to the steady-state solution; that is, proceed to p → 10,20…100 until the solution matrix [T] does not change. The results of the analysis are tabulated below. Column 1 labeled Tm(∞) is the steady-state distribution. Column 2, Tm(95%), is the 95% limit being sought as per the graph directly above. The third column is the temperature distribution at t = to = 248s, Tm(248s); at this elapsed time, Node 1 has reached its limit. Can you explain why this node was the first to reach this limit? Which nodes will be the last to reach their limits? Tm(∞) 55.80 49.93 47.67 59.03 51.72 49.19 63.89 52.98 50.14 62.84 53.35 50.46

Tm(95%) 54.51 48.93 46.78 57.58 50.63 48.23 62.20 51.83 49.13 61.20 52.18 49.43

Tm(248s) 54.51 48.64 46.38 57.64 50.32 47.79 62.42 51.52 48.68 61.35 51.86 48.98

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PROBLEM 5.127 KNOWN: Nodal network and operating conditions for a water-cooled plate. FIND: Transient temperature response. SCHEMATIC:

ASSUMPTIONS: (1) Steady-sate conditions, (2) Two-dimensional conduction. ANALYSIS: The energy balance method must be applied to each nodal region. Grouping similar regions, the following results are obtained. Nodes 1 and 5:  2α∆t 2α∆t  p+1 2α∆t p+1 2α∆t p+1 T2 − T6 = T1p + 1 +  T1 − 2 2 2 2   ∆x ∆y  ∆x ∆y   2α∆t 2α∆t  p+1 2α∆t p+1 2α∆t p+1 T4 − T10 = T5p + 1 +  T5 − 2 2 2 2   ∆x ∆y  ∆x ∆y  Nodes 2, 3, 4:  2α∆t 2α∆t  p+1 α∆t p+1 2α∆t p+1 α∆t p+1 p Tm-1,n − Tm+1,n − Tm,n-1 = Tm,n + 1 +  Tm,n − 2 2 2 2 2   ∆x ∆y  ∆x ∆x ∆y  Nodes 6 and 14:  2α∆t 2α∆t 2hα∆t  p+1 2α∆t p+1 2α∆t p+1 2hα∆t T1 − T7 = T∞ +T6p + + 1 +  T6 − 2 2 2 2  k∆y  k∆y ∆x ∆y ∆y ∆x   2α∆t 2α∆t 2hα∆t  p+1 2α∆t p+1 2α∆t p+1 2hα∆t p T15 − T19 = T∞ +T14 + + 1 +  T14 − 2 2 2 2  k∆y  k∆y ∆x ∆y ∆x ∆y  Continued …..

PROBLEM 5.127 (Cont.) Nodes 7 and 15:  2α∆t 2α∆t 2hα∆t  p+1 2α∆t p+1 α∆t p+1 α∆t p+1 2hα∆t p + + T2 − T6 − T8 = T∞ +T7  1 +  T7 − 2 2 2 2 2 k∆y  k∆y ∆y ∆y ∆x k∆x  ∆x  2α∆t 2α∆t 2hα∆t  p+1 α∆t p+1 α∆t p+1 2α∆t p+1 2hα∆t p + + T14 − T16 − T20 = T∞ +T15  1 +  T15 − 2 2 2 2 2 k∆y  k∆y ∆y ∆x ∆x ∆y  ∆x Nodes 8 and 16:  2α∆t 2α∆t 2 hα∆t 2 hα∆t  p+1 4 α∆t p+1 2 α∆t p+1 T T + + + − − 1 + T 2 2 3 k∆x 3 k∆y  8 2 3 2 7  3 3 x y y x ∆ ∆ ∆ ∆   4 α∆t p+1 2 α∆t p+1 2 hα∆t  1 1  T9 − T11 = T∞ + T8p − +   3 ∆x 2 3 ∆y 2 3 k  ∆x ∆y   2α∆t 2α∆t 2 hα∆t 2 hα∆t  p+1 2 α∆t p+1 2 α∆t p+1 T T + + + + − 1 +  T16 −  3 ∆y 2 11 3 ∆x 2 15 ∆x 2 ∆y 2 3 k∆x 3 k∆y   4 α∆t p+1 4 α∆t p+1 2 hα∆t  1 1  p T17 − T21 = − +   T∞ + T16 2 2 3 ∆x 3 ∆y 3 k  ∆x ∆y  Node 11:  2α∆t 2α∆t 2hα∆t  p+1 α∆t p+1 ∆t p+1 α∆t p+1 2hα∆t p T∞ +T11  1 + 2 + 2 +  T11 − 2 T8 − 2α 2 T12 − 2 T16 = k x k x ∆ ∆ ∆y ∆y ∆x ∆y  ∆x  Nodes 9, 12, 17, 20, 21, 22:  2α∆t 2α∆t  p+1 α∆t p+1 α∆t p+1 p+1 p+1 p Tm,n+1 + Tm,n-1 Tm-1,n + Tm+1,n + − = Tm,n 1 +  Tm,n −  ∆x 2 ∆y 2  ∆y 2 ∆x 2  Nodes 10, 13, 18, 23:  2α∆t 2α∆t  p+1 α∆t p+1 2α∆t p+1 p+1 p Tm,n+1 + Tm,n-1 Tm-1,n = Tm,n + − 1 +  Tm,n −  ∆x 2 ∆y 2  ∆y 2 ∆x 2  Node 19:  2α∆t 2α∆t  p+1 α∆t p+1 p+1 2α∆t p+1 p T14 + T24 T20 = T19 + − 1 +  T19 −  ∆x 2 ∆y 2  ∆y 2 ∆x 2  Nodes 24, 28:  2α∆t 2α∆t  p+1 2α∆t p+1 2α∆t p+1 2q′′ α∆t p T19 − T25 = o +T24 + 1 +  T24 − 2 2 2 2   k y ∆ ∆ ∆ ∆ ∆ x y y x    2α∆t 2α∆t  p+1 2α∆t p+1 2α∆t p+1 2q′′ α∆t p T23 − T27 = o +T28 + 1 +  T28 − 2 2 2 2   k y ∆ ∆ ∆ ∆ ∆ x y y x  

(

)

(

)

(

(

)

)

Continued …..

PROBLEM 5.127 (Cont.) Nodes 25, 26, 27:

)

(

 2α∆t 2α∆t  p+1 2α∆t p+1 2q′′ α∆t p+1 α∆t p+1 p+1 Tm,n+1 − Tm-1,n + Tm+1,n +Tm,n + = o 1 +  Tm,n − 2 2  2 2  k y ∆ x y y x ∆ ∆ ∆ ∆   The convection heat rate is q′conv = h [( ∆x/2 )( T6 − T∞ ) + ∆x ( T7 − T∞ ) + ( ∆x + ∆y )( T8 − T∞ ) / 2 + ∆y ( T11 − T∞ ) + ( ∆x +∆y )( T16 − T∞ ) / 2 + ∆x ( T15 − T∞ ) + ( ∆x/2 )( T14 − T∞ ) = q out.

The heat input is q′in = q′′o ( 4∆x ) and, on a percentage basis, the ratio is n ≡ (q′conv / q′in ) ×100. Results of the calculations (in °C) are as follows: Time: 5.00 sec;

n = 60.57%

19.612 19.446

19.712 19.597

24.217 25.658 27.581

24.074 25.608 27.554

19.974 20.105 21.370 23.558 25.485 27.493

20.206 20.490 21.647 23.494 25.417 27.446

Time: 15.0 sec;

n = 94.89%

23.228 22.896

23.363 23.096

28.294 30.063 32.095

28.155 30.018 32.072

23.716 23.761 25.142 27.652 29.908 32.021

Time: 23.00 sec; 23.663 23.311

23.802 23.516

28.782 30.591 32.636

28.644 30.546 32.613

24.042 24.317 25.594 27.694 29.867 31.987

Time: 10.00 sec; 20.292 20.609 21.730 23.483 25.396 27.429

22.269 22.394 22.723 21.981 22.167 22.791 24.143 27.216 27.075 26.569 28.898 28.851 28.738 30.901 30.877 30.823 Time: 20.00 sec;

24.165 24.491 25.733 27.719 29.857 31.976

n = 85.80%

23.574 23.226

23.712 23.430

28.682 30.483 32.525

28.543 30.438 32.502

23.025 23.302 24.548 26.583 28.690 30.786

23.137 23.461 24.673 26.598 28.677 30.773

n = 98.16% 24.073 24.110 25.502 28.042 30.330 32.452

24.409 24.682 25.970 28.094 30.291 32.419

24.535 24.861 26.115 28.122 30.282 32.409

n = 99.00% 24.165 24.200 25.595 28.143 30.438 32.563

24.503 24.776 26.067 28.198 30.400 32.531

24.630 24.957 26.214 28.226 30.392 32.520

COMMENTS: Temperatures at t = 23 s are everywhere within 0.13°C of the final steadystate values.

PROBLEM 5.128 KNOWN: Cubic-shaped furnace, with prescribed operating temperature and convection heat transfer on the exterior surfaces. FIND: Time required for the furnace to cool to a safe working temperature corresponding to an inner wall temperature of 35°C considering convection cooling on (a) the exterior surfaces and (b) on both the exterior and interior surfaces. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction through the furnace walls and (2) Constant properties. ANALYSIS: Assuming two-dimensional conduction through the walls and taking advantage of symmetry for the cubical shape, the analysis considers the quarter section shown in the schematic above. For part (a), with no cooling on the interior during the cool-down process, the inner surface boundary condition is adiabatic. For part (b), with cooling on both the exterior and interior, the boundary conditions are prescribed by the convection process. The boundaries through the centerline of the wall and the diagonal through the corner are symmetry planes and considered as adiabatic. We have chosen to use the finite-element software FEHT as the solution tool. Using FEHT, an outline of the symmetrical wall section is drawn, and the material properties are specified. To determine the initial conditions for the cool-down process, we will first find the temperature distribution for steady-state operation. As such, specify the boundary condition for the inner surface as a constant temperature of 900°C; the other boundaries are as earlier described. In the Setup menu, click on Steady-State, and then Run to obtain the steady-state temperature distribution. This distribution represents the initial temperature distribution, Ti (x, y, 0), for the wall at the onset of the cool-down process. Next, in the Setup menu, click on Transient; for the nodes on the inner surface, in the Specify | Boundary Conditions menu, deselect the Temperature box (900°C) and set the Flux box to zero for the adiabatic condition (part (a)); and, in the Run command, click on Continue (not Calculate). Be sure to change the integration time scale from seconds to hours. Because of the high ratio of wall section width (nearly 8.5 m) to the thickness (1 m), the conduction heat transfer through the section is nearly one-dimensional. We chose the x,y-section 1 m to the right of the centerline (1 m, y) as the location for examining the temperature-time history, and determining the cool-down time for the inner surface to reach the safe working temperature of 35°C. Continued …..

PROBLEM 5.128 (Cont.)

Time-to-cool, Part (a), Adiabatic inner surface. From the above temperature history, the cool-down time, ta, corresponds to the condition when Ta (1 m, 0, ta) = 35°C. As seen from the history, this location is the last to cool. From the View | Tabular Output, find that

<

t a = 1306 h = 54 days

Continued …..

PROBLEM 5.128 (Cont.) Time-to-cool, Part (b), Cooled inner surface. From the above temperature history, note that the center portion of the wall, and not the inner surface, is the last to cool. The inner surface cools to 35°C in approximately 175 h or 7 days. However, if the cooling process on the inner surface were discontinued, its temperature would increase and eventually exceed the desired safe working temperature. To assure the safe condition will be met, estimate the cool down time as, tb, corresponding to the condition when Tb (1 m, 0.75 m, tb) = 35°C. From the View | Tabular Output, find that

t b = 311 h = 13 days

<

COMMENTS: (1) Assuming the furnace can be approximated by a two-dimensional symmetrical section greatly simplifies our analysis by not having to deal with three-dimensional corner effects. We justify this assumption on the basis that the corners represent a much shorter heat path than the straight wall section. Considering corner effects would reduce the cool-down time estimates; hence, our analysis provides a conservative estimate. (2) For background information on the Continue option, see the Run menu in the FEHT Help section. Using the Run | Calculate command, the steady-state temperature distribution was determined for the normal operating condition of the furnace. Using the Run | Continue command (after clicking on Setup | Transient), this steady-state distribution automatically becomes the initial temperature distribution for the cool-down transient process. This feature allows for conveniently prescribing a non-uniform initial temperature distribution for a transient analysis (rather than specifying values on a node-by-node basis.

PROBLEM 5.129 KNOWN: Door panel with ribbed cross-section, initially at a uniform temperature of 275°C, is ejected from the hot extrusion press and experiences convection cooling with ambient air at 25°C and 2 a convection coefficient of 10 W/m ⋅K. FIND: (a) Using the FEHT View|Temperature vs. Time command, create a graph with temperaturetime histories of selected locations on the panel surface, T(x,0,t). Comment on whether you see noticeable differential cooling in the region above the rib that might explain the appearance defect; and Using the View|Temperature Contours command with the shaded-band option for the isotherm contours, select the From start to stop time option, and view the temperature contours as the panel cools. Describe the major features of the cooling process you have seen. Use other options of this command to create a 10-isotherm temperature distribution at some time that illustrates important features. How would you re-design the ribbed panel in order to reduce this thermally induced paint defect situation, yet retain the stiffening function required of the ribs? SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction in the panel, (2) Uniform convection coefficient over the upper and lower surfaces of the panel, (3) Constant properties. 3

PROPERTIES: Door panel material (given): ρ = 1050 kg/m , c = 800 J/kg⋅K, k = 0.5 W/m⋅K. ANALYSIS: (a) Using the Draw command, the shape of the symmetrical element of the panel (darkened region in schematic) was generated and elements formed as shown below. The symmetry lines represent adiabatic surfaces, while the boundary conditions for the exposed web and rib surfaces are characterized by (T∞, h). Continued …..

PROBLEM 5.129 (Cont.)

After running the calculation for the time period 0 to 400 s with a 1-second time step, the temperaturetime histories for three locations were obtained and the graph is shown below.

As expected, the region directly over the rib (0,0) cooled the slowest, while the extreme portion of the web (0, 13 mm) cooled the fastest. The largest temperature differences between these two locations occur during the time period 50 to 150 s. The maximum difference does not exceed 25°C. Continued …..

PROBLEM 5.129 (Cont.) (b) It is possible that the temperature gradients within the web-rib regions – rather than just the upper surface temperature differentials – might be important for understanding the panel’s response to cooling. Using the Temperature Contours command (with the From start to stop option), we saw that the center portion of the web and the end of the rib cooled quickly, but that the region on the rib centerline (0, 3-5 mm), was the hottest region. The isotherms corresponding to t = 100 s are shown below. For this condition, the temperature differential is about 21°C.

From our analyses, we have identified two possibilities to consider. First, there is a significant surface temperature distribution across the panel during the cooling process. Second, the web and the extended portion of the rib cool at about the same rate, and with only a modest normal temperature gradient. The last region to cool is at the location where the rib is thickest (0, 3-5 mm). The large temperature gradient along the centerline toward the surface may be the cause of microstructure variations, which could influence the adherence of paint. An obvious re-design consideration is to reduce the thickness of the rib at the web joint, thereby reducing the temperature gradients in that region. This fix comes at the expense of decreasing the spacing between the ribs.

PROBLEM 6.1 KNOWN: Variation of hx with x for laminar flow over a flat plate. FIND: Ratio of average coefficient, h x , to local coefficient, hx, at x. SCHEMATIC:

ANALYSIS: The average value of hx between 0 and x is 1 x C x ∫ h x dx = ∫ x -1/2dx x 0 x 0 C 1/2 = 2x = 2Cx -1/2 x = 2h x .

hx = hx hx Hence,

hx = 2. hx

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COMMENTS: Both the local and average coefficients decrease with increasing distance x from the leading edge, as shown in the sketch below.

PROBLEM 6.2 KNOWN: Variation of local convection coefficient with x for free convection from a vertical heated plate. FIND: Ratio of average to local convection coefficient. SCHEMATIC:

ANALYSIS: The average coefficient from 0 to x is 1 x C x -1/4 h x = ∫ h x dx = ∫ x dx x 0 x 0 4 C 3/4 4 4 hx = x = C x -1/4 = h x . 3 x 3 3 Hence,

hx 4 = . hx 3

<

The variations with distance of the local and average convection coefficients are shown in the sketch.

COMMENTS: Note that h x / h x = 4 / 3 is independent of x. Hence the average coefficient 4 for an entire plate of length L is h L = h L , where hL is the local coefficient at x = L. Note 3 also that the average exceeds the local. Why?

PROBLEM 6.3 KNOWN: Expression for the local heat transfer coefficient of a circular, hot gas jet at T∞ directed normal to a circular plate at Ts of radius ro. FIND: Heat transfer rate to the plate by convection. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Flow is axisymmetric about the plate, (3) For h(r), a and b are constants and n ≠ -2. ANALYSIS: The convective heat transfer rate to the plate follows from Newton’s law of cooling q conv = ∫ dq conv = ∫ h ( r ) ⋅ dA ⋅ ( T∞ − Ts ). A

A

The local heat transfer coefficient is known to have the form, h ( r ) = a + br n and the differential area on the plate surface is dA = 2π r dr.

Hence, the heat rate is q conv = ∫

ro 0

(a + brn ) ⋅ 2π r dr ⋅ (T∞ − Ts ) r

b n+2  o a q conv = 2π ( T∞ − Ts )  r 2 + r  n+2 2 0 b n+2  a q conv = 2π  ro2 + ro  ( T∞ − Ts ) . n+2 2 

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COMMENTS: Note the importance of the requirement, n ≠ -2. Typically, the radius of the jet is much smaller than that of the plate.

PROBLEM 6.4 KNOWN: Distribution of local convection coefficient for obstructed parallel flow over a flat plate. FIND: Average heat transfer coefficient and ratio of average to local at the trailing edge. SCHEMATIC:

ANALYSIS: The average convection coefficient is

(

)

1 L 1 L h dx 0.7 + 13.6x − 3.4x 2 dx = x L ∫0 L ∫0 1 hL = 0.7L + 6.8L2 − 1.13L3 = 0.7 + 6.8L − 1.13L2 L hL =

(

)

h L = 0.7 + 6.8 (3) − 1.13 (9 ) = 10.9 W/m 2 ⋅ K.

<

The local coefficient at x = 3m is h L = 0.7 + 13.6 (3) − 3.4 (9 ) = 10.9 W/m 2 ⋅ K. Hence,

<

h L / h L = 1.0. COMMENTS: The result h L / h L = 1.0 is unique to x = 3m and is a consequence of the existence of a maximum for h x x . The maximum occurs at x = 2m, where

(dh x / dx ) = 0 and

$

(d2h x / dx2 < 0.)

PROBLEM 6.5 KNOWN: Temperature distribution in boundary layer for air flow over a flat plate. FIND: Variation of local convection coefficient along the plate and value of average coefficient. SCHEMATIC:

ANALYSIS: From Eq. 6.17,

h=−

k ∂ T ∂ y y =0 k ( 70 × 600x ) =+ (Ts − T∞ ) (Ts − T∞ )

where Ts = T(x,0) = 90°C. Evaluating k at the arithmetic mean of the freestream and surface temperatures, T = (20 + 90)°C/2 = 55°C = 328 K, Table A.4 yields k = 0.0284 W/m⋅K. Hence, with Ts - T‡ = 70°C = 70 K,

h=

0.0284 W m ⋅ K ( 42, 000x ) K m 70 K

(

= 17x W m 2 ⋅ K

)

<

and the convection coefficient increases linearly with x.

The average coefficient over the range 0 ≤ x ≤ 5 m is 5 1 L 17 5 17 x 2 h = ∫ hdx = ∫ xdx = = 42.5 W m 2 ⋅ K 0 0 L 5 5 2 0

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PROBLEM 6.6 KNOWN: Variation of local convection coefficient with distance x from a heated plate with a uniform temperature Ts. FIND: (a) An expression for the average coefficient h12 for the section of length (x2 - x1) in terms of C, x1 and x2, and (b) An expression for h12 in terms of x1 and x2, and the average coefficients h1 and

h 2 , corresponding to lengths x1 and x2, respectively. SCHEMATIC:

ASSUMPTIONS: (1) Laminar flow over a plate with uniform surface temperature, Ts, and (2) Spatial variation of local coefficient is of the form h x = Cx −1/ 2 , where C is a constant. ANALYSIS: (a) The heat transfer rate per unit width from a longitudinal section, x2 - x1, can be expressed as

′ = h12 ( x 2 − x1 )( Ts − T∞ ) q12

(1)

where h12 is the average coefficient for the section of length (x2 - x1). The heat rate can also be written in terms of the local coefficient, Eq. (6.3), as x x ′ = 2 h x dx ( Ts − T∞ ) = ( Ts − T∞ ) 2 h x dx q12 x1 x1 Combining Eq. (1) and (2), x2 1 h12 = h dx ( x 2 − x1 ) x1 x

∫

∫

∫

and substituting for the form of the local coefficient, h x = Cx −1/ 2 , find that 2 − x1/ 2 1/ 2  x 2  x1/ x2 1 C x 1/ 2 − 2 1 h12 = Cx dx =   = 2C x x x x x 1/ 2 x x − − − 1 ( 2 1)  2 1  2 1 x1 (b) The heat rate, given as Eq. (1), can also be expressed as

∫

′ = h 2 x 2 ( Ts − T∞ ) − h1x1 ( Ts − T∞ ) q12

(2)

(3)

(4)

<

(5)

which is the difference between the heat rate for the plate over the section (0 - x2) and over the section (0 - x1). Combining Eqs. (1) and (5), find,

h x −h x h12 = 2 2 1 1 x 2 − x1

(6)

COMMENTS: (1) Note that, from Eq. 6.6, 1 x 1 x −1/ 2 hx = h x dx = Cx dx = 2Cx −1/ 2 2 0 x 0

∫

∫

or h x = 2hx. Substituting Eq. (7) into Eq. (6), see that the result is the same as Eq. (4).

<

(7)

PROBLEM 6.7 KNOWN: Radial distribution of local convection coefficient for flow normal to a circular disk. FIND: Expression for average Nusselt number. SCHEMATIC:

ASSUMPTIONS: Constant properties ANALYSIS: The average convection coefficient is 1 hdAs A s ∫ As 1 ro k n h= Nu o 1 + a ( r/ro )  2π rdr ∫   π ro2 0 D h=

r

o kNu o  r 2 ar n+2   +  h= ro3  2 ( n + 2 ) ron  0

where Nuo is the Nusselt number at the stagnation point (r = 0). Hence, r

n+2  o  r/r 2  r  ( ) hD a o  Nu D = = 2Nu o  +    2  k n+2 )  ro  (  0 Nu D = Nu o 1 + 2a/ ( n + 2 )

Nu D = 1+ 2a/ ( n + 2 ) 0.814Re1/2 Pr 0.36 . D

<

COMMENTS: The increase in h(r) with r may be explained in terms of the sharp turn which the boundary layer flow must make around the edge of the disk. The boundary layer accelerates and its thickness decreases as it makes the turn, causing the local convection coefficient to increase.

PROBLEM 6.8 KNOWN: Convection correlation and temperature of an impinging air jet. Dimensions and initial temperature of a heated copper disk. Properties of the air and copper. FIND: Effect of jet velocity on temperature decay of disk following jet impingement. SCHEMATIC:

ASSUMPTIONS: (1) Validity of lumped capacitance analysis, (2) Negligible heat transfer from sides and bottom of disk, (3) Constant properties. ANALYSIS: Performing an energy balance on the disk, it follows that E st = ρ Vc dT dt = − As (q′′conv + q′′rad ) . Hence, with V = AsL,

h (T − T∞ ) + h r (T − Tsur ) dT =− ρ cL dt

(

)

2 where, h r = εσ (T + Tsur ) T 2 + Tsur and, from the solution to Problem 6.7,

h=

k k 2a  1/ 2 0.36 Nu D = 1 +  0.814 ReD Pr D D n+2

With a = 0.30 and n = 2, it follows that 2 0.36 h = ( k D ) 0.936 Re1/ D Pr where ReD = VD/ν. Using the Lumped Capacitance Model of IHT, the following temperature histories were determined. Continued …..

PROBLEM 6.8 (Cont.) 1000

Temperature, T(K)

900 800 700 600 500 400 300 0

500

1000

1500

2000

2500

3000

Time, t(s) V = 4 m/s V = 20 m/s V = 50 m/s

The temperature decay becomes more pronounced with increasing V, and a final temperature of 400 K is reached at t = 2760, 1455 and 976s for V = 4, 20 and 50 m/s, respectively.

(

)

COMMENTS: The maximum Biot number, Bi = h + h r L k Cu , is associated with V = 50 m/s (maximum h of 169 W/m ⋅K) and t = 0 (maximum hr of 64 W/m2⋅K), in which case the maximum Biot number is Bi = (233 W/m2⋅K)(0.025 m)/(386 W/m⋅K) = 0.015 < 0.1. Hence, the lumped capacitance approximation is valid. 2

PROBLEM 6.9 KNOWN: Local convection coefficient on rotating disk. Radius and surface temperature of disk. Temperature of stagnant air. FIND: Local heat flux and total heat rate. Nature of boundary layer. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat transfer from back surface and edge of disk. ANALYSIS: If the local convection coefficient is independent of radius, the local heat flux at every point on the disk is

q′′ = h ( Ts − T∞ ) = 20 W / m 2 ⋅ K (50 − 20 ) °C = 600 W / m 2

<

Since h is independent of location, h = h = 20 W / m 2 ⋅ K and the total power requirement is

Pelec = q = hAs (Ts − T∞ ) = hπ ro2 (Ts − T∞ )

(

)

Pelec = 20 W / m 2 ⋅ K π (0.1m )

2

(50 − 20 ) °C = 18.9 W

<

If the convection coefficient is independent of radius, the boundary layer must be of uniform thickness δ. Within the boundary layer, air flow is principally in the circumferential direction. The circumferential velocity component uθ corresponds to the rotational velocity of the disk at the surface (y = 0) and increases with increasing r (uθ = Ωr). The velocity decreases with increasing distance y from the surface, approaching zero at the outer edge of the boundary layer (y → δ).

PROBLEM 6.10 KNOWN: Form of the velocity and temperature profiles for flow over a surface. FIND: Expressions for the friction and convection coefficients. SCHEMATIC:

ANALYSIS: The shear stress at the wall is

τs = µ

∂ u = µ  A + 2By − 3Cy 2  = Aµ .   y=0 ∂ y  y=0

Hence, the friction coefficient has the form, Cf = Cf =

τs 2 /2 ρ u∞

2Aν 2 u∞

=

2Aµ 2 ρ u∞

.

<

The convection coefficient is 2  − k f (∂ T/∂ y )y=0 − k f  E + 2Fy − 3Gy  y=0 h= = Ts − T∞ D − T∞

h=

−k f E . D − T∞

<

COMMENTS: It is a simple matter to obtain the important surface parameters from knowledge of the corresponding boundary layer profiles. However, it is rarely a simple matter to determine the form of the profile.

PROBLEM 6.11 KNOWN: Surface temperatures of a steel wall and temperature of water flowing over the wall. FIND: (a) Convection coefficient, (b) Temperature gradient in wall and in water at wall surface. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in x, (3) Constant properties. PROPERTIES: Table A-1, Steel Type AISI 1010 (70°C = 343K), ks = 61.7 W/m⋅K; Table A-6, Water (32.5°C = 305K), kf = 0.62 W/m⋅K. ANALYSIS: (a) Applying an energy balance to the control surface at x = 0, it follows that q′′x,cond − q′′x,conv = 0 and using the appropriate rate equations, Ts,2 − Ts,1 ks = h Ts,1 − T∞ . L Hence, k s Ts,2 − Ts,1 61.7 W/m ⋅ K 60$ C h= = = 705 W/m 2 ⋅ K. $ L Ts,1 − T∞ 0.35m 15 C

(

)

<

(b) The gradient in the wall at the surface is

(dT/dx )s = −

Ts,2 − Ts,1 L

=−

60$ C = −171.4$ C/m. 0.35m

In the water at x = 0, the definition of h gives

(

)

(dT/dx )f,x=0 = −

h Ts,1 − T∞ kf

(dT/dx )f,x=0 = −

705 W/m 2 ⋅ K $ 15 C = −17, 056$ C/m. 0.62 W/m ⋅ K

( )

COMMENTS: Note the relative magnitudes of the gradients. Why is there such a large difference?

<

PROBLEM 6.12 KNOWN: Boundary layer temperature distribution. FIND: Surface heat flux. SCHEMATIC:

PROPERTIES: Table A-4, Air (Ts = 300K): k = 0.0263 W/m⋅K. ANALYSIS: Applying Fourier’s law at y = 0, the heat flux is u y ∂ T  u   = − k ( T∞ − Ts )  Pr ∞  exp  − Pr ∞  ∂ y y=0 ν  y=0  ν   u = − k ( T∞ − Ts ) Pr ∞ ν = −0.0263 W/m ⋅ K (100K ) 0.7 × 5000 1/m.

q′′s = − k q′′s q′′s

q′′s = −9205 W/m 2 . COMMENTS: (1) Negative flux implies convection heat transfer to the surface. (2) Note use of k at Ts to evaluate q′′s from Fourier’s law.

<

PROBLEM 6.13 KNOWN: Air flow over a flat plate of length L = 1 m under conditions for which transition from laminar to turbulent flow occurs at xc = 0.5m based upon the critical Reynolds number, Re x,c = 5× 105. Forms for the local convection coefficients in the laminar and turbulent regions. FIND: (a) Velocity of the air flow using thermophysical properties evaluated at 350 K, (b) An expression for the average coefficient h lan ( x ) , as a function of distance from the leading edge, x, for the laminar

region, 0 ≤ x ≤ xc, (c) An expression for the average coefficient h turb ( x ) , as a function of distance from the leading edge, x, for the turbulent region, xc ≤ x ≤ L, and (d) Compute and plot the local and average convection coefficients, hx and h x , respectively, as a function of x for 0 ≤ x ≤ L. SCHEMATIC:

ASSUMPTIONS: (1) Forms for the local coefficients in the laminar and turbulent regions, hlam = Clamx-0.5 and htirb = Cturbx-0.2 where Clam = 8.845 W/m3/2⋅K, Cturb = 49.75 W/m2⋅K0.8, and x has units (m). PROPERTIES: Table A.4, Air (T = 350 K): k = 0.030 W/m⋅K, ν = 20.92 × 10-6 m2/s, Pr = 0.700. ANALYSIS: (a) Using air properties evaluated at 350 K with xc = 0.5 m, u x u ∞ = 5 × 105 ν x c = 5 × 105 × 20.92 × 10 −6 m 2 s 0.5 m = 20.9 m s Re x,c = ∞ c = 5 × 105 ν (b) From Eq. 6.5, the average coefficient in the laminar region, 0 ≤ x ≤ xc, is x 1 x 1 1 (1) hlam ( x ) = ∫ h lam ( x ) dx = Clam ∫ x −0.5dx = Clam x 0.5 = 2Clam x −0.5 = 2h lam ( x ) o x 0 x x (c) The average coefficient in the turbulent region, xc ≤ x ≤ L, is

<

<

x x   0.5 c 0.8 x 1  xc x x  h turb ( x ) = ∫ h lam ( x ) dx + ∫x c h turb ( x ) dx  = Clam 0.5 + C turb 0.8  x  0

h turb ( x ) =

)

(

1



0

0.8 2Clam x 0.5 − x c0.8  c + 1.25C turb x    x

x c 

(2)

<

Convection coefficient (W/m^2.K)

(d) The local and average coefficients, Eqs. (1) and (2) are plotted below as a function of x for the range 0 ≤ x ≤ L. 150

100

50

0 0

0.5 Distance from leading edge, x (m) Local - laminar, x xc Average - laminar, x xc

1

PROBLEM 6.14 KNOWN: Air speed and temperature in a wind tunnel. 8

FIND: (a) Minimum plate length to achieve a Reynolds number of 10 , (b) Distance from leading edge at which transition would occur. SCHEMATIC:

ASSUMPTIONS: (1) Isothermal conditions, Ts = T∞. -6 2

PROPERTIES: Table A-4, Air (25°C = 298K): ν = 15.71 × 10 m /s. ANALYSIS: (a) The Reynolds number is Re x =

ρ u∞x u∞x . = µ ν 8

To achieve a Reynolds number of 1 × 10 , the minimum plate length is then

(

−6 2 8 Re x ν 1×10 15.71× 10 m / s Lmin = = u∞ 50 m/s

) <

Lmin = 31.4 m. 5

(b) For a transition Reynolds number of 5 × 10 xc =

Re x,c ν u∞

=

(

5 × 105 15.71× 10-6 m 2 / s

x c = 0.157 m.

)

50 m/s

<

COMMENTS: Note that x c Re x,c = L ReL This expression may be used to quickly establish the location of transition from knowledge of Re x,c and Re L .

PROBLEM 6.15 KNOWN: Transition Reynolds number. Velocity and temperature of atmospheric air, water, engine oil and mercury flow over a flat plate. FIND: Distance from leading edge at which transition occurs for each fluid. SCHEMATIC:

ASSUMPTIONS: Transition Reynolds number is Re x,c = 5 × 105. PROPERTIES: For the fluids at T = 300K; Fluid

Table

2

v(m /s)

A-4

15.89 × 10

Water

A-6

0.858 × 10

Engine Oil

A-5

550 × 10

Mercury

A-5

0.113 × 10

-6

Air (1 atm)

-6

-6 -6

ANALYSIS: The point of transition is x c = Re x,c

5 × 105 ν ν. = u ∞ 1 m/s

Substituting appropriate viscosities, find Fluid Air Water Oil Mercury

xc(m) 7.95 0.43 275 0.06

<

COMMENTS: Due to the effect which viscous forces have on attenuating the instabilities which bring about transition, the distance required to achieve transition increases with increasing ν .

PROBLEM 6.16 KNOWN: Two-dimensional flow conditions for which v = 0 and T = T(y). FIND: (a) Verify that u = u(y), (b) Derive the x-momentum equation, (c) Derive the energy equation. SCHEMATIC:

Pressure & shear forces

Energy fluxes

ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3) Negligible body forces, (4) v = 0, (5) T = T(y) or ∂T/∂x = 0, (6) Thermal energy generation occurs only by viscous dissipation. ANALYSIS: (a) From the mass continuity equation, it follows from the prescribed conditions that ∂u/∂x = 0. Hence u = u(y). (b) From Newton’s second law of motion, ΣFx = (Rate of increase of fluid momentum)x,

{

}

 p −  p + ∂ p dx  dy ⋅ 1 +  −τ + τ + ∂ τ dy  dx ⋅ 1 = ρ u u + ∂ ρ u u dx dy ⋅ 1 − ρ u u dy ⋅ 1 ( ) ( ) [( ) ]   ∂ y    ∂ x  ∂ x   Hence, with τ = µ (∂ u/∂ y ) , it follows that

−

∂ p ∂ 2u . =µ ∂ x ∂ y2

∂ p ∂ τ ∂ ( ρ u ) u  = 0 + = ∂ x ∂ y ∂ x

<

(c) From the conservation of energy requirement and the prescribed conditions, it follows that

E in − E out = 0, or

 pu + ρ u e + u 2 / 2  dy ⋅1 +  −k ∂ T + τ u + ∂ (τ u ) dy  dx ⋅1     ∂ y  ∂ y 

(

{

− pu +

or,

∂ ∂ x

( pu ) dx + ρ u

)

(e + u / 2) + ∂∂ x  ρ u (e + u / 2) dx} dy ⋅1 − τ u − k ∂∂ Ty + ∂∂ y −k ∂∂ Ty  dy dx ⋅1 = 0 2

2

∂ (τ u ) ∂ ∂ ∂  ∂ T − ( pu ) −  ρ u e + u 2 / 2  + k  = 0 ∂ y ∂ x ∂ x ∂ y ∂ y

(

τ

)

∂ u ∂ τ ∂ p ∂ 2T +u −u +k = 0. ∂ y ∂ y ∂ x ∂ y2

Noting that the second and third terms cancel from the momentum equation, 2  ∂ 2T  ∂ u  k µ +   = 0.   ∂ y 2  ∂ y 

<

PROBLEM 6.17 KNOWN: Oil properties, journal and bearing temperatures, and journal speed for a lightly loaded journal bearing. FIND: Maximum oil temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3) Clearance is much less than journal radius and flow is Couette. ANALYSIS: The temperature distribution corresponds to the result obtained in the text Example on Couette flow, T(y) = T0 +

2 µ 2 y y  U  −   . 2k  L  L  

The position of maximum temperature is obtained from dT µ 2  1 2y  U  −  =0= dy 2k  L L2  or,

y = L/2.

The temperature is a maximum at this point since d 2 T/dy2 < 0. Hence, Tmax = T ( L/2 ) = T0 + $

Tmax = 40 C +

µ 2 1 1  µ U2 U  −  = T0 + 2k 8k 2 4

10-2kg/s ⋅ m (10m/s )

Tmax = 40.83$ C.

2

8 × 0.15 W/m ⋅ K

<

COMMENTS: Note that Tmax increases with increasing µ and U, decreases with increasing k, and is independent of L.

PROBLEM 6.18 KNOWN: Diameter, clearance, rotational speed and fluid properties of a lightly loaded journal bearing. Temperature of bearing. FIND: (a) Temperature distribution in the fluid, (b) Rate of heat transfer from bearing and operating power. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3) Couette flow. 3 -5 2 -3 PROPERTIES: Oil (Given): ρ = 800 kg/m , ν = 10 m /s, k = 0.13 W/m⋅K; µ = ρν = 8 × 10 kg/s⋅m.

ANALYSIS: (a) For Couette flow, the velocity distribution is linear, u(y) = U(y/L), and the energy equation and general form of the temperature distribution are 2 2 2  du  d 2T C µ U U k T = −   y 2 + 1 y + C2 . = −µ   = −µ   2k  L  k L  dy  dy2 2 Considering the boundary conditions dT/dy)y=L = 0 and T(0) = T0, find C2 = T0 and C1 = µU /L. Hence, 2 T = T0 + µ U 2 / k ( y/L ) − 1/ 2 ( y/L )  .  

( )



<



(b) Applying Fourier’s law at y = 0, the rate of heat transfer per unit length to the bearing is q ′ = − k (π D )

dT 

dy  y=0

= − (π D )

µ U2 L

(

3 = − π × 75 × 10 − m

) 8 ×10

−3

kg/s ⋅ m (14.14 m/s )

2

0.25 × 10 −3 m

= −1507.5 W/m

where the velocity is determined as

U = ( D/2 )ω = 0.0375m × 3600 rev/min ( 2π rad/rev ) / ( 60 s/min ) = 14.14 m/s.

The journal power requirement is

P′ = F(′ y=L )U = τ s( y=L ) ⋅ π D ⋅ U

(

)

P′ = 452.5kg/s 2 ⋅ m π × 75 ×10-3m 14.14m/s = 1507.5kg ⋅ m/s3 = 1507.5W/m

<

where the shear stress at y = L is

 14.14 m/s  U 2 τ s( y=L ) = µ (∂ u/∂ y )y=L = µ = 8 ×10−3 kg/s ⋅ m   = 452.5 kg/s ⋅ m. -3 L  0.25 × 10 m 

COMMENTS: Note that q′ = P′, which is consistent with the energy conservation requirement.

PROBLEM 6.19 KNOWN: Conditions associated with the Couette flow of air or water. FIND: (a) Force and power requirements per unit surface area, (b) Viscous dissipation, (c) Maximum fluid temperature. SCHEMATIC:

ASSUMPTIONS: (1) Fully-developed Couette flow, (2) Incompressible fluid with constant properties. -7 2 -3 PROPERTIES: Table A-4, Air (300K): µ = 184.6 × 10 N⋅s/m , k = 26.3 × 10 W/m⋅K; Table A-6, -6 2 Water (300K): µ = 855 × 10 N⋅s/m , k = 0.613 W/m⋅K. ANALYSIS: (a) The force per unit area is associated with the shear stress. Hence, with the linear velocity profile for Couette flow, τ = µ ( du/dy ) = µ ( U/L ) .

Air:

τ air = 184.6 ×10−7 N ⋅ s/m 2 ×

Air:

( P/A )air =

200 m/s = 0.738 N/m 2 0.005 m 200 m/s Water: τ water = 855 × 10−6 N ⋅ s/m 2 × = 34.2 N/m 2 . 0.005 m With the required power given by P/A = τ ⋅ U,

Water:

<

(0.738 N/m2 ) 200 m/s = 147.6 W/m2 ( P/A )water = (34.2 N/m 2 ) 200 m/s = 6840 W/m 2 .

<

2 2 (b) The viscous dissipation is µΦ = µ (du/dy ) = µ ( U/L ) . Hence, 2

N ⋅ s  200 m/s  4 3  0.005 m  = 2.95 ×10 W/m 2 m

Air:

( µΦ )air = 184.6 ×10−7

Water:

( µΦ )water = 855 ×10−6

2

N ⋅ s  200 m/s  6 3  0.005 m  = 1.37 × 10 W/m . 2 m

(c) From the solution to Part 4 of the text Example, the location of the maximum temperature corresponds to ymax = L/2. Hence, Tmax = T0 + µ U 2 / 8k and 2 -7 2

Air:

(Tmax )air = 27$ C +

Water:

(Tmax )water = 27

$

184.6 × 10

N ⋅ s/m

( 200 m/s )

= 30.5$ C

8 × 0.0263 W/m ⋅ K 855 × 10-6 N ⋅ s/m 2 ( 200 m/s )

2

C+

<

8 × 0.613 W/m ⋅ K

<

= 34.0$ C.

COMMENTS: (1) The viscous dissipation associated with the entire fluid layer, µΦ ( LA ) , must

 $

 $

equal the power, P. (2) Although µΦ water >> µΦ air , k water >> k air . Hence,

Tmax,water ≈ Tmax,air .

PROBLEM 6.20 KNOWN: Velocity and temperature difference of plates maintaining Couette flow. Mean temperature of air, water or oil between the plates. FIND: (a) Pr⋅Ec product for each fluid, (b) Pr⋅Ec product for air with plate at sonic velocity. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Couette flow, (3) Air is at 1 atm. PROPERTIES: Table A-4, Air (300K, 1atm), cp = 1007 J/kg⋅K, Pr = 0.707, γ = 1.4, R= 287.02 J/kg⋅K; Table A-6, Water (300K): cp = 4179 J/kg⋅K, Pr = 5.83; Table A-5, Engine oil (300K), cp = 1909 J/kg⋅K, Pr = 6400. ANALYSIS: The product of the Prandtl and Eckert numbers is dimensionless, Pr⋅ Ec = Pr

U2 m2 / s2 m 2 / s2 . ∩ ∩ cp ∆T ( J/kg ⋅ K ) K kg ⋅ m 2 / s 2 / kg

)

(

Substituting numerical values, find Air Pr⋅Ec 0.0028

Water 0.0056

Oil 13.41

<

(b) For an ideal gas, the speed of sound is c = (γ R T )

1/ 2

where R, the gas constant for air, is Ru/ Μ = 8.315 kJ/kmol⋅K/(28.97 kg/kmol) = 287.02 J/kg⋅K. Hence, at 300K for air, U = c = (1.4 × 287.02 J/kg ⋅ K × 300K )

1/ 2

= 347.2 m/s.

For sonic velocities, it follows that Pr ⋅ Ec = 0.707

(347.2 m/s )2 1007J / kg ⋅ K × 25K

<

= 3.38.

COMMENTS: From the above results it follows that viscous dissipation effects must be considered in the high speed flow of gases and in oil flows at moderate speeds. For Pr⋅Ec to be less than 0.1 in air with ∆T = 25°C, U should be < 60 m/s. ~

PROBLEM 6.21 KNOWN: Couette flow with moving plate isothermal and stationary plate insulated. FIND: Temperature of stationary plate and heat flux at the moving plate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3) Couette flow. ANALYSIS: The energy equation is given by 2  ∂ 2T  ∂ u 

0= k +µ   ∂ y 2  ∂ y 

Integrating twice find the general form of the temperature distribution, 2 2 ∂ 2T µ U ∂ T µ U =−   = −   y + C1 2

∂ y

∂ y

k L

T ( y) = −

µ U 2 y + C1y + C2 . 2k  L 

k L

2

Consider the boundary conditions to evaluate the constants,

∂ T/∂ y y=0 = 0 → C1 = 0 and T ( L ) = TL → C2 = TL +

µ 2 U . 2k

Hence, the temperature distribution is

 µ U2  T ( y ) = TL +    2k 

  y 2  1 −    .   L  

The temperature of the lower plate (y = 0) is  µ U2  T ( 0 ) = TL +  .

<

The heat flux to the upper plate (y = L) is ∂ T µ U2 q′′ ( L ) = − k . = y=L

<

 2k 

∂ y

L

COMMENTS: The heat flux at the top surface may also be obtained by integrating the viscous dissipation over the fluid layer height. For a control volume about a unit area of the fluid layer,

E ′′g = E ′′out

2

∂ u  ∫ µ  dy = q′′ ( L ) 0 ∂ y  L

µ U2 q′′ ( L ) = . L

PROBLEM 6.22 KNOWN: Couette flow with heat transfer. Lower (insulated) plate moves with speed U and upper plate is stationary with prescribed thermal conductivity and thickness. Outer surface of upper plate maintained at constant temperature, Tsp = 40°C. FIND: (a) On T-y coordinates, sketch the temperature distribution in the oil and the stationary plate, and (b) An expression for the temperature at the lower surface of the oil film, T(0) = To, in terms of the plate speed U, the stationary plate parameters (Tsp, ksp, Lsp) and the oil parameters (µ, ko, Lo). Determine this temperature for the prescribed conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed Couette flow and (3) Incompressible fluid with constant properties. ANALYSIS: (a) The temperature distribution is shown above with these key features: linear in plate, parabolic in oil film, discontinuity at plate-oil interface, and zero gradient at lower plate surface. (b) From Example 6.4, the general solution to the conservation equations for the temperature distribution in the oil film is 2 µ  U  2 To ( y ) = − Ay + C3 y + C4 where A=   2k o  Lo  and the boundary conditions are, At y = 0, insulated boundary

dTo  = 0;  dy  y = 0

At y = Lo, heat fluxes in oil and plate are equal,

C3 = 0

q′′o ( Lo ) = q′′sp ( Lo )

Continued...

PROBLEM 6.22 (Cont.) To ( Lo ) − Tsp dT  −k o  = dy  y = L R sp o

 dTo  = −2ALo   dy y=L  R = L k sp sp  sp

To ( L ) = − AL2o + C4

 k Lsp  C4 = Tsp + AL2o 1 + 2 o  Lo ksp   Hence, the temperature distribution at the lower surface is

To ( 0 ) = − A ⋅ 0 + C4 To ( 0 ) = Tsp +

 k µ U 2 1 + 2 o 2k o Lo 

Lsp   ksp 

<

Substituting numerical values, find

To ( 0 ) = 40$ C +

0.799 N ⋅ s m 2 0.145 3  ×  = 116.9$ C (5 m s )2 1 + 2 2 × 0.145 W m ⋅ K 5 1.5  

<

COMMENTS: (1) Give a physical explanation about why the maximum temperature occurs at the lower surface. (2) Sketch the temperature distribution if the upper plate moved with a speed U while the lower plate is stationary and all other conditions remain the same.

PROBLEM 6.23 KNOWN: Shaft of diameter 100 mm rotating at 9000 rpm in a journal bearing of 70 mm length. Uniform gap of 1 mm separates the shaft and bearing filled with lubricant. Outer surface of bearing is water-cooled and maintained at Twc = 30°C. FIND: (a) Viscous dissipation in the lubricant, µΦ(W/m3), (b) Heat transfer rate from the lubricant, assuming no heat lost through the shaft, and (c) Temperatures of the bearing and shaft, Tb and Ts. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed Couette flow, (3) Incompressible fluid with constant properties, and (4) Negligible heat lost through the shaft. ANALYSIS: (a) The viscous dissipation, µΦ, Eq. 6.40, for Couette flow from Example 6.4, is 2

2

2  47.1m s   du  U 7 3 µΦ = µ   = µ   = 0.03 N ⋅ s m 2   = 6.656 × 10 W m L  dy   0.001m 

<

where the velocity distribution is linear and the tangential velocity of the shaft is

U = π DN = π ( 0.100 m ) × 9000 rpm × ( min 60s ) = 47.1m s . (b) The heat transfer rate from the lubricant volume ∀ through the bearing is q = µΦ ⋅ ∀ = µΦ (π D ⋅ L ⋅ " ) = 6.65 × 107 W m3 (π × 0.100 m × 0.001m × 0.070 m ) = 1462 W

<

where  = 70 mm is the length of the bearing normal to the page. Continued...

PROBLEM 6.23 (Cont.) (c) From Fourier’s law, the heat rate through the bearing material of inner and outer diameters, Di and Do, and thermal conductivity kb is, from Eq. (3.27),

qr =

2π k b (Tb − Twc ) ln ( Do Di )

q ln ( Do Di ) Tb = Twc + r 2π k b Tb = 30$ C +

1462 W ln ( 200 100 ) 2π × 0.070 m × 45 W m ⋅ K

= 81.2$ C

<

To determine the temperature of the shaft, T(0) = Ts, first the temperature distribution must be found beginning with the general solution, Example 6.4,

µ  U 2 2 T ( y ) = −   y + C3 y + C4 2k  L  The boundary conditions are, at y = 0, the surface is adiabatic

dT  =0  dy  y = 0

C3 = 0

and at y = L, the temperature is that of the bearing, Tb

T ( L ) = Tb = −

µ  U 2 2   L + 0 + C4 2k  L 

C4 = Tb +

µ 2 U 2k

Hence, the temperature distribution is

µ 2  y2   T ( y ) = Tb + U 1 −  L2  2k   and the temperature at the shaft, y = 0, is

Ts = T ( 0 ) = Tb +

µ 2 0.03 N ⋅ s m 2 U = 81.3$ C + ( 47.1m s )2 = 303$ C 2k 2 × 0.15 W m ⋅ K

<

PROBLEM 6.24 KNOWN: Couette flow with heat transfer. FIND: (a) Dimensionless form of temperature distribution, (b) Conditions for which top plate is adiabatic, (c) Expression for heat transfer to lower plate when top plate is adiabatic. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) incompressible fluid with constant properties, (3) Negligible body forces, (4) Couette flow. ANALYSIS: (a) From Example 6.4, the temperature distribution is µ 2  y  y 2  y T = T0 + U  −    + ( TL − T0 )

2k

 L  L  

L

 y  y 2  y T − T0 µ U2  −   + = TL − T0 2k (TL − T0 )  L  L   L   or, with

θ ≡ (T − T0 ) TL − T0 ,

η ≡ y L,

Ec ≡ U 2 cp (TL − T0 ) Pr⋅ Ec  1  θ= η − η 2 + η = η 1 + Pr⋅ Ec (1 − η ) 2  2  Pr ≡ c p µ k ,

(

)

(1)

<

(2)

<

(b) For there to be zero heat transfer at the top plate, dT/dy)y=L = 0. Hence,

T −T dθ  Pr⋅ Ec Pr⋅ Ec ⋅ L 0= +1 = 0 (1 − 2η ) η =1 + 1 = −  dη η =1 L 2 2 There is no heat transfer at the top plate if, Ec⋅Pr = 2. (c) The heat transfer rate to the lower plate (per unit area) is

q′′0 = − k

( T − T ) dθ dT = −k L 0 dy y = 0 L dη η = 0

T −T q′′o = − k L 0 L

 Pr⋅ Ec   2 (1 − 2η ) η = 0 + 1

T − T  Pr⋅ Ec  q′′0 = − k L 0  + 1 = −2k ( TL − T0 ) L L  2 

< Continued...

PROBLEM 6.24 (Cont.) (d) Using Eq. (1), the dimensionless temperature distribution is plotted as a function of dimensionless distance, η = y/L. When Pr⋅Ec = 0, there is no dissipation and the temperature distribution is linear, so that heat transfer is by conduction only. As Pr ⋅Ec increases, viscous dissipation becomes more important. When Pr⋅Ec = 2, heat transfer to the upper plate is zero. When Pr⋅Ec > 2, the heat rate is out of the oil film at both surfaces.

theta = (T(y)-T0)/(TL-T0)

2

1.5

1

0.5

0 0

0.25

0.5 eta = y/L

Pr*Ec = 0, conduction Pr*Ec = 1 Pr*Ec = 2, adiabatic at y=L Pr*Ec = 4 Pr*Ec = 10

0.75

1

PROBLEM 6.25 KNOWN: Steady, incompressible, laminar flow between infinite parallel plates at different temperatures. FIND: (a) Form of continuity equation, (b) Form of momentum equations and velocity profile. Relationship of pressure gradient to maximum velocity, (c) Form of energy equation and temperature distribution. Heat flux at top surface. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional flow (no variations in z) between infinite, parallel plates, (2) Negligible body forces, (3) No internal energy generation, (4) Incompressible fluid with constant properties. ANALYSIS: (a) For two-dimensional, steady conditions, the continuity equation is

∂ (ρ u ) ∂ (ρ v) + = 0. ∂ x ∂ y

Hence, for an incompressible fluid (constant ρ) in parallel flow (v = 0),

∂ u = 0. ∂ x

<

The flow is fully developed in the sense that, irrespective of y, u is independent of x. (b) With the above result and the prescribed conditions, the momentum equations reduce to ∂ p ∂ 2u ∂ p

0=−

∂ x

+µ

0=−

∂ y2

∂ y

<

Since p is independent of y, ∂p/∂x = dp/dx is independent of y and

µ

∂ 2u ∂ y2

=µ

d 2u dy2

=

dp . dx

Since the left-hand side can, at most, depend only on y and the right-hand side is independent of y, both sides must equal the same constant C. That is,

µ

d 2u dy2

= C.

Hence, the velocity distribution has the form

u (y) =

C 2 y + C1y + C2 . 2µ

Using the boundary conditions to evaluate the constants,

u (0 ) = 0

→

C2 = 0 and u ( L ) = 0

→

C1 = −CL/2µ . Continued …..

PROBLEM 6.25 (Cont.) C 2 u (y) = y − Ly . 2µ

)

(

The velocity profile is

The profile is symmetric about the midplane, in which case the maximum velocity exists at y = L/2. Hence, C  L2  L2 dp u ( L/2 ) = u max = or u max = − . − 

2µ  

<

8µ dx

4  

(c) For fully developed thermal conditions, (∂T/∂x) = 0 and temperature depends only on y. Hence with v = 0, ∂u/∂x = 0, and the prescribed assumptions, the energy equation becomes 2 2

 du  ∂ i d T dp =k +u +µ  . 2 ∂ x dx  dy  dy ∂ i ∂ e 1 dp where With i = e + p/ρ, = + ∂ x ∂ x ρ dx

ρu

∂ e ∂ e∂ T ∂ e∂ ρ = + = 0. ∂ x ∂ T∂ x ∂ ρ∂ x 2

d 2T

 du  0=k +µ  . 2  dy  dy

Hence, the energy equation becomes

<

With du/dy = (C/2µ) (2y - L), it follows that d 2T C2 4y2 − 4Ly + L2 . =− 2 4kµ dy

)

(

Integrating twice,

T ( y) = −

C2 4kµ

 y 4 2Ly3 L2 y2  +  −  + C3 y + C4 3 2   3 

Using the boundary conditions to evaluate the constants,

T ( 0 ) = T2 Hence,

→

C4 = T2

and

T ( L ) = T1

C2 y T ( y ) = T2 +   ( T1 − T2 ) − 4kµ L

→

C3 =

C2 L3 ( T1 − T2 ) . + 24kµ L

 y4 2Ly3 L2 y2 L3 y  + −  − . 3 2 6   3 

<

From Fourier’s law,

q′′ ( L ) = − k q′′ ( L ) =

∂ T k C2 = ( T2 − T1 ) + ∂ y y=L L 4µ

4 3 L3   L − 2L3 + L3 −  6  3 

k C 2L3 . (T2 − T1 ) + L 24 µ

<

COMMENTS: The third and second terms on the right-hand sides of the temperature distribution and heat flux, respectively, represents the effects of viscous dissipation. If C is large (due to large µ or umax), viscous dissipation is significant. If C is small, conduction effects dominate.

PROBLEM 6.26 KNOWN: Pressure independence of µ, k and cp. FIND: Pressure dependence of ν and α for air at 350K and p = 1, 10 atm. ASSUMPTIONS: Perfect gas behavior for air. -6 2

-6 2

PROPERTIES: Table A-4, Air (350K, 1 atm): ν = 20.92 × 10 m /s, α = 29.9 × 10 m /s. ANALYSIS: The kinematic viscosity and thermal diffusivity are, respectively, ν =µ/ρ

α = k/ρ c p .

Hence, ν and α are inversely proportional to ρ. For an incompressible liquid, ρ is constant. Hence ν and α are independent of pressure.

<

For a perfect gas, ρ = p/RT. Hence, ρ is directly proportional to p, in which case ν and α vary inversely with pressure. It follows that ν and α are inversely proportional to pressure. To calculate ν or α for a perfect gas at p ≠ 1 atm, 1 p 1 α ( p ) = α (1 atm) ⋅ p

ν ( p ) = ν (1 atm ) ⋅

Hence, for air at 350K, p(atm) 1 10

2

ν(m /s) -6

20.92 × 10 -6 2.09 × 10

2

α(m /s) -6

29.9 × 10 -6 2.99 × 10

COMMENTS: For the incompressible liquid and the perfect gas, Pr = ν/α is independent of pressure.

<

PROBLEM 6.27 KNOWN: Characteristic length, surface temperature and average heat flux for an object placed in an airstream of prescribed temperature and velocity. FIND: Average convection coefficient if characteristic length of object is increased by a factor of five and air velocity is decreased by a factor of five. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties. ANALYSIS: For a particular geometry, Nu L = f ( ReL , Pr ). The Reynolds numbers for each case are Case 1:

(100m/s )1m 100 m 2 / s VL ReL,1 = 1 1 = = ν1 ν1 ν1

Case 2:

( 20m/s ) 5m = 100 m2 / s . V L ReL,2 = 2 2 = ν2 ν2 ν2

Hence, with ν1 = ν2, ReL,1 = ReL,2. Since Pr1 = Pr2, it follows that Nu L,2 = Nu L,1 . Hence, h 2L2 / k 2 = h1L1 / k1 L h 2 = h1 1 = 0.2 h1. L2 For Case 1, using the rate equation, the convection coefficient is q1 = h1A1 ( Ts − T∞ )1 h1 =

(q1 / A1 ) = q1′′ 20, 000 W/m 2 = (Ts − T∞ )1 (Ts − T∞ )1 ( 400 − 300 ) K

= 200 W/m 2 ⋅ K.

Hence, it follows that for Case 2 h 2 = 0.2 × 200 W/m 2 ⋅ K = 40 W/m 2 ⋅ K.

<

COMMENTS: If ReL,2 were not equal to ReL,1, it would be necessary to know the specific form of f(ReL, Pr) before h2 could be determined.

PROBLEM 6.28 KNOWN: Heat transfer rate from a turbine blade for prescribed operating conditions. FIND: Heat transfer rate from a larger blade operating under different conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Surface area A is directly proportional to characteristic length L, (4) Negligible radiation, (5) Blade shapes are geometrically similar. ANALYSIS: For a prescribed geometry, Nu =

hL = f ( Re L , Pr ). k

The Reynolds numbers for the blades are ReL,1 = ( V1L1 / ν ) = 15 / ν

Re L,2 = ( V2L 2 / ν ) = 15 / ν .

Hence, with constant properties, ReL,1 = ReL,2 . Also, Pr1 = Pr2 . Therefore, Nu 2 = Nu 1 h 2L 2 / k = h1L1 / k L L q1 h 2 = 1 h1 = 1 . L2 L 2 A1 Ts,1 − T∞

(

) (

)

(

)

Hence, the heat rate for the second blade is

( (

) )

L A 2 Ts,2 − T∞ q 2 = h 2 A 2 Ts,2 − T∞ = 1 q1 L2 A1 Ts,1 − T∞ Ts,2 − T∞ ( 400 − 35 ) 1500 W q2 = q1 = ( ) Ts,1 − T∞ (300 − 35)

(

)

q 2 = 2066 W.

<

COMMENTS: The slight variation of ν from Case 1 to Case 2 would cause ReL,2 to differ from ReL,1. However, for the prescribed conditions, this non-constant property effect is small.

PROBLEM 6.29 KNOWN: Experimental measurements of the heat transfer coefficient for a square bar in cross flow. FIND: (a) h for the condition when L = 1m and V = 15m/s, (b) h for the condition when L = 1m and V = 30m/s, (c) Effect of defining a side as the characteristic length. SCHEMATIC:

ASSUMPTIONS: (1) Functional form Nu = CRem Pr n applies with C, m, n being constants, (2) Constant properties. ANALYSIS: (a) For the experiments and the condition L = 1m and V = 15m/s, it follows that Pr as well as C, m, and n are constants. Hence hL α ( VL ) . m

Using the experimental results, find m. Substituting values h1L1  V1L1  =  h 2L 2  V2L 2 

m

50 × 0.5  20 × 0.5  = 40 × 0.5  15 × 0.5 

m

giving m = 0.782. It follows then for L = 1m and V = 15m/s, L  V⋅L  h = h1 1   L  V1 ⋅ L1 

m

= 50

0.5  15 × 1.0    m 2 ⋅ K 1.0  20 × 0.5  W

×

0.782

= 34.3W/m 2 ⋅ K.

<

= 59.0W/m 2 ⋅ K.

<

(b) For the condition L = 1m and V = 30m/s, find L  V⋅L  h = h1 1   L  V1 ⋅ L1 

m

= 50

0.5  30 × 1.0    m 2 ⋅ K 1.0  20 × 0.5  W

×

0.782

(c) If the characteristic length were chosen as a side rather than the diagonal, the value of C would change. However, the coefficients m and n would not change. COMMENTS: The foregoing Nusselt number relation is used frequently in heat transfer analysis, providing appropriate scaling for the effects of length, velocity, and fluid properties on the heat transfer coefficient.

PROBLEM 6.30 KNOWN: Local Nusselt number correlation for flow over a roughened surface. FIND: Ratio of average heat transfer coefficient to local coefficient. SCHEMATIC:

ANALYSIS: The local convection coefficient is obtained from the prescribed correlation, k k 1/3 = 0.04 Re0.9 x Pr x x 0.9 x 0.9 V h x = 0.04 k   Pr1/3 ≡ C1x -0.1. x ν  

h x = Nu x

To determine the average heat transfer coefficient for the length zero to x, x 1 x 1 ∫ h x dx = C1 ∫ x -0.1dx 0 x 0 x C1 x 0.9 hx = = 1.11 C1 x -0.1. x 0.9

hx ≡

Hence, the ratio of the average to local coefficient is h x 1.11 C1 x -0.1 = = 1.11. hx C1 x -0.1 COMMENTS: Note that Nu x / Nu x is also equal to 1.11. Note, however, that 1 x Nu x ≠ ∫ Nu x dx. x 0

<

PROBLEM 6.31 KNOWN: Freestream velocity and average convection heat transfer associated with fluid flow over a surface of prescribed characteristic length. FIND: Values of Nu L , Re L , Pr, jH for (a) air, (b) engine oil, (c) mercury, (d) water. SCHEMATIC:

PROPERTIES: For the fluids at 300K: Fluid

Table

Air

A.4

Engine Oil Mercury Water

2

ν(m /s) -6

15.89 × 10

0.0263

-6

550 × 10

A.5

0.145

-6

0.113 × 10

A.5

8.54

-6

0.858 × 10

A.6

2

α(m /s)

k(W/m⋅K)

0.613

Pr

-7

22.5 × 10

-7

0.859 × 10

0.71 6400

-7

45.30 × 10

0.025

-7

1.47 × 10

5.83

ANALYSIS: The appropriate relations required are Nu L =

hL k

Fluid

Air Engine Oil Mercury Water

Re L =

Nu L

3802 690 11.7 163

VL ν

Pr =

ν α

jH = StPr 2/3

ReL

St =

Pr

4

6.29 × 10

3

1.82 × 10

6

8.85 × 10

6

1.17 × 10

0.71 6403

Nu L ReL Pr

<

jH

0.068 0.0204 -6

0.025

4.52 × 10

5.84

7.74 × 10

COMMENTS: Note the wide range of Pr associated with the fluids.

-5

PROBLEM 6.32 KNOWN: Variation of hx with x for flow over a flat plate. FIND: Ratio of average Nusselt number for the entire plate to the local Nusselt number at x = L. SCHEMATIC:

ANALYSIS: The expressions for the local and average Nusselt numbers are Nu L Nu L

)

(

CL-1/2 L CL1/2 hLL = = = k k k h LL = k

where hL =

1 L C L 2C 1/2 L = 2 CL-1/2 . ∫ h x dx = ∫ x -1/2dx = 0 0 L L L

Hence, Nu L =

2 CL-1/2 ( L ) k

2 CL1/2 = k

and Nu L Nu L

= 2.

<

COMMENTS: Note the manner in which Nu L is defined in terms of h L . Also note that Nu L ≠

1 L ∫ Nu x dx. L 0

PROBLEM 6.33 KNOWN: Laminar boundary layer flow of air at 20°C and 1 atm having δ t = 1.13 δ . FIND: Ratio δ / δ t when fluid is ethylene glycol for same conditions. SCHEMATIC:

ASSUMPTIONS: (1) Laminar flow. PROPERTIES: Table A-4, Air (293K, 1 atm): Pr = 0.709; Table A-5, Ethylene glycol (293K): Pr = 211. ANALYSIS: The Prandtl number strongly influences relative growth of the velocity, δ , and thermal, δ t , boundary layers. For laminar flow, the approximate relationship is given by Pr n ≈

δ δt

where n is a positive coefficient. Substituting the values for air

(0.709 )n =

1 1.13

find that n = 0.355. Hence, for ethylene glycol it follows that

δ = Pr 0.355 = 2110.355 = 6.69. δt COMMENTS: (1) For laminar flow, generally we find n = 0.33. In which case, δ / δ t = 5.85.

(2) Recognize the physical importance of ν > α, which gives large values of the Prandtl number, and causes δ > δ t .

<

PROBLEM 6.34 KNOWN: Air, water, engine oil or mercury at 300K in laminar, parallel flow over a flat plate. FIND: Sketch of velocity and thermal boundary layer thickness. ASSUMPTIONS: (1) Laminar flow. PROPERTIES: For the fluids at 300K:

Fluid

Table

Air Water Engine Oil Mercury

A.4 A.6 A.5 A.5

Pr 0.71 5.83 6400 0.025

ANALYSIS: For laminar, boundary layer flow over a flat plate. δ ~ Pr n

δt

where n > 0. Hence, the boundary layers appear as shown below. Air:

Water:

Engine Oil:

Mercury:

COMMENTS: Although Pr strongly influences relative boundary layer development in laminar flow, its influence is weak for turbulent flow.

PROBLEM 6.35 KNOWN: Expression for the local heat transfer coefficient of air at prescribed velocity and temperature flowing over electronic elements on a circuit board and heat dissipation rate for a 4 × 4 mm chip located 120mm from the leading edge. FIND: Surface temperature of the chip surface, Ts. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Power dissipated within chip is lost by convection across the upper surface only, (3) Chip surface is isothermal, (4) The average heat transfer coefficient for the chip surface is equivalent to the local value at x = L. PROPERTIES: Table A-4, Air (assume Ts = 45°C, Tf = (45 + 25)/2 = 35°C = 308K, 1atm): ν = -6 2 -3 16.69 × 10 m /s, k = 26.9 × 10 W/m⋅K, Pr = 0.703. ANALYSIS: From an energy balance on the chip (see above),

qconv = E g = 30W.

(1)

Newton’s law of cooling for the upper chip surface can be written as

Ts = T∞ + q conv / h Achip

(2)

( )

where A chip =  2 . Assume that the average heat transfer coefficient h over the chip surface is equivalent to the local coefficient evaluated at x = L. That is, h chip ≈ h x ( L ) where the local coefficient can be evaluated from the special correlation for this situation, 0.85 hxx  Vx  Nu x = Pr1/ 3 = 0.04  

ν 

k

and substituting numerical values with x = L, find 0.85 k  VL  h x = 0.04  Pr1/ 3 

L ν 

 0.0269 W/m ⋅ K   10 m/s × 0.120 m  h x = 0.04     0.120 m  16.69 × 10-6 m 2 / s 

0.85

(0.703)1/ 3 = 107 W/m 2 ⋅ K.

The surface temperature of the chip is from Eq. (2), 2 Ts = 25$ C + 30 × 10-3 W/107 W/m2 ⋅ K × ( 0.004m ) = 42.5$ C.

<

COMMENTS: (1) Note that the estimated value for Tf used to evaluate the air properties was reasonable. (2) Alternatively, we could have evaluated h chip by performing the integration of the local value, h(x).

PROBLEM 6.36 KNOWN: Location and dimensions of computer chip on a circuit board. Form of the convection correlation. Maximum allowable chip temperature and surface emissivity. Temperature of cooling air and surroundings. FIND: Effect of air velocity on maximum power dissipation, first without and then with consideration of radiation effects. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Negligible temperature variations in chip, (3) Heat transfer exclusively from the top surface of the chip, (4) The local heat transfer coefficient at x = L provides a good approximation to the average heat transfer coefficient for the chip surface. PROPERTIES: Table A.4, air ( T = ( T∞ + Tc ) 2 = 328 K): ν = 18.71 × 10-6 m2/s, k = 0.0284 W/m⋅K, Pr = 0.703. ANALYSIS: Performing an energy balance for a control surface about the chip, we obtain Pc = qconv +

(

)

2 . With qrad, where qconv = hA s ( Tc − T∞ ) , qrad = h r A s ( Tc − Tsur ) , and h r = εσ ( Tc + Tsur ) Tc2 + Tsur

h ≈ h L , the convection coefficient may be determined from the correlation provided in Problem 6.35 1/3 (NuL = 0.04 Re0.85 L Pr ). Hence,

(

1/ 3 2 Pc = " 2 0.04 ( k L ) Re0.85 (Tc − T∞ ) + εσ (Tc + Tsur ) Tc2 + Tsur L Pr



)(Tc − Tsur )

where ReL = VL/ν. Computing the right side of this expression for ε = 0 and ε = 0.85, we obtain the following results. 0.3

Chip power, Pc(W)

0.25 0.2 0.15 0.1 0.05 0 0

5

10

15

20

25

Velocity, V(m/s) epsilon = 0.85 epsilon = 0

Since hL increases as V0.85, the chip power must increase with V in the same manner. Radiation exchange increases Pc by a fixed, but small (6 mW) amount. While hL varies from 14.5 to 223 W/m2⋅K over the prescribed velocity range, hr = 6.5 W/m2⋅K is a constant, independent of V. COMMENTS: Alternatively, h could have been evaluated by integrating hx over the range 118 ≤ x ≤ 122 mm to obtain the appropriate average. However, the value would be extremely close to hx=L.

PROBLEM 6.37 KNOWN: Form of Nusselt number for flow of air or a dielectric liquid over components of a circuit card. FIND: Ratios of time constants associated with intermittent heating and cooling. Fluid that provides faster thermal response. -5

2

PROPERTIES: Prescribed. Air: k = 0.026 W/m⋅K, ν = 2 × 10 m /s, Pr = 0.71. Dielectric liquid: k -6 2 = 0.064 W/m⋅K, ν = 10 m /s, Pr = 25. ANALYSIS: From Eq. 5.7, the thermal time constant is

τt =

ρ∀c hAs

Since the only variable that changes with the fluid is the convection coefficient, where

k k k  VL m n m n h = Nu L = CReL Pr = C  Pr L L L  ν  the desired ratio reduces to

τ t,air ( a )

m

h k  ν   Pr  = d = d a  d τ t,dielectric( d) ha ka  ν d   Pra  0.064  2 ×10 −5    = τ t,d 0.026  10 −6  τ t,a

0.8

0.33

 25   0.71   

n

= 88.6

Since its time constant is nearly two orders of magnitude smaller than that of the air, the dielectric liquid is clearly the fluid of choice. COMMENTS: The accelerated testing procedure suggested by this problem is commonly used to test the durability of electronic packages.

PROBLEM 6.38 KNOWN: Form of the Nusselt number correlation for forced convection and fluid properties. FIND: Expression for figure of merit FF and values for air, water and a dielectric liquid. -5

2

PROPERTIES: Prescribed. Air: k = 0.026 W/m⋅K, ν = 1.5 × 10 m /s, Pr = 0.70. Water: k = -6 2 -6 2 0.600 W/m⋅K, ν = 10 m /s, Pr = 5.0. Dielectric liquid: k = 0.064 W/m⋅K, ν = 10 m /s, Pr = 25 m n ANALYSIS: With Nu L ~ Re L Pr , the convection coefficient may be expressed as

h~

k  VL m n Vm  kPr n  Pr ~ L  ν  L1− m  ν m

   

The figure of merit is therefore

FF =

kPr n

<

νm

and for the three fluids, with m = 0.80 and n = 0.33,

(

FF W ⋅ s 0.8 / m 2.6 ⋅ K

)

Air 167

Water 64,400

Dielectric 11,700

<

Water is clearly the superior heat transfer fluid, while air is the least effective. COMMENTS: The figure of merit indicates that heat transfer is enhanced by fluids of large k, large Pr and small ν.

PROBLEM 6.39 KNOWN: Ambient, interior and dewpoint temperatures. Vehicle speed and dimensions of windshield. Heat transfer correlation for external flow. FIND: Minimum value of convection coefficient needed to prevent condensation on interior surface of windshield. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional heat transfer, (3) Constant properties. PROPERTIES: Table A-3, glass: kg = 1.4 W/m⋅K. Prescribed, air: k = 0.023 W/m⋅K, ν = 12.5 × -6 2 10 m /s, Pr = 0.70. ANALYSIS: From the prescribed thermal circuit, conservation of energy yields

T∞,i − Ts,i 1/ h i

=

Ts,i − T∞,o

t / k g + 1/ h o

where h o may be obtained from the correlation

h L 1/ 3 Nu L = o = 0.030 Re0.8 L Pr k -6

2

With V = (70 mph × 1585 m/mile)/3600 s/h = 30.8 m/s, ReD = (30.8 m/s × 0.800 m)/12.5 × 10 m /s 6 = 1.97 × 10 and 0.8 0.023 W / m ⋅ K ho = 0.030 1.97 × 106 (0.70 )1/ 3 = 83.1W / m2 ⋅ K

0.800 m

(

)

From the energy balance, with Ts,i = Tdp = 10°C −1 Ts,i − T∞,o )  t ( 1  hi = +   (T∞,i − Ts,i )  k g ho 

 (10 + 15 ) °C  0.006 m + 1  hi = (50 − 10 ) °C  1.4 W / m ⋅ K 83.1W / m 2 ⋅ K  hi = 38.3 W / m 2 ⋅ K

−1

<

COMMENTS: The output of the fan in the automobile’s heater/defroster system must maintain a velocity for flow over the inner surface that is large enough to provide the foregoing value of hi . In addition, the output of the heater must be sufficient to maintain the prescribed value of T∞,i at this velocity.

PROBLEM 6.40 KNOWN: Drag force and air flow conditions associated with a flat plate. FIND: Rate of heat transfer from the plate. SCHEMATIC:

ASSUMPTIONS: (1) Chilton-Colburn analogy is applicable. 3

PROPERTIES: Table A-4, Air (70°C,1 atm): ρ = 1.018 kg/m , cp = 1009 J/kg⋅K, Pr = 0.70, -6 2

ν = 20.22 × 10 m /s. ANALYSIS: The rate of heat transfer from the plate is

( ) (Ts − T∞ )

q = 2h L2

where h may be obtained from the Chilton-Colburn analogy, C h jH = f = St Pr 2/3 = Pr 2 / 3 2 ρ u ∞ cp 2 Cf 1 τ s 1 (0.075 N/2 ) / (0.2m ) = = = 5.76 ×10−4. 2 2 2 2 ρ u / 2 2 1.018 kg/m3 (40 m/s ) / 2 ∞ Hence, C h = f ρ u ∞ cp Pr -2/3 2 −2 / 3 h = 5.76 × 10-4 1.018kg/m3 40m/s (1009J/kg ⋅ K ) (0.70 )

(

h = 30 W/m 2 ⋅ K. The heat rate is

(

q = 2 30 W/m 2 ⋅ K

)

) (0.2m )2 (120 − 20) C $

<

q = 240 W.

COMMENTS: Although the flow is laminar over the entire surface (ReL = u∞L/ν = 40 m/s -6 2

5

× 0.2m/20.22 × 10 m /s = 4.0 × 10 ), the pressure gradient is zero and the Chilton-Colburn analogy is applicable to average, as well as local, surface conditions. Note that the only contribution to the drag force is made by the surface shear stress.

PROBLEM 6.41 KNOWN: Air flow conditions and drag force associated with a heater of prescribed surface temperature and area. FIND: Required heater power. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Reynolds analogy is applicable, (3) Bottom surface is adiabatic. 3

PROPERTIES: Table A-4, Air (Tf = 350K, 1atm): ρ = 0.995 kg/m , cp = 1009 J/kg⋅K, Pr = 0.700. ANALYSIS: The average shear stress and friction coefficient are F 0.25 N τs = D = = 1 N/m 2 2 A 0.25 m τs 1 N/m 2 Cf = = = 8.93 × 10−3. 2 2 3 ρ u ∞ / 2 0.995 kg/m (15m/s ) / 2 From the Reynolds analogy, St =

h C = f Pr −2 / 3 . 2 ρ u ∞ cp

Solving for h and substituting numerical values, find

(

h = 0.995 kg/m3 (15m/s ) 1009 J/kg ⋅ K 8.93 ×10-3 / 2 h = 85 W/m 2 ⋅ K.

) (0.7 )−2 / 3

Hence, the heat rate is

(

q = h A ( Ts − T∞ ) = 85W/m 2 ⋅ K 0.25m 2 q = 2.66 kW.

) (140 −15) C $

<

COMMENTS: Due to bottom heat losses, which have been assumed negligible, the actual power requirement would exceed 2.66 kW.

PROBLEM 6.42 KNOWN: Heat transfer correlation associated with parallel flow over a rough flat plate. Velocity and temperature of air flow over the plate. FIND: Surface shear stress l m from the leading edge. SCHEMATIC:

ASSUMPTIONS: (1) Modified Reynolds analogy is applicable, (2) Constant properties. -6 2

PROPERTIES: Table A-4, Air (300K, 1atm): ν = 15.89 × 10 m /s, Pr = 0.71, ρ = 1.16 3

kg/m . ANALYSIS: Applying the Chilton-Colburn analogy Cf Nu x 0.04 Re0.9 Pr1/ 3 2 / 3 2 / 3 2 / 3 x Pr Pr = St x Pr = = 2 Re x Pr Re x Pr Cf = 0.04 Re-0.1 x 2 where u x 50 m/s ×1m Re x = ∞ = = 3.15 ×106. -6 2 ν 15.89 × 10 m / s Hence, the friction coefficient is

(

Cf = 0.08 3.15 × 106

)

−0.1

(

2 /2 = 0.0179 = τ s / ρ u ∞

)

and the surface shear stress is

(

)

2 / 2 = 0.0179 × 1.16kg/m3 50 m/s / 2 τ s = Cf ρ u ∞ ( ) 2

τ s = 25.96 kg/m ⋅ s 2 = 25.96 N/m 2 . COMMENTS: Note that turbulent flow will exist at the designated location.

<

PROBLEM 6.43 KNOWN: Nominal operating conditions of aircraft and characteristic length and average friction coefficient of wing. FIND: Average heat flux needed to maintain prescribed surface temperature of wing. SCHEMATIC:

ASSUMPTIONS: (1) Applicability of modified Reynolds analogy, (2) Constant properties. -6

2

PROPERTIES: Prescribed, Air: ν = 16.3 × 10 m /s, k = 0.022 W/m⋅K, Pr = 0.72. ANALYSIS: The average heat flux that must be maintained over the surface of the air foil is q ′′ = h ( Ts − T∞ ) , where the average convection coefficient may be obtained from the modified Reynolds analogy.

Nu L Nu L Cf Pr 2 / 3 = = St Pr 2 / 3 = 2 Re L Pr ReL Pr1/ 3 Hence, with Re L = VL / ν = 100 m / s ( 2m ) / 16.3 × 10 −6 m 2 / s = 1.23 × 107 ,

Nu L = h=

(

)

0.0025 1/ 3 1.23 ×107 ( 0.72 ) = 13, 780 2

k 0.022 W / m ⋅ K Nu L = (13, 780 ) = 152 W / m2 ⋅ K L 2m

q′′ = 152 W / m 2 ⋅ K 5 − ( −23) °C = 4260 W / m 2

<

COMMENTS: If the flow is turbulent over the entire airfoil, the modified Reynolds analogy provides a good measure of the relationship between surface friction and heat transfer. The relation becomes more approximate with increasing laminar boundary layer development on the surface and increasing values of the magnitude of the pressure gradient.

PROBLEM 6.44 2

KNOWN: Average frictional shear stress of τ s = 0.0625 N/m on upper surface of circuit board with densely packed integrated circuits (ICs) FIND: Allowable power dissipation from the upper surface of the board if the average surface temperature of the ICs must not exceed a rise of 25°C above ambient air temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) The modified Reynolds analogy is applicable, (3) Negligible heat transfer from bottom side of the circuit board, and (4) Thermophysical properties required for the analysis evaluated at 300 K, 3

PROPERTIES: Table A-4, Air (Tf = 300 K, 1 atm): ρ = 1.161 kg/m , cp = 1007 J/kg⋅K, Pr = 0.707. ANALYSIS: The power dissipation from the circuit board can be calculated from the convection rate equation assuming an excess temperature (Ts - T∞) = 25°C. q = h As ( Ts − T∞ ) (1) The average convection coefficient can be estimated from the Reynolds analogy and the measured average frictional shear stress τ s .

Cf = St Pr 2 / 3 2

Cf =

τs ρ V2 / 2

St =

h ρ V cp

(2,3,4)

With V = u∞ and substituting numerical values, find h.

τs

ρ V2

=

h Pr 2 / 3 ρ V cp

h=

τ s cp −2 / 3 Pr V

h=

0.0625 N / m 2 ×1007 J / kg ⋅ K (0.707 )−2 / 3 = 39.7 W / m2 ⋅ K 2 m/s

Substituting this result into Eq. (1), the allowable power dissipation is q = 39.7 W / m 2 ⋅ K × 0.120 × 0.120 m 2 × 25 K = 14.3 W

(

)

COMMENTS: For this analyses using the modified or Chilton-Colburn analogy, we found Cf = 0.0269 and St = 0.0170. Using the Reynolds analogy, the results are slightly different with h = 31.5 W / m 2 ⋅ K and q = 11.3 W.

<

PROBLEM 6.45 KNOWN: Evaporation rate of water from a lake. FIND: The convection mass transfer coefficient, h m . SCHEMATIC:

ASSUMPTIONS: (1) Equilibrium at water vapor-liquid surface, (2) Isothermal conditions, (3) Perfect gas behavior of water vapor, (4) Air at standard atmospheric pressure. PROPERTIES: Table A-6, Saturated water vapor (300K): pA,sat = 0.03531 bar, ρA,sat = 3

1/vg = 0.02556 kg/m . ANALYSIS: The convection mass transfer (evaporation) rate equation can be written in the form hm =

n ′′A

( ρA,s − ρA,∞ )

where

ρ A,s = ρ A,sat , the saturation density at the temperature of the water and

ρ A,∞ = φρ A,sat which follows from the definition of the relative humidity, φ = pA/pA,sat and perfect gas behavior. Hence, n ′′A hm = ρ A,sat (1 − φ ) and substituting numerical values, find hm =

0.1 kg/m 2 ⋅ h ×1/3600 s/h 0.02556 kg/m3 (1 − 0.3)

= 1.55 × 10−3 m/s.

<

COMMENTS: (1) From knowledge of pA,sat, the perfect gas law could be used to obtain the saturation density. p A,sat ΜA 0.03531 bar × 18 kg/kmol ρ A,sat = = = 0.02548 kg/m3. -2 3 ℜT 8.314 ×10 m ⋅ bar/kmol ⋅ K (300K ) This value is within 0.3% of that obtained from Table A-6. (2) Note that psychrometric charts could also be used to obtain ρA,sat and ρA,∞.

PROBLEM 6.46 KNOWN: Evaporation rate from pan of water of prescribed diameter. Water temperature. Air temperature and relative humidity. FIND: (a) Convection mass transfer coefficient, (b) Evaporation rate for increased relative humidity, (c) Evaporation rate for increased temperature. SCHEMATIC:

ASSUMPTIONS: (1) Water vapor is saturated at liquid interface and may be approximated as a perfect gas. PROPERTIES: Table A-6, Saturated water vapor (Ts = 296K): ρ A,sat = vg-1 = (49.4 m3/kg)-1 = −1 3 0.0202 kg/m3; (Ts = 320 K): ρ A,sat = v-1 = 0.0715 kg/m3. g = 13.98 m / kg

)

(

ANALYSIS: (a) Since evaporation is a convection mass transfer process, the rate equation has the

(

)

 evap = h m A ρ A,s − ρ A,∞ and the mass transfer coefficient is form m  evap m 1.5 × 10−5 kg/s hm = = = 0.0179 m/s π D 2 / 4 ρ A,s − ρ A,∞ (π /4 )(0.23 m )2 0.0202 kg/m3

)(

(

)

<

with Ts = T∞ = 23°C and φ∞ = 0. (b) If the relative humidity of the ambient air is increased to 50%, the ratio of the evaporation rates is

 evap (φ∞ = 0.5 ) m  evap (φ∞ = 0 ) m

=

h m A  ρ A,s ( Ts ) − φ∞ ρ A,s ( T∞ ) h m A ρ A,s (Ts ) 

 evap (φ∞ = 0.5 ) = 1.5 ×10−5 kg/s 1 − 0.5 Hence, m 

= 1 − φ∞

ρ A,s (T∞ ) ρ A,s (Ts )

.

0.0202 kg/m3   = 0.75 ×10−5 kg/s. 3 0.0202 kg/m 

(c) If the temperature of the ambient air is increased from 23°C to 47°C, with φ∞ = 0 for both cases, the ratio of the evaporation rates is

) = hmAρA,s (47 C) = ρA,s (47 C) . (  evap (Ts = T∞ = 23 C ) h m Aρ A,s ( 23 C ) ρ A,s ( 23 C ) m kg/m Hence, m = 5.31 × 10 (T = T = 47 C ) = 1.5 ×10 kg/s 0.0715 0.0202 kg/m  evap Ts = T∞ = 47$ C m $



evap

s

∞

$

−5

$

$

$

$

3

3

−5

kg/s.

<

COMMENTS: Note the highly nonlinear dependence of the evaporation rate on the water  evap increases by 350%. temperature. For a 24°C rise in Ts , m

PROBLEM 6.47 KNOWN: Water temperature and air temperature and relative humidity. Surface recession rate. FIND: Mass evaporation rate per unit area. Convection mass transfer coefficient. SCHEMATIC:

ASSUMPTIONS: (1) Water vapor may be approximated as a perfect gas, (2) No water inflow; outflow is only due to evaporation. 3 PROPERTIES: Table A-6, Saturated water: Vapor (305K), ρg = v-1 g = 0.0336 kg/m ; 3 Liquid (305K), ρf = v-1 f = 995 kg/m .

ANALYSIS: Applying conservation of species to a control volume about the water,   −M A,out = M A,st d d dH  ′′evap A = ( ρf V ) = ( ρf AH ) = ρf A . −m dt dt dt Substituting numerical values, find  ′′evap = − ρf m

(

dH = −995kg/m3 −10−4 m/h dt

) (1/3600 s/h )

 ′′evap = 2.76 × 10−5 kg/s ⋅ m 2 . m

<

Because evaporation is a convection mass transfer process, it also follows that  ′′evap = n ′′A m or in terms of the rate equation,

(

)

 ′′evap = h m ρ A,s − ρ A,∞ = h m  ρ A,sat ( Ts ) − φ∞ ρA,sat (T∞ ) m  ′′evap = h m ρ A,sat (305K ) (1 − φ∞ ) , m and solving for the convection mass transfer coefficient, hm =

 ′′evap m

ρ A,sat (305K ) (1 − φ∞ )

h m = 1.37 × 10−3 m/s.

=

2.76 ×10−5 kg/s ⋅ m 2 0.0336 kg/m3 (1 − 0.4 )

<

COMMENTS: Conservation of species has been applied in exactly the same way as a conservation of energy. Note the sign convention.

PROBLEM 6.48 KNOWN: CO2 concentration in air and at the surface of a green leaf. Convection mass transfer coefficient. FIND: Rate of photosynthesis per unit area of leaf. SCHEMATIC:

ANALYSIS: Assuming that the CO2 (species A) is consumed as a reactant in photosynthesis at the same rate that it is transferred across the atmospheric boundary layer, the rate of photosynthesis per unit leaf surface area is given by the rate equation,

(

)

n′′A = h m ρ A,∞ − ρ A,s . Substituting numerical values, find

(

)

n′′A = 10−2 m/s 6 ×10-4 − 5 × 10−4 kg/m3 n′′A = 10−6 kg/s ⋅ m 2 .

<

COMMENTS: (1) It is recognized that CO2 transport is from the air to the leaf, and (ρA,s ρA,∞) in the rate equation has been replaced by (ρA,∞ - ρA,s). (2) The atmospheric concentration of CO2 is known to be increasing by approximately 0.3% per year. This increase in ρA,∞ will have the effect of increasing the photosynthesis rate and hence plant biomass production.

PROBLEM 6.49 KNOWN: Species concentration profile, CA(y), in a boundary layer at a particular location for flow over a surface. FIND: Expression for the mass transfer coefficient, hm, in terms of the profile constants, CA,∞ and DAB. Expression for the molar convection flux, N A ′′ . SCHEMATIC:

ASSUMPTIONS: (1) Parameters D, E, and F are constants at any location x, (2) DAB, the mass diffusion coefficient of A through B, is known. ANALYSIS: The convection mass transfer coefficient is defined in terms of the concentration gradient at the wall, h m ( x ) = − D AB

∂ CA / ∂ y )y=0 . (CA,s − CA,∞ )

The gradient at the surface follows from the profile, CA(y),

)

(

∂ CA  ∂ Dy2 + Ey + F = = + E.  ∂ y  y=0 ∂ y y=0 Hence, h m (x ) = −

D ABE

=

(CA,s − CA,∞ ) (

−D ABE . F − CA,∞

<

)

The molar flux follows from the rate equation,

(

)

N′′A = h m C A,s − CA,∞ = N′′A = − DABE.

(

− DABE ⋅ C A,s − C A,∞ . CA,s − CA,∞

)

(

)

<

COMMENTS: It is important to recognize that the influence of species B is present in the property DAB. Otherwise, all the parameters relate to species A.

PROBLEM 6.50 KNOWN: Steady, incompressible flow of binary mixture between infinite parallel plates with different species concentrations. FIND: Form of species continuity equation and concentration distribution. Species flux at upper surface. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional flow, (2) No chemical reactions, (3) Constant properties. ANALYSIS: For fully developed conditions, ∂CA/∂x = 0. Hence with v = 0, the species conservation equation reduces to d2 CA dy2

<

= 0.

Integrating twice, the general form of the species concentration distribution is CA ( y ) = C1y + C2. Using appropriate boundary conditions and evaluating the constants, CA ( 0) = CA,2 CA ( L ) = CA,1

→ →

C2 =CA,2 C1 = CA,1 − C A,2 /L,

(

)

the concentration distribution is

(

)

CA ( y ) = CA,2 + ( y/L ) CA,1 − CA,2 .

<

From Fick’s law, the species flux is N′′A ( L ) = − DAB N′′A ( L ) =

(

dCA dy y=L

)

DAB CA,2 − CA,1 . L

<

COMMENTS: An analogy between heat and mass transfer exists if viscous dissipation is 2 2 negligible. The energy equation is then d T/dy = 0. Hence, both heat and species transfer are influenced only by diffusion. Expressions for T(y) and q′′ (L ) are analogous to those for CA(y) and N′′A ( L ).

PROBLEM 6.51 KNOWN: Flow conditions between two parallel plates, across which vapor transfer occurs. FIND: (a) Variation of vapor molar concentration between the plates and mass rate of water production per unit area, (b) Heat required to sustain the process. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed, incompressible flow with constant properties, (3) Negligible body forces, (4) No chemical reactions, (5) All work interactions, including viscous dissipation, are negligible. ANALYSIS: (a) The flow will be fully developed in terms of the vapor concentration field, as well as the velocity and temperature fields. Hence

∂ CA =0 ∂ x

CA ( x,y ) = C A ( y ).

or

 = 0, v = 0 and constant DAB, the species conservation equation Also, with ∂CA/∂t = 0, N A reduces to d 2CA dy 2

= 0.

Separating and integrating twice, CA ( y ) = C1 ( y ) + C2 . Applying the boundary conditions, CA ( 0 ) = CA,0

→

C2 = CA,0

CA ( L ) = CA,L

→

CA,L = C1L + C2

C1 = −

CA,0 − CA,L L

find the species concentration distribution,

(

CA ( y ) = CA,0 − CA,0 − CA,L

) ( y/L ).

<

From Fick’s law, Eq. 6.19, the species transfer rate is N′′A = N′′A,s = − D AB

CA,0 − CA,L ∂ CA  = DAB .  ∂ y  y=0 L Continued …..

PROBLEM 6.51 (Cont.) Multiplying by the molecular weight of water vapor, Μ A, the mass rate of water production per unit area is n′′A = ΜA N′′A = ΜA D AB

CA,0 − CA,L L

.

<

(b) Heat must be supplied to the bottom surface in an amount equal to the latent and sensible heat transfer from the surface, q′′ = q′′lat + q′′sen

 dT  q′′ = n′′A,s h fg +  − k  .  dy  y=0 The temperature distribution may be obtained by solving the energy equation, which, for the prescribed conditions, reduces to d 2T dy2

= 0.

Separating and integrating twice, T ( y ) = C1y + C2 . Applying the boundary conditions, T ( 0 ) = T0 T ( L ) = TL

→ →

C2 = T0 C1 = ( T1 − T0 ) / L

find the temperature distribution, T ( y ) = T0 − ( T0 − TL ) y/L. Hence, −k

(T − T ) dT  =k 0 L .  dy  y=0 L

Accordingly, q′′ = ΜA DAB

CA,0 − CA,L L

h fg + k

(T0 − TL ) . L

<

COMMENTS: Despite the existence of the flow, species and energy transfer across the air are uninfluenced by advection and transfer is only by diffusion. If the flow were not fully developed, advection would have a significant influence on the species concentration and temperature fields and hence on the rate of species and energy transfer. The foregoing results would, of course, apply in the case of no air flow. The physical condition is an example of Poiseuille flow with heat and mass transfer.

PROBLEM 6.52 KNOWN: The conservation equations, Eqs. E.24 and E.31. FIND: (a) Describe physical significance of terms in these equations, (b) Identify approximations and special conditions used to reduce these equations to the boundary layer equations, Eqs. 6.33 and 6.34, (c) Identify the conditions under which these two boundary layer equations have the same form and, hence, an analogy will exist. ANALYSIS: (a) The energy conservation equation, Eq. E.24, has the form ∂ i ∂ i ∂  ∂ T ∂  ∂ T   ∂ p ∂ p & ρu +ρ v = k + k + u +v + µ Φ + q.     ∂ x ∂ y ∂ x  ∂ x ∂ y  ∂ y  ∂ x ∂ y  1a 1b 2a 2b 3 4 5 The terms, as identified, have the following phnysical significance: 1. Change of enthalpy (thermal + flow work) advected in x and y directions, 2. Change of conduction flux in x and y directions, 3. Work done by static pressure forces, 4. Word done by viscous stresses, 5. Rate of energy generation. The species mass conservation equation for a constant total concentration has the form ∂ CA ∂ CA ∂  ∂ CA  ∂  ∂ CA  & u +v = D AB + DAB + NA    ∂ x ∂ y ∂ x ∂ x  ∂ y ∂ y  1a 1b 2a 2b 3

<

1. Change in species transport due to advection in x and y directions, 2. Change in species transport by diffusion in x and y directions, and 3. Rate of species generation.

<

(b) The special conditions used to reduce the above equations to the boundary layer equations are: & A = 0) , without internal constant properties, incompressible flow, non-reacting species ( N

heat generation (q& = 0 ), species diffusion has negligible effect on the thermal boundary layer, u(∂ p/∂ x) is negligible. The approximations are,  ∂ u ∂ u ∂ v >> , ,  u >> v ∂ y ∂ x ∂ y  ∂ T ∂ T Thermal b.1.: >> Concentration b.1.:  ∂ y ∂ x  The resulting simplified boundary layer equations are Velocity boundary layer

∂ v ∂ x ∂ C A ∂ CA >> .  ∂ x  ∂ y

2

∂ T ∂ T ∂ 2T ν  ∂ u  ∂ CA ∂ CA ∂ 2 CA u +v =α +  u +v = D AB  ∂ x ∂ y ∂ x ∂ y ∂ y2 c  ∂ y  ∂ y2 1a 1b 2a 3 1c 1d 2b where the terms are: 1. Advective transport, 2. Diffusion, and 3. Viscous dissipation.

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(c) When viscous dissipation effects are negligible, the two boundary layer equations have identical form. If the boundary conditions for each equation are of the same form, an analogy between heat and mass (species) transfer exists.

PROBLEM 6.53 KNOWN: Thickness and inclination of a liquid film. Mass density of gas in solution at free surface of liquid. FIND: (a) Liquid momentum equation and velocity distribution for the x-direction. Maximum velocity, (b) Continuity equation and density distribution of the gas in the liquid, (c) Expression for the local Sherwood number, (d) Total gas absorption rate for the film, (e) Mass rate of NH3 removal by a water film for prescribed conditions.

SCHEMATIC: NH3 (A) – Water (B) L = 2m δ = 1 mm D = 0.05m W = πD = 0.157m 3 ρA,o = 25 kg/m -9 2 DAB = 2 × 10 m /s φ = 0° ASSUMPTIONS: (1) Steady-state conditions, (2) The film is in fully developed, laminar flow, (3) Negligible shear stress at the liquid-gas interface, (4) Constant properties, (5) Negligible gas concentration at x = 0 and y = δ, (6) No chemical reactions in the liquid, (7) Total mass density is constant, (8) Liquid may be approximated as semi-infinite to gas transport. 3 -6 2 PROPERTIES: Table A-6, Water, liquid (300K): ρf = 1/vf = 997 kg/m , µ = 855 × 10 N⋅s/m , ν -6 2 = µ/ρf = 0.855 × 10 m /s. ANALYSIS: (a) For fully developed flow (v = w = 0, ∂u/∂x = 0), the x-momentum equation is 0 = ∂τ yx / ∂ y + X where τ yx = µ (∂ u/∂ y ) and X = ( ρ g ) cos φ . That is, the momentum equation reduces to a balance between gravitational and shear forces. Hence,

(

)

µ ∂ 2 u/∂ y 2 = − ( ρ g ) cos φ . Integrating, ∂ u/∂ y = − ( g cos φ /ν ) y + C1 Applying the boundary conditions, ∂ u/∂ y )y=0 = 0 → C1 = 0 u (δ ) = 0

C2 = g cos φ

→

u = − ( g cos φ /2ν ) y 2 + C1y + C 2 .

δ2 . 2ν

)

(

g cos φ δ 2  2 1 − ( y/δ )  2ν 2ν and the maximum velocity exists at y = 0, Hence,

u=

g cos φ

(

<

δ 2 − y2 =

)

u max = u (0 ) = g cos φ δ 2 / 2ν .

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(b) Species transport within the liquid is influenced by diffusion in the y-direction and convection in the x-direction. Hence, the species continuity equation with u assumed equal to umax throughout the region of gas penetration is

Continued …..

PROBLEM 6.53 (Cont.) 2

∂ρ ∂ ρA u A = DAB ∂ x ∂ y2

∂ 2 ρ A u max ∂ρ A = . D AB ∂ x ∂ y2

Appropriate boundary conditions are: ρA(x,0) = ρA,o and ρA(x,∞) = 0 and the entrance condition is: ρA(0,y) = 0. The problem is therefore analogous to transient conduction in a semi-infinite medium due to a sudden change in surface temperature. From Section 5.7, the solution is then ρ A − ρ A,o y y ρ A = ρ A,o erfc = erf 1/ 2 1/ 2 0 − ρ A,o 2 ( D AB x/u max ) 2 ( D AB x/u max )

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(c) The Sherwood number is defined as Sh x = ∂ρ A ∂ y y=0

h m,x x

where

D AB

= − ρ A,o

2

(π )1/ 2

h m,x ≡

n ′′A,x

ρ A,o

=

−D AB∂ρ A / ∂ y )y=0 ρ A,o

 y 2 u max  1   4 D AB x  2 ( D AB x/u max )1/ 2

1/ 2

exp  −

= − ρ A,o y=0

 u max  π D x   AB 

.

Hence, h m,x

1/ 2 D u  = max AB  π x 

Sh x =

1

(π )1/ 2

1/ 2

 u max x   D   AB 

=

1

(π )1/ 2

1/ 2

 u max x   ν 

1/ 2

 ν  D   AB 

and with Rex ≡ umax x/ν, Sh x = 1/ (π )



1/ 2 

<

1/2 Re1/2 Sc1/2 = 0.564 Re1/2 x Sc .  x

(d) The total gas absorption rate may be expressed as

n A = h m,x ( W ⋅ L ) ρ A,o where the average mass transfer convection coefficient is 1/ 2 1/ 2 1 L 1  u max D AB   4u max D AB  L dx h m,x = ∫ h m,x dx =   ∫ 0 1/2 =   . π π L L 0 L x Hence, the absorption rate per unit width is

ρ A,o . n A / W = ( 4u max D AB L / π ) (e) From the foregoing results, it follows that the ammonia absorption rate is 1/ 2

<

1/ 2

1/ 2  4 g cosφδ 2 D L  D L  4u AB n A =  max AB  W ρ A,o =   π 2πν    

W ρ A,o .

Substituting numerical values,

(

) (

)

1/ 2

2   −3 2 -9 2 2 × 10 m /s 2m   4 × 9.8 m/s × 1 × 10 m nA =   -6 2 2π × 0.855 × 10 m / s    

( 0.157m ) 25 kg/m3 = 6.71 × 10−4 kg/s.

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COMMENTS: Note that ρA,o ≠ ρA,∞, where ρA,∞ is the mass density of the gas phase. The value of ρA,o depends upon the pressure of the gas and the solubility of the gas in the liquid.

PROBLEM 6.54 KNOWN: Cross flow of gas X over object with prescribed characteristic length L, Reynolds number, and average heat transfer coefficient. Thermophysical properties of gas X, liquid Y, and vapor Y. FIND: Average mass transfer coefficient for same object when impregnated with liquid Y and subjected to same flow conditions. SCHEMATIC:

ASSUMPTIONS: (1) Heat and mass transfer analogy is applicable, (2) Vapor Y behaves as perfect gas PROPERTIES:

2

(Given)

ν(m /s)

Gas X

21 × 10

-6

k(W/m⋅K)

2

α(m /s) -6

0.030

29 × 10

0.665

1.65 × 10

Vapor Y 4.25 × 10 0.023 Mixture of gas X - vapor Y: Sc = 0.72

4.55 × 10

Liquid Y

-7

3.75 × 10

-5

-7 -5

ANALYSIS: The heat-mass transfer analogy may be written as h m,L L h L Nu L = L = f ( ReL , Pr ) Sh L = = f ( ReL ,Sc ) k DAB The flow conditions are the same for both situations. Check values of Pr and Sc. For Pr, the properties are those for gas X (B). 21× 10−6 m 2 / s ν Pr = B = = 0.72 α B 29 ×10-6m 2 / s while Sc = 0.72 for the gas X (B) - vapor Y (A) mixture. It follows for this situation h m,L L h L D Nu L = L = Sh L = or h m,L = h L AB . k DAB k Recognizing that DAB = ν B / Sc = 21.6 × 10-6m 2 / s (0.72 ) = 30.0 × 10−6 m 2 / s and substituting numerical values, find 30.0 ×10-6m 2 / s 2 h m,L = 25 W/m ⋅ K × = 0.0250 m/s. 0.030 W/m ⋅ K

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COMMENTS: Note that none of the thermophysical properties of liquid or vapor Y are required for the solution. Only the gas X properties and the Schmidt number (gas X - vapor Y) are required.

PROBLEM 6.55 KNOWN: Free stream velocity and average convection mass transfer coefficient for fluid flow over a surface of prescribed characteristic length. FIND: Values of Sh L , Re L , Sc and jm for (a) air flow over water, (b) air flow over naphthalene, and (c) warm glycerol over ice. SCHEMATIC:

PROPERTIES: For the fluids at 300K: Table

2

-6

ν(m /s)×10

Fluid(s)

2

DAB(m /s)

A-4 A-5 A-8

Air Glycerin Water vapor - Air

15.89 634 -

-4 0.26 × 10

A-8

Naphthalene - Air

-

0.62 × 10

A-8

Water - Glycerol

-

0.94 × 10

-5 -9

ANALYSIS: (a) Water (νapor) - Air: h L (0.01m/s )1m Sh L = m = = 385 D AB 0.26 × 10-4 m2 / s VL (1 m/s )1m Re L = = = 6.29 × 104 -6 2 ν 15.89 × 10 m / s 0.16 × 10−6 m 2 / s ν Sc = = = 0.62 DAB 0.26 × 10-6 m 2 / s h 0.01 m/s jm = St mSc 2/3 = m Sc 2/3 = ( 0.62 )2 / 3 = 0.0073. V 1 m/s

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(b) Naphthalene (νapor) - Air: Sh L = 1613

Re L = 6.29 × 104

Sc = 2.56

jm = 0.0187.

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(c) Water (1iquid) - Glycerol: Sh L = 1.06 × 107

Re L = 1577

Sc = 6.74 × 105

jm = 76.9.

COMMENTS: Note the association of ν with the freestream fluid B.

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PROBLEM 6.56 KNOWN: Characteristic length, surface temperature, average heat flux and airstream conditions associated with an object of irregular shape. FIND: Whether similar behavior exists for alternative conditions, and average convection coefficient for similar cases. SCHEMATIC: Case:

1 1 100 1 275 325

L,m V, m/s p, atm T∞, K Ts, K q ′′, W/m

2

2

h, W/m ⋅ K D AB × 10

+4

12,000

-

-

-

-

240

-

-

-

-

-

-

1.12

1.12

2

, m /s -

2 2 50 1 275 325

3 2 50 0.2 275 325

4 2 50 1 300 300

5 2 250 0.2 300 300

ASSUMPTIONS: (1) Heat and mass transfer analogy is applicable; that is, f(ReL,Pr) = f(ReL,Sc), see Eqs. 6.57 and 6.61. PROPERTIES: Table A-4, Air (300K, 1 atm): ν1 = 15.89 × 10 −6 m 2 / s, Pr1 = 0.71, k1 = 0.0263 W/m ⋅ K. ANALYSIS: Re L,1 = V1L1 / ν1 = (100 m/s × 1m ) / 15.89 × 10−6 m 2 / s = 6.29 × 106 and Pr1 = 0.71. Case 2:

Re L,2 =

V2 L 2

ν2

50 m/s × 2m

=

-6

6

= 6.29 × 10 ,

2

Pr2 = 0.71.

15.89 × 10 m / s

From Eq. 6.57 it follows that Case 2 is analogous to Case 1. Hence Nu 2 = Nu 1 and h L k2 L W 1m 2 = h1 1 = 240 = 120 W/m ⋅ K. h2 = 1 1 2 k1 L 2 L2 m ⋅ K 2m Case 3: With p = 0.2 atm, ν 3 = 79.45 × 10

−6

V3 L 3

2

m / s and Re L,3 =

ν3

< 50 m/s × 2m

=

-6

79.45 × 10 m / s

Case 4:

Re L,4 = Re L,1 , Sc 4 =

=

15.89 × 10

−6

-4

2

= 0.142 ≠ Pr1 .

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Hence, Case 4 is not analogous to Case 1. Case 5:

Re L,5 =

Sc5 =

V5 L5

ν5 ν5

D AB,5

=

250 m/s × 2m -6

2

−6

2

6

79.45 × 10 m / s =

79.45 × 10

m /s

-4

Pr3 = 0.71.

2

m /s

1.12 × 10 m / s

D AB,4

6

= 1.26 × 10 ,

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Since Re L,3 ≠ Re L,1 , Case 3 is not analogous to Case 1. ν4

2

2

1.12 × 10 m / s

= 6.29 × 10 = Re L,1

= 0.71 = Pr1 .

Hence, conditions are analogous to Case 1, and with Sh 5 = Nu 1 , h m,5 = h1

L1

D AB,5

L5

k1

= 240

W 2

m ⋅K

×

1m 2m

×

1.12 × 10

−4

2

m /s

0.0263 W/m ⋅ K

= 0.51 m/s.

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COMMENTS: Note that Pr, k and Sc are independent of pressure, while ν and DAB vary inversely with pressure.

PROBLEM 6.57 KNOWN: Surface temperature and heat loss from a runner’s body on a cool, spring day. Surface temperature and ambient air-conditions for a warm summer day. FIND: (a) Water loss on summer day, (b) Total heat loss on summer day. SCHEMATIC:

ASSUMPTIONS: (1) Heat and mass transfer analogy is applicable. Hence, from Eqs. 6.57 and 6.61, f(ReL,Pr) is of same form as f(ReL,Sc), (2) Negligible surface evaporation for Case 1, (3) Constant properties, (4) Water vapor is saturated for Case 2 surface and may be approximated as a perfect gas. -5

2

PROPERTIES: Air (given): ν = 1.6×10 m /s, k = 0.026 W/m⋅K, Pr = 0.70; Water vapor -5 2

air (given): DAB = 2.3×10 m /s; Table A-6, Saturated water vapor (T∞ = 303K):

ρ A,sat = vg-1 = 0.030 kg/m3 ; ( Ts = 308K ) : ρ A,sat = vg-1 = 0.039 kg/m3 , h fg = 2419 kJ/kg.

ANALYSIS: (a) With Re L,2 = Re L,1 and Sc=ν /D AB = 1.6 × 10 −5 m 2 / s/2.3 × 10-5 m 2 / s=0.70=Pr, it follows that Sh L = Nu L . Hence

h m L/DAB = hL/k D q1 DAB 500 W 2.3 × 10-5m 2 / s  0.0221  h m = h AB = = =  m/s. k As (Ts − T∞ )1 k As ( 20K ) 0.026 W/m ⋅ K  As  Hence, from the rate equation, with As as the wetted surface  0.0221  m n A = h m As ρ A,s − ρ A,∞ =  As  ρ A,sat Ts,2 − φ∞ ρ A,sat T∞,2     A s s  

(

)

(

)

(

)

n A = 0.0221 m3 / s ( 0.039 − 0.6 × 0.030 ) kg/m3 = 4.64 ×10−4 kg/s.

<

(b) The total heat loss for Case 2 is comprised of sensible and latent contributions, where

(

)

q 2 = qsen + q lat = hAs Ts,2 − T∞,2 + n A h fg .

(

)

Hence, with hAs = q1 / Ts,1 − T∞,1 = 25 W/K, q 2 = 25 W/K (35 − 30 ) C + 4.64 × 10-4 kg/s × 2.419 × 106 J/kg $

q 2 = 125 W + 1122 W = 1247 W. COMMENTS: Note the significance of the evaporative cooling effect.

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PROBLEM 6.58 KNOWN: Heat transfer results for an irregularly shaped object. FIND: (a) The concentration, C A, and partial pressure, p A, of vapor in an airstream for a

(

)

drying process of an object of similar shape, (b) Average mass transfer flux, n ′′A kg/s ⋅ m 2 . SCHEMATIC:

Case 1:

Heat Transfer

Case 2:

Mass Transfer

ASSUMPTIONS: (1) Heat-mass transfer analogy applies, (b) Perfect gas behavior. -6 2

PROPERTIES: Table A-4, Air (323K, 1 atm): ν = 18.20×10 m /s, Pr = 0.703, k = -3

28.0×10 W/m⋅K; Plastic vapor (given): -5 2 2.6× m /s.

M A=

82 kg/kmol, psat(50°C) = 0.0323 atm, DAB =

ANALYSIS: (a) Calculate Reynolds numbers VL 120 m/s × 1m Re1 = 1 1 = = 6.59 ×10 6 -6 2 ν 18.2 × 10 m / s 60 m/s × 2m Re 2 = = 6.59 ×10 6. -6 2 18.2 ×10 m / s Note that ν 18.2 × 10−6 m 2 / s Pr1 = 0.703 Sc 2 = = = 0.700. D AB 2.6 ×10-5m 2 / s Since Re 1 = Re2 and Pr1 = Sc2, the dimensionless solutions to the energy and species equations are identical. That is, from Eqs. 6.54 and 6.58,

(

T∗ x∗ , y∗

) = C∗A ( x∗, y∗ )

C − CA,s T − Ts = A T∞ − Ts CA,∞ − CA,s

(1)

where T∗ and C∗A are defined by Eqs. 6.37 and 6.38, respectively. Now, determine p CA,s = A,sat = 0.0323 atm/8.205 × 10-2m 3 ⋅ atm/kmol ⋅ K × ( 273 + 50 ) K ℜT CA,s = 1.219 ×10 −3 kmol/kg.

(

)

Continued …..

PROBLEM 6.58 (Cont.) Substituting numerical values in Eq. (1),

(

CA = CA,s + C A,∞ − C A,s

) TT −−TTs ∞

s

80 −100 )o C ( − 3 3 − 3 3 CA = 1.219 × 10 kmol/m + 0 − 1.219 ×10 kmol/m o

(

)

( 0 −100 )

C

CA = 0.975 ×10−3 kmol/m 3.

<

The vapor pressure is then

<

p A = CA ℜT = 0.0258 atm. (b) For case 1, q′′ = 2000 W/m 2 . The rate equations are q′′ = h ( Ts − T∞ )

(2)

(

n ′′A = h m CA,s − CA,∞

) M A.

(3)

From the analogy Nu L = Sh L

→

h L1 h m L2 = k DAB

h L k = 2 . h m L1 DAB

or

(4)

Combining Eqs. (2) - (4), h n ′′A = q ′′ m h

( CA,s − CA,∞ ) M A = q ′′ L1DAB ( CA,s − CA,∞ ) M A ( Ts − T∞ )

( Ts − T∞ )

L2 k

which numerically gives n ′′A = 2000 W/m

2

( 2m ( 28 × 10

) (1.219 ×10 W/m ⋅ K )

1m 2.6 × 10 -5m2 / s

n ′′A = 9.28 ×10 −4 kg/s ⋅ m2 .

-3

-3

)

− 0 kmol/m3 ( 82 kg/kmol )

(100-0) K

<

COMMENTS: Recognize that the analogy between heat and mass transfer applies when the conservation equations and boundary conditions are of the same form.

PROBLEM 6.59 KNOWN: Convection heat transfer correlation for flow over a contoured surface. FIND: (a) Evaporation rate from a water film on the surface, (b) Steady-state film temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (b) Constant properties, (c) Negligible radiation, (d) Heat and mass transfer analogy is applicable. -6 2 PROPERTIES: Table A-4, Air (300K, 1 atm): k = 0.0263 W/m⋅K, ν = 15.89×10 m /s, Pr = 0.707; 3 Table A-6, Water (Ts ≈ 280K): vg = 130.4 m /kg, hfg = 2485 kJ/kg; Table A-8, Water-air (T ≈ -4 2 298K): DAB = 0.26×10 m /s. ANALYSIS: (a) The mass evaporation rate is

 evap = n A = h m A  ρ A,sat ( Ts ) − φ∞ ρ A,sat (T∞ ) = h m A ρ A,sat (Ts ). m

From the heat and mass transfer analogy:

ReL =

(10 m/s ) 1m = 6.29 ×105 VL = ν 15.89 × 10-6 m 2 / s

(

Sh L = 0.43 6.29 ×105

)

0.58

Sh L = 0.43 Re0.58 Sc0.4 L Sc =

ν 15.89 × 10−6 m 2 / s = = 0.61 DAB 26 × 10-6 m 2 / s

(0.61)0.4 = 814

DAB 0.26 × 10−4 m 2 / s hm = Sh L = (814 ) = 0.0212 m/s L 1m −1 ρ A,sat (Ts ) = vg (Ts ) = 0.0077 kg/m3. Hence,

 evap = 0.0212m/s × 1m 2 × 0.0077kg/m3 = 1.63 × 10−4 kg/s. m

<

(b) From a surface energy balance, q′′conv = q′′evap , or

 ′′evap h fg h L ( T∞ − Ts ) = m With

(

Nu L = 0.43 6.29 × 105 hL =

Hence,

Ts = T∞ −

)

0.58

(m ′′evaphfg ) . hL

(0.707 )0.4 = 864

k 0.0263 W/m ⋅ K Nu L = 864 = 22.7 W/m 2 ⋅ K. L 1m

Ts = 300K −

(

1.63 × 10-4 kg/s ⋅ m 2 2.485 × 106 J/kg

) = 282.2K.

<

22.7 W/m 2 ⋅ K COMMENTS: The saturated vapor density, ρA,sat, is strongly temperature dependent, and if the

initial guess of Ts needed for its evaluation differed from the above result by more than a few degrees, the density would have to be evaluated at the new temperature and the calculations repeated.

PROBLEM 6.60 KNOWN: Surface area and temperature of a coated turbine blade. Temperature and pressure of air flow over the blade. Molecular weight and saturation vapor pressure of the naphthalene coating. Duration of air flow and corresponding mass loss of naphthalene due to sublimation. FIND: Average convection heat transfer coefficient. SCHEMATIC:

ASSUMPTIONS: (1) Applicability of heat and mass transfer analogy, (2) Negligible change in As due to mass loss, (3) Naphthalene vapor behaves as an ideal gas, (4) Solid/vapor equilibrium at surface of coating, (5) Negligible vapor density in freestream of air flow. 3

-6

PROPERTIES: Table A-4, Air (T = 300K): ρ = 1.161 kg/m , cp = 1007 J/kg⋅K, α = 22.5 × 10 2 -5 2 m /s. Table A-8, Naphthalene vapor/air (T = 300K): DAB = 0.62 × 10 m /s.

ANALYSIS: From the rate equation for convection mass transfer, the average convection mass transfer coefficient may be expressed as

hm =

(

na

As ρ A,s − ρ A,∞

)

=

∆m / ∆t As ρ A,s

where

ρ A,s = ρ A,sat ( Ts ) =

M a p A,sat

ℜ Ts

=

(128.16 kg / kmol )1.33 × 10−4 bar = 6.83 × 10 −4 kg / m3 3 0.08314 m ⋅ bar / kmol ⋅ K (300K )

Hence,

hm =

0.008 kg / (30 min× 60s / min )

(

0.05m 2 6.83 × 10−4 kg / m3

)

= 0.13m / s

Using the heat and mass transfer analogy with n = 1/3, we then obtain

 α  h = h m ρ cp Le2 / 3 = h m ρ cp    DAB 

(

(

2/3

)

= 0.130 m / s 1.161kg / m3 ×

1007 J / kg ⋅ K 22.5 × 10−6 / 0.62 ×10−5

)

2/3

= 359 W / m 2 ⋅ K

COMMENTS: The naphthalene sublimation technique has been used extensively to determine convection coefficients associated with complex flows and geometries.

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PROBLEM 6.61 KNOWN: Mass transfer experimental results on a half-sized model representing an engine strut. 1/3 FIND: (a) The coefficients C and m of the correlation Sh L = CRem L Sc

n! r!( n − r )!

for the mass

transfer results, (b) Average heat transfer coefficient, h, for the full-sized strut with prescribed operating conditions, (c) Change in total heat rate if characteristic length LH is doubled. SCHEMATIC:

Mass transfer

Heat transfer

ASSUMPTIONS: Analogy exists between heat and mass transfer.

(

)

PROPERTIES: Table A-4, Air T = (T∞ + Ts ) / 2 = 400K, 1 atm : ν =26.41× 10-6 m 2 / s, k = 0.0338 W/m⋅K, Pr = 0.690; T = 300K : ν B = 15.89 × 10−6 m 2 / s; Table A-8, Naphthalene-air

(

)

(300K, 1 atm): D AB = 0.62 × 10 −5 m 2 / s, Sc = ν B / D AB = 15.89 × 10 −6 m 2 / s/0.62 × 10-5 m 2 / s=2.56. ANALYSIS: (a) The correlation for the mass transfer experimental results is of the form 1/3 Sh L = CRe m L Sc . The constants C,m may be evaluated from two data sets of Sh L and ReL ; choosing the middle sets (2,3):

(Sh L )2 (ReL )m2 = (Sh L )3 (ReL )3m

or m =

Then, using set 2, find C =

log Sh L )2 / Sh L ) 3 ] log [491/568] = = 0.80. log  Re L )2 / Re L )3  log [120,000/144,000]   Sh L )2

)

=

491

= 0.031.

1/3 Re m (120, 000 )0.8 2.561/3 L 2 Sc (b) For the heat transfer analysis of the strut, the correlation will be of the form 1/ 3 where Re = V L / ν and the constants C,m were Nu L = h L ⋅ L H / k = 0.031 Re0.8 L H L Pr determined in Part (a). Substituting numerical values,

 60 m/s × 0.06 m  h L = Nu L ⋅ = 0.031   LH  26.41 × 10-6 m2 / s  k

0.8

0.6901/ 3

0.0338 W/m ⋅ K 0.06 m

= 198 W/m 2 ⋅ K.

< <

<

(c) The total heat rate for the strut of characteristic length L H is q=h As ( Ts − T∞ ) , where As = 2.2 LH⋅l and 0.8 -1 0.8 -1 -0.2 A ~ L h~Nu L ⋅ L-1 s H H ~ RE L ⋅ L H ~ L H ⋅ L H ~ L H Hence, q~h ⋅ As ~ L-0.2 ( LH ) ~ L0.8 H H . If the characteristic length were doubled, the heat rate 0.8 would be increased by a factor of (2) = 1.74.

( )

<

PROBLEM 6.62 KNOWN: Boundary layer temperature distribution for flow of dry air over water film. FIND: Evaporative mass flux and whether net energy transfer is to or from the water. SCHEMATIC:

ASSUMPTIONS: (1) Heat and mass transfer analogy is applicable, (2) Water is well insulated from below. PROPERTIES: Table A-4, Air (Ts = 300K, 1 atm): k = 0.0263 W/m⋅K; Table A-6, Water 3 6 vapor (Ts = 300K): ρ A,s = v-1 g = 0.0256 kg/m , h fg = 2.438 ×10 J/kg; Table A-8, Air-water

vapor ( Ts = 300K ) : DAB = 0.26 × 10−4 m 2 / s. ANALYSIS: From the heat and mass transfer analogy,

ρ A − ρ A,s

u y  = 1 − exp  −Sc ∞  . ρ A,∞ − ρ A,s ν   Using Fick’s law at the surface (y = 0), the species flux is u ∂ ρA = + ρ A,s D AB Sc ∞ ∂ y y=0 ν n′′A = 0.0256 kg/m3 × 0.26 ×10-4 m 2 / s × ( 0.6 ) 5000 m-1 = 2.00 ×10−3 kg/s ⋅ m 2 . n′′A = − DAB

The net heat flux to the water has the form q′′net = q′′conv − q′′evap = + k

u ∂ T − n′′A h fg = k ( T∞ − Ts ) Pr ∞ − n′′A h fg ∂ y y=0 ν

and substituting numerical values, find W kg (100K ) 0.7 × 5000 m-1 − 2 × 10−3 2 × 2.438 × 106 J/kg m⋅K s⋅m q′′net = 9205 W/m 2 − 4876 W/m 2 = 4329 W/m 2 . q′′net = 0.0263

Since q′′net > 0, the net heat transfer is to the water. COMMENTS: Note use of properties (DAB and k) evaluated at Ts to determine surface fluxes.

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PROBLEM 6.63 KNOWN: Distribution of local convection heat transfer coefficient for obstructed flow over a flat plate with surface and air temperatures of 310K and 290K, respectively. FIND: Average convection mass transfer coefficient. SCHEMATIC:

ASSUMPTIONS: Heat and mass transfer analogy is applicable. PROPERTIES: Table A-4, Air ( Tf = ( Ts + T∞ ) / 2 = (310 + 290 ) K/2 = 300 K, 1 atm ) : k = 0.0263 W/m ⋅ K, ν = 15.89 × 10-6 m2 / s, Pr = 0.707. Table A-8, Air-napthalene (300K, 1 atm): DAB = 0.62 × 10−5 m 2 / s, Sc = ν /DAB = 2.56. ANALYSIS: The average heat transfer coefficient is

(

)

1 L 1 L h x dx = ∫ 0.7 + 13.6x − 3.4x 2 dx ∫ L 0 L 0 h L = 0.7 + 6.8L − 1.13L2 = 10.9 W/m 2 ⋅ K. hL =

Applying the heat and mass transfer analogy with n = 1/3, Equation 6.66 yields Nu L Pr1/3

=

Sh L Sc1/3

Hence, h L Sc1/3 = L DAB k Pr1/3 1/ 3 DAB Sc1/3 0.62 × 10-5m 2 / s  2.56  2 h m,L = h L = 10.9 W/m ⋅ K k Pr1/3 0.0263 W/m ⋅ K  0.707  h m,L L

h m,L = 0.00395 m/s.

<

COMMENTS: The napthalene sublimation method provides a useful tool for determining local convection coefficients.

PROBLEM 6.64 KNOWN: Radial distribution of local Sherwood number for uniform flow normal to a circular disk. FIND: (a) Expression for average Nusselt number. (b) Heat rate for prescribed conditions. SCHEMATIC:

ASSUMPTIONS: (1) Constant properties, (2) Applicability of heat and mass transfer analogy.

)

(

PROPERTIES: Table A-4, Air T = 75$C = 348 K : k = 0.0299 W/m ⋅ K, Pr = 0.70. ANALYSIS: (a) From the heat and mass transfer analogy, Equation 6.66, Nu D Pr 0.36

=

Sh D Sc0.36

where r Sh 1 n ∫ As Sh D ( r ) dA s = o 2π ∫ o 1 + a ( r/ro ) rdr  2 0   As π ro ro 2Sh o  r 2 ar n + 2   +  = Sh o 1 + 2a/ ( n + 2 ) Sh D = ro2  2 ( n + 2 ) ron  0

Sh D =

Hence,

<

Nu D=0.814 1+ 2a/ ( n + 2 ) Re1/2 Pr 0.36 . D

(b) The heat rate for these conditions is

(

2 k 1/2 0.36 π D q = hA ( Ts − T∞ ) = 0.814 1 + 2a/ ( n + 2 ) ReD Pr D 4

(

q = 0.814 (1 + 2.4/7.5 ) 0.0299 W/m ⋅ K (π 0.02 m/4 ) 5 × 104 q = 9.92 W.

)

1/ 2

) (Ts − T∞ ) (0.07 )0.36

(100 C) $

<

COMMENTS: The increase in h(r) with r may be explained in terms of the sharp turn which the boundary layer flow must make around the edge of the disk. The boundary layer accelerates and its thickness decreases as it makes the turn, causing the local convection coefficient to increase.

PROBLEM 6.65 KNOWN: Convection heat transfer correlation for wetted surface of a sand grouse. Initial water content of surface. Velocity of bird and ambient air conditions. FIND: Flight distance for depletion of 50% of initial water content. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Vapor behaves as a perfect gas, (3) Constant properties, (4) Applicability of heat and mass transfer analogy. PROPERTIES: Air (given): v = 16.7 × 10-6 m2 / s; Air-water vapor (given): DAB = 0.26 × 10−4 m 2 / s; Table A-6, Water vapor (Ts = 305 K): vg = 29.74 m /kg; (Ts = 310 3

3

K), vg = 22.93 m /kg. ANALYSIS: The maximum flight distance is X max = Vt max where the time to deplete 50% of the initial water content ∆M is t max =

∆M ∆M . =  evap h m As ρ A,s − ρ A,∞ m

(

)

The mass transfer coefficient is DAB D = 0.034Re4/5 Sc1/3 AB L L L 1/ 2 Sc = ν /D AB = 0.642, L = ( As ) = 0.2 m VL 30 m/s × 0.2 m ReL = = = 3.59 × 105 -6 2 ν 16.7 × 10 m / s h m = Sh L

(

h m = 0.034 3.59 × 105

)

4/5

(0.642 )1/ 3

(0.26 ×10−4 m2 / s/0.2 m) = 0.106 m/s.

Hence, t max =

(

) 

0.025 kg

−1 −1 0.106 m/s 0.04 m 2 ( 29.74 ) − 0.25 ( 22.93)  kg/m3

= 259 s



X max = 30 m/s ( 259 s ) = 7785 m = 7.78 km. COMMENTS: Evaporative heat loss is balanced by convection heat transfer from air. Hence, Ts < T∞.

<

PROBLEM 6.66 KNOWN: Water-soaked paper towel experiences simultaneous heat and mass transfer while subjected to parallel flow of air, irradiation from a radiant lamp bank, and radiation exchange with surroundings. Average convection coefficient estimated as h = 28.7 W/m2⋅K. FIND: (a) Rate at which water evaporates from the towel, nA (kg/s), and (b) The net rate of radiation transfer, qrad (W), to the towel. Determine the irradiation G (W/m2). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Vapor behaves as an ideal gas, (3) Constant properties, (4) Towel experiences radiation exchange with the large surroundings as well as irradiation from the lamps, (5) Negligible heat transfer from the bottom side of the towel, and (6) Applicability of the heat-mass transfer analogy. PROPERTIES: Table A.4, Air (Tf = 300 K): ρ = 1.1614 kg/m3, cp = 1007 J/kg⋅K, α = 22.5 × 10-6 m2/s; Table A.6, Water (310 K): ρA,s = ρg = 1/νg = 1/22.93 = 0.0436 kg/m3, hfg = 2414 kJ/kg. ANALYSIS: (a) The evaporation rate from the towel is

(

n A = h m As ρ A,s − ρ A,∞

)

where h m can be determined from the heat-mass transfer analogy, Eq. 6.92, with n = 1/3, h hm

= ρ c p Le

2/3

 α  = ρ cp    DAB 

2/3

 22.5 × 10−6  = 1.614 kg m × 1007 J kg ⋅ K    0.26 × 10−4    3

2/3

= 1062 J m3 ⋅ K

h m = 28.7 W m 2 ⋅ K 1062 J m3 ⋅ K = 0.0270 m s The evaporation rate is

n A = 0.0270 m s × ( 0.0925 × 0.0925 ) m 2 ( 0.0436 − 0 ) kg m3 = 1.00 ×10−5 kg s

<

(b) Performing an energy balance on the towel considering processes of evaporation, convection and radiation, find

E in − E out = q conv − q evap + q rad = 0 hAs ( T∞ − Ts ) − n A h fg + q rad = 0

q rad = 1.00 × 10−5 kg s × 2414 × 103 J kg − 27.8 W m 2 ( 0.0925 m )

2

q rad = 2414 W + 4.76 W = 28.9 W

( 290 − 310 ) K

< Continued...

PROBLEM 6.66 (Cont.) The net radiation heat transfer to the towel is comprised of the absorbed irradiation and the net exchange between the surroundings and the towel,

(

4 − T4 q rad = α GAs + ε Asσ Tsur s

) (

)

28.9 W = 0.96G ( 0.0925 m ) + 0.96 × ( 0.0925 m ) 5.67 × 10−8 W m 2 ⋅ K 4 300 4 − 3104 K 4 2

2

Solving, find the irradiation from the lamps, G = 3583 W/m2. COMMENTS: (1) From the energy balance in Part (b), note that the heat rate by convection is considerably smaller than that by evaporation. (2) As we’ll learn in Chapter 12, the lamp irradiation found in Part (c) is nearly 3 times that of solar irradiation to the earth’s surface.

<

PROBLEM 6.67 KNOWN: Thin layer of water on concrete surface experiences evaporation, convection with ambient air, and radiation exchange with the sky. Average convection coefficient estimated as h = 53 W/m2⋅K. FIND: (a) Heat fluxes associated with convection, q′′conv , evaporation, q′′evap , and radiation exchange with the sky, q′′rad , (b) Use results to explain why the concrete is wet instead of dry, and (c) Direction of heat flow and the heat flux by conduction into or out of the concrete. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Vapor behaves as an ideal gas, (3) Constant properties, (4) Water surface is small compare to large, isothermal surroundings (sky), and (4) Applicability of the heat-mass transfer analogy. PROPERTIES: Table A.4, Air (Tf = (T∞ + Ts)/2 = 282.5 K): ρ = 1.243 kg/m3, cp = 1007 J/kg⋅K, α = 2.019 ×105 m2/s; Table A.8, Water-air (Tf = 282.5 K): DAB = 0.26 × 10-4 m2/s (282.5/298)3/2 = 0.24 × 10-4 m2/s; Table A.6, Water (Ts = 275 K): ρA,s = ρg = 1/νg = 1/181.7 = 0.0055 kg/m3, hfg = 2497 kJ/kg; Table A.6, Water ( T∞ = 290 K): ρA,s = 1/69.7 = 0.0143 kg/m3. ANALYSIS: (a) The heat fluxes associated with the processes shown on the schematic are Convection:

q′′conv = h ( T∞ − Ts ) = 53 W m 2 ⋅ K ( 290 − 275 ) K = +795 W m 2 Radiation Exchange:

(

)

(

)

<

q′′rad = εσ Ts4 − Tsky = 0.96 × 5.76 × 10−8 W m 2 ⋅ K 4 2754 − 2404 K 4 = +131W m 2

<

Evaporation:

q′′evap = n′′A h fg = −2.255 × 10−4 kg s ⋅ m 2 × 2497 × 103 J kg = −563.1W m 2

<

where the evaporation rate from the surface is

(

)

n ′′A = h m ρ A,s − ρ A,∞ = 0.050 m s ( 0.0055 − 0.7 × 0.0143 ) kg m3 = −2.255 × 10−4 kg s ⋅ m 2 Continued...

PROBLEM 6.67 (Cont.) and where the mass transfer coefficient is evaluated from the heat-mass transfer analogy, Eq. 6.92, with n = 1/3,

 α  h = ρ cp Le2 / 3 = ρ cp   hm  D AB 

2/3

2/3  2.019 × 10−5  3  = 1.243kg m × 1007 J kg ⋅ K   0.26 10−4 



×



h = 1058 J m3 ⋅ K hm h m = 53 W m 2 ⋅ K 1058 J m3 ⋅ K = 0.050 m s (b) From the foregoing evaporation calculations, note that water vapor from the air is condensing on the liquid water layer. That is, vapor is being transported to the surface, explaining why the concrete surface is wet, even without rain. (c) From an overall energy balance on the water film considering conduction in the concrete as shown in the schematic,

E in − E out = 0 q′′conv − q′′evap − q′′rad − q′′cond = 0 q′′cond = q′′conv − q′′evap − q′′rad

(

)(

)

q′′cond = 1795 W m3 − −563.1W m 2 − +131W m 2 = 1227 W m 2 The heat flux by conduction is into the concrete.

<

PROBLEM 6.68 KNOWN: Heater power required to maintain wetted (water) plate at 27°C, and average convection coefficient for specified dry air temperature, case (a). FIND: Heater power required to maintain the plate at 37°C for the same dry air temperature if the convection coefficients remain unchanged, case (b). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Convection coefficients unchanged for different plate temperatures, (3) Air stream is dry at atmospheric pressure, and (4) Negligible heat transfer from the bottom side of the plate. 3

PROPERTIES: Table A-6, Water (Ts,a = 27°C = 300 K): ρA,s = 1/vg = 0.02556 kg/m , hfg = 2.438 × 6 3 6 10 J/kg; Water (Ts,b = 37°C = 310 K): ρA,s = 1/vg = 0.04361 kg/m , hfg = 2.414 × 10 J/kg. ANALYSIS: For case (a) with Ts = 27°C and Pe = 432 W, perform an energy balance on the plate to determine the mass transfer coefficient hm.

E in − E out = 0

(

)

Pe,a − q′′evap + q′′cv As = 0 Substituting the rate equations and appropriate properties,

(

)

(

)

Pe,a −  h m ρ A,s − ρ A,∞ h fg + h Ts,a − T∞  As = 0   432 W +  h m 0.0256 kg / m3 − 0 × 2.438 × 106 J / kg +  20 W / m 2 ⋅ K ( 27 − 32 ) K  × 0.2 m 2 = 0 

(

)

where ρA,s and hfg are evaluated at Ts = 27°C = 300 K. Find,

h m = 0.0363 m / s For case (b), with Ts = 37°C and the same values for h and h m , perform an energy balance to determine the heater power required to maintain this condition.

(

)

(

)

Pe,b −  h m ρ A,s − 0 h fg + h Ts,b − T∞  As = 0   Pe,b − 0.0363 m / s ( 0.04361 − 0 ) kg / m3 × 2.414 × 106 J / kg +  20 W / m 2 ⋅ K (37 − 32 ) × 0.2 m 2 = 0  Pe,b = 784 W where ρA,a and hfg are evaluated at Ts = 37°C = 310 K.

<

PROBLEM 6.69 KNOWN: Dry air at 32°C flows over a wetted plate of width 1 m maintained at a surface temperature of 27°C by an embedded heater supplying 432 W. FIND: (a) The evaporation rate of water from the plate, nA (kg/h) and (b) The plate temperature Ts when all the water is evaporated, but the heater power remains the same. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Vapor behaves as an ideal gas, (3) Constant properties, and (4) Applicability of the heat-mass transfer analogy. PROPERTIES: Table A.4, Air (Tf = (32 + 27)°C/2 = 302.5 K): ρ = 1.153 kg/m3, cp = 1007 J/kg⋅K, α = 2.287 ×105 m2/s; Table A.8, Water-air (Tf ≈ 300 K): DAB = 0.26 × 10-4 m2/s; Table A.6, Water (Ts = 27°C = 300 K): ρA,s = 1/νg = 1/39.13 = 0.0256 kg/m3, hfg = 2438 kJ/kg. ANALYSIS: (a) Perform an energy balance on the wetted plate to obtain the evaporation rate, nA.

E in − E out = 0

Pe + qconv − q evap = 0

Pe + hAs ( T∞ − Ts ) − n A h fg = 0

(1)

In order to find h , invoke the heat-mass transfer analogy, Eq. (6.92) with n = 1/3, h hm

= ρ c p Le

2/3

 α  = ρ cp    D AB 

The evaporation rate equation

(

n A = h m As ρ A,s − ρ A,∞

2/3

 2.287 × 10−5  = 1.153 kg m × 1007 J kg ⋅ K    0.26 × 10−4    3

2/3

)

Substituting Eqs. (2) and (3) into Eq. (1), find h m Pe + 1066 J m3 ⋅ K h m As (T∞ − Ts ) − h m As ρ A,s − ρ A,∞ h fg = 0

)

(

3

= 1066 J m ⋅ K (2)

(

)

(4)

432 W + 1066 J m3 ⋅ K (32 − 27 ) K − ( 0.0256 − 0 ) kg m3 × 2438 × 103 J kg  (0.200 × 1) m 2 ⋅ h m = 0





432 + [5330 - 62,413] × 0.20 hm = 0 hm = 0.0378 m/s Using Eq. (3), find n A = 0.0378 m s (0.200 × 1) m 2 (0.0256 − 0 ) kg m3 = 1.94 × 10−4 kg s = 0.70 kg h (b) When the plate is dry, all the power must be removed by convection, Pe = qconv = h As(Ts - T∞ )

<

Assuming h is the same as for conditions with the wetted plate,

Ts = T∞ + Pe h As = T∞ + Pe (1066h m ) As

(

)

Ts = 32$ C + 432 W 1066 × 0.0378 W m 2 ⋅ K × 0.200 m 2 = 85.6$ C

<

PROBLEM 6.70 KNOWN: Surface temperature of a 20-mm diameter sphere is 32°C when dissipating 2.51 W in a dry air stream at 22°C. FIND: Power required by the imbedded heater to maintain the sphere at 32°C if its outer surface has a thin porous covering saturated with water for the same dry air temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Heat and mass transfer analogy is applicable, (3) Heat transfer convection coefficient is the same for the dry and wet condition, and (3) Properties of air and the diffusion coefficient of the air-water vapor mixture evaluated at 300 K. 3

-6

PROPERTIES: Table A-4, Air (300 K, 1 atm): ρ = 1.1614 kg/m , cp = 1007 J/kg⋅K, α = 22.5 × 10 2 -4 2 m /s; Table A-8, Water-air mixture (300 K, 1 atm): DA-B = 0.26 × 10 m /s; Table A-4, Water (305 3 6 K, 1 atm): ρA,s = 1/vg = 0.03362 kg/m , hfg = 2.426 × 10 J/kg. ANALYSIS: For the dry case (d), perform an energy balance on the sphere and calculate the heat transfer convection coefficient. E in − E out = Pe,d − q cv = 0 Pe,d − h As (Ts − T∞ ) = 0 2.51 W − hπ ( 0.020 m ) × (32 − 22 ) K = 0 2

h = 200 W / m 2 ⋅ K

Use the heat-mass analogy, Eq. (6.67) with n = 1/3, to determine h m . h hm

 α    DAB 

2/3

= ρ cp 

200 W / m2 ⋅ K hm

 22.5 × 106 m 2 / s  = 1.1614 kg / m × 1007 J / kg ⋅ K    0.26 × 106 m 2 / s   

2/3

3

h m = 0.188 m / s For the wet case (w), perform an energy balance on the wetted sphere using values for h and h m to determine the power required to maintain the same surface temperature. E in − E out = Pe,w − q cv − q evap = 0

(

)

Pe,w −  h ( Ts − T∞ ) + h m ρ A,s − ρ A,∞ h fg  As = 0 Pe,w −  200 W / m 2 ⋅ K (32 − 22 ) K +



0.188 m / s ( 0.03362 − 0 ) kg / m3 × 2.426 × 106 J / kg  π ( 0.020 m ) = 0



Pe,w = 21.8 W

2

<

COMMENTS: Note that ρA,s and hfg for the mass transfer rate equation are evaluated at Ts = 32°C = 305 K, not 300 K. The effect of evaporation is to require nearly 8.5 times more power to maintain the same surface temperature.

PROBLEM 6.71 KNOWN: Operating temperature, ambient air conditions and make-up water requirements for a hot tub. FIND: Heater power required to maintain prescribed conditions. SCHEMATIC:

ASSUMPTIONS: (1) Side wall and bottom are adiabatic, (2) Heat and mass transfer analogy is applicable. PROPERTIES: Table A-4, Air ( T = 300K, 1 atm ) : ρ = 1.161 kg/m 3 , cp = 1007 J/kg⋅K, α -6 2

= 22.5× 10 m /s; Table A-6, Sat. water vapor (T = 310K): hfg = 2414 kJ/kg, ρA,sat(T) = 1/vg = 3

-1

3

3

(22.93m /kg) = 0.0436 kg/m ; (T∞ = 290K): ρA,sat(T∞) = 1/vg = (69.7 m /kg) 3 -6 2 kg/m ; Table A-8, Air-water vapor (298K): DAB = 26 × 10 m /s.

-1

= 0.0143

ANALYSIS: Applying an energy balance to the control volume, & evap h fg ( T ) . q elec = qconv + qevap = h A ( T − T∞ ) + m Obtain h A from Eq. 6.67 with n = 1/3, h h = A = ρ cp Le2/3 h m h mA h A = h mA ρ c p Le2/3 =

& evap m

ρ A,sat ( T ) − φ∞ ρ A,sat ( T∞ )

ρ cp Le2/3 .

Substituting numerical values,

(

)

Le = α /D AB = 22.5 ×10 −6 m 2 / s /26 ×10−6 m 2 / s = 0.865 hA =

10-3 kg/s

[ 0.0436 − 0.3× 0.0143] kg/m3

1.161

kg m3

× 1007

J ( 0.865)2 / 3 kg ⋅ K

hA = 27.0 W/K.

Hence, the required heater power is q elec = 27.0W/K ( 310 − 290 ) K + 10-3kg/s × 2414kJ/kg × 1000J/kJ q elec = ( 540 + 2414 ) W = 2954 W. COMMENTS: The evaporative heat loss is dominant.

<

PROBLEM 6.72 KNOWN: Water freezing under conditions for which the air temperature exceeds 0°C. FIND: (a) Lowest air temperature, T∞, before freezing occurs, neglecting evaporation, (b) The mass transfer coefficient, hm, for the evaporation process, (c) Lowest air temperature, T∞, before freezing occurs, including evaporation. SCHEMATIC:

No evaporation

With evaporation

ASSUMPTIONS: (1) Steady-state conditions, (2) Water insulated from ground, (3) Water surface has ε = 1, (4) Heat-mass transfer analogy applies, (5) Ambient air is dry. 3

PROPERTIES: Table A-4, Air (Tf ≈ 2.5°C ≈ 276K, 1 atm): ρ = 1.2734 kg/m , cp = 1006 -6 2 J/kg⋅K, α = 19.3 × 10 m /s; Table A-6, Water vapor (273.15K): hfg = 2502 kJ/kg, ρg = 1/vg = 4.847 × 10 kg/m ; Table A-8, Water vapor - air (298K): DAB = 0.26 × 10 −4 m 2 / s. -3

3

ANALYSIS: (a) Neglecting evaporation and performing an energy balance,

q′′conv − q′′rad = 0 4 =0 h ( T∞ − Ts ) − εσ Ts4 − Tsky

(

T∞ = 0o C +

)

or

(

4 T∞ =Ts + ( εσ / h ) Ts4 − Tsky

)

1 × 5.667 ×10-8 W/m 2 ⋅ K 4  0 + 273) 4 − ( −30 + 273) 4  = 4.69o C. (   25 W/m2 ⋅ K

<

(b) Invoking the heat-mass transfer analogy in the form of Eq. 6.67 with n = 1/3, h = ρ cp Le 2/3 or h m = h/ρ c p Le 2/3 where Le = α /DAB hm

 19.3 × 10-6 m2 / s  h m = 25 W/m2 ⋅ K /1.273 kg/m3 (1006 J/kg ⋅ K )    0.26 ×10 -4m 2 / s 

(

)

2/3

= 0.0238 m/s.

<

(c) Including evaporation effects and performing an energy balance gives q′′conv − q′′rad − q′′evap = 0

(

)

& ′′ h fg = h m ρA,s − ρA,∞ h fg , ρ A,s = ρ g and ρ A, ∞ = 0. Hence, where q′′evap = m

(

)

(

)

4 + h /h ρ −0 h T∞ = Ts + ( εσ / h ) T s4 − Tsky ( m ) g fg 0.0238 m/s T∞ = 4.69o C + × 4.847 × 10−3 kg/m 3 × 2.502 ×10 6 J/kg 2 25 W/m ⋅ K

T∞ = 4.69o C+ 11.5 oC = 16.2oC.

<

PROBLEM 6.73 KNOWN: Wet-bulb and dry-bulb temperature for water vapor-air mixture. FIND: (a) Partial pressure, pA, and relative humidity, φ, using Carrier’s equation, (b) pA and φ using psychrometric chart, (c) Difference between air stream, T∞, and wet bulb temperatures based upon evaporative cooling considerations. SCHEMATIC:

ASSUMPTIONS: (1) Evaporative cooling occurs at interface, (2) Heat-mass transfer analogy applies, (3) Species A and B are perfect gases. PROPERTIES: Table A-6, Water vapor: pA,sat (21.1°C) = 0.02512 bar, pA,sat (37.8°C) = 0.06603 bar, hfg (21.1°C) = 2451 kJ/kg; Table A-4, Air (Tam = [TWB + TDB]/2 ≅ 300K, 1 atm): -6 2 α = 22.5 × 10 m /s, cp = 1007 J/kg⋅K; Table A-8, Air-water vapor (298K): DAB = 0.26 × 10 4 2 m /s. ANALYSIS: (a) Carrier’s equation has the form p − p gw ( TDB − TWB ) p v = pgw − 1810 − TWB

(

)

where pv = partial pressure of vapor in air stream, bar pgw = sat. pressure at TWB = 21.1°C, 0.02512 bar p = total pressure of mixture, 1.033 bar TDB = dry bulb temperature, 37.8°C TWB = wet bulb temperature, 21.1°C. Hence, p v = 0.02512 bar −

(1.013 − 0.02512 ) bar × (37.8 − 21.1)o C = 0.0142 bar. 1810 − ( 21.1 + 273.1) K

The relative humidity, φ, is then p pv 0.0142 bar φ≡ A = = = 0.214. pA,sat p 37.8o C 0.06603 bar A

(

<

)

(b) Using a psychrometric chart TWB = 21.1o C = 70oF   TDB = 37.8o C = 100o F

<

φ ≈ 0.225

<

p v = φ p sat = 0.225 × 0.06603 bar = 0.0149 bar. Continued …..

PROBLEM 6.73 (Cont.) (c) An application of the heat-mass transfer analogy is the process of evaporative cooling which occurs when air flows over water. The change in temperature is estimated by Eq. 6.73.

( T∞ − Ts ) =

( M A / M B ) h fg  pA,sat ( Ts ) c p Le 2/3

 

p

p  − A,∞  p 

where cp and Le are evaluated at Tam = (T∞ + Ts )/2 and pA,∞ = pv, as determined in Part (a). Substituting numerical values, using Le = α/DAB,

( T∞ − Ts ) =

J 0.0149 bar  kg  0.02491 bar −  1.013 bar 2/3 1.013 bar   22.5 × 10-6m 2 / s  1007 J/kg ⋅ K   -4 2  0.26 × 10 m / s 

(18 kg/kmol/29 kg/kmol)

( T∞ − Ts )

× 2451 ×103

= 17.6oC.

<

Note that cp and α are associated with the air. COMMENTS: The following table compares results from the two calculation methods. Carrier’s Eq.

Psychrometric Chart

pv (bar)

0.0142

0.0149

φ

0.214

0.225

Evaporative Cooling T∞ - Ts = 17.6°C Observed Difference TDB - TWB = 16.7°C 17.6 −16.7 % Difference: × 100 = 5.4%. 16.7

PROBLEM 6.74 KNOWN: Wet and dry bulb temperatures. FIND: Relative humidity of air. SCHEMATIC:

ASSUMPTIONS: (1) Perfect gas behavior for vapor, (2) Steady-state conditions, (3) Negligible radiation, (4) Negligible conduction along thermometer. 3

PROPERTIES: Table A-4, Air (308K, 1 atm): ρ = 1.135 kg/m , cp = 1007 J/kg⋅K, α = 23.7 × -6 2 3 10 m /s; Table A-6, Saturated water vapor (298K): vg = 44.25 m /kg, hfg = 2443 kJ/kg; (318K): 3 -4 2 vg = 15.52 m /kg; Table A-8, Air-vapor (1 atm, 298K): DAB = 0.26 × 10 m /s, DAB (308K) = -4 2 3/2 -4 2 0.26 × 10 m /s × (308/298) = 0.27 × 10 m /s, Le = α/DAB = 0.88. ANALYSIS: From an energy balance on the wick, Eq. 6.71 follows from Eq. 6.68. Dividing Eq. 6.71 by ρA,sat(T∞), T∞ − Ts h  = h fg  m  ρA,sat ( T∞ )  h 

 ρ A,sat ( Ts ) ρ A,∞  −  .  ρA,sat ( T∞ ) ρA,sat ( T∞ ) 

With  ρ A,∞ / ρ A,sat ( T∞ )  ≈ φ∞ for a perfect gas and h/hm given by Eq. 6.67, φ∞ =

ρA,sat ( Ts )

ρA,sat ( T∞ )

−

ρ cp Le2/3 ρA,sat ( T∞ ) h fg

( T∞ − Ts ) .

Using the property values, evaluate ρA,sat ( Ts )

ρA,sat ( T∞ )

=

vg T∞

vg ( Ts )

=

(

15.52 = 0.351 44.25

ρ A,sat ( T∞ ) = 15.52 m3 /kg

)

−1

= 0.064 kg/m 3 .

Hence, φ ∞ = 0.351 −

1.135 kg/m3 (1007 J/kg ⋅ K )

( 0.88)

2/3

(

0.064 kg/m 3 2.443 ×106 J/kg

φ ∞ = 0.351 − 0.159 = 0.192.

)

( 45 − 25 ) K

<

COMMENTS: Note that latent heat must be evaluated at the surface temperature (evaporation occurs at the surface).

PROBLEM 6.75 KNOWN: Heat transfer correlation for a contoured surface heated from below while experiencing air flow across it. Flow conditions and steady-state temperature when surface experiences evaporation from a thin water film. FIND: (a) Heat transfer coefficient and convection heat rate, (b) Mass transfer coefficient and evaporation rate (kg/h) of the water, (c) Rate at which heat must be supplied to surface for these conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Heat-mass transfer analogy applies, (3) Correlation requires properties evaluated at Tf = (Ts + T∞)/2. PROPERTIES: Table A-4, Air (Tf = (Ts + T∞)/2 = (290 + 310)K/2 = 300 K, 1 atm): ν = -6 2 15.89 × 10 m /s, k = 0.0263 W/m⋅K, Pr = 0.707; Table A-8, Air-water mixture (300 K, 1 atm): -4 2 3 DAB = 0.26 × 10 m /s; Table A-6, Sat. water (Ts = 310 K): ρA,sat = 1/vg = 1/22.93 m /kg = 3 0.04361 kg/m , hfg = 2414 kJ/kg. ANALYSIS: (a) To characterize the flow, evaluate ReL at Tf VL 10 m/s ×1 m Re L = = = 6.293 ×10 5 -6 2 ν 15.89 ×10 m / s and substituting into the prescribed correlation for this surface, find

(

NuL = 0.43 6.293 ×105

)

0.58

( 0.707 )0.4 = 864.1

Nu L ⋅ k 864.1× 0.0263 W/m ⋅ K = = 22.7 W/m2 ⋅ K. L 1m Hence, the convection heat rate is

<

hL =

q conv = h LA s ( Ts − T∞ ) q conv = 22.7 W/m 2 ⋅ K × 1 m 2 ( 310 − 290 ) K = 454 W

<

(b) Invoking the heat-mass transfer analogy h L Sh L = m = 0.43Re0.58 Sc0.4 L DAB where ν 15.89 × 10−6 m 2 / s = = 0.611 DAB 0.26 × 10-4m2 / s and ν is evaluated at Tf. Substituting numerical values, find Sc =

Continued …..

PROBLEM 6.75 (Cont.)

(

Sh L = 0.43 6.293 ×105 hm =

)

0.58

( 0.611)0.4 = 815.2

Sh L ⋅ D AB 815.2 × 0.26 ×10−4 m 2 / s = = 2.12× 10−2 m/s. L 1m

<

The evaporation rate, with ρ A,s = ρ A,sat ( Ts ) , is

(

)

& = h mA s ρA,s − ρA,∞ m & = 2.12 ×10-2 m/s ×1 m 2 ( 0.04361 − 0 ) kg/m 3 m & = 9.243 ×10-4kg/s = 3.32 kg/h. m

<

(c) The rate at which heat must be supplied to the plate to maintain these conditions follows from an energy balance.

E& in − E& out = 0 q in − q conv − q evap = 0 where qin is the heat supplied to sustain the losses by convection and evaporation. q in = q conv + q evap & fg q in = h LAs ( Ts − T∞ ) + mh

q in = 454 W + 9.243 ×10-4 kg/s × 2414 ×103 J/kg q in = ( 254 + 2231) W = 2685 W.

<

COMMENTS: Note that the loss from the surface by evaporation is nearly 5 times that due to convection.

PROBLEM 6.76 KNOWN: Thickness, temperature and evaporative flux of a water layer. Temperature of air flow and surroundings. FIND: (a) Convection mass transfer coefficient and time to completely evaporate the water, (b) Convection heat transfer coefficient, (c) Heater power requirement per surface area, (d) Temperature of dry surface if heater power is maintained. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Applicability of heat and mass transfer analogy with n = 1/3, (3) Radiation exchange at surface of water may be approximated as exchange between a small surface and large surroundings, (4) Air is dry (ρA,∞ = 0), (5) Negligible heat transfer from unwetted surface of the plate. PROPERTIES: Table A-6, Water (Tw = 340K): ρf = 979 kg/m , ρ A,sat = v g−1 = 0.174 kg / m3 , 3

3

h fg = 2342 kJ / kg. Prescribed, Air: ρ = 1.08 kg/m , cp = 1008 J/kg⋅K, k = 0.028 W/m⋅K. Vapor/Air: -4

2

DAB = 0.29 × 10 m /s. Water: εw = 0.95. Plate: εp = 0.60. ANALYSIS: (a) The convection mass transfer coefficient may be determined from the rate equation n′′A = h m ( ρ A,s − ρ A, ∞ ) , where ρ A,s = ρ A,sat ( Tw ) and ρ A,∞ = 0. Hence,

hm =

n ′′A

ρ A,sat

=

0.03kg / s ⋅ m 2 0.174 kg / m3

<

= 0.172 m / s

The time required to completely evaporate the water is obtained from a mass balance of the form − n ′′A = ρ f dδ / dt, in which case

ρf

0

t

∫ δ i dδ = −n′′A ∫ 0 dt

3 ρf δ i 979 kg / m (0.002m ) t= = = 65.3s n ′′A 0.03kg / s ⋅ m 2

< 3

(b) With n = 1/3 and Le = α/DAB = k/ρcp DAB = 0.028 W/m⋅K/(1.08 kg/m × 1008 J/kg⋅K × 0.29 × -4 2 10 m /s) = 0.887, the heat and mass transfer analogy yields

h=

k hm DAB

Le1/ 3

=

0.028 W / m ⋅ K ( 0.172 m / s )

1/ 3 0.29 ×10−4 m 2 / s (0.887 )

= 173W / m 2 ⋅ K

<

The electrical power requirement per unit area corresponds to the rate of heat loss from the water. Hence, Continued …..

PROBLEM 6.76 (Cont.)

(

4 − T4 ′′ = q′′evap + q′′conv + q′′rad = n ′′A h fg + h ( Tw − T∞ ) + ε wσ Tw Pelec sur ′′ = 0.03 kg / s ⋅ m Pelec

2

(

)

2.342 × 10 J / kg + 173 W / m ⋅ K ( 40K ) + 0.95 × 5.67 × 10 6

2

−8

2

W/m ⋅K

4

)

(

4

340 − 300

4

)

′′ = 70, 260 W / m 2 + 6920 W / m 2 + 284 W / m 2 = 77, 464 W / m 2 Pelec

<

(c) After complete evaporation, the steady-state temperature of the plate is determined from the requirement that

(

)

(

4 ′′ = h Tp − T∞ + ε pσ Tp4 − Tsur Pelec

(

) )

(

77, 464 W / m 2 = 173 W / m 2 ⋅ K Tp − 300 + 0.60 × 5.67 × 10−8 W / m 2 ⋅ K 4 Tp4 − 3004

Tp = 702K = 429°C

) <

COMMENTS: The evaporative heat flux is the dominant contributor to heat transfer from the water layer, with convection of sensible energy being an order of magnitude smaller and radiation exchange being negligible. Without evaporation (a dry surface), convection dominates and is approximately an order of magnitude larger than radiation.

PROBLEM 6.77 KNOWN: Heater power required to maintain water film at prescribed temperature in dry ambient air and evaporation rate. FIND: (a) Average mass transfer convection coefficient h m , (b) Average heat transfer convection coefficient h, (c) Whether values of hm and h satisfy the heat-mass analogy, and (d) Effect on evaporation rate and disc temperature if relative humidity of the ambient air were increased from 0 to 0.5 but with heater power maintained at the same value. SCHEMATIC:

ASSUMPTIONS: (1) Water film and disc are at same temperature; (2) Mass and heat transfer coefficient are independent of ambient air relative humidity, (3) Constant properties. 3

3

PROPERTIES: Table A-6, Saturated water (305 K): vg = 29.74 m /kg, hfg = 2426 × 10 J/kg; -6 2 Table A-4, Air ( T = 300 K, 1 atm ) : k = 0.0263 W/m⋅K, α = 22.5 × 10 m /s, Table A-8, Air-4

water vapor (300 K, 1 atm): DAB = 0.26 × 10

2

m /s.

ANALYSIS: (a) Using the mass transfer convection rate equation,

(

)

n A = h mAs ρA,s − ρA, ∞ = h mAs ρA,sat (1 − φ∞ ) and evaluating ρA,s = ρA,sat (305 K) = 1/vg (305 K) with φ ∞ ~ ρA,∞ = 0, find hm =

hm =

(

nA

As ρ A,s − ρA,∞

)

2.55 ×10−4 kg/hr/ ( 3600s/hr )

(π (0.020 m ) ) 2

/ 4 (1/29.74 − 0 ) kg/m3

= 6.71× 10−3 m/s.

<

(b) Perform an overall energy balance on the disc, q = q conv + q evap = hAs ( Ts − T∞ ) + n Ah fg and substituting numerical values with hfg evaluated at Ts , find h: 200 ×10−3 W = hπ ( 0.020 m ) / 4( 305 − 295 ) K + 7.083 ×10-8 kg/s × 2426 ×103 J/kg 2

h = 8.97 W/m 2 ⋅ K.

< Continued …..

PROBLEM 6.77 (Cont.) (c) The heat-mass transfer analogy, Eq. 6.67, requires that h ? k  DAB 1/3 =   . h m DAB  α  Evaluating k and DAB at T =

( Ts + T∞ ) / 2 = 300 K

and substituting numerical values, 1/3

0.0263 W/m ⋅ K  0.26 ×10− 4 m 2 / s    = 1337 ≠ 6.71 ×10 -3 m/s 0.26 ×10 -4 m 2 / s  22.5 ×10 -6 m 2 / s  8.97 W/m 2 ⋅ K

= 1061

Since the equality is not satisfied, we conclude that, for this situation, the analogy is only approximately met (≈ 30%). (d) If φ ∞ = 0.5 instead of 0.0 and q is unchanged, nA will decrease by nearly a factor of two, as will nAhfg = qevap. Hence, since qconv must increase and h remains nearly constant, Ts - T∞ must increase. Hence, Ts will increase. COMMENTS: Note that in part (d), with an increase in Ts , hfg decreases, but only slightly, and ρA,sat increases. From a trial-and-error solution assuming constant values for hm and h, the disc temperature is 315 K for φ ∞ = 0.5.

PROBLEM 6.78 KNOWN: Power-time history required to completely evaporate a droplet of fixed diameter maintained at 37°C. FIND: (a) Average mass transfer convection coefficient when droplet, heater and dry ambient air are at 37°C and (b) Energy required to evaporate droplet if the dry ambient air temperature is 27°C. SCHEMATIC:

ASSUMPTIONS: (1) Wetted surface area of droplet is of fixed diameter D, (2) Heat-mass transfer analogy is applicable, (3) Heater controlled to operate at constant temperature, Ts = 37°C, (4) Mass of droplet same for part (a) and (b), (5) Mass transfer coefficients for parts (a) and (b) are the same. PROPERTIES: Table A-6, Saturated water (37°C = 310 K): hfg = 2414 kJ/kg, ρA,sat = 1/vg = 3 1/22.93 = 0.04361 kg/m ; Table A-8, Air-water vapor (Ts = 37°C = 310 K, 1 atm): DAB = 0.26 -6 2 3/2 -6 2 × 10 m /s(310/289) = 0.276 × 10 m /s; Table A-4, Air (T = (27 + 37)°C/2 = 305 K, 1 3 -6 2 atm): ρ = 1.1448 kg/m , cp = 1008 J/kg⋅K, ν = 16.39 × 10 m /s, Pr = 0.706. ANALYSIS: (a) For the isothermal conditions (37°C), the electrical energy Q required to evaporate the droplet during the interval of time ∆t = te follows from the area under the P-t curve above, te

Pdt =  20 ×10−3 W × ( 50 × 60 ) s + 0.5 × 20 ×10-3 W (100 − 50 ) × 60s    Q = 90 J. Q =

∫0

From an overall energy balance during the interval of time ∆t = te, the mass loss due to evaporation is Q = Mh fg or M = Q/h fg M = 90 J/2414 ×103 J/kg = 3.728 ×10 -5 kg. To obtain the average mass transfer coefficient, write the rate equation for an interval of time ∆t = te, & ⋅ t e = h mAs ρ A,s − ρ A,∞ ⋅ t e = h mAs ρ A,s (1 − φ∞ ) ⋅ t e M = m

(

)

Substituting numerical values with φ ∞ = 0, find

(

)

3.278 × 10−5 kg = h m π ( 0.004 m ) 2 / 4 0.04361 kg/m 3 × (100 × 60 ) s Continued …..

PROBLEM 6.78 (Cont.)

<

hm = 0.0113 m/s. -5

(b) The energy required to evaporate the droplet of mass M = 3.728 × 10 overall energy balance,

kg follows from an

Q = Mh fg + hAs ( Ts − T∞ ) where h is obtained from the heat-mass transfer analogy, Eq. 6.67, using n = 1/3, h k = = ρ cp Le 2/3 n h m DAB Le where ν 16.39 × 10−6 m2 / s = = 0.594 DAB 0.276 ×10-4 m 2 / s Sc 0.594 Le = = = 0.841. Pr 0.706 Sc =

Hence, h = 0.0113 m/s × 1.1448 kg/m3 × 1008 J/kg ⋅ K ( 0.841)

2/3

= 11.62 W/m 2 ⋅ K.

and the energy requirement is

(

)

Q = 3.728 × 10-5 kg × 2414 kJ/kg + 11.62 W/m 2 ⋅ K π ( 0.004 m ) 2 / 4 ( 37 − 27 )o C Q =

( 90.00 + 0.00145 ) J

= 90 J.

The energy required to meet the convection heat loss is very small compared to that required to sustain the evaporative loss.

<

PROBLEM 6.79 KNOWN: Initial plate temperature Tp (0) and saturated air temperature (T∞) in a dishwasher at the 2

start of the dry cycle. Thermal mass per unit area of the plate Mc/As = 1600 J/m ⋅K. FIND: (a) Differential equation to predict plate temperature as a function of time during the dry cycle and (b) Rate of change in plate temperature at the start of the dry cycle assuming the average convection heat transfer coefficient is 3.5 W/m2⋅K. SCHEMATIC:

ASSUMPTIONS: (1) Plate is spacewise isothermal, (2) Negligible thermal resistance of water film on plate, (3) Heat-mass transfer analogy applies. 3

PROPERTIES: Table A-4, Air ( T =(55 + 65)°C/2 = 333 K, 1 atm): ρ = 1.0516 kg/m , cp = 1008 -6 2 J/kg⋅K, Pr = 0.703, ν = 19.24× 10 m /s; Table A-6, Saturated water vapor, (Ts = 65°C = 338 K): ρ A 3

3

= 1/vg = 0.1592 kg/m , hfg = 2347 kJ/kg; (Ts = 55°C = 328 K): ρ A = 1/vg = 0.1029 kg/m ; Table A-8, -4 2 3/2 4 Air-water vapor (Ts = 65°C = 338 K, 1 atm): DAB = 0.26 × 10 m /s (338/298) = 0.314 × 102

m /s. ANALYSIS: (a) Perform an energy balance on a rate basis on the plate,

E& in − E& out = E& st

(

)

′′ ′′ q conv − q evap = ( Mc/A s ) dTp /dt .

Using the rate equations for the heat and mass transfer fluxes, find

h  T∞ − Tp ( t )  − hm  ρ A,s ( Ts ) − ρ A,∞ ( T∞ )  h fg = ( Mc/As )( dT/dt ) .

<

(b) To evaluate the change in plate temperature at t = 0, the start of the drying process when Tp (0) = 65°C and T∞ = 55°C, evaluate hm from knowledge of h = 3.5 W/m 2 ⋅ K using the heat-mass transfer analogy, Eq. 6.67, with n = 1/3, 2/3 2/3 h  Sc   ν / D AB  = ρ cp Le 2/3 = ρ cp   = ρ cp   hm  Pr   Pr  and evaluating thermophysical properties at their appropriate temperatures, find  19.24 × 10-6m2 /s/0.314 × 10 -4m 2 /s  3.5 W/m2 ⋅ K = 1.0516 kg/m3 × 1008 J/kg ⋅ K     hm 0.703  

2/3

Substituting numerical values into the conservation expression of part (a), find

h m = 3.619 × 10 −3 m/s.

(

) <

3.5 W/m2 ⋅ K (55 − 65 ) C − 3.619 × 10-3 m/s ( 0.1592 − 0.1029 ) kg/m3 × 2347 ×10 3 J/kg = 1600 J/m 2 ⋅ K dTp /dt o

dTp /dt = −[ 35.0 + 478.2 ] W/m 2 ⋅ K/1600 J/m 2 ⋅ K = −0.32 K/s. COMMENTS: This rate of temperature change will not be sustained for long, since, as the plate cools, the rate of evaporation (which dominates the cooling process) will diminish.

PROBLEM 7.1 KNOWN: Temperature and velocity of fluids in parallel flow over a flat plate. FIND: (a) Velocity and thermal boundary layer thicknesses at a prescribed distance from the leading edge, and (b) For each fluid plot the boundary layer thicknesses as a function of distance. SCHEMATIC:

ASSUMPTIONS: (1) Transition Reynolds number is 5 × 105. PROPERTIES: Table A.4, Air (300 K, 1 atm): ν = 15.89 × 10-6 m2/s, Pr = 0.707; Table A.6, Water (300 K): ν = µ/ρ = 855 × 10-6 N⋅s/m2/997 kg/m3 = 0.858 × 10-6 m2/s, Pr = 5.83; Table A.5, Engine Oil (300 K): ν = 550 × 10-6 m2/s, Pr = 6400; Table A.5, Mercury (300 K): ν = 0.113 × 10-6 m2/s, Pr = 0.0248. ANALYSIS: (a) If the flow is laminar, the following expressions may be used to compute δ and δt, respectively,

δ=

5x

δt =

2 Re1/ x

δ

Fluid

Pr1/ 3

Air Water Oil Mercury

where u x 1m s ( 0.04 m ) 0.04 m 2 s Re x = ∞ = = ν ν ν

Rex

δ (mm)

δt (mm)

2517 4.66 × 104 72.7 3.54 × 105

3.99 0.93 23.5 0.34

4.48 0.52 1.27 1.17

(b) Using IHT with the foregoing equations, the boundary layer thicknesses are plotted as a function of distance from the leading edge, x. 10

5

BL thickness, deltat (mm)

BL thickness, delta (mm)

8 6 4 2 0

4 3 2 1 0

0

10

20

30

Distance from leading edge, x (mm) Air Water Oil Mercury

40

0

10

20

30

40

Distance from leading edge, x (mm) Air Water Oil Mercury

COMMENTS: (1) Note that δ ≈ δt for air, δ > δt for water, δ >> δt for oil, and δ < δt for mercury. As expected, the boundary layer thicknesses increase with increasing distance from the leading edge. (2) The value of δt for mercury should be viewed as a rough approximation since the expression for δ/δt was derived subject to the approximation that Pr > 0.6.

<

PROBLEM 7.2 KNOWN: Temperature and velocity of engine oil. Temperature and length of flat plate. FIND: (a) Velocity and thermal boundary layer thickness at trailing edge, (b) Heat flux and surface shear stress at trailing edge, (c) Total drag force and heat transfer per unit plate width, and (d) Plot the boundary layer thickness and local values of the shear stress, convection coefficient, and heat flux as a function of x for 0 ≤ x ≤ 1 m. SCHEMATIC:

ASSUMPTIONS: (1) Critical Reynolds number is 5 × 105, (2) Flow over top and bottom surfaces. PROPERTIES: Table A.5, Engine Oil (Tf = 333 K): ρ = 864 kg/m3, ν = 86.1 × 10-6 m2/s, k = 0.140 W/m⋅K, Pr = 1081. ANALYSIS: (a) Calculate the Reynolds number to determine nature of the flow,

u L 0.1m s × 1m ReL = ∞ = = 1161 ν 86.1×10−6 m 2 s Hence the flow is laminar at x = L, from Eqs. 7.19 and 7.24, and −1/ 2 = 5 1m 1161 −1/ 2 = 0.147 m δ = 5L ReL ( )( ) −1/ 3 δ t = δ Pr −1/ 3 = 0.147 m (1081) = 0.0143m

< <

(b) The local convection coefficient, Eq. 7.23, and heat flux at x = L are

hL =

k 2 Pr1/ 3 = 0.140 W m ⋅ K 0.332 1161 1/ 2 1081 1/ 3 = 16.25 W m 2 ⋅ K 0.332 Re1/ ( ) ( ) L L 1m

q′′x = h L ( Ts − T∞ ) = 16.25 W m 2 ⋅ K ( 20 − 100 ) C = −1300 W m 2 $

<

Also, the local shear stress is, from Eq. 7.20, ρu2 864 kg m3 τ s,L = ∞ 0.664 Re−L1/ 2 = (0.1m s )2 0.664 (1161)−1/ 2

2

2

<

τ s,L = 0.0842 kg m ⋅ s 2 = 0.0842 N m 2 (c) With the drag force per unit width given by D′ = 2Lτ s,L where the factor of 2 is included to account for both sides of the plate, it follows that

(

)

−1/ 2

2 −1/ 2 = (1m ) 864 kg m3 (0.1m s ) / 2 1.328 (1161) D′ = 2L ρ u ∞ 2 1.328 Re L 2

= 0.337 N m

For laminar flow, the average value h L over the distance 0 to L is twice the local value, hL, h L = 2h L = 32.5 W m 2 ⋅ K The total heat transfer rate per unit width of the plate is $ q′ = 2Lh L ( Ts − T∞ ) = 2 (1m ) 32.5 W m 2 ⋅ K ( 20 − 100 ) C = −5200 W m

<

< Continued...

PROBLEM 7.2 (Cont.) (c) Using IHT with the foregoing equations, the boundary layer thickness, and local values of the convection coefficient and heat flux were calculated and plotted as a function of x.

deltax*10, hx*100, -q''x

5000

4000

3000

2000

1000

0 0

0.2

0.4

0.6

0.8

1

Distance from leading edge, x (m) BL thickness, deltax * 10 (mm) Convection coefficient, hx * 100 (N/m^2) Heat flux, - q''x (W/m^2)

COMMENTS: (1) Note that since Pr >> 1, δ >> δt. That is, for the high Prandtl liquids, the velocity boundary layer will be much thicker than the thermal boundary layer. (2) A copy of the IHT Workspace used to generate the above plot is shown below. // Boundary layer thickness, delta delta = 5 * x * Rex ^-0.5 delta_mm = delta * 1000 delta_plot = delta_mm * 10 // Scaling parameter for convenience in plotting // Convection coefficient and heat flux, q''x q''x = hx * (Ts - Tinf) Nux = 0.332 * Rex^0.5 * Pr^(1/3) Nux = hx * x / k hx_plot = 100 * hx // Scaling parameter for convenience in plotting q''x_plot = ( -1 ) * q''x // Scaling parameter for convenience in plotting // Reynolds number Rex = uinf * x / nu // Properties Tool: Engine oil // Engine Oil property functions : From Table A.5 // Units: T(K) rho = rho_T("Engine Oil",Tf) // Density, kg/m^3 cp = cp_T("Engine Oil",Tf) // Specific heat, J/kg·K nu = nu_T("Engine Oil",Tf) // Kinematic viscosity, m^2/s k = k_T("Engine Oil",Tf) // Thermal conductivity, W/m·K Pr = Pr_T("Engine Oil",Tf) // Prandtl number

// Assigned variables Tf = (Ts + Tinf) / 2 Tinf = 100 + 273 Ts = 20 + 273 uinf = 0.1 x=1

// Film temperature, K // Freestream temperature, K // Surface temperature, K // Freestream velocity, m/s // Plate length, m

PROBLEM 7.3 KNOWN: Velocity and temperature of air in parallel flow over a flat plate. FIND: (a) Velocity boundary layer thickness at selected stations. Distance at which boundary layers merge for plates separated by H = 3 mm. (b) Surface shear stress and v(δ) at selected stations. SCHEMATIC:

ASSUMPTIONS: (1) Steady flow, (2) Boundary layer approximations are valid, (3) Flow is laminar. 3

-6

PROPERTIES: Table A-4, Air (300 K, 1 atm): ρ = 1.161 kg/m , ν = 15.89 × 10

2

m /s.

ANALYSIS: (a) For laminar flow,

5x

δ =

5

=

Re1/2 x

(u ∞ / ν )

1/2

x ( m) δ ( mm )

0.001 0.126

x1/2 = 0.01 0.399

5x1/2

= 3.99 ×10 −3 x1/2 .

(25 m/s/15,89×10-6 m 2 / s )

1/2

0.1 1.262

Boundary layer merger occurs at x = xm when δ = 1.5 mm. Hence 0.0015 m x1/2 = 0.376 m1/2 x m = 141 mm. m = -3 1/2 3.99 ×10 m

<

(b) The shear stress is

τ s,x = 0.664 x ( m)

(

τ s,x N/m 2

)

ρu 2∞ / 2 Re1/2 x

=

0.001 6.07

ρu 2∞ / 2

( u∞ /ν )1 / 2 x1/2 0.01 1.92

0.1

0.664 ×1.161 kg/m3 ( 25 m/s ) / 2 2

=

(

)

1 / 2 1/2 25 m/s/15.89 ×10-6 m 2 / s x

=

0.192 x1/2

(N/m2 ) .

0.61 1/2

The velocity distribution in the boundary layer is v = (1/2) (νu∞/x) (ηdf/dη - f). At y = δ, η ≈ 5.0, f ≈ 3.24, df/dη ≈ 0.991. 1/2 0.5 v= 15.89 × 10−6 m2 / s × 25 m/s ( 5.0 × 0.991 − 3.28) = 0.0167/x1/2 m/s. 1/2 x

x ( m) v ( m/s )

(

)

0.001 0.528

0.01 0.167

(

)

0.1 0.053

COMMENTS: (1) v L = 0.25 m, the air flow is laminar over the entire heater. For the first strip, q1 = h1 (∆L × w)(Ts - T‡ ) where h1 is obtained from

h1 =

k 2 1/ 3 0.664 Re1/ x Pr ∆L 1/ 2

 2 m s × 0.01m  0.0429 W m ⋅ K  h1 = × 0.664   43.54 ×10−6 m 2 s  0.01m  

(0.683)1/ 3 = 53.8 W

q1 = 53.8 W m 2 ⋅ K ( 0.01m × 0.2 m )(500 − 25 ) C = 51.1W $

m2 ⋅ K

<

For the fifth strip, q5 = q 0 −5 − q 0 − 4 ,

q5 = h 0 −5 (5∆L × w )( Ts − T∞ ) − h 0− 4 ( 4∆L × w )( Ts − T∞ ) q5 = (5h 0−5 − 4h 0− 4 ) ( ∆L × w )( Ts − T∞ )

Hence, with x5 = 5∆L = 0.05 m and x4 = 4∆L = 0.04 m, it follows that h 0 − 5 = 24.1 W/m2⋅K and h 0 − 4 =

26.9 W/m2⋅K and

q5 = (5 × 24.1 − 4 × 26.9 ) W m 2 ⋅ K ( 0.01× 0.2 ) m 2 (500 − 25) K = 12.2 W .

<

Similarly, where h 0 −10 = 17.00 W/m2⋅K and h 0 − 9 = 17.92 W/m2⋅K.

q10 = (10h 0 −10 − 9h0 −9 ) ( ∆L × w )( Ts − T∞ )

q10 = (10 × 17.00 − 9 × 17.92 ) W m 2 ⋅ K ( 0.01× 0.2 ) m 2 (500 − 25) K = 8.3 W

< Continued...

PROBLEM 7.9 (Cont.) For the entire heater, h0 − 25 =

k

 2 1/ 3 0.0429 0.664 Re1/ = × 0.664  L Pr

L and the heat rate over all 25 strips is

1/ 2

   43.54 × 10−6 

0.25

2 × 0.25

(0.683)1/ 3 = 10.75 W

m2 ⋅ K

q 0 − 25 = h0 − 25 ( L × w )( Ts − T∞ ) = 10.75 W m 2 ⋅ K ( 0.25 × 0.2 ) m 2 (500 − 25 ) C = 255.3 W (b,c) Using the IHT Correlations Tool, External Flow, for Laminar or Mixed Flow Conditions, and following the same method of solution as above, the heat rates for the first, fifth, tenth and all the strips were calculated for air velocities of 2, 5 and 10 m/s. To evaluate the heat rates for fully turbulent conditions, the analysis was performed setting Res,c = 1 × 10-6. The results are tabulated below.

<

$

Flow conditions Laminar

Fully turbulent

u ‡ (m/s) 2 5 10 2 5 10

q1 (W) 51.1 80.9 114 17.9 37.3 64.9

q5 (W) 12.1 19.1 27.0 10.6 22.1 38.5

q10 (W) 8.3 13.1 18.6 9.1 19.0 33.1

q0-25 (W) 256 404 572 235 490 853

COMMENTS: (1) An alternative approach to evaluating the heat loss from a single strip, for example, strip 5, would take the form q 5 = h5 ( ∆L × w )( Ts − T∞ ) , where h 5 ≈ h x = 4.5∆L or h5 ≈ ( h x = 5∆L + h x = 4∆L ) 2 . (2) From the tabulated results, note that for both flow conditions, the heat rate for each strip and the entire heater, increases with increasing air velocity. For both flow conditions and for any specified velocity, the strip heat rates decrease with increasing distance from the leading edge. (3) The effect of flow conditions, laminar vs. fully turbulent flow, on strip heat rates shows some unexpected behavior. For the u ‡ = 5 m/s condition, the effect of turbulent flow is to increase the heat rates for the entire heater and the tenth and fifth strips. For the u ‡ = 10 m/s, the effect of turbulent flow is to increase the heat rates at all locations. This behavior is a consequence of low Reynolds number (Rex = 2.3 × 104) at x = 0.25 m with u ‡ = 10 m/s. (4) To more fully appreciate the effects due to laminar vs. turbulent flow conditions and air velocity, it is useful to examine the local coefficient as a function of distance from the leading edge. How would you use the results plotted below to explain heat rate behavior evident in the summary table above? Local coefficient, hx (W/m^2.K)

100 80 60 40 20 0 0

0.02

0.04

0.06

0.08

Distance from the leading edge, x (m) uinf = 2 m/s, laminar flow uinf = 5 m/s, laminar flow uinf = 10 m/s, laminar flow uinf = 10 m/s, fully turbulent flow

0.1

PROBLEM 7.10 KNOWN: Speed and temperature of atmospheric air flowing over a flat plate of prescribed length and temperature. 5

5

6

FIND: Rate of heat transfer corresponding to Rex,c = 10 , 5 × 10 and 10 . SCHEMATIC:

ASSUMPTIONS: (1) Flow over top and bottom surfaces. 3

-6

PROPERTIES: Table A-4, Air (Tf = 348K, 1 atm): ρ = 1.00 kg/m , ν = 20.72 × 10 0.0299 W/m⋅K, Pr = 0.700.

2

m /s, k =

ANALYSIS: With u L 25 m/s × 1m Re L = ∞ = = 1.21 ×106 -6 2 ν 20.72 ×10 m / s the flow becomes turbulent for each of the three values of Rex,c. Hence,

)

(

1/3 Nu L = 0.037 Re 4/5 L − A Pr 1/2 A = 0.037 Re 4/5 x,c − 0.664 Re x,c 5

Rex,c

5

10

6

5×10

10

__________________________________________________________________

A Nu L

(

hL W/m 2 ⋅ K q′ ( W/m )

)

160 2272 67.9

13,580

871 1641 49.1 9820

1671 931 27.8 5560

where q′ = 2 h LL ( Ts − T∞ ) is the total heat loss per unit width of plate. COMMENTS: Note that h L decreases with increasing Rex,c , as more of the surface becomes covered with a laminar boundary layer.

PROBLEM 7.11 KNOWN: Velocity and temperature of air in parallel flow over a flat plate of 1-m length. FIND: (a) Calculate and plot the variation of the local convection coefficient, hx(x), with distance for flow conditions corresponding to transition Reynolds numbers of 5 × 105, 2.5 × 105 and 0 (fully turbulent), (b) Plot the variation of the average convection coefficient, h x ( x ) , for the three flow conditions of part (a), and (c) Determine the average convection coefficients for the entire plate, h L , for the three flow conditions of part (a). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant surface temperature, and (3) Critical Reynolds depends upon prescribed flow conditions. PROPERTIES: Table A.4, Air (Tf = 300 K, 1 atm): ν = 15.89 × 10-6 m2/s, k = 0.0263 W/m⋅K, Pr = 0.707. ANALYSIS: (a) The Reynolds number for the plate (L = 1 m) is u L 10 m s × 1m ReL = ∞ = = 6.29 × 105 . 6 2 − ν 15.89 × 10 m s Hence, the boundary layer conditions are mixed with Rex,c = 5 × 105, 5 × 105 x c = L Re x,c ReL = 1m = 0.795 m 5

(

)

6.29 × 10

Using the IHT Correlation Tool, External Flow, Local coefficients for Laminar or Turbulent Flow, hx(x) was evaluated and plotted with critical Reynolds numbers of 5 × 105, 2.5 × 105 and 0 (fully turbulent). Note the location of the laminar-turbulent transition for the first two flow conditions.

Local coefficient, hx (W/m^2.K)

100

80

60

40

20

0 0

0.2

0.4

0.6

0.8

1

Distance from the leading edge, x (m) Rexc = 5.0e5, Mixed flow Rexc = 2.5e5, Mixed flow Rexc = 0, Fully turbulent flow

Continued...

PROBLEM 7.11 (Cont.) (b) Using the IHT Correlation Tool, External Flow, Average coefficient for Laminar or Mixed Flow, h x ( x ) was evaluated and plotted for the three flow conditions. Note that the change in h x ( x ) at the

Average coefficient, hLbar (W/m^2.K)

critical length, xc, is rather gradual, compared to the abrupt change for the local coefficient, hx(x). 100

80

60

40

20

0 0

0.2

0.4

0.6

0.8

1

Distance from the leading edge, x (m) Rexc = 5.0e5, Mixed flow Rexc = 2.5e5, Mixed flow Rexc = 0, Fully turbulent flow

(c) The average convection coefficients for the plate can be determined from the above plot since h L = h x ( L ) . The values for the three flow conditions are, respectively,

h L = 17.4, 27.5 and 37.8 W m 2 ⋅ K COMMENTS: A copy of the IHT Workspace used to generate the above plots is shown below. // Method of Solution: Use the Correlation Tools, External Flow, Flat Plate, for (i) Local, laminar or turbulent flow and (ii) Average, laminar or mixed flow, to evaluate the local and average convection coefficients as a function of position on the plate. In each of these tools, the value of the critical Reynolds number, Rexc, can be set corresponding to the special flow conditions. // Correlation Tool: External Flow, Plate Plate, Local, laminar or turbulent flow. Nux = Nux_EF_FP_LT(Rex,Rexc,Pr) // Eq 7.23,37 Nux = hx * x / k Rex = uinf * x / nu Rexc =1e-10 // Evaluate properties at the film temperature, Tf. //Tf = (Tinf + Ts) / 2 /* Correlation description: Parallel external flow (EF) over a flat plate (FP), local coefficient; laminar flow (L) for RexRexc, Eq 7.37; 0.6 Tm) is appropriate, k 0.634 W/m ⋅ K 0.4 h i = 0.023Re4/5 = 0.023 (12,611 )4 / 5 ( 4.16 )0.4 = 1230 W/m 2 ⋅K. D Pr D 0.040 m For the external flow, ReD = VD/ν = 100 m/s × 0.040 m/38.79 × 10 -6 m 2 / s = 1.031 ×105 and from Table 7.4, C = 0.26 and m = 0.6; Pr ≤ 10, n = 0.37, and Pr ≈Prs , n h o = ( k/D ) CRem D Pr ( Pr/Prs )

1/4

ho =

40.7 × 10−3 W/m ⋅ K

(

× 0.26 1.031 × 105

)

0.6

( 0.684) 0.37 ( 1)1 / 4 = 234 W/m2 ⋅ K.

0.040 m 2 Substituting numerical values into Eq. 8.46 with P = πD and U = 197 W/m ⋅K, T∞ − Tm,o & p = exp − PLU/mc T∞ − Tm,i

(

)

(1)

 π × 0.040 m × 4 m  o Tm,o = 225o C − ( 225 − 30 ) Cexp  − ×197 W/m 2 ⋅ K  = 47.6o C  0.25 kg/s × 4179 J/kg ⋅ K  COMMENTS: Note the assumed value of Tm,o to evaluate water properties was reasonable. Using Eq. (1), replacing T∞ and U with Ts and hi, respectively, find Ts = 63.2°C; hence, Prs (Ts ) ≈ 0.687. The assumption that Pr ≈ Prs in the Zhukauskas relation is reasonable.

<

PROBLEM 8.68 KNOWN: Single tube heat exchanger for cooling blood. FIND: (a) Temperature at which properties are evaluated in estimating h, (b) Prandtl number for the blood, Pr, (c) Flow condition: laminar or turbulent, (d) Average heat transfer coefficient, h, for blood flow, (e) Total heat rate, q, (f) Required length of tube, L, when U is known. SCHEMATIC:

(

)

3

-7

PROPERTIES: Blood Given, Tm : ρ = 1000 kg/m , ν = 7 × 10 4000 J/kg⋅K.

2

m /s, k = 0.5 W/m⋅K, cp =

ASSUMPTIONS: (1) Flow and thermal conditions fully developed, (2) Thermal resistance of tube material is negligible, (3) Overall heat transfer coefficient between blood and water-ice mixture is U = 2

300 W/m ⋅K, (4) Constant properties, (5) Negligible heat transfer enhancement associated with coiling. o ANALYSIS: (a) Evaluate properties at Tm = ( To +Ti ) / 2 = ( 40 + 30) C/2 = 35oC. (b) The Prandtl number is

Pr =

c pµ k

=

c pνρ k

4000 J/kg ⋅ K × 7 × 10−7 m 2 / s ×1000 kg/m3 ) ( = = 5.60. 0.5 W/m ⋅ K

< <

(c) Calculate Reynolds number as

&ρ & & 4 ×10−4 m 3 /min (1 min/60s ) 4m 4V 4V Re D = = = = = 1213 π D µ π Dνρ π Dν π × 2.5 × 10−3 m × 7 ×10−7 m 2 / s

<

Hence, the flow is laminar, (d) For laminar and fully developed conditions, Eq. 8.55 is the proper correlation,

NuD = hD/k = 3.66

h = 3.66 × 0.5 W/m ⋅ K/2.5 × 10−3 m = 732 W/m 2 ⋅ K,

<

(e) The total heat rate follows from an overall energy balance, Eq. 8.37,

& p ( To − Ti ) & c p ( To − Ti ) = ρVc q=m

(

)

q = 1000 kg/m 3 10−4 m 3 /min/60 s/min 4000 J/kg ⋅ K × ( 30 − 40 )o C = −66.7 W.

<

(f) Using the rate equation, Eq. 8.47, solve for L,

L=

q 66.7 W = = 0.81 m Uπ D ∆Tlm 300 W/m2 ⋅ K π × 2.5 × 10−3 m × 34.8o C

(

)

where A = πDL and ∆Tl m = [(40 - 0)°C - (30 - 0)°C]/ l n(40/30) = 34.8°C.

<

PROBLEM 8.69 KNOWN: Flow conditions associated with water passing through a pipe and air flowing over the pipe. FIND: (a) Differential equation which determines the variation of the mixed-mean temperature of the water, (b) Heat transfer per unit length of pipe at the inlet and outlet temperature of the water. SCHEMATIC:

ASSUMPTIONS: (1) Negligible temperature drop across the pipe wall, (2) Negligible radiation exchange between outer surface of insulation and surroundings, (3) Fully developed flow throughout pipe, (4) Negligible potential and kinetic energy and flow work effects. -6

PROPERTIES: Table A-6, Water (Tm,i = 200°C): cp,w = 4500 J/kg⋅K, µw = 134 × 10 -6

kw = 0.665 W/m⋅K, Prw = 0.91; Table A-4, Air (T∞ = - 10°C): νa = 12.6 × 10 W/m⋅K, Pra = 0.71, Prs ≈ 0.7.

2

N⋅s/m ,

2

m /s, ka = 0.023

ANALYSIS: (a) Following the development of Section 8.3.1 and applying an energy balance to a differential element in the water, we obtain

& c p,w Tm − dq − m & c p,w ( Tm + dTm ) = 0. m Hence

& c p,w dTm dq = −m

where

dq = UidAi ( Tm − T∞ ) = U iπ D dx ( Tm − T∞ ) .

Substituting into the energy balance, it follows that

d Tm Uπ D =− i ( Tm − T∞ ) . & cp dx m

(1)

<

The overall heat transfer coefficient based on the inside surface area may be evaluated from Eq. 3.30 which, for the present conditions, reduces to

Ui =

1 . 1 D  D + 2t  D 1 + ln + h i 2k  D  D + 2t h o

(2)

For the inner water flow, Eq. 8.6 gives

Re D =

& 4m 4 × 2 kg/s = = 19,004. π Dµ w π (1 m ) ×134 ×10 −6 kg/s ⋅ m Continued …..

PROBLEM 8.69 (Cont.) Hence, the flow is turbulent. With the assumption of fully developed conditions, it follows from Eq. 8.60 that k 0.3. h i = w × 0.023 Re 4/5 Prw (3) D D For the external air flow

Re D =

V ( D+2t ) ν

=

4 m/s (1.3m )

12.6 ×10−6 m 2 / s

= 4.13 ×105.

Using Eq. 7.31 to obtain the outside convection coefficient,

ho =

ka 1/4 × 0.076 Re0.7 Pra0.37 ( Pra / P rs ) . D ( D + 2t )

(4)

(b) The heat transfer per unit length of pipe at the inlet is

(

)

q′ = π D Ui Tm,i − T∞ .

(5)

From Eqs. (3 and 4),

hi =

0.665 W/m ⋅ K × 0.023 (19,004 ) 4 / 5 ( 0.91)0.3 = 39.4 W/m 2 ⋅ K 1m

ho =

0.7 0.023 W/m ⋅ K × 0.076 4.13 ×105 ( 0.71)0.37 (1)1 / 4 = 10.1 W/m2 ⋅ K. (1.3 m )

(

)

Hence, from Eq. (2) −1

  1 1m 1  1.3  1 Ui =  + ln   + ×   39.4 W/m 2 ⋅ K 0.1 W/m ⋅ K  1  1.3 10.1 W/m 2 ⋅ K 

= 0.37 W/m 2 ⋅ K

and from Eq. (5)

(

q′ = π (1 m ) 0.37 W/m 2 ⋅ K

) ( 200 +10)o C = 244 W/m.

<

Since Ui is a constant, independent of x, Eq. (1) may be integrated from x = 0 to x = L. The result is analogous to Eq. 8.42b and may be expressed as

 π DL    π ×1m × 500m = exp  − Ui  = exp  − × 0.37 W/m 2 ⋅ K   m  T∞ − Tm,i  2 kg/s × 4500 J/kg ⋅ K   & c p,w  T∞ − Tm,o Hence = 0.937. T∞ − Tm,i T∞ − Tm,o

(

)

Tm,o = T∞ + 0.937 Tm,i − T∞ = 187o C.

<

COMMENTS: The largest contribution to the denominator on the right-hand side of Eq. (2) is made by the conduction term (the insulation provides 96% of the total resistance to heat transfer). For this reason the assumption of fully developed conditions throughout the pipe has a negligible effect on the calculations. Since the reduction in Tm is small (13°C), little error is incurred by evaluating all properties of water at Tm,i.

PROBLEM 8.70 KNOWN: Inner and outer radii and thermal conductivity of a teflon tube. Flowrate and temperature of confined water. Heat flux at outer surface and temperature and convection coefficient of ambient air. FIND: Fraction of heat transfer to water and temperature of tube outer surface. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Fully-developed flow, (3) One-dimensional conduction, (4) Negligible tape contact and conduction resistances. -6

PROPERTIES: Table A-6, Water (Tm = 290K): µ = 1080 × 10 7.56.

kg/s⋅m, k = 0.598 W/m⋅K, Pr =

ANALYSIS: The outer surface temperature follows from a surface energy balance

( 2π

ro L) q′′ =

Ts,o − T∞

( h o2π r oL )−1 ( ln ( ro / ri ) / 2π

(

Lk ) + (1 / 2π ri Lh i )

−T

) ( r / k ) ln ( rs,o/ r ) +m( r / r ) / h . o o i o i i T

q′′ = h o Ts,o − T∞ + With

Ts,o − Tm

+

& ( π Dµ ) = 4 ( 0.2kg/s ) / π ( 0.02 m )1080 ×10 −6 kg/s ⋅ m  = 11,789 Re D = 4 m/  

the flow is turbulent and Eq. 8.60 yields h i = ( k/Di ) 0.023ReD Pr 4/5

0.4

= (0.598 W/m ⋅ K/0.02 m )(0.023 )(11,789 )

4/5

( 7.56)0.4 = 2792 W/m 2 ⋅ K.

Hence 2

2

(

)

2000 W/m = 25 W/m ⋅ K Ts,o − 300K +

and solving for Ts,o, The heat flux to the air is

(

Ts,o − 290 K

(0.013 m/0.35 W/m ⋅ K ) ln (1.3 ) + (1.3 ) /

(2792 W/m ⋅ K )

Ts,o = 308.3 K.

2

<

)

q′′o = h o Ts,o − T∞ = 25 W/m 2 ⋅ K ( 308.3 − 300 ) K = 207.5 W/m 2 . Hence,

q′′i / q′′ = ( 2000 − 207.5) W/m 2 /2000 W/m 2 = 0.90.

<

COMMENTS: The resistance to heat transfer by convection to the air substantially exceeds that due to conduction in the teflon and convection in the water. Hence, most of the heat is transferred to the water.

PROBLEM 8.71 KNOWN: Temperature recorded by a thermocouple inserted in a stack containing flue gases with a prescribed flow rate. Diameters and emissivities of thermocouple tube and gas stack. Conditions associated with stack surroundings. FIND: Equations for predicting thermocouple error and error associated with prescribed conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Flue gas has properties of air at Tg ≈ 327°C, (3) Stack forms a large enclosure about the thermocouple tube and surroundings form a large enclosure around the stack, (4) Stack surface energy balance is unaffected by heat loss to tube, (5) Gas flow is fully developed, (6) Negligible conduction along thermocouple tube, (7) Stack wall is thin. 3

-7

PROPERTIES: Table A-4, Air (Tg ≈ 600K, pg = 1 atm): ρ = 0.58 kg/m , µ = 305.8 × 10 -6

ν = 52.7× 10

2

N⋅s/m ,

2

m /s, k = 0.0469 W/m⋅K, Pr = 0.685.

ANALYSIS: Determination of the thermocouple error necessitates determining the gas temperature Tg and relating it to the thermocouple temperature Tt. From an energy balance applied to a control surface about the thermocouple, q conv = q rad or h t A t Tg − Tt = ε tσ A t Tt4 − Ts4 .

)

(

(

)

εσ Tg = Tt + t Tt4 − Ts4 . ht

Hence

(

)

(1)

<

However, Ts is unknown and must be determined from an energy balance on the stack wall.

q conv,i = qconv,o + qrad

(

)

(

4 h i As Tg − Ts = h o As ( Ts − T∞ ) + εsσ A s Ts4 − Tsur or

(

)

h εσ 4 . Tg = Ts + o ( Ts − T∞ ) + s Ts4 − Tsur hi hi

) (2)

<

Tg and Ts may be determined by simultaneously solving Eqs. (1) and (2). For the prescribed conditions & g / ρπ Ds2 / 4 Dt 4 m & gD t ρ VDt ρ m 4 ×1 kg/s × 0.01 m ReDt = = = = = 1157. 2 µ µ πµ Ds π × 305.8 ×10-7 N ⋅ s/m2 ( 0.6 m )2 Continued …..

(

)

PROBLEM 8.71 (Cont.) Assuming (Pr/Prs ) = 1, it follows from the Zhukauskus correlation 0.37 Nu D = 0.26Re0.6 Dt Pr where C = 0.26 and m = 0.6 from Table 7.4. Hence

ht =

0.0469 W/m ⋅ K (1157 ) 0.6 ( 0.685 ) 0.87 × 0.26 = 73 W/m2 ⋅ K. 0.01 m Tg = 573 K +

Hence, from Eq. (1)

0.8 × 5.67 × 10−8 W/m 2 ⋅ K 4 73 W/m2 ⋅ K

(5734 −Ts4 )K 4

Tg = 573 K + 67 K − 6.214 ×10 −10 Ts4 = 640 − 6.214 ×10 −10 Ts4. Re Ds =

Also,

&g 4m π Ds µ

=

4 × 1 kg/s π ( 0.6 m ) 305.8 ×10−7 N ⋅ s/m2

(1a)

= 6.94 ×10 4

and the gas flow is turbulent. Hence from the Dittus-Boelter correlation,

hi =

(

)

k 0.3 = 0.0469 W/m ⋅ K × 0.023 6.94 ×104 4 / 5 × 0.685 0.3 = 12 W/m 2 ⋅ K. 0.023Re 4/5 Pr ( ) Ds Ds 0.6 m

Hence from Eq. (2)

Tg = Ts +

25 0.8 × 5.67 × 10−8 W/m 2 ⋅ K 4  4 Ts − 300 4  K4 ( Ts − 300 K ) +  2   12 12 W/m ⋅ K

Tg = Ts + 2.083Ts − 625 K + 3.78 × 10 −9 Ts4 − 30.6 K = −655.6 K + 3.083Ts + 3.78 × 10−9 Ts4 .

(2a)

Solve Eqs. (1a) and (2a) by trial-and-error. Assume values for Ts and determine Tg from (1a) and (2a). Continue until values of Tg agree.

Ts (K) 400 375 387 388 Hence

Tg (K) → (1a) 624 628 626 626

Tg (K) → (2a) 674 575 622 626

Ts = 388 k, Tg = 626 K

and the thermocouple error is

Tg − Tt = 626 K − 573 K = 53oC.

<

COMMENTS: The thermocouple error results from radiation exchange between the thermocouple tube and the cooler stack wall. Anything done to ↑ Ts would ↓ this error (e.g., ↓ ho or ↑ T∞ and Tsur). The error also ↓ with ↑ ht. The error could be reduced by installing a radiation shield around the tube.

PROBLEM 8.72 KNOWN: Platen heated by hot ethylene glycol flowing through tubing arrangement with spacing S soldered to lower surface. Top surface exposed to convection process. FIND: Tube spacing S and heating fluid temperature Tm which will maintain the top surface at 45 ± 0.25°C. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions; (2) Lower surface is insulated, all heat transfer from hot fluid is into platen; (3) Copper tube is thick-walled such that interface between solder and platen is isothermal; (4) Fully developed flow conditions in tube. PROPERTIES: Table A.4, Ethylene glycol (Tm = 60°C): µ = 0.00522 N⋅s/m2, k = 0.2603 W/m⋅K. ANALYSIS: Begin the analysis by setting up a nodal mesh (9 ×6) to represent the platen experiencing convection on the top surface ( T∞ , h) while the two side boundaries are symmetry adiabats. On the lower surface, nodes 46 and 47 represent the isothermal platen-solder interface maintained at To by the hot fluid. The remaining nodes (49-54) are insulated on their lower boundary.

The heat rate supplied by the tube to the platen can be expressed as

q′cv = 0.5h o (π Di )( Tm − To )

(1)

From energy balances about nodes 46 and 47, the heat rate into the platen by conduction can be expressed as

q′cd = q′a + q′b + q′c

(2)

q′a = k ( ∆x 2 )( T46 − T37 ) ∆y

(3) Continued...

PROBLEM 8.72 (Cont.) q′b = k ( ∆x )( T47 − T38 ) ∆y

(4)

q′c = k ( ∆y 2 )( T47 − T48 ) ∆x

(5)

and we require that

q′cd = q′cv

(6)

The convection coefficient for internal flow can be estimated from a correlation assuming fully developed flow. First, characterize the flow with

ReD =

 4m 4 × 0.06 kg s = = 1829 π Di µ π (0.008 m ) 0.00522 N ⋅ s m 2

and since it is laminar,

h D Nu D = o i = 3.66 k h o = 3.66 × 0.2603W m ⋅ K 0.008 m = 119.1W m ⋅ K where properties are evaluated at Tm. Using the IHT Finite-Difference Tool for Two-Dimensional Steady-State Conditions and the Properties Tool for Ethylene Glycol, along with the foregoing rate equations and energy balances, Eqs. (1-6), a model was developed to solve for the temperature distribution in the platen. In the solution, we determined what hot fluid temperature was required to maintain T1 = 45°C. Two trials were run. In the first, the nodal arrangement was as shown above (9 × 6) for which S/2 = (9 - 1)∆x = 42.67 mm with ∆x = 2Di/3 = 5.33 mm and ∆y = w/5 = 5 mm. In the second trial, we repositioned the right-hand symmetry adiabat to pass vertically through the nodes 6-51 so that now the nodal mesh is (6 × 6) and S/2 = (6 - 1)∆x = 26.65 mm with ∆x and ∆y remaining the same. The results of the trials are tabulated below. Trial

Mesh

T1 (°C)

T6 (°C)

T9 (°C)

Tm (°C)

q′cv (W/m)

1 2

9×6 6×6

45.0 45.0

43.5 44.5

43.0 ---

105 85

80.5 52.6

From the trial 2 results, the surface temperature uniformity is (T1 - T6) = 0.5°C which satisfies the ±0.25°C requirement. So that suitable tube spacing and fluid temperature are S = 53 mm

Tm = 85°C

<

COMMENTS: (1) Recognize that the grid spacing is quite coarse and good practice demands that we repeat the analysis decreasing the nodal spacing until no further changes are seen in Tm. (2) In the first trial, note that Tm = 105°C which of course, is not possible.

PROBLEM 8.73 KNOWN: Features of tubing used in a ground source heat pump. Temperature of surrounding soil. Fluid inlet temperature and flowrate. FIND: (a) Effect of tube length on outlet temperature, (b) Recommended tube length and the effect of variations in the flowrate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible conduction resistance in soil, (4) Negligible KE, PE and flow work changes, (5) Fluid properties correspond to those of water. PROPERTIES: Table A.6 (assume Tm = 277 K): cp = 4206 J/kg⋅K, µ = 1560 × 10-6 N⋅s/m2, k = 0.577 W/m⋅K, Pr = 11.44.

 π Di µ = 4 ( 0.03kg s ) π ( 0.025m )1560 ANALYSIS: (a) For the prescribed conditions, ReD = 4m 6 2 – 10 N ¹ s m = 980 and the flow is laminar. Assuming thermal entry length conditions, which is reasonable for Pr = 11.44, Eq. 8.56 may be used to determine the average convection coefficient Nu D = 3.66 +

0.0668 ( D L ) ReD Pr 1 + 0.04 ( D L ) ReD Pr 

2/3

With Ts used in lieu of T‡ , Eq. 8.46b may be used to determine Tm,o,

 L = exp  −   ′ Ts − Tm,i  mcp R tot

Ts − Tm,o

   

where R ′tot accounts for the convection and tube wall conduction resistances,

R ′tot = R ′cnv + R ′cnd = (1 π Di h ) + ln ( Do Di ) 2π k t and

Do = Di + 2t = 41mm . Using the Correlations and Properties Toolpads of IHT, the following results were obtained for the effect of the tube length L on Tm,o.

Continued...

PROBLEM 8.73 (Cont.)

Outlet temperature, Tmo(C)

10

8

6

4

2

0 10

15

20

25

30

35

40

45

50

Tube length, L(m) mdot = 0.015 kg/s mdot = 0.030 kg/s mdot = 0.045 kg/s

 = 0.030 kg/s, the The longer the tube the larger the rate of heat extraction from the soil, and for m temperature rise of ∆T = (Tm,o - Tm,i) ≈ 6°C is well below the maximum possible value of ∆Tmax = 10°C.  = 0.015 kg/s), the (b) The length should be at least 50 m long. If the flowrate were reduced by 50% ( m corresponding temperature rise would be close to ∆Tmax and L = 50 m would be close to optimal. However, for the nominal flowrate and a 50% increase from the nominal, the length should exceed 50 m to recover more heat and provide a heat pump inlet temperature which is closer to the maximum possible value.

COMMENTS: In practice, the tube surface temperature would be less than 10°C (if the temperature of the soil well removed from the tube were at 10°C), thereby reducing the heat extraction rate and Tm,o.

PROBLEM 8.74 KNOWN: Reynolds numbers for fully developed turbulent flow of water in a smooth circular tube. FIND: Nusselt numbers based on the Colburn, Petukhov and Gnielinski correlations. SCHEMATIC:

ANALYSIS: The correlations are Colburn, Equation 8.59, 1/3 Nu D = 0.023Re4/5 D Pr

Petukhov, Equation 8.62, NuD =

( f/8) Re D Pr 1/2 1.07 + 12.7 ( f/8 ) Pr 2 / 3 −1

(

)

−2

f = (1.82log10 Re D − 1.64 ) Gnielinski, Equation 8.63, Nu D =

( f/8)( ReD − 1000 ) Pr 1/2 1 +12.7 ( f/8 ) Pr 2 / 3 −1

(

f = ( 0.79lnReD − 1.64 )

)

−2

It follows that: ReD ___________________________________________________________________________________________________________

Correlation

4

4000

5

10

10

___________________________________________________________________________________________________________

Colburn

31.8

66.2

417.9

Petukhov

40.0

81.2

549.5

Gnielinski

30.0

75.0

560.0 4

COMMENTS: The Colburn equation does well in the transitional region (ReD ≤ 10 ), where the Gnielinski equation provides the best predictions, but can significantly underpredict the Nusselt number under fully turbulent conditions.

PROBLEM 8.75 KNOWN: Effect of entry length on average Nusselt number for turbulent flow in a tube. FIND: Ratio of average to fully developed Nusselt numbers for prescribed conditions. SCHEMATIC:

ASSUMPTIONS: (1) Sharp edged inlet, (2) Combined entry region. ANALYSIS: From Eq. 8.64, Nu D C = 1+ Nu D,fd ( x/D ) m 0.23 and with C = 24Re− and m = 0.815 − 2.08 ×10 −6 Re D , D

Nu D = 1+ Nu D,fd

( x/D )(

0.23 24Re − D

0.815− 2.08×10 −6 ReD

)

.

It follows that _______________________________________________________________

( Nu D / Nu D,fd )

Re D

b x / Dg

_______________________________________________________________ 4

1.464

10

1.112

10

1.420

10

1.142

10

4 5 5

10 60 10 60

COMMENTS: The assumption Nu D ≈ Nu fd for x/D = 10 would result in underprediction of Nu D by approximately 45%. The underprediction is only approximately 10% for x/D = 60.

PROBLEM 8.76 KNOWN: Fluid enters a thin-walled tube of 5-mm diameter and 2-m length with a flow rate of 0.04 kg/s and temperature of Tm,i = 85°C; tube surface temperature is maintained at Ts = 25°C; and, for this base operating condition, the outlet temperature is Tm,o = 31.1°C. FIND: The outlet temperature if the flow rate is doubled? SCHEMATIC:

ASSUMPTIONS: (1) Flow is fully developed and turbulent, (2) Fluid properties are independent of temperature, and (3) Constant surface temperature cooling conditions.

 cp , the ANALYSIS: For the base operating condition (b), the rate equation, Eq. 8.42b, with C = m capacity rate, is

Ts − Tm,o

)b

Ts − Tm,i

 PL h b  = exp  −   Cb 

(1)

Substituting numerical values, with P = πD, find the ratio, h b / Cb ,

25 − 31.1 = exp π × 0.005 m × 2 m ( h b / Cb ) 25 − 85

h b / Cb = 72.77 m −2 For the new operating condition (n), the flow rate is doubled, Cn = 2Cb, and the convection coefficient scales according to the Dittus-Boelter relation, Eq. 8.60,  0.8 h Re0.8 m D

(

)

(

h n = 20.8 h b and h n / C n = 20.8 / 2 h b / Cb

)

(2)

Using the rate equation for the new operating condition, find

Ts − Tm,o

)n

Ts − Tm,i 25 − Tm,o 25 − 85 Tm,o

)n

 PL h n = exp  −  Cn

  = exp  − PL × 0.871 h b / Cb  

(

)

(3)

= exp  −π × 0.005 m × 2 m × 0.871× 72.77 m −2   

)n = 33.2°C

<

PROBLEM 8.77 KNOWN: Flow rate and inlet temperature of air passing through a rectangular duct of prescribed dimensions and surface heat flux. FIND: Air and duct surface temperatures at outlet. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform surface heat flux, (3) Constant properties, (4) Atmospheric pressure, (3) Fully developed conditions at duct exit, (6) Negligible KE, PE and flow work effects.

(

)

-7

PROPERTIES: Table A-4, Air Tm ≈ 300K, 1 atm : cp = 1007 J/kg⋅K, µ = 184.6 × 10 k = 0.0263 W/m⋅K, Pr = 0.707.

2

N⋅s/m ,

ANALYSIS: For this uniform heat flux condition, the heat rate is

q = q ′′s As = q′′s  2 ( L × W ) + 2 ( L × H ) 

q = 600 W/m 2  2 (1m × 0.016m ) + 2 (1m × 0.004m )  = 24 W. From an overall energy balance

Tm,o =Tm,i +

q 24 W = 300K + = 379 K. & cp m 3 ×10 −4 kg/s × 1007 J/kg ⋅ K

<

The surface temperature at the outlet may be determined from Newton’s law of cooling, where

Ts,o = Tm,o + q ′′/h. From Eqs. 8.67 and 8.1

Dh =

4 A c 4 ( 0.016m × 0.004m ) = = 0.0064 m P 2 ( 0.016m + 0.004m )

& Dh 3 ×10 −4 kg/s ( 0.0064m ) ρ u m Dh m Re D = = = = 1625. µ Acµ 64 ×10−6m 2 184.6 ×10−7 N ⋅ s/m2

(

)

Hence the flow is laminar, and from Table 8.1

h=

k 0.0263 W/m ⋅ K 5.33 = 5.33 = 22 W/m 2 ⋅ K Dh 0.0064 m

Ts,o = 379 K +

600 W/m 2 22 W/m 2 ⋅ K

= 406 K.

<

COMMENTS: The calculations should be reperformed with properties evaluated at Tm = 340 K. The change in Tm,o would be negligible, and Ts,o would decrease slightly.

PROBLEM 8.78 KNOWN: Flow rate and temperature of air entering a triangular duct of prescribed dimensions and surface temperature. FIND: Air outlet temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform surface temperature, (4) Fully developed conditions throughout, (5) Air is at atmospheric pressure, (6) Negligible potential and kinetic energy and flow work effects. PROPERTIES: Table A-4, Air (assume Tm ≈ 325K, 1 atm): cp = 1008 J/kg⋅K, µ = 196.4 × -7

2

10 N⋅s/m , k = 0.0282 W/m⋅K, Pr = 0.707. ANALYSIS: From Eqs. 8.67 and 8.1

(

)

−4 2 4 A c 4 1.73 × 10 m Dh = = = 0.0115 m P 3 ( 0.02m )

& Dh 4 × 10−4 kg/s ( 0.0115 m ) ρ u mD h m Re D = = = = 1354. µ Ac µ 1.73 ×10 −4m 2 196.4 ×10 −7 N ⋅ s/m 2

(

)

Hence the flow is laminar and from Table 8.1, h=

k 0.0282 W/m ⋅ K 2.47 = 2.47 = 6.1 W/m 2 ⋅ K. Dh 0.0115 m

From Eq. 8.42b it follows that, with P = 3 W,  PL  Tm,o =Ts − Ts − Tm,i exp  − h  m   & cp 

(

)

 3 × 0.02m × 2m × 6.1 W/m 2 ⋅ K   Tm,o = 100o C − 100oC − 27o C exp  −  4 ×10−4 kg/s ×1008 J/kg ⋅ K   

(

)

Tm,o = 88o C. COMMENTS: With Tm,o = 88°C, Tm = 330K and there is no need to re-evaluate the properties.

<

PROBLEM 8.79 KNOWN: Temperature and velocity of gas flow between parallel plates of prescribed surface temperature and separation. Thickness and location of plate insert. FIND: Heat flux to the plates (a) without and (b) with the insert. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible radiation, (3) Gas has properties of atmospheric air, (4) Plates are of infinite width W, (5) Fully developed flow. 3

-7

PROPERTIES: Table A-4, Air (1 atm, Tm = 1000K): ρ = 0.348 kg/m , µ = 424.4 × 10 = 0.0667 W/m⋅K, Pr = 0.726.

kg/s⋅m, k

ANALYSIS: (a) Based upon the hydraulic diameter Dh, the Reynolds number is

Dh = 4 Ac / P = 4 ( H ⋅ W ) / 2 ( H + W ) = 2H = 80 mm Re Dh =

ρ u m D h 0.348 kg/m3 ( 60 m/s ) 0.08 m = = 39,360. µ 424.4 × 10−7 kg/s ⋅ m

Since the flow is fully developed and turbulent, use the Dittus-Boelter correlation, 0.3 = 0.023 39,360 4 / 5 0.726 0.3 = 99.1 NuD = 0.023 Re 4/5 ( ) ( ) D Pr k 0.0667 W/m ⋅ K h= Nu D = 99.1 = 82.6 W/m 2 ⋅ K Dh 0.08 m q′′ = h ( Tm − Ts ) = 82.6 W/m 2 ⋅ K (1000 − 350 ) K = 53,700 W/m 2 . (b) From continuity, & = ( ρ u mA ) = ( ρ u mA ) m a b

<

u m )b = um )a ( ρ A ) a / ( ρ A )b = 60 m/s ( 40/20 ) = 120 m/s.

For each of the resulting channels, Dh = 0.02 m and ρ u m D h 0.348 kg/m3 (120 m/s ) 0.02 m Re Dh = = = 19,680. −7

µ

424.4 × 10

kg/s ⋅ m

Since the flow is still turbulent,

Nu D = 0.023(19,680 )4 / 5 ( 0.726 )0.3 = 56.9

h=

56.9 ( 0.0667 W/m ⋅ K ) 0.02 m

= 189.8 W/m 2 ⋅ K

q′′ = 189.8 W/m ⋅ K (1000 − 350 ) K = 123,400 W/m . 2

2

COMMENTS: From the Dittus-Boelter equation, 0.8 0.2 h b / ha = u m,b / u m,a Dh,a / Dh,b = ( 2 )0.8 ( 4 ) 0.2 = 1.74 ×1.32 = 2.30.

(

) (

)

Hence, heat transfer enhancement due to the insert is primarily a result of the increase in um and secondarily a result of the decrease in Dh.

<

PROBLEM 8.80 KNOWN: Temperature, pressure and flow rate of air entering a rectangular duct of prescribed dimensions and surface temperature. FIND: Air outlet temperature and duct heat transfer rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform surface temperature, (4) Fully developed flow throughout, (5) Negligible potential and kinetic energy and flow work effects. -7

PROPERTIES: Table A-4, Air (assume Tm ≈ 325K, 1 atm): cp = 1008 J/kg⋅K, µ = 196.4 × 10 2

N⋅s/m , k = 0.0282 W/m⋅K, Pr = 0.707. ANALYSIS: From Eqs. 8.67 and 8.1,

Dh =

4 A c 4 × ( 0.15 × 0.075) m 2 = = 0.10 m P 2 ( 0.15 + 0.075 ) m

Re D =

0.1 kg/s ( 0.1m ) & Dh ρ u mD h m = = = 45,260. −7 N ⋅ s/m 2 µ A cµ 0.15m × 0.075m 196.4 × 10 ( )

Hence the flow is turbulent, and from Eq. 8.60

h=

k 0.0282 W/m ⋅ K 0.023 Re 4/5 Pr 0.4 = 0.023 ( 45,260 ) 4 / 5 ( 0.707 ) 0.4 = 30 W/m 2 ⋅ K. D Dh 0.10 m

From Eq. 8.42b, with P = 2(W + H),

 PL  Tm,o = Ts − Ts − Tm,i exp  − h  m  & c p    2 ( 0.15m + 0.075m ) 2m 30W/m2 ⋅ K  Tm,o = 400 K − ( 400 − 285 ) K exp  − 0.1 kg/s × 1008 J/kg ⋅ K 

(

)

(

(

 

<

Tm,o = 312 K and from Eq. 8.37

) 

)

& cp Tm,o − Tm,i = 0.1 kg/s × 1008 J/kg ⋅ K ( 312 − 285) K = 2724 W. q=m

<

COMMENTS: (1) The calculations may be checked by determining q from Eqs. 8.44 and 8.45. We obtain ∆Tl m = 101o C and q = 2724 W. (2) Tm has been over-estimated. The calculations should be repeated with properties evaluated at Tm = 299 K.

PROBLEM 8.81 KNOWN: Dimensions of semi-circular copper tubes in contact at plane surfaces. Thermal contact resistance. Tube flow conditions. FIND: (a) Heat rate per unit tube length, and (b) The effect on the heat rate when the fluids are ethylene glycol, the exchanger tube is fabricated from an aluminum alloy, or the exchanger tube thickness is increased. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Adiabatic outer surface, (4) Fully developed flow, (5) Negligible heat loss to surroundings. PROPERTIES: Table A.1, Copper (T ≈ 300 K): k = 400 W/m⋅K; Table A.6, Water (290 K): µ = 1080 × 10-6 N⋅s/m2, k = 0.598 W/m⋅K, Pr = 7.56; (330 K): µ = 489 × 10-6 N⋅s/m2, k = 0.65 W/m⋅K, Pr = 3.15; (given): µ = 800 × 10-6 kg/s⋅m, k = 0.625 W/m⋅K, Pr = 5.35. ANALYSIS: (a,b) Heat transfer from the hot to cold fluids is enhanced by conduction through the semicircular portions of the tube walls. The walls may be approximated as straight fins with an insulated tip, and the thermal circuit is shown below.

Note that, since each semi-circular surface is insulated on one side, surfaces may be combined to yield a single fin of thickness 2t with convection on both sides. Also, due to the equivalent geometry and the assumption of constant properties, there is symmetry on opposite sides of the contact resistance. From the thermal circuit, the heat rate is

q′ =

Th,m − Tc,m R ′tot

(1)

For flow through the semi-circular tube,

 c 4m  h 4mA   ρ u m Dh mD 4m = = = = µ Ac µ Ac Pµ Pµ ( 2ri + π ri ) µ 4 × 0.2 kg s ReD = = 9725 ( 2 + π ) 0.02 m × 800 ×10−6 kg s ⋅ m

ReD =

the flow is turbulent. Using the Colburn correlation, 4 / 5 Pr1/ 3 = 0.023 9725 4 / 5 5.35 1/ 3 = 62.4 Nu D = 0.023ReD ( ) ( )

(2)

(3) Continued...

)

(

PROBLEM 8.81 (Cont.)

2 4Ac 4 π ri 2 2π Dh = 0.02 m = 0.0244 m = = P (π + 2 ) ri π + 2

h = Nu D

(4)

k 0.625 = 62.4 = 1600 W m 2 ⋅ K . Dh 0.0244

(5)

Find now values for the thermal resistance of the circuit.

R ′conv =

1 1 = = 0.0156 m ⋅ K W 2ri h ( 0.04 m )1600 W m 2 ⋅ K

(6)

1 θ R ′fin = b = q′f ( hP′kA′ )1/ 2 tanh ( hP kA ) L c c L = π ri 2 = π ( 0.01m ) = 0.0314 m

Ac = 2t ⋅1m = 0.006 m 2

(7)

P ≈ 2.1 m (8,9,10)

(1600 W m2 ⋅ K × 2 m m × 400 W m ⋅ K × 0.006 m2 s ) = 87.6 W K ⋅ m 1/ 2 ( hP kAc )1/ 2 L = (1600 W m 2 ⋅ K × 2 m 400 W m ⋅ K × 0.006 m2 ) 0.0314 m = 1.15 1/ 2

( hP′kA′c )1/ 2 =

1 = 0.0140 m ⋅ K W 87.6 W m ⋅ K ( 0.817 ) t 0.003m R ′cond = = = 1.875 × 10−4 m ⋅ K W 2kri 2 ( 400 W m ⋅ K )(0.02 m )

R ′fin =

R ′t,c =

R ′′t,c 2ri

=

10−5 m 2 ⋅ K W = 2.5 × 10−4 m ⋅ K W 2 ( 0.02 m )

(11) (12)

(13)

The equivalent resistance of the parallel circuit is −1 + R ′−1 −1 = 71.4 W m ⋅ K + 64.1W m ⋅ K −1 = 7.38 × 10−3 m ⋅ K W (14) R ′eq = R ′fin ( ) conv Hence

)

(

(

)

R ′tot = 2 R ′eq + R ′cond + R ′t,c

(15)

)

(

R ′tot =  2 7.38 × 10−3 + 1.875 × 10−4 + 2.50 × 10−4  m ⋅ K W = 0.0154 m ⋅ K W   q′ =

(330 − 290 ) K 0.0154 m ⋅ K W

= 2600 W m .

<

(c) Using the IHT Workspace with the foregoing equations, analyses were performed and the results summarized in the table below. The “Conditions” are described below; the “Change” is relative to the base case condition. Continued …..

PROBLEM 8.81 (Cont.) Condition* Base case Ethylene glycol Aluminum alloy Thicker tube

R ′conv × 10

(m⋅K/W) 156.8 1382 156.8 156.8

4

R ′fin × 104

R ′cond × 104

R ′tot × 104

R ′eq × 104

q′

Change

(m⋅K/W) 140.1 923.1 183.2 130.6

(m⋅K/W) 1.88 1.88 4.24 2.50

(m⋅K/W) 154.2 1113.0 180.0 150.0

(m⋅K/W) 73.96 553.5 84.49 71.25

(W/m) 2594 359 2223 2667

(%) --86.2 -14.3 +2.8

*Conditions: change from base case Base case - water, copper (k = 400 W/m⋅K), t = 3 mm Ethylene glycol - ethylene glycol instead of water Aluminum alloy - alloy (k = 177 W/m⋅K) instead of copper Thicker tube - t = 4 mm instead of 3 mm As expected, using ethylene glycol as the working fluid would decrease the heat rate. Note that R ′conv is the dominate resistance since the convection coefficient is considerably reduced compared to that with water. Using aluminum alloy, rather than copper, as the tube material reduces the heat rate by 14%. Conduction-convection (fin) in the tube wall is important as can be seen by examining the change in R ′fin relative to the base condition. Increasing the tube wall thickness for the copper tube exchanger from 3 to 4 mm had only a marginal positive effect on the heat rate. COMMENTS: A more accurate calculation would account for the absence of symmetry about the contact plane. Evaluation of water properties at Th,m = 330 K and Tc,m = 290 K yields hh = 2060 W/m2⋅K and hc = W/m2⋅K.

PROBLEM 8.82 KNOWN: Heat exchanger to warm blood from a storage temperature 10°C to 37° at 200 ml/min. Tubing has rectangular cross-section 6.4 mm × 1.6 mm sandwiched between plates maintained at 40°C. FIND: (a) Length of tubing and (b) Assessment of assumptions to indicate whether analysis under- or over-estimates length. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes, (3) Blood flow is fully developed, (4) Blood has properties of water, and (5) Negligible tube wall and contact resistance. 3

PROPERTIES: Table A-6, Water ( Tm ≈ 300 K): cp,f = 4179 J/kg⋅K, ρ f = 1/vf = 997 kg/m , νf = -7

µfvf = 8.58 × 10

2

m /s, k = 0.613 W/m⋅K, Pr = 5.83.

ANALYSIS: (a) From an overall energy balance and the rate equation,

(

)

& c p Tm,o − Tm,i = hA s∆TLMTD q=m

(1)

where

∆TLMTD =

( 40 −15 ) − ( 40 − 37 ) ∆T1 − ∆T2 = = 10o C. ln ( ∆T1 / ∆T2 ) ln ( 2 5 / 3)

To estimate h, find the Reynolds number for the rectangular tube,

u D 0.326 m/s × 0.00256 m Re D = m h = = 973 ν 8.58 × 10−7 m2 / s where

Dh = 4 Ac / P = 4 ( 6.4 mm ×1.6 mm ) / 2 ( 6.4 +1.6 ) mm = 2.56 mm Ac = ( 6.4 mm × 1.6 mm ) = 1.024 ×10 −5m 2

(

)

& /Ac = 200 m l/60 s 10 −6m 3 / ml /1.024 ×10 −5 m 2 = 0.326 m/s. & ρ Ac = ∀ u m = m/ Hence the flow is laminar, but assuming fully developed flow with an isothermal surface from Table 8.1 with b/a = 6.4/1.6 = 4, hDh 4.4 × 0.613 W/m ⋅ K Nu D = = 4.4 h= = 1054 W/m2 ⋅ K.

k

0.00256 m

Continued …..

PROBLEM 8.82 (Cont.) From Eq. (1) with

As = PL = 2 ( 6.4 + 1.6 ) × 10−3m × L=1.6 ×10−2 LZ & = ρ Acu m = 997 kg/m 3 ×1.024 ×10 −5 m2 × 0.326 m/s = 3.328 ×10− 3 kg/s m the length of the rectangular tubing can be found as

3.328 × 10−3 kg/s × 4179 J/kg ⋅ K ( 37 − 15) K = 1054 W/m 2 ⋅ K ×1.6 × 10-2 Lm 2 × 10 K L = 1.8 m.

<

(b) Consider these comments with regard to whether the analysis under- or over-estimates the length, ⇒ fully-developed flow - L/Dh = 1.8 m/0.00256 = 700; not likely to have any effect, ⇒ negligible tube wall resistance - depends upon materials of construction; if plastic, analysis under predicts length, ⇒ negligible thermal contact resistance between tube and heating plate - if present, analysis under predicts length.

PROBLEM 8.83 KNOWN: Coolant flowing through a rectangular channel (gallery) within the body of a mold. FIND: Convection coefficient when the coolant is process water or ethylene glycol. SCHEMATIC:

ASSUMPTIONS: (1) Gallery can be approximated as a rectangular channel with a uniform surface temperature, (2) Fully developed flow conditions. PROPERTIES: Table A.6, Water ( Tm = (140 + 15)°C/2 = 350 K): ρ = 974 kg/m3, µ = 365 × 10-6 n⋅s/m2, ν = µ/ρ = 3.749 × 10-7 m2/s, k = 0.668 W/m⋅K, Pr = 2.29; Table A.5, Ethylene glycol ( Tm = 350 K): ρ = 1079 kg/m3, ν = 3.17 × 10-6 m2/s, k = 0.261 W/m⋅K, Pr = 34.6. ANALYSIS: The characteristic length of the channel, the hydraulic diameter, Eq. 8.67, is Dh = 4Ac P where Ac is the cross-sectional flow area and P is the wetted perimeter. For our channel,

Dh =

4 (a × b )

2 (a + b )

=

4 × 0.090 m × 0.0095 m = 0.0172 m 2 ( 0.090 + 0.0095 ) m

For the water coolant, from the continuity equation, find the Reynolds number to characterize the flow  V 1.3 × 10−3 m3 s um = = = 1.52 m s Ac 0.090 m × 0.0095 m

u D 1.52 m s × 0.0172 m ReDh = m h = = 69, 736 ν 3.749 × 10−7 m 2 s Since the flow is turbulent, and assuming fully developed conditions, use the Dittus-Boelter correlation, Eq. 8.60, to estimate the convection coefficient, hDh 0.8 0.4 Nu Dh = Pr 0.4 = 0.023 ( 69, 736 ) ( 2.29 ) = 240 = 0.023Re0.8 Dh k

hw =

0.668 W m ⋅ K × 240 = 9326 W m 2 ⋅ K 0.0172 m

<

Repeating the calculations using properties for the ethylene glycol coolant, find

ReDh = 8, 247

Nu Dh = 128

h eg = 1957 W m 2 ⋅ K

< Continued...

PROBLEM 8.83 (Cont.) COMMENTS: (1) The convection coefficient for the water coolant is more than 4 times greater than that with the ethylene glycol coolant. The corrosion protection afforded by the latter coolant greatly compromises the thermal performance of the gallery. In such situations, it is useful to explore a compromise between corrosion protection and thermal performance by using an aqueous solution of ethylene glycol (50%-50%, for example). (2) Recognize that for the ethylene glycol coolant calculation the Reynolds number is slightly below the lower limit of applicability of the Dittus-Boelter correlation.

PROBLEM 8.84 KNOWN: Dimensions, surface temperature and thermal conductivity of a cold plate. Velocity, inlet temperature, and properties of coolant. FIND: (a) Model for determining the heat rate q and outlet temperature, Tm,o, (b) Values of q and Tm,o for prescribed conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible KE, PE and flow work changes, (3) Constant properties, (4) Symmetry about the midplane (horizontal) of the cold plate and the midplane (vertical) of each cooling channel, (5) Negligible heat transfer at sidewalls of cold plate, (6) Onedimensional conduction from outer surface of cold plate to base surface of channel and within the channel side walls, which act as extended surfaces. PROPERTIES: Water (prescribed): ρ = 984 kg/m3, cp = 4184 J/kg⋅K, µ = 489 × 10-6 N⋅s/m2, k = 0.65 W/m⋅K, Pr = 3.15. ANALYSIS: (a) The outlet temperature, Tm,o, may be determined from the energy balance prescribed by Eq. 8.46b,

 1 = exp  −  m  1cp R tot Ts − Tm,i 

Ts − Tm,o

   

 1 = ρu m A c is the flowrate for a single channel and Rtot is the total resistance to heat transfer where m between the cold plate surface and the coolant for a particular channel. This resistance may be determined from the symmetrical section shown schematically, which represents one-half of the cell associated with a full channel. With the number of channels (and cells) corresponding to N = W/S, there are 2N = 2(W/S) symmetrical sections, and the total resistance Rtot of a cell is one-half that of a symmetrical section. Hence, Rtot = Rss/2, where the resistance of the symmetrical section includes the effect of conduction through the outer wall of the cold plate and convection from the inner surfaces. Hence,

R ss =

(H − h ) 2 + 1 k cp (SW ) ηo hA t

where At = Af + Ab = 2(h/2 × W) + (w × W), h is the average convection coefficient for the channel flow, and ηo is the overall surface efficiency.

A ηo = 1 − f (1 − ηf ) At Continued...

PROBLEM 8.84 (Cont.) The efficiency ηf corresponds to that of a straight, rectangular fin with an adiabatic tip, Eq. 3.89, and Lc = w/2. With D h = 4A c P = 4w 2 4w = w = 0.006 m , Re D = ρ u m D h µ = 984 kg/m3 × 2 m/s × 0.006 h

m/489 × 10-6 N⋅s/m2 = 24,150 and the channel flow is turbulent. Assuming fully-developed flow throughout the channel, the Dittus-Boelter correlation, Eq. 8.60, may therefore be used to evaluate h , where Nu D ≈ Nu D,fd = 0.023Re 4D/ 5 Pr 0.4 The total heat rate for the cold plate may be expressed as

(

 1cp Tm,o − Tm,i q = Nq1 = Nm

)

(b) For the prescribed conditions, 3  = ρ u A = 984 kg m ( 2 m s )( 0.006 m ) = 0.0708 kg s m 1 m c 2

Nu D = 0.023 ( 24,150 )

(3.15 )0.4 = 116.8 h = 116.8 k D h = 116.8 ( 0.65 W m ⋅ K ) ( 0.006 m ) = 12, 650 W m 2 ⋅ K A f = 2 ( h 2 × W ) = 2 ( 0.003 m × 0.1m ) = 6 × 10−4 m 2 A t = Af + A b = 6 × 10−4 m 2 + ( 0.006 m × 0.1m ) = 1.2 × 10−3 m 2 4/5

(

)1/ 2

1/ 2

=  h ( 2δ + 2W ) k cp (δ W ) = [12,650 W/m2⋅K(0.008 + 0.200)m/400 With m = hPf k cp Acf W/m⋅K(0.004 × 0.100)m2]1/2 = 128.2 m-1. tanh m ( h 2 ) tanh (128.2 × 0.003 ) 0.366 = = = 0.952 ηf = m (h 2 ) 128.2 × 0.003 0.385

ηo = 1 − 0.5 (1 − 0.952 ) = 0.976 R ss =

( 0.010 − 0.006 ) m 2 + 400 W m ⋅ K (0.01m × 0.1m ) 0.976

1

(12650 W m2 ⋅ K )1.2 ×10−3 m2

R ss = (0.005 + 0.0675 ) K W = 0.0725 K W With Rtot = Rss/2 = 0.0362 K/W, Ts − Tm,o   1 = exp  −  = 0.911 Ts − Tm,i  0.0708 kg s × 4184 J kg ⋅ K × 0.0362 K W 

(

)

Tm,o = Ts − 0.911 Ts − Tm,i = 360 K − 0.911(360 − 300 ) K = 305.3 K The total heat rate is

(

)

 c q = Nm 1 p Tm,o − Tm,i = 10 × 0.0708 kg s × 4184 J kg ⋅ K (305.3 − 300 ) K = 15, 700 W

< <

COMMENTS: The prescribed properties correspond to a value of Tm which significantly exceeds that obtained from the foregoing solution ( Tm = 302.6 K). Hence, the calculations should be repeated using more appropriate thermophysical properties (see next problem). From Eq. 3.85, the effectiveness of the extended surface is −1

ε = R t,b R t,f = ( hδ W )

( hA f ηf )−1 = ( A f ηf

(

δ W ) = 6 × 10

−4

2

m × 0.954

) (0.004 m × 0.10 m ) = 1.43.

Hence, the ribs are only marginally effective in enhancing heat transfer to the coolant.

PROBLEM 8.85 KNOWN: Geometry, surface temperature and thermal conductivity of a cold plate. Velocity and inlet temperature of coolant. FIND: Effect of channel width on total heat rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible KE, PE and flow work changes, (3) Constant properties, (4) Symmetry about midplane (horizontal) of the cold plate and the midplane (vertical) of each channel, (5) Negligible heat transfer at sidewalls of cold plate, (6) One-dimensional conduction from outer surface of cold plate to base surface of channel and within the channel side walls, which act as extended surfaces. PROPERTIES: Water: Evaluated at Tm using the Properties Toolpad of IHT. ANALYSIS: The model developed for the preceding problem was entered into the workspace of IHT, with the Dittus-Boelter equation and exponential relation accessed from the Correlations Toolpad and modified to account for the hydraulic diameter and the total resistance to heat transfer. Calculations were performed for Case 1: Case 2: Case 3: Case 4: Case 5:

w = 96 mm, N = 1, S = W = 100 mm w = 46 mm, N = 2, S = 50 mm w = 21 mm, N = 4, S = 25 mm w = 6 mm, N = 10, S = 10 mm w = 1 mm, N = 20, S = 5 mm

and the results are tabulated as follows.

(

Case

N

Dh (m)

ReD

h W m ⋅K

1 2 3 4 5

1 2 4 10 20

0.01129 0.01062 0.00933 0.00600 0.00171

24,820 25,320 22,340 14,620 4761

8269 8895 9142 10,070 13,740

2

)

Tm,o (K)

q (W)

300.7 302.3 302.6 304.3 317.2

3164 10,370 10,960 12,940 17,160

It is clearly beneficial to increase the number of channels, with the total heat rate increasing by approximately a factor of 5 as N increases from 1 to 20. The heat rate may be increased further by increasing um, and hence the flowrate per channel, although an upper limit would be associated with the pressure drop, which would increase with decreasing Dh. Could additional heat transfer enhancement be achieved by altering the thickness δ of the channel walls? COMMENTS: Note that results obtained for Case 4 differ from those of the preceding problem due to different fluid properties. In this case the properties were evaluated at the actual value of Tm = 302.6 K, rather than at an assumed (significantly larger) value.

PROBLEM 8.86 KNOWN: Heat sink with 24 passages for air flow removes power dissipation from circuit board. FIND: Operating temperature of the board and pressure drop across the sink. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes, (3) Negligible thermal resistance between the circuit boards and air passages, (4) Sink surface and board are isothermal at Ts . 3

PROPERTIES: Table A-4, Air ( T ≈ 310 K,1 atm): ρ = 1.1281 kg/m , cp = 1008 J/kg⋅K, ν = -6 2 16.89 × 10 m /s, k = 0.0270 W/m⋅K, Pr = 0.706. ANALYSIS: The air outlet temperature follows from Eq. 8.43,  PLh  = exp  − .  m & Ts − Tm,i c p  

Ts − Tm,o

The mass flow rate for the entire sink is & = 1.1281 kg/m 3 × 0.060 m 3 / s = 6.77 ×10−2 kg/s & = ρ∀ m and the Reynolds number for a rectangular passage is u D Re D = m h ν where Dh = 4Ac / P = 4 ( 6 mm × 25 mm ) / 2 ( 6 + 25 ) mm = 9.68 mm um =

& m/24 6.77 ×10 −2 kg/s/24 = = 16.7 m/s ρ Ac 1.1281 kg/m3 (6 × 25) ×10− 6 m 2

giving Re D =

16.7 m/s × 9.68 × 10−3 m 16.89 × 10−6 m 2 / s

= 9571.

Assume the flow is turbulent and fully developed and using the Dittus-Boelter correlation find NuD = 0.023Re 4/5 Pr 0.4 = 0.023 ( 9571) h=

4/5

( 0.706) 0.4 = 30.6

Nu ⋅ k 30.6 × 0.027 W/m ⋅ K = = 85.4 W/m 2 ⋅ K. Dh 0.00968 m Continued …..

PROBLEM 8.86 (Cont.) From an overall energy balance on the sink,

(

& c p Tm,o − Tm,i q=m

)

& cp T m,o =Tm,i + q/m

Tm,o = 27o C + 50 W/6.77 ×10 −2 kg/s × 1008 J/kg ⋅ K = 27.73oC Hence, the operating temperature of the circuit board for these conditions is   −3 2 Ts − 27.73  2 ( 6 + 25 ) ×10 m × 0.150m ×85.4 W/m ⋅ K  = exp  −  Ts −27 6.77 × 10−2 kg/s/24 × 1008 J/kg ⋅ K  

)

(

Ts = 30o C.

<

The pressure drop in the rectangular passage for the smooth surface condition follows from Eqs. 8.22 and 8.20 ∆p = f

ρ u 2m L Dh

where −1/4

f = 0.31Re1/4 D = 0.316 ( 9554 ) ∆p = 0.0320

= 0.0320.

1.1281 kg/m 3 (16.7 m/s ) 2 0.00968 m

× 0.150 m = 156 N/m 2 .

<

COMMENTS: (1) The analysis has been simplified by assuming the channel is rectangular and all four sides experience heat transfer. Since the insulated surface is a small portion of the total passage surface area, the effect can’t be very large. & ∆p = 0.060 m3/s × 156 (2) The power required to move the air through the heat sink is Pelec = ∀ 2 N/m = 9.4 W.

PROBLEM 8.87 KNOWN: Channel formed between metallic blades dissipating heat by internal volumetric generation. FIND: (a) The heat removal rate per blade for the prescribed thermal conditions and (b) Time required to slow a train of mass 106 kg from 120 km/h to 50 km/h. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions for channel blades and air flow, (2) The blades form a channel of rectangular (a × b) cross-section and length L, (3) Negligible kinetic energy changes and axial conduction inside the channel, and (4) Fully developed flow conditions in the channel. PROPERTIES: Table A.4, Air ( Tm ≈ 350 K, 1 atm): ρ = 0.995 kg/m3, cp = 1009 J/kg ⋅K, ν = 20.92 × 10-6 m2/s, k = 0.030 W/m⋅K, Pr = 0.700. ANALYSIS: (a) The heat removal rate by the air from a single channel (one blade) follows from an overall energy balance,

(

 p Tm,o − Tm,i q = mc

)

(1)

where the outlet temperature can be determined from Eq. 8.42b,

 PL = exp  −  mc  p Ts − Tm,i 

Ts − Tm,o

 h  

(2)

The hydraulic diameter, Dh, flows from Eq. 8.67, 2 4Ac 4 ( a × b ) 4 ( 0.220 × 0.004 ) m = = = 0.0079 m Dh = P 2 ( a + b ) 2 ( 0.220 + 0.004 ) m

(3)

Assuming Tm = 350K , the Reynolds number is

u D 50 m s × 0.0079m ReDh = m h = = 18, 779 ν 20.92 × 10−6 m 2 s

(4)

Using the Dittus-Boelter correlation, Eq. 8.60,

Nu Dh =

hDh 0.8 0.4 Pr0.4 = 0.023 (18, 779 ) (0.700 ) = 52.37 = 0.023 Re0.8 Dh k

(5) Continued...

PROBLEM 8.87 (Cont.) h=

0.030 W m ⋅ K × 52.37 = 199 W m 2 ⋅ K 0.0079m

Hence, the outlet temperature is

 2 (0.220 + 0.004 ) m × 0.070m  199 W m 2 ⋅ K  = exp   0.0438 kg s × 1009 J kg ⋅ K  (600 − 25)$ C 600 − Tm,o

Tm,o = 100.7$ C where

 = ρAcum = 0.995 kg/m3 × (0.220 × 0.004) m2 × 50 m/s = 0.0438 kg/s m and the rate of heat removal per blade, from Eq. (1), is

q = 0.0438 kg s ×1009 J kg ⋅ K (100.7 − 25 ) C = 3.346 kW $

<

(b) From an energy balance on the locomotive over an interval of time, ∆t, the heat energy transferred to the air stream results in a change in kinetic energy of the train,

− Eout = ∆E = KEf − KEi − (q × N ) × ∆t =

(

1 M Vf2 − Vi2 2

(6)

)

1 −3346 W blade × 2000blades × ∆t (s ) = × 10−6 kg 2

 50, 000 2  120, 000 2  2 2   −  m s  3600   3600  

∆t = 69s

<

COMMENTS: (1) For the channel, L/Dh = 0.070 m/0.0079 m = 8.9 < 10 so that the assumption of fully developed conditions may not be satisfied. Recognize that the flow at the channel entrance may be highly turbulent because of the upstream fan swirl and ducting. (2) What benefits could be realized by increasing the heat transfer coefficient? Aside from increasing velocity, what design changes would you make to increase h? (3) Our assumption for Tm = 350 K at which to evaluate properties is reasonable considering Tm = (100.7 + 25)°C/2 = 335 K.

PROBLEM 8.88 KNOWN: Chip and cooling channel dimensions. Channel flowrate and inlet temperature. Chip temperature. FIND: Water outlet temperature and chip power. SCHEMATIC:

ASSUMPTIONS: (1) Negligible kinetic and potential energy changes for water, (2) Uniform channel surface temperature, (3) Tm = 300 K, (4) Fully developed flow. -6

PROPERTIES: Table A-6, Water ( Tm = 300 K): cp = 4179 J/kg⋅K, µ = 855 × 10 0.613 W/m⋅K, Pr = 583.

kg/s⋅m, k =

ANALYSIS: Using the hydraulic diameter, find the Reynolds number, 4 ( H × W ) 2 ( 50 × 200 ) µ m 2 −6 Dh = = 10 m/ µ m = 8 ×10−5 m

2 (H + W )

250 µ m

(

)

10−4 kg/s 8×10−5 m & 1D h ρ u mD h m Re D = = = = 936. µ A cµ ( 50 × 200 )10−12 m2 855 ×10 −6 kg/s ⋅ m

(

)

Hence, the flow is laminar and, from Table 8.1, NuD = 4.44, so that

h = Nu D

4.44 ( 0.613 W/m ⋅ K ) k = = 34,022 W/m 2 ⋅ K. − 5 Dh 8 × 10 m -6

With P = 2(H + W) = 2(250 µm) 10

Ts − Tm,o Ts − Tm,i

=

350K − Tm,o 60 K

-4

m/µm = 5 ×10

 PL = exp  −  m  & 1 cp

m, Eq. 8.42b yields

 5 × 10−6 m2 × 34,022 W/m 2 ⋅ K    h  = exp  −  −4 kg/s × 4179 J/kg ⋅ K   10   

Tm,o = 350K − 60 K exp ( −0.407 ) = 310 K.

<

Hence, from Eq. 8.37,

(

)

(

)

& c p Tm,o − Tm,i = Nm & 1c p Tm,o − Tm,i = 50 × 10−4 kg/s ( 4179 J/kg ⋅ K )( 20 K ) = 418 W. q=m

<

2

COMMENTS: (1) The chip heat flux of 418 W/cm is extremely large and the method provides a very efficient means of heat removal from high power chips. However, clogging of the microchannels is a potential problem which could seriously compromise reliability. (2) L/Dh = 125 and 0.05 ReDPr = 272. Hence, fully developed conditions are not realized and h > 34,022. The actual power dissipation is therefore greater than 418 W.

PROBLEM 8.89 KNOWN: Chip and cooling channel dimensions. Channel flow rate and inlet temperature. Temperature of chip at base of channel. FIND: (a) Water outlet temperature and chip power, (b) Effect of channel width and pitch on power dissipation. SCHEMATIC:

ASSUMPTIONS: (1) Negligible flow work and kinetic and potential energy changes for water, (2) Flow may be approximated as fully developed and channel walls as isothermal for purposes of estimating the convection coefficient, (3) One-dimensional conduction along channel side walls, (4) Adiabatic condition at end of side walls, (5) Heat dissipation is exclusively through fluid flow in channels, (6) Constant properties. -6

PROPERTIES: Table A-6, Water ( Tm = 300K): cp = 4179 J/kg⋅K, µ = 855 × 10 kg/s⋅m, k = 0.613 W/m⋅K, Pr = 5.83. ANALYSIS: (a) The channel sidewalls act as fins, and a unit channel/sidewall combination is shown in schematic (a), where the total number of unit cells corresponds to N = L/S. With N = 50 and L = 10 mm, S = 200 µm and δ = S – W = 150 µm. Alternatively, the unit cell may be represented in terms of a single fin of thickness δ, as shown in schematic (b). The thermal resistance of the unit cell may -1 be obtained from the expression for a fin array, Eq. (3.103), Rt,o = (ηohAt) , where At = Af + Ab = L -4 -4 -6 2 (2 H + W) = 0.01m (4 × 10 + 0.5 × 10 ) m = 4.5 × 10 m . With Dh = 4 (H × W)/2 (H + W) = 4 (2 -4 -4 -4 -5  × 10 m × 0.5 × 10 m)/2 (2.5 × 10 m) = 8 × 10 m, the Reynolds number is ReD = ρum Dh/µ = m 1 -4

-5

-4

-4

-6

Dh/Acµ = 10 kg/s × 8 × 10 m/(2 × 10 m × 0.5 × 10 m) 855 × 10 kg/s⋅m = 936. Hence, the flow is laminar, and assuming fully developed conditions throughout a channel with uniform surface temperature, Table 8.1 yields NuD = 4.44. Hence,

h=

k 0.613 W / m ⋅ K × 4.44 Nu D = = 34, 022 W / m 2 ⋅ K 5 − Dh 8 × 10 m

With m = (2h/kchδ) the fin efficiency is

ηf =

1/2

2

-4

= (68,044 W/m ⋅K/140 W/m⋅K × 1.5 × 10 m)

1/2

-1

= 1800 m and mH = 0.36,

tanh mH 0.345 = = 0.958 mH 0.36

and the overall surface efficiency is

A 4.0 × 10−6 ηo = l − f (l − ηf ) = l − (l − 0.958) = 0.963 At 4.5 × 10−6 The thermal resistance of the unit cell is then Continued …..

PROBLEM 8.89 (Cont.) −1

R t,o = (ηo h A t )

(

= 0.963 × 34, 022 W / m 2 ⋅ K × 4.5 × 10−6 m 2

)

−1

= 6.78 K / W

The outlet temperature follows from Eq. (8.46b),

(

)



l



Tm,o = Ts − Ts − Tm,i exp  −  = 350K − ( 60K ) ×  m  cp R t,o      1  = 307.8K exp  −  10−4 kg / s × 4179 J / kg ⋅ K × 6.78K / W   

<

The heat rate per channel is then

(

)

 1 cp Tm,o − Tm,i = 10−4 kg / s × 4179 J / kg ⋅ K (17.8K ) = 7.46 W q1 = m and the chip power dissipation is

q = Nq l = 50 × 7.46 W = 373 W

<

(b) The foregoing result indicates significant heat transfer from the channel side walls due to the large value of ηf. If the pitch is reduced by a factor of 2 (S = 100 µm), we obtain

S = 100 µ m, W = 50 µ m, δ = 50 µ m, N = 100 : q1 = 7.04 W, q = 704 W

<

Hence, although there is a reduction in ηf due to the reduction in δ (ηf = 0.89) and therefore a slight reduction in the value of ql, the effect is more than compensated by the increase in the number of channels. Additional benefit may be derived by further reducing the pitch to whatever minimum value of δ is imposed by manufacturing or structural limitations. There would also be an advantage to increasing the channel hydraulic diameter and or flowrate, such that turbulent flow is achieved with a correspondingly larger value of h. COMMENTS: (1) Because electronic devices fail by contact with a polar fluid such as water, great care would have to be taken to hermetically seal the devices from the coolant channels. In lieu of water, a dielectric fluid could be used, thereby permitting contact between the fluid and the electronics. However, all such fluids, such as air, are less effective as coolants. (2) With L/Dh = 125 and L/Dh)fd ≈ 0.05 ReD Pr = 273, fully developed flow is not achieved and the value of h = hfd underestimates the actual value of h in the channel. The coefficient is also underestimated by using a Nusselt number that presumes heat transfer from all four (rather than three) surfaces of a channel.

PROBLEM 8.90 KNOWN: Chip and cooling channel dimensions. Channel flow rate and inlet temperature. Temperature of chip at base of channel. FIND: (a) Outlet temperature and chip power dissipation for dielectric liquid, (b) Outlet temperature and chip power dissipation for air. SCHEMATIC:

ASSUMPTIONS: (1) Negligible flow work and kinetic and potential energy changes, (2) Flow may be approximated as fully developed and channel walls as isothermal for purposes of estimating the convection coefficient, (3) One-dimensional conduction along the channel side walls, (4) Adiabatic condition at end of side walls, (5) Heat dissipation is exclusively through fluid flow in channels, (6) Constant properties. PROPERTIES: Prescribed. Dielectric liquid: cp = 1050 J/kg⋅K, k = 0.065 W/m⋅K, µ = 0.0012 2 -7 2 N⋅s/m , Pr = 15. Air: cp = 1007 J/kg⋅K, k = 0.0263 W/m⋅K, µ = 185 × 10 N⋅s/m , Pr = 0.707. ANALYSIS: (a) The channel side walls act as fins, and a unit channel/sidewall combination is shown in schematic (a), where δ = S – W = 150 µm. Alternatively, the unit cell may be represented in terms of a single fin of thickness δ, as shown in schematic (b). The thermal resistance of the unit cell -1 may be obtained from the expression for a fin array, Eq. (3.103), Rt,o = (ηo h At) , where At = Af + -6 2 -8 2 Ab = L (2 H + W) = 4.5 × 10 m . With Ac = H × W = 10 m and Dh = 4 Ac/2(H + W) = 8 × 10 5  m, the Reynolds number is ReD = ρumDh/µ = m 1 Dh/Acµ = 667. Hence, the flow is laminar, and assuming fully developed conditions throughout a channel with uniform surface temperature, Table k 0.065 W / m ⋅ K × 4.44 Nu D = 8.1 yields NuD = 4.44. Hence, h = = 3608 W / m 2 ⋅ K 5 − Dh 8 × 10 m With m = (2 h/kch δ)

ηf =

1/2

-1

= 586 m and mH = 0.117, the fin efficiency is

tanh mH 0.1167 = = 0.995 mH 0.117

and the overall surface efficiency is

A 4.0 × 10−6 ηo = l − f (1 − ηf ) = l − (l − 0.995 ) = 0.996. At 4.5 × 10−6 The thermal resistance of the unit cell is then −1

R t,o = (ηo h A t )

(

= 0.996 × 3608 W / m 2 ⋅ K × 4.5 × 10−6 m 2

)

−1

= 61.9 K / W

The outlet temperature follows from Eq. (8.46b),

 l Tm,o = Ts − Ts − Tm,i exp  −  m  1 cp R t,o 

(

)

  = 350K   Continued …..

PROBLEM 8.90 (Cont.)   1 − (60K ) exp  −  = 298.6K  10−4 kg / s × 1050 J / kg ⋅ K × 61.9 K / W   

<

The heat rate per channel is then

(

)

 1 cp Tm,o − Tm,i = 10−4 kg / s × 1050 J / kg ⋅ K × 8.6 K = 0.899 W q1 = m and the chip power dissipation is

<

q = Nq1 = 50 × 0.899 W = 45.0 W

(b) With m 1 = 10 −6 kg / s, Re D = m l D h / A c µ = 432 and the flow is laminar. Hence, with NuD = 4.44,

h=

k 0.0263 W / m ⋅ K × 4.44 Nu D = = 1460 W / m 2 ⋅ K 5 − Dh 8 × 10 m

With m = (2 h/kchδ)

ηf =

1/2

-1

= 373 m and mH = 0.0746, the fin efficiency is

tanh mH 0.0744 = = 0.998 mH 0.0746

and the overall surface efficiency is

A 4.0 × 10−6 ηo = l − f (1 − ηf ) = l − (l − 0.998 ) = 0.998 At 4.5 × 10−6 Hence,

−1

R t,o = (ηo h A t )

(

= 0.998 × 1460 W / m 2 ⋅ K × 4.5 × 10−6 m 2

)

−1

= 153K / W

The outlet temperature is then

 l Tm,o = Ts − Ts − Tm,i exp  −  m  1 cp R t,o 

(

)

  = 350K  

  1  = 349.9 K − (60K ) exp  −  10−6 kg / s × 1007 J / kg ⋅ K × 153K / W   

(

<

)

 1 cp Tm,o − Tm,i = 10−6 kg / s × 1007 J / kg ⋅ K × 59.9 K = 0.060 W q1 = m q = Nq1 = 3.02 W

<

COMMENTS: (1) For laminar flow in the channels, there is a clear advantage to using the dielectric liquid instead of air. (2) The prescribed channel geometry is by no means optimized, and the number of fins should be increased by reducing S. Also, channel dimensions and/or flow rates could be increased to achieve turbulent flow and hence much larger values of h. (3) With L/Dh = 125 and L/Dh)fd ≈ 0.05 ReD Pr = 500 for the dielectric liquid, fully developed flow is not achieved and its assumption yields a conservative (under) estimate of the convection coefficient. The coefficient is also underestimated by using a Nusselt number that presumes heat transfer from all four (rather than three) surfaces of a channel.

PROBLEM 8.91 KNOWN: Arrangement of chips and cooling channels for a substrate. Contact and conduction resistances. Coolant velocity and inlet temperature. FIND: (a) Coolant temperature rise, (b) Chip and substrate temperatures. SCHEMATIC:

ASSUMPTIONS: (1) Constant properties, (2) Fully-developed flow, (3) Negligible kinetic and potential energy changes, (4) Heat transfer exclusively to water, (5) Steady-state conditions. 3

PROPERTIES: Water (given): ρ = 1000 kg/m , cp = 4180 J/kg⋅K, k = 0.610 W/m⋅K, Pr = 5.8, µ = -6

855 × 10

kg/s⋅m.

ANALYSIS: (a) For a single flow channel, the overall energy balance yields Tm,o − Tm,i =

q & cp m

=

NL P ρ u mA cc p

10 × 5 W

= 1000 kg/m

From the thermal circuit, q=

To − Tm R t,c + R cond + R conv

3

o

(1 m/s )( 0.005 m )

2

4180 J/kg ⋅ K

(

= 0.48 C.

<

)

R t,c = R ′′t,c / A s = 0.5 ×10−4 m2 ⋅ K/W /10 ( 0.005 m )2 = 0.2 K/W. 2

With Dh = 4Ac /P = 4(0.005 m) /4(0.005 m) = 0.005 m, ρ u m Dh 1000 kg/m (1 m/s ) 0.005 m = = 5848. −6 µ 855 ×10 kg/s ⋅ m 3

ReD =

With turbulent flow, the Dittus-Boelter correlation yields k 0.4  0.61 W/m ⋅ K  4/5 h = 0.023Re 4/5 = (5.8 )0.4 = 5849 W/m 2 ⋅ K D Pr  0.023 ( 5848 ) D  0.005 m  R conv = ( hAs )

−1

(

= 5849 W/m2 ⋅ K × 4 × 0.005 m × 0.2 m

)

−1

= 0.043 K/W.

Approximating Tm as (Tm,i + Tm,o)/2 = 25.24°C,

(

)

Tc = Tm + q R t,c + R cond + R conv = 25.24o C + 50 W ( 0.2 + 0.12 + 0.043 ) K/W = 43.3oC.

<

Similarly, from the thermal circuits, Ts = Tm + q × R conv = 25.24 o C + 50W × 0.043K/W = 27.4 oC

<

COMMENTS: (1) Since the coolant temperature rise is less than 0.5°C, all chip temperatures will be within 0.5°C of each other. (2) The channel surface temperature may also be obtained from Eq. 8.42b, yielding the same result.

PROBLEM 8.92 KNOWN: Power dissipation of components on each side of a hollow core PCB. Dimensions of PCB. Inlet temperature and flow rate of air. FIND: Outlet air temperature and inlet and outlet surface temperatures for prescribed flow rates. SCHEMATIC:

ASSUMPTIONS: (1) Steady flow, (2) Negligible flow work and potential and kinetic energy changes, (3) Channel may be approximated as infinite parallel plates, (4) Uniform surface heat flux, (5) Fully developed flow at exit, (6) Constant properties. 3

PROPERTIES: Table A-4, Air ( Tm ≈ 310K): ρ = 1.128 kg/m , cp = 1007 J/kg⋅K, µ = 189.3 × 10 2 N⋅s/m , k = 0.0270 W/m⋅K, Pr = 0.706.

-7

ANALYSIS: Performing an energy balance for a control surface about the hollow core, 2q = m c p

(Tm,o − Tm,i ) , in which case Tm,o =

2q 80 W + Tm,i = + 20°C = 59.7°C  cp m 0.002 kg / s × 1007 J / kg ⋅ K

<

The surface temperatures may be obtained from Newton’s law of cooling, q s′′ = h ( Ts − Tm ). Hence, with h → ∞ at the entrance, where the thermal boundary layer thickness is zero,

<

Ts,i = Tm,i = 20°C

With ReD = ρ umDh/µ = m Dh/Ac µ, where Dh = 2H = 0.008m and Ac = H × W = 0.004m × 0.3m = 2 2 -7 2 0.0012m , ReD = (0.002 kg/s × 0.008m)/(0.0012m × 189.3 × 10 N⋅s/m ) = 704 and the flow is 2 2 laminar. With a uniform surface heat flux, q′′s = q/(W × L) = 40 W/(0.3m) = 444 W/m , Table 8.3 yields NuD = 8.23. Hence,

h=

Nu D k Dh

=

8.23 × 0.027 W / m ⋅ K = 27.8 W / m 2 ⋅ K 0.008m

q′′s 444 W / m 2 Ts,o = Tm,o + = 59.7°C + = 75.7°C h 27.8 W / m 2 ⋅ K

<

If the flowrate is increased by a factor of 5,

Tm,o =

2q 80 W + Tm,i = + 20°C = 27.9°C  cp m 0.01kg / s × 1007 J / kg ⋅ K

<

The surface temperature at the inlet is unchanged, Continued …..

PROBLEM 8.92 (Cont.) Ts,i = 20°C

<

but with ReD = 3520, flow in the channel is now turbulent. Using Eq. (8.60) as a first approximation,

 k  4/5 0.4 4 / 5 0.4  0.027 W / m ⋅ K  2  0.023 (3520 ) ( 0.706 ) = 46.4 W / m ⋅ K  0.023 Re D Pr =  0.008m    Dh 

h =

q′′ 444 W / m 2 Ts,o = Tm,o + s = 27.9°C + = 37.5°C h 46.4 W / m 2 ⋅ K

<

COMMENTS: (1) With L/Dh = 37.5 and L/Dh)fd ≈ 0.05 ReD Pr = 25 for the laminar flow, it is reasonable to assume fully developed conditions at the exit. The same may be said for the turbulent flow condition. (2) The temperature difference, Ts – Tm, increases from approximately 0 at the entrance to a maximum value associated with fully developed conditions.

PROBLEM 8.93 KNOWN: Printed-circuit board (PCB) with uniform temperature Ts cooled by laminar, fully developed flow in a parallel-plate channel. The air flow with an inlet temperature of Tm,i is driven by a pressure difference, ∆p.

(

)

FIND: The average heat removal rate per unit area, qs′′ W / m 2 , from the PCB. SCHEMATIC:

ASSUMPTIONS: (1) Laminar, fully developed flow, (2) Upper and lower walls of the channel are insulated and of infinite extent in the transverse direction, (3) PCB has uniform surface temperature, (4) Constant properties, (5) Negligible kinetic and potential energy changes and flow work. 3

PROPERTIES: Table A-4, Air (Tm = 293 K, 1 atm): ρ = 1.192 kg/m , cp = 1007 J/kg⋅K, ν = 1.531 -5 2 × 10 m /s, k = 0.0258 W/m⋅K, Pr = 0.709. ANALYSIS: The energy equations for determining the heat rate from one surface of the board are Eqs. 8.37 and 8.42b

(

)

 cp Tm,o − Tm,i = qs′′ As q=m

(1)

 PL h  = exp  −   m  cp  Ts − Tm,i  

(2)

Ts − Tm,o

where As = Lw and P = 2(w + a) where w is the width in the transverse direction. For the fully developed flow condition, the velocity is estimated from the friction pressure drop relation, Eq. 8.22a,

(

)

∆p = f ρ u 2m / 2 ( L / Dh )

(3)

where the hydraulic diameter for the channel cross section is

Dh =

4 (w a ) 4 Ac = = 2a P 2 (w + a )

a T∞ (hot surface facing upward) 7 and RaL > 10 , (3) Constant properties. -6

2

PROPERTIES: Table A-4, air (p = 1 atm, Tf ≈ 310K): ν = 16.9 × 10 m /s, k = 0.0270 W/m⋅K, Pr -6 2 -1 = 0.706, α = ν/Pr = 23.9 × 10 m /s, β = 0.00323 K . ANALYSIS: From an energy balance for the outer surface, Ts,o − Ts,i αSG S − q ′′conv − E = q ′′cond = R ′′tot

(

)

4

αSG S − h Ts,o − T∞ − εσ Ts,o =

Ts,o − Ts,i 2R ′′p + R ′′i

where R ′′p = ( t1 / k p ) = 2.78 × 10 −5 m 2 ⋅ K / W and R i′′ = ( t 2 / k i ) = 1.923 m 2 ⋅ K / W. For a hot surface facing upward and Ra L = gβ ( Ts,o − T∞ ) L3 / αν > 107 , h is obtained from Eq. 9.31. Hence, with cancellation of L, h=

k L

 9.8 m / s 2 × 0.00323 K −1 1/ 3 0.15 Ra L = 0.15 × 0.0270 W / m ⋅ K   −12 4 2  16.9 × 23.9 × 10

2

= 1.73 W / m ⋅ K

4/3

1/ 3

   m /s 

(Ts,o − T∞ )1/ 3

(Ts,o − 305 K )4 / 3

Hence,

(

2

0.5 750 W / m ⋅ K

)

2

− 1.73 W / m ⋅ K

Solving, we obtain

4/3

( Ts, o − 305 )4 / 3 − 0.5 × 5.67 × 10 −8 W / m 2 ⋅ K 4 Ts,4o Ts,o = 318.3K = 45.3°C

Hence, the heat load is q = ( W ⋅ L t ) q′′cond = (3.5m ×10m )

=

(

Ts, o − 263K 5.56 × 10

−5

)

2

+ 1.923 m ⋅ K / W

< ( 45.3 + 10 ) °C 1.923m 2 ⋅ K / W

<

= 1007 W

COMMENTS: (1) The thermal resistance of the aluminum panels is negligible compared to that of the insulation. (2) The value of the convection coefficient is h = 1.73 ( Ts,o − T∞ )

1/ 3

2

= 4.10 W / m ⋅ K.

PROBLEM 9.43 KNOWN: Inner surface temperature and composition of a furnace roof. Emissivity of outer surface and temperature of surroundings. FIND: (a) Heat loss through roof with no insulation, (b) Heat loss with insulation and inner surface temperature of insulation, and (c) Thickness of fire clay brick which would reduce the insulation temperature, Tins,i, to 1350 K. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through the composite wall, (3) Negligible contact resistance, (4) Constant properties. PROPERTIES: Table A-4, Air (Tf ≈ 400 K, 1 atm): k = 0.0338 W/m⋅K, ν = 26.4 × 10-6 m2/s, α = 38.3 × 10-6 m2/s, Pr = 0.69, β = (400 K)-1 = 0.0025 K-1; Table A-1, Steel 1010 (600 K): k = 48.8 W/m⋅K; Table A-3 Alumina-Silica blanket (64 kg/m3, 750 K): k = 0.125 W/m⋅K; Table A-3, Fire clay brick (1478 K): k = 1.8 W/m⋅K. ANALYSIS: (a) Without the insulation, the thermal circuit is

Performing an energy balance at the outer surface, it follows that

Ts,i − Ts,o

qcond = qconv + q rad

L1 k1A + L3 k 3A

(

)

)

(

4 − T 4 (1,2) = hA Ts,o − T∞ + εσ A Ts,o sur

where the radiation term is evaluated from Eq. 1.7. The characteristic length associated with free convection from the roof is, from Eq. 9.29 L = As P = 16m 2 16 m = 1m . From Eq. 9.25, with an assumed value for the film temperature, Tf = 400 K, Ra L =

(

)

gβ Ts,o − T∞ L3

να

=

(

)(

9.8 m s 2 0.0025 K −1 Ts,o − T∞ 26.4 × 10

−6

m

2

s × 38.3 × 10

−6

) (l m )3 2

m s

Hence, from Eq. 9.31 1/ 3 k 1/ 3 3 0.0338 W m ⋅ K h = 0.15Ra1/ 0.15 2.42 × 107 Ts,o − T∞ = 1.47 Ts,o − T∞ L = L 1m

(

)

(

)

(

= 2.42 × 107 Ts,o − T∞

%

1/ 3

)

W m 2 ⋅ K .(3) Continued...

PROBLEM 9.43 (Cont.) The energy balance can now be written

(0.08m 1.8 W

(1700 − Ts,o ) K

m ⋅ K + 0.005m 48.8 W m ⋅ K )

(

= 1.47 Ts,o − 298 K

)

4/3

4 4 4 +0.3 × 5.67 × 10−8 W m 2 ⋅ K  Ts,o − ( 298 K )   

and from iteration, find Ts,o ≈ 895 K. Hence,

{

q = 16m 2 1.47 (895 − 298 )

4/3

}

4 4 W m 2 + 0.3 × 5.67 × 10−8 W m 2 ⋅ K 4 (895 K ) − ( 298 K ) 





q = 16m 2 {7, 389 + 10, 780} W m 2 = 2.91 × 105 W . (b) With the insulation, an additional conduction resistance is provided and the energy balance at the outer surface becomes Ts,i − Ts,o 4 4 (4) = hA Ts,o − T∞ + εσ A Ts,o − Tsur L1 k1A + L 2 k 2 A + L3 k 3A

<

(

(1700 − Ts,o ) K (0.08m 1.8 + 0.02 0.125 + 0.005 48.8 ) m

2

⋅K W

)

(

)

(

= 1.47 Ts,o − 298K

)4 / 3

4 4 +0.3 × 5.67 × 10−8 W m 2 ⋅ K 4 Ts,o − ( 298 K )  .

From an iterative solution, it follows that Ts,o ≈ 610 K. Hence,

{

q = 16m 2 1.47 ( 610 − 298 )

4/3





}

4 4 W m 2 + 0.3 × 5.67 × 10−8 W m 2 ⋅ K 4 ( 610 K ) − ( 298 K ) 





q = 16m 2 {3111 + 2221} W m 2 = 8.53 × 104 W . The insulation inner surface temperature is given by Ts,i − Tins,i q= . L1 k1A Hence L 0.08 m Tins,i = −q 1 + Ts,i = −8.53 × 104 W + 1700 K = 1463 K . k1A 1.8 W m ⋅ K × 16m 2 (c) To determine the required thickness L1 of the fire clay brick to reduce Tins,i = 1350 K, we keyboarded Eq. (4) into the IHT Workspace and found

<

<

L1 = 0.13 m.

<

COMMENTS: (1) The accuracy of the calculations could be improved by re-evaluating thermophysical properties at more appropriate temperatures. (2) Convection and radiation heat losses from the roof are comparable. The relative contribution of radiation increases with increasing Ts,o, and hence decreases with the addition of insulation. (3) Note that with the insulation, Tins,i = 1463 K exceeds the melting point of aluminum (933 K). Hence, molten aluminum, which can seep through the refractory, would penetrate, and thereby degrade the insulation, under the specified conditions.

PROBLEM 9.44 KNOWN: Dimensions and emissivity of top surface of amplifier. Temperature of ambient air and large surroundings. FIND: Effect of surface temperature on convection, radiation and total heat transfer from the surface. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Quiescent air. PROPERTIES: Table A.4, air (Obtained from Properties Tool Pad of IHT). ANALYSIS: The total heat rate from the surface is q = qconv + qrad. Hence,

(

4 q = hAs ( Ts − T∞ ) + εσ As Ts4 − Tsur

)

where As = L2 = 0.25 m2. Using the Correlations and Properties Tool Pads of IHT to evaluate the average convection coefficient for the upper surface of a heated, horizontal plate, the following results are obtained. 160 140

Heat rate (W)

120 100 80 60 40 20 50

55

60

65

70

75

Surface temperature, Ts(C) Total heat rate, q Convection heat rate, qconv Radiation heat rate, qrad

Over the prescribed temperature range, the radiation and convection heat rates are virtually identical and the heat rate increases from approximately 66 to 153 W. COMMENTS: A surface temperature above 50°C would be excessive and would accelerate electronic failure mechanisms. If operation involves large power dissipation (> 100 W), the receiver should be vented.

PROBLEM 9.45 KNOWN: Diameter, thickness, emissivity and initial temperature of silicon wafer. Temperature of air and surrounding. FIND: (a) Initial cooling rate, (b) Time required to achieve prescribed final temperature. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat transfer from side of wafer, (2) Large surroundings, (3) Wafer may be treated as a lumped capacitance, (4) Constant properties, (5) Quiescent air. 3

PROPERTIES: Table A-1, Silicon ( T = 187°C = 460K): ρ = 2330 kg/m , cp = 813 J/kg⋅K, k = 87.8 -6 2 W/m⋅K. Table A-4, Air (Tf,i = 175°C = 448K): ν = 32.15 × 10 m /s, k = 0.0372 W/m⋅K, α = 46.8 × -6 2 -1 10 m /s, Pr = 0.686, β = 0.00223 K . SOLUTION: (a) Heat transfer is by natural convection and net radiation exchange from top and 2 2 bottom surfaces. Hence, with As = πD /4 = 0.0177 m ,

(

)

)

(

4  q = As  h t + h b ( Ti − T∞ ) + 2 εσ Ti4 − Tsur   where the radiation flux is obtained from Eq. 1.7, and with L = As/P = 0.0375m and RaL = gβ (Ti 3

5

T∞) L /αν = 2.30 × 10 , the convection coefficients are obtained from Eqs. 9.30 and 9.32. Hence,

) ( k 4 = 0.0372 W / m ⋅ K × 5.9 = 5.9 W / m 2 ⋅ K h b = ( 0.27 Ra1/ ) L L 0.0375m

ht =

q = 0.0177 m

2

k 4 = 0.0372 W / m ⋅ K × 11.8 = 11.7 W / m 2 ⋅ K 0.54 Ra1/ L L 0.0375m

(

)

(11.7 + 5.9 ) W / m 2 ⋅ K (300K ) + 2 × 0.65 × 5.67 × 10 −8 W / m 2 ⋅ K 4 5984 − 2984 K 4   

q = 0.0177 m 2 (5280 + 8845 ) W / m 2  = 250 W  

<

(b) From the generalized lumped capacitance model, Eq. 5.15,

ρ cAsδ

∫

(

)

dT 4 A = − ( h t + h b ) ( T − T∞ ) + 2εσ T 4 − Tsur   s  dt

(

 T t  h t + hb dT = −  Ti 0

∫



4  ) (T − T∞ ) + 2εσ (T 4 − Tsur )

ρ cδ

 dt  Continued …..

PROBLEM 9.45 (Cont.) Using the DER function of IHT to perform the integration, thereby accounting for variations in h t and hb with T, the time tf to reach a wafer temperature of 50°C is found to be

t f ( T = 320K ) = 181s

S u rfa ce h e a t flu xe s , q ''(W /m ^2 )

<

325

W a fe r te m p e ra tu re , T(C )

275 225 175 125

4500 4000 3500 3000 2500 2000 1500 1000 500 0 0

75

20

40

60

80 100 120 140 160 180 200 Tim e , t(s )

25 0

20

40

60

N a tu ra l co n ve ctio n fro m to p s u rfa ce N a tu ra l co n ve ctio n fro m b o tto m s u rfa ce R a d ia tio n fro m to p o r b o tto m s u rfa ce

80 100 120 140 160 180 200 Tim e , t(s )

As shown above, the rate at which the wafer temperature decays with increasing time decreases due to reductions in the convection and radiation heat fluxes. Initially, the surface radiative flux (top or bottom) exceeds the heat flux due to natural convection from the top surface, which is twice the flux due to natural convection from the bottom surface. However, because q ′′rad and q′′cnv decay 4

5/4

, respectively, the reduction in q ′′rad with decreasing T is more pronounced, and at t = 181s, q ′′rad is well below q ′′cnv,t and only slightly larger than q ′′cnv,b .

approximately as T and T

(

)

2 2 COMMENTS: With hr,i = εσ ( Ti + Tsur ) Ti2 + Tsur = 14.7 W / m ⋅ K, the largest cumulative

coefficient of h tot = hr,i + h t,i = 26.4 W / m 2 ⋅ K corresponds to the top surface. If this coefficient is used to estimate a Biot number, it follows that Bi = h tot (δ / 2 ) / k = 1.5 × 10 −4 capacitance approximation is excellent.

1 and the lumped

PROBLEM 9.46 KNOWN: Pyrex tile, initially at a uniform temperature Ti = 140°C, experiences cooling by convection with ambient air and radiation exchange with surroundings. FIND: (a) Time required for tile to reach the safe-to-touch temperature of Tf = 40°C with free convection and radiation exchange; use T = ( Ti + Tf ) 2 to estimate the average free convection and linearized radiation coefficients; comment on how sensitive result is to this estimate, and (b) Time-tocool if ambient air is blown in parallel flow over the tile with a velocity of 10 m/s. SCHEMATIC:

ASSUMPTIONS: (1) Tile behaves as spacewise isothermal object, (2) Backside of tile is perfectly insulated, (3) Surroundings are large compared to the tile, (4) For forced convection situation, part (b), assume flow is fully turbulent. PROPERTIES: Table A.3, Pyrex (300 K): ρ = 2225 kg/m3, cp = 835 J/kg⋅K, k = 1.4 W/m⋅K, ε = 0.80 (given); Table A.4, Air ( Tf = ( Ts + T∞ ) 2 = 330.5 K, 1 atm ) : ν = 18.96 × 10-6 m2/s, k = 0.0286 W/m⋅K, α = 27.01 × 10-6 m2/s, Pr = 0.7027, β = 1/Tf. ANALYSIS: (a) For the lumped capacitance system with a constant coefficient, from Eq. 5.6, Ts ( t ) − T∞   hAs   = exp  −   t Ti − T∞   ρ Vc   where h is the combined coefficient for the convection and radiation processes, h = hcv + h rad and

As = L2

V = L2d

(1)

(2) (3,4)

The linearized radiation coefficient based upon the average temperature, Ts , is Ts = ( Ti + Tf

)

2 = (140 + 40 ) C 2 = 90$ C = 363 K $

(

2 h rad = εσ ( Ts + Tsur ) Ts2 + Tsur

)

(5) (6)

(

)

h rad = 0.8 × 5.67 × 10−8 W m 2⋅ K 4 (363 + 298 ) 3632 + 2982 K 3 = 6.61 W m 2⋅ K The free convection coefficient can be estimated from the correlation for the flat plate, Eq. 9.30, with Ra L =

gβ∆TL3

να

L = As P = L2 4L = 0.25L

(7,8) Continued...

PROBLEM 9.46 (Cont.) Ra L =

9.8 m s

(1 330 K )(363 − 298 ) K (0.25 × 0.200 m )3

2

18.96 × 10

−6

m

2

s × 27.01 × 10

(

14

Nu L = 0.54Ra L = 0.54 4.712 × 10

5

)

14

−6

m

2

= 4.712 × 10

5

s

= 14.18 2

hcv = Nu L k L = 14.18 × 0.0286 W m⋅ K 0.25 × 0.200 m = 8.09 W m ⋅ K

From Eq. (2), it follows h = ( 6.61 + 8.09 ) W m 2⋅ K = 14.7 W m 2⋅ K From Eq. (1), with As/V = 1/d, where d is the tile thickness, the time-to-cool is found as 40 − 25 140 − 25



  3  2225 kg m × 0.010 m × 835 J kg ⋅ K  2

14.7 W m ⋅ K × t f

= exp  −

<

t f = 2574s = 42.9 min

Using the IHT Lumped Capacitance Model with the Correlations Tool, Free Convection, Flat Plate, we can perform the analysis where both hcv and hrad are evaluated as a function of the tile temperature. The time-to-cool is

<

t f = 2860s = 47.7 min

which is 10% higher than the approximate value. (b) Considering parallel flow with a velocity, u ∞ = 10 m s over the tile, the Reynolds number is u L 10 m s × 0.200m 5 = 1.055 × 10 Re L = ∞ = −6 2 ν 18.96 × 10 m s

but, assuming the flow is turbulent at the upstream edge, use Eq. 7.41 to estimate h cv , 4 5

Nu L = 0.037 Re L

13

Pr

(

= 0.037 1.055 × 10

5

)

4 5

(0.7027 )1 3 = 343.3 2

hcv = Nu L k L = 343.3 × 0.0286 W m ⋅ K 0.200m = 49.1 W m ⋅ K

Hence, using Eqs. (2) and (1), find

<

t f = 661s = 11.0 min

COMMENTS: (1) For the conditions of part (a), Bi = hd/k= 14.7 W/m2 ⋅K × 0.01m / 1.4 W/m⋅K = 0.105. We conclude that the lumped capacitance analysis is marginally applicable. For the condition of part (b), Bi = 0.4 and, hence, we need to consider spatial effects as explained in Section 5.4. If we considered spatial effects, would our estimates for the time-to-cool be greater or less than those from the foregoing analysis? (2) For the conditions of part (a), the convection and radiation coefficients are shown in the plot below as a function of cooling time. Can you use this information to explain the relative magnitudes of the tf estimates?

20

Coefficients, (W/m^2.K)

2

h = 57.2 W m ⋅ K

15

10

5

0 0

1000

2000

Elapsed cooling time, t (s) Average coefficient, part (a) Variable coefficient, hcv + hrad, part (b) Convection coefficient, hcv Radiation coefficient, hrad

3000

PROBLEM 9.47 KNOWN: Stacked IC boards within a duct dissipating 500 W with prescribed air flow inlet temperature, flow rate, and internal convection coefficient. Outer surface has emissivity of 0.5 and is exposed to ambient air and large surroundings at 25°C. FIND: Develop a model to estimate outlet temperature of the air, Tm,o, and the average surface temperature of the duct, Ts , following these steps: (a) Estimate the average free convection for the outer surface, ho , assuming an average surface temperature of 37°C; (b) Estimate the average (linearized) radiation coefficient for the outer surface, h rad , assuming an average surface temperature of 37°C; (c) Perform an overall energy balance on the duct considering (i) advection of the air flow, (ii) dissipation of electrical power in the ICs, and (iii) heat transfer from the fluid to the ambient air and surroundings. Express the last process in terms of thermal resistances between the fluid and the mean fluid temperature, Tm , and the outer temperatures T∞ and Tsur; (d) Substituting numerical values into the expression of part (c), calculate Tm,o and Ts ; comment on your results and the assumptions required to develop your model. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible potential energy, kinetic energy and flow work changes for the internal flow, (3) Constant properties, (4) Power dissipated in IC boards nearly uniform in longitudinal direction, (5) Ambient air is quiescent, and (5) Surroundings are isothermal and large relative to the duct. -5

2

PROPERTIES: Table A-4, Air (Tf = ( Ts + T∞)/2 = 304 K): ν = 1.629 × 10 m /s, α = 2.309 × 10 5

2

-1

m /s, k = 0.0266 W/m⋅K, β = 0.003289 K , Pr = 0.706.

ANALYSIS: (a) Average, free-convection coefficient over the duct. Heat loss by free convection occurs on the vertical sides and horizontal top and bottom. The methodology for estimating the average coefficient assuming the average duct surface temperature Ts = 37°C follows that of Example 9.3. For the vertical sides, from Eq. 9.25 with L = H, find gβ Ts − T∞ H3 Ra L =

(

να

)

9.8 m / s 2 × 0.003289 K −1 (37 − 25 ) K × (0.150 m )

3

Ra L =

1.629 ×10−5 m 2 / s × 2.309 ×10−5 m 2 / s

= 3.47 × 106

The free convection is laminar, and from Eq. 9.27, 4 0.670 Ra1/ L Nu L = 0.68 + 9 /16 4 / 9

1 + ( 0.492 / Pr ) 

 

Continued …..

-

PROBLEM 9.47 (Cont.)

(

0.670 × 3.47 ×106

)

1/ 4

h H Nu L = v = 0.68 + = 23.2 4/9 k 1 + ( 0.492 / 0.706 )9 /16    h v = 4.11 W / m 2 ⋅ K For the top and bottom surfaces, Lc = (As/P) = (w × L)/(2w + 2L) = 0.0577 m, hence, RaL = 1.974 × 5 10 and with Eqs. 9.30 and 9.32, respectively,

h L 4 Nu L = t c = 0.54 Ra1/ L ; k h L 4; Nu L = b c = 0.27 Ra1/ L k

Top surface: Bottom surface:

h t = 5.25 W / m 2 ⋅ K h b = 2.62 W / m 2 ⋅ K

The average coefficient for the entire duct is

(

)

h cv,o = 2h v + h t + h b / 2 = ( 2 × 4.11 + 5.25 + 2.62 ) W / m 2 ⋅ K = 4.02 W / m 2 ⋅ K

<

(b) Average (linearized) radiation coefficient over the duct. Heat loss by radiation exchange between the duct outer surface and the surroundings on the vertical sides and horizontal top and bottom. With Ts = 37°C, from Eq. 1.9,

(

2 h rad = εσ ( Ts + Tsur ) Ts2 + Tsur

)

(

)

h rad = 0.5 × 5.67 × 10−8 W / m 2 ⋅ K 4 (310 + 298 ) 3102 + 2982 K 3 = 3.2 W / m 2 ⋅ K

<

(c) Overall energy balance on the fluid in the duct. The control volume is shown in the schematic below and the energy balance is

E in − E out + E gen = 0 −q adv + Pelec − q out = 0

(1)

 ρ,  =∀ The advection term has the form, with m

(

 cp Tm,o − Tm,i q adv = m

)

(2)

and the heat rate qout is represented by the thermal circuit shown below and has the form, with Tsur = T∞,

q out =

(

Tm − T∞

R cv,i + 1/ R cv,o + 1/ R rad

(3)

−1

)

where Tm is the average mean temperature of the fluid, (Tm,i + Tm,o)/2. The thermal resistances are evaluated with As = 2(w + H) L as

R cv,i = 1/ hi As

(4)

R cv,o = 1/ h cv,o As

(5)

R rad = 1/ h rad As

(6) Continued …..

PROBLEM 9.47 (Cont.) Using this energy balance, the outlet temperature of the air can be calculated. From the thermal circuit, the average surface temperature can be calculated from the relation

q out = ( Tm − Ts ) / R cv,i

(7)

(d) Calculating Tm,o and Ts . Substituting numerical values into the expressions of Part (c), find

Tm,o = 45.7°C

<

Ts = 34.0°C

The heat rates and thermal resistance results are

qadv = 480.5 W

q out = 19.5 W

R cv,i = 0.020 K / W

R cv,o = 0.250 K / W

R rad = 0.313 K / W

COMMENTS: (1) We assumed Ts = 37°C for estimating hcv,o and h rad , whereas from the energy balance we found the value was 34.0°C. Performing an interative solution, with different assumed Ts we would find that the results are not sensitive to the Ts value, and that the foregoing results are satisfactory. (2) From the results of Part (d) for the heat rates, note that about 4% of the electrical power is transferred from the duct outer surface. The present arrangement does not provide a practical means to cool the IC boards. (3) Note that Tm,i < Ts < Tm,o. As such, we can’t utilize the usual log-mean temperature (LMTD) expression, Eq. 8.45, in the rate equation for the internal flow analysis. It is for this reason we used the overall coefficient approach representing the heat transfer by the thermal circuit. The average surface temperature of the duct, Ts, is only used for the purposes of estimating hcv,o and h rad . We represented the effective temperature difference between the fluid and the ambient/surroundings as Tm − T∞ . Because the fluid temperature rise is not very large, this assumption is a reasonable one.

PROBLEM 9.48 KNOWN: Parallel flow of air over a highly polished aluminum plate flat plate maintained at a uniform temperature Ts = 47°C by a series of segmented heaters. FIND: (a) Electrical power required to maintain the heater segment covering the section between x1 = 0.2 m and x2 = 0.3m and (b) Temperature that the surface would reach if the air blower malfunctions and heat transfer occurs by free, rather than forced, convection. SCHEMATIC:

ASSUMPTIONS : (l) Steady-state conditions, (2) Backside of plate is perfectly insulated, (3) Flow is turbulent over the entire length of plate, part (a), (4) Ambient air is extensive, quiescent at 23°C for part (b). PROPERTIES: Table A.4, Air (Tf = (Ts + T‡ )/2 = 308K): υ = 16.69 × 10-6 m2/s, k = 0.02689 W/m⋅K, α = 23.68 × 10-6 m2/s, Pr = 0.7059, β = 1/Tf ; Table A.12, Aluminum, highly polished: ε = 0.03. ANALYSIS: (a) The power required to maintain the segmented heater (x1 - x2) is

Pe = h x1− x2 ( x 2 − x1 ) w (Ts − T∞ )

(1)

where h x1− x2 the average coefficient for the section between x1 and x2 can be evaluated as the average of the local values at x1 and x2,

h x1− x2 = ( h ( x1 ) + h ( x 2 )) / 2

(2)

Using Eq. 7.37 appropriate for fully turbulent flow, with Rex = u ‡ x /k, Nu x1 = 0.0296 Re 4x / 5 Pr1/ 3 4/5  10m s × 0.2m  1/ 3  Nu X1 = 0.0296  0.7059 ) = 304.6 (  16.69 × 10−6 m 2 s 





h x1 = Nu x1 k x1 = 304.6 × 0.02689 W m ⋅ K 0.2m = 40.9 W m 2⋅ K Nu x2 = 421.3

h x2 = 37.8 W m 2 ⋅ K

Hence, from Eq (2) to obtain h x1− x2 and Eq. (1) to obtain Pe, h x1− x2 = ( 40.9 + 37.8) W m 2⋅ K 2 = 39.4 W m 2⋅ K

Pe = 39.4 W m 2 ⋅ K (0.3 − 0.2 ) m × 0.2m ( 47 − 23) C = 18.9 W $

< Continued...

PROBLEM 9.48 (Cont.) (b) Without the airstream flow, the heater segment experiences free convection and radiation exchange with the surroundings,

(

)

4  x −x w Pe =  h cv (Ts − T∞ ) + εσ Ts4 − Tsur   ( 2 1 )

(3)

We will assume that the free convection coefficient, h cv , for the segment is the same as that for the entire plate. Using the correlation for a flat plate, Eq. 9.30, with

Ra L =

gβ∆TL3c να

Lc =

As 0.2 × 0.5m 2 = = 0.0714 m P 2 (0.2 + 0.5) m

and evaluating properties at Tf = 308 K,

Ra L =

3 9.8 m s 2 (1 308K )( 47 − 23)( 0.0714m )

16.69 ×10−6 m s 2 × 23.68 ×10−6 m 2 s

(

4 5 Nu L = 0.54Ra1/ L = 0.54 7.033 × 10

)

14

= 7.033 × 105

= 15.64

h cv = Nu L k Lc = 15.64 × 0.02689 W m ⋅ K 0.0714m = 5.89 W m 2⋅ K Substituting numerical values into Eq. (3), 18.9W = 5.89 W m ⋅ K ( Ts − 296 ) + 0.03 × 5.67 × 10



2

Ts = 447 K = 174$ C

−8

2

W m ⋅K

4

(T

4 4 s − 296

) (0.3 − 0.2 ) m × 0.2m <

COMMENTS: Recognize that in part (b), the assumed value for Tf = 308 K is a poor approximation. Using the above relations in the IHT work space with the Properties Tool, find that Ts = 406 K = 133 °C using the properly evaluated film temperature (Tf) and temperature difference (∆T) in the correlation. From this analysis, hcv = 8.29 W m 2 ⋅ K and hrad = 0.3 W/m2⋅K. Because of the low emissivity of the plate, the radiation exchange process is not significant.

PROBLEM 9.49 KNOWN: Correlation for estimating the average free convection coefficient for the exterior surface of a long horizontal rectangular cylinder (duct) exposed to a quiescent fluid. Consider a horizontal 0.15 m-square duct with a surface temperature of 35°C passing through ambient air at 15°C. FIND: (a) Calculate the average convection coefficient and the heat rate per unit length using the HD correlation, (b) Calculate the average convection coefficient and the heat rate per unit length considering the duct as formed by vertical plates (sides) and horizontal plates (top and bottom), and (c) Using an appropriate correlation, calculate the average convection coefficient and the heat rate per unit length for a duct of circular cross-section having a diameter equal to the wetted perimeter of the rectangular duct of part (a). Do you expect the estimates for parts (b) and (c) to be lower or higher than those obtained with the H-D correlation? Explain the differences, if any. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Ambient air is quiescent, (3) Duct surface has uniform temperature. -5

2

PROPERTIES: Table A-4, air (Tf = (Ts + T∞)/2 = 298 K, 1 atm): ν = 1.571 × 10 m /s, k = 0.0261 -5 2 W/m⋅K, α = 2.22 × 10 m /s, Pr = 0.708. ANALYSIS: (a) The Hahn-Didion (H-D) correlation [ASHRAE Proceedings, Part 1, pp 262-67, 1972] has the form 1/ 8 H 1/ 4 Nu p = 0.55 Ra p   p



Ra p ≤ 107



where the characteristic length is the half-perimeter, p = (w + H), and w and H are the horizontal width and vertical height, respectively, of the duct. The thermophysical properties are evaluated at the film temperature. Using IHT, with the correlation and thermophysical properties, the following results were obtained. Rap 5.08 × 10

7

(

Nu D

hp W / m2 ⋅ K

42.6

3.71

)

q′p ( W / m ) 44.5

<

where the heat rate per unit length of the duct is

q′p = h p 2 ( H + w )( Ts − T∞ ) .

(b) Treating the duct as a combination of horizontal (top: hot-side up and bottom: hot-side down) and two vertical plates (v) as considered in Example 9.3, the following results were obtained ht 2

(W/m ⋅K) 5.62

hb

hv

h hv

2

2

2

(W/m ⋅K) 2.81

(W/m ⋅K) 4.78

(W/m ⋅K) 4.50

q′hv (W/m) 54.0

< Continued

PROBLEM 9.49 (Cont.) where the average coefficient and heat rate per unit length for the horizontal-vertical plate duct are

h hv = ( h t + h b + 2 h v ) / 4 q′hv = h hv 2 ( H + w )( Ts − T∞ ). (c) Consider a circular duct having a wetted perimeter equal to that of the rectangular duct, for which the diameter is

π D = 2 (H + w )

D = 0.191 m

Using the Churchill-Chu correlation, Eq. 9.34, the following results are obtained.

Nu D

RaD 1.31 × 10

7

30.6

(

h D W / m2 ⋅ K 4.19

)

q′D ( W / m ) 50.3

<

where the heat rate per unit length for the circular duct is

q′D = π D h D ( Ts − T∞ ) . COMMENTS: (1) The H-D correlation, based upon experimental measurements, provided the lowest estimate for h and q′. The circular duct analysis results are in closer agreement than are those for the horizontal-vertical plate duct. (2) An explanation for the relative difference in h and q′ values can be drawn from consideration of the boundary layers and induced flows around the surfaces. Viewing the cross-section of the square duct, recognize that flow induced by the bottom surface flows around the vertical sides, joining the vertical plume formed on the top surface. The flow over the vertical sides is quite different than would occur if the vertical surface were modeled as an isolated vertical surface. Also, flow from the top surface is likewise modified by flow rising from the sides, and doesn’t behave as an isolated horizontal surface. It follows that treating the duct as a combination of horizontal-vertical plates (hv results), each considered as isolated, would over estimate the average coefficient and heat rate. (3) It follows that flow over the horizontal cylinder more closely approximates the situation of the square duct. However, the flow is more streamlined; thinnest along the bottom, and of increasing thickness as the flow rises and eventually breaks away from the upper surface. The edges of the duct disrupt the rising flow, lowering the convection coefficient. As such, we expect the horizontal cylinder results to be systematically higher than for the H-D correlation that accounts for the edges.

PROBLEM 9.50 KNOWN: Straight, rectangular cross-sectioned fin with prescribed geometry, base temperature, and environmental conditions. FIND: (a) Effectiveness considering only free convection with average coefficient, (b) Effectiveness considering also radiative exchange, (c) Finite-difference equations suitable for considering local, rather than average, values. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional conduction in fin, (4) Width of fin much larger than length, w >> L, (5) Uniform heat transfer coefficient over length for Parts (a) and (b). PROPERTIES: Table A-1, Aluminum alloy 2024-T6 (T ≈ (45 + 25) / 2 = 35 ° C ≈ 300 K), k = 177 W/m⋅K; Table A-11, Aluminum alloy 2024-T6 (Given), ε = 0.82; Table A-4. Air (Tf ≈ 300 K), ν = 15.89 × 10 -6 m2/s, k = 26.3 × 10-3 W/m⋅K, α = 22.5 × 10-6 m2/s, β = 1/Tf = 33.3 × 10-3 K-1. ANALYSIS: (a) The effectiveness of a fin is determined from Eq. 3.81

ε = q f hA c,bθ b where h is the average heat transfer coefficient. The fin heat transfer follows from Eq. 3.72 sinh mL + (h / mk) cosh mL qf = M cosh mL + (h / mk) sinh mL

(1)

(2)

where

M = ( hPkA c )

1/ 2

θb

and

m = ( hP kAc )

1/ 2

.

(3,4)

Horizontal, flat plate correlations assuming Tf = (Tb + T∞ ) / 2 ≈ 300 K may be used to estimate h , Eqs. 9.30 to 9.32. Calculate first the Rayleigh number gβ Ts − T∞ L3c Ra Lc = (5)

(

να

)

where Ts is the average temperature of the fin surface and Lc is the characteristic length from Eq. 9.29,

Lc ≡

As L× w L = ≈ ⋅ P 2L + 2w 2

Substituting numerical values,

Ra Lc =

(6)

(

)

3 9.8 m s 2 × 1/ 300K × (310 − 298 ) K 100 × 10−3 / 2 m3

22.5 × 10−6 m 2 s × 15.89 × 10−6 m 2 s

= 1.37 ×105

(7) Continued...

PROBLEM 9.50 (Cont.) where Ts ≈ ( Tb + Tf ) / 2 = 310K. Recognize the importance of this assumption which must be justified for a precise result. Using Eq. 9.30 and 9.32 for the upper and lower surfaces, respectively, 1/ 4 k 0.0263 W m ⋅ K h u = Nu L × = = 5.47 W m 2 ⋅ K = 10.4, Nu L = 0.54 1.37 × 105 c L c c 100 × 10−8 2 m

(

)

(

)

Nu L = 0.27 1.37 × 105 c

1/ 4

)

(

h" = 2.73 W m 2 ⋅ K

= 5.20,

The average value is estimated as hc = ( h u + h" ) 2 = 4.10 W m 2⋅ K . Using this value in Eqs. (3) and (4), find

(

)

1/ 2

M =  4.10 W m 2⋅ K ( 2w ) m × 177 W m ⋅ K w × 2 × 10−3 m 2 



m = ( hc P / kA c )

1/ 2

(



( 45 − 25 )$ C = 34.1w W

)



Substituting these values into Eq. (2), with mL = 0.481 and qf /w = q′f . q 'f = 34.1W m ×

1/ 2

=  4.1W / m 2⋅ K ( 2w ) m 177W / m ⋅ K w × 2 × 10−3 m 2 

(

)



sinh 0.481 + 4.1W m 2 ⋅ K 4.81m −1 × 177 W m ⋅ K cosh 0.481

(

cosh 0.481 + 4.86 × 10

and then from Eq. (1), the effectiveness is

)

(

−3

)sinh 0.481

= 4.81m −1 .

= 15.2 W m

ε = 15.2 W m × w 4.1W m 2 ⋅ K w × 2 × 10−3 m ( 45 − 25 ) C = 92.7. $

<

(b) If radiation exchange with the surroundings is considered, use Eq. 1.9 to determine

(

)

h r = εσ ( Ts + Tsur ) Ts + Tsur = 0.82 × 5.67 × 10 2

2

−8

2

W m ⋅K

4

(310 + 298 ) (3102 + 2982 )K 3 = 5.23 W

2

m ⋅K.

This assumes the fin surface is gray-diffuse and small compared to the surroundings. Using h = hc + h r where h c is the convection parameter from part (a), find h = ( 4.10 + 5.23 ) W m 2⋅ K = 9.33 W m 2⋅ K, M = 51.4w W, m = 7.26m −1, q′f = 31.8 W m giving

<

ε = 85.2 (c) To perform the numerical method, we used the IHT Finite Difference Equation Tool for 1-D, SS, extended surfaces. The convection coefficient for each node was expressed as

h tot,m = h u (Tm ) + h " ( Tm ) 2 + h r ( Tm )

The effectiveness was calculated from Eq. (1) where the fin heat rate is determined from an energy balance on the base node. q f = q cond + q cv + q rad q b = q cond = kA c ( Tb − T1 ) / ∆x q a = q cv + q rad = h tot,b ( P ⋅ ∆ w 2 )( Tb − Tinf

)

h tot,b = ( h u (Tb ) + h" ( Tb )) 2 + h r ( Tb ) Continued...

PROBLEM 9.50 (Cont.) The results of the analysis (15 nodes, ∆x = L/15)

qf = 33.6 W m

<

ε = 83.2

COMMENTS: (1) From the analytical treatments, parts (a) and (b), considering radiation exchange nearly doubles the fin heat rate (31.8 vs. 15.2 W/m) and reduces the effectiveness from 92.7 to 85.2. The numerical method, part (c) considering local variations for hc and hrad, provides results for q ′f and ε which are in close agreement with the analytical method, part (b). (2) The IHT Finite Difference Equation Tool provides a powerful approach to solving a problem as tedious as this one. Portions of the work space are copied below to illustrate the general logic of the analysis. // Method of Solution: The Finite-Difference Equation tool for One-Dimensional, Steady-State Conditions for an extended surface was used to write 15 nodal equations. The convection and linearized radiation coefficient for each node was separately calculated by a User-Defined Function. */ // User-Defined Function - Upper surface convection coefficients: /* FUNCTION h_up ( Ts ) h_up = 0.0263 / 0.05 * NuLcu NuLcu = 0.54 * (11,421 * (Ts - 298) )^0.25 RETURN h_up END */ // User-Defined Function - Linearized radiation coefficients: /* FUNCTION h_rad ( Ts ) h_rad = 0.82 * sigma * (Ts + 298 ) * (Ts^2 + 298^2 ) sigma = 5.67e-8 RETURN h_rad END */ /* Node 1: extended surface interior node;e and w labeled 2 and b. */ 0.0 = fd_1d_xsur_i(T1,T2,Tb,k,qdot,Ac,P,deltax,Tinf,htot1,q''a) q''a = 0 // Applied heat flux, W/m^2; zero flux shown qdot = 0 htot1 = ( h_up(T1) + h_do(T1) ) / 2 + h_rad(T1) /* Node 2: extended surface interior node;e and w labeled 2 and b. */ 0.0 = fd_1d_xsur_i(T2,T3,T1,k,qdot,Ac,P,deltax,Tinf,htot2,q''a) htot2 = ( h_up(T2) + h_do(T2) ) / 2 + h_rad(T2) /* Node 15: extended surface end node, e-orientation; w labeled inf. */ 0.0 = fd_1d_xend_e(T15,T14,k,qdot,Ac,P,deltax,Tinf,htot15,q''a,Tinf,htot15,q''a) htot15 = ( h_up(T15) + h_do(T15) ) / 2 + h_rad(T15) // Assigned Variables: Tb = 45 + 273 Tinf = 25 + 273 Tsur = 25 + 273 L = 0.1 deltax = L / 15 k = 177 Ac = t * w t = 0.002 w=1 P=2*w Lc = L / 2

// Base temperature, K // Ambient temperature, K // Surroundings temperature, K // Length of fin, m // Space increment, m // Thermal conductivity, W/m.K; fin material // Cross-sectional area, m^2 // Fin thickness, m // Fin width, m; unity value selected // Perimeter, m // Characteristic length, convection correlation, m

// Fin heat rate and effectiveness qf = qcond + qcvrad qcond = k * Ac * (Tb - T1 ) / deltax qcvrad = htotb * P * deltax / 2 * ( Tb - Tinf ) htotb = ( h_up(Tb) + h_do(Tb) ) / 2 + h_rad(Tb) eff = qf / ( htotb * Ac * (Tb - Tinf) )

// Heat rate from the fin base, W // Heat rate, conduction, W // Heat rate, combined radiation convection, W // Total heat transfer coefficient, W/m^2.K // Effectivenss

PROBLEM 9.51 KNOWN: Dimensions, emissivity and operating temperatures of a wood burning stove. Temperature of ambient air and surroundings. FIND: Rate of heat transfer. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Quiescent air, (3) Negligible heat transfer from pipe elbow, (4) Free convection from pipe corresponds to that from a vertical plate. PROPERTIES: Table A.4, air (Tf = 400 K): ν = 26.41 × 10-6 m2/s, k = 0.0338 W/m⋅K, α = 38.3 × 10-6 m2/s, β = 0.0025 K-1, Pr = 0.69. Table A.4, air (Tf = 350 K): ν = 20.92 × 10-6 m2/s, k = 0.030 W/m⋅K, α = 29.9 × 10-6 m2/s, Pr = 0.70, β = 0.00286 K-1. ANALYSIS: Three distinct contributions to the heat rate are made by the 4 side walls, the top surface, and the pipe surface. Hence qt = 4qs + qt + qp, where each contribution includes transport due to convection and radiation. qs = hs L2s Ts,s − T∞ + h rad,s L2s Ts,s − Tsur q t = h t L2s Ts,s − T∞ + h rad,s L2s Ts,s − Tsur

(

( (

)(

) )

)

( (

(

) )

)(

q p = h p π Dp Lp Ts,p − T∞ + h rad,p π Dp Lp Ts,p − Tsur The radiation coefficients are

(

h rad,s = εσ Ts,s + Tsur

(

2 = 12.3 W ) (Ts,s2 + Tsur )

h rad,p = εσ Ts,p + Tsur

2 + T 2 = 7.9 W )(Ts,p sur )

(

)

m2 ⋅ K m2 ⋅ K

)

For the stove side walls, RaL,s = gβ Ts,s − T∞ L3s αν = 4.84 × 109. Similarly, with (As/P) = L2s 4 L s = 0.25 m, RaL,t = 7.57 × 107 for the top surface, and with Lp = 2 m, RaL,p = 3.59 × 1010 for the stove pipe. For the side walls and the pipe, the average convection coefficient may be determined from Eq. 9.26, 2

  1/ 6   0.387Ra L Nu L = 0.825 + 8 / 27    1 + ( 0.492 Pr )9 /16     

Continued...

PROBLEM 9.51 (Cont.) which yields Nu L,s = 199.9 and Nu L,p = 377.6. For the top surface, the average coefficient may be obtained from Eq. 9.31, 3 Nu L = 0.15Ra1/ L which yields Nu L,t = 63.5. With h = Nu ( k L ) , the convection coefficients are

hs = 6.8 W m 2 ⋅ K , Hence,

h t = 8.6 W m 2 ⋅ K ,

h p = 5.7 W m 2 ⋅ K

( ) q t = ( h t + h rad,s ) L2s ( Ts,s − 300 K ) = 20.9 W m 2 ⋅ K (1m 2 ) ( 200 K ) = 4180 W (

) (

)

qs = hs + h rad,s L2s Ts,s − 300 K = 19.1W m 2 ⋅ K 1m 2 ( 200 K ) = 3820 W

(

q p = h p + h rad,p

)(π Dp Lp )(Ts,p − 300 K ) = 13.6 W

m ⋅ K (π × 0.25 m × 2 m )(100 K ) = 2140 W 2

and the total heat rate is

q tot = 4qs + q t + q p = 21, 600 W

<

COMMENTS: The amount of heat transfer is significant, and the stove would be capable of maintaining comfortable conditions in a large, living space under harsh (cold) environmental conditions.

PROBLEM 9.52 KNOWN: Plate, 1m × 1m, inclined at 45° from the vertical is exposed to a net radiation heat flux of 2

300 W/m ; backside of plate is insulated and ambient air is at 0°C. FIND: Temperature plate reaches for the prescribed conditions. SCHEMATIC:

2

ASSUMPTIONS: (1) Net radiation heat flux (300 W/m ) includes exchange with surroundings, (2) Ambient air is quiescent, (3) No heat losses from backside of plate, (4) Steady-state conditions. PROPERTIES: Table A-4, Air (assuming Ts = 84°C, Tf = (Ts + T∞)/2 = (84 + 0)°C/2 = 315K, 1 -6 2 -6 2 atm): ν = 17.40 × 10 m /s, k = 0.0274 W/m⋅K, α = 24.7 × 10 m /s, Pr = 0.705, β = 1/Tf. ANALYSIS: From an energy balance on the plate, it follows that q′′rad = q′′conv . That is, the net radiation heat flux into the plate is equal to the free convection heat flux to the ambient air. The temperature of the surface can be expressed as

Ts = T∞ + q′′rad / h L (1) where hL must be evaluated from an appropriate correlation. Since this is the bottom surface of a heated inclined plate, “g” may be replaced by “g cos θ”; hence using Eq. 9.25, find

Ra L =

gcos θβ ( Ts − T∞ ) L3 9.8m/s2 × cos45° (1/315K )( 84 − 0) K (1m )3 = = 4.30 × 109. − 6 2 − 6 2 να 17.40 × 10 m / s × 24.7 × 10 m / s 9

Since RaL > 10 , conditions are turbulent and Eq. 9.26 is appropriate for estimating Nu L

  0.387Ra1L/ 6  Nu L = 0.825 + 8/27  1 + ( 0.492/Pr )9/16    

(

)

     

2

 1/6  0.387 4.30 ×109  Nu L = 0.825 + 8/27  1 + ( 0.492/0.705 )9/16    

(2)

2     = 193.2  

hL = Nu L k / L = 193.2 × 0.0274W/m ⋅ K/1m = 5.29W/m2 ⋅ K. Substituting hL from Eq. (3) into Eq. (1), the plate temperature is Ts = 0 °C + 300W/m 2 /5.29W/m 2 ⋅K = 57 °C.

(3)

<

COMMENTS: Note that the resulting value of Ts ≈ 57°C is substantially lower than the assumed value of 84°C. The calculation should be repeated with a new estimate of Ts, say, 60°C. An alternate approach is to write Eq. (2) in terms of Ts, the unknown surface temperature and then combine with Eq. (1) to obtain an expression which can be solved, by trial-and-error, for Ts.

PROBLEM 9.53 KNOWN: Horizontal rod immersed in water maintained at a prescribed temperature. FIND: Free convection heat transfer rate per unit length of the rod,

q′conv

SCHEMATIC:

ASSUMPTIONS: (1) Water is extensive, quiescent medium. 3

PROPERTIES: Table A-6, Water (Tf = (Ts + T∞)/2 = 310K): ρ = 1/vf = 993.0 kg/m , ν = µ/ρ = -6 2 3 -7 2 3 695 × 10 N⋅s/m /993.0 kg/m = 6.999 × 10 m /s, α = k/ρc = 0.628 W/m⋅K/993.0 kg/m × 4178 -7 2 -6 -1 J/kg⋅K = 1.514 × 10 m /s, Pr = 4.62, β = 361.9 × 10 K . ANALYSIS: The heat rate per unit length by free convection is given as

q′conv = h D ⋅π D ( Ts − T∞ ). (1) To estimate hD , first find the Rayleigh number, Eq. 9.25, 2 −6 −1 ( 56 − 18) K ( 0.005m )3 g β (Ts − T∞ ) D3 9.8m/s 361.9 × 10 K Ra D = = = 1.587 ×105 − 7 2 − 7 2 να 6.999 ×10 m /s ×1.514 ×10 m / s

(

)

and use Eq. 9.34 for a horizontal cylinder,

  0.387Ra1D/ 6  Nu D =  0.60 + 8/27  1 + ( 0.599/Pr )9/16    

(

)

     

 1/6  0.387 1.587 × 105  Nu D =  0.60 + 8/27  1 + ( 0.599/4.62 )9/16    

2

2     = 10.40  

hD = Nu D k / D = 10.40 × 0.628W/m ⋅ K/0.005m = 1306W/m 2 ⋅ K. (2) Substituting for hD from Eq. (2) into Eq. (1), q′conv = 1306W/m 2 ⋅ K ×π ( 0.005m )( 56 − 18 ) K = 780W/m. < COMMENTS: (1) Note the relatively large value of hD ; if the rod were immersed in air, the heat transfer coefficient 2

would be close to 5 W/m ⋅K. (2) Eq. 9.33 with appropriate values of C and n from Table 9.1 could also be used to estimate

(

)

hD . Find

0.25 Nu D = CRa nD = 0.48 1.587 ×105 = 9.58 hD = Nu D k / D = 9.58× 0.628W/m ⋅ K/0.005m = 1203W/m2 ⋅ K. By comparison with the result of Eq. (2), the disparity of the estimates is ~8%.

PROBLEM 9.54 KNOWN: Horizontal, uninsulated steam pipe passing through a room. FIND: Heat loss per unit length from the pipe. SCHEMATIC:

ASSUMPTIONS: (1) Pipe surface is at uniform temperature, (2) Air is quiescent medium, (3) Surroundings are large compared to pipe. -6

PROPERTIES: Table A-4, Air (Tf = (Ts + T∞)/2 = 350K, 1 atm): ν = 20.92 × 10 2

-6

-3

W/m⋅K, α = 29.9 × 10 m /s, Pr = 0.700, β = 1/Tf = 2.857 × 10

2

m /s, k = 0.030

-1

K .

ANALYSIS: Recognizing that the heat loss from the pipe will be by free convection to the air and by radiation exchange with the surroundings, we can write

(

)

4 . q′ = q′conv + q′rad = π D  hD (Ts − T∞ ) + εσ Ts4 − Tsur   To estimate hD , first find RaL, Eq. 9.25, and then use the correlation for a horizontal cylinder, Eq. 9.34, Ra L =

g β (Ts − T∞ ) D

3 =

9.8m/s

2

(1/350K ) ( 400 − 300) K ( 0.150m) 3

να 20.92 ×10 −6 m2 / s × 29.9 ×10 −6 m 2 / s 2     0.387Ra1L/ 6 Nu D = 0.60 + 8/27    1 + ( 0.559/Pr )9/16 







7



)

(

= 1.511 × 10

(1)

2

1/6   0.387 1.511 × 10 7   Nu D = 0.60 + = 31.88 8/27  9/16   1 + ( 0.559/0.700 )     

hD = Nu D ⋅ k / D = 31.88 × 0.030W/m ⋅ K/0.15m = 6.38W/m2 ⋅ K. (2) Substituting for hD from Eq. (2) into Eq. (1), find q ′ = π ( 0.150m )  6.38W/m2 ⋅ K ( 400 − 300 ) K + 0.85 × 5.67 ×10 −8 W / m2 ⋅ K4 400 4 − 300 4 K4   

(

q′ = 301W/m + 3 9 7 W / m = 698W/m.

)

<

COMMENTS: (1) Note that for this situation, heat transfer by radiation and free convection are of equal importance. (2) Using Eq. 9.33 with constants C,n from Table 9.1, the estimate for hD is 0.333 Nu D = CRanL = 0.125 1.511× 107 = 30.73

(

)

h D = Nu D k / D = 30.73 × 0.030W/m⋅ K/0.150m = 6.15W/m2 ⋅ K. The agreement is within 4% of the Eq. 9.34 result.

PROBLEM 9.55 KNOWN: Diameter and outer surface temperature of steam pipe. Diameter, thermal conductivity, and emissivity of insulation. Temperature of ambient air and surroundings. FIND: Effect of insulation thickness and emissivity on outer surface temperature of insulation and heat loss. SCHEMATIC: See Example 9.4, Comment 2. ASSUMPTIONS: (1) Pipe surface is small compared to surroundings, (2) Room air is quiescent. PROPERTIES: Table A.4, air (evaluated using Properties Tool Pad of IHT). ANALYSIS: The appropriate model is provided in Comment 2 of Example 9.4 and includes use of the following energy balance to evaluate Ts,2,

q′cond = q′conv + q′rad ≡ q′

(

2π ki Ts,1 − Ts,2 ln ( r2 r1 )

) = h2π r

(

(

)

4 4 2 Ts,2 − T∞ + ε 2π r2σ Ts,2 − Tsur

)

from which the total heat rate q′ can then be determined. Using the IHT Correlations and Properties Tool Pads, the following results are obtained for the effect of the insulation thickness, with ε = 0.85. 170

800 700

140

Heat loss, q'(W/m)

Surface temperature, Ts,2(C)

600 110

80

500 400 300 200

50 100 20

0 0

0.01

0.02

0.03

0.04

Insulation thickness, t(m)

0.05

0

0.01

0.02

0.03

0.04

0.05

Insulation thickness, t(m)

The insulation significantly reduces Ts,2 and q′ , and little additional benefits are derived by increasing t above 25 mm. For t = 25 mm, the effect of the emissivity is as follows.

Continued...

44

53

42

52

Heat loss, q'(W/m)

Surface temperature, Ts,2(C)

PROBLEM 9.55 (Cont.)

40

38

51

50

36

49

34

48 0.1 0.2 0.3 0.4

0.5 0.6 0.7 0.8 0.9

Emissivity, eps

1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1

Emissivity, eps

Although the surface temperature decreases with increasing emissivity, the heat loss increases due to an increase in net radiation to the surroundings.

PROBLEM 9.56 KNOWN: Dimensions and temperature of beer can in refrigerator compartment. FIND: Orientation which maximizes cooling rate. SCHEMATIC:

ASSUMPTIONS: (1) End effects are negligible, (2) Compartment air is quiescent, (3) Constant properties. -6

PROPERTIES: Table A-4, Air (Tf = 288.5K, 1 atm): ν = 14.87 × 10 -6

21.0 × 10

2

-3

m /s, Pr = 0.71, β = 1/Tf = 3.47 × 10

2

m /s, k =0.0254 W/m⋅K, α =

-1

K .

ANALYSIS: The ratio of cooling rates may be expressed as

q v h v π DL ( Ts − T∞ ) = = qh h h π DL ( Ts − T∞ )

hv . hh

For the vertical surface, find

Ra L =

gβ ( Ts − T∞ ) 3 9.8m/s 2 × 3.47 ×10−3 K−1 ( 23°C ) 3 L = L = 2.5 ×109 L3 − 6 2 − 6 2 να 14.87 ×10 m /s 21 ×10 m / s

(

)(

)

Ra L = 2.5 ×10 9 ( 0.15 ) = 8.44 ×10 6 , 3

(

)

2

  6 1/6 0.387 8.44 × 10   and using the correlation of Eq. 9.26, Nu L = 0.825 +  = 29.7. 8/27 9/16   1 + ( 0.492/0.71 )      k 0.0254W/m ⋅ K Hence hL = hv = Nu L = 29.7 = 5.03W/m 2 ⋅ K. L 0.15m gβ ( Ts − T∞ ) 3 For the horizontal surface, find Ra D = D = 2.5 ×10 9 ( 0.06 )3 = 5.4 × 105 να

(

)

2

1/6   0.387 5.4 × 105   and using the correlation of Eq. 9.34, Nu D =  0.60 +  = 12.24 8/27 9/16   1 + ( 0.559/0.71)      k 0.0254W/m ⋅ K hD = h h = Nu D = 12.24 = 5.18W/m 2 ⋅ K. D 0.06m q v 5.03 Hence = = 0.97. < qh 5.18

COMMENTS: In view of the uncertainties associated with Eqs. 9.26 and 9.34 and the neglect of end effects, the above result is inconclusive. The cooling rates are approximately the same.

PROBLEM 9.57 KNOWN: Length and diameter of tube submerged in paraffin of prescribed dimensions. Properties of paraffin. Inlet temperature, flow rate and properties of water in the tube. FIND: (a) Water outlet temperature, (b) Heat rate, (c) Time for complete melting. SCHEMATIC:

ASSUMPTIONS: (1) Negligible k.e. and p.e. changes for water, (2) Constant properties for water and paraffin, (3) Negligible tube wall conduction resistance, (4) Free convection at outer surface associated with horizontal cylinder in an infinite quiescent medium, (5) Negligible heat loss to surroundings, (6) Fully developed flow in tube. -6

PROPERTIES: Water (given): cp = 4185 J/kg⋅K, k = 0.653 W/m⋅K, µ = 467 × 10

kg/s⋅m, Pr =

2.99; Paraffin (given): Tmp = 27.4°C, hsf = 244 kJ/kg, k = 0.15 W/m⋅K, β = 8 × 10

K , ρ = 770

-4

3

-6

kg/m , ν = 5 × 10

2

-8

m /s, α = 8.85 × 10

-1

2

m /s.

ANALYSIS: (a) The overall heat transfer coefficient is

1 1 1 = + . U hi ho To estimate h i , find

Re D =

& 4m 4 × 0.1kg/s = =10,906 π Dµ π × 0.025m × 467 ×10 −6 kg/s ⋅ m

and noting the flow is turbulent, use the Dittus-Boelter correlation 4/5 NuD = 0.023Re4D/ 5 Pr 0.3 = 0.023 (10,906 ) ( 2.99) 0.3 = 54.3

Nu D k 54.3× 0.653W/m ⋅ K = = 1418W/m 2 ⋅ K. D 0.025m To estimate h o , find hi =

Ra D =

gβ ( Ts − T∞ ) D3 να

9.8m/s2 ) 8 ×10−4 K −1 ( 55 − 27.4 ) K ( 0.025m )3 ( = 5 × 10−6 m2 / s × 8.85 × 10−8 m2 / s

Ra D = 7.64 ×10 6 2

and using the correlation of Eq. 9.34,

h o = Nu D

Nu D

  0.387Ra1D/ 6   =  0.60 +  = 35.0 8/27 9/16   1 + ( 0.559/Pr )     

k 0.15W/m ⋅ K = 35.0 = 210W/m 2 ⋅ K. D 0.025m

Alternatively, using the correlation of Eq. 9.33, Continued …..

PROBLEM 9.57 (Cont.) NuD = CRa nD with C = 0.48, n = 0.25 0.15W/m ⋅ K h o = 25.2 = 151W/m 2 ⋅ K. 0.025m

NuD = 25.2

The significant difference in ho values for the two correlations may be due to difficulties associated with high Pr applications of one or both correlations. Continuing with the result from Eq. 9.34, 1 1 1 1 1 = + = + = 5.467 ×10 −3 m 2 ⋅ K / W

U

hi

ho

1418

210

U = 183W/m 2 ⋅ K. Using Eq. 8.46, find

 π DL   π × 0.025m × 3m W  = exp  − U  = exp  − 183   mc  T∞ − Tm,i  0.1kg/s × 4185J/kg ⋅ K m2 ⋅ K   & p 

T∞ − Tm,o

(

)

Tm,o = T∞ − T∞ − Tm,i 0.902 = 27.4 − ( 27.4 − 60 ) 0.902  °C

<

Tm,o = 56.8°C. (b) From an energy balance, the heat rate is

(

)

& p Tm,i − Tm,o = 0.1kg/s × 4185J/kg ⋅ K ( 60 − 56.8 ) K = 1335W q = mc

<

or using the rate equation,

q = U A ∆ Tl m = 183W/m 2 ⋅ Kπ ( 0.025m ) 3m

( 60 − 27.4) K − ( 56.8 − 27.4) K ln

60 − 27.4 56.8 − 27.4

q = 1335W. (c) Applying an energy balance to a control volume about the paraffin,

E in = ∆E st q ⋅ t = ρ V hsf = ρL  WH − π D 2 / 4  h sf   t=

770kg/m 3 × 3m  π ( 0.25m) 2 − ( 0.025m )2  2.44× 105 J/kg  1335W 4  

t = 2.618 × 104 s = 7.27h.

<

COMMENTS: (1) The value of ho is overestimated by assuming an infinite quiescent medium. The fact that the paraffin is enclosed will increase the resistance due to free convection and hence decrease q and increase t. 2 (2) Using ho = 151W/m 2 ⋅ K results in U = 1 3 6 W / m ⋅ K,Tm,o = 57.6°C, q = 1009 W and t = 9.62 h.

PROBLEM 9.58 KNOWN: A long uninsulated steam line with a diameter of 89 mm and surface emissivity of 0.8 transports steam at 200°C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20°C. FIND: (a) The heat loss per unit length for a calm day when the ambient air temperature is 20°C; (b) The heat loss on a breezy day when the wind speed is 8 m/s; and (c) For the conditions of part (a), calculate the heat loss with 20-mm thickness of insulation (k = 0.08 W/m⋅K). Would the heat loss change significantly with an appreciable wind speed? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Calm day corresponds to quiescent ambient conditions, (3) Breeze is in crossflow over the steam line, (4) Atmospheric air and large surroundings are at the same temperature; and (5) Emissivity of the insulation surface is 0.8. -5

2

PROPERTIES: Table A-4, Air (Tf = (Ts + T∞)/2 = 383 K, 1 atm): ν = 2.454 × 10 m /s, k = -5 2 0.03251 W/m⋅K, α = 3.544 × 10 m /s, Pr = 0.693. ANALYSIS: (a) The heat loss per unit length from the pipe by convection and radiation exchange with the surroundings is

q′b = q′cv + q′rad

(

(

)

4 − T4 q′b = h D Pb Ts,b − T∞ + ε Pbσ Ts,b ∞

)

Pb = π Db

(1,2)

where Db is the diameter of the bare pipe. Using the Churchill-Chu correlation, Eq. 9.34, for free convection from a horizontal cylinder, estimate h D

  1/ 6   0.387 Ra h Db  D Nu D = = 0.60 + 8 / 27  k   1 + (0.559 / Pr )9 /16     

2 (3)

where properties are evaluated at the film temperature, Tf = (Ts + T∞)/2 and gβ (Ts − T∞ ) D3b Ra D =

(4)

να

Substituting numerical values, find for the bare steam line Ra D 3.73 × 10

6

Nu D

hD (W/m ⋅K)

q′cv ( W / m )

q′rad ( W / m )

q′b ( W / m )

21.1

7.71

388

541

929

2

Continued …..

<

PROBLEM 9.58 (Cont.) (b) For forced convection conditions with V = 8 m/s, use the Churchill-Bernstein correlation, Eq. 7.56, 2 Pr1/ 3 0.62 Re1/ h D Db D Nu D = = 0.3 + 1/ 4 k 1 + ( 0.4 / Pr )2 / 3   

5/8    Re  D 1 +      282, 000    

4/5

where ReD = VD/ν. Substituting numerical values, find Re D 2.17×10

Nu D

h D,b (W/m ⋅K)

q′cv ( W / m )

q′rad ( W / m )

q′b ( W / m )

82.5

30.1

1517

541

2058

4

2

<

(c) With 20-mm thickness insulation, and for the calm-day condition, the heat loss per unit length is

(

)

q′ins = Ts,o − T∞ / R ′tot

(1)

−1 R ′t = R ′ins + [1/ R ′cv + 1/ R ′rad ]

(2)

where the thermal resistance of the insulation from Eq. 3.28 is

R ′ins = n ( Do / Db ) / [2π k ]

(3)

and the convection and radiation thermal resistances are

( ) R ′rad = 1/ ( h rad π Do ) R ′cv = 1/ h D,oπ Do

(

)(

2 + T2 h rad,o = εσ Ts,o + T∞ Ts,o ∞

(4)

)

(5,6)

The outer surface temperature on the insulation, Ts,o, can be determined by an energy balance on the surface node of the thermal circuit.

Ts,b − Ts,o R ′ins

=

Ts,o − T∞

[1/ R ′cv + 1/ R ′rad ]−1

Substituting numerical values with Db,o = 129 mm, find the following results.

R ′ins = 0.7384 m ⋅ K / W

h D,o = 5.30 W / m 2 ⋅ K

R ′cv = 0.4655 K / W

h rad = 5.65 W / m 2 ⋅ K

R ′rad = 0.4371 K / W Ts,o = 62.1°C

q′ins = 187 W / m

< Continued …..

PROBLEM 9.58 (Cont.) COMMENTS: (1) For the calm-day conditions, the heat loss by radiation exchange is 58% of the total loss. Using a reflective shield (say, ε = 0.1) on the outer surface could reduce the heat loss by 50%. (2) The effect of a 8-m/s breeze over the steam line is to increase the heat loss by more than a factor of two above that for a calm day. The heat loss by radiation exchange is approximately 25% of the total loss. (3) The effect of the 20-mm thickness insulation is to reduce the heat loss to 20% the rate by free convection or to 9% the rate on the breezy day. From the results of part (c), note that the insulation resistance is nearly 3 times that due to the combination of the convection and radiation process ′ is the thermal resistances. The effect of increased wind speed is to reduce R ′cv , but since R ins dominant resistance, the effect will not be very significant. (4) Comparing the free convection coefficients for part (a), Db = 89 mm with Ts,b = 200°C, and part (b), Db,o = 129 mm with Ts,o = 62.1°C, it follows that h D,o is less than h D,b since, for the former, the steam line diameter is larger and the diameter smaller. (5) The convection correlation models in IHT are especially useful for applications such as the present one to eliminate the tediousness of evaluating properties and performing the calculations. However, it is essential that you have experiences in hand calculations with the correlations before using the software.

PROBLEM 9.59 KNOWN: Horizontal tube, 12.5mm diameter, with surface temperature 240°C located in room with an air temperature 20°C. FIND: Heat transfer rate per unit length of tube due to convection. SCHEMATIC:

ASSUMPTIONS: (1) Ambient air is quiescent, (2) Surface radiation effects are not considered. -6

2

PROPERTIES: Table A-4, Air (Tf = 400K, 1 atm): ν = 26.41 × 10 m /s, k= 0.0338 W/m⋅K, α = -6 2 -3 -1 38.3 × 10 m /s, Pr = 0.690, β = 1/Tf = 2.5 × 10 K . ANALYSIS: The heat rate from the tube, per unit length of the tube, is

q′ = h π D ( Ts − T∞ )

where h can be estimated from the correlation, Eq. 9.34, 2     0.387Ra1/6 D Nu D = 0.60 + . 8/27  9/16    

1 + ( 0.559/Pr ) 







From Eq. 9.25,

(

2 −3 −1 −3 gβ ( Ts − T∞ ) D3 9.8m/s × 2.5 ×10 K ( 240 − 20 ) K × 12.5 ×10 m Ra D = = να 26.41× 10−6 m 2 / s × 38.3 × 10−6 m 2 / s

)

3

= 10,410.

2

  1/6   0.387 (10,410 ) Hence, Nu D = 0.60 + = 4.40 8/27    1 + ( 0.559/0.690 )9/16      k 0.0338W/m ⋅ K h = Nu D = × 4.40 = 11.9W/m 2 ⋅ K. − 3 D 12.5 ×10 m The heat rate is

(

)

q′ = 11.9W/m 2 ⋅ K ×π 12.5 ×10 −3m ( 240 − 20 ) K = 103W/m.

<

COMMENTS: Heat loss rate by radiation, assuming an emissivity of 1.0 for the surface, is

(

)

(

)

q ′rad = ε Pσ Ts4 − T∞4 =1 × π 12.5 ×10 −3 m × 5.67 ×10− 8

q′rad = 138W/m.

( 240 + 273 )4 − ( 20 + 273 )4  K4   m ⋅K  W

2

4

Note that P = π D. Note also this estimate assumes the surroundings are at ambient air temperature. In this instance, q′rad > q′conv .

PROBLEM 9.60 KNOWN: Insulated steam tube exposed to atmospheric air and surroundings at 25°C. FIND: (a) Heat transfer rate by free convection to the room, per unit length of the tube; effect on quality, x, at outlet of 30 m length of tube; (b) Effect of radiation on heat transfer and quality of outlet flow; (c) Effect of emissivity and insulation thickness on heat rate. SCHEMATIC:

ASSUMPTIONS: (1) Ambient air is quiescent, (2) Negligible surface radiation (part a), (3) Tube wall resistance negligible. PROPERTIES: Steam tables, steam (sat., 4 bar): hf = 566 kJ/kg, Tsat = 416 K, hg = 2727 kJ/kg, hfg = 2160 kJ/kg, vg = 0.476 × 103 m3/kg; Table A.3, magnesia, 85% (310 K): km = 0.051 W/m⋅K; Table A.4, air (assume Ts = 60°C, Tf = (60 + 25)°C/2 = 315 K, 1 atm): ν = 17.4 × 10-6 m2/s, k = 0.0274 W/m⋅K, α = 24.7 × 10-6 m2/s, Pr = 0.705, Tf = 1/315 K = 3.17 × 10-3 K-1. ANALYSIS: (a) The heat rate per unit length of the tube (see sketch) is given as,

T −T q′ = i ∞ R ′t

where

D 1  1 1 1  ln 3 + = +  R ′t  h oπ D3 2π k m D 2 h iπ D1 

−1 (1,2)

To estimate h o , we have assumed Ts ≈ 60°C in order to calculate RaL from Eq. 9.25, 3 gβ ( Ts − T∞ ) D33 9.8 m s 2 × 3.17 × 10−3 K −1 ( 60 − 25 ) K ( 0.115 m ) Ra D = = = 3.85 × 106 . 6 2 6 2 − − να 17.4 × 10 m s × 24.7 × 10 m s

The appropriate correlation is Eq. 9.34; find 2

(

)

2

1/ 6     6 1/ 6 0.387 3.85 10 ×     0.387 ( Ra D )   Nu D = 0.60 + = 0.60 + = 21.4  8 / 27 8 / 27      1 + ( 0.559 Pr )9 /16  1 + (0.559 0.705 )9 /16         

ho =

k 0.0274 W m ⋅ K Nu D = × 21.4 = 5.09 W m 2 ⋅ K . D3 0.115 m

Substituting numerical values into Eq. (2), find

  1 1 115 1 = + + ln  R ′t  5.09 W m 2 ⋅ K × π 0.115 m 2π × 0.051 W m ⋅ K 65 11, 000 W m 2 ⋅ Kπ × 0.055 m  1

and from Eq. (1),

q′ = 0.430 W m ⋅ K ( 416 − 298 ) K = 50.8 W m

−1

= 0.430 W m ⋅ K

< Continued...

PROBLEM 9.60 (Cont.) We need to verify that the assumption of Ts = 60°C is reasonable. From the thermal circuit,

)

(

Ts = T∞ + q′ h o π D3 = 25$ C + 50.8 W m 5.09 W m 2 ⋅ K × π × 0.115 m = 53$ C . Another calculation using Ts = 53°C would be appropriate for a more precise result. Assuming q′ is constant, the enthalpy of the steam at the outlet (L = 30 m), h2, is

 = 2727 kJ kg − 50.8 W m × 30 m 14.97 kg s = 2625 kJ kg h 2 = h1 − q′ ⋅ L m  = ρ g Ac u m with ρg = 1 vg and Ac = π D12 4 . For negligible pressure drop, where m x = ( h 2 − h f ) h fg = ( 2625 − 566 ) kJ kg ( 2160 kJ kg ) = 0.953.

<

(b) With radiation, we first determine Ts by performing an energy balance at the outer surface, where

q′i = q′conv,o + q′rad

(

Ti − Ts 4 = h oπ D3 ( Ts − T∞ ) + π D3εσ Ts4 − Tsur R ′i

)

and

R ′i =

D 1 1 ln 3 + hiπ D1 2π k m D2

From knowledge of Ts, q′i = ( Ti − Ts ) R ′i may then be determined. Using the Correlations and Properties Tool Pads of IHT to determine h o and the properties of air evaluated at Tf = (Ts + T‡ )/2, the following results are obtained. Condition

Ts (°C)

q′i (W/m)

ε = 0.8, D3 = 115 mm ε = 0.8, D3 = 165 mm ε = 0.2, D3 = 115 mm ε = 0.2, D3 = 165 mm

41.8 33.7 49.4 38.7

56.9 37.6 52.6 35.9

COMMENTS: Clearly, a significant reduction in heat loss may be realized by increasing the insulation thickness. Although Ts, and hence q′conv,o , increases with decreasing ε, the reduction in q′rad is more than sufficient to reduce the heat loss.

PROBLEM 9.61 KNOWN: Dissipation rate of an electrical cable suspended in air. FIND: Surface temperature of the cable, Ts. SCHEMATIC:

ASSUMPTIONS: (1) Quiescent air, (2) Cable in horizontal position, (3) Negligible radiation exchange.

PROPERTIES: Table A-4, Air (Tf = (Ts + T∞)/2 = 325K, based upon initial estimate for Ts, 1 atm): -6 2 -6 2 ν = 18.41 × 10 m /s, k = 0.0282 W/m⋅K, α = 26.2 × 10 m /s, Pr = 0.704. ANALYSIS: From the rate equation on a unit length basis, the surface temperature is

Ts = T∞ + q ′ / π Dh where h is estimated by an appropriate correlation. Since such a calculation requires knowledge of Ts, an iteration procedure is required. Begin by assuming Ts = 77°C and calculated RaD, Ra D = g β∆T D3 / να where ∆T = Ts − T∞ and Tf = ( Ts + T∞ ) / 2 For air, β = 1/Tf, and substituting numerical values, Ra D = 9.8

m 2

(1/325K ) ( 77 − 27 ) K ( 0.025m )

3

/18.41 ×10

−6 m

2

s

s Using the Churchill-Chu relation, find h.

× 26.2 × 10

  1/6   0.387Ra D hD  Nu D = = 0.60 + 8/27  k   1 + ( 0.559/Pr ) 9/16     

(

)

1/6  0.387 4.884 × 10 4 0.0282W/m ⋅ K   h= 0.60 + 8/27 0.025m  1 + ( 0.559/0.704 )9/16    

−6m

2

s

(1,2,3)

= 4.884 × 10 4.

2 (4)

2

  2  = 7.28W/m ⋅ K.  

Substituting numerical values into Eq. (1), the calculated value for Ts is Ts = 27°C + (3 0 W / m ) / π × 0.025m × 7.28W/m 2 ⋅ K = 79.5°C. This value is very close to the assumed value (77°C), but an iteration with a new value of 79°C is warranted. Using the same property values, find for this iteration: Ra D = 5.08 ×10 4 h = 7.35W/m 2 ⋅ K Ts = 79°C. We conclude that Ts = 79°C is a good estimate for the surface temperature. COMMENTS: Recognize that radiative exchange is likely to be significant and would have the effect of reducing the estimate for Ts.

<

PROBLEM 9.62 KNOWN: Dissipation rate of an immersion heater in a large tank of water. FIND: Surface temperature in water and, if accidentally operated, in air. SCHEMATIC:

ASSUMPTIONS: (1) Quiescent ambient fluid, (2) Negligible radiative exchange. PROPERTIES: Table A-6, Water and Table A-4, Air: T(K) Water 315 Air 1500

3

k⋅10 (W/m⋅K) 634 100

7

2

ν⋅10 (µ/ρ,m /s) 6.25 2400

7

2

α⋅10 (k/ρcp ,m /s) 1.531 3500

Pr 4.16 0685

6

-1

β⋅10 (K ) 400.4 666.7

ANALYSIS: From the rate equation, the surface temperature, Ts, is

Ts = T∞ + q / ( π D L h )

(1)

where h is estimated by an appropriate correlation. Since such a calculation requires knowledge of Ts, an iteration procedure is required. Begin by assuming for water that Ts = 64°C such that Tf = 315K. Calculate the Rayleigh number,

Ra D =

3 2 −6 −1 gβ∆TD3 9.8m/s × 400.4 × 10 K ( 64 − 20 ) K ( 0.010m) = = 1.804 ×10 6. − 7 2 − 7 2 να 6.25 × 10 m / s ×1.531× 10 m / s

(2)

Using the Churchill-Chu relation, find   1/6   0.387Ra D hD  Nu D = = 0.60 + 8/27  k   1 + ( 0.559/Pr ) 9/16     

(

)

2

(3)

2

 6 1/6   0.387 1.804 × 10  0.634W/m ⋅ K  h= 0.60 + = 1301W/m 2 ⋅ K.  8/27  0.01m   1 + ( 0.559/4.16 ) 9/16      Substituting numerical values into Eq. (1), the calculated value for Ts in water is

Ts = 20° C + 550W/ π × 0.010m × 0.30m ×1301W/m 2 ⋅ K = 64.8°C.

< Continued …..

PROBLEM 9.62 (Cont.) Our initial assumption of Ts = 64°C is in excellent agreement with the calculated value. With accidental operation in air, the heat transfer coefficient will be nearly a factor of 100 less. Suppose h ≈ 2 5 W / m 2 ⋅ K, then from Eq. (1), T ≈ 2360°C. Very likely the heater will burn out. s

2

Using air properties at Tf ≈ 1500K and Eq. (2), find RaD = 1.815 × 10 . Using Eq. 9.33, Nu D = CRa nD with C= 0.85 and n = 0.188 from Table 9.1, find h = 22.6W/m 2 ⋅ K. Hence, our first estimate for the surface temperature in air was reasonable,

Ts ≈ 2300°C.

<

However, radiation exchange will be the dominant mode, and would reduce the estimate for Ts. Generally such heaters could not withstand operating temperatures above 1000°C and safe operation in air is not possible.

PROBLEM 9.63 KNOWN: Motor shaft of 20-mm diameter operating in ambient air at T∞ = 27°C with surface temperature Ts ≤ 87°C. FIND: Convection coefficients and/or heat removal rates for different heat transfer processes: (a) For a rotating horizontal cylinder as a function of rotational speed 5000 to 15,000 rpm using the recommended correlation, (b) For free convection from a horizontal stationary shaft; investigate whether mixed free and forced convection effects for the range of rotational speeds in part (a) are significant using the recommended criterion; (c) For radiation exchange between the shaft having an emissivity of 0.8 and the surroundings also at ambient temperature, Tsur = T∞ ; and (d) For cross flow of ambient air over the stationary shaft, required air velocities to remove the heat rates determined in part (a). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Shaft is horizontal with isothermal surface. PROPERTIES: Table A.4, Air (Tf = (Ts + T∞ )/2 = 330 K, 1 atm): ν = 18.91 × 10-6 m2/s , k = 0.02852 W/m⋅K, α = 26.94 × 10-6 m2/s, Pr = 0.7028, β = 1/Tf . ANALYSIS: (a) The recommended correlation for the a horizontal rotating shaft is 2 / 3 Pr1/ 3 Nu D = 0.133Re D ReD < 4.3 ×105 0.7 < Pr < 670 where the Reynolds number is ReD = ΩD2 ν and Ω ( rad s ) is the rotational velocity. Evaluating properties at Tf = (Ts + T∞ )/2, find for ω = 5000 rpm, 2 ReD = (5000rpm × 2π rad rev / 60s min )( 0.020m ) 18.91× 10−6 m2 s = 11, 076

Nu D = 0.133 (11, 076 )

23

(0.7028 )1/ 3 = 58.75

h D,rot = Nu D k D = 58.75 × 0.02852 W m ⋅ K 0.020m = 83.8 W m 2 ⋅ K

<

The heat rate per unit shaft length is

q′rot = h D,rot (π D )( Ts − T∞ ) = 83.8 W m 2 ⋅ K (π × 0.020m )(87 − 27 ) C = 316 W m $

<

The convection coefficient and heat rate as a function of rotational speed are shown in a plot below. (b) For the stationary shaft condition, the free convection coefficient can be estimated from the Churchill-Chu correlation, Eq. (9.34) with Continued...

PROBLEM 9.63 (Cont.) Ra D = Ra D =

gβ∆TD3 να 3 9.8 m s 2 (1 330K )(87 − 27 ) K ( 0.020m )

18.91 × 10−6 m 2 s × 26.94 × 10−6 m 2 s

  6 0.387Ra1/   D Nu D = 0.60 + 8 / 27    1 + ( 0.559 Pr )9 /16     

= 27, 981

2

2

  1/ 6   0.387 ( 27, 981) Nu D = 0.60 + = 5.61 8 / 27  9 /16   1 + ( 0.559 0.7028 )      h D,fc = Nu D k D = 5.61× 0.02852 W m ⋅ K 0.020m = 8.00 W m 2⋅ K q′fc = 8.00 W m 2⋅ K (π × 0.020m )(87 − 27 ) C = 30.2 W m $

<

Mixed free and forced convection effects may be significant if 3 Pr 0.137 ReD < 4.7 GrD

(

)

where GrD = RaD/Pr, find using results from above and in part (a) for ω = 5000 rpm, 0.137 3

11, 076 ? < ? 4.7 ( 27,981 0.7028 ) 0.7018  

= 383

We conclude that free convection effects are not significant for rotational speeds above 5000 rpm. (c) Considering radiation exchange between the shaft and the surroundings, 4 h rad = εσ (Ts + Tsur ) Ts2 + Tsur

)

(

)

<

q′rad = 6.57 W m 2 ⋅ K (π × 0.020m )(87 − 27 ) K = 24.8 W m

<

(

h rad = 0.8 × 5.67 × 10−8 W m 2⋅ K (360 + 300 ) 3602 + 3002 K 3 = 6.57 W m 2⋅ K and the heat rate by radiation exchange is

q′rad = h rad (π D )( Ts − Tsur )

(d) For cross flow of ambient air at a velocity V over the shaft, the convection coefficient can be estimated using the Churchill-Bernstein correlation, Eq. 7.57, with Re D ,cf =

VD ν

Nu D,cf = h D,cf D k = 0.3 +

0.62 Re1D/ ,2cf Pr1/ 3

0

1 + 0.4 Pr

5

2 / 3 1/ 4

1 +  Re    !  282,000  D , cf

"# #$

5/ 8 4 / 5

Continued...

PROBLEM 9.63 (Cont.) From the plot below (left) for the rotating shaft condition of part (a), h D,rot vs. rpm, note that the convection coefficient varies from approximately 75 to 175 W/m2 ⋅ K. Using the IHT Correlations Tool, Forced Convection, Cylinder, which is based upon the above relations, the range of air velocities V required to achieve h D,cf in the range 75 to 175 W/m2 ⋅ K was computed and is plotted below

200

50

150

40

Air velocity, V (m/s)

Coefficient or heat rate

(right).

100

50

30

20

10 0 5000

10000 Rotational speed, rpm (rev/min)

15000 0 75

Convection coefficient, h (W/m^2.K) Heat rate, q'*10^-1 (W/m)

100

125

150

175

Convection coefficient, hDbar (W/m^2.K)

Note that the air cross-flow velocities are quite substantial in order to remove similar heat rates for the rotating shaft condition. COMMENTS: We conclude for the rotational speed and surface temperature conditions, free convection effects are not significant. Further, radiation exchange, part (c) result, is less than 10% of the convection heat loss for the lowest rotational speed condition.

PROBLEM 9.64 KNOWN: Horizontal pin fin of 6-mm diameter and 60-mm length fabricated from plain carbon steel (k = 57 W/m⋅K, ε = 0.5). Fin base maintained at Tb = 150°C. Ambient air and surroundings at 25°C. FIND: Fin heat rate, qf, by two methods: (a) Analytical solution using average fin surface temperature of

Ts = 125$ C to estimate the free convection and linearized radiation coefficients; comment on sensitivity of fin heat rate to choice of Ts ; and, (b) Finite-difference method when coefficients are based upon local temperatures, rather than an average fin surface temperature; compare result of the two solution methods. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in the pin fin, (3) Ambient air is quiescent and extensive, (4) Surroundings are large compared to the pin fin, and (5) Fin tip is adiabatic.

(

)

PROPERTIES: Table A.4, Air (Tf = Ts + T∞ 2 = 348 K): ν = 20.72 × 10-6 m2/s, k = 0.02985 W/m⋅K, α = 29.60 × 10-6 m2/s, Pr = 0.7003, β = 1/Tf. ANALYSIS: (a) The heat rate for the pin fin with an adiabatic tip condition is, Eq. 3.76, q f = M tanh ( mL ) M = ( h tot PkA c )

1/ 2

P = πD

(1) m = ( hP kAc )

1/ 2

θb

Ac = π D2 4

θ b = Tb − T∞

(2,3) (4-6)

and the average coefficient is the sum of the convection and linearized radiation processes, respectively, h tot = h fc + h rad

(7)

(

evaluated at Ts = 125$ C with Tf = Ts + T∞

)

2 = 75$ C = 348 K .

Estimating h fc : For the horizontal cylinder, Eq. 9.34, with Ra D =

gβ∆TD3

να Continued …..

PROBLEM 9.64 (Cont.) Ra D =

3 9.8 m s 2 (1 348K )(125 − 25 )( 0.006m )

20.72 × 10−6 m 2 s × 29.60 × 10−6 m 2 s 2

= 991.79

  1/ 6   0.387 Ra D Nu D = 0.60 + 8 / 27    1 + ( 0.559 Pr )9 /16     

2

  1/ 6   0.387 (991.79 ) Nu D = 0.60 + = 2.603 8 / 27    1 + ( 0.559 0.7003)9 /16      h fc = Nu D k D = 2.603 × 0.02985 W m ⋅ K 0.006m = 12.95 W m 2⋅ K Calculating h rad : The linearized radiation coefficient is

(

2 h rad = εσ ( Ts + Tsur ) Ts2 + Tsur

)

(8)

)

(

h rad = 0.5 × 5.67 × 10−8 W m 2 ⋅ K 4 (398 + 298 ) 3982 + 2982 K3 = 4.88 W m 2⋅ K Substituting numerical values into Eqs. (1-7) , find

<

qfin = 2.04W

with θ b = 125 K, A c = 2.827 × 10−5 m 2 , P = 0.01885 m, m = 2.603m −1, M = 2.909 W, and h tot = 17.83 W m 2⋅ K . Using the IHT Model, Extended Surfaces, Rectangular Pin Fin, with the Correlations Tool for Free Convection and the Properties Tool for Air, the above analysis was repeated to obtain the following results.

( C) $

115

120

125

130

135

qf ( W )

1.989

2.012

2.035

2.057

2.079

-2.3

-1.1

0

+1.1

+2.2

Ts

(qf − qf ,o )

q fo (%)

The fin heat rate is not very sensitive to the choice of Ts for the range Ts = 125 ± 10 °C. For the base case condition, the fin tip temperature is T(L) = 114 °C so that Ts ≈ (T(L) + Tb ) /2 = 132°C would be consistent assumed value. Continued …..

PROBLEM 9.64 (Cont.) (b) Using the IHT Tool, Finite-Difference Equation, Steady- State, Extended Surfaces, the temperature distribution was determined for a 15-node system from which the fin heat rate was determined. The local free convection and linearized radiation coefficients h tot = h fc + h rad, were evaluated at local temperatures, Tm , using IHT with the Correlations Tool, Free Convection, Horizontal Cylinder, and the Properties Tool for Air, and Eq. (8). The local coefficient htot vs. Ts is nearly a linear function for the range 114 ≤ Ts ≤ 150°C so that it was reasonable to represent htot (Ts) as a Lookup Table Function. The fin heat rate follows from an energy balance on the base node, (see schematic next page)

qf = qa + q b = ( 0.08949 + 1.879 ) W = 1.97 W

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qa = h b ( P∆x 2 )( Tb − T∞ ) q b = kAc ( Tb − T1 ) ∆x where Tb = 150°C, T1 = 418.3 K = 145.3°C, and hb = htot (Tb) = l8.99 W m2 ⋅ K .

Considering variable coefficients, the fin heat rate is -3.3% lower than for the analytical solution with the assumed Ts = 125°C. COMMENTS: (1) To validate the FDE model for part (b), we compared the temperature distribution and fin heat rate using a constant htot with the analytical solution ( Ts = 125°C). The results were identical indicating that the 15-node mesh is sufficiently fine. (2) The fin temperature distribution (K) for the IHT finite-difference model of part (b) is Tb 423

T01 418.3`

T02 414.1

T03 410.3

T04 406.8

T05 403.7

T06 401

T07 398.6

T08 396.6

T09 394.9

T10 393.5

T11 392.4

T12 391.7

T13 391.2

T14 391

T15 390.9

PROBLEM 9.65 KNOWN: Diameter, thickness, emissivity and thermal conductivity of steel pipe. Temperature of water flow in pipe. Cost of producing hot water. FIND: Cost of daily heat loss from an uninsulated pipe. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Negligible convection resistance for water flow, (3) Negligible radiation from pipe surroundings, (4) Quiescent air, (5) Constant properties. -6

2

PROPERTIES: Table A-4, air (p = 1 atm, Tf ≈ 295K): ka = 0.0259 W/m⋅K. ν = 15.45 × 10 m /s, -6 2 -3 -1 α = 21.8 × 10 m /s, Pr = 0.708, β = 3.39 × 10 K . ANALYSIS: Performing an energy balance for a control surface about the outer surface, q ′cond = q ′conv + q ′rad , it follows that

T − Ts,o R ′cond

(

)

4 = hπ Do Ts,o − T∞ + ε pπ Doσ Ts,o

(1)

where R ′cond = "n ( D o / Di ) / 2π k p = "n (100 / 84 ) / 2π ( 60 W / m ⋅ K ) = 4.62 × 10 −4 m ⋅ K / W. The convection coefficient may be obtained from the Churchill and Chu correlation. Hence, with RaD =

gβ (Ts,o - T∞) D3o / αν = 9.8 m / s 2 × 3.39 × 10−3 K −1 ( 0.1m )3 ( Ts,o − 268K ) / = 98, 637

(Ts,o − 268) ,

(21.8 × 15.45 × 10

−12

4

m /s

2

)

2

Nu D

h=

  1/ 6   0.387 Ra D 1/ 6 = 0.60 +  = 0.60 + 2.182 ( Ts,o − 268 ) 8 / 27   1 + ( 0.559 / Pr )9 /16     

ka Do

{

2

{

(

Nu D = 0.259 W / m ⋅ K 0.60 + 2.182 Ts,o − 268

}

2

)1/ 6 }

2

Substituting the foregoing expression for h , as well as values of R ′cond , D o , ε p and σ into Eq. (1), an iterative solution yields

Ts,o = 322.9 K = 49.9°C

It follows that h = 6.10 W / m 2 ⋅ K, and the heat loss per unit length of pipe is 2 2 4 4 −8 q ′ = q ′conv + q ′rad = 6.10 W / m ⋅ K (π × 0.1m ) 54.9K + 0.6 (π × 0.1m ) 5.67 × 10 W / m ⋅ K (322.9K )

= (105.2 + 116.2 ) W / m = 221.4 W / m

The corresponding daily energy loss is Q′ = 0.221kW / m × 24 h / d = 5.3kW ⋅ h / m ⋅ d

C′ = (5.3kW ⋅ h / m ⋅ d )($0.05 / kW ⋅ h ) = $0.265 / m ⋅ d and the associated cost is COMMENTS: (1) The heat loss is significant, and the pipe should be insulated. (2) The conduction resistance of the pipe wall is negligible relative to the combined convection and radiation resistance at the outer surface. Hence, the temperature of the outer surface is only slightly less than that of the water.

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PROBLEM 9.66 KNOWN: Insulated, horizontal pipe with aluminum foil having emissivity which varies from 0.12 to 0.36 during service. Pipe diameter is 300 mm and its surface temperature is 90°C. FIND: Effect of emissivity degradation on heat loss with ambient air at 25°C and (a) quiescent conditions and (b) cross-wind velocity of 5 m/s. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Surroundings are large compared to pipe, (3) Pipe has uniform temperature. -6

PROPERTIES: Table A-4, Air (Tf = (90 + 25)°C/2 = 330K, 1 atm): ν = 18.9 × 10 -3 -6 2 28.5 × 10 W/m⋅K, α = 26.9 × 10 m /s, Pr = 0.703.

2

m /s, k =

ANALYSIS: The heat loss per unit length from the pipe is

(

4 q′ = hP ( Ts − T∞ ) + εσ P Ts4 − Tsur

)

where P = πD and h needs to be evaluated for the two ambient air conditions. (a) Quiescent air. Treating the pipe as a horizontal cylinder, find gβ ( Ts − T∞ ) D3 9.8m/s2 (1/330K )( 90 − 25 ) K ( 0.30m )3 Ra D = = = 1.025 ×108 − 6 2 − 6 2 να 18.9 × 10 m / s × 26.9 × 10 m / s -5 12 and using the Churchill-Chu correlation for 10 < ReD < 10 .   1/6   0.387Ra D Nu D = 0.60 + 8/27    1 + ( 0.559/Pr )9/16     

(

)

2

2

 1/6   0.387 1.025 ×108  Nu D =  0.60 + = 56.93 8/27    1 + ( 0.559/0.703 )9/16      hD = Nu D k / D = 56.93× 0.0285 W / m ⋅ K/0.300m = 5.4W/m 2 ⋅ K. Continued …..

PROBLEM 9.66 (Cont.) Hence, the heat loss is

(

)

q ′ = 5.4W/m 2 ⋅ K ( π 0.30m )( 90 − 25 ) K + ε × 5.67 ×10 −8 W / m 2 ⋅ K ( π 0.300m ) 363 4 − 2984 K 4

ε = 0.12 → q′ = ( 331 + 61) = 3 9 2 W / m q′ = 331 + 506ε  ε = 0.36 → q′ = ( 331 + 182 ) = 5 1 3 W / m

< <

The radiation effect accounts for 16 and 35%, respectively, of the heat rate. (b) Cross-wind condition. With a cross-wind, find Re D =

VD 1 0 m / s × 0.30m = = 1.587 ×10 5 − 6 2 ν 18.9 × 10 m / s

and using the Hilpert correlation where C = 0.027 and m = 0.805 from Table 7.2,

(

1/3 = 0.027 1.587 ×105 Nu D = CRem D Pr

)

0.805

( 0.703)1/3 = 368.9

hD = NuD ⋅ k / D = 368.9 × 0.0285W/m ⋅ K/0.300m = 3 5 W / m 2 ⋅ K. Recognizing that combined free and forced convection conditions may exist, from Eq. 9.64 with n = 3, Nu3m = Nu3F + Nu3N

(

hm = 5.43 + 353

)

1/3

= 3 5 W / m2 ⋅ K

we find forced convection dominates. Hence, the heat loss is

(

)

q ′ = 3 5 W / m2 ⋅ K ( π 0.300m)( 90 − 25 ) K + ε × 5.67 ×10 − 8 W / m2 ⋅ K (π 0.300m) 393 4 − 298 4 K 4

{

0.12 → q ′ = 2144 + 102 = 2246W/m q′ = 2144 + 853ε εε = = 0.36 → q ′ = 2144 + 307 = 2451W/m

< <

The radiation effect accounts for 5 and 13%, respectively, of the heat rate. COMMENTS: (1) For high velocity wind conditions, radiation losses are quite low and the degradation of the foil is not important. However, for low velocity and quiescent air conditions, radiation effects are significant and the degradation of the foil can account for a nearly 25% change in heat loss. 2

(2) The radiation coefficient is in the range 0.83 to 2.48 W/m ⋅K for ε = 0.12 and 0.36, respectively. Compare these values with those for convection.

PROBLEM 9.67 KNOWN: Diameter, emissivity, and power dissipation of cylindrical heater. Temperature of ambient air and surroundings. FIND: Steady-state temperature of heater and time required to come within 10°C of this temperature. SCHEMATIC:

ASSUMPTIONS: (1) Air is quiescent, (2) Duct wall forms large surroundings about heater, (3) Heater may be approximated as a lumped capacitance. PROPERTIES: Table A.4, air (Obtained from Properties Tool Pad of IHT). ANALYSIS: Performing an energy balance on the heater, the final (steady-state) temperature may be obtained from the requirement that q′ = q′conv + q′rad , or

q′ = h (π D )( T − T∞ ) + h r (π D )( T − Tsur )

(

)

2 . Using the Correlations Tool where h is obtained from Eq. 9.34 and hr = εσ ( T + Tsur ) T 2 + Tsur Pad of IHT to evaluate h , this expression may be solved to obtain T = 854 K = 581°C

<

 ′ = q′ − q′ Under transient conditions, the energy balance is of the form, E st conv − q′rad , or

(

)

ρ cp π D2 4 dT dt = q′ − h (π D )(T − T∞ ) − h r (π D )(T − Tsur ) Using the IHT Lumped Capacitance model with the Correlations Tool Pad, the above expression is integrated from t = 0, for which Ti = 562.4 K, to the time for which T = 844 K. The integration yields t = 183s

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The value of Ti = 562.4 K corresponds to the steady-state temperature for which the power dissipation is balanced by convection and radiation (see solution to Problem 7.44). COMMENTS: The forced convection coefficient (Problems 7.43 and 7.44) of 105 W/m2⋅K is much larger than that associated with free convection for the steady-state conditions of this problem (14.6 W/m2⋅K). However, because of the correspondingly larger heater temperature, the radiation coefficient with free convection (42.9 W/m2⋅K) is much larger than that associated with forced convection (15.9 W/m2⋅K).

PROBLEM 9.68 KNOWN: Cylindrical sensor of 12.5 mm diameter positioned horizontally in quiescent air at 27°C. FIND: An expression for the free convection coefficient as a function of only ∆T = Ts - T∞ where Ts is the sensor temperature. ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform temperature over cylindrically shaped sensor, (3) Ambient air extensive and quiescent. PROPERTIES: Table A-4, Air (Tf, 1 atm): β = 1/Tf and Ts (°C) 30 55 80

Tf (K) 302 314 327

6

2

ν × 10 m /s

6

2

α × 10 m /s

16.09 17.30 18.61

3

k × 10 W/m⋅K

22.8 24.6 26.5

Pr

26.5 27.3 28.3

0.707 0.705 0.703

ANALYSIS: For the cylindrical sensor, using Eqs. 9.25 and 9.34,

gβ∆TD3 Ra D = να

  1/6   0.387Ra hD D  D Nu D = = 0.60 + 8/17  k   1 + ( 0.559/Pr ) 9/16     

2 (1,2)

where properties are evaluated at (Tf = Ts + T∞)/2. With 30 ≤ Ts ≤ 80°C and T∞ = 27°C, 302 ≤ Tf ≤ 326 K. Using properties evaluated at the mid-range of Tf, Tf = 314K, find

Ra D =

9.8m/s2 ( 1/314K) ∆ T ( 0.0125m )3

17.30 ×10−6 m 2 / s × 24.6 ×10−6 m 2 / s

 0.0273W/m ⋅ K  hD =  0.60 + 0.0125m  

{

= 143.2 ∆T

 1/6  0.387 (143∆T ) 8/27   1 + ( 0.559/0.705 )9/16    

hD = 2.184 0.60 + 0.734∆ T1/6

2

}. 2

(3)

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COMMENTS: (1) The effect of using a fixed film temperature, Tf = 314K = 41°C, for the full range 30 ≤ Ts ≤ 80°C can be seen by comparing results from the approximate Eq. (3) and the correlation, Eq. (2), with the proper film temperature. The results are summarized in the table. Correlation _____________________________

(

Ts (°C)

∆T = Ts - T∞ (°C)

RaD

Nu D

h D W / m2 ⋅ K

30 55

3 28

518 4011

2.281 3.534

4.83 7.72

)

Eq. (3)

(

h D W / m2 ⋅ K

)

4.80 7.71

The approximate expression for h D is in excellent agreement with the correlation. (2) In calculating heat rates it may be important to consider radiation exchange with the surroundings.

PROBLEM 9.69 KNOWN: Thin-walled tube mounted horizontally in quiescent air and wrapped with an electrical tape passing hot fluid in an experimental loop. FIND: (a) Heat flux q′′e from the heating tape required to prevent heat loss from the hot fluid when (a) neglecting and (b) including radiation exchange with the surroundings, (c) Effect of insulation on q′′e and convection/radiation rates. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Ambient air is quiescent and extensive, (3) Surroundings are large compared to the tube. PROPERTIES: Table A.4, Air (Tf = (Ts + T∞ )/2 = (45 + 15)°C/2 = 303 K, 1 atm): ν = 16.19 × 10-6 m2/s, α = 22.9 × 10-6 m2/s, k = 26.5 × 10-3 W/m⋅K, Pr = 0.707, β = 1/Tf. ANALYSIS: (a,b) To prevent heat losses from the hot fluid, the heating tape temperature must be maintained at Tm; hence Ts,i = Tm. From a surface energy balance,

(

q′′e = q′′conv + q′′rad = hDi + h r

) (Ts,i − T∞ )

(

where the linearized radiation coefficient, Eq. 1.9, is h r = εσ Ts,i + T∞

(

)

) (Ts,i2 + T∞2 ) , or

h r = 0.95 × 5.67 × 10−8 W m 2 ⋅ K 4 (318 + 288 ) 3182 + 2882 K 3 = 6.01W m 2 ⋅ K . Neglecting radiation: For the horizontal cylinder, Eq. 9.34 yields

(

)

gβ Ts,i − T∞ Di3 9.8 m s 2 (1 303 K )( 45 − 15 ) K ( 0.020 m )3 Ra D = = = 20, 900 να 16.19 × 10−6 m 2 s × 22.9 × 10−6 m 2 s

  1/ 6 h D Di   0.386Ra  D i Nu D = = 0.60 + 8 / 27  k   1 + ( 0.559 Pr )9 /16     

2

Continued ….

PROBLEM 9.69 (Cont.) 2

  1/ 6   0.386 ( 20, 900 ) 0.0265 W m ⋅ K  2 hD = 0.60 +   = 6.90 W m ⋅ K i 8 / 27 0.020 m   1 + ( 0.559 0.707 )9 /16      Hence, neglecting radiation, the required heat flux is q′′e = 6.90 W m 2 ⋅ K ( 45 − 15 ) K = 207 W m 2 ⋅ K

<

Considering radiation: The required heat flux considering radiation is q′′e = ( 6.90 + 6.01) W m 2 ⋅ K ( 45 − 15 ) K = 387 W m 2 ⋅ K

<

(c) With insulation, the surface energy balance must be modified to account for an increase in the outer diameter from Di to Do = Di + 2t and for the attendant thermal resistance associated with conduction across the insulation. From an energy balance at the inner surface of the insulation,

q′′e (π Di ) = q′cond =

(

2π k i Tm − Ts,o ln ( Do Di )

)

and from an energy balance at the outer surface,

(

q′cond = q′conv + q′rad = π Do h Do + h r

)(Ts,o − T∞ )

The foregoing expressions may be used to determine Ts,o and q′′e as a function of t, with the IHT Correlations and Properties Tool Pads used to evaluate h D . The desired results are plotted as follows. o

350

25

20 Heat rate (W/m)

Required heat flux, qe''(W/m^2)

300

250

200

15

10

5 150

0 0

0.004

0.008

0.012

0.016

0.02

Insulation thickness, t(m)

100 0

0.004

0.008

0.012

Insulation thickness, t(m)

0.016

0.02

Total Convection Radiation

By adding 20 mm of insulation, the required power dissipation is reduced by a factor of approximately 3. Convection and radiation heat rates at the outer surface are comparable. COMMENTS: Over the range of insulation thickness, Ts,o decreases from 45°C to 20°C, while h D o and hr decrease from 6.9 to 3.5 W/m2⋅K and from 3.8 to 3.3 W/m2⋅K, respectively.

PROBLEM 9.70 KNOWN: A billet of stainless steel AISI 316 with a diameter of 150 mm and length 500 mm emerges from a heat treatment process at 200°C and is placed into an unstirred oil bath maintained at 20°C. FIND: (a) Determine whether it is advisable to position the billet in the bath with its centerline horizontal or vertical in order decrease the cooling time, and (b) Estimate the time for the billet to cool to 30°C for the better positioning arrangement. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions for part (a), (2) Oil bath approximates a quiescent fluid, (3) Consider only convection from the lateral surface of the cylindrical billet; and (4) For part (b), the billet has a uniform initial temperature. PROPERTIES: Table A-5, Engine oil (Tf = (Ts + T∞)/2): see Comment 1. Table A-1, AISI 316 3 (400 K): ρ = 8238 kg/m , cp = 468 J/kg⋅K, k = 15 W/m⋅K. ANALYSIS: (a) For the purpose of determining whether the horizontal or vertical position is preferred for faster cooling, consider only free convection from the lateral surface. The heat loss from the lateral surface follows from the rate equation

q = h As ( Ts − T∞ )

Vertical position. The lateral surface of the cylindrical billet can be considered as a vertical surface of height L, width P = πD, and area As = PL. The Churchill-Chu correlation, Eq. 9.26, is appropriate to estimate h L , 2

  1/ 6   0.387 Ra L h L  Nu L = L = 0.825 + 8 / 27  k   1 + ( 0.492 / Pr )9 /16     

Ra L =

gβ ( Ts − T∞ ) L3

να

with properties evaluated at Tf = (Ts + T∞)/2. Horizontal position. In this position, the billet is considered as a long horizontal cylinder of diameter D for which the Churchill-Chu correlation of Eq. 9.34 is appropriate to estimate h D , 2

  1/ 6   0.387 Ra h D  D Nu L = D = 0.60 +  8 / 27 k   1 + ( 0.55 / Pr )9 /16     

Continued …..

PROBLEM 9.70 (Cont.) Ra D =

gβ (Ts − T∞ ) D3

να

with properties evaluated at Tf. The heat transfer area is also As = PL. Using the foregoing relations in IHT with the thermophysical properties library as shown in Comment 1, the analysis results are tabulated below.

Ra L = 1.36 ×1011

Nu L = 801

h L = 218 W / m 2 ⋅ K

(vertical)

Ra D = 3.67 ×109

Nu D = 245

h D = 221 W / m 2 ⋅ K

(horizontal)

Recognize that the orientation has a small effect on the convection coefficient for these conditions, but we’ll select the horizontal orientation as the preferred one. (b) Evaluate first the Biot number to determine if the lumped capacitance method is valid.

h ( D / 2 ) 221 W / m 2 ⋅ K (0.150 m / 2 ) Bi = D o = = 1.1 k 15 W / m ⋅ K Since Bi >> 0.1, the spatial effects are important and we should use the one-term series approximation for the infinite cylinder, Eq. 5.49. Since h D will decrease as the billet cools, we need to estimate an average value for the cooling process from 200°C to 30°C. Based upon the analysis summarized in Comment 1, use h D = 119 W / m 2 ⋅ K. Using the transient model for the infinite cylinder in IHT, (see Comment 2) find for T(ro, to) = 30°C,

<

t o = 3845 s = 1.1 h

COMMENTS: (1) The IHT code using the convection correlation functions to estimate the coefficients is shown below. This same code was used to calculate h D for the range 30 ≤ Ts ≤ 200°C 2

and determine that an average value for the cooling period of part (b) is 119 W/m ⋅K. /* Results - convection coefficients, Ts = 200 C hDbar hLbar D L Tinf_C Ts_C 221.4 217.5 0.15 0.5 20 200

*/

/* Results - correlation parameters, Ts = 200 C NuDbar NuLbar Pr RaD RaL 244.7 801.3 219.2 3.665E9 1.357E11 */ /* Results - properties, Ts = 200 C; Tf = 383 K Pr alpha beta deltaT k nu Tf 219.2 7.188E-8 0.0007 180 0.1357 1.582E-5 383 /* Correlation description: Free convection (FC), long horizontal cylinder (HC), 10^-5 3 × 10 ), k k h = Nu L ⋅ =  0.069Ra1/3 Pr 0.074  L   L L 0.0272W/m ⋅ K  5 1/3 0.705 0.074  = 2.13W/m2 ⋅ K h= ( )  0.069 3.417 × 10  0.060m  

(

(4)

)

q = 2.13W/m 2 ⋅ K × 3m 2 ( 50 − 29 ) K = 134W.

< 3

5

(b) For separation distance L = 10mm, from Eq. (3) it follows that RaL = (10/60) × 3.417 × 10 = 1582. Since RaL < 1700, heat transfer occurs by conduction only, such that hL 0.0272W/m ⋅ K Nu L = =1 or h = 1× = 2.72W/m 2 ⋅ K k 0.010m q = 2.72W/m 2 ⋅ K × 3m 2 ( 50 − 29 ) K = 171W.

<

COMMENTS: Note that as L increases, from 10mm to 60mm, the heat rate decreases. However, For L ≥ 60mm, the heat rate will not change. This follows from Eq. 4 which, for Ra > 3 × 5 10 , h is independent of separation distance L.

PROBLEM 9.93 KNOWN: Rectangular cavity of two parallel, 0.5m square plates with insulated inter-connecting sides and with prescribed separation distance and surface temperatures. FIND: Heat flux between surfaces for three orientations of the cavity: (a) Vertical τ = 90°C, (b) Horizontal with τ = 0°, and (c) Horizontal with τ = 180°. SCHEMATIC:

ASSUMPTIONS: (1) Radiation exchange is negligible, (2) Air is at atmospheric pressure. -6

PROPERTIES: Table A-4, Air (Tf = (T1 + T2)/2 = 300K, 1 atm): ν = 15.89 × 10 -6

0.0263 W/m⋅K, α = 22.5 × 10

2

-3

m /s, Pr = 0.707, β = 1/Tf = 3.333 × 10

2

m /s, k =

-1

K .

ANALYSIS: The convective heat flux between the two cavity plates is

q′′conv = h ( T1 − T2 )

where h is estimated from the appropriate enclosure correlation which will depend upon the Rayleigh number. From Eq. 9.25, find g β (T1 −T 2 ) L3 9.8m/s 2 × 3.333 × 10−3 K− 1 ( 325 − 275 ) K ( 0.05m )3 Ra L = = = 5.710 ×105. − 6 2 − 6 2 να 15.89 × 10 m / s × 22.5 × 10 m / s Note that H/L = 0.5/0.05 = 10, a factor which is important in selecting correlations. (a) With τ = 90°, for a vertical cavity, Eq. 9.50, is appropriate, 0.28

−1 / 4

 Pr Ra   H   0.707 × 5.71 × 10 5  = 0.22     L  0.22 + Pr  L   0.22 + 0.707  k 0.0263W/m ⋅ K hL = Nu L = × 4.692 = 2.47W/m 2 ⋅ K L 0.05m q′′conv = 2.47W/m 2 ⋅ K ( 325 − 275 ) K = 123W/m 2.

Nu L = 0.22 

0.28

( 10 ) −1 / 4 = 4.692

<

(b) With τ = 0* for a horizontal cavity heated from below, Eq. 9.49 is appropriate.

h=

(

)

1/3 k k 0.0263W/m ⋅ K Nu L = 0.069 Ra1/3 Pr 0.074 = 0.069 5.710 ×105 ( 0.707 )0.074 L L L 0.05m 2 h = 2.92W/m ⋅ K

q′′conv = 2.92W/m 2 ⋅ K ( 325 − 275) K = 146W/m 2 .

<

(c) For τ = 180° corresponding to the horizontal orientation with the heated plate on the top, heat transfer will be by conduction. That is, k Nu L = 1 or h L = Nu L ⋅ = 1 × 0.0263W/m ⋅ K / ( 0.05m ) = 0.526W/m 2 ⋅ K.

L

q′′conv = 0.526W/m ⋅ K ( 325 − 275 ) K = 26.3W/m 2 . 2

COMMENTS: Compare the heat fluxes for the various orientations and explain physically their relative magnitudes.

<

PROBLEM 9.94 KNOWN: Horizontal flat roof and vertical wall sections of same dimensions exposed to identical temperature differences. FIND: (a) Ratio of convection heat rate for horizontal section to that of the vertical section and (b) Effect of inserting a baffle at the mid-height of the vertical wall section on the convection heat rate. SCHEMATIC:

ASSUMPTIONS: (1) Ends of sections and baffle adiabatic, (2) Steady-state conditions. PROPERTIES: Table A-4, Air ( T = (T1 + T2 ) / 2 = 277K,1atm ) : ν = 13.84 × 10 m /s, k = -6

-6

2

2

0.0245 W/m⋅K, α = 19.5 × 10 m /s, Pr = 0.713. ANALYSIS: (a) The ratio of the convection heat rates is

q hor h hor As ∆T h hor = = . q vert h vert As∆ T h vert

(1)

To estimate coefficients, recognizing both sections have the same characteristics length, L = 0.1m, with 3 RaL = gβ∆TL /να find

Ra L =

9.8m/s2 × (1/277K ) (18 − ( −10 ) ) K ( 0.1m )3 13.84 × 10−6 m 2 / s ×19.5 × 10−6 m 2 / s

= 3.67 ×106.

The appropriate correlations for the sections are Eqs. 9.49 and 9.52 (with H/L = 30), Nu L hor = 0.069Ra1L/ 3 Pr0.074

Nu L vert = 0.42Ra1L/ 4 Pr0.012 ( H / L )

− 0.3

.

(3,4)

Using Eqs. (3) and (4), the ratio of Eq. (1) becomes,

(

)

6 1/3 1 / 3 0.074 0.069 3.67 × 10 ( 0.713)0.074 q hor 0.069RaL Pr = = = 1.57. 1/4 q vert 0.42Ra1 / 4 Pr 0.012 ( H / L ) −0.3 6 0.012 − 0.3 L 0.42 3.67 × 10 ( 0.713) ( 30)

(

)

<

(b) The effect of the baffle in the vertical wall section is to reduce H/L from 30 to 15. Using Eq. 9.52, it follows, −0.3 −0.3 q baf h baf ( H / L )baf  15  = = = = 1.23. −0.3  30  q h

(H / L )

<

That is, the effect of the baffle is to increase the convection heat rate. COMMENTS: (1) Note that the heat rate for the horizontal section is 57% larger than that for the vertical section for the same (T1 – T2). This indicates the importance of heat losses from the ceiling or roofs in house construction. (2) Recognize that for Eq. 9.52, the Pr > 1 requirement is not completely satisfied. (3) What is the physical explanation for the result of part (b)?

PROBLEM 9.95 KNOWN: Double-glazed window of variable spacing L between panes filled with either air or carbon dioxide. FIND: Heat transfer across window for variable spacing when filled with either gas. Consider these conditions (outside, T1; inside, T2): winter (-10, 20°C) and summer (35°C, 25°C). SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange is negligible, (3) Gases are at atmospheric pressure, (4) Perfect gas behavior. PROPERTIES: Table A-4: Winter, T = ( − 10 + 20 ) ° C / 2 = 288K, Summer, T = ( 35 + 25 ) °C / 2 = 3 0 3 K :

α 2 6 (m /s × 10 )

3

ν 2 6 (m /s × 10 )

k × 10 (W/m⋅K)

Gas (1 atm)

T (K)

Air Air CO2 CO2

288 303

20.5 22.9

14.82 16.19

24.9 26.5

288

10.2

7.78

15.74

303

11.2

8.55

16.78

ANALYSIS: The heat flux by convection across the window is q ′′ = h ( T1 − T2 ) where the convection coefficient is estimated from the correlation of Eq. 9.53 for large aspect ratios 10< H/L < 40, 1/4

Nu L = hL/k = 0.046RaL . Substituting numerical values for winter (w) and summer (s) conditions, Ra L,w,air =

9.8m/s

2

20.5 × 10

(1/288K ) ( 20 − ( −10) ) KL3

−6

8 3

Ra L,s,air = 8.724 × 10 L

2

m / s × 14.82 ×10

−6

2

= 3.360 ×109 L3

m /s 10 3

Ra L,w,CO = 1.286 × 10

the heat transfer coefficients are

(

h w,air = ( 0.0249W/m ⋅ K / L) × 0.046 3.360 × 109 L3 h s,air = 0.209L−1/4

h w,CO2 = 0.244L−1/4

9 3

Ra L,s,CO = 3.378 × 10 L

L

2

)

1/4

2

= 0.276L− 1/4

h s,CO2 = 0.186L−1/4.

For a separation distance such that H/L = 40, the maximum aspect ratio for the correlation, with H = 1.5 m, L = 37.5 mm find q ′′w,air = 18.8W/m2

q ′′s,air = 4.7W/m2

q ′′w,CO = 16.6W/m2 2

q ′′s,CO = 4.2W/m2 . 2

Using CO2 rather than air reduces the heat loss/gain by approximately 12%. Note the winter heat rate for this window is nearly 4 times that for summer.

PROBLEM 9.96 KNOWN: Dimensions of double pane window. Thickness of air gap. Temperatures of room and ambient air. FIND: (a) Temperatures of glass panes and heat rate through window, (b) Resistance of glass pane relative to smallest convection resistance. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Negligible glass pane thermal resistance, (3) Constant properties. PROPERTIES: Table A-3, Plate glass: kp = 1.4 W/m⋅K. Table A-4, Air (p = 1 atm). Tf,i = 287.6K: -6 2 -6 2 -1 νi = 14.8 × 10 m /s, ki = 0.0253 W/m⋅K, αi = 20.9 × 10 m /s, Pri = 0.710, β i = 0.00348 K . T = -6 2 -6 2 (Ts,i + Ts,o)/2 = 272.8K: ν = 13.49 × 10 m /s, k = 0.0241 W/m⋅K, α = 18.9 × 10 m /s, Pr = 0.714, -1 -6 2 -6 2 β = 0.00367 K . Tf,o = 258.2K: νo = 12.2 × 10 m /s, ko = 0.0230W/m⋅K, α = 17.0 × 10 m /s, Pr -1 = 0.718, β o = 0.00387 K . ANALYSIS: (a) The heat rate may be expressed as q = qo = ho H 2 Ts,o − T∞,o

(1)

(

(2)

(

)

q = qg = hg H 2 Ts,i − Ts,o

(

q = qi = h i H 2 T∞,i − Ts,i

)

)

(3)

where ho and hi may be obtained from Eq. (9.26),

  1/ 6   0.387 Ra H Nu H = 0.825 + 8 / 27    1 + ( 0.492 / Pr )9 /16     

2

with Ra H = gβ o ( Ts,o − T∞,o ) H 3 / α oν o and Ra H = gβ i ( T∞,i − Ts,i ) H 3 / α iν i , respectively. Assuming 4

7

10 < Ra L < 10 , hg is obtained from

−0.3

4 0.012 H / L Nu L = 0.42 Ra1/ ( ) L Pr

where Ra L = gβ ( Ts,i − Ts,o ) L3 / αν . A simultaneous solution to Eqs. (1) – (3) for the three unknowns yields Continued …..

PROBLEM 9.96 (Cont.)

Ts,i = 9.1°C,

<

Ts,o = −9.6°C, q = 35.7 W

where hi = 3.29 W / m 2 ⋅ K, ho = 3.45 W / m 2 ⋅ K and hg = 1.90 W / m 2 ⋅ K. (b) The unit conduction resistance of a glass pane is R ′′cond = L p / k p = 0.00429 m 2 ⋅ K / W, and the 2 smallest convection resistance is R ′′ = 1 / h = 0.290 m ⋅ K / W. Hence, conv,o

(

o)

R ′′cond τ*, the critical tilt angle, Eq. 9.56 is appropriate for estimating h. 1/4 1/4

Nu L = Nu L (τ = 90 ) ⋅ ( sin τ )

= 4.72 ( sin75° )

= 4.68

h = Nu L k / L = 4.68 × 0.0263W/m ⋅ K/0.05m = 2.46W/m 2 ⋅ K. q′′ = 2.46W/m 2 ⋅ K ( 325 − 275 ) K =123W/m 2 .

<

COMMENTS: Note that Nu L (τ = 0 ) > Nu L (τ = 90 °) . For the cavity conditions there is little change in h for tilt angles, τ, from 45° to 90°.

PROBLEM 9.100 KNOWN: Dimensions and surface temperatures of a flat-plate solar collector. FIND: (a) Heat loss across collector cavity, (b) Effect of plate spacing on the heat loss. SCHEMATIC:

ASSUMPTIONS: Negligible radiation. PROPERTIES: Table A.4, Air ( T = (T1 + T2)/2 = 323 K): ν = 18.2 × 10-6 m2/s, k = 0.028 W/m⋅K, α = 25.9 × 10-6 m2/s, β = 0.0031 K-1. ANALYSIS: (a) Since H/L = 2 m/0.03 m = 66.7 > 12, τ < τ* and Eq. 9.54 may be used to evaluate the convection coefficient associated with the air space. Hence, q = h As(T1 - T2), where h = (k/L) Nu L and • 1.6 1/ 3  1708   1708 (sin 1.8τ )   Ra L cos τ   Nu L = 1 + 1.44 1 − 1 1 − + −        Ra L cos τ  Ra L cos τ      5830 

For L = 30 mm, the Rayleigh number is gβ ( T1 − T2 ) L

3

Ra L =

αν

9.8 m s =

2

(0.0031K )(40 C )(0.03 m )

25.9 × 10

−1

−6

m

2

$

•

3

s × 18.2 × 10

−6

m

2

= 6.96 × 10

4

s

and RaL cosτ = 3.48 × 10 . It follows that Nu L = 3.12 and h = (0.028 W/m⋅K/0.03 m)3.12 = 2.91 W/m2⋅K. Hence, 4

2

( )(40 C ) = 466 W

q = 2.91 W m ⋅ K 4 m

<

$

2

500

500

480

480

460

Heat loss, q(W)

Heat loss, q(W)

(b) The foregoing model was entered into the workspace of IHT, and results of the calculations are plotted as follows.

440

420

440

420

400 0.017

460

400

0.028

0.039

Plate spacing, L(m)

0.05

0.011

0.012

0.013

0.014

0.015

0.016

Plate spacing, L(m)

Continued...

PROBLEM 9.100 (Cont.) 900

Heat loss, q(W)

800

700

600

500

400 0.005

0.006

0.007

0.008

0.009

0.01

Plate spacing, L(m)

The plots are influenced by the fact that the third and second terms on the right-hand side of the correlation are set to zero at L ≈ 0.017 m and L ≈ 0.011 m, respectively. For the range of conditions, minima in the heat loss of q ≈ 410 W and q = 397 W are achieved at L ≈ 0.012 m and L = 0.05 m, respectively. Operation at L ≈ 0.02 m corresponds to a maximum and is clearly undesirable, as is operation at L < 0.011 m, for which conditions are conduction dominated. COMMENTS: Because the convection coefficient is low, radiation effects would be significant.

PROBLEM 9.101 KNOWN: Cylindrical 120-mm diameter radiation shield of Example 9.5 installed concentric with a 100-mm diameter tube carrying steam; spacing provides for an air gap of L = 10 mm. FIND: (a) Heat loss per unit length of the tube by convection when a second shield of diameter 140 m is installed; compare the result to that for the single shield calculation of the example; and (b) The heat loss per unit length if the gap dimension is made L = 15 mm (rather than 10 mm). Do you expect the heat loss to increase or decrease? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, and (b) Constant properties. -6

2

PROPERTIES: Table A-4, Air (Tf = (Ts + T∞)/2 = 350 K, 1 atm): ν = 20.92 × 10 m /s, k = 0.030 W/m⋅K, Pr = 0.700. ANALYSIS: (a) The thermal circuit representing the tube with two concentric cylindrical radiation shields having gap spacings L = 10 mm is shown above. The heat loss per unit length by convection is

q′ =

Ti − T2 R ′g1 + R ′g2

(1)

where the R ′g represents the thermal resistance of the annular gap (spacing). From Eq. 9.58, 59 and 60, find

R ′g =

n ( D o / D i ) 2π k eff

(2)

( )

(3)

Ra L

(4)

1/ 4

k eff Pr   = 0.386   k  0.861 + Pr  Ra ∗c =

 n ( Do / Di )

Ra ∗c

1/ 4

4

(

)

5 L3 Di−3/ 5 + Do−3/ 5

Ra L = gβ (To − Ti ) L3 / αν

(5)

where the properties are evaluated at the average temperature of the bounding surfaces, Tf = (Ti + To)/2. Recognize that the above system of equations needs to be solved iteratively by initial guess values of T1, or solved simultaneously using equation-solving software with a properties library. The results are tabulated below. Continued …..

PROBLEM 9.101 (Cont.) (b) Using the foregoing relations, the analyses can be repeated with L = 15 mm, so that Di = 130 mm and D2 = 160 mm. The results are tabulated below along with those from Example 9.5 for the singleshield configuration. Shields

L(mm)

R ′g1 (m⋅K/W)

R ′g2 (m⋅K/W)

R ′tot (m⋅K/W)

T1(°C)

q′ (W/m)

1 2 2

10 10 15

0.7658 1.008 0.9773

--0.8855 0.8396

0.76 1.89 1.82

--74.8 74.3

100 44.9 46.8

COMMENTS: (1) The effect of adding the second shield is to more than double the thermal resistance of the shields to convection heat transfer. (2) The effect of gap increase from 10 to 15 mm for the two-shield configuration is slight. Increasing L allows for greater circulation in the annular space, thereby reducing the thermal resistance. (3) Note the difference in thermal resistances for the annular spaces R ′g1 of the one-and two-shield configurations with L = 10 mm. Why are they so different (0.7658 vs. 1.008 m⋅K/W, respectively)? (4) See Example 9.5 for details on how to evaluate the properties for use with the correlation.

PROBLEM 9.102 KNOWN: Operating conditions of a concentric tube solar collector. FIND: Convection heat transfer per unit length across air space between tubes. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Long tubes. -6

PROPERTIES: Table A-4, Air (T = 50°C, 1 atm): ν = 18.2 × 10 -6 2 -1 = 25.9 × 10 m /s, Pr = 0.71, β = 0.0031 K .

2

m /s, k = 0.028 W/m⋅K, α

ANALYSIS: For the annular region Ra L =

g β (Ts − T ∞ ) L3 να

9.8m/s2 )(0.0031K−1 ) ( 70 − 30) °C ( 0.025m ) 3 ( = (18.2× 10−6 m2 / s)( 25.9× 10−6 m2 / s)

Ra L = 4.03 ×10 4. Hence, from Eq. 9.60, Ra *c =

ln ( 0.15/0.10 ) 

4

( 0.025m)3 ( 0.10) −3/5 + ( 0.15)

−3 / 5  5

× 4.03× 10 4 = 3857.



Accordingly, Eq. 9.59 may be used, in which case

( )

1/4 1/4 Pr   k eff = 0.386k  Ra *c   0.861 + Pr  1/4 0.71   k eff = 0.386 ( 0.028W/m ⋅ K )  ( 3857 )1/4 = 0.07W/m ⋅ K.   0.861 + 0.71 

From Eq. 9.58, it then follows that q′ =

2 π ( 0.07W/m ⋅ K ) 2π keff ( Ti − To ) = ( 70 − 30 ) ° C = 43.4W/m. ln ( Do / Di ) ln ( 0.15/0.10 )

<

COMMENTS: An additional heat loss is related to thermal radiation exchange between the inner and outer surfaces.

PROBLEM 9.103 KNOWN: Annulus formed by two concentric, horizontal tubes with prescribed diameters and surface temperatures is filled with water. FIND: Convective heat transfer rate per unit length of the tubes. SCHEMATIC:

ASSUMPTIONS: Steady-state conditions. -3

PROPERTIES: Table A-6, Water (Tf = 325K): ρ = (1/1.013 × 10 -6

528 × 10

2

-6

N⋅s/m , k = 0.645 W/m⋅K, Pr = 3.42, β = 471.2 × 10

-1

3

m /kg), cp = 4182 J/kg⋅K, µ =

K .

ANALYSIS: From Eqs. 9.58 and 9.59,

q′ =

( ) 4 3 / 5 + D−3 / 5 5 . Ra *c = ln ( Do / Di )  ⋅ Ra L / L3 ( D− o ) i

1/4 k eff Pr   * 1/4 = 0.386  Ra .  c k  0.861 + Pr 

2 π k eff (T − T ) ln ( Do / D i ) i o

From Eq. 9.60,

(1,2) (3)

The Rayleigh number follows from Eq. 9.25 using ν = µ/ρ and α = k/ρ cp , g β ( To − Ti ) L = να

2

9.8m/s × 471.2 × 10

3

Ra L =

−6

K

−1

( 350 − 300) K ×

(

−3

12.5 × 10

m

)

3

3   −3 3  0.645W/m ⋅ K × 1.013 × 10 m /kg −6 2 −3 m  528 × 10 N ⋅ s / m × 1.013 ×10  ×  kg  4182J/kg ⋅ K   

  

Ra L = 5.396 ×10 5. Using this value in Eq. (3), find *

(

)

5

4   75  × 5.396 × 10 5 / 12.5 × 10 −3 m 3  50 × 10 −3 m  −3 / 5 + 75 ×10 −3 m  −3 / 5  = 5.164 × 10 4      50      

Ra c = ln 

and then evaluating Eq. (2), find

(

)

1/4 k eff 3.42   4 1/4 = 5.50 = 0.386  5.164 × 10  k  0.861 + 3.42 

k eff = 5.50 ×0.645W/m ⋅ K = 3.55W/m ⋅ K. The heat rate from Eq. (1) is then,

q′ =

2π × 3.55W/m ⋅ K ( 350 − 300 ) K = 2.75kW/m. ln ( 75/50 )

<

COMMENTS: Note that the Ra*c value is within prescribed limits for Eq. 9.50 or Eq. (3). Note also the characteristic length in RaL is L = (Do – Di)/2, the annulus gap.

PROBLEM 9.104 KNOWN: Annulus formed by two concentric, horizontal tubes with prescribed diameters and surface temperatures is filled with nitrogen at 5 atm. FIND: Convective heat transfer rate per unit length of the tubes. SCHEMATIC:

ASSUMPTIONS: (1) Thermophysical properties k, µ, and Pr, are independent of pressure, (2) Density is proportional to pressure, (3) Perfect gas behavior. PROPERTIES: Table A-4, Nitrogen ( T = ( Ti + To ) / 2 = 350K, 5 atm) : k = 0.0293 W/m⋅K, µ = -7

2

3

3

200 × 10 N⋅s/m , ρ(5 atm) = 5 ρ (1 atm) = 5 × 0.9625 kg/m = 4.813 kg/m , Pr = 0.711, ν = µ/ρ = -6 2 3 -6 2 4.155 × 10 m /s, α = k/ρc = 0.0293 W/m⋅K/(4.813 kg/m × 1042 J/kg⋅K) = 5.842 × 10 m /s. ANALYSIS: From Eqs. 9.58 and 9.59

q′ =

( )

1/4 k eff Pr   * 1/4 = 0.386  Ra .  c k  0.861 + Pr 

2 π k eff (T − T ) ln ( Do / D i ) o i

From Eq. 9.60,

(

3 / 5 + D− 3/5 Ra *c = ln ( Do / Di )  Ra L / L3 D− o i 4

(1,2)

). 5

(3) The Rayleigh number, RaL, follows from Eq. 9.25, and Ra *c from Eq. (3), gβ ( To − Ti ) L3 9.8m/s 2 (1/350K )( 400 − 300 ) K ( 0.025m )3 Ra L = = = 1.802 ×10 6. − 6 2 − 6 2 αν 5.842 ×10 m / s × 4.155 ×10 m / s

(

)

4 5  250  Ra *c = ln ×1.802 ×10 6 / ( 0.025m )3 0.20− 3/5 + 0.25− 3 / 5 m 3 = 98,791   200  and then evaluating Eq. (2), 1/4 k eff 0.711   = 0.386  ( 98,791)1/4 = 5.61.  k  0.861 + 0.711  Hence, the heat rate, Eq. (1), becomes

q′ =

2π × 5.61× 0.0293W/m ⋅ K ( 400 − 300 ) K = 463W/m. ln ( 250/200 )

COMMENTS: Note that the heat loss by convection is nearly six times that for conduction. Radiation transfer is likely to be important for this situation. The effect of nitrogen pressure is to decrease ν which in turn increases RaL; that is, free convection heat transfer will increase with increase in pressure.

<

PROBLEM 9.105 KNOWN: Diameters and temperatures of concentric spheres. FIND: Rate at which stored nitrogen is vented. SCHEMATIC:

ASSUMPTIONS: (1) Negligible radiation. 5

PROPERTIES: Liquid nitrogen (given): hfg = 2 × 10 J/kg; Table A-4, Helium ( T = (Ti + To)/2 = -6 2 2 -1 180K, 1 atm): ν = 51.3 × 10 m /s, k = 0.107 W/m⋅K, α = 76.2 m /s, Pr = 0.673, β = 0.00556 K . ANALYSIS: Performing an energy balance for a control surface about the liquid nitrogen, it follows that

& fg. q = q conv = mh From the Raithby and Hollands expressions for free convection between concentric spheres,

q conv = k eff π ( DiDo / L )( To − Ti ) 1/4

k eff = 0.74k  Pr/ ( 0.861 + Pr ) 

( )

1/4

Ra *s

    L RaL Ra *s =    ( Do D i )4 D−7 / 5 + D−7 / 5 5  o i  

where

(

Ra L =

Ra *s =

gβ ( To − Ti ) L3 να

(

0.05m

)

=

)

(

(51.3×10

−6 2

)(

−6 2

m / s 76.2 ×10

3.59 ×10 5

4 5 1.10m2 1 + (1.1) −7 / 5  m −7



)

9.8m/s2 0.00556K −1 ( 206K )( 0.05m)3

)

= 529



1/4

k eff = 0.74 ( 0.107W/m ⋅ K )  0.673/ ( 0.861 + 0.673 ) Hence,

m /s

= 3.59 ×105

(

)

(529 )1/4 = 0.309W/ m ⋅ K.

q conv = ( 0.309W/m ⋅ K ) π 1.10m 2 /0.05m 206 K = 4399 W.

The rate at which nitrogen is lost from the system is therefore & = q conv / hfg = 4399W/2 ×105 J / k g = 0.022 kg/s. m COMMENTS: The heat gain and mass loss are large. Helium should be replaced by a noncondensing gas of smaller k, or the cavity should be evacuated.

<

PROBLEM 9.106 KNOWN: Concentric spheres with prescribed surface temperatures. FIND: Convection heat transfer rate. SCHEMATIC:

ASSUMPTIONS: (1) Quiescent air in void space, (2) Density ~ pressure; other properties independent, (3) Perfect gas behavior. -3

-1

-6

PROPERTIES: Table A-4, Air (Tf = 300K, 3 atm): β = 3.33 × 10 K , ν = 1/3 × 15.89 × 10 2 -6 2 m /s, k = 0.263 W/m⋅K, α = 1/3 × 22.5 × 10 m /s, Pr = 0.707. ANALYSIS: The heat transfer rate due to free convection is

q = k eff π ( D iDo / L )( Ti − To )

where

(9.61)

( )

1/4 k eff Pr   * 1/4 = 0.74  Ra s  k  0.861 + Pr      L Ra L Ra *s =    ( Do D i ) 4 D−7 / 5 + D−7 / 5 5  o i  

(

(9.63)

)

g β ( Ti − To ) L3

Ra L =

(9.62)

να

.

(9.25)

Substituting numerical values in the above expressions, find that Ra L =

(

9.8m/s2 × 3.333 × 10 −3 K −1 ( 325 − 375 ) K 12.5 ×10 −3 × 10− 6 m2 / s (1 / 3 ) 22.5 × 10− 6 m2 / s (1/315.89 )

)m 3

3

= 80,928

    −3 12.5 × 10 m 80,928  *  Ra s =  ⋅ = 330.1 4 4 5 − 3 − 3 − 7 / 5 − 7 / 5    100 ×10 × 75 ×10 m  75 × 10− 3 m + 100 × 10− 3 m        

(

k eff

(

)

)

(

1/4

0.707     0.861 + 0.707 

= 0.74 

k Hence, the heat rate becomes

)

(

)

( 330.1)1 / 4 = 2.58.

W   75 × 10−3 m × 100 × 10−3 m    ( 325 − 275) K = 64.0W. q =  2.58 × 0.263 π  m ⋅ K    12.5 ×10 −3m 

<

COMMENTS: Note the manner in which the thermophysical properties vary with pressure. Assuming perfect gas behavior, ρ ~ p. Also, k, µ and cp are independent of pressure. Hence, Pr is -1 -1 independent of pressure, but ν = µ/ρ ~ p and α = k/ρc ~ p .

PROBLEM 9.107 KNOWN: Cross flow over a cylinder with prescribed surface temperature and free stream conditions. FIND: Whether free convection will be significant if the fluid is water or air. SCHEMATIC:

ASSUMPTIONS: (1) Constant properties, (2) Combined free and forced heat transfer. -6

2

PROPERTIES: Table A-6, Water (Tf = (T∞ + Ts)/2 = 300K): ν = µ vf = 855 × 10 N⋅s/m × 1.003 -3 3 -7 2 -6 -1 × 10 m /kg = 8.576 × 10 m /s, β = 276.1 × 10 K ; Table A-4, Air (300K, 1 atm): ν = 15.89 × 10 6 2 -3 -1 m /s, β = 1/Tf = 3.333 × 10 K . ANALYSIS: Following the discussion of Section 9.9, the general criterion for delineating the relative 2 significance of free and forced convection depends upon the value of Gr/Re . If free convection is significant.

GrD /Re 2D ≥ 1 where

GrD = g β ( T∞ − Ts )D 3 / ν 2 and

(1)

Re D = VD/ ν .

(2,3)

(a) When the surrounding fluid is water, find

(

GrD = 9.8m/s 2 × 276.1 ×10−6 K −1 ( 35 − 20 ) K ( 0.05m )3 / 8.576 ×10−7 m 2 /s Re D = 0.05m/s × 0.05m/8.576 ×10−7 m 2 / s = 2915

)

2

= 68, 980

GrD / Re 2D = 68,980/29152 = 0.00812.

<

We conclude that since GrD /Re2D > 1, free convection dominates the heat transfer process. COMMENTS: Note also that for the air flow situation, surface radiation exchange is likely to be significant.

<

PROBLEM 9.108 KNOWN: Parallel air flow over a uniform temperature, heated vertical plate; the effect of free convection on the heat transfer coefficient will be 5% when GrL /Re2L = 0.08. FIND: Minimum vertical velocity required of air flow such that free convection effects will be less than 5% of the heat rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Criterion for combined free-forced convection determined from experimental results. -6

2

PROPERTIES: Table A-4, Air (Tf = (Ts + T∞)/2 = 315K, 1 atm): ν = 17.40 × 10 m /s, β = 1/Tf. ANALYSIS: To delineate flow regimes, according to Section 9.9, the general criterion for predominately forced convection is that GrL /Re2L 350°C.

PROBLEM 10.29 KNOWN: Horizontal, stainless steel bar submerged in water at 25°C. FIND: Heat rate per unit length of the bar. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Film pool boiling, (3) Water at 1 atm. PROPERTIES: Table A-6, Water, liquid (1 atm, Tsat = 100°C): ρ l = 957.9 kg/m3, hfg = 2257 3 kJ/kg; Table A-6, Water, vapor, (Tf = (Ts + Tsat )/2 ≈ 450K): ρ v = 4.81 kg/m , cp,v = 2560 J/kg⋅K, µv -6 2 = 14.85 × 10 N⋅s/m , kv = 0.0331 W/m⋅K. ANALYSIS: The heat rate per unit length is

q′s = q s / l = q′′π D = hπ D ( Ts − Tsat ) = h π D ∆Te

where ∆Te = (250-100)°C = 150°C. Note from the boiling curve of Figure 10.4, that film boiling will occur. From Eq. 10.10, 3 4 / 3 + h h1/3 h 4 / 3 = hconv or h = hconv + h rad if hconv > h rad . rad

4

(

)

To estimate the convection coefficient, use Eq. 10.9, 1/4

 g ( ρ − ρ ) h′ D3  h conv D l v fg  Nu D = = C kv  ν v k v ∆Te   

where C = 0.62 for the horizontal cylinder and h ′fg = h fg + 0.8cp,v ( Ts − Tsat ) . Find 1/4

h conv =

0.0331W/m ⋅ K 0.050m

 9.8m/s 2 ( 957.9 − 4.81 ) k g / m 3 2257 × 10 3 + 0.8 × 2560J/kg ⋅ K × 150 K  ( 0.050m)3     0.62  −6 2   14.85 × 10 /4.81 m / s × 0.0331 W / m ⋅ K × 150K  

(

)

hconv = 273 W / m 2 ⋅ K. To estimate the radiation coefficient, use Eq. 10.11, 4 ε σ Ts4 − Tsat 0.50 × 5.67 ×10 −8 W / m 2 ⋅ K 4 5234 − 3734 K 4 hrad = = = 1 1 W / m 2 ⋅ K. Ts − Tsat 150K Since hconv > hrad, the simpler form of Eq. 10.10 is appropriate. Find, h =  273 + 3 / 4 ×11 W / m2 ⋅ K = 2 8 1 W / m2 ⋅ K.

(



)

(

(

)

)



Using the rate equation, find q′s = 281W/m 2 ⋅ K × π × ( 0.050m ) ×150K = 6.62kW/m. COMMENTS: The effect of the water being subcooled (T = 25°C < Tsat ) is considered to be negligible.

<

PROBLEM 10.30 KNOWN: Heater element of 5-mm diameter maintained at a surface temperature of 350°C when immersed in water under atmospheric pressure; element sheath is stainless steel with a mechanically polished finish having an emissivity of 0.25. FIND: (a) The electrical power dissipation and the rate of evaporation per unit length; (b) If the heater element were operated at the same power dissipation rate in the nucleate boiling regime, what temperature would the surface achieve? Calculate the rate of evaporation per unit length for this operating condition; and (c) Make a sketch of the boiling curve and represent the two operating conditions of parts (a) and (b). Compare the results of your analysis. If the heater element is operated in the power-controlled mode, explain how you would achieve these two operating conditions beginning with a cold element. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, and (2) Water exposed to standard atmospheric pressure and uniform temperature, Tsat. PROPERTIES: Table A-6, Saturated water, liquid (100°C): ρ" = 957.9 kg / m3 , c p," = 4217 J/kg⋅K, µ" = 279 × 10 −6 N ⋅ s / m 2 , Pr" = 1.76, hfg = 2257 kJ/kg, h ′fg = h fg + 0.80 c p,v ( Ts − Tsat ) = −3

3

2905 kJ/kg, σ = 58.9 × 10 N/m; Saturated water, vapor (100°C): ρv = 0.5955 kg/m ; Water vapor -6 (Tf = 498 K): ρv = 1/vv = 12.54 kg/m3, cp,v = 3236 J/kg⋅K, kv = 0.04186 W/m⋅K, ηv = 1.317 × 10 2 m /s. ANALYSIS: (a) Since ∆Te > 120°C, the element is operating in the film-boiling (FB) regime. The electrical power dissipation per unit length is q′s = h (π D )( Ts − Tsat ) (1) where the total heat transfer coefficient is 4/3 + h 1/ 3 h 4 / 3 = h conv rad h The convection coefficient is given by the correlation, Eq. 10.9, with C = 0.62,

(2)

1/ 4

 g ( ρ − ρ ) h′ D3  h conv D " v fg  = C kv  ηv k v ( Ts − Tsat )   

(3) 1/ 4

 9.8 m / s 2 (833.9 − 12.54 ) kg / m3 × 2.905 × 106 J / kg ⋅ K (0.005 m )3   h conv = 0.62  6 2 −   1.31× 10 m / s × 0.04186 W / m ⋅ K (350 − 100 ) K   h conv = 626 W / m 2 ⋅ K −8

The radiation coefficient, Eq. (10.11), with σ = 5.67 × 10

2

4

W/m ⋅K , is Continued …..

h rad =

h rad =

(

4 εσ Ts4 − Tsat

(Ts − Tsat )

PROBLEM 10.30 (Cont.)

)

(

)

0.25σ 6234 − 3734 K 4

(350 − 100 ) K

= 4.5 W / m 2 ⋅ K

Substituting numerical values into Eq. (2) for h, and into Eq. (1) for q′s , find

h = 630 W / m 2 ⋅ K q′s = 630 W / m 2 ⋅ K (π × 0.005 m )(350 − 100 ) K = 2473 W / m

<

q′′s = q′s / π D = 0.157 MW / m 2 The evaporation rate per unit length is

 ′b = q′s / h fg = 3.94 kg / h ⋅ m m

<

(b) For the same heat flux, q′′s = 0.157 MW / m 2 , using the Rohsenow correlation for the nucleate boiling (NB) regime, find ∆Te, and hence Ts. 1/ 2 

 g ( ρ" − ρ v )  q′′s = µ" h fg   σ  

3

c ∆T   p," e   C h Pr n   s,f fg " 

where, from Table 10.1, for stainless steel mechanically polished finish with water, Cs,f = 0.013 and n = 1.0.

0.157 × 106 W / m 2 = 279 × 10−6 N ⋅ s / m 2 × 2.257 × 106 J / kg 1/ 2

 9.8 m / s 2 (957.9 − 0.5955 ) kg / m3   × −3 N / m   58.9 10 ×     4217 J / kg ⋅ K × ∆Te ×   0.013 × 2.257 ×106 J / kg ×1.76    ∆Te = Ts − Tsat = 10.5 K

3

<

Ts = 110.5°C

The evaporation rate per unit length is

 ′b = q′′s (π D ) h fg = 3.94 kg / h ⋅ m m

< Continued …..

PROBLEM 10.30 (Cont.) (c) The two operating conditions are shown on the boiling curve, which is fashioned after Figure 10.4. For FB the surface temperature is Ts = 350°C (∆Te = 250°C). The element can be operated at NB 2 with the same heat flux, qs′′ = 0.157 MW/m , with a surface temperature of Ts = 110°C (∆Te = 10°C). Since the heat fluxes are the same for both conditions, the evaporation rates are the same.

If the element is cold, and operated in a power-controlled mode, the element would be brought to the NB condition following the arrow shown next to the boiling curve near ∆Te = 0. If the power is increased beyond that for the NB point, the element will approach the critical heat flux (CHF) condition. If q′′s is increased beyond q′′max , the temperature of the element will increase abruptly, and the burnout condition will likely occur. If burnout does not occur, reducing the heat flux would allow the element to reach the FB point.

PROBLEM 10.31 KNOWN: Inner and outer diameters, outer surface temperature and thermal conductivity of a tube. Saturation pressure of surrounding water and convection coefficient associated with gas flow through the tube. FIND: (a) Heat rate per unit tube length, (b) Mean temperature of gas flow through tube. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Uniform surface temperature, (3) Water is at saturation temperature, (4) Tube is horizontal. PROPERTIES: Table A-6, saturated water, liquid (p = 1.523 bars): Tsat = 385 K, ρ " = 950 kg / m3 , 3

hfg = 2225 kJ/kg. Table A-6, saturated water, vapor (Tf = 480 K): ρv = 9.01 kg/m , cp,v = 2940 J/kg⋅K. µv = 15.9 × 10 N⋅s/m , kv = 0.0381 W/m⋅K, ν v = 1.77 × 10−6 m 2 / s. -6

2

ANALYSIS: (a) The heat rate per unit length is q ′ = h oπ Do ( Ts,o − Tsat ) , where ho includes contributions due to convection and radiation in film boiling. With C = 0.62 and h ′fg = h fg + 0.80

(

)

6

c p,v Ts,o − Tsat = 2.67 × 10 J / kg, Eq. 10.9 yields 1/ 4

hconv,o

 9.8 m / s 2 ( 950 − 9 ) kg / m3 × 2.67 × 106 J / kg ( 0.03m )3  0.0381 W / m ⋅ K   =   0.62  −6 2   0.030m 1.77 × 10 m / s × 0.0381 W / m ⋅ K × 190 K  

From Eq. 10.11, the radiation coefficient is h rad,o =

0.30 × 5.67 × 10

−8

2

W/m ⋅K

4

(575

4

− 385

4

(575 − 385 ) K

)K

2

= 376 W / m ⋅ K

4 2

= 7.8 W / m ⋅ K

From Eq. 10.10b, it follows that 2

ho = hconv,o + 0.75 hrad,o = 382 W / m ⋅ K

and the heat rate is

(

)

q ′ = hoπ D o Ts,o − Tsat = 382 W / m ⋅ K (π × 0.03m )190 K = 6840 W

<

2

(

(b) From the thermal circuit, with R ′conv,i = ( h iπ Di )−1 = 500 W / m 2 ⋅ K × π × 0.026m

)

−1

= 0.0245 m ⋅ K / W

and R ′cond = "n ( Do / Di ) / 2π k s = "n ( 0.030 / 0.026 ) / 2π (15 W / m ⋅ K ) = 0.00152 m ⋅ K / W, q′ =

Tm − Ts,o Tm − 575 K = R conv,i + R ′cond ( 0.0245 + 0.00152 ) m ⋅ K / W

Tm = 575 K + 6840 W / m ( 0.0260 m ⋅ K / W ) = 753 K

<

COMMENTS: Despite the large temperature of the gas, the rate of heat transfer is limited by the large thermal resistances associated with convection from the gas and film boiling. The resistance due to film boiling is R ′fb = (π D o ho )−1 = 0.0278 m ⋅ K / W.

PROBLEM 10.32 KNOWN: Cylinder of 120 mm diameter at 1000K quenched in saturated water at 1 atm FIND: Describe the quenching process and estimate the maximum heat removal rate per unit length during cooling. SCHEMATIC:

ASSUMPTIONS: Water exposed to 1 atm pressure, Tsat = 100°C. ANALYSIS: At the start of the quenching process, the surface temperature is Ts(0) = 1000K. Hence, ∆Te = Ts – Tsat = 1000K – 373K = 627K, and from the typical boiling curve of Figure 10.4, film boiling occurs. As the cylinder temperature decreases, ∆Te decreases, and the cooling process follows the boiling curve sketched above. The cylinder boiling process passes through the Leidenfrost point D, into the transition or unstable boiling regime (D → C). 2

At point C, the boiling heat flux has reached a maximum, q ′′max = 1.26 MW/m (see Example 10.1). Hence, the heat rate per unit length of the cylinder is

q′s = q′max = q′′max (π D ) = 1.26MW/m 2 π ( 0.120m ) = 0.475MW/m.

<

As the cylinder cools further, nucleate boiling occurs (C → A) and the heat rate drops rapidly. Finally, at point A, boiling no longer is present and the cylinder is cooled by free convection. COMMENTS: Why doesn’t the quenching process follow the cooling curve of Figure 10.3?

PROBLEM 10.33 KNOWN: Horizontal platinum wire of diameter of 1 mm, emissivity of 0.25, and surface temperature of 800 K in saturated water at 1 atm pressure. FIND: (a) Surface heat flux, q′′s , when the surface temperature is Ts = 800 K and (b) Compute and plot on log-log coordinates the heat flux as a function of the excess temperature, ∆Te = Ts - Tsat, for the range 150 ≤ ∆Te ≤ 550 K for emissivities of 0.1, 0.25, and 0.95; separately plot the percentage contribution of radiation as a function of ∆Te. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Film pool boiling. PROPERTIES: Table A.6, Saturated water, liquid (Tsat = 100°C, 1 atm): ρ" = 957.9 kg/m3, hfg = 2257 kJ/kg; Table A.6, Water, vapor (Tf = (Ts + Tsat)/2 = (800 + 373)K/2 = 587 K): ρv = 58.14 kg/m3, cp,v = 7065 J/kg⋅K, µv = 21.1 × 10-6 N⋅s/m2, kv = 81.9 × 10-3 W/m⋅K. ANALYSIS: (a) The heat flux is

q′′s = h ( Ts − Tsat ) = h∆Te where ∆Te = (800 - 373)K = 427 indicative of film boiling. From 10.10, 4 / 3 + h h −1/ 3 h 4 / 3 = h conv or h = h conv + (3 4 ) h rad rad if hrad < hconv. Use Eq. 10.9 with C = 0.62 for a horizontal cylinder, 3 1/ 4  ′ ρ ρ g h D − ( ) h D " v fg  Nu D = conv = C  kv  ν v k v ( Ts − Tsat )    1/ 4

  3 9.8 m s 2 (957.9 − 58.14 ) kg m3 × 4670 k J kg ( 0.001m ) hconv × 0.001m   = 0.62   −3 6 2 3 − 81.9 × 10 W m ⋅ K 21.1 × 10 N ⋅ s m 58.14 kg m × 0.0819 W m ⋅ K (800 − 373 ) K   

)

(

h conv = 2155 W m 2 ⋅ K where h ′fg = h fg + 0.8c p,v ( Ts − Tsat ) = 2257 kJ kg + 0.8 × 7065 J kg ⋅ K (800 − 373 ) K = 4670 kJ kg. To estimate the radiation coefficient, use Eq. 10.11, 4 εσ Ts4 − Tsat 0.25σ 8004 − 3734 K 4 h rad = = = 13.0 W m 2 ⋅ K . Ts − Tsat 800 373 K − ( ) Since h rad < h conv , use the simpler expression,

(

)

(

)

h = 2155 W m 2⋅ K + (3 4 )13.0 W m 2⋅ K = 2165 W m 2⋅ K. Using the rate equation, find Continued...

PROBLEM 10.33 (Cont.) q′′s = 2165 W m 2 ⋅ K (800 − 373) K = 0.924 MW m 2 .

<

(b) Using the IHT Correlations Tool, Boiling, Film Pool Boiling, combined with the Properties Tool for Water, the heat flux, q′′s , was calculated as a function of the excess temperature, ∆Te for emissivities of

0.1, 0.25 and 0.95. Also plotted is the ratio (%) of q′′rad q′′ as a function of ∆Te . 3

8E6

1E6 600000

2 q''rad/q''s (%)

Heat flux, q'' (W/m^2)

4E6

200000 80000 40000

1

10000 100

400 Temperature excess, deltaTe (K) eps = 0.1 eps = 0.25 eps = 0.95 Critical heat flux, q''max (W/m^2) Minimum heat flux, q''min (W/m^2)

800

0 100

200

300

400

500

600

Temperature excess, deltaTe (K) eps = 0.1 eps = 0.25 eps = 0.95

From the q′′s vs. ∆Te plot, note that the heat rate increases markedly with increasing excess temperature. On the plot scale, the curves for the three emissivity values, 0.1, 0.25 and 0.95, overlap indicating that the overall effect of emissivity change on the total heat flux is slight. Also shown on the plot are the critical heat flux, q′′max = 1.26 MW/m2, and the minimum heat flux, q′′min = 18.9kW/m2, at the Leidenfrost point. These values are computed in Example 10.1. Note that only for the extreme value of ∆Te is the heat flux in film pool boiling in excess of the critical heat flux. The relative contribution of the radiation mode is evident from the q′′rad qs′′ vs. ∆Te plot. The maximum contribution by radiation is less than

3% and surprisingly doesn’t occur at the maximum excess temperature. By examining a plot of q′′rad vs. ∆Te , we’d see that indeed q′′rad increases markedly with increasing ∆Te ; however, q′′conv increases even more markedly so that the relative contribution of the radiation mode actually decreases with increasing temperature for ∆Te > 250 K. Note that, as expected, the radiation heat flux, q′′rad , is proportional to the emissivity. COMMENTS: Since q′′s < q′′max = 1.26 MW m 2 , the prescribed condition can only be achieved in power-controlled heating by first exceeding q′′max and then decreasing the flux to 0.924 MW/m2.

PROBLEM 10.34 KNOWN: Surface temperature and emissivity of strip steel. FIND: Heat flux across vapor blanket. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Vapor/jet interface is at Tsat for p = 1 atm, (3) Negligible effect of jet and strip motion. 3

PROPERTIES: Table A-6, Saturated water (100°C): ρ l = 957.9 kg/m , hfg = 2257 kJ/kg; 3

-6

2

Saturated water vapor (Tf = 640K): ρv = 175.4 kg/m , cp,v = 42 kJ/kg⋅K, µv = 32 × 10 N⋅s/m , -6 2 k = 0.155 W/m⋅K, ν v = 0.182 × 10 m /s. ANALYSIS: The heat flux is

q′′s = h∆Te where ∆Te = 907 K − 373 K = 534 K h = h conv + ( 3 / 4 ) h rad .

and

4/3 h 4 / 3 = hconv + h rad h1/3

With

h ′fg = h fg + 0.80cp,v ( Ts − Tsat ) = 2.02 ×107 J/kg

or

Equation 10.9 yields

(

)

1/4

 9.8m/s 2 ( 957.9 −175.4 ) k g / m 3 2.02 ×10 7 J / k g (1 m )3    Nu D = 0.62   − 6 2  0.182 ×10 m / s ( 0.155 W / m⋅ K )( 907 − 373) K 

= 6243.

Hence,

hconv = Nu D k v / D = 6243W/m 2 ⋅ K ( 0.155 W / m ⋅K / 1 m ) = 968 W / m 2 ⋅ K hrad =

(

4 ε σ Ts4 − Tsat

Ts − Tsat

) = 0.35 × 5.67 ×10−8 W / m 2 ⋅ K 4 (9074 − 3734 ) K 4 ( 907 − 373) K

hrad = 24 W / m 2 ⋅ K

(

)

Hence,

h = 968 W / m2 ⋅ K + ( 3 / 4) 24 W / m 2 ⋅ K = 986 W / m 2 ⋅ K

And

q′′s = 986 W / m 2 ⋅ K (907 − 373 ) K = 5.265 ×105 W / m 2 .

<

COMMENTS: The foregoing analysis is a very rough approximation to a complex problem. A more rigorous treatment is provided by Zumbrunnen et al. In ASME Paper 87-WA/HT-5.

PROBLEM 10.35 KNOWN: Copper sphere, 10 mm diameter, initially at a prescribed elevated temperature is quenched in a saturated (1 atm) water bath. FIND: The time for the sphere to cool (a) from Ti = 130 to 110°C and (b) from Ti = 550°C to 220°C . SCHEMATIC:

ASSUMPTIONS: (1) Sphere approximates lumped capacitance, (2) Water saturated at 1 atm. PROPERTIES: Table A-1, Copper: ρ = 8933 kg/m3; Table A.11, Copper (polished) : ε = 0.04, typical value; Table A.4, Water: as required for the pool boiling correlations. ANALYSIS: Treating the sphere as a lumped capacitance and performing an energy balance, see Eq. 5.14,

E in − E out = E st

− qs′′ ⋅ As = ρ c∀

dT dt

(1,2)

For the sphere, V = πD3 / 6 and As = πD2 . Using the IHT Lumped Capacitance Model to solve this differential equation, we need to specify (1) the specific heat of the copper sphere as a function sphere temperature; use IHT Properties Tool, Copper; and (2) the heat flux, qs′′ , associated with the pool boiling processes; use IHT Correlations Tool, Boiling: (a) Cooling from Ti =130° to 110°: Nucleate pool boiling, Rohsenhow correlation, Eq. 10.5, (b) Cooling from Ti = 550 to 220°C : Film Pool Boiling, Eq. 10.9 with C = 0.67 (sphere). The thermophysical properties for water required of the correlations are provided by the IHT Tool, Properties-Water. The specific heat of copper as a function of sphere temperature is provided by the IHT Tool, Properties-Copper. The temperature-time histories for each of the cooling processes are plotted below. Nucleate pool boiling quench process

Film pool boiling quench process

130

500

Sphere temperature, T (s)

Sphere temperature, Ts (C)

400 120

110

100

300

200

100 0

1

2

3

Elapsed time, t (s)

4

5

0

10

20

30

Elapsed time, t (s)

Continued...

PROBLEM 10.35 (Cont.) Using the Explore feature in the IHT Plot Window, the elapsed times for the quench process were found as: Quench process Ti - Tf (°C) ∆t(s) Nucleate pool boiling 130-110 0.76 Film pool boiling 550-220 13.5 COMMENTS: (1) Comparing the elapsed times for the two processes, the nucleate pool boiling process cools 20°C in 0.76s (26.3°C/s) vs. 330°C in 13.5s (24.4°C/s) for the film pool boiling process. (2) The IHT Workspace used to generate the temperature-time history for the nucleate pool boiling process is shown below. // Correlations Tool - Boiling, Nucleate Pool Boiling, Heat flux qs'' = qs_dprime_NPB(Csf,n,rhol,rhov,hfg,cpl,mul,Prl,sigma,deltaTe,g) // Eq 10.5 g = 9.8 // Gravitational constant, m/s^2 deltaTe = Ts - Tsat // Excess temperature, K Ts = Ts_C + 273 // Surface temperature, K //Ts_C = 130 Tsat = 100 + 273 // Saturation temperature, K /* Evaluate liquid(l) and vapor(v) properties at Tsat. From Table 10.1 (Fill in as required), */ // fluid-surface combination: Csf = 0.013 // Polished copper-water combination, Table 10.1 n = 1.0 /* Correlation description: Heat flux for nucleate pool boiling (NPB), water-surface combination (Cf,n), Eq 10.5, Table 10.1 . See boiling curve, Fig 10.4 . */ // Properties Tool- Water: // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); xv = 1 // Quality (0=sat liquid or 1=sat vapor) rhov = rho_Tx("Water",Tsat,xv) // Density, kg/m^3 hfg = hfg_T("Water",Tsat) // Heat of vaporization, J/kg sigma = sigma_T("Water",Tsat) // Surface tension, N/m (liquid-vapor) // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); xl = 0 // Quality (0=sat liquid or 1=sat vapor) rhol = rho_Tx("Water",Tsat,xl) // Density, kg/m^3 cpl = cp_Tx("Water",Tsat,xl) // Specific heat, J/kg·K mul = mu_Tx("Water",Tsat,xl) // Viscosity, N·s/m^2 Prl = Pr_Tx("Water",Tsat,xl) // Prandtl number // Lumped Capacitance Model: /* Conservation of energy requirement on the control volume, CV. */ Edotin - Edotout = Edotst Edotin = 0 Edotout = As * ( + qs'' ) Edotst = rho * vol * cp * Der(Ts,t) /* The independent variables for this system and their assigned numerical values are */ As = pi * D^2 / 4 // surface area, m^2 vol = pi * D^3 / 6 // volume, m^3 D = 0.01 rho = 8933 // density, kg/m^3 // Properties Tool - Copper // Copper (pure) property functions : From Table A.1 // Units: T(K) cp = cp_T("Copper",Ts) // Specific heat, J/kg·K

PROBLEM 10.36 KNOWN: Saturated water at 1 atm is heated in cross flow with velocities 0 – 2 m/s over a 2 mmdiameter tube. FIND: Plot the critical heat flux as a function of water velocity; identify the pool boiling and transition regions between the low and high velocity ranges. SCHEMATIC:

ASSUMPTIONS: Nucleate boiling in the presence of external forced convection. 3

3

PROPERTIES: Table A-6, Water (1 atm): Tsat = 100°C, ρ l = 957.9 kg/m , ρ v = 0.5955 kg/m , hfg -3 = 2257 kJ/kg, σ = 58.9 × 10 N/m. ANALYSIS: The Lienhard-Eichhorn correlations for forced convection boiling with cross flow over a cylinder are appropriate for estimating q ′′max , Eqs. 10.12 and 10.13. Low Velocity 1/3   ρ v h fg   4σ   q′′max = 1 +  ρ V 2D   V π     v 1/3   1 kg J   4 × 58.9 ×10 −3 N / m   3 q′′max = 0.5955 × 2257 ×10 1+  V π kg   0.5955kg/m3 V2 0.002m   m3  

q′′max = 4.2782 ×105 V + 2.921×10 6 V1/3. High Velocity 1/3  1/2   ρ v h fg  1  ρl 3 / 4 1  ρl   σ q′′max = +  V      π 169  ρ v  19.2  ρ v   ρv V2 D    

q′′max =

1 kg J  1  957.9 3 / 4 0.5955 × 2257 ×103  +  0.5955  3 π kg 169    m  1/3 

 1  957.9 1/2  58.9 ×10−3 N / m      19.2  0.5955   0.5955kg/m3 V2 0.002m 

V  

q′′max = 6.4299 ×105 V + 3.280 ×10 6 V1/3 Continued …..

PROBLEM 10.36 (Cont.) The transition between the low and high velocity regions occurs when 1/2   0.275  ρl   ′′ q max = ρ v h fg V + 1    π  ρv    

q′′max = 0.5955

J  0.275  957.9 1/2  3 × 2257 ×10 V + 1 = 6.0627 ×106 V.   3

kg m

kg

 π

 0.5955 



(3)

For pool boiling conditions when the velocity is zero, the critical heat flux must be estimated according to the correlation for the small horizontal cylinder as introduced in Problem 10.22. If the cylinder were 2 “large,” the critical heat flux would be 1.26 MW/m as given by the Zuber-Kutateladze correlation, Eq. 10.7. Following the analysis of Problem 10.22, find Bo = 0.40 and the critical heat flux for the “small” 2 mm cylinder is

q′′max ) pool = 1.18 ×1.26 M W / m 2 = 1.49 W / m 2. The graph below identifies four regions: pool boiling where q ′′max = 1.49 M W / m2 from V = 0 to 0.15 m/s and the low velocity, transition and high velocity regimes.

PROBLEM 10.37 KNOWN: Saturated water at 1 atm and velocity 2 m/s in cross flow over a heater element of 5 mm diameter. FIND: Maximum heating rate, q ′[ W / m] . SCHEMATIC:

ASSUMPTIONS: Nucleate boiling in the presence of external forced convection. 3

3

PROPERTIES: Table A-6, Water (1 atm): Tsat = 100°C, ρ l = 957.9 kg/m , ρ v = 0.5955 kg/m , hfg -3 = 2257 kJ/kg, σ = 58.9 × 10 N/m. ANALYSIS: The Lienhard-Eichhorn correlation for forced convection with cross flow over a cylinder is appropriate for estimating q ′′max . Assuming high-velocity region flow, Eq. 10.13 with Eq. 10.14 can be written as 1/3  1/2   ρ v h fg V  1  ρl 3 / 4 1  ρl   σ q′′max = +  .     169  ρ  π 19.2  ρ v   ρ v V 2D    v   





Substituting numerical values, find

q′′max =

 1  957.9 3 / 4 1 0.5955kg/m3 × 2257 ×103 J / k g × 2 m / s  +   π 169  0.5955  1/2 

1  957.9  19.2  0.5955 

1/3 

    0.5955kg/m 3 ( 2 m / s ) 2 0.005m    58.9 × 10− 3 N / m

   

q′′max = 4.331MW/m 2 . The high-velocity region assumption is satisfied if 1/2 q′′max ? 0.275  ρl  +1

ρ v h fg V < π

   ρv 

4.331×106 W / m2 0.5955kg/m 3 × 2257 ×10 3 J / k g × 2 m / s

? 0.275 957.9 1/2  

= 1.61<

π

   0.5955 

The inequality is satisfied. Using the q ′′max estimate, the maximum heating rate is q′max = q′′max ⋅π D = 4.331MW/m 2 ×π ( 0.005m ) = 68.0kW/m.

+ 1 = 4.51.

<

COMMENTS: Note that the effect of the forced convection is to increase the critical heat flux by 4.33/1.26 = 3.4 over the pool boiling case.

PROBLEM 10.38 KNOWN: Correlation for forced-convection local boiling inside a vertical tube. FIND: Boiling heat transfer rate per unit length of the tube. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Local boiling occurs when tube wall is 15°C above the saturation temperature. ANALYSIS: From experimental results, the heat transfer coefficient can be estimated by the correlation

p  h = 2.54 ( ∆Te )3 exp    15.3 

 W / m2 ⋅ K   

where ∆Te is the excess temperature, Ts − Tsat [ K ] , and p is the pressure [bar]. The heat transfer rate per unit length is

q′ = π D h ∆Te . Evaluating the heat transfer coefficient, find

h = 2.54 (15K ) exp (4 bar/15.3) = 11,134 W / m 2 ⋅ K. 3

The heat rate is then.

q′ = π ( 0.050m ) ×11,134W/m 2 ⋅ K ×15K = 26.2 kW/m. COMMENTS: The saturation temperature at 4 bar is Tsat = 406.5K according to Table A-6.

<

PROBLEM 10.39 KNOWN: Forced convection and boiling processes occur in a smooth tube with prescribed water velocity and surface temperature. FIND: Heat transfer rate per unit length of the tube. SCHEMATIC:

ASSUMPTIONS: (1) Fully-developed flow, (2) Nucleate boiling conditions occur on inner wall of tube, (3) Forced convection and boiling effects can be separately estimated. 3

PROPERTIES: Table A-6, Water (Tm = 95°C = 368K): ρ l = 1/vf = 962 kg/m , ρ v = 1/vg = 0.500 3 -6 2 kg/m , hfg = 2270 kJ/kg, cp,l = 4212 J/kg⋅K, µl = 296 × 10 N⋅s/m , k l = 0.678 W/m⋅K, Prl = 1.86, -3

-7

2

σ = 60 × 10 N/m, ν l = 3.08 × 10 m /s. ANALYSIS: Experimentation has indicated that the heat transfer rate can be estimated as the sum of the separate effects due to forced convection and boiling. On a per unit length basis, q′ = q′fc + q ′boil. For forced convection, ReD = u m D / ν l = 1.5m/s × 0.015m/3.08 × 10 −7 m2 / s = 73,052. Since Re > 2300, flow is turbulent and since fully developed, use the Dittus-Boelter correlation but with the 0.023 coefficient replaced by 0.019 and n = 0.4, NuD = h D / k = 0.019Re 4D/ 5 Pr n k 0.678W/m ⋅ K h = Nu D = × 0.019 ( 73,052 ) 4 / 5 (1.86 )0.4 = 8563W/m 2 ⋅ K.

D

0.015m

q′fc = h π D ( Ts − Tm ) = 8,563W/m 2 ⋅ K ⋅ π ( 0.015m )(110 − 95 ) ° C = 6052W/m. For boiling, ∆Te = (110 – 100)°C = 10°C and hence nucleate boiling occurs. From the Rohsenow equation, with Csf = 0.006 and n = 1.0, 3 1/2  c p,l ∆Te   g ( ρl − ρ v )    q′′boil = µl h fg   σ  Csf hfg Pr n    l   q ′boil = 296 × 10

−6 N ⋅ s 3 J × 2270 × 10 2 kg

m

q ′′boil = 1.22 ×106 W / m2

1/2 

 9 . 8 m / s2 ( 962 − 0.5) k g / m3    −3   60 × 10 N / m

   4212J/kg ⋅ K × 10K   J  0.006 × 2270 × 103 × ( 1.86 )1.0   kg 

3

′ = qboil ′′ ( π D ) = 1.22 × 106 W / m2 (π × 0.015m ) = 57,670W/m. q boil

The total heat rate for both processes is

q′ = ( 6052 + 57,670 ) W / m = 6.37 ×104 W/m.

<

COMMENTS: Recognize that this method provides only an estimate since the processes are surely coupled.

PROBLEM 10.40 KNOWN: Saturated steam condensing on the outside of a brass tube and water flowing on the inside of the tube; convection coefficients are prescribed. FIND: Steam condensation rate per unit length of the tube. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions. 3

PROPERTIES: Table A-6, Water, vapor (0.1 bar): Tsat ≈ 320K, hfg = 2390 × 10 J/kg; Table A-1, Brass ( T = ( Tm + Tsat ) / 2 ≈ 300K ) : k = 110 W / m ⋅ K ANALYSIS: The condensation rate per unit length follows from Eq. 10.33 written as

& ′ = q ′/ h ′fg m

(1)

where the heat rate follows from Eq. 10.32 using an overall heat transfer coefficient

q′ = Uo ⋅π Do ( Tsat − Tm )

and from Eq. 3.31,

 1 D /2 D D 1 Uo =  + o ln o + o  k Di Di h i   ho

(2) −1 (3)

  1 0.0095m 19 19 1 Uo =  + ln +   6800W/m 2 ⋅ K 1 1 0 W / m ⋅ K 16.5 16.5 5200W/m 2 ⋅ K  Uo = 147.1× 10−6 + 12.18× 10−6 + 192.3× 10−6   

−1

−1

W / m 2 ⋅ K = 2627W/m 2 ⋅ K.

Combining Eqs. (1) and (2) and substituting numerical values (see below for h ′fg ), find

& ′ = U oπD o ( Tsat − Tm ) / h ′fg m & ′ = 2627W/m 2 ⋅ Kπ ( 0.019m )( 320 − 303 ) K/2410 ×103 J / k g = 1.11×10 −3 kg/s. m

<

COMMENTS: (1) Note from evaluation of Eq. (3) that the thermal resistance of the brass tube is not negligible. (2) From Eq. 10.26, with Ja = c p, l ( Tsat − T s ) / h fg , h ′fg = h fg [1 + 0.68Ja ] . Note from expression for Uo, that the internal resistance is the largest. Hence, estimate Ts,o ≈ To – (Ro/ΣR) (To – Tm) ≈ 313K. Hence, h ′fg ≈ 2390 ×103 J / k g 1 + 0.68 × 4179J/kg ⋅ K (320 − 313 )K / 2390 ×103J /kg   

h ′fg = 2410kJ/kg

where c p, l for water (liquid) is evaluated at Tf = (Ts,o + To)/2 ≈ 317K.

PROBLEM 10.41 KNOWN: Insulated container having cold bottom surface and exposed to saturated vapor. FIND: Expression for growth rate of liquid layer, δ(t); thickness formed for prescribed conditions; compare with vertical plate condensate for same conditions. SCHEMATIC:

ASSUMPTIONS: (1) Side wall effects are negligible and, (2) Vapor-liquid interface is at Tsat , (3) Temperature distribution in liquid is linear, (4) Constant properties. 3

PROPERTIES: Table A-6, Saturated vapor (p = 1.0133 bar): Tsat = 100°C, ρ v = 0.596 kg/m , hfg = 3 -6 2257 kJ/kg; Table A-6, Saturated liquid (Tf = 90°C = 363K): ρl = 1000 kg/m , µl = 313 × 10 2

N⋅s/m , kl = 0.676 W/m⋅K, c p, l = 4207 J/kg⋅K. ANALYSIS: Perform a surface energy balance on the interface (see above) recognizing that & l / A = ρl dδ /dt from an overall mass rate balance on the liquid to obtain m

& m T −T dδ T −T E& ′′in − E& ′′out = q′′conds − q′′cond = h fg − k l sat s = ρl h fg − k l sat s = 0 (1) A δ dt δ where q ′′conds is the condensation heat flux and q ′′cond is the conduction heat flux into the liquid layer of thickness δ with linear temperature distribution. Eq. (1) can be rewritten as

ρl h fg

dδ T −T = k l sat s . dt δ

Separate variables and integrate with limits shown to obtain the liquid layer growth rate, δ t kl ( Tsat − Ts ) δ dδ = dt 0 0 ρl hfg

∫

∫

1/2

or

 2k ( T − T )  δ =  l sat s t  ρl hfg  

.

(2)

<

For the prescribed conditions, the liquid layer thickness and condensate formed in one hour are 1/2

 W kg J  δ (1hr ) =  2 × 0.676 (100 − 80 ) °C × 3600s/1000 3 × 2257 × 103  m⋅K kg   m

= 6.57mm

<

M (1hr ) = ρl Aδ = 1000kg/m 3 × 200 ×10 −6 m 2 × 6.57 ×10−3m = 1.314 ×10 −3 kg.

<

Continued …..

PROBLEM 10.41 (Cont.) The condensate formed on a vertical plate with the same conditions follows from Eq. 10.33,

& ⋅ t = h L A ( Tsat − Ts ) ⋅ t / h ′fg M vp = m where h ′fg and h L follow from Eqs. 10.26 and 10.30, respectively.

(

h ′fg = h fg (1 + 0.68Ja ) = h fg 1 + 0.68cp,l ∆T / hfg

)

  J h ′fg = 2257 ×103 J / k g  1+ 0.68 × 4207 (100 − 80 ) °C/2257 ×103 J / k g  = 2314kJ/kg kg ⋅ K   1/4

hL = 0.943  g ρl ( ρl − ρv ) k3l h′fg / µl ( Tsat − Ts ) L   

hL = 0.943 9.8m/s2 × 1000kg/m3 (1000 − 0.596) k g / m3 ( 0.676W/m ⋅ K )3  1/4

×2314 ×103 J/kg/313 ×10 −6 N ⋅ s / m 2 (100 − 80 ) °C × 0.2m   hL = 8155 W / m 2 ⋅ K. Hence,

M vp = 8155W/m 2 ⋅ K × 200 ×10 −6 m 2 (100 − 80 ) °C × 3600s/2314 × 103 J/kg M vp = 5.08 × 10−2 kg.

<

COMMENTS: (1) Note that the condensate formed by the vertical plate is an order of magnitude larger. For the vertical plate the rate of condensate formation is constant. For the container bottom surface, the rate decreases with increasing time since the conduction resistance increases as the liquid layer thickness increases. (2) For the vertical plate, assumed to be square 14.1 × 14.1 mm, the Reynolds number, Eq. 10.35 and 10.33, is

Reδ =

Reδ =

& 4m 4 hL A ( Tsat − Ts ) = µl b µl b h′fg

(

)

4

8155W/m 2 ⋅ K 200 × 10−6 m2 (100 − 80) ° C

313 ×10 −6 N ⋅ s / m 2 ×14.1×10 −3 m

2314 kJ/kg

Reδ = 12.8. Hence, using Eq. 10.30 to estimate h L is correct since, in fact, the film is laminar.

PROBLEM 10.42 KNOWN: Vertical tube experiencing condensation of steam on its outer surface. FIND: Heat transfer and condensation rates. SCHEMATIC:

ASSUMPTIONS: (1) Film condensation, (2) Negligible non-condensibles, (3) D/2 >> δ, vertical plate behavior. 3

PROPERTIES: Table A-6, Water, vapor (1.0133 bar): Tsat = 100°C, ρ v = 0.596 kg/m , hfg = 2257 3 -6 2 kJ/kg; Table A-6, Water, liquid (Tf = 97°C): ρl = 960.6 kg/m , µl = 289 × 10 N⋅s/m , c p, l = 4214 J/kg⋅K, kl = 0.679 W/m⋅K. ANALYSIS: The heat transfer and condensation rates are

q = h L (π D L )( Tsat − Ts )

& = q / h ′fg m

where h ′fg = h fg (1 + 0.68Ja ) and Ja = c p,l ( Tsat − Ts) / h fg. Hence Ja = 4214 J/kg⋅K (100 3

94)K/2257 × 10 J/kg = 0.0112 and h ′fg = 2274 kJ/kg. Assume laminar film condensation and use Eq. 10.31 to estimate h L , 1/4

Nu L =

hL =

hLL kl

0.679W/m ⋅ K

Hence,

1.0m

 ρ l g ( ρ l − ρ v ) h ′fg L3   = 0.943   µ l k l ( Tsat − Ts ) 

1/4

 960.6kg/m 3 × 9 . 8 m / s 2 ( 960.6 − 0.596 ) k g / m 3 × 2274 × 103J / kg × (1m ) 3  × 0.943   −6 2  289 × 10 N ⋅ s / m × 0.679W/m ⋅ K ( 100 − 94 ) K 

q = 7360W/m 2 ⋅ K ( π × 0.100m × 1m )(100 − 94 ) K = 13.87kW. & = 13.9 ×103 W/2274 ×10 3 J / k g = 0.00610kg/s. m -6

2

Check the laminar film assumption: Reδ = 4 m& / µ lb = 4 × 0.00610 kg/s/289 × 10 N⋅s/m × (π × 0.100m) = 269. Since 30 < Reδ < 1800, the flow is wavy, not laminar. By combining Eqs. 10.33 and 10.35 with 10.38 (see Example 10.3), find Reδ, Reδ µ l b h ′fg 4 A s ( Tsat − Ts )

=

Reδ 1.22

1.08Reδ

− 5.2

⋅

kl

(ν / g ) 2 l

1/3

Continued …..

2

= 7 3 6 0 W / m ⋅ K.

PROBLEM 10.42 (Cont.) −6

289 × 10

( π 0.10m ) × 2274 × 103J / kg 1 = ⋅ 1.22 4 ( π × 0.10m × 1m ) (100 − 94 ) K 1.08Reδ − 5.2  N⋅s / m

2

(



0.679W/m ⋅ K −6

289 × 10

/960.6

)

2

4

2

1/3 2

m / s /9.8m/s



Solving, we obtain Reδ = 311. Using Eq. 10.38, find

(

2

hL ν l / g

)

1/3

=

kl

Reδ 1.22 1.08Reδ − 5.2

2

hL = 8 5 0 7 W / m ⋅ K

q = 8 5 0 7 W / m ⋅ K ( π × 0.10m × 1m )(100 − 94 ) K = 16.0kW

<

& = 16.0 ×10 3 W / 2 2 7 4 ×10 3 J / k g = 7.05 ×10 −3 kg/s. m

<

2

COMMENTS: To determine whether the assumption D/2 >> δ is satisfied, use Eq. 10.25 to estimate δ(L) ≈ 0.12mm. Despite the laminar film assumption, clearly the assumption is justified and the vertical plate correlation is applicable.

PROBLEM 10.43 KNOWN: Vertical tube experiencing condensation of steam on its outer surface. FIND: Heat transfer and condensation rates. SCHEMATIC:

ASSUMPTIONS: (1) Film condensation, (2) Negligible condensibles in steam, (3) D/2 >> δ, vertical plate behavior. 3

PROPERTIES: Table A-6, Water, vapor (1.5 bar): Tsat ≈ 385K, ρ v = 0.876 kg/m , hfg = 2225 3 -6 kJ/kg; Table A-6, Water, (liquid Tf = 376K): ρl = 956.2 kg/m , c p, l = 4220 J/kg⋅K, µl = 271 × 10 2

N⋅s/m , kl = 0.681 W/m⋅K. ANALYSIS: The heat transfer and condensation rates are

q = h L (π D L )( Tsat − Ts )

& = q / h ′fg m

where h ′fg = h fg (1 + 0.68Ja ) and Ja = cp, l ( Tsat − Ts ) / h fg . Hence, Ja = 4220 J/kg⋅K (385 – 3 367)K/2225 × 10 J/kg = 0.0171 and h ′fg = 2277 kJ/kg. Assume the flow is wavy. Combine Eqs.

10.33 and 10.35 with 10.38, find Reδ.

Reδ µ l b h ′fg Reδ = ⋅ 4 As ( Tsat − Ts ) 1.08Re1.22 − 5.2 δ

(

kl

ν2 /g

)

1/3

271× 10−6 N ⋅ s / m 2 (π × 0.10m ) × 2277 × 103 J/kg 4 × ( π × 0.10m ×1m )( 385 − 367 ) K =

1

⋅

0.681W/m ⋅ K

1/3 1.08Re1.22 δ − 5.2  271×10 −6 /956.2 2 m 4 / s 2 /9.8m/s2     

Reδ = 832. Using Eq. 10.38, find

(

h L ν l2 / g kl

)

(

)

1/3

=

Reδ 1.08Re1.22 δ − 5.2

h L = 7,127W/m 2 ⋅ K.

q = 7127W/m 2 ⋅ K (π × 0.1m ×1m )( 385 − 367 ) K = 40.3kW

< <

& = 40.3 ×103 W/2277 ×103 J / k g = 0.0177kg/s. m Continued …..

PROBLEM 10.43 (Cont.) COMMENTS: Since 30 < Reδ < 1800, the wavy flow film assumption is justified. By comparing these results with those of Problem 10.42, the effect of increased pressure on condensation can be seen. p (bar)

Tsat(K)

Tsat-Ts(K)

373 385

6 18

(

2

hL W / m ⋅ K

)

q (kW)

& ⋅ 10 3 ( k g / s ) m 1.01 1.5

8507 7127

16.0 40.3

7.05 17.7

The effect of increasing the pressure from 1.01 to 1.5 bar is to increase the excess temperature threefold, to decrease hL by 16%, and to increase the rates by a factor of 2.5.

PROBLEM 10.44 KNOWN: Saturated steam at one atmosphere condenses on the outer surface of a vertical tube; water flow within tube experiences 4°C temperature rise. FIND: Required flow rate to maintain tube wall at 94°C. SCHEMATIC:

ASSUMPTIONS: (1) Laminar wavy film condensation on a vertical surface, (2) Negligible concentration of non-condensible gases in the stream, (3) Thermal resistance of tube wall is negligible, (4) Water flow is fully developed, (5) Tube wall surface is at uniform temperature Ts. -6

2

PROPERTIES: Table A-6, Water (Assume Tm ≈ 300K): cp = 4179 J/kg⋅K, µ = 855 × 10 N⋅s/m , k = 0.613 W/m⋅K, Pr = 5.83. ANALYSIS: From the results of Problem 10.42, the heat rate for laminar wavy condensation on the outside surface of the tube was found to be q = 16.0 kW. From an energy balance on the water flowing within the tube, the flow rate is & = q / c p Tm,o − Tm,i = 16.0 ×103 W/4179J/kg ⋅ K × 4K = 0.957kg/s. m (1) To determine the inlet temperature of the water, the rate equation is required.

(

)

q = U A ∆Tlm

<

U=

1 1 / hi + 1 / ho

∆Tlm =

∆T1 − ∆ T2 . ln ( ∆T1 / ∆T2 )

(2,3,4)

2

From Problem 10.42, ho = 8507 W/m ⋅K. Evaluate Re for the water flow using Eq. 8.6

& π Dµ = 4 × 0.957kg/s/ π × 0.092m × 855 ×10 −6 N ⋅ s / m 2 = 15,493. Re = 4m/ The flow is turbulent and since fully-developed, Eq. 8.60 is an appropriate correlation.

Nu = h i D i / k = 0.023Re4D/ 5 Pr 0.4 = 0.023 (15,493)

4/5

( 5.83 )0.4 = 104.7

h i = Nu ⋅ k / Di = 104.7 × 0.613W/m⋅ K/0.092m = 6 9 8 W / m 2 ⋅ K. 2

Hence, U = 1/[1/698 + 1/8507] = 645 W/m ⋅K. Substituting numerical values into the rate equation, Eq. (2), with A = π Di L, find

∆Tlm = q / U A = 16.0 × 103 W / 6 4 5 W / m 2 ⋅ K × (π 0.092m × 1m ) = 85.6K. Recalling now Eq. (4), note that ∆T1 - ∆T2 = 4K and that Tm,o – Tm,i = 4K, hence,

85.6K = 4K/ln

94 − Tm,i

(

94 − Tm,i + 4

)

giving

Tm,i = 6.3°C.

COMMENTS: Note that the Tm = 300K assumption is not reasonable and an iteration should be made. Also, it is likely that the thermal resistance of the tube wall is not negligible.

PROBLEM 10.45 KNOWN: Cooled vertical plate 500-mm high and 200-mm wide condensing saturated steam at 1 atm.

 = 25 kg/h, (b) FIND: (a) Surface temperature, Ts, required to achieve a condensation rate of m  ≤ 50 kg h , and (c) Compute and plot Ts as a function of the condensation rate for the range 15 ≤ m  , but if the plate is 200 mm high and 500 mm wide (vs. 500 Compute and plot Ts for the same range of m mm high and 200 mm wide for parts (a) and (b)). SCHEMATIC:

ASSUMPTIONS: (1) Film condensation, (2) Negligible non-condensables in steam. PROPERTIES: Table A-6, Water, vapor (1.0133 bar): Tsat = 100°C, ρv = 0.5963 kg m3 , hfg = 2257 kJ/kg; Table A-6, Water, liquid (Tf ≈ (74 + 100)°C/2 ≈ 360 K): ρ" = 967.1 kg/m3 , cp," = 4203 J/kg⋅K, µ" = 324 × 10-6 N⋅s/m2 , k " = 0.674 W/m⋅K. ANALYSIS: (a) The surface temperature can be determined from the rate equation, Eq. 10.32, written as

 ′fg h L As Ts = Tsat − q h L As = Tsat − mh where h ′fg = hfg (1 + 0.68 Ja) and Ja = c p," (Tsat − Ts )/hfg . To evaluate Ts, we need values of hL and h ′fg , both of which require knowledge of Ts . Hence, we need to assume a value of Ts and iterate the solution until good agreement with calculated Ts value is achieved. Assume Ts = 74°C and evaluate h ′fg and Reδ .

(

)

h ′fg = 2257 kJ kg 1 + 0.68  4203J kg ⋅ K (100 − 74 ) K 2257 × 103 J kg  = 2331kJ kg





−6  µ b = 4 × ( 25 3600 ) kg s 324 × 10 Reδ = 4m N ⋅ s m 2 × 0.2m = 429 . "

Since 30 < Reδ < 1800, the flow is wavy-laminar and Eq. 10.38 is appropriate, Reδ k" hL = ⋅ 1/ 3 1.08 Re1.22 δ − 5.2 ν 2 g

( ) "

hL =

429 108 ( 429 )

1.22

⋅

− 5.2 

0.674 W m ⋅ K

(

)

1/ 3

= 7312 W m 2 ⋅ K

2 4 2 2 −6  324 × 10 967.1 m s 9.8 m s    kg  J  W 7312 Hence, Ts = 100$ C − ( 25 3600 ) × 2331 × 103 × ( 0.2 × 0.5 ) m 2  = 78$ C .  2 kg  s  m ⋅K

<

This value is to be compared to the assumed value of 74°C. See comment 1. (b,c) Using the IHT Correlations Tool, Film Condensation, Vertical Plate for laminar, wavy-laminar and turbulent regions, combined with the Properties Tool for Water, the surface temperature Ts was Continued...

PROBLEM 10.45 (Cont.)  , considering the two plate configurations as calculated as a function of the condensation rate, m indicated in the plot below.

Plate temperature, Ts (C)

100

80

60

40 15

25

35

45

Condensation rate, mdot (kg/h) 500 mm high x 200 mm wide 200 mm high x 500 mm wide

As expected the condensation rate increases with decreasing surface temperature. The plate with the shorter height (L = 200 mm vs 500 mm) will have the thinner boundary layer and, hence, the higher average convection coefficient. Since both plate configurations have the same total surface area, the 200mm height plate will have the larger heat transfer and condensation rates. For the range of conditions examined, the condensate flow is in the wavy-laminar region. COMMENTS: (1) With the IHT model developed for parts (b) and (c), the result for the part (a) 2  = 25 kg/h is Ts = 78.2°C (Reδ = 439 and h L = 7403 W/m ⋅ K) . Hence, the assumed conditions with m value (Ts = 74°C) required to initiate the analysis was a good one. (2) A copy of the IHT Workspace model used to generate the above plot is shown below. /* Correlations Tool - Film Condensation, Vertical Plate, Laminar, wavy-laminar and turbulent regions: */ NuLbar = NuL_bar_FCO_VP(Redelta,Prl) // Eq 10.37, 38, 39 NuLbar = hLbar * (nul^2 / g)^(1/3) / kl g = 9.8 // Gravitational constant, m/s^2 Ts = Ts_C + 273 // Surface temperature, K Ts_C = 78 // Initial guess value used to solve the model Tsat = 100 + 273 // Saturation temperature, K // The liquid properties are evaluated at the film temperature, Tf, Tf = Tfluid_avg(Ts,Tsat) // The condensation and heat rates are q = hLbar * As * (Tsat - Ts) // Eq 10.32 As = L * b // Surface Area, m^2 mdot = q / h'fg // Eq 10.33 h'fg = hfg + 0.68 * cpl * (Tsat - Ts) // Eq 10.26 // The Reynolds number based upon film thickness is Redelta = 4 * mdot / (mul * b) // Eq 10.35 // Assigned Variables: L = 0.5 // Vertical height, m b = 0.2 // Width, m mdot_h = mdot * 3600 // Condensation rate, kg/h //mdot_h = 25 // Design value, part (a) // Properties Tool - Water: // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); xl = 0 // Quality (0=sat liquid or 1=sat vapor) rhol = rho_Tx("Water",Tf,xl) // Density, kg/m^3 hfg = hfg_T("Water",Tsat) // Heat of vaporization, J/kg cpl = cp_Tx("Water",Tf,xl) // Specific heat, J/kg·K mul = mu_Tx("Water",Tf,xl) // Viscosity, N·s/m^2 nul = nu_Tx("Water",Tf,xl) // Kinematic viscosity, m^2/s kl = k_Tx("Water",Tf,xl) // Thermal conductivity, W/m·K Prl = Pr_Tx("Water",Tf,xl) // Prandtl number

PROBLEM 10.46 KNOWN: Plate dimensions, temperature and inclination. Pressure of saturated steam. FIND: (a) Heat transfer and condensation rates for vertical plate, (b) Heat transfer and condensation rates for inclined plate. SCHEMATIC:

ASSUMPTIONS: (1) Conditions correspond to the turbulent film region, (2) Constant properties. 3

PROPERTIES: Table A-6, saturated vapor (p=1.0133 bars): Tsat = 100°C, ρv = 0.596 kg/m , hfg = 2257 kJ/kg. Table A-6, saturated liquid (Tf = 75°C): ρ " = 975 kg / m3 , µ" = 375 × 10−6 N ⋅ s / m 2 , k " = 0.668 W / m ⋅ K, c p," = 4193 J / kg ⋅ K.

ANALYSIS: (a) Expressing hL in terms of Reδ by combining Eqs. (10.33) and (10.35) and substituting into Eq. (10.38), it follows that

Reδ µ" h′fg

4 L (Tsat − Ts )

=

Reδ

⋅

(

k"

)

1/ 3 1.08 Re1.22 δ − 5.2 ν "2 / g

(1)

where, with Ja = c p," ( Tsat − Ts ) / h fg = 0.0929, h ′fg = h fg (1 + 0.68 Ja ) = 2400 kJ / kg. From an iterative solution to Eq. (1), we obtain Reδ = 2370, and the assumption of a turbulent film is justified. From Eqs. (10.35) and (10.33) the condensation and heat rates are then

 = m

µ" b Reδ = 0.444 kg / s 4

<

 h′fg = 0.444 kg / s × 2.4 × 106 J / kg = 1.065 × 106 W q=m

<

From Eq. (10.32), we also obtain hL = q / [( bL )( Tsat − Ts )] = 5325 W / m 2 ⋅ K. (b) With hL(incl ) ≈ ( cos θ )1/ 4 hL , we obtain hL(incl ) ≈ 0.917 × 5325 W / m 2 ⋅ K = 4880 W / m 2 ⋅ K. If the inclination reduces hL by 8.73%, the heat and condensation rates are reduced by equivalent amounts. Hence,

 = 0.407 kg / s, m

q = 0.977 × 106 W

<

COMMENTS: The initial guess of a turbulent film region was motivated by the value of L = 2m, which was believed to be large enough for transition to turbulence. Note that the solution could also have been obtained by accessing the Film Condensation correlations of IHT, implementation of which does not require an assumption of flow conditions.

PROBLEM 10.47 KNOWN: Saturated ethylene glycol (1 atm) condensing on a vertical plate at 420K. FIND: Heat transfer rate to the plate and condensation rate. SCHEMATIC:

ASSUMPTIONS: (1) Film condensation, (2) Negligible non-condensible gases in vapor. 3

PROPERTIES: Table A-5, Ethylene glycol vapor (1 atm): Tsat = 470K, ρ v ≈ 0 kg/m , h fg = 812 kJ/kg; Table A-5, Ethylene glycol, liquid (Tf = (Ts + Tsat )/2 ≈ 445K; use properties at upper limit of table 373K): ρl = 1058.5 3

-2

kg/m , c p, l = 2742 J/kg⋅K, µl = 0.215 × 10

2

N⋅s/m , kl = 0.263, W/m⋅K.

ANALYSIS: The heat transfer and condensation rates are given by Eqs. 10.32 and 10.33. q = h L A s ( Tsat − Ts )

& = q / h ′fg , m

where h ′fg = h fg (1 + 0.68 Ja) and Ja = c p, l ( Tsat − Ts ) / h fg . Substituting property values at Tf = (Ts + 3 Tsat )/2, find h ′fg = 812 kJ/kg (1 + 0.68 [2742 J/kg⋅K (470 – 420)K/812 × 10 J/kg]) = 905 kJ/kg.

Assuming the flow is laminar, use Eq. 10.30 to evaluate hL . hL

 g ρ l ( ρl − ρv ) k 3l h ′fg  = 0.943    µ l ( Tsat − Ts ) L 

1/4

1/4

 9 . 8 m / s 2 × 1 0 5 8 . 5 k g / m 3 ( 1058.5 − 0 ) k g / m 3 ( 0 . 2 6 3 W / m ⋅ K )3 × 905 × 10 3J / k g    −2 2 0.215 × 10 N ⋅ s / m ( 470 − 420 ) K × 0.3m  

2

find hL = 1451 W/m ⋅K. Using the rate equations, find q = 1451 W / m ⋅ K ( 0.3 × 0.1 ) m 2

2

( 470 − 420 ) K = 2.18kW

& = 2.18 × 10 3 W / 9 0 5 ×10 3J / k g = 0.002405kg/s = 8.66kg/h. m

Determine whether the flow is indeed laminar:

& / µlb = 4 × 0.002405 Re δ = 4 m

-2

2

kg/s/0.215 × 10 N⋅s/m ×

0.1m = 44.7. Since 30 < Reδ < 1800, the flow is in the wavy-laminar region. Hence, the correlation of Eq. 10.38 is more appropriate. Combining Eq. 10.33 and 10.35 with 10.38 (see Example 10.3), Reδ µ l b h ′fg 4 A s ( Tsat − Ts ) −2

0.215 × 10

=

Reδ 1.22 1.08Reδ − 5.2

2

3

N ⋅ s / m × 0.1m × 905 × 10 J / k g

4 × ( 0.3 × 0.1 ) m

2

( 470 − 420 ) K

⋅

kl

(ν / g ) 2 l

1/3

0.263W/m ⋅ K

1

=

1.22

1.08Reδ

− 5.2

(

 

−2

0.215 × 10

)

4

2

2

/1058.5 m / s / 9 . 8 m / s

 

1/3

find Reδ = 45 and using Eq. 10.38, find hL =

Hence,

Reδ 1.22 1.08Reδ − 5.2

q = 2.21kW

⋅

kl

(ν / g ) 2 l

1/3

2

=1 4 7 0 W / m ⋅ K.

& = 2.44 × 10 −3 kg/s. m

COMMENTS: Note the wavy-laminar value of Reδ is within 1.3% of the laminar value.

<

PROBLEM 10.48 KNOWN: Vertical plate 2.5 m high at a surface temperature Ts = 54°C exposed to steam at atmospheric pressure. FIND: (a) Condensation and heat transfer rates, (b) Whether turbulent flow would still exist if the height were halved, and (c) Compute and plot the condensation rates for the two plate heights (2.5 m and 1.25 m) as a function of surface temperature for the range, 54 ≤ Ts ≤ 90°C. SCHEMATIC:

ASSUMPTIONS: (1) Film condensation, (2) Negligible non-condensables in steam. PROPERTIES: Table A-6, Water, vapor (1 atm): Tsat = 100°C, ρv = 0.596 kg/m3, hfg = 2257 kJ/kg; Table A-6, Water, liquid (Tf = (100 + 54)°C/2 = 350 K): ρ" = 973.7 kg/m3, k " = 0.668 W/m⋅K, µ" = 365 × 10-6 N⋅s/m2 , cp," = 4195 J/kg⋅K, Pr" = 2.29.

ANALYSIS: (a) The heat transfer and condensation rates are given by Eqs. 10.32 and 10.33,

q′ = h L L ( Tsat − Ts )

 ′ = q′ h ′fg m

(1,2)

where, from Eq. 10.26, with Ja = c p," (Tsat − Ts)/hfg ,

(

h′fg = h fg 1 + 0.68 c p," (Tsat − Ts ) h fg  h′fg = 2257

)

 4195 J kg ⋅ K (100 − 54 ) K   kJ  1 + 0.68    = 2388 kJ kg . 3 J kg kg     2257 10 ×  

Assuming turbulent flow conditions, Eq. 10.39 is the appropriate correlation, 1/ 3 2

(

hL ν " g

)

k"

=

Reδ

(

8750 + 58 Pr −0.5 Reδ0.75 − 253

Reδ > 1800

)

(3)

Not knowing Reδ or h L , another relation is required. Combine Eq. 10.33 and 10.35,

h ′fg  Re µ b  = δ "  . A ( Tsat − T )  4  A ( Tsat − T ) Substitute Eq. (4) for h L into Eq. (3), with A = bL, Reδ µ" bh′fg Reδ k" . = ⋅ 1/ 3 4 ( bL )(Tsat − T ) 8750 + 58 Pr −0.5 Re0.75 − 253 2 ν" g " δ hL =

 ′fg mh

(

)(

)

(4)

(5)

Using appropriate properties with L = 2.5 m, find Continued...

PROBLEM 10.48 (Cont.) 365 × 10−6 N ⋅ s m 2 × 2388 × 103 J kg (6) 4 × 2.5m (100 − 54 ) K 1 0.668 W m ⋅ K = ⋅ 1/ 3 −0.5 2 4 2 8750 + 58 ( 2.29 ) Reδ0.75 − 253   6 2 −  365 × 10 973.7 m s 9.8m s    Reδ = 2979 .

) (

(

)

Note that Reδ > 1800, so indeed the flow is turbulent, and using Eq. (4) or (3), find h L = 5645 W m 2 ⋅ K . From the rate equations (1) and (2), the heat transfer and condensation rates are q′ = 5645 W m 2 ⋅ K × 2.5m 100 − 54 K = 649k W m

(

)

< <

 ′ = 649 ×103 W m 2388 ×103 J kg = 0.272 kg s ⋅ m . m (b) If the height of the plate were halved, L = 1.25 m, Eq. (6) would only need to be modified for this new value. Using the calculated values for the LHS and the last term on the RHS, Eq. (6) becomes,

3.78960 =

1 −0.5

8750 + 58 ( 2.29 )

(

Reδ0.75 − 253

)

× 27, 493

(7)

and after some manipulation , find Reδ = 1280 .

Since 1800 > Reδ , the flow is not turbulent, but wavy-laminar. Now the procedure follows that of Example 10.3. For L = 1.25 m with wavy-laminar flow, Eq. 10.38 is the appropriate correlation. The calculations yield these results: Reδ = 1372 h L = 5199 W m 2 ⋅ K

q′ = 299 kW m

<

 ′ = 0.125 kg s ⋅ m . m

Note that the height was decreased by a factor of 2 while the rates decreased by a factor of 2.2! Would you have expected this result? (c) Using the IHT Correlation Tool, Film Condensation, Vertical Plate for laminar, wavy-laminar, and turbulent regions, combined with the Properties Tool for Water, the condensation rates were calculated as a function of the surface temperature considering the two plate heights indicated.

Condensation rate, mdot' (kg/s.m)

0.5

0.375

0.25

0.125

0 50

60

70

80

90

Plate temperature, Ts (C) 2.5 m high plate 1.25 m high plate

Continued...

PROBLEM 10.48 (Cont.) The condensation rate decreases nearly linearly with increasing surface temperature. The inflection in the upper curve (L = 2.5 m) corresponds to the flow transition at Reδ = 1800 between wavy-laminar and turbulent. For surface temperature lower than 76°C, the flow is turbulent over the 2.5 m plate. The flow over the 1.25 m plate is always in the wavy-laminar region. The fact that the 2.5 m plate experiences turbulent flow explains the height-rate relationship mentioned in the closing sentences of part (b). COMMENTS: A copy of the IHT model used to generate the above plot is shown below. /* Correlations Tool - Film Condensation, Vertical Plate, Laminar, wavy-laminar and turbulent regions: */ NuLbar = NuL_bar_FCO_VP(Redelta,Prl) // Eq 10.37, 38, 39 NuLbar = hLbar * (nul^2 / g)^(1/3) / kl g = 9.8 // Gravitational constant, m/s^2 Ts = Ts_C + 273 // Surface temperature, K Ts_C = 54 // Part (a) design condition Tsat = 100 + 273 // Saturation temperature, K // The liquid properties are evaluated at the film temperature, Tf, Tf = Tfluid_avg(Ts,Tsat) // The condensation and heat rates are q = hLbar * As * (Tsat - Ts) // Eq 10.32 As = L * b // Surface Area, m^2 mdot = q / h'fg // Eq 10.33 h'fg = hfg + 0.68 * cpl * (Tsat - Ts) // Eq 10.26 // The Reynolds number based upon film thickness is Redelta = 4 * mdot / (mul * b) // Eq 10.35 // Assigned Variables: L = 1.25 // Height, m b=1 // Width, m // Properties Tool - Water: // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); xl = 0 // Quality (0=sat liquid or 1=sat vapor) rhol = rho_Tx("Water",Tf,xl) // Density, kg/m^3 hfg = hfg_T("Water",Tsat) // Heat of vaporization, J/kg cpl = cp_Tx("Water",Tf,xl) // Specific heat, J/kg·K mul = mu_Tx("Water",Tf,xl) // Viscosity, N·s/m^2 nul = nu_Tx("Water",Tf,xl) // Kinematic viscosity, m^2/s kl = k_Tx("Water",Tf,xl) // Thermal conductivity, W/m·K Prl = Pr_Tx("Water",Tf,xl) // Prandtl number

PROBLEM 10.49 KNOWN: Two vertical plate configurations maintained at 90°C for condensing saturated steam at 1 atm: single plate L × w and two plates each L/2 × w where L and w are the vertical and horizontal dimensions, respectively. FIND: Which case will provide the larger heat transfer or condensation rate. SCHEMATIC:

ASSUMPTIONS: (1) Negligible concentration of non-condensible gases in the steam. 3

PROPERTIES: Table A-6, Saturated water vapor (1 atm): Tsat = 100°C, ρ v = (1/vg) = 0.596 kg/m , hfg = 2257 kJ/kg; Saturated water (Tf = (Ts + Tsat )/2 = (90 + 100)°C/2 = 95°C = 368K): ρ l = (1/vf) = 3

-6

2

962 kg/m , µl = 296 × 10 N⋅s/m , k l = 0.678 W/m⋅K, cp,l = 4212 J/kg⋅K. ANALYSIS: The heat transfer and condensation rates are

q = h LA s ( Tsat − Ts )

& = q / h ′fg m

where, for the two cases,

hL,1As,1 = hL,1 ( L) [ L × w]

hL,2 As,2 = hL,2 ( L / 2)  2 ( L / 2× w ) 

and the average convection coefficients are evaluated at L and L/2, respectively. Hence,

h L,1 ( L ) [ L × w ] h L,1 ( L ) & q1 m = 1 = = . & 2 hL,2 ( L / 2)  2 ( L / 2 × w )  hL,2 ( L / 2) q2 m -1/4

For laminar film condensation on both plates, using the correlation of Eq. 10.30, with hL α L

q1 / q 2 = ( L / [ L / 2 ])

−1/4

= 0.84.

<

Hence, case 2 is preferred and provides 16% more heat transfer. When Reδ = 30 for case 1 with the given conditions, find from Eq. 10.37

(

h L ν l2 / g kl

)

1/3

1/3 2   − 6 2 3 2 h L  296 × 10 N ⋅ s / m /962kg/m /9.8m/s    = 0.678W/m ⋅ K

(

,

)

Continued …..

PROBLEM 10.49 (Cont.)

(

h L ν l2 / g kl

)

1/3 −1/3

= 1.47Reδ−1/3 = 1.47 ( 30 )

hL = 15,061W/m 2 ⋅ K and then from Eq. 10.30, 1/4

g ρ ( ρ − ρ )k 3 h ′  l l v l fg  hL = 0.943   µ l ( Tsat − Ts ) L    where

h ′fg = h fg + 0.68c p,l ( Tsat − Ts ) h ′fg = 2257kJ/kg + 0.68 × 4212J/kg ⋅ K (100 − 90 ) K = 2286kJ/kg, 15,061W/m2 ⋅ K = 1/4

 9.8m/s2 × 962kg/m 3 ( 962 − 0.596) k g / m3 ( 0.678W/m ⋅ K )3  0.943  2286kJ/kg    296 ×10−6 N ⋅ s / m 2 (100 − 90 ) K L   L = 34 mm.

We can anticipate for other, larger values of L that the comparison of hL values cannot be so easily made. However, according to Figure 10.15, we expect the same behavior of hL in the wavy region and anticipate that indeed case 2 will provide the greater condensation rate. Note that in the turbulent region with the increase in hL with Reδ, we cannot conclude with certainty which case is preferred. COMMENTS: In dealing with single-phase, forced or free convection, we associate thin thermal boundary layers with higher heat transfer rates. For vertical plates, we would expect the shorter plate to have the higher convection heat transfer coefficient. The results of this problem suggest the same is true for condensation on the vertical plate.

PROBLEM 10.50 KNOWN: Number, diameter and wall temperature of condenser tubes in a square array. Pressure of saturated steam around tubes. FIND: Rates of heat transfer and condensation per unit length of the array. SCHEMATIC:

ASSUMPTIONS: (1) Laminar film condensation on tubes, (2) Negligible concentration of noncondensable gases in steam. PROPERTIES: Table A-6, saturated vapor (psat = 0.105 bar): Tsat = 320 K = 47°C, ρv = 0.0715 3

kg/m , hfg = 2390 kJ/kg. Table A-6, saturated liquid (Tf = 32°C = 305 K): ρ " = 995 kg / m3 , µ" = 769 × 10

−6

2

N ⋅ s / m , k " = 0.620 W / m ⋅ K, c p," = 4178 J / kg ⋅ K.

ANALYSIS: The average heat rate per unit length for a single tube is q1′ = hD,N (π D )( Tsat − Ts ) , where hD,N is obtained from Eq. 10.41. With Ja = c p," ( Tsat − Ts ) / h fg = 0.052 and h ′fg = h fg (l + 6

6

0.68 Ja) = 1.04 (2.390 × 10 J/kg) = 2.48 × 10 J/kg, 1/ 4

h D,N

 g ρ " ( ρ " − ρ v ) k 3" h ′fg   = 0.729   N µ" ( Tsat − Ts ) D 

1/ 4

hD,N

 9.8 m / s 2 × 995 kg / m3 (995 − 0.0715 ) kg / m3 ( 0.62 W / m ⋅ K )3 2.48 × 106 J / kg  = 0.729   2 −6 25 × 769 × 10 N ⋅ s / m ( 30°C ) 0.025m  

2 = 3260 W / m ⋅ K

The heat rate per unit length of the array is q ′ = N 2q ′l . Hence, q′ = N 2 h D,N (π D )( Tsat − Ts ) = 625 × 3260 W / m 2 ⋅ K (π × 0.025m ) 30°C = 4.79 × 106 W / m

<

The corresponding condensation rate is

′= m

q′ 4.79 × 106 W / m = = 1.93kg / s ⋅ m h ′fg 2.48 × 106 J / kg

<

COMMENTS: Because of turbulence generation due to splashing from one tube to another in a vertical column, the foregoing value of hD,N is expected to underestimate the actual value of hD,N and hence to underpredict the heat and condensation rates.

PROBLEM 10.51 KNOWN: Tube wall diameters and thermal conductivity. Mean temperature and flow rate of water flow through tube. Pressure of saturated steam around tube. FIND: (a) Rates of heat transfer and condensation per unit length, (b) Effect of flow rate on heat transfer. SCHEMATIC:

ASSUMPTIONS: (1) Negligible concentration of noncondensible gases in the steam, (2) Uniform tube surface temperatures, (3) Laminar film condensation, (4) Fully-developed internal flow, (5) Constant properties. 2

PROPERTIES: Table A-6, water (Tm = 290 K): µ = 0.00108 N⋅s/m , k = 0.598 W/m⋅K, Pr = 7.56. 3 Table A-6, saturated vapor (p = 0.135 bar): Tsat = 325 K = 52°C, ρv = 0.0904 kg/m , hfg = 2378 kJ/kg. Table A-6, saturated liquid (Tf ≈ Tsat): ρ " = 987 kg / m3 , c p," = 4182 J / kg ⋅ K, µ" = 528 × 10

−6

2

N ⋅ s / m , k " = 0.645 W / m ⋅ K.

ANALYSIS: (a) From the thermal circuit, the heat rate may be expressed as

q′ = where,

Tsat − Tm R ′fc + R ′cond + R ′conv

(1)

R ′cond = n ( Do / Di ) / 2π k s = 0.00152 m ⋅ K / W

 / π D µ = 11, 336, the flow is The convection resistance is R ′conv = (π Di h i )−1 . With Re D = 4m i turbulent and the Dittus-Boelter correlation yields  k  4/5 0.4 4 / 5 0.4  0.598 W / m ⋅ K  2 hi =   0.023 (11, 336 ) ( 7.56 ) = 2082 W / m ⋅ K  0.023 Re D Pr =  0.026m    Di  The convection resistance is then −1 −1 R ′conv = (π Di hi ) = π × 0.026m × 2082 W / m 2 ⋅ K = 0.00588 m ⋅ K / W

)

(

The resistance associated with the condensate film is R ′fc = (π D o ho ) , where ho is given by Eq. 10.40. With C = 0.729, 1/ 4

ho

1/ 4

 gρ " ( ρ " − ρ v ) k 3" h fg ′  =C   µ" ( Tsat − Ts,o ) Do 

 9.8 m / s 2 × 987 (987 − 0.09 ) kg 2 / m 6 ( 0.645 W / m ⋅ K )3 h ′fg  = 0.729   2 −6   528 × 10 N ⋅ s / m (325 − Ts,o ) × 0.030m 1/ 4

 W3 ⋅ kg   h o = 462   m8 ⋅ K3 ⋅ s   

(

1/ 4

  h′fg   325 T − s,o  

)

(

)

where h ′fg = h fg + 0.68c p," Tsat − Ts,o = 2.38 × 106 J / kg + 2844 J / kg ⋅ K 325 − Ts,o The unknown surface temperature may be determined from an additional rate equation, such as Continued …..

q′ =

PROBLEM 10.51 (Cont.)

Ts,o − Tm

(2)

R ′cond + R ′conv

Substituting the thermal resistances into Eqs. (1) and (2), an iterative solution yields

q′ = 4270 W / m

Ts,o = 321.6 K = 48.6°C

<

The condensation rate is then

 ′cond = m

q′ 4270 W / m = = 0.00179 kg / s ⋅ m h ′fg 2.39 ×106 J / kg

<

The corresponding values of the condensate convection coefficient and resistance are

h o = 13,380 W / m 2 ⋅ K R ′fc = 0.000793m ⋅ K / W

and

Because R ′conv is much larger than R ′cond and R ′fc , attention should be paid to reducing the convection resistance in order to increase the heat rate. The resistance to heat flow by convection is the limiting factor. (b) The effects of varying the flow rate are shown below 50

10000

Su rfa ce te m p e ra ture , Ts ,o (C )

H e a t ra te (W /m )

9000 8000 7000 6000 5000

48

46

44

42

40

4000 0

0 .2 0 .4 0 .6 0 .8

1

1 .2 1 .4 1 .6 1 .8

Tu b e flo w ra te (kg /s )

2

0

0 .2 0 .4 0 .6 0 .8

1

1 .2 1 .4 1 .6 1 .8

Tu be flo w ra te (kg /s )

The effect of increasing m on q ′ is significant and is accompanied by a reduction in Ts,o. COMMENTS: (1) Use of the IHT convection and condensation correlations, as well as its temperature-dependent properties of water facilitated the numerical solution. (2) Evaluation of the film properties at Tsat is reasonable for part (a), since Tf = (Ts,o + Tsat)/2 = 50.3°C ≈ Tsat. However, with increasing m and hence decreasing Ts,o, the approximation would become inappropriate.

2

PROBLEM 10.52 KNOWN: Inner surface of a vertical thin-walled container of length L and diameter D experiences condensation of a saturated vapor. Container wall maintained at a uniform surface temperature by flowing cold water across its outer surface. FIND: Expression for the time, tf , required to fill the container with condensate assuming the condensate film is laminar. Express your result in terms of D, L, (Tsat − Ts), g and appropriate fluid properties. SCHEMATIC:

ASSUMPTIONS: (1) Laminar film condensation on a vertical surface, (2) Uniform temperature container wall surface, and (3) Mass of liquid condensate in the laminar film negligible compared to liquid mass on bottom of container. ANALYSIS: From an instantaneous mass balance on the container,

 m(t) =

dM dt

(1)

 (t) is the condensate rate and the liquid mass in the container, M, is Where m M = ρ π D2 4 L − x "

(

)(

)

(2)

The condensate rate from Eq. 10.33 can be expressed as

 (t ) = m

h A ( T − Ts ) q = s s sat h′fg h′fg

where the average film coefficient over the height 0 to x from Eq. 10.30 is, 1/ 4  gρ ( ρ − ρ ) k 3h ′  " " v fg "  hs = 0.943   µ" (Tsat − Ts ) x   

(3)

(4)

and the surface area over which condensation occurs is

As = π Dx

(5) Continued...

PROBLEM 10.52 (Cont.) Substituting Eqs (2-5) into Eq. (1), 1/ 4

 gρ ( ρ − ρ ) k 3h ′  " " ν " fg  0.943   µ" ( Tsat − Ts ) L   

L1/ 4 x1/ 4

(π Dx )

(

h ′fg = − ρ" π D2 4

) dxdt

(6)

Separate variables and identify the limits of integration, 1/ 4    gρ ( ρ − ρ ) k 3h ′  t 0   " " fg ν 1/ 4 2 "  L (π D )  h′fg ρ " π D 4   ∫ f dt = − ∫ x −3/ 4dx 0.943    x =L   0  µ" ( Tsat − Ts ) L      

(

)

(7)

The RHS integrates to 0

−  x1/ 4 (1 4 ) = 4L1/ 4  L

(8)

and solving for tf,

      2 ρ" π D 4 Lh ′fg   tf = 4   1/ 4  gρ ( ρ − ρ ) k 3h ′    " " v " fg   0.943  π DL T T − ( )( sat s )   µ" ( Tsat − Ts ) L       

(

)

<

COMMENTS: The numerator and denominator in the bracketed expression are of special significance. The numerator is product of the mass in the filled container and the latent heat of vaporization; that is, the total energy removed by the cold water. What is physical significance of the denominator? Can you interpret the time-to-fill, tf , expression in light of these terms?

PROBLEM 10.53 KNOWN: Tube of Problem 10.42 in horizontal position experiences condensation on its outer surface. FIND: Heat transfer and condensation rates. SCHEMATIC:

ASSUMPTIONS: (1) Laminar film condensation, (2) End effects negligible, (3) Negligible concentration of non-condensible gases in steam. 3

PROPERTIES: Table A-6, Water, vapor (1 atm): Tsat = 100°C, ρ v = 0.596 kg/m , hfg = 2257 kJ/kg; 3 Table A-6, Water, liquid (Tf = (Ts + Tsat )/2 = 370K): ρl = 960.6 kg/m , c p, l = 4214 J/kg⋅K, µl = -6

2

289 × 10 N⋅s/m , kl =0.679 W/m⋅K. ANALYSIS: From Eq. 10.32 with A = π D L and Eq. 10.33, the heat transfer and condensation rates are

& = q / h ′fg q = h L (π D L ) ( Tsat − Ts ) m where from Eq. 10.26 with Ja = c p, l ( Tsat − Ts ) / h fg, find

h ′fg = h fg [1 + 0.68Ja ] = 2257kJ/kg 1 + 0.68  4214J/kg ⋅ K (100 − 94 ) K / 2 257 × 10 J / k g  = 2274    3

kJ

.

kg

For laminar film condensation, Eq. 10.40 is the appropriate correlation for a cylinder with C = 0.729, 1/4  g ρ ( ρ − ρ ) k3 h′  l l v fg l  . hD = 0.729   µ l ( Tsat − Ts ) D 





1/4

 9 . 8 m / s 2 × 960.6kg/m 3 ( 960.6 − 0.596 ) k g / m 3 ( 0.679W/m ⋅ K )3 ×2274 ×10 3 J / k g  h D = 0.729   −6 2 289 × 10 N ⋅ s / m ( 100 − 94 ) K × 0.1m   hD = 10,120 W / m 2 ⋅ K. Hence, the heat transfer and condensation rates are 2

q = 10,120 W / m ⋅ K (π × 0.1m ×1m )(100 − 94 ) K = 19.1kW

< <

& = 19.1 ×103 W/2274 ×10 3 J / k g = 8.39 ×10−3 kg/s. m

COMMENTS: A comparison of the above results for the horizontal tube with those for a vertical tube (Problem 10.42) follows: Position Vertical Horizontal

(

2

h W / m ⋅K

)

8,507 10,120

The rates are higher for the horizontal case. Why?

& ⋅ 10 3 ( k g / s ) m

q(kW) 16.0 19.1

7.05 8.39

PROBLEM 10.54 KNOWN: Horizontal pipe passing through an air space with prescribed temperature and relative humidity. FIND: Water condensation rate per unit length of the pipe. SCHEMATIC:

ASSUMPTIONS: (1) Film condensation occurs on horizontal tube. PROPERTIES: Table A-6, Water, vapor (T∞ = 37°C = 310K): pA,sat = 0.06221 bar; Table A-6, 3 Water, vapor (pA = φ ⋅ pA,sat = 0.04666 bar): TA,sat ≈ 305K, ρ v = 0.04 kg/m , hfg = 2426 kJ/kg; Table 3 A-6, Water, liquid (Tf = (Ts + TA,sat )/2 = 297K): ρl = 997.2 kg/m , c p, l = 4180 J/kg⋅K, µl = 917 × -6

2

10 N⋅s/m , kl = 0.609 W/m⋅K. ANALYSIS: From Eq. 10.33, the condensate rate per unit length is

&′= m

h (π D) ( Tsat − Ts ) q′ = L h ′fg h ′fg

where, from Eq. 10.26, with Ja = cp, l ( Tsat − Ts ) / h fg , h ′fg = h fg 1 + 0.68c p, l ( Tsat − T s ) / h fg  = 2426

1 + 0.68 × 4180 J 305 − 288 K /2 4 2 6 × 10 3 J  ( ) kg  kg ⋅ K kg  kJ

h ′fg = 2474 kJ/kg. Note that Tsat = TA,sat is the saturation temperature of the water vapor in air at 37°C having a relative humidity φ = 0.75. That is, Tsat = 305K while Ts = 15°C = 288K. Assuming laminar film condensation on the horizontal pipe, it follows from Eq. 10.40 that, 1/4  g ρ ( ρ − ρ ) k3 h′  l l v fg l  hD = 0.729   µl ( Tsat − Ts ) D 





1/4

 9 . 8 m / s 2 × 997.2kg/m 3 ( 997.2 − 0.04 ) k g / m 3 ( 0.609W/m ⋅ K )3 × 2474 × 10 3 J / k g  h D = 0.729   −6 2 917 × 10 N ⋅ s / m ( 305 − 288 ) K × 0.025m   hD = 7925 W / m 2 ⋅ K. Hence, the condensate rate is, 2 &′

m = 7925 W / m ⋅ K (π × 0.025m )( 305 − 288 ) K/2474 ×103 J/kg & ′ = 4.28 ×10−3 kg/s ⋅ m. m

COMMENTS: The actual dropwise condensation rate exceeds the foregoing estimate.

<

PROBLEM 10.55 KNOWN: Horizontal tube, 50mm diameter, with surface temperature of 34°C is exposed to steam at 0.2 bar. FIND: Estimate the heat transfer and condensation rates per unit length of the tube. SCHEMATIC:

ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in steam. 3

PROPERTIES: Table A-6, Saturated steam (0.2 bar): Tsat = 333K, ρ v = 0.129 kg/m , hfg = 2358 3 kJ/kg; Table A-6, Water, liquid (Tf = (Ts + Tsat )/2 = 320K): ρl = 989.1 kg/m , c p, l = 4180 J/kg⋅K, -6

2

µl = 577 × 10 N⋅s/m , kl = 0.640 W/m⋅K.

ANALYSIS: From Eqs. 10.32 and 10.33, the heat transfer and condensate rates per unit length of the tube are

q′ = h D (π D )( Tsat − Ts )

& ′ = q′ / h ′fg m

where from Eq. 10.26 with Ja = c p,l ( Tsat − Ts ) / h fg,

h ′fg = h fg [1 + 0.68 Ja ] = 2358

kJ  1 + 0.68 × 4180J/kg ⋅ K ( 333 − 307 ) K/2358 ×103 J/kg     kg

h ′fg = 2432 kJ/kg. For laminar film condensation, Eq. 10.40 is appropriate for estimating h D with C = 0.729, 1/4

 g ρ ( ρ − ρ ) k3 h′  l l v l fg  hD = 0.729   µl ( T − T ) D  sat s   h D = 0.729

 9 . 8 m / s 2 × 9 8 9 . 1 k g / m 3 ( 989.1 − 0.129 ) k g / m 3 ( 0 . 6 4 0 W / m ⋅ K )3 × 2432 × 103 J / k g    −6 2 577 × 10 N ⋅ s / m ( 333 − 307 ) K × 0.050m  

hD = 6926 W / m 2 ⋅ K. Hence, the heat transfer and condensation rates are

q′ = 6926 W / m 2 ⋅ K (π × 0.050m )( 333 − 307 ) K = 28.3kW/m

<

& ′ = 28.3 ×103 W / m / 2 4 3 2 ×10 3 J / k g = 1.16 ×10 − 2 kg/s ⋅ m. m

<

PROBLEM 10.56 KNOWN: Horizontal tube 1m long with surface temperature of 70°C used to condense steam at 1 bar. FIND: Diameter required for condensation rate of 125 kg/h. SCHEMATIC:

ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in steam. 3

PROPERTIES: Table A-6, Water, vapor (1 atm): Tsat = 100°C, ρ v = 0.596 kg/m , hfg = 2257 kJ/kg; 3 Table A-6, Water, liquid (Tf = (Ts + Tsat )/2 = 358K): ρl = 968.6 kg/m , c p, l = 4201 J/kg⋅K, µl = -6

2

332 × 10 N⋅s/m , kl = 0.673 W/m⋅K. ANALYSIS: From the rate equation, Eq. 10.33, with A = π D L, the required diameter is

& h ′fg / π L h D ( Tsat − Ts ) D =m where from Eq. 10.26 with Ja = c p,l ( Tsat − Ts ) / h fg,

(1)

4201J/kg ⋅ K × (100 − 70 ) K  kJ   = 2343kJ/kg. h ′fg = h fg (1 + 0.68Ja ) = 2257  1 + 0.68  kg  2257 ×103 J/kg  Substituting numerical values, Eq. (1) becomes 125 kg J D= × 2343 ×103 / π ×1m × h

3600 s

kg

(2)

−1 D (100 − 70 ) K = 863.2h D .

(3)

The appropriate correlation for hD is Eq. 10.40 with C = 0.729, 1/4  g ρ ( ρ − ρ ) k3 h′  l l v fg l  . hD = 0.729   µ l ( Tsat − Ts ) D 



(4)



Substitute Eq. (4) for hD into Eq. (3) and use numerical values, 863.2 D−1 = 0.729 × 1/4

 9.8m/s2 × 968.6kg/m3 (968.6 − 0.596 ) k g / m3 ( 0.673W/m ⋅ K )3 × 2343 × 103 J/kg      332 ×10 −6 N ⋅ s / m2 (100 − 70 ) K× D 863.2 D−1 = 3693.4 D−1 / 4 D = 0.144m = 144mm. COMMENTS: Note for this situation Ja = 0.06.

<

PROBLEM 10.57 KNOWN: Saturated R-12 vapor at 1 atm condensing on the outside of a horizontal tube. FIND: Tube surface temperature required for condensation rate of 50 kg/h. SCHEMATIC:

ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in vapor. 3

PROPERTIES: Table A-5, R-12 Saturated vapor (1 atm): Tsat = 243K, ρ v = 6.32 kg/m , hfg = 165 3 kJ/kg; Table A-5, R-12 Saturated liquid (Tf ≈ 240K): ρl = 1498 kg/m , c p, l = 892.3 J/kg⋅K, µl = -2

2

0.0385 × 10 N⋅s/m , kl = 0.069 W/m⋅K. ANALYSIS: The surface temperature or temperature difference can be written as follows from Eq. 10.33, & h ′fg / h D π D L ∆T = Tsat − Ts = m (1) where A = π D L. To evaluate h ′fg and hD , we require knowledge of Ts or ∆T. Assume a ∆T = 10°C, then Ts = 233K and Tf = (Ts + Tsat )/2 = 240K. From Eq. 10.26 with Ja = c p, l ∆T / h fg , find h ′fg = h fg (1 + 0.68Ja ) = 165

kJ  J J  1 + 0.68 × 892.3 × 10K/165× 10 3  = 171kJ/kg.  kg  kg ⋅ K kg 

(2)

The appropriate correlation for hD is Eq. 10.40 with C = 0.729; substitute properties and find hD in terms of ∆T. 1/4  g ρ ( ρ − ρ ) k3 h′  l l v l fg  hD = 0.729   µl ( Tsat − Ts ) D 





1/4

 9.8m/s2 × 1498kg/m3 (1498 − 6.32 ) k g / m3 ( 0.069W/m ⋅ K )3 × 171 ×103 J/kg   h D = 0.729    0.0385 × 10− 2 N ⋅ s / m2 × ∆T × 0.010m hD = 3082 ∆ T1/4 .

(3)

Substitute Eq. (3) into Eq. (1) for hD , and solve for ∆T,

∆T =

(

)

50 k g / s ×171× 103 J / k g / 3082 ∆T1/4 π ( 0.010m ) ×1m 3600

∆T = 12.9K

or

Ts = 230K.

<

COMMENTS: We used the assumed value of Ts or ∆T only to evaluate properties. Our estimate for Tf = 240K is to be compared to the calculated value of Tf ≈ 236K. An iteration is probably not necessary.

PROBLEM 10.58 KNOWN: Saturation temperature and inlet flow rate of R-12. Diameter, length and temperature of tube. FIND: Rate of condensation and outlet flow rate. SCHEMATIC:

ASSUMPTIONS: (1) Negligible concentration of noncondensables in vapor. 3

PROPERTIES: Given, R-12, saturated vapor: ρv = 6 kg/m , hfg = 160 kJ/kg, µv = 150 × 10

-7

2

N⋅s/m . Table A-5, R-12, saturated liquid (Tf = 300 K): ρ " = 1306 kg / m3 , c p," = 978 J / kg ⋅ K, 2

µ" = 0.0254 N ⋅ s / m , k " = 0.072 W / m ⋅ K.

ANALYSIS: The Reynolds number associated with the inlet vapor flow is Re v,i = 4 m v,i / π Dµ v = 0.04 kg / s / π × 0.025m × 150 × 10

−7

2

N ⋅ s / m = 33, 950 < 35, 000. Hence, the average convection

coefficient may be obtained from Eq. 10.42, where h ′fg = h fg + 0.375 c p," ( Tsat − Ts ) = (1.6 × 10 + 5

5

0.375 × 978 × 20) J/kg = 1.67 × 10 J/kg.

 gρ " ( ρ " − ρ v ) k 3" h ′fg

hD = 0.555 



µ " ( Tsat − Ts ) D

1/ 4

  

 2  9.8 m / s ≈ 0.555   

(

1306 kg / m

3

)( 2

0.072 W / m ⋅ K ) 1.67 × 10 3

2

0.0254 N ⋅ s / m × 20 K × 0.025m

1/ 4

5

 J / kg    

h D = 297 W / m 2 ⋅ K The heat rate is then

q = π DL h D ( Tsat − Ts ) = π × 0.025m × 2m × 297 W / m 2 ⋅ K × 20 K = 933 W and the condensation rate is

 cond = m

q 933 W = = 0.0056 kg / s h ′fg 1.67 × 105

<

The flow rate of vapor leaving the tube is then

 v,o = m  v,i − m  cond = ( 0.0100 − 0.0056 ) kg / s = 0.0044 kg / s m

<

PROBLEM 10.59 KNOWN: Array of condenser tubes exposed to saturated steam at 0.1 bar. FIND: (a) Condensation rate per unit length of square array, (b) Options for increasing the condensation rate. SCHEMATIC:

ASSUMPTIONS: (1) Film condensation on tubes, (2) Negligible non-condensable gases in steam. PROPERTIES: Table A.6, Saturated water vapor (0.1 bar): Tsat ≈ 320 K, ρv = 0.072 kg/m3, hfg = 2390 kJ/kg; Table A.6, Water, liquid (Tf = (Ts + Tsat)/2 = 310 K): ρ" = 993.1 kg/m3, cp," = 4178 J/kg⋅K, µ" = 695 × 10-6 N⋅s/m2, k " = 0.628 W/m⋅K. ANALYSIS: (a) From Eq. 10.33, the condensation rate for a N ×N square array is

′=m  L = h D,N ⋅ N t (π D )(Tsat − Ts ) h′fg m

where h D,N is the average coefficient for the tubes in a vertical array of N tubes. With Ja = cp," ∆T/hfg = 4178 J/kg⋅K × (320 - 300)K/2390 × 103 J/kg = 0.035, Eq. 10.26 yields h ′fg = hfg(1 + 0.68 Ja) = 2390 kJ/kg(1 + 0.68 × 0.035) = 2447 kJ/kg. For a vertical tier of N = 10 horizontal tubes, the average coefficient is given by Eq. 10.41, 1/ 4  gρ ( ρ − ρ ) k 3 h ′  " " v fg "  h D,N = 0.729   Nµ" (Tsat − Ts ) D    1/ 4

h D,N

 9.8 m s 2 × 993.1kg m3 (993.1 − 0.072 ) kg m3 ( 0.628 W m⋅ K )3 × 2447 × 103 J kg  = 0.729   −6 2 10 × 695 × 10 N ⋅ s m (320 − 300 ) K × 0.008 m  

h D,N = 6210 W m 2⋅ K . Hence, the condensation rate for the entire array per unit tube length is  ′ = 6210 W m 2⋅ K (100 )π × 0.008 m (320 − 300 ) K 2447 ×103 J kg m

 ′ = 0.128 kg s ⋅ m = 459 kg h ⋅ m . m

<

(b) Options for increasing the condensation rate include reducing the surface temperature and/or the number of tubes in a vertical tier. By varying the temperature of cold water flowing through the tubes, it is feasible to maintain surface temperatures in the range 280 ≤ Ts ≤ 300 K. Using the Correlations and Properties Toolpads of IHT, the following results were obtained for N = 10, 5 and 2, with Nt = 100 in each case. The results are based on properties evaluated at p = 0.1 bar, for which the Properties Toolpad yielded Tsat = 318.9 K. Continued...

PROBLEM 10.59 (Cont.)

Condensation rate, mdot'(kg/s.m)

0.3

0.2

0.1 280

290

300

Surface temperature, Ts(K) N = 10 N=5 N=2

Clearly, there are significant benefits associated with reducing both Ts and N. COMMENTS: Note that, since h D,N ∝ N-1/4, the average coefficient decreases with increasing N due to a corresponding increase in the condensate film thickness. From the result of part (a), the coefficient for the topmost tube is h D = 6210 W/m2⋅K(10)1/4 = 11,043 W/m2⋅K.

PROBLEM 10.60 KNOWN: Thin-walled concentric tube arrangement for heating deionized water by condensation of steam. FIND: Estimates for convection coefficients on both sides of the inner tube. Inner tube wall outlet temperature. Whether condensation provides fairly uniform inner tube wall temperature approximately equal to the steam saturation temperature. SCHEMATIC:

ASSUMPTIONS: (1) Negligible thermal resistance of inner tube wall, (2) Internal flow is fully developed. 3

PROPERTIES: Deionized water (given): ρ = 982.3 kg/m , cp = 4181 J/kg⋅K, k = 0.643 W/m⋅K, µ = -6 2 548 × 10 N⋅s/m , Pr = 3.56; Table A-6, Saturated vapor (1 atm): Tsat = 100°C, ρ v = (1/vg) = 0.596 3 kg/m , hfg = 2265 kJ/kg; Table A-6, Saturated water (assume Ts ≈ 75°C, Tf = (75 + 100)°C/2 = 360K): 3 -6 2 ρl = (1/vf) = 967 kg/m , µl = 324 × 10 N⋅s/m , kl = 0.674 W/m⋅K, c p, l = 4203 J/kg⋅K. ANALYSIS: From an energy balance on the inner tube assuming a constant wall temperature,

(

)

(

hc Tsat − Ts,o = hi Ts,o − Tm,o

)

where h c and hi are, respectively, the heat transfer coefficients for condensation (c) on a horizontal cylinder and internal (i) flow in a tube. Condensation. From Eq. 10.40, for the horizontal tube, 1/4

 g ρ ( ρ − ρ ) k 3h ′  l l v l fg  hc = 0.729   µl ( Tsat − Ts ) D   

{

where h ′fg = h fg 1 + 0.68c p,l ( Tsat − Ts ) / h fg

}

{ h ′fg = 2265 kJ/kg {1+1.262 ×10 −3 (100 − Ts )}

h ′fg = 2265 kJ/kg 1 + 0.68 × 4203 J / k g ⋅ K (100 − Ts ) /2265 ×103 J/kg

}

hc = 0.729  9.8m/s2 × 967kg/m3 ( 067 − 0.596 ) k g / m3 ( 0.674W/m ⋅ K )3 × 

{

}

1/4

2265 1 + 1.262 ×10 −3 (100 − Ts ) kJ/kg/324 ×10 −6 N ⋅ s / m 2 (100 − Ts ) 0.030 m  

Continued …..

PROBLEM 10.60 (Cont.) 1/4

1 +1.262 ×10 −3 (100 − T )  s  hc = 2.843× 104  100 − Ts  

.

Internal flow. From Eq. 8.6, evaluating properties at Tm , find

Re D =

& 4m 4 × 5 kg/s = = 3.872 ×105 − 6 2 πµ D π × 548 ×10 N ⋅ s / m × 0.030 m

and for turbulent flow use the Colburn equation,

hD 1/3 Nu D = i = 0.023Re0.8 D Pr k hi =

)

(

0.8 0.023 × 0.643 W / m ⋅ K 3.872 ×105 ( 3.56 )1/3 = 2.22 × 104 W / m 2 ⋅ K. 0.03 m

<

Substituting numerical values into the energy balance relation,

(

 1 + 1.262 ×10−3 100 − T s,o 4 2.843 ×10   100 − Ts,o 

1/4

)   

(100 − Ts,o ) K (

)

= 2.22 × 104 W / m2 ⋅ K Ts,o − 60 K and by trial-and-error, find

Ts,o ≈ 75°C. With this value of Ts, find that

hc = 1.29 × 104 W / m2 ⋅ K

<

which is approximately half that for the internal flow. Hence, the tube wall cannot be at a uniform temperature. This could only be achieved if h c ? h i .

PROBLEM 10.61 KNOWN: Heat dissipation from multichip module to saturated liquid of prescribed temperature and properties. Diameter and inlet and outlet water temperatures for a condenser coil. FIND: (a) Condensation and water flow rates. (b) Tube surface inlet and outlet temperatures. (c) Coil length. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions since rate of heat transfer from the module is balanced by rate of heat transfer to coil, (2) Fully developed flow in tube, (3) Negligible changes in potential and kinetic energy for tube flow. PROPERTIES: Saturated fluorocarbon (Tsat = 57°C, given): kl = 0.0537 W/m⋅K, c p, l = 1100 3 3 -3 2 J/kg⋅K, h ′fg ≈ h fg = 84,400 J/kg. ρl = 1619.2 kg/m , ρ v = 13.4 kg/m , σ = 8.1 × 10 kg/s , µl = 440 -6

3

× 10 kg/m⋅s, Prl = 9; Table A-6, Water, sat. liquid ( Tm = 300K ) : ρ = 997 kg/m , cp = 4179 J/kg⋅K, -6

2

µ = 855 × 10 N⋅s/m , k = 0.613 W/m⋅K, Pr = 5.83. ANALYSIS: (a) With

q = (q ′′ × A )module = 105 W / m 2 ( 0.1m )2 = 10 3 W the condensation rate is

& con = m

q 103 W = = 0.0118 kg/s h′fg 84,400 J / k g

<

and the required water flow rate is

& = m

(

q

c p Tm,o − Tm,i

)

=

1000 W = 7.98 ×10 −3 kg/s. 4179 J / k g ⋅ K ( 30K )

<

(b) The Reynolds number for flow through the tube is

Re D =

& 4m 4 × 7.98 ×10 −3 k g / s = = 1188. π Dµ π ( 0.01m ) 855 ×10−6 N ⋅ s / m 2

Hence, the flow is laminar. Assuming a uniform wall temperature,

h i = Nu D k / D = 3.66 ( 0.613 W / m ⋅ K/0.01m ) = 224 W / m 2 ⋅ K. Continued …..

PROBLEM 10.61 (Cont.) For film condensation on the outer surface, Eq. 10.40 yields

(

)(

)

1/4

 9.8m/s 2 1619.2 k g / m 3 1605.8kg/m3 ( 0.0537 W / m ⋅ K )3 84,400J/kg    h o = 0.729   − 6 440 × 10 k g / m ⋅ s × 0.01m ( Tsat − Ts )   h o = 2150 (57 − Ts )

−1/4

.

From an energy balance on a portion of the tube surface,

h o ( Tsat − Ts ) = h i ( Ts − Tm ) or

2150 ( 57 − Ts )

3/4

= 224 ( Ts − Tm )

At the entrance where ( Tm,i = 285K ) , trial-and-error yields:

<

Ts,i = 50.6°C and at the exit where ( Tm,o = 315K ) ,

<

Ts,o = 55.4°C (c) From Eqs. 8.44 and 8.45,

L=

q hiπ D∆Tlm

where

∆Tl m = L=

(

( Ts − Tm,i ) − ( Ts − Tm,o ) = 41 −11 = 22.8°C ln  ( Ts − Tm,i ) / ( Ts − Tm,o ) ln ( 41/11)   1000 W

)

224 W / m2 ⋅ K π ( 0.01m ) 22.8°C

= 6.23m.

<

COMMENTS: Some control over system performance may be exercised by adjusting the water & ( Tm,o − Tm,i ) is reduced for a prescribed q. The value of h i is increased flow rate. By increasing m, substantially if the internal flow is turbulent.

PROBLEM 10.62 KNOWN: Saturated ethylene glycol vapor at 1 atm condensing on a sphere of 100 mm diameter having surface temperature of 150°C. FIND: Condensation rate. SCHEMATIC:

ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in vapor. 3

PROPERTIES: Table A-5, Saturated ethylene glycol, vapor (1 atm): Tsat = 470K, ρ v ≈ 0 kg/m , hfg = 812 kJ/kg; Table A-5, Ethylene glycol, liquid (Tf = 423K, but use values at 373K, limit of data in 3 -2 2 table): ρl = 1058.5 kg/m , c p, l = 2742 J/kg⋅K, µl = 0.215 × 10 N⋅s/m , kl = 0.263 W/m⋅K. ANALYSIS: The condensation rate is given by Eq. 10.33 as

(

)

h L π D2 ( Tsat − Ts ) q & = m = h′fg h′fg 2

where A = π D for the sphere and h ′fg , with Ja = c p, l ∆T / h fg , is given by Eq. 10.26 as h ′fg = h fg (1 + 0.68Ja ) = 812

kJ  J  1 + 0.68 × 2742 ( 470 − 423) K/812 ×103 J/kg  = 900kJ/kg.  kg  kg ⋅ K 

The average heat transfer coefficient for the sphere follows from Eq. 10.40with C = 0.815, 1/4

g ρ ( ρ − ρ )k 3 h ′  l l v l fg  hD = 0.815   µ l ( Tsat − Ts ) D   

 9 . 8 m / s 2 × 1058.5kg/m 3 (1058.5 − 0 ) k g / m 3 ( 0.263W/m ⋅ K )3× 900 × 10 3 J / k g  h D = 0.815   −2 2 0.215 × 10 N ⋅ s / m ( 470 − 423 ) K × 0.100m  

1/4

hD = 1674 W / m 2 ⋅ K. Hence, the condensation rate is

& = 1674W/m 2 ⋅ K ×π ( 0.100m ) 2 ( 470 − 423 ) K/900 ×103 J/kg m & = 2.75 ×10 −3 kg/s. m

<

COMMENTS: Recognize this estimate is likely to be a poor one since properties were not evaluated at the proper Tf which was beyond the limit of the table.

PROBLEM 10.63 KNOWN: Copper sphere of 10 mm diameter, initially at 50°C, is placed in a large container filled with saturated steam at 1 atm. FIND: Time required for sphere to reach equilibrium and the condensate formed during this period. SCHEMATIC:

ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in vapor, (3) Sphere is spacewise isothermal, (4) Sphere experiences heat gain by condensation only. 3

PROPERTIES: Table A-6, Saturated water vapor (1 atm): Tsat = 100°C, ρ v = 0.596 kg/m , hfg = 3 2257 kJ/kg; Table A-6, Water, liquid (Tf ≈ (75 + 100)°C/2 = 360K): ρl = 967.1 kg/m , c p, l = 4203 -6

2

J/kg⋅K, µl = 324 × 10 N⋅s/m , kl = 0.674 W/m⋅K; Table A-1, Copper, pure ( T = 75° C) : ρ sp = 3

8933 kg/m , cp,sp = 389 J/kg⋅K. ANALYSIS: Using the lumped capacitance approach, an energy balance on the sphere provides,

E& in − E& out = E& st & h ′fg = h D A s ( Tsat − Ts ) = ρsp c p,sp Vs m

dTs . dt

(1)

Properties of the sphere, ρ sp and cp ,sp ,. Will be evaluated at Ts = ( 50 +100 ) ° C / 2 = 75° C, while water (liquid) properties will be evaluated at Tf = ( Ts + Tsat ) / 2 = 87.5 °C ≈ 360K. From Eq. 10.26 with Ja = c p, l ∆T / h fg where ∆T = Tsat - Ts , find

1 + 0.68 4203 J × 100 − 75 K /2 2 5 7 × 10 3 J / k g   = 2328 kJ . (2) ( )     kg  kg ⋅ K kg kJ

h ′fg = h fg (1 + 0.68Ja ) = 2257

To estimate the time required to reach equilibrium, we need to integrate Eq. (1) with appropriate limits. However, to perform the integration, an appropriate relation for the temperature dependence of hD needs to be found. Using Eq. 10.40 with C = 0.815, 1/4  g ρ ( ρ − ρ ) k3 h′  l l v l fg  hD = 0.815  .  µ l ( Tsat − Ts ) D 





Substitute numerical values and find, 1/4

 9 . 8 m / s 2 × 967.1kg/m 3 ( 967.1 − 0.596 ) k g / m 3 ( 0 . 6 7 4 W / m ⋅ K )3 × 2328 × 10 3 J / k g  h D = 0.815   −6 2 324 × 10 N ⋅ s / m ( Tsat − Ts ) × 0.010m   −1/4

hD = B ( Tsat − Ts )

where

B = 30,707W/m 2 ⋅ ( K )

3/4

. (3)

Continued …..

PROBLEM 10.63 (Cont.) 1

Substitute Eq. (3) into Eq. (1) for hD and recognize Vs / A s = π D 3 / π D 2 = D / 6 , 6

dT B ( Tsat − Ts )−1/4 ( Tsat − Ts ) = ρsp c p,sp ( D / 6 ) s . dt

(4)

Note that d(Ts) = - d(Tsat – Ts); letting ∆T ≡ Tsat – Ts and separating variables, the energy balance relation has the form

ρsp c p,sp ( D / 6 ) ∆T d ( ∆T ) t dt = − (5) 0 ∆To ∆T3 / 4 B where the limits of integration have been identified, with ∆ To = Tsat − Ti and Ti = Ts(0). Performing the integration, find

∫

t=−

∫

ρsp cp,sp ( D / 6) B

⋅

1  1/4  ∆T − ∆ T1/4 o  .   1− 3 / 4

Substituting numerical values with the limits, ∆T = 0 and ∆To = 100-50 = 50°C,

t=−

8933kg/m3 × 389J/kg ⋅ K ( 0.010m/6) 30,707 W / m 2 ⋅ K 3 / 4

× 4  01/4 − 501 / 4  K1/4  

<

t = 2.0s.

To determine the total amount of condensate formed during this period, perform an energy balance on a time interval basis,

E in − E out = ∆E = E final − E initial Ein = ρsp cp,sp V ( Tfinal − Tinitial ]

(6)

where Tfinal = Tsat and Tinitial = Ti = Ts(0). Recognize that

E in = M h ′fg

(7)

where M is the total mass of vapor that condenses. Combining Eqs. (6) and (7),

M=

M=

ρsp cp,sp V h ′fg

[Tsat − Ti ]

8933kg/m3 × 389J/kg ⋅ K (π / 6)( 0.010m )3 2328 × 103 J/kg

[100 − 50 ] K

M = 3.91 ×10 −5 kg.

<

COMMENTS: The total amount of condensate could have been evaluated from the integral, t 0

t q t h D A s ( Tsat − Ts ) dt dt = 0 h ′fg 0 h ′fg

& =∫ M = ∫ mdt

∫

giving the same result, but with more effort.

PROBLEM 10.64 KNOWN: Saturated steam condensing on the inside of a horizontal pipe. FIND: Heat transfer coefficient and the condensation rate per unit length of the pipe. SCHEMATIC:

ASSUMPTIONS: (1) Film condensation with low vapor velocities. 3

PROPERTIES: Table A-6, Saturated water vapor (1.5 bar): Tsat ≈ 385K, ρ v = 0.88 kg/m , hfg = 3 2225 kJ/kg; Table A-6, Saturated water (Tf = (Tsat + Ts)/2 ≈ 380K): ρl = 953.3 kg/m , c p, l = 4226 -6

2

J/kg⋅K, µl = 260 × 10 N⋅s/m , kl = 0.683 W/m⋅K. ANALYSIS: The condensation rate per unit length follows from Eq. 10.33 with A = π D L and has the form

&′= m

& m = h D (π D )( Tsat − Ts ) / h ′fg L

where hD is estimated from the correlation of Eq. 10.42 with Eq. 10.43, 1/4

g ρ ( ρ − ρ )k 3 h ′  l l v l fg  hD = 0.555   µ l ( Tsat − Ts ) D    where

3 J 3 J h ′fg = h fg + cp,l ( Tsat − Ts ) = 2225 ×103 + × 4226 ( 385 − 373 ) K 8 kg 8 kg ⋅ K h ′fg = 2244kJ/kg. Hence, 1/4

kg kg  3  2 3  9.8m/s × 953.3 3 ( 953.3 − 0.88 ) 3 ( 0.683W/m ⋅ K ) 2244 ×10 J/kg  m m hD = 0.555   − 6 2   260 × 10 N ⋅ s / m ( 385 − 373) K × 0.075m   hD = 7127W/m 2 ⋅ K. It follows that the condensate rate per unit length of the tube is

& ′ = 7127W/m 2 ⋅ K (π × 0.075m )(385 − 373)K / 2225 ×103 J / k g = 9.06 ×10−3 kg / s ⋅ m. m

<

PROBLEM 10.65 KNOWN: Horizontal pipe passing through an air space with prescribed temperature and relative humidity. FIND: Water condensation rate per unit length of pipe. SCHEMATIC:

ASSUMPTIONS: (1) Drop-wise condensation, (2) Copper tube approximates well promoted surface. PROPERTIES: Table A-6, Water vapor (T∞ = 37°C = 310K): pA,sat = 0.06221 bar; Table A-6, Water vapor (pA = φ⋅pa,sat = 0.04666 bar): Tsat = 305K = 32°C, hfg = 2426 kJ/kg; Table A-6, Water, liquid (Tf = (Ts + Tsat )/2 = 297K): c p, l = 4180 J / kg ⋅ K. ANALYSIS: From Eq. 10.33, the condensate rate per unit length is

&′= m

q ′ h L (π D )(Tsat − Ts ) = h′fg h′fg

where from Eq. 10.26, with Ja = c p, l ( Tsat − T s ) / h fg ,

h ′fg = h fg [1 + 0.68Ja ] = 2426

kJ 1 + 0.68 × 4180J/kg ⋅ K ( 305 − 288 ) K/2426k J/kg  kg 

h ′fg = 2474kJ/kg. Note that Tsat is the saturation temperature of the water vapor in air at 37°C having a relative humidity, φ = 0.75. That is, Tsat = 305K and Ts = 15°C + 288K. For drop-wise condensation, the correlation of Eq. 10.44 yields

hdc = 51,104 + 2044Tsat

22° C < Tsat < 100°C 2

where the units of h dc and Tsat are W/m ⋅K and °C.

hdc = 51,104 + 2044 ( 32° C) = 116,510 W / m2 ⋅ K. Hence, the condensation rate is

& ′ = 116,510 W / m 2 ⋅ K ( π × 0.025m )( 305 − 288 ) K/2474× 10 3 J/kg m & ′ = 6.288 ×10−2 kg/s ⋅ m m

<

COMMENTS: From the result of Problem 10.54 assuming laminar film condensation, the condensation rate was m& ′film = 4.28 ×10 −3 k g / s ⋅ m which is an order of magnitude less than for the rate assuming drop-wise condensation.

PROBLEM 10.66 KNOWN: Beverage can at 5°C is placed in a room with ambient air temperature of 32°C and relative humidity of 75%. FIND: The condensate rate for (a) drop-wise and (b) film condensation. SCHEMATIC:

ASSUMPTIONS: (1) Condensation on top and bottom surface of can neglected, (2) Negligible noncondensibles in water vapor-air, and (b) For film condensation, film thickness is small compared to diameter of can. PROPERTIES: Table A-6, Water vapor (T∞ = 32°C = 305 K): pA,sat = 0.04712 bar; Water vapor (pA = φ⋅pA,sat = 0.03534 bar): Tsat ≈ 300 K = 27°C, hfg = 2438 kJ/kg; Water, liquid (Tf = (Ts + Tsat)/2 = 289 K): c p," = 4185 J/kg⋅K. ANALYSIS: From Eq. 10.33, the condensate rate is

 = m

h (π DL )( Tsat − Ts ) q = h ′fg h ′fg

where from Eq. 10.26, with Ja = c p," (Tsat – Ts)/hfg,

h ′fg = h fg [1 + 0.68 Ja ] h ′fg = 2438 kJ / kg 1 + 0.68 × 4185 J / kg ⋅ K (300 − 278 ) K / 2438 kJ / kg  h ′fg = 2501 kJ / kg Note that Tsat is the saturation temperature of the water vapor in air at 32°C having a relative humidity of φ∞ = 0.75. (a) For drop-wise condensation, the correlation of Eq. 10.44 with Tsat = 300 K = 27°C yields

h = h dc = 51,104 + 2044 Tsat

22°C < Tsat ≤ 100°C

2

where the units of h dc are W/m ⋅K and Tsat are °C,

h dc = 51,104 + 2044 × 27 = 106, 292 W / m 2 ⋅ K Hence, the condensation rate is  = 1.063 × 105 W / m 2 ⋅ K (π × 0.065 m × 0.125 m )( 27 − 5 ) K / 2501 kJ / kg m

<

 = 0.0229 kg / s m Continued …..

PROBLEM 10.66 (Cont.) (b) For film condensation, we used the IHT tool Correlations, Film Condensation, which is based upon Eqs. 10.37, 10.38 or 10.39 depending upon the flow regime. The code is shown in the Comments section, and the results are

Reδ = 24, flow is laminar

<

 = 0.00136 kg / s m

Note that the film condensation rate estimate is nearly 20 times less than for drop-wise condensation. COMMENTS: The IHT code identified in part (b) follows: /* Results, NuLbar 0.5093

Part (b) - input variables and rate parameters Redelta hLbar mdot D L Ts 24.05 6063 0.001362 0.065 0.125 278

Tsat 300

*/

/* Thermophysical properties evaluated at Tf; hfg at Tsat Prl Tf cpl h'fg hfg kl mul nul 7.81 289 4185 2.501E6 2.438E6 0.5964 0.001109 1.11E-6*/ // Other input variables required in the correlation L = 0.125 b = pi * D D = 0.065 /* Correlation description: Film condensation (FCO) on a vertical plate (VP). If Redelta> m & h. COMMENTS: (1) Be sure to recognize why Ch → ∞. Note also that m (2) Note that Tc = (Tc,i + Tc,o)/2 = (15 + 48.2)°C/2 ≈ 305 K. This compares favorably with the value of 300 K at which properties of the cold fluid were evaluated.

PROBLEM 11.63 KNOWN: Concentric tube heat exchanger with prescribed conditions. FIND: (a) Maximum possible heat transfer, (b) Effectiveness, (c) Whether heat exchanger should be run in PF or CF to minimize size or weight; determine ratio of required areas for the two flow conditions. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties, (4) Overall heat transfer coefficient remains unchanged for PF or CF conditions. PROPERTIES: Hot fluid (given): c = 2100 J/kg⋅K; Cold fluid (given): c = 4200 J/kg⋅K. ANALYSIS: (a) The maximum possible heat transfer rate is given by Eq. 11.19.

(

)

q max = Cmin Th,i − Tc,o . & h ch, giving The minimum capacity fluid is the hot fluid with Cmin = m

(

)

& h c h Th,i − Tc,o = 0.125 q max = m

kg J × 2100 ( 210 − 40 ) K = 44,625W. s kg ⋅ K

<

(b) The effectiveness is defined by Eq. 11.20 and the heat rate, q, can be determined from an energy balance on the cold fluid.

(

)

& c c c Tc,o − Tc,i / q max ε = q / q max = m ε = 0.125 kg/s× 4200 J / k g ⋅ K (95 − 40 ) K/44,625W = 0.65.

<

(c) Operating the heat exchanger under CF conditions will require a smaller heat transfer area than for PF conditions. The ratio of the areas is

ACF q / U ∆ Tl m,CF ∆ Tl m,PF = = . A PF q / U ∆ Tl m,PF ∆ Tlm,CF To calculate the LMTD, first find Th,o from overall energy balances on the two fluids.

Th,o = Th,i −

& c cc m 0.125 × 4200 Tc,o − Tc,i = 210°C − ( 95 − 40 ) ° C = 100°C. & h ch m 0.125 × 2100

(

)

Using Eq. 11.15 with ∆T1 and ∆T2 as shown below, find ∆ Tlm = (∆T1 - ∆T2)/ l n (∆T1/∆T2). Substituting values, find

ACF  ( 210 − 40 ) − (100 − 95 ) / ln (170/5) 46.8°C = = = 0.55. A PF ( 210 − 95 ) − (100 − 40)  ln (115/60 ) 84.5°C

<

COMMENTS: In solving part (c), it is also possible to use Figs. 11.15 and 11.16 to evaluate NTU values for corresponding ε and Cmin/Cmax values. With knowledge of NTU it is then possible to find ACF/APF.

PROBLEM 11.64 KNOWN: Concentric tube HXer with prescribed inlet fluid temperatures, fluid flow rates and overall coefficient. FIND: (a) Maximum heat transfer rate, qmax; (b) Outlet fluid temperatures when area is 0.33 m2 with CF operation; (c) Compute and plot the effectiveness, ε, and fluid outlet temperatures, Tc,o and Th,o, as a function of UA for the range 50 ≤ UA ≤ 1000 W/K for CF operation with all other conditions remaining the same; as UA becomes very large, find asymptotic value for Th,o; (d) Largest heat transfer rate which could be achieved if HXer is very long with PF operation; effectiveness for this arrangement; and (e) Compute and plot ε, Tc,o and Th,o as a function of UA for the range 50 ≤ UA ≤ 1000 W/K for PF operation with all other conditions remaining the same; as UA becomes very large, find asymptotic value for Tc,o and Th,o. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties. PROPERTIES: Table A-6, Water (Assume Tc,o ≈ 85°C, Tc ≈ 335 K): cc = 4186 J/kg⋅K, (Assume Th,o ≈ 100°C, Th ≈ 100°C, Th ≈ 420 K): ch = 4302 J/kg⋅K. ANALYSIS: (a) With Cmin = Ch, the maximum heat transfer rate from Eq. 11.19 is

(

)

(

)

q max = C min Th,i − Tc,i = C h Th,i − Tc,i =

42 kg 3600 s

× 4302 ×

J kg ⋅ K

( 200 − 35 ) K = 8281 W .

<

(b) Using the ε - NTU method, find ε from values of Cmin, Cmin/Cmax, and NTU. C min = 42 3600 kg s × 4302 J kg ⋅ K = 50.19 W K ,

C min C max =

42 kg h × 4302 J kg ⋅ K 84 kg h × 4186 J kg ⋅ K

= 0.514

NTU = UA Cmin = 180 W m 2 ⋅ K × 0.33m 2 50.19 W K = 1.184 . Using Eq. 11.30 for counter flow operation, with Cr = Cmin/Cmax, find that 1 − exp [− NTU (1 − Cr )] 1 − exp [−1.18 (1 − 0.514 )] ε= = = 0.616 . 1 − Cr exp [− NTU (1 − Cr )] 1 − 0.514 exp [−1.18 (1 − 0.514 )] From the definition of effectiveness, ε = Ch (Th,i - Th,o)/Cmin (Th,i - Tc,i), it follows that

(

)

Th,o = Th,i − ε Th,i − Tc,i = 200$ C − 0.62 ( 200 − 35 ) C = 98.4$ C . $

<

Equating the energy balances on both fluids, Ch (Th,i - Th,o ) = Cc (Tc,o - Tc,i ), find

(

)

Tc,o = ( Ch Ce ) Th,i − Th,o + Tc,i = 0.514 ( 200 − 98.4 ) C + 35$ C = 87.2$ C . $

< Continued...

PROBLEM 11.64 (Cont.) (c) Using the IHT Heat Exchanger Tool, Concentric Tube, counter flow operation and the Properties Tool for Water, a model was developed using the effectiveness NTU method employed in the previous analysis to compute ε, Tc,o and Th,o as a function of UA for CF operation. The results are plotted and discussed below. (d) For PF with same prescribed inlet conditions, the temperature distributions appear as shown above when A→ ∞ . At the outlet, Tc,o = Th,o, and from the sketch δTh,max + δTc,max = (200 - 35)°C = 165°C. From the energy balance, find

Chδ Th,max = Ccδ Tc,max and solving simultaneously, find

δ Th,max = 109.0$ C

Th,o = Th,i − δ Th,max = 200 − 109.0 = 91.0$ C .

The heat rate and effectiveness are

< <

q = Ch ⋅ δ Th,max = 50.19 W K × 109.0K = 5471W

ε = q q max = 5471W 8, 281W = 0.661 .

Counterflow operation

Tco, Tho (C), eps*100

Tco , Tho (C), eps*100

(e) Using the IHT model from part (c), but for PF operation, the effectiveness, Tc,o and Th,o were computed and plotted as a function of UA. 130 110 90 70 50 30 0

200

400

600

800

1000

Parallel flow operation 130 110 90 70 50 30 100

200

300

400

UA (W/K)

UA (W/K)

Tho (C), minumum fluid Tco (C), maximum fluid Effectiveness, eps*100

Tho (C), minumum fluid Tco (C), maximum fluid Effectiveness, eps*100

500

COMMENTS: (1) From the plot for CF operation as UA increases, the minimum (hot) fluid outlet temperature, Th,o, decreases to the cold fluid temperature, Tc,i. That is when UA → ∞ , Th,o →Tc,i. As UA → ∞ , the effectiveness approaches unity as expected since a very large CF heat exchanger has a heat rate qmax and ε = 1. (2) From the plot for PF operation, as UA increases, Th,o and Tc,o approach an asymptotic value, 91.0°C. Also, as UA → ∞ , the effectiveness increases, approaching 0.661, rather than unity as would be the case for CF operation.

PROBLEM 11.65 KNOWN: Flow rates and inlet temperatures of water and glycol in counterflow heat exchanger. Desired glycol outlet temperature. Heat exchanger diameter and overall heat transfer coefficient without and with spherical inserts. FIND: (a) Required length without spheres, (b) Required length with spheres, (c) Explanation for reduction in fouling and pump power associated with using spheres. SCHEMATIC:

ASSUMPTIONS: (1) Negligible kinetic energy, potential energy and flow work changes, (2) Negligible heat loss to surroundings, (3) Constant properties, (4) Negligible tube wall thickness.

(

)

PROPERTIES: Table A-5, Ethylene glycol Th = 70°C : cp,h = 2606 J/kg⋅K; Table A-6, Water

(Tc ≈ 35°C ) : cp,c = 4178 J/kg⋅K.

ANALYSIS: (a) With Ch = Cmin = 1303 W/K and Cc = Cmax = 2089 W/K, Cr = 0.624. With actual and maximum possible heat rates of

(

)

q = Ch Th,i − Th,o = 1303 W / K (100 − 40 ) °C = 78,180 W

(

)

q max = Cmin Th,i − Tc,i = 1303 W / K (100 − 15 ) °C = 110, 755 W the effectiveness is ε = q/qmax = 0.706. From Eq. 11.30b,

NTU =

 ε −1  1  0.294  ln   = −2.66 ln   = 1.71 Cr − 1  ε Cr − 1   0.559 

Hence, with A = πDL and NTU = UA/Cmin,

C NTU 1303 W / K × 1.71 L = min = = 9.46m π Di U π (0.075m )1000 W / m 2 ⋅ K

<

 c, m  h, Th,i, Th,o and Tc,i are unchanged, Cr, ε and NTU are unchanged. Hence, with U (b) Since m 2

= 2000 W/m ⋅K,

<

L = 4.73m

(c) Because the spheres induce mixing of the flows, the potential for contaminant build-up on the surfaces, and hence fouling, is reduced. Although the obstruction to flow imposed by the spheres acts to increase the pressure drop, the reduction in the heat exchanger length reduces the pressure drop. The second effect may exceed that of the first, thereby reducing pump power requirements. COMMENTS: The water outlet temperature is Tc,o = Tc,i + q/Cc = 15°C + 78,180 W/2089 W/K =

(

)

52.4°C. The mean temperature Tc = 33.7°C is close to that used to evaluate the specific heat of water.

PROBLEM 11.66 KNOWN: Concentric tube, counter-flow heat exchanger. FIND: Total heat transfer rate and outlet temperatures of both fluids. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties. PROPERTIES: Table A-6, Water ( Th ≈ 68°C ≈ 340 K): ch = 4188 J/kg⋅K; Table A-6, Water ( Tc ≈ 37°C = 310 K): cc = 4178 J/kg⋅K. ANALYSIS: Using the ε-NTU method, begin by evaluating the capacity rates.

& h ch = 2.5kg/s × 4188 J / k g ⋅ K = 10,470 W / K Ch = m & ccc = 5.0kg/s × 4178 J / k g ⋅ K = 20,890 W / K Cc = m Hence, Cmin = Ch and Cmin/Cmax = 0.50 From the definition, Eq. 11.25,

NTU = U A / Cmin = 1000W/m 2 ⋅ K × 23m 2 / (10,470W/K ) = 2.20. Using values of NTU and Cmin/Cmax, find from Fig. 11.15, that

ε ≈ 0.80. From the definition of ε, Eq. 11.20, it follows that

(

)

q = ε qmax = ε Cmin Th,i − Tc,i = 0.80 ×10,470W/K (100 − 20 ) K = 670kW.

<

Performing energy balances on both fluids, find

Tc,o = Tc,i + q / Cc = 20 °C + 670kW/20,890W/K = 52.1°C

<

Th,o = Th,i − q / C h = 100°C − 670kW/10,470W/K = 36.0°C.

<

COMMENTS: (1) Note that Tc = (20 + 52.1)°C/2 ≈ 310 K and Th = (100 + 36)°C/2 = 341 K and that these values agree well with those used to evaluate the properties. (2) Eq. 11.30 could be used to evaluate ε; the result gives ε = 0.800.

PROBLEM 11.67 KNOWN: Shell and tube heat exchanger for cooling exhaust gases with water. FIND: Required surface area using ε-NTU method. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible changes in kinetic and potential energies, (3) Constant properties, (4) Gases have properties of air. PROPERTIES: Table A-6, Water, liquid ( Tc = (85 + 35)°C/2 = 333 K): cp = 4185 J/kg⋅K. ANALYSIS: Using the ε-NTU method, the area can be expressed as

A = NTU ⋅ Cmin / U

(1)

where NTU must be found from knowledge of ε and Cmin/Cmax = Cr. The capacity rates are:

& c cp,c = 2.5kg/s × 4185 J / k g ⋅ K = 10,463W/K Cc = m

Equating the energy balance relation for each fluid,

(

)(

)

Ch = Cc Tc,o − Tc,i / Th,i − Th,o = 10,463W/K ( 85 − 35 ) / ( 200 − 93) = 4889W/K. Hence,

Cr = Cmin / Cmax = Ch / Cc = 4889/10,463 = 0.467. The effectiveness of the exchanger, with qmax = Cmin (Th,i – Tc,i) and Cmin = Ch, is

(

)

(

)

ε = q / q max = C h Th,i − Th,o / Ch Th,i − Tc,i = ( 200 − 93 ) / ( 200 − 35 ) = 0.648. Considering the HXer to be a single shell with 2,4….tube passes, Eqs. 11.31b,c are appropriate to evaluate NTU. −1/2 E −1 2 / ε1 − (1 + Cr ) NTU = − 1 + C2r ln E= . 1/2 E +1 2 1 + Cr

(

)

(

Substituting numerical values,

E=

2/0.648 − (1 + 0.467 )

(1 + 0.4672 )

1/2

(

= 1.467 NTU = − 1 + ( 0.467 )2

)

)

−1/2

ln

1.467 − 1 = 1.51. 1.467 + 1

Using the appropriate numerical values in Eq. (1), the required area is

A = 1.51 × 4889W/K/180W/m 2 ⋅ K = 40.9m 2. COMMENTS: Figure 11.16 could also have been used with Cr and ε to find NTU.

<

PROBLEM 11.68 KNOWN: Dimensions, fluid flow rates, and fluid temperatures for a counterflow heat exchanger used to heat blood. FIND: (a) Outlet temperature of the blood, (b) Effect of water flowrate and inlet temperature on heat rate and blood outlet temperature. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties. PROPERTIES: Table A.6, Water ( Tm ≈ 55°C): cp = 4183 J/kg⋅K.  c ANALYSIS: (a) Using the ε - NTU method, we first obtain Ch = ( m h p,h ) = (0.10 kg/s × 4183 J/kg⋅K)  c = 418.3 W/K and Cc = ( m c p,c ) = (0.05 kg/s × 3500 J/kg⋅K) = 175 W/K = Cmin. Hence, (Cmin/Cmax) =

0.418 and NTU =

UA Cmin

500 W m 2⋅ K )π ( 0.055 m )( 0.5 m ) ( = = 0.247 . 175 W K

From Eq. 11.30, ε = 0.21. Hence, from Eq. 11.23

(

)

q = ε Cmin Th,i − Tc,i = 0.21(175 W K )( 60 − 18 ) C = 1544 W . From Eq. 11.7, q 1544 W Tc,o = Tc,i + = 18$ C + = 26.8$ C Cc 175 W K $

<

 h does not have a significant effect on ε for the prescribed (b) Because the variation of Cmin/Cmax with m h. NTU, Tc,o and q increase only slightly with increasing m 2000

29

1800

28

Heat rate, q(W)

Blood outlet temperature, Tco(C)

30

27 26

1600

1400

1200

25 24

1000 0.05

0.1

0.15

Water mass flowrate, mdoth(kg/s) Thi = 70 C Thi = 60 C Thi = 50 C

0.2

0.05

0.1

0.15

0.2

Water mass flowrate, mdoth(kg/s) Tci = 70 C Tci = 60 C Tci = 50 C

However, the water inlet temperature does have a significant effect, and accelerated heating is achieved with Th,i = 70°C.

 h = 0.2 kg/s and Th,i = 70°C, the outlet temperature of the blood is still below COMMENTS: With m the desired level of Tc,o ≈ 37°C. This value of Tc,o could be increased by increasing L or Th,i.

PROBLEM 11.69 KNOWN: Inlet temperatures and flow rates of water (c) and ethylene glycol (h) in a shell-and-tube heat exchanger (one shell pass and two tube passes) of prescribed area and overall heat transfer coefficient. FIND: (a) Heat transfer rate and fluid outlet temperatures and (b) Compute and plot the effectiveness, ε,  h , for the and fluid outlet temperatures, Tc,o and Th,o as a function of the flow rate of ethylene glycol, m  h ≤ 5 kg/s. range 0.5 ≤ m SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties, and (4) Overall coefficient remains unchanged. PROPERTIES: Table A-5, Ethylene glycol ( Tm ≈ 40°C): cp = 2474 J/kg⋅K; Table A-6, Water ( Tm ≈ 15°C): cp = 4186 J/kg⋅K. ANALYSIS: (a) Using the ε-NTU method we first obtain  c Ch = m h p,h = ( 2kg s × 2474J kg ⋅ K ) = 4948 W K

( ) Cc = ( mc cp,c ) = (5kg s × 4186J kg ⋅ K ) = 20, 930 W K . 

Hence with Cmin = Ch = 4948 W/K and Cr = Cmin/Cmax = 0.236, NTU =

UA Cmin

800 W m 2 ⋅ K )15m 2 ( = = 2.43 . 4948 W K

From Fig. 11.16, ε = 0.81 and from Eq. 11.23

(

)

q = ε Cmin Th,i − Tc,i = 0.81( 4948 W K )( 60 − 10 ) K = 2 × 105 W . From Eqs. 11.6 and 11.7, energy balances on the fluids,

Tc,o = Tc,i +

Ch q Cc

$

= 60 C − = 10$ C +

2 × 105 W 4948 W K

<

= 19.6$ C

2 × 105 W 20, 930 W K

<

= 19.6$ C .

(b) Using the IHT Heat Exchanger Tool, Shell and Tube, and the Properties Tool for Water and Ethylene Glycol, Tc,o, Th,o, and ε as a function of  h were computed and plotted. m

 h ) note that ε → 1 while At very low Cmin, (low m  h increases, both fluid outlet Th,o → Tc,i. As m temperatures increase and the effectiveness decreases.

Tco, Tho (C), eps*100

Th,o = Th,i −

q

<

100 80 60 40 20 0 0

1

2

3

4

Hot fluid flow rate, mdoth (kg/s) Tco (C), max fluid Tho (C), min fluid Effectiveness, eps*100

5

PROBLEM 11.70 KNOWN: Flow rate, specific heat and inlet temperature of gas in cross-flow heat exchanger. Flow rate and temperature of water which enters as saturated liquid and leaves as saturated vapor. Number of tubes, tube diameter and overall heat transfer coefficient. FIND: Required tube length. SCHEMATIC:

ASSUMPTIONS: (1) Negligible kinetic and potential energy changes, (2) Negligible heat loss to surroundings, (3) Constant gas specific heat. 6

PROPERTIES: Table A-6, Saturated Water, (T = 450 K): hfg = 2.024 ×10 J/kg. ANALYSIS: Use effectiveness-NTU method

ε=

q q max

=

(

q

Cmin Th,i − Tc,i

)

=

(

q

& h c p,h Th,i − Tc,i m

)

& c h fg = 3 k g / s × 2.024 ×10 6 J / k g = 6.072 × 106 W q=m ε=

6.072 × 106 W = 0.571 10 k g / s ×1120J/kg ⋅ K (1400 − 450 ) K

Cmin / Cmax = 0.

From Fig. 11.19, find

NTU ≈ 0.8 ≈ Uo Nπ D o L / Cmin L≈

0.8 × 10 kg/s ×1120J/kg ⋅ K 50 W / m 2 ⋅ K × 500π × 0.025m

<

= 4.56m.

COMMENTS: (1) The gas outlet temperature is

& h c p,h = 1400K − 6.072 × 106 W/10kg/s ×1120J/kg ⋅ K = 857.9 K. Th,o = Th,i − q / m (2) Using the LMTD method,

∆ Tlm,CF =  (1400 − 450 ) − ( 858 − 450 ) / ln  (1400 − 450 ) / (858 − 450 )  = 641K.

From Fig. 11.13, find F = 1, so the area and length are Ao = q / Uo F ∆Tlm,CF = 6.072 ×10 6 W / 50 W / m 2 ⋅ K ×1 × 641K = 189m 2

(

L = A / Nπ Do = 189m 2 /500π ( 0.025m ) = 4.82m.

)

PROBLEM 11.71 KNOWN: Gas flow conditions upstream of a tube bank of prescribed geometry. Flow rate and inlet temperature of water passing through the tubes. FIND: (a) Overall heat transfer coefficient, (b) Water and gas outlet temperatures, (c) Effect of water flow rate on heat recovery and outlet temperatures. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible heat loss to the surroundings and kinetic and potential energy changes, (4) Negligible tube fouling and wall thermal resistance, (5) Fully developed water flow, (6) Gas properties are those of air. PROPERTIES: Table A.6, Water (Assume Tm ≈ 340 K): cp = 4188 J/kg⋅K, µ = 420 × 10-6 N⋅s/m2, k = 0.660 W/m⋅K, Pr = 2.66; Table A.4, Air (Assume Tm ≈ 600 K): cp = 1051 J/kg⋅K, ν = 52.7 × 10-6 m2/s, k = 0.047 W/m⋅K, Pr = 0.69.

 c,1 = ANALYSIS: (a) For the prescribed conditions, U = (1/hi + 1/ho)-1. For the internal flow, with m 0.025 kg/s,  4m 4 × 0.025 kg s c,1 Re D = = = 3032 . π Dµ π ( 0.025 m ) 420 × 10−6 N ⋅ s m 2 Hence, assuming turbulent flow, 4 / 5 0.4 Nu D = 0.023 Re D Pr = 0.023 (3032 )

4/5

hi =

k D

Nu D =

0.660 W m⋅ K 0.025 m

For the external flow, Vmax =

( 2.66 )0.4 = 20.8

20.8 = 548 .

0.05 m

(0.05 − 0.025 ) m

5.0 m s = 10.0 m s . Hence

V D 10 m s × 0.025 Re D,max = max = = 4744 ν 52.7 × 10−6 m 2 s 0.36 From the Zhukauskas correlation and Tables 7.7 and 7.8, Nu D = ( 0.97 ) 0.27 Re0.63 ( Pr Prs )1/ 4 . D,max Pr Neglecting the Prandtl number ratio,

Nu D = ( 0.97 ) 0.27 ( 4744 )

0.63

ho =

k D

Nu D =

0.047 W m⋅ K 0.025 m

(0.69 )0.36 = 47.4 47.4 = 89.1W m 2⋅ K . Continued...

PROBLEM 11.71 (Cont.)

<

Hence, U = (1/548 + 1/89.1)-1 = 76.7 W/m2⋅K.  c = 2.5 kg/s, Cc = (b) The fluid outlet temperatures may be determined from the ε-NTU method. With m  c cp,c = 2.5 kg/s × 4188 J/kg⋅K = 10,470 W/K. With Ch = m  h c p,h = 2.25 kg/s × 1051 J/kg⋅K = 2365 m W/K, Cmin/Cmax = Cmixed/Cunmixed = 2365/10,470 = 0.23. Hence, with A = NπDL = 100π × 0.025 m × 4 m = 31.4 m2,

(

2

NTU =

UA

=

C min

76.7 W m ⋅ K 31.4 m

2

2365 W K

) = 0.95

From Fig. 11.19, ε ≈ 0.61. From Eq. 11.19, qmax = Cmin(Th,i - Tc,i) = 2365 W/K(800 - 300)K = 1.18 × 106 W. Hence, q = εqmax = 0.72 × 106 W. From Eq. 11.6b,

(Th,i − Th,o ) = C

q

(

=

h

0.72 × 106 W 2365 W K

)

q

=

<

Th,o = 496 K

0.72 × 106 W

<

Tc,o = 369 K = 69 K Cc 10, 470 W K (c) Using the appropriate Heat Exchangers, Correlations and Properties Toolpads of IHT, the following results were obtained. From Eq. 11.7b,

Tc,o − Tc,i =

= 304 K

540

800000

460 750000 Heat rate, q(W)

Outlet temperatures (K)

500

420 380 340

700000

300 2

4

6

8

10

12

14

16

Water flow rate, mdotc(kg/s)

18

20 650000 2

Gas outlet temperature, Tho(K) Water outlet temperature, Tco(K)

4

6

8

10

12

14

16

18

20

Water flow rate, mdotc(kg/s)

 c (and m  c,1 ), hi increases, thereby increasing U and q. However, because the total With increasing m  c = 20 kg/s only yields U = 83.9 W/m2⋅K, despite resistance is dominated by the gas-side condition, m

 c is much the fact that hi = 2180 W/m2⋅K. Because the extent to which q increases with increasing m  c itself, Tc,o decreases with increasing m  c . Hence, there is a trade-off smaller than the increase in m between the amount of hot water and the temperature at which it is delivered. If, for example, the  c cannot exceed 8 kg/s. To maintain an acceptable temperature must exceed 50°C (Tc,o > 323 K), m  c, m  h (and V) should be increased, thereby increasing ho, and hence U value of Tc,o, while increasing m and q. COMMENTS: If the air and water property functions of IHT are used to evaluate properties at appropriate mean values of the inlet and outlet fluid temperatures, the following, more accurate, results would be obtained for Parts (a) and (b): ε = 0.582, q = 0.697 × 106 W, Tc,o = 366.6 K, Th,o = 508.8 K, hi = 523 W/m2⋅K, ho = 86.5 W/m2⋅K and U = 74.2 W/m2⋅K.

PROBLEM 11.72  c and inlet KNOWN: Tube arrangement in steam-to-air, cross-flow heat exchanger. Flow rate m temperature of air. Condensing temperature of steam.  = 12 kg/s, (b) Effect of m  c on air outlet temperature, heat rate FIND: (a) Air outlet temperature for m c and condensation rate.

SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Negligible steam side convection and tube wall conduction resistance, (4) Mean air temperature is 350 K.

(

)

PROPERTIES: Table A.4, Air (Assume Tc ≡ Tc,i + Tc,o 2 ≈ 350 K, 1 atm): ρ = 0.995 kg/m3, cp = 1009 J/kg⋅K, ν = 20.92 × 10-6 m2/s, k = 0.030 W/m⋅K, Pr = 0.700; Ts = 400 K: Pr = 0.690. ANALYSIS: (a) For a single-pass, cross-flow heat exchanger with one fluid mixed and the other unmixed, Fig. 11.19 can be used to obtain ε, where Cmin/Cmax = Cmixed/Cunmixed = 0 and NTU = UA/Cmin =  c cp. From Eq. 11.5, U = h o , and the Zhukauskas correlation may be used to estimate h o . U(πDL)N/ m

 c = ρVA ≈ ρVNTLST. Hence, The upstream velocity may be obtained from m  m 12 kg s c V= = = 1.44 m s . ρ N T LST 0.995 kg m3 × 30 × 2 m × 0.14 m For aligned tubes, ST 0.14 m Vmax = V= 1.44 m s = 2.88 m s ST − D ( 0.14 − 0.07 ) m V D 2.88 m s × 0.07 m Re D,max = max = = 9637 . −6 2 ν 20.92 × 10 m s From Table 7.7, select values of C = 0.27 and m = 0.63. Hence, 0.36 Nu D = 0.27 Re0.63 ( Pr Prs ) D,max Pr

0.25

Nu D = 0.27 (9637 )

0.63

ho = Nu D

k D

= 77.1

(0.70 )0.36 ( 0.70

0.030 W m⋅ K 0.07 m

0.69 )

0.25

= 77.1

= 33.0 W m 2⋅ K .

Hence, NTU =

h oπ DLN  c m c p

=

33.0 W m 2⋅ K × π ( 0.07 m ) 2 m (1200 ) 12 kg s × 1009 J kg ⋅ K

= 1.44 .

From Fig. 11.19, find ε ≈ 0.77 and then determine Continued...

PROBLEM 11.72 (Cont.)

( ) = Tc,o − Tc,i  c (T − T ) q max Ts − Tc,i m c p s c,i Tc,o = Tc,i + ε ( Ts − Tc,i ) = 300 K + 0.77 ( 400 − 300 ) K = 377 K = 104$ C

ε=

q

=

 c m c p Tc,o − Tc,i

<

(b) With q = εqmax = εCc(Ts - Tc,i) and the condensation rate given by Eqs. 10.33 and 10.26, q q  m ≈ cd = h ′fg h fg the foregoing model may be used with the Heat Exchangers, Correlations and Properties Toolpads of IHT   c on Tc,o, q and m to determine the effect of m cd . 380

3E6

370 Heat rate, q(W)

Air outlet temperature, Tco(K)

2.5E6

360

2E6

1.5E6

1E6

350

500000

10

20

30

40

50

10

20

Air flow rate, mdotc(kg/s)

30

40

50

Air flow rate, mdotc(kg/s)

 c , q must also increase. However, since the increase in q is Since h o increases with increasing m  c , Tc,o decreases with increasing m  c. proportionally less than the increase in m

Condensation rate, mdotcd(kg/s)

1.4

1.2

1

0.8

0.6

0.4

0.2 10

20

30

40

50

Air flow rate, mdotc(kg/s)

The condensation rate increases proportionally with the increase in q, and if the objective is to maximize  c should be maintained. the condensation rate, the largest value of m COMMENTS: If the objective is to heat the air, there is obviously a trade-off between maintaining elevated values of the flowrate and outlet temperature.

PROBLEM 11.73 KNOWN: Heat exchanger operating in parallel-flow configuration. FIND: Expression for Rlm/Rt which doesn’t involve temperatures. Plot result. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible change in kinetic and potential energy. ANALYSIS: (a) For the exchanger, the rate equation is

q = UA∆Tlm and we can define thermal resistances as

(

)

R t = Th,i − Tc,i / q

or

R lm = ( ∆Tlm ) / q = 1/UA.

Using the rate equation and the definition of effectiveness, find the thermal resistance based upon the inlet temperatures of the hot and cold fluids as

(

)

R t = C min T h,i −Tc,i / C min ⋅q = 1/ ε C min. The ratio of these resistances is

R lm 1/UA ε ε = = = Rt 1/ ε Cmin U A / Cmin NTU and for the parallel flow, concentric tube configuration using Eq. 11.29a,

R lm 1 − exp  − NTU ( 1 + Cr )  1 − exp ( −B ) = = Rt NTU (1 + Cr ) B

<

where B = NTU(1 + Cr). Evaluating the ratio for various values of B, find B 0.1 0.5 1.0 3.0 5.0 10.0

<

Rlm/Rt 0.95 0.79 0.63 0.32 0.20 0.10

COMMENTS: (1) For Cmax → ∞, Cr → 0; hence B → NTU. (2) For Cmax ≈ Cmin, B → 2NTU or 1 B ~ C−min . (3) For B > 1, Rlm/Rt → B . (5) We conclude that care must be taken in representing heat exchangers with a thermal resistance, recognizing that the resistance will depend on flow rates for wide ranges of conditions. -1

PROBLEM 11.74 KNOWN: Heat exchanger condensing steam at 100°C with cooling water supplied at 15°C. FIND: (a) Thermal resistance of the exchanger, (b) Change in thermal resistance if fouling is 0.0002 2 m ⋅K/W on each of the inner and outer tube surfaces, and (c) Plot the thermal resistance as a function of tube water inlet rate assuming all other conditions remain unchanged; comment on whether UA will remain constant if the flow rate changes. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes. PROPERTIES: Table A-6, Water ( Tm = 42°C=315 K): cp = 4179 J/kg⋅K. ANALYSIS: (a) For an exchanger, using the rate equation,

(

)

q = UA ∆Tlm = Th,i − Tc,i / Rt , the thermal resistance of the exchanger can be expressed as

(

)

Cmin Th,i − Tc,i T −T 1 q 1 R t = h,i c,i = = ⋅ max = . UA∆ Tlm Cmin ⋅ q C min q ε Cmin For the present exchanger with Cr = 0, use Eq. 11.36a with

& c cp,c = 0.5kg/s × 4179 J / k g ⋅ K = 2090W/K Cmin = m NTU = U A / Cmin = 2000 W / m 2 ⋅ K × 0.5m 2 /2090 W / K = 0.478 ε = 1 − exp ( − NTU ) = 0.380. Hence, the thermal resistance is

R t = 1/0.380 × 2090 W / K = 1.258 ×10−3 K/W.

<

(b) With fouling present, the overall heat transfer coefficient will decrease. No fouling:

With fouling:

1 UoA

=

1 1 + h h A h h c Ac

R′′ R′′ 1 1 1 2R ′′f = + f,c + f,h = + U f A U oA A c Ah Uo A A Continued …..

PROBLEM 11.74 (Cont.) 1 1 2 × 0.0002 m 2 ⋅ K / W = + Uf A 2000 W / m 2 ⋅K × 0.5m 2 0.5m 2 Uf A = 555.6 W / K . It follows that NTU = UfA/Cmin = 0.266 and ε f = 0.233 giving

R t,f = 1/ ε f C min = 2.050 ×10−3 K / W and hence the increase in thermal resistance due to fouling is

( R t,f − R t ) / R t = ( 2.050 − 1.258) /1.258 = 63%.

<

(c) With no fouling, the thermal resistance, when all other conditions (UoA = 1000 W/K) remain unchanged, depends on Cmin only as NTU = UoA/Cmin,

1 1 Rt = = ε Cmin Cmin

  UA   1 − exp  −   Cmin   

Cmin (W/K) 3 Rt (K/W × 10 )

200 4.967

400 2.723

−1

600 2.055

  1000 W / K   = 1 − exp  −  Cmin  Cmin    1

1000 1.582

1500 1.370

2000 1.271

−1

3000 1.176

From the plot note that Rt is a weak function of Cmin above Cmin > 1000 W/K, from which we conclude that using a constant Rt would be reasonable.

Concerning the variability of UA with changing 0.8 & c0.8. Cmin: if most of the resistance is on the water side and the flow is turbulent, hc ≈ Re0.8 D ≈ u m ≈m It follows that hc will depend significantly on changes in Cmin. However, if hc and hh are of similar magnitude, the effect of Cmin on U may not be significant.

PROBLEM 11.75 KNOWN: Air conditioner modeled as a reversed Carnot heat engine, with refrigerant as the working fluid, operating between indoor and outdoor temperatures of 23 and 43°C, respectively, removing 5 kW from a building. Compressor and fan motor efficiency of 80%. FIND: (a) Required motor power assuming negligible thermal resistances between the refrigerant in the condenser and the outside air and between the refrigerant in the evaporator and the inside air, and -3 (b) Required power if thermal resistances are each 3 × 10 K/W. SCHEMATIC:

ASSUMPTIONS: (1) Ideal heat exchanger with no losses, (2) Air conditioner behaves as reversed Carnot engine. ANALYSIS: (a) With negligible thermal resistances, the Carnot cycle and reversed heat engine can be represented as shown above. Hence,

w & ideal = qH − q L = qL ( TH / TL ) − 1 = 5kW ( 316 K / 2 9 6 K) − 1 = 0.3378 kW.

Considering the fan power requirement and the efficiency of the motor,

& act = ( w & ideal + w& fan ) / ηc = ( 0.3378 + 0.200 ) kW/0.8 = 0.672 kW. w

<

-3

(b) Consider now thermal resistances of Rt = 3 × 10 K/W on the high temperature (condenser) and low temperature (evaporator) sides. Low side: in order to remove heat from the room, TC < Ti. That is

(

)

Ti − TC = qR t = 5 kW 3×10−3 K / W = 15 K TC = Ti − 1 5 K = 23°C − 15 K = 8°C. High side: in order to reject heat from the condenser to the outside air, TH > To,

TH − T o = q HR t = q c (T H / Tc ) R t

TH − ( 43 + 273) K = 5 kW  TH / ( 8+ 273) 3 ×10 −3 K / W

TH = 333.9 K = 61 °C.

The work required for this cycle is

w & ideal = qH − q L = qL ( TH / TL ) − 1 = 5 kW ( 61+ 273) K /( 8 + 273) K − 1 = 0.943kW & act = ( w & ideal + w& fan ) / ηc = ( 0.943 + 0.2 ) kW/0.8 = 1.43 kW. w

<

The effect of finite thermal resistances in the evaporator and condenser is to increase the power by a factor of two.

PROBLEM 11.76 KNOWN: Flow rate and pressure of saturated vapor entering a condenser. Number and diameter of condenser tubes. Water flow rate and inlet temperature. Tube outside convection coefficient. FIND: (a) Water outlet temperature, (b) Total tube length, (c) Effect of fouling on mass condensation, (d) Effect of water flow rate and inlet temperature on condenser performance. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss to surroundings and potential and kinetic energy changes, (2) Constant properties, (3) Negligible wall conduction resistance and fouling (initially). PROPERTIES: Water (given): cp = 4178 J/kg⋅K, µ = 700 × 10-6 kg/s⋅m, k = 0.628 W/m⋅K, Pr = 4.6; Table A.6, Sat. steam (355 K): hfg = 2.304 × 106 J/kg; With fouling: R′′f = 0.0003 m2⋅K/W.

(

)

(

)

  h  c ANALYSIS: (a) From an energy balance, qh = m h i h,i − i h,o = m h fg = q c = m c p,c Tc,o − Tc,i , or  c m h c,p

Tc,o = Tc,i +

 c m c p,c

= 280 K +

1.5 kg s × 2.304 × 106 J kg 15 kg s × 4178 J kg ⋅ K

<

= 335.1K .

(b) Since Cr = 0, NTU = -ln(1 - ε), where

ε=

q q max

=

( ) = (335.1 − 280 ) K = 0.735  c m (355 − 280 ) K c p,c (Th,i − Tc,i )

 c m c p,c Tc,o − Tc,i

Hence, NTU = -ln(1 - 0.735) = 1.327 = UA/Cmin. The overall heat transfer coefficient is given by 1/U = 1/ h i + 1/ h o . For the internal tube flow, Re D =

 4m c,1

4 × 15 kg s 100

= 27, 284 π ( 0.01m ) 700 × 10−6 kg s ⋅ m Hence, assuming fully developed flow with the Dittus-Boelter correlation,

π Dµ

=

Nu D = 0.023 Re4D/ 5 Pr n = 0.023 ( 27, 284 )

4/5

hi = ( k D ) Nu D =

0.628 W m⋅ K 0.01m −1

and U = [(1 9408 ) + (1 5000 )]

( 4.6 )0.4 = 149.8

149.8 = 9408 W m 2⋅ K

W m 2⋅ K = 3265 W m 2⋅ K . Hence, the heat transfer area is

(

)

2 2  c A=m c p,c ( NTU U ) = 15 kg s ( 4178 J kg ⋅ K ) 1.327 3265 W m ⋅ K = 25.5 m

and the tube length is L = A/NπD = 25.5 m2/100π(0.01 m) = 8.11 m.

<

(c) With fouling, the overall heat transfer coefficient is 1/Uw = 1/Uwo + R′′f . Hence, Continued...

PROBLEM 11.76 (Cont.)

(

)

1 U w = 3.063 × 10−4 + 3 × 10 −4 m 2 ⋅ K W Uw = 1649 W/m2⋅K.

(

NTU = UA Cmin = 1649 W m 2⋅ K × 25.5 m 2

) (15 kg s × 4178 J kg ⋅ K ) = 0.671

From Eq. 11.36a, ε = 1 - exp(-NTU) = 1 - exp(-0.671) = 0.489. Hence, q = εqmax = 0.489 × 15 kg/s × 4178 J/kg⋅K(355 - 280)K = 2.30 × 106 W. Without fouling the heat rate was 6 6  h q=m h fg = 1.5 kg s × 2.304 × 10 J kg = 3.46 × 10 W .

<

6 6   Hence, m h,w m h,wo = 2.30 × 10 3.46 × 10 = 0.666 .  The condensation rate with fouling is then m h,w = 0.666 × 1.5 kg s = 0.998 kg s .

(d) The prescribed water inlet temperature of Tc,i = 280 K is already at the lower limit of available sources, and it would not be feasible to consider smaller values. In addition, with h i already quite large,

 c is not likely to provide a significant improvement in performance. Using the Heat an increase in m

 c ≤ 25 kg/s. Exchanger and Correlations Tools from IHT, the following results were obtained for 15 ≤ m

Condensation rate, mdoth(kg/s)

1.2

1.1

1

0.9 15

17

19

21

23

25

Water mass flow rate, mdotc(kg/s)

 c , there is approximately an 18% increase in the heat rate, and hence in the Over the specified range of m condensation rate. This increase is, in part, due to the increase in h i from 9408 to 14,160 W/m2⋅K, which increases U from 1649 to 1752 W/m2⋅K, as well as to a reduction in Tc,o from 316.6 to 306.0 K, which increases the mean driving potential for heat transfer. COMMENTS: There is a significant reduction in performance due to fouling, which can not be restored  c . The desired performance could be achieved by oversizing the condenser, that is, by by increasing m increasing the number of tubes and/or the tube length.

PROBLEM 11.77 KNOWN: Rankine cycle with saturated steam leaving the boiler at 2 MPa and a condenser pressure of 10 kPa. Net reversible work of 0.5 MW. FIND: (a) Thermal efficiency of ideal Rankine cycle, (b) Required cooling water flow rate to condenser at 15°C with allowable temperature rise of 10°C, and (c) Design of a shell and tube heat exchanger (one shell and multiple tube passes) to satisfy condenser flow rate and temperature rise. SCHEMATIC:

ASSUMPTIONS: (1) Negligible loss from condenser to surroundings, (2) Negligible kinetic and potential energy changes in heat exchanger, (3) Ideal Rankine cycle, and (4) Negligible thermal resistance on condensate side of exchanger tubes. PROPERTIES: Steam Tables, (Wark, 4th Edition): (1) p1 = p4 = 10 kPa = 0.10 bar, Tsat = 45.8°C = -3 3 319 K, vf = 1.0102 × 10 m /kg, hf = 191.83 kJ/kg; (3) p2 = p3 = 2 Mpa = 20 bar, hg = 2799.5 kJ/kg, sg = 6.3409 kJ/kg⋅K; (4) s4 = s3 = 6.3409 kJ/kg⋅K, p4 = 0.10 bar, sf = 0.6493 kJ/kg⋅K, sg = 8.1502 kJ/kg⋅K, hf = 191.83 kJ/kg⋅K, hfg = 2392.8 kJ/kg; Table A-6, Water (Tsat = 293 K): cp,c = 4182 J/kg⋅K, µ = 1007 -6 2 5 2 5 × 10 N⋅s/m , k = 0.603 W/m⋅K, Pr = 7.0. Note: 1 bar = 10 N/m = 10 Pa. ANALYSIS: (a) Referring to your thermodynamics text, find that

η=

wnet w t − wp ( h3 − h4 ) − v1 ( p2 − p1 ) = = QH QH h3 − h 2

where the net work is the turbine minus the pump work. Assuming the liquid in the pump is incompressible,

(

)

wp = v1 ( p2 − p1 ) = 1.0102 ×10−3 m3 /kg 2× 106 − 10 × 103 N / m 2 = 2.01kJ/kg. To find the enthalpies at states 2, 3, and 4, consider the individual processes. For the pump,

h 2 = h 1 + w p = (191.83 + 2.01) kJ/kg = 193.84 kJ/kg.

Since the exit state of the boiler is saturated at p3 = 2 Mpa,

h 3 = h g = 2799.5 kJ/kg.

QH = h3 − h2 = ( 2799.5 −193.84 ) kJ/kg = 2605.7 kJ/kg. Since the process from 3 to 4 is isentropic, s4 – s3, hence

(

)

x 4 = ( s4 − sf ) / s g − s f = ( 6.3409 − 0.6493) / ( 8.1502 − 0.6493) = 0.759

h 4 = h f + xh fg = 191.83 + 0.759 ( 2392.8 ) kJ/kg = 2007.5kJ/kg. Continued …..

PROBLEM 11.77 (Cont.) w t = h3 − h4 = ( 2799.5 − 2007.5) kJ/kg = 792.0 kJ/kg. Substituting appropriate values, the thermal efficiency is

η=

( 792.0 − 2.01) kJ/kg 2605.7 kJ/kg

<

= 0.303 = 30.3%.

(b) From an overall balance on the cycle, the heat rejected to the condenser is

Qc = QH − w net =  2605.7 − ( 792.0 − 2.01) kJ/kg = 1815.7 kJ/kg.

Since the net reversible power is 0.5 MW, the required steam rate (h) is 6 & &h =W m net / w net = 0.5 ×10 W / ( 792.0 − 2.01) kJ/kg = 0.6329 kg/s. Hence, the heat rate to be removed by the cold water passing through the condenser is

(

&h=m & c c p,c Tc,out − Tc,in q c = Q c ⋅m

)

& c × 4182 J / k g ⋅ K ( 25 − 15) K 1815.7 kJ/kg × 0.6329 k g / s = 1.149 × 106 W = m

<

& c = 27.47 k g / s m

where cp,c = cp,f is evaluated at T2, Tc,in = 15°C and Tc,out – Tc,in = 10°C, the specified allowable rise. (c) To design the heat exchanger we need to evaluate UA. Considering the shell-tube configuration and since Cr = Cmin/Cmax = 0,

ε = 1 − exp ( − NTU ) = 1 − exp − ( U A / Cmin )  ε=

ε=

q q max

=

qc

(

& c cp,c Th − Tc,i m

)

1.149 × 106 W = 0.326 27.47kg/s × 4182 J / k g ⋅ K ( 45.7 − 15) K

 UA  0.326 = 1 − exp  −   27.47 k g / s × 4182 J / k g ⋅ K  UAs = 45,372 W / K & c c p,c. Our design process will involve the following steps: select tube diameter, D = where Cmin = m 15 mm; set um = 2 m/s in each tube and find number of tubes; perform internal flow calculation to estimate h c and then determine the length.

(

& c = ρ Ac Nu m = 1.010 ×10 −3 m 3 / k g m N = 78.5 ≈ 79.

) (π (0.015m )2 / 4) 2 m / s × N = 27.47 k g / s −1

Continued …..

PROBLEM 11.77 (Cont.) For flow in a single tube,

Re D =

4 ( 27.47 kg/s/79 ) &t 4m = = 29,310. π Dµ π ( 0.015m ) 1007 × 10−6 N ⋅ s / m 2

Assuming the flow is fully developed and using the Dittus-Boelter correlation,

Nu =

hD = 0.023Re0.8 Pr 0.4 = 0.023 ( 29,310 )0.8 ( 7.00 )0.4 = 187.7 D k

h = 0.603W/m ⋅ K ×187.7/0.015m = 7544 W / m 2 ⋅ K. Hence, the tube length is

UAs = h (π DL) N = 45,372 W / K L = 45,372 W / K / 7 5 4 4 W / m 2 ⋅ K ×π ( 0.015m ) 79 = 1.6m and our design has the following parameters:

N = 79 tubes

L =1.6m

D = 15 mm.

<

COMMENTS: (1) The selection of the tube diameter and water velocity values (15 mm, 2 m/s) was based upon prior experience; they seemed reasonable. We could, however, establish other requirements which would influence these choices such as allowable pressure drop and standard tube sizes.

PROBLEM 11.78 KNOWN: Rankine cycle with saturated steam leaving the boiler at 2 Mpa and a condenser pressure of 10 kPa. Heat rejected to the condenser of 2.3 MW. Condenser supplied with cooling water at rate of 70 kg/s at 15°C. FIND: (a) Size of the condenser as determined by the parameter, UA, and (b) Reduction in thermal efficiency of the cycle if U decreases by 10% due to fouling assuming water flow rate and inlet temperature and the condenser steam pressure remain fixed. SCHEMATIC:

ASSUMPTIONS: (1) Negligible loss from condenser to surroundings, (2) Negligible kinetic and potential energy changes in exchanger, (3) Ideal Rankine cycle, (4) For fouled operating condition, m& c, Tc,i and p4 remain the same. PROPERTIES: Steam Tables (Wark, 4th Edition): See previous problem for calculations to obtain cycle enthalpies; h1 = 191.83 kJ/kg, h4 = 2007.5 kJ/kg. ANALYSIS: (a) For the condenser, recognize that Cmin = Cc, and Cr = Cmin/Cmax = 0,

ε=

q q max

= 1 − exp ( − NTU ) = 1 − exp ( − U A / C min )

& c cp,c = 70 kg/s × 4182 J / k g ⋅ K = 292,740 W / K Cmin = m

(

)

q max = Cmin Th − Tc,i = 292,740 W / K ( 45.7 −15 ) K = 8.987 ×106 W. q = q h = 2.30 ×106 W 2.30 × 106 W

  UA = 0.256 = 1 − exp  −  8.987 ×106 W  292,740 W / K 

<

UA = 86,538 W / K. (b) In the fouled condition, U is reduced 10%, hence

Uf A = 0.9UA = 77,884 W / K and

NTUf =

Uf A 77,884 W / K = = 0.266 Cmin 292,740 W / K

ε f = 1 − exp ( −NTU f ) = 1 − exp ( −0.266 ) = 0.234. Continued …..

PROBLEM 11.78 (Cont.) If we operate the cycle at the same back pressure p4 = 10 kPa so that Th = 45.7°C, the heat removal rate must decrease,

q h = ε q max = 0.234 × 8.987 ×10 6 W = 2.103 × 10 6 W since qmax = Cmin (Th – Tc,i) remains the same. From the previous problem, we found the heat rejected as

h 4 − h1 = ( 2007.5 −191.83 ) kJ/kg = 1815.7 kJ/kg and hence the cycle steam rate through the fouled condenser is

& h,f = qh / ( h4 − h1 ) = 2.103 ×106 W/1815.7 kJ/kg =1.158 kg/s. m For the unfouled condenser of part (a), the steam rate was

& h = 2.3MW/1815.7 kJ/kg = 1.267 kg/s. m Hence, we see that fouling reduces the steam rate by 8.5% when U is decreased 10%. Since p4 remains the same, the thermal efficiency remains unchanged,

<

η = 30.3% as calculated in the previous problem. However, the net work of the cycle will decrease 8.5%.

COMMENTS: Fouling of the condenser heat exchanger has no effect on the thermal efficiency of the cycle since the back pressure at the condenser is maintained constant. The effect is, however, to reduce the heat rejection rate while maintaining exchanger flow rate and inlet temperature fixed. Comparing the conditions: Parameter UA, W/K ε qh, MW & net w

Clean 86,538 0.256 2.300 --

Fouled 77,884 0.234 2.103 --

Change (%) -10.0 -8.6 -8.6 -8.6

PROBLEM 11.79 KNOWN: Compact heat exchanger (see Example 11.6) after extended use has prescribed fouling factors on water and gas sides. FIND: Gas-side overall heat transfer coefficient. SCHEMATIC:

ASSUMPTIONS: (1) Heat transfer coefficients on the inside and outside (cold- and hot-sides) are the same as for the unfouled condition, (2) Temperature effectiveness of the finned hot side surface is the same as for the unfouled condition. ANALYSIS: The overall heat transfer coefficient follows from Eq. 11.1 as

R′′f,c R′′f,h 1 1 1 = + +Rw + + Uh Ah (ηo hA )c (ηo A)c (ηo A )h (ηoh A )h

where Rw and R ′′f are the wall resistance and fouling factors, respectively. Multiply both sides by Ah and recognizing that ηo,c = 1, obtain

R′′f,c R′′ 1 1 1 = + + Ah R w + f,h + . Uh hc ( Ac / Ah ) ( Ac / Ah ) ηo,h ηo hh Substitute numerical values from Example 11.6 results (hh, ηo,h, Ah Rw, Ac/Ah) and those from the problem statement R′′f,h , R′′f,c , hc to find,

(

)

1 1 = Uh 1500 W / m2 ⋅ K ( 0.143) +

0.0005m 2 ⋅ K / W 0.001m 2 ⋅ K / W 1 + 3.51× 10−5 m 2 ⋅ K / W + + 0.91 ( 0.143) 0.91× 1 8 3 W / m 2 ⋅ K

)

(

1 = 4.662 ×10−3 + 6.993× 10−3 + 3.51 ×10−5 +5.495 ×10 −4 +6.005 ×10−3 m 2 ⋅ K / W Uh Uh = 65.4 W / m 2 ⋅ K.

< 2

COMMENTS: For the unfouled condition, we found Uh = 93.4 W/m ⋅K from Example 11.6. Note that the thermal resistance of the tube-fin material is negligible and that fouling has a significant effect, reducing Uh by 41%.

PROBLEM 11.80 KNOWN: Compact heat exchanger with prescribed core geometry and operating parameters. FIND: Required heat exchanger volume; number of tubes in the longitudinal and transverse directions, NL and NT ; required tube length. SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible KE and PE changes, (3) Single pass operation, (4) Gas properties are those of air. 3

PROPERTIES: Table A-6, Water ( Tc = 325 K): ρ = 987.2 kg/m , cp = 4182 J/kg⋅K; Table A-4, Air (Assume Th,o ≈ 400 K, Th ≈ 550 K, 1 atm): cp = 1040 J/kg⋅K. ANALYSIS: To find the Hxer volume, first find Ah using the ε-NTU method. By definition,

V =A h /α

A h = NTU ⋅ Cmin / U h.

and

(1,2)

Find the capacity rates, q, qmax and ε:

& c c p,c = 2 k g / s × 4182 J / k g ⋅ K = 8364 W / K Cc = m & h c p,h = 1.25 kg/s ×1040 J / k g ⋅ K = 1300 W / K ← Cmin Ch = m Hence,

Cr =

Cmin = 0.155. Cmax

It follows that

ε=

q q max

=

(

Cc Tc,o − Tc,i

(

)

Cmin Th,i − Tc,i

)

=

8364 W / K ( 350 − 300 ) K = 0.804. 1300 W / K ( 700 − 300) K

With ε = 0.804 and Cr = 0.155, find NTU ≈ 1.7 from Fig. 11.18 for a single-pass, cross flow Hxer with both fluids unmixed. Using Eqs. (2) and (1), find

A h = 1.7 ×1300 W / K / 9 3 . 4 W / m 2 ⋅ K = 23.7m2 V = 23.7m 2 /269m 2 / m 3 = 0.0880m 3. Continued …..

PROBLEM 11.80 (Cont.) To determine the number of tubes in the longitudinal direction, consider the tubular arrangement in the sketch. The Hxer volume can be written as

V = A fr × l L

(3)

l L = ( NL −1) l + Df

(4)

where

and NL is the number of tubes in the longitudinal direction. Combining Eqs. (3) and (4) and substituting numerical values, find

NL = ( V / Afr − Df ) / l + 1

(5)

where Df is the overall diameter of the finned tube, and NL = 0.0880m3 /0.20 m 2 − 0.0285m /0.0343 + 1 = 13.0 ≈ 13.

(

)

<

&c To determine the number of tubes in the transverse direction, compare the overall water flow rate m & t. That is, with that for a single tube, m

& t = ρcA t Vi m

(

)

(6)

where At is the tube inner cross-sectional area π Di2 / 4 and Vi the internal velocity. Hence,

& c /m & t = ( 2 k g / s ) /987.2 k g / m 3 × N=m

π ( 0.0138m) 2 × 0.100 m / s = 135.4 ≈ 135. 4

The total number of tubes required, N, is 135; the number in the transverse direction is

<

NT = N / NL = 135/13 = 10.4 ≈ 11.

To determine the water tube length, recognize that the total area (Ah), less that of the finned surfaces (Af), will be that of the water tube surface area. That is,

A h − A f = π D ol T ⋅ N.

From specification of the core geometry, we know Af/Ah = 0.830; solve for l T to obtain

l T = Ah (1 −A f / A h ) / π Do ⋅ N

(7)

l T = 23.7m 2 (1 − 0.830 ) / π ( 0.0164m ) ×135 = 0.58m.

<

COMMENTS: In summary we find that Total number of tubes, N (NT × NL) Tubes in longitudinal direction, NL Tubes in transverse direction, NT

143 13 11

2

with a total surface area of 27.3 m . The length of the exchanger is Length in longitudinal direction, l L Length in transverse direction, l T

0.44 m 0.58 m.

PROBLEM 11.81 KNOWN: Compact heat exchanger geometry, gas-side flow rate and inlet temperature, water-side convection coefficient, water flow rate, and water inlet and outlet temperatures. FIND: Gas-side overall heat transfer coefficient. Required heat exchanger volume. SCHEMATIC:

ASSUMPTIONS: (1) Gas has properties of atmospheric air at an assumed mean temperature of 700 K, (2) Negligible fouling, (3) Negligible heat exchange with the surroundings and negligible kinetic and potential energy and flow work changes. PROPERTIES: Table A-1, aluminum (T ≈ 300 K): k = 237 W/m⋅K. Table A-4, air (p = 1 atm, T = -7 2 700 K): cp = 1075 J/kg⋅K, µ = 338.8 × 10 N⋅s/m , Pr = 0.695. Table A-6, water ( T = 330 K): cp = 4184 J/kg⋅K. ANALYSIS: For the prescribed heat exchanger core,

1 1 1 = + Ah R w + ηo,h h h U h h c ( Ac / A h ) where

A c Di  Af ,h  8.2 ≈ (1 − 0.913) = 0.070 1 − = A h Do  A h  10.2 The product of Ah and the wall conduction resistance is

Ah R w =

ln ( Do / Di ) 2π kL / A h

=

Di ln ( Do / Di ) 2 k ( Ac / A h )

=

0.0082m × ln (10.2 / 8.2 )

2 × 237 W / m ⋅ K ( 0.070 )

= 5.39 × 10−4 m 2 ⋅ K / W 2

2

 h / σ A fr = 1.25 kg/s/0.534 × 0.20 m = 11.7 kg/s⋅m , With a gas-side mass velocity of G = m Re =

G D h 11.7 kg / s ⋅ m 2 × 0.00363m = = 1254 µ 338.8 × 10−7 N ⋅ s / m 2

and Fig. 11.21 yields jH ≈ 0.0096. Hence,

hh ≈

0.0096 G c p Pr 2 / 3

=

)

(

0.0096 11.7 kg / s ⋅ m 2 (1075 J / kg ⋅ K )

(0.695 )

2/3

= 154 W / m 2 ⋅ K

Continued …..

PROBLEM 11.81 (Cont.) With r2c = r2 + t/2 = 15.8 mm + 0.330 mm/2 = 15.97 mm, r2c/r1 = 15.97/5.1 = 3.13, L = r2 – r1 = 10.7 -6 2 1/2 2 mm, Lc = L + t/2 = 10.87 mm = 0.0109m, Ap = Lct = 3.59 × 10 m , and L3/ c (hh/kAp) = 0.484, Fig. 3.19 yields ηf ≈ 0.77. Hence,

A ηo,h = 1 − f (1 − ηf ) = 1 − 0.913 (1 − 0.77 ) = 0.790 A −1

(

2

U h = 1500 W / m ⋅ K × 0.07

)

−1

+ 5.39 × 10

−4

2

(

2

m ⋅ K / W + 0.79 × 154 W / m ⋅ K

)

−1

2

= 0.0183 m ⋅ K / W

U h = 54.7 W / m 2 ⋅ K

< 5

With q = Cc (Tc,o – Tc,i) = 4184 W/K × 80 K = 3.35 × 10 W, qmax = Cmin (Th,i – Tc,i) = 1344 W/K × 5 535 K = 7.19 × 10 W, ε = 0.466 and Cr = 0.321. From Figure 11.18, we then obtain NTU ≈ 0.65. The required gas-side surface area is then

Ah =

NTU × Cmin 0.65 × 1344 W / K = = 16.0 m 2 2 Uh 54.7 W / m ⋅ K 2

3

With α = 587 m /m , the required volume is

V=

Ah 16 m 2 = = 0.0273m3 2 3 α 587 m / m

<

COMMENTS: (1) Although Uh is small and Ah larger for the continuous fins than for the circular fins of Example 11.6, the much larger value of α, renders the volume requirement smaller. (2) The heat exchanger length is L = V/Afr = 0.137 m, and the number of tube rows is

NL ≈

L + 1 = 7.23 ≈ 7. SL

(3) The hypothetical fin radius (r2 = 15.8 mm) was estimated to be the arithmetic mean of one-half the center-to-center spacing between one tube and its six neighbors.

PROBLEM 11.82 KNOWN: Cooling coil geometry. Air flow rate and inlet and outlet temperatures. Freon pressure and convection coefficient. FIND: Required number of tube rows. SCHEMATIC:

ASSUMPTIONS: (1) Negligible fouling, (2) Constant properties, (3) Negligible heat loss to surroundings. -7

2

PROPERTIES: Table A-4, Air ( Th = 300 K, 1 atm): cp = 1007 J/kg⋅K, µ = 184.6 × 10 N⋅s/m , k = 0.0263 W/m⋅K, Pr = 0.707; Table A-5, Sat. R-12 (1 atm): Tsat = Tc = 243 K, hfg = 165 kJ/kg. ANALYSIS: The required number of tube rows is

NL = ( L − Df ) /SL + 1

where

L = V / Afr

A h = NTU ( Cmin / U h )

V =A h / α

1 / Uh = 1 / hc ( Ac / Ah ) + Ah Rw + 1/ ηo,h h h . -5

2

From Ex. 11.6, (Ac/Ah) = 0.143 and AhRw = 3.51 × 10 m ⋅K/W. With

G=

&h m 1.50 k g / s = = 20.9 k g / s ⋅ m 2 2 σ Afr 0.449 × 0.16m

Re =

GD h 20.9 kg/s ⋅ m 2 × 6.68 ×10 −3 m = = 7563 µ 184.6 × 10−7 N ⋅ s / m 2

and Fig. 11.20 gives jH ≈ 0.0068. Hence,

h h = jh

Gcp Pr

2/3

= 0.0068

20.9 k g / s ⋅ m2 × 1007 J / k g ⋅ K

( 0.707 )

2/3

= 180 W / m 2 ⋅ K.

(

)

1/2 -6 2 With Lc = 6.18 mm and Ap = 1.57 × 10 m from Ex. 11.6, L3c / 2 hh /kA p = 0.338 and, from

Fig. 3.19, ηf ≈ 0.89 for r2c/r1 = 1.75. Hence, as in Ex. 11.6, ηo,h = 0.91 and

(

)

(

1 / Uh = 1/ 5000 W / m 2 ⋅ K 0.143 + 3.51×10− 5 m 2 ⋅ K / W + 1/ 0.91 ×180 W / m 2 ⋅ K Uh = 133 W / m 2 ⋅ K. Continued …..

)

PROBLEM 11.82 (Cont.) & h cp,h = 1511 W/K, With Cmin/Cmax = 0 and Cmin = m

ε=

q qmax

=

( ) = 20 K = 0.30 Ch ( Th,i − Tc,i ) 67 K

Ch Th,i − Th,o

NTU = − ln (1 − ε ) = 0.355 and

C 1511W/K A h = NTU min = 0.355 = 4.03m 2 . 2 Uh 133 W / m ⋅ K Hence,

Ah 4.03m2 L= = = 0.0937m α Afr 269 m 2 / m 3 0.16m2

(

)

and

NL =

L − Df 0.0652 + 1= + 1 = 2.9. SL 0.0343m

Hence, three or more rows must be used. COMMENTS: For the prescribed operating conditions, the heat rate would be

(

)

q = Ch Th,i − Th,o = 1 5 1 1 W / K ( 20 K ) = 30,220 W. If R-12 enters the tubes as saturated liquid, a flow rate of at least

&c = m

q 30,220 W = = 0.183 kg/s h fg 165,000 J / k g

would be needed to maintain saturated conditions in the tubes.

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PROBLEM 11.83 KNOWN: Cooling coil geometry. Air flow rate and inlet temperature. Freon pressure and convection coefficient. FIND: Air outlet temperature. SCHEMATIC:

ASSUMPTIONS: (1) Negligible fouling, (2) Constant properties, (3) Negligible heat loss to surroundings. -7

2

PROPERTIES: Table A-4, Air ( Th ≈ 300 K, 1 atm): cp = 1007 J/kg⋅K, µ = 184.6 × 10 N⋅s/m , k = 0.0263 W/m⋅K, Pr = 0.707; Table A-5, Sat. R-12 (1 atm): Tsat = Tc = 243 K, hfg = 165 kJ/kg. ANALYSIS: To obtain the air outlet temperature, we must first obtain the heat rate from the ε-NTU method. To find Ah, first find the heat exchanger length,

L ≈ ( NL −1) SL + D f = 3 ( 0.0343m ) + 0.0285m = 0.131m.

Hence,

V = Afr L = 0.16m 2 ( 0.131m ) = 0.021m3

(

)

A h = α V = 269m 2 / m 3 0.021m 3 = 5.65m 2 . The overall coefficient is

1 1 1 = + Ah R w + Uh hc ( Ac / Ah ) ηo,h h h -5

2

where Ex. 11.6 yields (Ac/Ah) = 0.143 and AhRw = 3.51 × 10 m ⋅K/W. With

G=

&h m 1.50 k g / s = = 20.9 kg/s ⋅ m 2 σ Afr 0.449 × 0.16 m 2

Re =

GD h 20.9 kg/s ⋅ m 2 × 6.68 ×10 −3 m = = 7563. µ 184.6 × 10−7 N ⋅ s / m 2

Fig. 11.20 gives jH ≈ 0.0068. Hence,

h h = jh

Gc p Pr 2 / 3

= 0.0068

20.9 kg/s ⋅ m 2 ×1007 J / k g ⋅ K

( 0.707 ) 2 / 3

h h = 180 W / m 2 ⋅ K. Continued …..

PROBLEM 11.83 (Cont.)

(

)

1/2 -6 2 With Lc = 6.18 mm and Ap = 1.57 × 10 m from Ex. 11.6, L3c / 2 hh /kA p = 0.338 and, from

Fig. 3.19, ηf ≈ 0.89 for r2c/r1 = 1.75. Hence, as in Ex. 11.6, ηo,h = 0.91 and

1 1 1 = + 3.51 ×10−5 m 2 ⋅ K / W + Uh 5000 W / m 2 ⋅ K 0.143 0.91 180 W / m 2 ⋅ K

(

)

(

)

Uh = 133 W / m 2 ⋅ K. With

& h cp,h = 1.5 kg/s (1007 J / k g ⋅ K ) = 1 5 1 1 W / K Cmin = C h = m NTU =

Uh Ah 1 3 3 W / m 2 ⋅K × 5.65m 2 = = 0.497. Cmin 1511W/K

With Cmin/Cmax = 0, Eq. 11.36a yields

ε = 1 − exp ( − NTU ) = 1 − exp ( −0.497 ) = 0.392. Hence,

(

)

q = ε q max = ε Cmin Th,i − Tc,i = 0.392 (1 5 1 1 W / K ) 67 K q = 39,685 W. The air outlet temperature is

Th,o = Th,i −

q 39,685 W = 310 K − = 283.7 K. Ch 1511W/K

COMMENTS: If R-12 enters the tubes as saturated liquid, a flow rate of at least

&c = m

q 39,685 W = = 0.241kg/s h fg 165,000 J / k g

would be needed to maintain saturated conditions in the tubes.

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PROBLEM 11.84 KNOWN: Cooling coil geometry. Gas flow rate and inlet temperature. Water pressure, flow rate and convection coefficient. FIND: Required number of tube rows. SCHEMATIC:

ASSUMPTIONS: (1) Negligible fouling, (2) Constant properties, (3) Negligible heat loss to surroundings. -7

PROPERTIES: Table A-4, Air ( Th ≈ 725 K, 1 atm): cp = 1081 J/kg⋅K, µ = 346.7 × 10 2

N⋅s/m , k = 0.0536 W/m⋅K, Pr = 0.698; Table A-6, Sat. water (2.455 bar): Tsat = Tc = 400 K, hfg = 2183 kJ/kg. ANALYSIS: The required number of tube rows is

NL =

L − Df +1 SL

where

L=

V Afr

V=

Ah α

C A h = NTU min Uh

1 1 1 = + Ah R w + . Uh hc ( Ac / Ah ) ηo,h hh From Ex. 11.6, (Ac/Ah) ≈ 0.143 and

A hR w =

Di ln ( D o / D i ) ( 0.0138m ) ln (16.4/13.8 ) = = 5.55 ×10 −4 m 2 ⋅ K/W. 2k ( Ac / A h ) 2 (15 W / m ⋅ K)( 0.143)

With

G=

&h m 3.0 k g / s = = 18.6 k g / s ⋅ m 2 σ Afr 0.449 × 0.36m2

Re =

GDh 18.6 k g / s ⋅ m2 × 6.68× 10−3 m = = 3576 µ 346.7 × 10−7 N ⋅ s / m 2

and Fig. 11.20 gives jh ≈ 0.009. Hence,

h h = jh

Gc p Pr

2/3

= 0.009

18.6 kg/s ⋅ m 2 ×1081J/kg ⋅ K

( 0.698)

2/3

= 230 W / m 2 ⋅ K.

Continued …..

PROBLEM 11.84 (Cont.)

(

)

1/2 -6 2 With r2c/r1 = 1.75, Lc = 6.18 mm and Ap = 1.57 × 10 m from Ex. 11.6, L3c / 2 hh /kA p = 1.52

and Fig. 3.19 gives ηf ≈ 0.40. Hence,

ηo,h = 1 −

Af (1 − ηf ) = 1 − 0.83 (1 − 0.4) = 0.50. A

Hence,

1 1 1 = + 5.55 ×10−4 m 2 ⋅ K / W + Uh 104 W / m2 ⋅ K 0.143 0.50 230 W / m2 ⋅ K

(

)

(

)

Uh = 100.5 W / m 2 ⋅ K. With

(

)

q=m & c hfg = 0.5 kg/s 2.183 ×106 J / k g = 1.092 ×106 W Cmin = Ch = 3.0 kg/s (1081J/kg ⋅ K ) = 3243 W / K

(

)

q max = Cmin Th,i − Tc,i = 3 2 4 3 W / K (500 K ) = 1.622 ×106 W find

ε=

q q max

=

1.092 × 106 W 1.622 × 106 W

= 0.674.

From Eq. 11.36b

NTU = − ln (1 − ε ) = − ln (1 − 0.674 ) = 1.121. Hence,

C 3243 W / K A h = NTU min = 1.121 = 36.17m 2 2 Uh 100.5 W / m ⋅ K L=

Ah 36.17m 2 = = 0.373m Afrα 0.36m 2 269m 2 / m 3

NL =

(

)

L − Df 373 − 28.5 + 1= + 1 = 11.06 ≈ 11. SL 34.3

COMMENTS: The gas outlet temperature is

Th,o = Th,i −

q Cmin

= 900 K −

1.092 ×106 W = 564 K. 3243 W / K

Hence Th = (900 K + 564 K)/2 = 732 K is in good agreement with the assumed value.

<

PROBLEM 11.85 KNOWN: Cooling coil geometry. Gas flow rate and inlet temperature. Water pressure and convection coefficient. FIND: Gas outlet temperature. SCHEMATIC:

ASSUMPTIONS: (1) Negligible fouling, (2) Constant properties, (3) Negligible heat loss to surroundings. -7

2

PROPERTIES: Table A-4, Air ( Th ≈ 725 K, 1 atm): cp = 1081 J/kg⋅K, µ = 346.7 × 10 N⋅s/m , k = 0.0536 W/m⋅K, Pr = 0.698; Table A-6, Sat. water (2.455 bar): Tsat = Tc = 400 K, hfg = 2183 kJ/kg. ANALYSIS: To obtain Th,o, first obtain q from the ε-NTU method. To determine NTU, Ah must be found from knowledge of L.

L ≈ ( NL −1) SL + D f = 10 ( 0.0343m ) + 0.0285m = 0.372m. Hence,

V = Afr L = 0.36m 2 ( 0.372m ) = 0.134m3

(

)

A h = α V = 269m 2 / m 3 0.134m 3 = 36.05m 2 . The overall coefficient is

1 1 1 = + Ah R w + . Uh hc ( Ac / Ah ) ηo,h hh From Ex. 11.6, (Ac/Ah) ≈ 0.143 and

A hR w =

Di ln ( D o / D i ) ( 0.0138m ) ln (16.4/13.8 ) = = 5.55 ×10 −4 m 2 ⋅ K/W. 2k ( Ac / A h ) 2 (15 W / m ⋅ K)( 0.143)

With

G=

&h m 3.0 k g / s = = 18.6 k g / s ⋅ m 2 2 σ Afr 0.449 × 0.36m

Re =

GDh 18.6 k g / s ⋅ m2 × 6.68× 10−3 m = = 3576 µ 346.7 × 10−7 N ⋅ s / m 2

and Fig. 11.20 gives jH ≈ 0.009. Hence, Continued …..

PROBLEM 11.85 (Cont.) h h = jh

Gcp Pr 2 / 3

18.6 k g / s ⋅ m 2 × 1081J/kg ⋅ K

= 0.009

( 0.698 )2 / 3

h h = 230 W / m 2 ⋅ K.

(

)

1/2 -6 2 With r2c/r1 = 1.75, Lc = 6.18 mm and Ap = 1.57 × 10 m from Ex. 11.6, L3c / 2 hh /kA p = 1.52

and Fig. 3.19 gives ηf ≈ 0.40. Hence,

ηo,h = 1 −

Af (1 − ηf ) = 1 − 0.83 (1 − 0.4) = 0.50. A

Hence,

1 1 1 = + 5.55 × 10−4 m 2 ⋅ K / W + Uh 104 W / m2 ⋅ K 0.143 0.50 230 W / m 2 ⋅ K

(

)

(

)

Uh = 100.5 W / m 2 ⋅ K. With

Cmin = Ch = 3 k g / s (1081J/kg ⋅ K ) = 3243 W / K

(

)

2 2 Uh Ah 100.5 W / m ⋅ K 36.05m NTU = = = 1.117. Cmin 3243 W / K Since Cmin/Cmax = 0, Eq. 11.36a gives

ε = 1 − exp ( − NTU ) = 1 − exp ( −1.117 ) = 0.673. Hence,

(

)

q = ε Cmin Th,i − Tc,i = 0.673 (3243 W / K )( 500 K ) = 1.091×10 6 W and

1.091× 106 W Th,o = Th,i − = 900 K − = 564 K. Cmin 3243 W / K q

COMMENTS: (1) The assumption of Th = 725 K is good. (2) If water enters the tubes as saturated liquid, a flow rate of at least

&c = m

q 1.091× 106 W = = 0.50 kg/s h fg 2.183 ×10 6 J/kg

would be need to maintain saturated conditions in the tubes.

<

PROBLEM 12.1 KNOWN: Rate at which radiation is intercepted by each of three surfaces (see (Example 12.1). 2

FIND: Irradiation, G[W/m ], at each of the three surfaces. SCHEMATIC:

ANALYSIS: The irradiation at a surface is the rate at which radiation is incident on a surface per unit area of the surface. The irradiation at surface j due to emission from surface 1 is

Gj =

q1− j Aj

. -3

2

With A1 = A2 = A3 = A4 = 10 m and the incident radiation rates q1-j from the results of Example 12.1, find 12.1 ×10−3 W G2 = = 1 2 . 1 W / m2 −3 2

10

G3 =

G4 =

<

m

28.0 ×10−3 W 10−3 m2

19.8 ×10 −3 W 10 −3 m 2

= 28.0 W / m 2

<

= 19.8 W / m 2 .

<

COMMENTS: The irradiation could also be computed from Eq. 12.15, which, for the present situation, takes the form

G j = I1 cosθ j ω1 − j 2

where I1 = I = 7000 W/m ⋅sr and ω1-j is the solid angle subtended by surface 1 with respect to j. For example,

G 2 = I1 cosθ 2 ω1 −2 G 2 = 7000 W / m 2 ⋅ sr × cos 30°

10−3 m2 × cos60 °

( 0.5m )2

G 2 = 12.1W/m 2 . Note that, since A1 is a diffuse radiator, the intensity I is independent of direction.

PROBLEM 12.2 KNOWN: A diffuse surface of area A1 = 10-4m2 emits diffusely with total emissive power E = 5 × 104 W/m2 . FIND: (a) Rate this emission is intercepted by small surface of area A2 = 5 × 10-4 m2 at a prescribed location and orientation, (b) Irradiation G2 on A2, and (c) Compute and plot G2 as a function of the separation distance r2 for the range 0.25 ≤ r2 ≤ 1.0 m for zenith angles θ2 = 0, 30 and 60°. SCHEMATIC:

ASSUMPTIONS: (1) Surface A1 emits diffusely, (2) A1 may be approximated as a differential surface area and that A 2 r22 D 2 . assuming that re-S S COMMENTS: Can you verify that the direct solar intensity, Idir, is a reasonable value, assuming that 4 the solar disk emits as a black body at 5800 K? I b,S = σTS4 / π = σ 5800 K / π

9



1

6

= 2.04 × 107 W / m2 ⋅ sr . Because of local cloud formations, it is possible to have an appreciable diffuse component. But it is not likely to have such a high direct component as given in the problem statement.

PROBLEM 12.8 KNOWN: Directional distribution of solar radiation intensity incident at earth’s surface on an overcast day. FIND: Solar irradiation at earth’s surface. SCHEMATIC:

ASSUMPTIONS: (1) Intensity is independent of azimuthal angle θ. ANALYSIS: Applying Eq. 12.17 to the total intensity 2π π / 2 Ii (θ ) cosθ sinθ dθ d φ 0 0

G=∫

∫

π /2 cos 2 θ sinθ d θ 0

G = 2 π In ∫

 1  π /2 G = ( 2 π sr ) × 8 0 W / m 2 ⋅ sr  − cos3 θ   3  0 π G = −167.6W/m 2 ⋅ sr  cos3 − cos3 0  2   G = 167.6W/m 2.

<

PROBLEM 12.9 KNOWN: Emissive power of a diffuse surface. FIND: Fraction of emissive power that leaves surface in the directions π/4 ≤ θ ≤ π/2 and 0 ≤ φ ≤ π. SCHEMATIC:

ASSUMPTIONS: (1) Diffuse emitting surface. ANALYSIS: According to Eq. 12.12, the total, hemispherical emissive power is ∞ 2π π / 2 I λ ,e ( λ, θ , φ ) cos θ sin θ d θ d φ dλ. 0 0 0

E=∫

∫ ∫

For a diffuse surface Iλ,e (λ, θ, φ) is independent of direction, and as given by Eq. 12.14,

E = π I e. The emissive power, which has directions prescribed by the limits on θ and φ, is ∞ π π /2 I ( λ ) dλ  dφ   cos θ sin θ d θ  0 λ ,e 0 / 4 π   

∆E = ∫

∫

∫

π /2 2 π  sin θ  1  ∆E = I e [φ ]0 ×   = I e [π ]  1 − 0.707 2  2   2 π / 4

(

)

∆E = 0.25 π I e . It follows that

∆E 0.25 π Ie = = 0.25. E π Ie COMMENTS: The diffuse surface is an important concept in radiation heat transfer, and the directional independence of the intensity should be noted.

<

PROBLEM 12.10 KNOWN: Spectral distribution of Eλ for a diffuse surface. FIND: (a) Total emissive power E, (b) Total intensity associated with directions θ = 0o and θ = 30o, and (c) Fraction of emissive power leaving the surface in directions π/4 ≤ θ ≤ π/2. SCHEMATIC:

ASSUMPTIONS: (1) Diffuse emission. ANALYSIS: (a) From Eq. 12.11 it follows that ∞

5

10

15

∞

20

E = ∫ E λ (λ ) dλ = ∫ (0) dλ + ∫ (100) dλ + ∫ (200) dλ + ∫ (100) dλ + ∫ (0) dλ 0 0 5 10 15 20 E = 100 W/m2 ⋅µm (10 − 5) µm + 200W/m2 ⋅µm (15 − 10) µm + 100 W/m2 ⋅µm (20−15) µm

<

E = 2000 W/m2 (b) For a diffuse emitter, Ie is independent of θ and Eq. 12.14 gives Ie =

E

π

=

2000 W m 2

π sr

<

Ie = 637 W m 2⋅ sr (c) Since the surface is diffuse, use Eqs. 12.10 and 12.14, E(π 4 → π 2) E E(π 4 → π 2) E E(π 4 → π 2) E

2π

π /2

∫ ∫π / 4 = 0

Ie cos θ sin θ dθ dφ

π Ie

π /2

2π

∫ cosθ sin θ dθ ∫0 = π /4 π

=

dφ

π /2

=

1  sin 2 θ 

 π 

2

2π

φ 0  π / 4

1 1 2  (1 − 0.707 2 )(2π − 0)  = 0.50 π  2 

COMMENTS: (1) Note how a spectral integration may be performed in parts. (2) In performing the integration of part (c), recognize the significance of the diffuse emission assumption for which the intensity is uniform in all directions.

<

PROBLEM 12.11 KNOWN: Diffuse surface ∆Ao, 5-mm square, with total emissive power Eo = 4000 W/m2. FIND: (a) Rate at which radiant energy is emitted by ∆Ao, qemit; (b) Intensity Io,e of the radiation field emitted from the surface ∆Ao; (c) Expression for qemit presuming knowledge of the intensity Io,e beginning with Eq. 12.10; (d) Rate at which radiant energy is incident on the hemispherical surface, r = R1 = 0.5 m, due to emission from ∆Ao; (e) Rate at which radiant energy leaving ∆Ao is intercepted by the small area ∆A2 located in the direction (40o, φ) on the hemispherical surface using Eq. 12.5; also determine the irradiation on ∆A2; (f) Repeat part (e), for the location (0o, φ); are the irradiations at the two locations equal? and (g) Irradiation G1 on the hemispherical surface at r = R1 using Eq. 12.5. SCHEMATIC:

ASSUMPTIONS: (1) Diffuse surface, ∆Ao, (2) Medium above ∆Ao is also non-participating, (3) R12 >> ∆A o , ∆A2. ANALYSIS: (a) The radiant power leaving ∆Ao by emission is

<

qemit = Eo⋅∆Ao = 4000 W/m2 (0.005 m × 0.005 m) = 0.10 W (b) The emitted intensity is Io,e and is independent of direction since ∆Ao is a diffuser emitter,

<

Io,e = E o π = 1273 W m 2⋅ sr

The intensities at points P1 and P2 are also Io,e and the intensity in the directions shown in the schematic above will remain constant no matter how far the point is from the surface ∆Ao since the space is nonparticipating. (c) From knowledge of Io,e, the radiant power leaving ∆Ao from Eq. 12.10 is, 2π

π /2

q emit = ∫ Io,e ∆Ao cos θ sin θ dθ dφ = Io,e ∆A o ∫ cos θ sin θ dθ dφ = π Io,e ∆A o = 0.10 W h φ = 0 ∫θ = 0

<

(d) Defining control surfaces above ∆Ao and on A1, the radiant power leaving ∆Ao must pass through A1. That is,

<

q1,inc = E o ∆A o = 0.10 W

Recognize that the average irradiation on the hemisphere, A1, where A1 = 2π R12 , based upon the definition, Section 12.2.3, G1 = q1,inc A1 = E o ∆ A o 2π R12 = 63.7 mW m 2 where q1,inc is the radiant power incident on surface A1. Continued...

PROBLEM 12.11 (Cont.) (e) The radiant power leaving ∆Ao intercepted by ∆A2, where ∆A2 = 4×10-6 m2, located at (θ = 45o, φ) as per the schematic, follows from Eq. 12.5, q ∆A →∆A = Io,e ∆Ao cos θ o ∆ω 2 − o o 2 where θo = 45o and the solid angle ∆A2 subtends with respect to ∆Ao is ∆ω 2 − o = ∆A 2 cos θ 2 R12 = 4 × 10−6 m 2 ⋅ 1 (0.5m)2 = 1.60 × 10−5 sr where θ2 = 0o, the direction normal to ∆A2, q ∆A →∆A = 1273 W m 2 ⋅ sr × 25 × 10−6 m 2 cos 45o × 1.60 × 10−5 sr = 3.60 × 10−7 W o 2

<

From the definition of irradiation, Section 12.2.3, G 2 = q ∆A →∆A ∆A 2 = 90 mW m 2 o 2 (f) With ∆A2, located at (θ = 0o, φ), where cosθo = 1, cosθ2 = 1, find ∆ω 2 − o = 1.60 × 10−5 sr

q ∆A →∆A = 5.09 × 10−7 W o 2

G 2 = 127 mW m 2

<

Note that the irradiation on ∆A2 when it is located at (0o, φ) is larger than when ∆A2 is located at (45o, φ); that is, 127 mW/m2 > 90 W/m2. Is this intuitively satisfying? (g) Using Eq. 12.15, based upon Figure 12.10, find G1 = ∫ I1,i dA1 ⋅ dω 0 −1 A1 = π Io,e ∆Ao ∆A1 = 63.7 mW m 2 h

<

where the elemental area on the hemispherical surface A1 and the solid angle ∆Ao subtends with respect to ∆A1 are, respectively, dA1 = R12 sin θ dθ dφ

dω o −1 = ∆A o cos θ R12

From this calculation you found that the average irradiation on the hemisphere surface, r = R1, is G1 = 63.7 mW m 2 . From parts (e) and (f), you found irradiations, G2 on ∆A2 at (0o, φ) and (45o, φ) as 127 mW/m2 and 90 mW/m2, respectively. Did you expect G1 to be less than either value for G2? How do you explain this? COMMENTS: (1) Note that from Parts (e) and (f) that the irradiation on A1 is not uniform. Parts (d) and (g) give an average value. (2) What conclusions would you reach regarding G1 if ∆Ao were a sphere?

PROBLEM 12.12 KNOWN: Hemispherical and spherical arrangements for radiant heat treatment of a thin-film material. Heater emits diffusely with intensity Ie,h = 169,000 W/ m2⋅sr and has an area 0.0052 m2. FIND: (a) Expressions for the irradiation on the film as a function of the zenith angle, θ, and (b) Identify arrangement which provides the more uniform irradiation, and hence better quality control for the treatment process. SCHEMATIC:

ASSUMPTIONS: (1) Heater emits diffusely, (2) All radiation leaving the heater is absorbed by the thin film. ANALYSIS: (a) The irradiation on any differential area, dAs, due to emission from the heater, Ah , follows from its definition, Section 12.2.3,

q G = h →s dAs

(1)

Where qh→s is the radiant heat rate leaving Ah and intercepted by dAs. From Eq. 12.5,

q h →s = Ie,h ⋅ dA h cos θ h ⋅ ωs − h

(2)

where ωs-h is the solid angle dAs subtends with respect to any point on Ah. From the definition, Eq. 12.2,

ω=

dA n

(3)

r2

where dAn is normal to the viewing direction and r is the separation distance. For the hemisphere: Referring to the schematic above, the solid angle is

ωs − h =

dAs R2

and the irradiation distribution on the hemispheric surface as a function of θh is G = Ie,h A h cosθ h R 2

(1)

<

For the sphere: From the schematic, the solid angle is

ωs,h =

dAs cos θs R o2

=

dAs 4R 2 cos θ h

where Ro, from the geometry of sphere cord and radii with θs = θh, is Continued...

PROBLEM 12.12 (Cont.) R o = 2R cos θ h and the irradiation distribution on the spherical surface as a function of θh is G = Ie,h d A h 4R 2

(2)

<

(b) The spherical shape provides more uniform irradiation as can be seen by comparing Eqs. (1) and (2). In fact, for the spherical shape, the irradiation on the thin film is uniform and therefore provides for better quality control for the treatment process. Substituting numerical values, the irradiations are:

G hem = 169, 000 W m 2⋅ sr × 0.0052m 2 cosθ h

(2m )2 = 219.7 cosθ h W

m2

2

Gsph = 169, 000 W m 2⋅ sr × 0.0052 m 2 4 ( 2m ) = 54.9 W m 2

(3) (4)

COMMENTS: (1) The radiant heat rate leaving the diffuse heater surface by emission is

q tot = π Ie,h A h = 276.1W The average irradiation on the spherical surface, Asph = 4πR2,

Gsph = q tot Asph = 276.1W 4π ( 2m ) = 54.9 W m 2 2

while the average irradiation on the hemispherical surface, Ahem = 2πR2 is

G hem = 276.1W 2π ( 2m ) = 109.9 W m 2 2

(2) Note from the foregoing analyses for the sphere that the result for G sph is identical to that found as Eq. (4). That follows since the irradiation is uniform. (3) Note that G hem > G sph since the surface area of the hemisphere is half that of the sphere. Recognize that for the hemisphere thin film arrangement, the distribution of the irradiation is quite variable with a maximum at θ = 0° (top) and half the maximum value at θ = 30°.

PROBLEM 12.13 KNOWN: Hot part, ∆Ap, located a distance x1 from an origin directly beneath a motion sensor at a distance Ld = 1 m. FIND: (a) Location x1 at which sensor signal S1 will be 75% that corresponding to x = 0, directly beneath the sensor, So, and (b) Compute and plot the signal ratio, S/So, as a function of the part position x1 for the range 0.2 ≤ S/So ≤ 1 for Ld = 0.8, 1.0 and 1.2 m; compare the x-location for each value of Ld at which S/So = 0.75. SCHEMATIC:

ASSUMPTIONS: (1) Hot part is diffuse emitter, (2) L2d >> ∆Ap, ∆Ao. ANALYSIS: (a) The sensor signal, S, is proportional to the radiant power leaving ∆Ap and intercepted by ∆Ad, S ~ q p →d = I p,e ∆A p cos θ p ∆ω d − p

(1)

L cos θ p = cos θ d = d = Ld (L2d + x12 )1/ 2 R

(2)

when

∆ω d − p =

∆A d ⋅ cos θ d R

2

= ∆Ad ⋅ Ld (L2d + x12 )3 / 2

(3)

Hence, q p → d = I p,e ∆A p ∆A d

L2d

(4)

(L2d + x12 )2

It follows that, with So occurring when x= 0 and Ld = 1 m,

 L2  d = =  2 2 2 2 2 2 So L (L + 0 )  Ld + x1  d d S

L2d (L2d + x12 ) 2

2

(5)

so that when S/So = 0.75, find,

<

x1 = 0.393 m (b) Using Eq. (5) in the IHT workspace, the signal ratio, S/So, has been computed and plotted as a function of the part position x for selected Ld values.

Continued...

PROBLEM 12.13 (Cont.) 1

Signal ratio, S/So

0.8

0.6

0.4

0.2

0 0

1

2

Part position, x (m) Sensor position, Ld = 0.8 m Ld = 1 m Ld = 1.2 m

When the part is directly under the sensor, x = 0, S/So = 1 for all values of Ld. With increasing x, S/So decreases most rapidly with the smallest Ld. From the IHT model we found the part position x corresponding to S/So = 0.75 as follows. S/So 0.75 0.75 0.75

Ld (m) 0.8 1.0 1.2

x1 (m) 0.315 0.393 0.472

If the sensor system is set so that when S/So reaches 0.75 a process is initiated, the technician can use the above plot and table to determine at what position the part will begin to experience the treatment process.

PROBLEM 12.14 KNOWN: Diameter and temperature of burner. Temperature of ambient air. Burner efficiency. FIND: (a) Radiation and convection heat rates, and wavelength corresponding to maximum spectral emission. Rate of electric energy consumption. (b) Effect of burner temperature on convection and radiation rates. SCHEMATIC:

ASSUMPTIONS: (1) Burner emits as a blackbody, (2) Negligible irradiation of burner from surrounding, (3) Ambient air is quiescent, (4) Constant properties. -6

2

PROPERTIES: Table A-4, air (Tf = 408 K): k = 0.0344 W/m⋅K, ν = 27.4 × 10 m /s, α = 39.7 × -6 2 -1 10 m /s, Pr = 0.70, β = 0.00245 K . ANALYSIS: (a) For emission from a black body

(

)

q rad = As E b = π D 2 / 4 σ T 4 = π ( 0.2m ) / 4  5.67 × 10−8 W / m 2 ⋅ K 4 (523 K ) = 133 W   2

4

3

2

<

-1

With L = As/P = D/4 = 0.05m and RaL = gβ (Ts - T∞) L /αν = 9.8 m/s × 0.00245 K (230 K) 3 -12 4 2 5 (0.05m) /(27.4 × 39.7 × 10 m /s ) = 6.35 × 10 , Eq. (9.30) yields

h=

)

(

k k 4 =  0.0344 W / m ⋅ K  0.54 6.35 × 105 1/ 4 = 10.5 W / m 2 ⋅ K Nu L =   0.54 Ra1/   L L 0.05m L   2 qconv = h As ( Ts − T∞ ) = 19.4 W / m 2 ⋅ K π ( 0.2m ) / 4  230 K = 75.7 W  

<

The electric power requirement is then

q + qconv (133 + 75.7 ) W Pelec = rad = = 232 W η 0.9

<

The wavelength corresponding to peak emission is obtained from Wien’s law, Eq. (12.27)

λmax = 2898µ m ⋅ K / 523K = 5.54 µ m

<

(b) As shown below, and as expected, the radiation rate increases more rapidly with temperature than

(

)

the convection rate due to its stronger temperature dependence Ts4 vs. Ts5 / 4 . Continued …..

PROBLEM 12.14(Cont.) 500

Heat rate (W)

400 300 200 100 0 100

150

200

250

300

350

Surface temperature (C) qconv qrad Pelec

COMMENTS: If the surroundings are treated as a large enclosure with isothermal walls at Tsur = T∞ 2

4 = 293 K, irradiation of the burner would be G = σ Tsur = 418 W/m and the corresponding heat rate would be As G = 13 W. This input is much smaller than the energy outflows due to convection and radiation and is justifiably neglected.

PROBLEM 12.15 KNOWN: Evacuated, aluminum sphere (D = 2m) serving as a radiation test chamber. FIND: Irradiation on a small test object when the inner surface is lined with carbon black and at 600K. What effect will surface coating have? SCHEMATIC:

ASSUMPTIONS: (1) Sphere walls are isothermal, (2) Test surface area is small compared to the enclosure surface. ANALYSIS: It follows from the discussion of Section 13.3 that this isothermal sphere is an enclosure behaving as a blackbody. For such a condition, see Fig. 12.12(c), the irradiation on a small surface within the enclosure is equal to the blackbody emissive power at the temperature of the enclosure. That is

G1 = Eb ( Ts ) = σ Ts4 G1 = 5.67 ×10 −8 W / m 2 ⋅K 4 ( 600K ) = 7348W/m 2 . 4

<

The irradiation is independent of the nature of the enclosure surface coating properties. COMMENTS: (1) The irradiation depends only upon the enclosure surface temperature and is independent of the enclosure surface properties. (2) Note that the test surface area must be small compared to the enclosure surface area. This allows for inter-reflections to occur such that the radiation field, within the enclosure will be uniform (diffuse) or isotropic. (3) The irradiation level would be the same if the enclosure were not evacuated since, in general, air would be a non-participating medium.

PROBLEM 12.16 KNOWN: Isothermal enclosure of surface area, As, and small opening, Ao, through which 70W emerges. FIND: (a) Temperature of the interior enclosure wall if the surface is black, (b) Temperature of the wall surface having ε = 0.15. SCHEMATIC:

ASSUMPTIONS: (1) Enclosure is isothermal, (2) Ao lambda) (%)

80

60

40

20

0 0.1

0.4

0.8

2

6

10

40

80

Wavelength, lambda (mum) Solar spectrum, TbS = 5800 K (left) Blackbody, Tbs = 300 K (right)

The left-hand curve in the plot represents the percentage of solar flux approximated as the 5800 Kblackbody spectrum in the spectral region less than λ. The right-hand curve represents the percentage of 300 K-blackbody flux in the spectral region less than λ. Referring to upper abscissa scale of Figure 12.23, for the solar flux, 75% of the solar flux is at wavelengths shorter than 1 µm. For the blackbody flux (300 K), 75% of the blackbody flux is at wavelengths shorter than 20 µm. These values are in agreement with points on the solar and 300K-blackbody curves, respectively, in the above plot.

PROBLEM 12.24 KNOWN: Thermal imagers operating in the spectral regions 3 to 5 µm and 8 to 14 µm. FIND: (a) Band-emission factors for each of the spectral regions, 3 to 5 µm and 8 to 14 µm, for temperatures of 300 and 900 K, (b) Calculate and plot the band-emission factors for each of the spectral regions for the temperature range 300 to 1000 K; identify the maxima, and draw conclusions concerning the choice of an imager for an application; and (c) Considering imagers operating at the maximum-fraction temperatures found from the graph of part (b), determine the sensitivity (%) required of the radiation detector to provide a noise-equivalent temperature (NET) of 5°C. ASSUMPTIONS: The sensitivity of the imager’s radiation detector within the operating spectral region is uniform. ANALYSIS: (a) From Eqs. 12.30 and 12.31, the band-emission fraction F(λ1 → λ2, T) for blackbody emission in the spectral range λ1 to λ2 for a temperature T is

F( λ1→λ 2, T ) = F( 0→λ 2, T) − F( 0→λ 1, T) Using the IHT Radiation | Band Emission tool (or Table 12.1), evaluate F(0-λT) at appropriate λ⋅T products: 3 to 5 µm region

F( λ1− λ 2, 300 K ) = 0.1375 − 0.00017 = 0.01359

<

F( λ1− λ 2, 900 K ) = 0.5640 − 0.2055 = 0.3585

<

8 to 14µm region

F( λ1− λ 2, 300 K ) = 0.5160 − 0.1403 = 0.3758

<

F( λ1− λ 2, 900 K ) = 0.9511 − 0.8192 = 0.1319

<

(b) Using the IHT Radiation | Band Emission tool, the band-emission fractions for each of the spectral regions is calculated and plotted below as a function of temperature. Band fractions for thermal imaging spectral regions 0.4

Band fraction for range

0.3

0.2

0.1

0 300

400

500

600

700

800

900

1000

Temperature, T (K) 3 to 5 um region 8 to 14 um region

Continued …..

PROBLEM 12.24 (Cont.) For the 3 to 5 µm imager, the band-emission factor increases with increasing temperature. For low temperature applications, not only is the radiant power σ T 4 , T ≈ 300 K low, but the band fraction

(

)

is small. However, for high temperature applications, the imager operating conditions are more favorable with a large band-emission factor, as well as much higher radiant power σ T4 , T → 900 K .

(

)

For the 8 to 14 µm imager, the band-emission factor decreases with increasing temperature. This is a more favorable instrumentation feature, since the band-emission factor (proportionally more power) becomes larger as the radiant power decreases. This imager would be preferred over the 3 to 5 µm imager at lower temperatures since the band-emission factor is 8 to 10 times higher. Recognizing that from Wien’s law, the peaks of the blackbody curves for 300 and 900 K are approximately 10 and 3.3 µm, respectively, it follows that the imagers will receive the most radiant power when the peak target spectral distributions are close to the operating spectral region. It is good application practice to chose an imager having a spectral operating range close to the peak of the blackbody curve (or shorter than, if possible) corresponding to the target temperature. The maxima band fractions for the 3 to 5 µm and 8 to 14 µm spectral regions correspond to temperatures of 960 and 355 K, respectively. Other application factors not considered (like smoke, water vapor, etc), the former imager is better suited with higher temperature scenes, and the latter with lower temperature scenes. (c) Consider the 3 to 5 µm and 8 to 14 µm imagers operating at their band-emission peak temperatures, 355 and 960 K, respectively. The sensitivity S (% units) of the imager to resolve an NET of 5°C can be expressed as

S (%) =

F( λ1−λ 2, T1) − F(λ1−λ 2, T2 ) F( λ1− λ 2, T1)

× 100

where T1 = 355 or 960 K and T2 = 360 or 965 K, respectively. Using this relation in the IHT workspace, find

S3−5 = 0.035%

S8 −14 = 0.023%

<

That is, we require the radiation detector (with its signal-processing system) to resolve the output signal with the foregoing precision in order to indicate a 5°C change in the scene temperature.

PROBLEM 12.25 KNOWN: Tube furnace maintained at Tf = 2000 K used to calibrate a heat flux gage of sensitive 2 area 5 mm mounted coaxial with the furnace centerline, and positioned 60 mm from the opening of the furnace. 2

FIND: (a) Heat flux (kW/m ) on the gage, (b) Radiant flux in the spectral region 0.4 to 2.5 µm, the sensitive spectral region of a solid-state (photoconductive type) heat-flux gage, and (c) Calculate and plot the heat fluxes for each of the gages as a function of the furnace temperature for the range 2000 ≤ Tf ≤ 3000 K. Compare the values for the two types of gages; explain why the solid-state gage will always indicate systematically low values; does the solid-state gage performance improve, or become worse as the source temperature increases? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Graphite tube furnace behaves as a blackbody, (3) Areas of gage and furnace opening are small relative to separation distance squared, and (4) Extension tube is cold relative to the furnace. ANALYSIS: (a) The heat flux to the gage is equal to the irradiation, Gg, on the gage and can be expressed as (see Section 12.2.3)

G g = I f ⋅ cos θ g ⋅ ∆ω f − g where ∆ωf - g is the solid angle that the furnace opening subtends relative to the gage. From Eq. 12.2, with θg = 0°

∆ω f − g ≡

dA n r2

=

A f cos θ g L2

=

1

6

π 0.0125 m 2 / 4 × 1

10.060 m6

2

= 3.409 × 10−2 sr

The intensity of the radiation from the furnace is

1 6

1

6

I f = E b,f Tf / π = σTf4 / π = 5.67 × 10−8 W / m2 ⋅ K4 2000 K 4 / π = 2.888 × 105 W / m2 ⋅ sr Substituting numerical values,

G g = 2.888 × 105 W / m2 ⋅ sr × 1 × 3.409 × 10-2 sr = 9.84 kW / m2

<

(b) The solid-state detector gage, sensitive only in the spectral region λ1 = 0.4 µm to λ2 = 2.5 µm, will receive the band irradiation.

G g, λ1−λ 2 = F(λ1→λ 2, Tf ) ⋅ G g,b =  F(0→λ 2, Tf ) − F( 0→λ1, Tf )  G g,b   Continued …..

PROBLEM 12.25 (Cont.) where for λ1 Tf = 0.4 µm × 2000 K = 800 µm⋅K, F(0 - λ1) = 0.0000 and for λ2 ⋅ Tf = 2.5 µm × 2000 K = 5000 µm⋅K, F(0 - λ2) = 0.6337. Hence,

G g,λ1− λ 2 = 0.6337 − 0.0000 × 9.84 kW / m2 = 6.24 kW / m2

<

(c) Using the foregoing equation in the IHT workspace, the heat fluxes for each of the gage types are calculated and plotted as a function of the furnace temperature.

H e a t flu x , G g (k W /m ^ 2 )

60

H e a t h e a t flu x g a g e ca lib ra tio n s

40

20

0 2000

2200

2400

2600

2800

3000

Fu rn a c e te m p e ra tu re , Tf (K ) B la c k h e a t flu x g a g e S o lid -s ta te g a g e , 0 .4 to 2 .5 u m

For the black gage, the irradiation received by the gage, Gg, increases as the fourth power of the furnace temperature. For the solid-state gage, the irradiation increases slightly greater than the fourth power of the furnace temperature since the band-emission factor for the spectral region, F(λ1 - λ2, Tf), increases with increasing temperature. The solid-state gage will always indicate systematic low readings since its band-emission factor never approaches unity. However, the error will decrease with increasing temperature as a consequence of the corresponding increase in the band-emission factor. COMMENTS: For this furnace-gage geometrical arrangement, evaluating the solid angle, ∆ωf - g, and the areas on a differential basis leads to results that are systematically high by 1%. Using the view factor concept introduced in Chapter 13 and Eq. 13.8, the results for the black and solid-state 2 gages are 9.74 and 6.17 kW/m , respectively.

PROBLEM 12.26 KNOWN: Geometry and temperature of a ring-shaped radiator. Area of irradiated part and distance from radiator. FIND: Rate at which radiant energy is incident on the part. SCHEMATIC:

ASSUMPTIONS: (1) Heater emits as a blackbody. ANALYSIS: Expressing Eq. 12.5 on the basis of the total radiation, dq = Ie dAh cosθ dω, the rate at which radiation is incident on the part is

q h − p = ∫ dq = Ie ∫ ∫ cos θ dω p − h dA h ≈ Ie cos θ ⋅ ω p − h ⋅ A h Since radiation leaving the heater in the direction of the part is oriented normal to the heater surface, θ = 0 and cos θ = 1. The solid angle subtended by the part with respect to the heater is ωp-h = Ap cos θ1/L2, while the area of the heater is Ah ≈ 2πrhW = 2π(L sin θ1)W. Hence, with Ie = Eb/π = σ Th4 π ,

(

)

$ 2 4 5.67 × 10−8 W m 2⋅ K 4 (3000 K ) 0.007 m cos 30 qh −p ≈ × × 2π (1.5 m ) 0.03m π (3m )2

q h − p ≈ 278.4 W

<

COMMENTS: The foregoing representation for the double integral is an excellent approximation since W 1, Wien’s law and (b) C2/λT > 1 and C2/λT > 1 (or λT > 1. Hence, the –1 term in the denominator of the Planck law is insignificant, giving E λ ,b ( λ, T ) ≈ C1 / λ 5 exp ( −C 2 / λT ) .

(

)

<

This approximate relation is known as Wien’s law. The ratio of the emissive power by Wien’s law to that by the Planck law is,

Eλ ,b,Wien Eλ ,b,Planck

=

1/exp ( C2 / λ T)

1/  exp ( C2 / λ T) − 1

.

For the condition λT = λmax T = 2898 µm⋅K, C2/λT =

14388 µm ⋅ K 2898 µ m ⋅ K

= 4.966 and

Eλ ,b Wien 1/exp ( 4.966 ) = = 0.9930. Eλ ,b Planck 1/ exp ( 4.966 ) − 1

<

That is, for λT ≤ 2898 µm⋅K, Wien’s law is a good approximation to the Planck distribution. (b) If C2/λT > C2), the exponential term may be expressed as a series that can be approximated by the first two terms. That is,

ex = 1 + x +

x2 x3 + + ... ≈ 1 + x 2! 3!

when

x tfix?

PROBLEM 12.30 KNOWN: Spectral distribution of emissivity for zirconia and tungsten filaments. Filament temperature. FIND: (a) Total emissivity of zirconia, (b) Total emissivity of tungsten and comparative power requirement, (c) Efficiency of the two filaments. SCHEMATIC:

ASSUMPTIONS: (1) Negligible reflection of radiation from bulb back to filament, (2) Equivalent surface areas for the two filaments, (3) Negligible radiation emission from bulb to filament. ANALYSIS: (a) From Eq. (12.38), the emissivity of the zirconia is ∞

ε = ∫ ε λ ( E λ / E b ) dλ = ε1 F(0→0.4 µ m ) + ε 2 F(0.4→0.7 µ m ) + ε 3 F(0.7 µ m →∞ ) o

) (

(

ε = ε1 F(0→0.4µ m ) + ε 2 F(0→0.7 µ m ) − F(0→0.4µ m ) + ε 3 1 − F(0→0.7 µ m )

)

From Table 12.1, with T = 3000 K

λ T = 0.4µ m × 3000 ≡ 1200µ m ⋅ K :

F(0→0.4 µ m ) = 0.0021

λ T = 0.7 µ m × 3000 K = 2100µ m ⋅ K : F(0→0.7 µ m ) = 0.0838 ε = 0.2 × 0.0021 + 0.8 (0.0838 − 0.0021) + 0.2 × (1 − 0.0838 ) = 0.249

<

(b) For the tungsten filament,

(

ε = ε1 F(0→ 2µ m ) + ε 2 1 − F(0→ 2µ m )

)

With λT = 6000µm⋅K, F(0 → 2µm) = 0.738

ε = 0.45 × 0.738 + 0.1(1 − 0.738) = 0.358

<

Assuming, no reflection of radiation from the bulb back to the filament and with no losses due to ′′ = εσ T 4 . natural convection, the power consumption per unit surface area of filament is Pelec Continued …..

PROBLEM 12.30 (Cont.) Zirconia:

′′ = 0.249 × 5.67 ×10−8 W / m 2 ⋅ K 4 (3000 K ) = 1.14 × 106 W / m 2 Pelec

Tungsten:

′′ = 0.358 × 5.67 × 10−8 W / m 2 ⋅ K 4 (3000 K ) = 1.64 × 106 W / m 2 Pelec

4

4

Hence, for an equivalent surface area and temperature, the tungsten filament has the largest power

<

consumption. (c) Efficiency with respect to the production of visible radiation may be defined as

∫ ε λ Eλ ,b dλ = ∫0.4 ε λ (Eλ ,b / E b ) = ε vis F = 0.4 0.7

ηvis

0.7

E

ε

ε

(0.4→0.7 µ m )

With F(0.4 → 0.7 µm) = 0.0817 for T = 3000 K, Zirconia: Tungsten:

ηvis = (0.8 / 0.249 ) 0.0817 = 0.263

ηvis = (0.45 / 0.358 ) 0.0817 = 0.103

Hence, the zirconia filament is the more efficient. COMMENTS: The production of visible radiation per unit filament surface area is Evis = ηvis ′′ . Hence, Pelec Zirconia:

E vis = 0.263 × 1.14 × 106 W / m 2 = 3.00 ×105 W / m 2

Tungsten:

E vis = 0.103 × 1.64 × 106 W / m 2 = 1.69 × 105 W / m 2

Hence, not only is the zirconia filament more efficient, but it also produces more visible radiation with less power consumption. This problem illustrates the benefits associated with carefully considering spectral surface characteristics in radiative applications.

<

PROBLEM 12.31 KNOWN: Variation of spectral, hemispherical emissivity with wavelength for two materials. FIND: Nature of the variation with temperature of the total, hemispherical emissivity. SCHEMATIC:

ASSUMPTIONS: (1) ε λ is independent of temperature. ANALYSIS: The total, hemispherical emissivity may be obtained from knowledge of the spectral, hemispherical emissivity by using Eq. 12.38 ∞ ε λ ( λ ) Eλ ,b ( λ , T ) dλ E λ ,b ( λ, T ) ∞ 0 ε (T ) = = ελ ( λ ) d λ. 0 E b (T ) E b (T )

∫

∫

We also know that the spectral emissive power of a blackbody becomes more concentrated at lower wavelengths with increasing temperature (Fig. 12.13). That is, the weighting factor, Eλ,b (λ,T)/Eb (T) increases at lower wavelengths and decreases at longer wavelengths with increasing T. Accordingly, Material A:

ε(T) increases with increasing T

Material B:

ε(T) decreases with increasing T.

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PROBLEM 12.32 KNOWN: Metallic surface with prescribed spectral, directional emissivity at 2000 K and 1 µm (see Example 12.6) and additional measurements of the spectral, hemispherical emissivity. FIND: (a) Total hemispherical emissivity, ε, and the emissive power, E, at 2000 K, (b) Effect of temperature on the emissivity. SCHEMATIC:

ANALYSIS: (a) The total, hemispherical emissivity, ε, may be determined from knowledge of the spectral, hemispherical emissivity, ε λ , using Eq. 12.38. 2 µ m E λ ,b (λ , T)dλ 4 µ m E λ ,b (λ , T)dλ ∞ ε λ (λ )E λ ,b (λ , T) dλ E b (T) = ε1 + ε2 0 0 2µ m E b (T) E b (T)

∫

ε (T) = ∫

∫

or from Eqs. 12.28 and 12.30,

ε (T) = ε1F(0→ λ ) + ε 2  F(0→ λ ) − F(0→ λ )  1 2 1 From Table 12.1, λ1 = 2 µ m,

T = 2000 K : λ1T = 4000 µ m ⋅ K,

λ2 = 4 µ m, T = 2000 K : λ2T = 8000 µ m ⋅ K,

F(0 →λ ) = 0.481 1 F(0→λ ) = 0.856 2

Hence,

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ε (T) = 0.36 × 0.481 + 0.20(0.856 − 0.481) = 0.25 From Eqs. 12.28 and 12.37, the total emissive power at 2000 K is E(2000 K) = ε (2000 K) ⋅ Eb (2000 K) E(2000 K) = 0.25 × 5.67 × 10 −8 W m 2⋅ K 4 × (2000 K) 4 = 2.27 × 105 W m 2 .

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(b) Using the Radiation Toolpad of IHT, the following result was generated. 0.4

Emissivity, eps

0.3

0.2

0.1

0 500

1000

1500

2000

2500

3000

Surface temperature, T(K)

Continued...

PROBLEM 12.32 (Cont.) At T ≈ 500 K, most of the radiation is emitted in the far infrared region (λ > 4 µm), in which case ε ≈ 0. With increasing T, emission is shifted to lower wavelengths, causing ε to increase. As T → ∞, ε → 0.36. COMMENTS: Note that the value of ε λ for 0 < λ ≤ 2 µm cannot be read directly from the ε λ distribution provided in the problem statement. This value is calculated from knowledge of ε λ ,θ (θ ) in Example 12.6.

PROBLEM 12.33 KNOWN: Relationship for determining total, hemispherical emissivity, ε, by integration of the spectral emissivity distribution, ελ (Eq. 12.38). FIND: Evaluate ε from ελ for the following cases: (a) Ex. 12.5, use the result to benchmark your code, (b) tungsten at 2800 K, and (c) aluminum oxide at 1400 K. Use the intrinsic function INTEGRAL of IHT as your solution tool. SCHEMATIC:

ASSUMPTIONS: (1) Surfaces are diffuse emitters. ANALYSIS: (a) Using IHT as the solution tool, Eq. 12.38 is entered into the workspace, and a lookup table created to represent the spectral emissivity distribution. See Comment 1 for the IHT annotated code. The result is ε = 0.558, in agreement with the analysis of Ex. 12.5 using the bandemission factors. (b, c) Using the same code as for the benchmarking exercise in Part (a), but with new look-up table files (*.lut) representing the spectral distributions tabulated below, the total hemispherical emissivities for the tungsten at 2800 K and aluminum oxide at 1400 K are:

ε W = 0.31

ε Al2O3 = 0.38

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These results compare favorably with values of 0.29 and 0.41, respectively, from Fig. 12.19. See Comment 2.

Tungsten, 2800 K

Aluminum oxide, 1400 K

λ (µm)

ελ

λ (µm)

ελ

λ (µm)

ελ

λ (µm)

ελ

0.3

0.47

2.0

0.26

0.6

0.19

4.5

0.50

0.4

0.48

4.0

0.17

0.8

0.18

5

0.70

0.5

0.47

6.0

0.05

1.0

0.175

6

0.88

0.6

0.44

8.0

0.03

1.5

0.175

10

0.96

1.0

0.38

10

0.03

2

0.19

12.5

0.9

3

0.29

15

0.53

4

0.4

20

0.39

Continued …..

PROBLEM 12.33 (Cont.) COMMENTS: (1) The IHT code to obtain ε from ελ for the case of Ex. 12.5 spectral distribution is shown below. // Benchmarking use of INTEGRAL and LOOKUPVAL functions // Calculating total emissivity from spectral distribution /* Results: integration from 0.05 to 15 by steps of 0.02, tabulated every 10 lLb eps eps_t T lambda LLB 198.2 0.001 0.5579 1600 14.85 0.1982 */ // Emissivity integral, Eq. 12.38 eps_t = pi * INTEGRAL(lL,lambda) / (sigma * T^4) sigma = 5.67 e-8 // Blackbody Spectral intensity, Tools | Radiation /* From Planck’s law, the blackbody spectral intensity is */ lL = eps *lLb lLb = l_lambda_b(lambda, T, C1, C2) // Eq. 12.25 // where units are lLb(W/m^2.sr.mum), lambda (mum) and T (K) with C1 = 3.7420e8 // First radiation constant, W⋅mum^4/m^2 C2 = 1.4388e4 // Second radiation constant, mum⋅K // and (mum) represents (micrometers). // Emissivity function eps = LOOKUPVAL(eps_L, 1, lambda, 2) /* The table file name is eps_L.lut, with 2 columns and 6 rows. See Help | Solver | Lookup Tables | Lookupval 0.05 0.4 1.99 0.4 2 0.8 4.99 0.8 5 0.001 100 0.001 */ // Input variable T = 1600

(2) For tungsten at 2800 K, the spectral limits for 98% of the blackbody spectrum are 0.51 and 8.3 µm. For aluminum at 1400 K, the spectral limits for 98% of the blackbody spectrum are 1.0 and 16.7 µm. For both cases, the foregoing tabulated spectral emissivity distributions are adequately represented for integration within the 98% limits.

PROBLEM 12.34 KNOWN: Spectral directional emissivity of a diffuse material at 2000K. FIND: (a) Total, hemispherical emissivity, (b) Emissive power over the spectral range 0.8 to 2.5 µm and for directions 0 ≤ θ ≤ π/6. SCHEMATIC:

ASSUMPTIONS: (1) Surface is diffuse emitter. ANALYSIS: (a) Since the surface is diffuse, ε λ,θ is independent of direction; from Eq. 12.36, ε λ,θ = ε λ. Using Eq. 12.38, ∞ ε ( T) = ε ( λ ) Eλ ,b ( λ ,T ) dλ / Eb ( T) 0 λ 1.5 ∞ E (T ) = ε1 E λ ,b ( λ,2000 ) dλ / E b + ε E ( λ ,2000 ) dλ / E b. 0 1.5 2 λ ,b Written now in terms of F(0 → λ), with F(0 → 1.5) = 0.2732 at λT = 1.5 × 2000 = 3000 µm⋅K, (Table 12.1) find,

∫

∫

∫

ε ( 2000K ) = ε1 × F( 0→1.5) + ε 2 1 − F( 0→1.5 )  = 0.2 × 0.2732 + 0.8[1 − 0.2732 ] = 0.636.  

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(b) For the prescribed spectral and geometric limits, from Eq. 12.12, 2.5 2π π / 6 ∆E = ε λ ,θ Iλ ,b ( λ , T ) cosθ sin θ d θ d φ d λ 0.8 0 0 where Iλ,e (λ, θ, φ) = ε λ,θ Iλ,b (λ,T). Since the surface is diffuse, ε λ,θ = ε λ, and noting Iλ,b is

∫

∫

∫

independent of direction and equal to Eλ,b/π, we can write 2.5  1.5   2π π / 6  E b ( T )  0.8 ε1 E λ ,b ( λ, T ) dλ 1.5 ε2 E λ ,b ( λ, T ) dλ  ∆E =  cosθ sin θ d θ dφ  +   0 Eb ( T) Eb ( T )  0  π  

∫

∫

∫

∫





or in terms F(0 → λ) values,

 2π sin2 θ π / 6  σ T4 ∆E = φ × {ε1[ F0→1.5 − F0→0.8] + ε 2 [ F0→2.5 −F 0→1.5]}.  0 0 2 π   From Table 12.1:



∆E = 2π ×



λT = 0.8 × 2000 = 1600 µm⋅K

F(0 → 0.8)= 0.0197

λT = 2.5 × 2000 = 5000 µm⋅K

F(0 → 2.5) = 0.6337

sin 2 π / 6   5.67 × 10 −8 × 2000 4 W 2

 

(

π

m2

)

⋅ {0.2 [0.2732 − 0.0197 ] + 0.8 [ 0.6337 − 0.2732]}

∆E = 0.25 × 5.67× 10−8 × 2000 4 W / m 2 × 0.339 = 76.89 k W / m2 .

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PROBLEM 12.35 KNOWN: Directional emissivity, ε θ, of a selective surface. FIND: Ratio of the normal emissivity, ε n, to the hemispherical emissivity, ε. SCHEMATIC:

ASSUMPTIONS: Surface is isotropic in φ direction. ANALYSIS: From Eq. 12.36 written on a total, rather than spectral, basis, the hemispherical emissivity is π /2 εθ (θ ) cos θ sin θ d θ. 0

ε = 2∫

Recognizing that the integral can be expressed in two parts, find π /4 π /2 ε = 2  ∫ ε (θ ) cos θ sin θ dθ + ∫ ε (θ ) cos θ sin θ dθ  0 π / 4   π /4 π /2  cosθ sin θ dθ + 0.3 ∫ cos θ sin θ dθ  ε = 2  0.8 ∫ 0 / 4 π  

 sin 2 θ π / 4 sin 2 θ π / 2    ε = 2 0.8 +0.3 0 2 2 π /4    1 1 ε = 2  0.8 ( 0.50 − 0) + 0.3 × (1 − 0.50)  = 0.550. 2  2  The ratio of the normal emissivity (ε n) to the hemispherical emissivity is

εn 0.8 = = 1.45. ε 0.550 COMMENTS: Note that Eq. 12.36 assumes the directional emissivity is independent of the φ coordinate. If this is not the case, then Eq. 12.35 is appropriate.

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PROBLEM 12.35 KNOWN: Directional emissivity, ε θ, of a selective surface. FIND: Ratio of the normal emissivity, ε n, to the hemispherical emissivity, ε. SCHEMATIC:

ASSUMPTIONS: Surface is isotropic in φ direction. ANALYSIS: From Eq. 12.36 written on a total, rather than spectral, basis, the hemispherical emissivity is π /2 εθ (θ ) cos θ sin θ d θ. 0

ε = 2∫

Recognizing that the integral can be expressed in two parts, find π /4 π /2 ε = 2  ∫ ε (θ ) cos θ sin θ dθ + ∫ ε (θ ) cos θ sin θ dθ  0 π / 4   π /4 π /2  cosθ sin θ dθ + 0.3 ∫ cos θ sin θ dθ  ε = 2  0.8 ∫ 0 / 4 π  

 sin 2 θ π / 4 sin 2 θ π / 2    ε = 2 0.8 +0.3 0 2 2 π /4    1 1 ε = 2  0.8 ( 0.50 − 0) + 0.3 × (1 − 0.50)  = 0.550. 2  2  The ratio of the normal emissivity (ε n) to the hemispherical emissivity is

εn 0.8 = = 1.45. ε 0.550 COMMENTS: Note that Eq. 12.36 assumes the directional emissivity is independent of the φ coordinate. If this is not the case, then Eq. 12.35 is appropriate.

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PROBLEM 12.36 KNOWN: The total directional emissivity of non-metallic materials may be approximated as εθ = εn cos θ where εn is the total normal emissivity. FIND: Show that for such materials, the total hemispherical emissivity, ε, is 2/3 the total normal emissivity. SCHEMATIC:

ANALYSIS: From Eq. 12.36, written on a total rather than spectral basis, the hemispherical emissivity ε can be determined from the directional emissivity εθ as

ε=2

1 0π / 2 εθ cos θ sin θ dθ

With ε θ = ε n cos θ , find

ε = 2 εn

1 0π /2 cos2θ sin θ dθ "

'0

ε = −2 ε n cos3 θ / 3 |

π /2

= 2 / 3 εn

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COMMENTS: (1) Refer to Fig. 12.17 illustrating on cartesian coordinates representative directional distributions of the total, directional emissivity for nonmetallic and metallic materials. In the schematic above, we’ve shown ε θ vs. θ on a polar plot for both types of materials, in comparison with a diffuse surface. (2) See Section 12.4 for discussion on other characteristics of emissivity for materials.

PROBLEM 12.37 KNOWN: Incandescent sphere suspended in air within a darkened room exhibiting these characteristics: initially: brighter around the rim after some time: brighter in the center FIND: Plausible explanation for these observations. ASSUMPTIONS: (1) The sphere is at a uniform surface temperature, Ts. ANALYSIS: Recognize that in observing the sphere by eye, emission from the central region is in a nearly normal direction. Emission from the rim region, however, has a large angle from the normal to the surface.

Note now the directional behavior, ε θ, for conductors and non-conductors as represented in Fig. 12.17. Assume that the sphere is fabricated from a metallic material. Then, the rim would appear brighter than the central region. This follows since ε θ is higher at higher angles of emission. If the metallic sphere oxidizes with time, then the ε θ characteristics change. Then ε θ at small angles of θ become larger than at higher angles. This would cause the sphere to appear brighter at the center portion of the sphere. COMMENTS: Since the emissivity of non-conductors is generally larger than for metallic materials, you would also expect the oxidized sphere to appear brighter for the same surface temperature.

PROBLEM 12.38 KNOWN: Detector surface area. Area and temperature of heated surface. Radiant power measured by the detector for two orientations relative to the heated surface. FIND: (a) Normal emissivity of heated surface, (b) Whether surface is a diffuse emitter. SCHEMATIC:

ASSUMPTIONS: (1) Detector intercepts negligible radiation from surroundings, (2) A1 and A2 are differential surfaces. ANALYSIS: The radiant power leaving the heated surface and intercepting the detector is

q12 (θ ) = I1 (θ ) A1 cos θ ω2 −1 I1 (θ ) = ε1 (θ ) Ib,1 = ε1 (θ ) σ T14 / π

ω 2−1 =

A2 cos θ

( L/cos θ ) 2

.

Hence,

q12 (θ ) = ε1 (θ )

σ T14 π

A1 cos θ

q (θ ) π L2 ε1 (θ ) = 12 σ T14 A1A 2 ( cos θ ) 4

A 2 cos θ

( L/cos θ ) 2

(a) For the normal condition, θ = 0, cosθ = 1, and ε 1 (θ) ≡ ε 1,n is

ε1,n =

1.155 ×10−6 W π

( 0.5m ) 2

5.67 ×10−8 W / m 2 ⋅ K 4 (1000K ) 4 5 × 10−6 m2 × 4 × 10−6 m 2

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ε1,n = 0.80. (b) For the orientation with θ = 60° and cosθ = 0.5, so that ε 1 (θ = 60°) is

ε1 ( 60° ) =

5.415 × 10−8 W π

( 0.5m )2

5.67 ×10−8 W / m 2 ⋅ K 4 (1000K ) 4 5 ×10−6 m2 × 4 ×10−6 m 2 ( 0.5) 4

ε1 ( 60° ) = 0.60. Since ε 1,n ≠ ε 1(60°), the surface is not a diffuse emitter. COMMENTS: Even if ε 1 (60°) were equal to ε 1,n, it could not be concluded there was diffuse emission until results were obtained for a wider range of θ.

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PROBLEM 12.39 KNOWN: Radiation thermometer responding to radiant power within a prescribed spectral interval and calibrated to indicate the temperature of a blackbody. FIND: (a) Whether radiation thermometer will indicate temperature greater than, less than, or equal to Ts when surface has ε < 1, (b) Expression for Ts in terms of spectral radiance temperature and spectral emissivity, (c) Indicated temperature for prescribed conditions of Ts and ε λ. SCHEMATIC:

ASSUMPTIONS: (1) Surface is a diffuse emitter, (2) Thermometer responds to radiant flux over interval dλ about λ. ANALYSIS: (a) The radiant power which reaches the radiation thermometer is

q λ = ε λI λ ,b ( λ ,Ts ) ⋅ A t ⋅ ωt

(1)

where At is the area of the surface viewed by the thermometer (referred to as the target) and ωt the solid angle through which At is viewed. The thermometer responds as if it were viewing a blackbody at Tλ, the spectral radiance temperature,

q λ = I λ ,b ( λ , Tλ ) ⋅ At ⋅ ω t .

(2)

By equating the two relations, Eqs. (1) and (2), find

I λ ,b ( λ ,Tλ ) = ε λ I λ ,b ( λ , Ts ) .

(3)

Since ε λ < 1, it follows that Iλ,b(λ, Tλ) < Iλ,b(λ, Ts) or that Tλ < Ts. That is, the thermometer will always indicate a temperature lower than the true or actual temperature for a surface with ε < 1. (b) Using Wien’s law in Eq. (3), find 1 I λ ( λ, T ) = C1λ −5 exp ( −C2 / λT )

π

1 1 C1λ −5 exp ( −C2 / λ Tλ ) = ε λ ⋅ C1λ −5 exp ( −C2 / λ Ts ) . π π -5

Canceling terms (C1λ /π), taking natural logs of both sides of the equation and rearranging, the desired expression is

1 1 λ = + lnε λ . Ts Tλ C2

(4)

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(c) For Ts = 1000K and ε = 0.9, from Eq. (4), the indicated temperature is

1 1 λ 1 0.65 µ m = − ln ε λ = − ln ( 0.9 ) Tλ Ts C 2 1000K 14,388 µ m ⋅ K That is, the thermometer indicates 5K less than the true temperature.

Tλ = 995.3K.

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PROBLEM 12.40 KNOWN: Spectral distribution of emission from a blackbody. Uncertainty in measurement of intensity. FIND: Corresponding uncertainities in using the intensity measurement to determine (a) the surface temperature or (b) the emissivity. ASSUMPTIONS: Diffuse surface behavior. ANALYSIS: From Eq. 12.25, the spectral intensity associated with emission may be expressed as

ε λ C1 / π I λ ,e = ε λ I λ ,b = λ 5 exp ( C2 / λ T ) −1 (a) To determine the effect of temperature on intensity, we evaluate the derivative,

−C 2 / λ T 2 ) ( =− 2 ∂T {λ 5 exp ( C2 / λT ) −1} 2 ∂ I λ ,e (C2 / λ T ) exp ( C2 /λ T) = Iλ ,e

∂ I λ ,e

( ε λ C1 / π ) λ 5 exp ( C2 /λ T )

exp ( C2 / λT ) − 1

∂T

Hence,

dT 1 − exp ( −C2 / λ T) d Iλ ,e = T Iλ ,e ( C2 / λ T )

(

)

With d Iλ ,e / Iλ ,e = 0.1, C2 = 1.439 ×104 µ m ⋅ K and λ = 10 µ m,

dT  1 − exp ( −1439K/T)  =  × 0.1 T  1439K/T  T = 500 K:

dT/T = 0.033 → 3.3% uncertainty

T = 1000 K:

dT/T = 0.053 → 5.5% uncertainty

< <

(b) To determine the effect of the emissivity on intensity, we evaluate

∂ I λ ,e

I = I λ ,b = λ ,e ∂ ελ ελ d ε λ d I λ ,e Hence, = = 0.10 → 10% uncertainty ελ Iλ ,e

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COMMENTS: The uncertainty in the temperature is less than that of the intensity, but increases with increasing temperature (and wavelength). In the limit as C2/λT → 0, exp (- C2/λT) → 1 – C2/λT and dT/T → d Iλ,e/Iλ,e. The uncertainty in temperature then corresponds to that of the intensity measurement. The same is true for the uncertainty in the emissivity, irrespective of the value of T or λ.

PROBLEM 12.41 KNOWN: Temperature, thickness and spectral emissivity of steel strip emerging from a hot roller. Temperature dependence of total, hemispherical emissivity. FIND: (a) Initial total, hemispherical emissivity, (b) Initial cooling rate, (c) Time to cool to prescribed final temperature. SCHEMATIC:

ASSUMPTIONS: (1) Negligible conduction (in longitudinal direction), convection and radiation from surroundings, (2) Negligible transverse temperature gradients. 3

PROPERTIES: Steel (given): ρ = 7900 kg/m , c = 640 J/kg⋅K, ε = 1200ε i/T (K). ANALYSIS: (a) The initial total hemispherical emissivity is ∞ εi = ε E (1200 ) / E b (1200 )  dλ 0 λ  λb and integrating by parts using values from Table 12.1, find

∫

λ T = 1200 µm ⋅ K → F(0 −1µ m) = 0.002; λ T = 7200 µm ⋅ K → F(0 − 6 µ m ) = 0.819 εi = 0.6 × 0.002 + 0.4 ( 0.819 − 0.002 ) + 0.25 (1 −0.819 ) = 0.373.

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(b) From an energy balance on a unit surface area of strip (top and bottom), − E& out = dEst /dt − 2εσ T4 = d( ρδ cT ) /dt dT 

2εiσ Ti4

dt i

ρδ c

 =−

=

−2 ( 0.373 ) 5.67× 10 − 8 W / m 2 ⋅ K 4 (1200 K )4 7900 k g / m3 ( 0.003 m) ( 640 J / k g ⋅ K )

= −5.78 K / s.

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(c) From the energy balance,

2ε (1200/T ) σ T 4 Tf dT dT 2400ε iσ t ρδ c  1 1  =− i ,∫ =− dt, t = − Ti T3 dt ρδ c ρδ c ∫0 4800ε iσ  T 2 T2  i   f t=

7900 k g / m 3 ( 0.003m ) 640 J / k g ⋅ K

 1 1  −2 −   K = 311s. 4800 K × 0.373 × 5.67 ×10−8 W / m 2 ⋅ K4  6002 12002  2

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COMMENTS: Initially, from Eq. 1.9, hr = εiσ Ti3 = 36.6 W/m ⋅K. Assuming a plate width of W = 3 1m, the Rayleigh number may be evaluated from RaL = gβ(Ti - T∞) (W/2) /να. Assuming T∞ = 300 8 K and evaluating properties at Tf = 750 K, RaL = 2.4 × 10 . From Eq. 9.31, NuL = 93, giving h = 10 2 W/m ⋅K. Hence heat loss by radiation exceeds that associated with free convection. To check the 2 validity of neglecting transverse temperature gradients, compute Bi = h(δ/2)/k. With h = 36.6 W/m ⋅K and k = 28 W/m⋅K, Bi = 0.002 10 µm, find that

(

)

Gabs = 1.0 0.5 × 600 W / m 2 ⋅ µ m (5 − 0 ) µ m + 600 W / m 2 ⋅ µ m ( 0.5 × 0.5 )(10 − 5) µm + 0 Gabs = 2250 W / m 2 . (d) The total, hemispherical absorptivity is defined as the fraction of the total irradiation that is absorbed. From Eq. 12.45, G 2250 W / m 2 α = abs = = 0.30. G 7500 W / m 2

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COMMENTS: Recognize that the total, hemispherical absorptivity, α = 0.3, is for the given spectral irradiation. For a different Gλ, one would then expect a different value for α.

PROBLEM 12.44 KNOWN: Temperature and spectral emissivity of small object suspended in large furnace of prescribed temperature and total emissivity. FIND: (a) Total surface emissivity and absorptivity, (b) Reflected radiative flux and net radiative flux to surface, (c) Spectral emissive power at λ = 2 µm, (d) Wavelength λ1/2 for which one-half of total emissive power is in spectral region λ ≥ λ1/2. SCHEMATIC:

ASSUMPTIONS: (1) Surface is opaque and diffuse, (2) Walls of furnace are much larger than object. ANALYSIS: (a) The emissivity of the object may be obtained from Eq. 12.38, which is expressed as ∞

∫ ε (T ) = o s

ε λ ( λ ) E λ ,b ( λ , Ts ) dλ Eb (T )

= ε1  F( 0→3µ m ) − F(0→1µ m )  + ε 2 1 − F( 0→3µ m ) 









where, with λ1Ts = 400 µm⋅K and λ2Ts = 1200 µm⋅K, F(0→1µm) = 0 and F( 0→3µ m ) = 0.002. Hence,

ε ( Ts ) = 0.7 ( 0.002 ) + 0.5 (0.998 ) = 0.500

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The absorptivity of the surface is determined by Eq. 12.46, ∞

∞

∫ αλ (λ ) G λ (λ ) dλ = ∫o αλ (λ ) E λ ,b (λ , Tf ) dλ α= o ∞

∫o

E b ( Tf )

G λ ( λ ) dλ

Hence, with λ1Tf = 2000 µm⋅K and λ2Tf = 6000 µm⋅K, F(0→1µm) = 0.067 and F( 0→3µ m ) = 0.738. It follows that

α = α1  F(0 →3µ m ) − F( 0 →1µ m )  + α 2 1 − F(0 →3µ m )  = 0.7 × 0.671 + 0.5 × 0.262 = 0.601    

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(b) The reflected radiative flux is G ref = ρ G = (1 − α ) E b ( Tf ) = 0.399 × 5.67 × 10−8 W m 2⋅ K 4 ( 2000 K ) = 3.620 × 105 W m 2 4

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The net radiative flux to the surface is q′′rad = G − ρ G − ε E b ( Ts ) = α E b ( Tf ) − ε E b ( Ts ) 4 4 q′′rad = 5.67 × 10−8 W m 2⋅ K 4  0.601( 2000 K ) − 0.500 ( 400 K )  = 5.438 × 105 W m 2





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(c) At λ = 2 µm, λTs = 800 K and, from Table 12.1, Iλ,b(λ,T)/σT5 = 0.991 × 10-7 (µm⋅K⋅sr)-1. Hence, Continued...

PROBLEM 12.44 (Cont.) Iλ ,b = 0.991 × 10−7 × 5.67 × 10−8

W m 2⋅ K 4

µ m ⋅ K ⋅ sr

× ( 400 K ) = 0.0575 5

W m 2 ⋅ µ m ⋅ sr

Hence, with Eλ = ελEλ,b = ελπIλ,b, E λ = 0.7 (π sr ) 0.0575 W m 2 ⋅ µ m ⋅ sr = 0.126 W m 2 ⋅ µ m

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(d) From Table 12.1, F(0→λ) = 0.5 corresponds to λTs ≈ 4100 µm⋅K, in which case,

λ1/ 2 ≈ 4100 µ m ⋅ K 400 K ≈ 10.3 µ m

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COMMENTS: Because of the significant difference between Tf and Ts, α ≠ ε. With increasing Ts → Tf, ε would increase and approach a value of 0.601.

PROBLEM 12.45 KNOWN: Small flat plate maintained at 400 K coated with white paint having spectral absorptivity distribution (Figure 12.23) approximated as a stairstep function. Enclosure surface maintained at 3000 K with prescribed spectral emissivity distribution. FIND: (a) Total emissivity of the enclosure surface, εes, and (b) Total emissivity, ε, and absorptivity, α, of the surface. SCHEMATIC:

ASSUMPTIONS: (1) Coated plate with white paint is diffuse and opaque, so that αλ = ελ, (2) Plate is small compared to the enclosure surface, and (3) Enclosure surface is isothermal, diffuse and opaque. ANALYSIS: (a) The total emissivity of the enclosure surface at Tes = 3000 K follows from Eq. 12.38 which can be expressed in terms of the bond emission factor, F(0-λT), Eq. 12.30,

ε e,s = ε1F(0 − λ T ) + ε 2 1 − F(0 − λ T )  = 0.2 × 0.738 + 0.9 [1 − 0.738] = 0.383 1 es 1 es  

<

where, from Table 12.1, with λ1Tes = 2 µm × 3000 K = 6000 µm⋅K, F(0-λT) = 0.738. (b) The total emissivity of the coated plate at T = 400 K can be expressed as

ε = α1F(0 − λ T ) + α 2  F(0 − λ T ) − F(0 − λ T )  + α3 1 − F(0 − λ T )  1 s 2 s 1 s  2 s    ε = 0.75 × 0 + 0.15 [0.002134 − 0.000 ] + 0.96 [1 − 0.002134] = 0.958

<

where, from Table 12.1, the band emission factors are: for λ1Ts = 0.4 × 400 = 160 µm⋅K, find F( 0 − λ T ) = 0.000; for λ2Tes = 3.0 × 400 = 1200 µm⋅K, find F( 0 − λ T ) = 0.002134. The total 1 s

2 s

absorptivity for the irradiation due to the enclosure surface at Tes = 3000 K is

α = α1F( 0 − λ T ) + α 2  F( 0 − λ T ) − F( 0 − λ T )  + α 3 1 − F(0 − λ T )  1 es 2 es 2 es  2 es    α = 0.75 × 0.002134 + 0.15 [0.8900 − 0.002134] + 0.96 [1 − 0.8900] = 0.240

<

where, from Table 12.1, the band emission factors are: for λ1Tes = 0.4 × 3000 = 1200 µm⋅K, find F( 0 − λ T ) = 0.002134; for λ2Tes = 3.0 × 3000 = 9000 µm⋅K, find F( 0 − λ T ) = 0.8900. 1 es

2 es



$

COMMENTS: (1) In evaluating the total emissivity and absorptivity, remember that ε = ε ε λ ,Ts and α = α(αλ, Gλ) where Ts is the temperature of the surface and Gλ is the spectral irradiation, which if the surroundings are large and isothermal, Gλ = Eb,λ(Tsur). Hence, α = α(αλ ,Tsur ). For the opaque, diffuse surface, αλ = ε λ. (2) Note that the coated plate (white paint) has an absorptivity for the 3000 K-enclosure surface irradiation of α = 0.240. You would expect it to be a low value since the coating appears visually “white”. (3) The emissivity of the coated plate is quite high, ε = 0.958. Would you have expected this of a “white paint”? Most paints are oxide systems (high absorptivity at long wavelengths) with pigmentation (controls the “color” and hence absorptivity in the visible and near infrared regions).

PROBLEM 12.46 KNOWN: Area, temperature, irradiation and spectral absorptivity of a surface. FIND: Absorbed irradiation, emissive power, radiosity and net radiation transfer from the surface. SCHEMATIC:

ASSUMPTIONS: (1) Opaque, diffuse surface behavior, (2) Spectral distribution of solar radiation corresponds to emission from a blackbody at 5800 K. ANALYSIS: The absorptivity to solar irradiation is ∞ ∞ αλ Gλ d λ αλ Eλ b ( 5800 K) dλ αs = 0 = 0 = α1F(0.5 →1 µm ) + α 2 F(2→∞ ). G Eb

∫

∫

From Table 12.1,

λT = 2900 µm⋅K:

F(0 → 0.5 µm) = 0.250

λT = 5800 µm⋅K:

F(0 → 1 µm) = 0.720

λT = 11,600 µm⋅K:

F(0 → 2 µm) = 0.941

αs = 0.8 ( 0.720 − 0.250 ) + 0.9 ( 1− 0.941) = 0.429.

(

)

Gabs = αSGS = 0.429 1200 W / m 2 = 515 W / m 2.

Hence,

<

The emissivity is ∞ ε= ε E ( 400 K ) dλ / E b = ε1F( 0.5→1 µm ) + ε 2F( 2→∞ ). 0 λ λb

∫

From Table 12.1,

λT = 200 µm⋅K:

F(0 → 0.5 µm) = 0

λT = 400 µm⋅K:

F(0 → 1 µm) = 0

λT = 800 µm⋅K

F(0 → 2 µm) = 0.

Hence, ε = ε 2 = 0.9, 4 E = εσ Ts4 = 0.9 × 5.67 ×10 −8 W / m 2 ⋅ K 4 ( 400 K ) = 1306 W / m 2 .

<

The radiosity is

J = E + ρSGS = E + (1 −α s ) G S = [1306 + 0.571 ×1200 ] W / m 2 = 1991W/m 2 .

<

The net radiation transfer from the surface is

q net = ( E − αSGS ) As = (1306 − 515 ) W / m 2 × 4 m 2 = 3164 W. COMMENTS: Unless 3164 W are supplied to the surface by other means (for example, by convection), the surface temperature will decrease with time.

<

PROBLEM 12.47 KNOWN: Temperature and spectral emissivity of a receiving surface. Direction and spectral distribution of incident flux. Distance and aperture of surface radiation detector. FIND: Radiant power received by the detector. SCHEMATIC:

2

ASSUMPTIONS: (1) Target surface is diffuse, (2) Ad/L 2 µ m = α λ > 2 µ m = 1 and h is obtained from Eq. (9.35)

Nu D = 2 +

4 0.589 Ra1/ D 9 /16  4 / 9

1 + ( 0.469 / Pr ) 

=

hD k

(2)



with Ra D = gβ ( Ts − T∞ ) D3 / να . Radiation absorption at the inner surface of the bulb may be expressed as

(

q′′rad,i = α G = α Pelec / π D2

)

(3)

where, from Eq. (12.46), 0.4

α = α1 ∫ 0

2.0

∞

(G λ / G ) dλ + α 2 ∫0.4 (G λ / G ) dλ + α3 ∫2 (G λ / G ) dλ Continued …..

PROBLEM 12.52 (Cont.) The irradiation is due to emission from the filament, in which case (Gλ/G) ~ (Eλ/E)f = (εf,λEλ,b/εfEb). Hence, α = (α1 / ε f

) ∫0

0.4

(

)

ε f , λ E λ ,b / E b dλ + (α 2 / ε f

∞

) ∫0.4 ε f , λ ( E λ ,b / E b ) dλ + (α 3 / ε f ) ∫2 2.0

(

)

ε f , λ E λ ,b / E b dλ (4)

where, from the spectral distribution of Problem 12.25, εf,λ ≡ ε1 = 0.45 for λ < 2µm and εf,λ ≡ ε2 = 0.10 for λ > 2µm. From Eq. (12.38)

(

∞

ε f = ∫ ε f ,λ ( E λ ,b / E b ) dλ = ε1 F(0→ 2µ m ) + ε 2 1 − F(0→ 2µ m ) 0

)

With λ Tf = 2 µ m × 3000 K = 6000 µ m ⋅ K, F( 0→ 2 µ m ) = 0.738 from Table 12.1. Hence,

ε f = 0.45 × 0.738 + 0.1(1 − 0.738 ) = 0.358 Equation (4) may now be expressed as

(

)

(

α = (α1 / ε f ) ε1 F( 0 → 0.4 µ m ) + (α 2 / ε f ) ε1 F( 0 → 2 µ m ) − F( 0 → 0.4 µ m ) + (α 3 / ε f ) ε 2 1 − F( 0 → 2 µ m )

)

where, with λT = 0.4µm × 3000 K = 1200 µm⋅K, F(0→0.4µm) = 0.0021. Hence, α = (1 / 0.358 ) 0.45 × 0.0021 + ( 0.1 / 0.358 ) 0.45 × ( 0.738 − 0.0021) + (1 / 0.358 ) 0.1 (1 − 0.738 ) = 0.168

Substituting Eqs. (2) and (3) into Eq. (1), as well as values of εb = 1 and α = 0.168, an iterative solution yields

Ts = 348.1 K

<

COMMENTS: For the prescribed conditions, q′′rad,i = 713 W / m 2 , q′′rad,o = 385.5 W / m 2 and q′′conv = 327.5 W / m 2 .

PROBLEM 12.53 KNOWN: Spectral emissivity of an opaque, diffuse surface. FIND: (a) Total, hemispherical emissivity of the surface when maintained at 1000 K, (b) Total, hemispherical absorptivity when irradiated by large surroundings of emissivity 0.8 and temperature 1500 K, (c) Radiosity when maintained at 1000 K and irradiated as prescribed in part (b), (d) Net radiation flux into surface for conditions of part (c), and (e) Compute and plot each of the parameters of parts (a)(c) as a function of the surface temperature Ts for the range 750 < Ts ≤ 2000 K. SCHEMATIC:

ASSUMPTIONS: (1) Surface is opaque, diffuse, and (2) Surroundings are large compared to the surface. ANALYSIS: (a) When the surface is maintained at 1000 K, the total, hemispherical emissivity is evaluated from Eq. 12.38 written as λ1 ∞ E λ ,b (T) dλ E b (T) + ε λ ,2 E (T) dλ E b (T) 0 λ1 λ ,b

∞ 0

ε = ∫ ε λ E λ ,b (T) dλ E b (T) = ε λ ,1 ∫

∫

ε = ε λ ,1F(0 − λ T) + ε λ ,2 (1 − F(0 − λ T) ) 1 1 where for λT = 6µm × 1000 K = 6000µm⋅K, from Table 12.1, find F0 − λ T = 0.738 . Hence,

<

ε = 0.8 × 0.738 + 0.3(1 − 0.738) = 0.669. (b) When the surface is irradiated by large surroundings at Tsur = 1500 K, G = Eb(Tsur). From Eq. 12.46, ∞

α = ∫ α λ G λ dλ 0

∞

∫0

∞

G λ dλ = ∫ ε λ E λ ,b (Tsur ) dλ E b (Tsur ) 0

α = ε λ ,1F(0 − λ T ) + ε λ ,2 (1 − F(0 − λ T ) ) 1 sur 1 sur where for λ1Tsur = 6 µm × 1500 K = 9000 µm⋅K, from Table 12.1, find F(0 − λ T) = 0.890 . Hence,

<

α = 0.8 × 0.890 + 0.3 (1 − 0.890) = 0.745. Note that α λ = ε λ for all conditions and the emissivity of the surroundings is irrelevant. (c) The radiosity for the surface maintained at 1000 K and irradiated as in part (b) is J = εEb (T) + ρG = εEb (T) + (1 − α)Eb (Tsur) J = 0.669 × 5.67 × 10-8 W/m2 ⋅K4 (1000 K)4 + (1 − 0.745) 5.67 × 10-8 W/m2 ⋅K4 (1500 K)4 J = (37,932 + 73,196) W/m2 = 111,128 W/m2

< Continued...

PROBLEM 12.53 (Cont.) 4 (d) The net radiation flux into the surface with G = σTsur is

q″rad,in = αG − εE b (T) = G − J q″rad,in = 5.67 × 10-8 W/m2 ⋅K (1500 K)4 − 111,128 W/m2

<

q″rad,in = 175,915 W/m2.

(e) The foregoing equations were entered into the IHT workspace along with the IHT Radiaton Tool, Band Emission Factor, to evaluate F( 0−λT ) values and the respective parameters for parts (a)-(d) were computed and are plotted below. 1

eps or alpha

0.9 0.8 0.7 0.6 0.5 500

1000

1500

2000

Surface temperature, Ts (K) Emissivity, eps Absorptivity, alpha; Tsur = 1500K

Note that the absorptivity, α = α (α λ , Tsur ) , remains constant as Ts changes since it is a function of

α λ (or ε λ ) and Tsur only. The emissivity ε = ε (ε λ , Ts ) is a function of Ts and increases as Ts increases. Could you have surmised as much by looking at the spectral emissivity distribution? At what condition is ε = α?

J or q''radin (W/m^2)

1E6

500000

0

-5E5 500

1000

1500

2000

Surface temperature, Ts (K) Radiosity, J (W/m^2) Net radiation flux in, q''radin (W/m^2)

The radiosity, J1 increases with increasing Ts since Eb(T) increases markedly with temperature; the reflected irradiation, (1 - α)Eb(Tsur) decreases only slightly as Ts increases compared to Eb(T). Since G is independent of Ts, it follows that the variation of q′′rad,in will be due to the radiosity change; note the sign difference. COMMENTS: We didn’t use the emissivity of the surroundings (ε = 0.8) to determine the irradiation onto the surface. Why?

PROBLEM 12.54 KNOWN: Furnace wall temperature and aperture diameter. Distance of detector from aperture and orientation of detector relative to aperture. FIND: (a) Rate at which radiation from the furnace is intercepted by the detector, (b) Effect of aperture window of prescribed spectral transmissivity on the radiation interception rate. SCHEMATIC:

ASSUMPTIONS: (1) Radiation emerging from aperture has characteristics of emission from a blackbody, (2) Cover material is diffuse, (3) Aperture and detector surface may be approximated as infinitesimally small. ANALYSIS: (a) From Eq. 12.5, the heat rate leaving the furnace aperture and intercepted by the detector is

q = IeAa cos θ1 ωs −a . From Eqs. 12.14 and 12.28

E b (Tf ) σ Tf4 5.67 ×10 −8 (1500 ) 4 Ie = = = = 9.14 ×10 4 W / m 2 ⋅sr. π π π From Eq. 12.2,

ωs −a =

An r

2

=

As ⋅ cos θ 2 r

2

=

10−5 m2 × cos45 °

(1m )

2

= 0.707 × 10−5 sr.

Hence

q = 9.14 ×104 W / m 2 ⋅ sr π ( 0.02m ) 2 / 4  cos30°× 0.707 ×10 −5 sr = 1.76 ×10 −4 W.  

<

(b) With the window, the heat rate is

q = τ ( I e Aa cos θ1 ωs −a ) where τ is the transmissivity of the window to radiation emitted by the furnace wall. From Eq. 12.55, ∞ ∞ τ λ Gλ dλ τ λ Eλ ,b ( Tf ) dλ 2 τ= 0 = 0 = 0.8 E / E dλ = 0.8F( 0→ 2 µm) . ∞ ∞ 0 λ ,b b G λ dλ E λ ,b dλ 0 0

∫

∫

∫

∫

∫ (

)

With λT = 2 µm × 1500K = 3000 µm⋅K, Table 12.1 gives F(0 → 2 µm) = 0.273. Hence, with τ = 0.273 × 0.8 = 0.218, find

q = 0.218 ×1.76 × 10−4 W = 0.384 × 10−4 W.

<

PROBLEM 12.55 KNOWN: Thermocouple is irradiated by a blackbody furnace at 1500 K with 25 mm2 aperture. Optical fiber of prescribed spectral transmissivity in sight path. FIND: (a) Distance L from the furnace detector should be positioned such that its irradiation is G = 50 W/m2 and, (b) Compute and plot irradiation, G, vs separation distance L for the range 100 ≤ L ≤ 400 mm for blackbody furnace temperatures of Tf = 1000, 1500 and 2000 K. SCHEMATIC:

ASSUMPTIONS: (1) Furnace aperture emits diffusely, (2) Ad 1.38 µm. (b) The transmissivity to solar irradiation, GS, follows from Eq. 12.55, ∞ ∞ τ S = τ λ G λ ,S dλ / G S = τ λ E λ ,b ( λ,5800K ) dλ / E b ( 5800K ) 0 0

∫

∫

1.38 E λ ,b ( λ,5800K ) dλ / E b ( 5800K ) = τ λ ,1F(0 →λ ) = 0.7 × 8.56 = 0.599 1 0

τ S = τ λ ,b ∫

<

where λ1 TS = 1.38 × 5800 = 8000 µm⋅K and from Table 12.1, F( 0 → λ ) = 0.856. From Eqs. 12.52 1 and 12.57, ∞

ρS = ∫ ρλ Gλ ,S dλ / GS = ρ λ ,1F( 0 → λ ) = 0.1× 0.856 = 0.086 1 0

<

α S = 1 − ρ S − τ S = 1 − 0.086 − 0.599 = 0.315.

<

(c) For the surface at Ts = 350K, the emissivity can be determined from Eq. 12.38. Since the surface is diffuse, according to Eq. 12.65, α λ = ε λ, the expression has the form ∞ 0

∞ 0

ε = ∫ ελ E λ ,b ( Ts ) d λ / E b ( Ts ) = ∫ α λ Eλ ,b ( 350K ) dλ / E b ( 350K ) ε = α λ ,1F( 0 −1.38 µ m ) + αλ ,2 1 − F( 0 −1.38 µ m )  = α λ ,2 = 1  

<

where from Table 12.1 with λ1 TS = 1.38 × 350 = 483 µm⋅K, F( 0 − λ T ) ≈ 0. (d) The net heat flux by radiation to the surface is determined by a radiation balance

q′′rad = GS − ρSG S − τ SGS − E q′′rad = αSGS − E

q′′rad = 0.315 × 750 W / m 2 − 1.0 × 5.67 ×10−8 W / m 2 ⋅ K 4 ( 350K ) = −615 W / m 2 . 4

<

PROBLEM 12.60 KNOWN: Large furnace with diffuse, opaque walls (Tf, ε f) and a small diffuse, spectrally selective object (To, τλ, ρ λ). FIND: For points on the furnace wall and the object, find ε, α, E, G and J. SCHEMATIC:

ASSUMPTIONS: (1) Furnace walls are isothermal, diffuse, and gray, (2) Object is isothermal and diffuse. ANALYSIS: Consider first the furnace wall (A). Since the wall material is diffuse and gray, it follows that

<

ε A = ε f = α A = 0.85. The emissive power is E A = εA E b ( Tf ) = εAσ Tf = 0.85 × 5.67 × 10 − 8 W / m 2 ⋅ K 4 ( 3000 K ) = 3.904 ×10 6 W / m 2 . Since the furnace is an isothermal enclosure, blackbody conditions exist such that

<

G A = J A = E b ( Tf ) = σ Tf4 = 5.67 ×10− 8 W / m 2 ⋅ K 4 ( 3000K ) 4 = 4.593 ×10 6 W / m 2 .

<

4

Considering now the semitransparent, diffuse, spectrally selective object at To = 300 K. From the radiation balance requirement, find

α λ = 1 − ρ λ −τ λ or α1 = 1 − 0.6 − 0.3 = 0.1 and α 2 = 1 − 0.7 − 0.0 = 0.3 ∞ 0

α B = ∫ α λ Gλ dλ / G = F0 − λ T ⋅α1 + (1 − F0 −λ T ) ⋅ α 2 = 0.970 × 0.1 + (1 − 0.970 ) × 0.3 = 0.106

<

where F0 - λT = 0.970 at λT = 5 µm × 3000 K = 15,000 µm⋅K since G = Eb(Tf). Since the object is diffuse, ε λ = α λ, hence εB =

∞

∫0

ε λ E λ ,b ( To ) dλ / E b,o = F0− λ Tα1 + (1 − F0− λ T ) ⋅ α 2 = 0.0138 × 0.1 + (1 − 0.0138 ) × 0.3 = 0.297

<

where F0-λT = 0.0138 at λT = 5 µm × 300 K = 1500 µmK. The emissive power is

E B = ε BE b,B (To ) = 0.297 × 5.67 ×10 −8 W / m 2 ⋅ K 4 ( 300 K ) = 136.5 W / m 2 . 4

<

The irradiation is that due to the large furnace for which blackbody conditions exist, G B = G A = σ Tf4 = 4.593 ×106 W / m 2 .

<

The radiosity leaving point B is due to emission and reflected irradiation,

J B = EB + ρBG B = 136.5 W / m 2 + 0.3 × 4.593 ×10 6 W / m 2 = 1.378 ×10 6 W / m 2 . < 6

2

If we include transmitted irradiation, JB = EB + (ρ B + τB) GB = EB + (1 - α B) GB = 4.106 × 10 W/m . In the first calculation, note how we set ρ B ≈ ρ λ (λ < 5 µm).

PROBLEM 12.61 KNOWN: Spectral characteristics of four diffuse surfaces exposed to solar radiation. FIND: Surfaces which may be assumed to be gray. SCHEMATIC:

ASSUMPTIONS: (1) Diffuse surface behavior. ANALYSIS: A gray surface is one for which α λ and ε λ are constant over the spectral regions of the irradiation and the surface emission. For λ = 3 µm and T = 5800K, λT = 17,400 µm⋅K and from Table 12.1, find F(0 → λ) = 0.984. Hence, 98.4% of the solar radiation is in the spectral region below 3 µm. For λ = 6 µm and T = 300K, λT = 1800 µm⋅K and from Table 12.1, find F(0 → λ) = 0.039. Hence, 96.1% of the surface emission is in the spectral region above 6 µm. Hence:

Surface A is gray:

α S ≈ ε = 0.8

Surface B is not gray:

α S ≈ 0.8, ε ≈ 0.3

Surface C is not gray:

α S ≈ 0.3, ε ≈ 0.7

Surface D is gray:

α S ≈ ε = 0.3.

< < < <

PROBLEM 12.62 KNOWN: A gray, but directionally selective, material with α (θ, φ) = 0.5(1 - cosφ). FIND: (a) Hemispherical absorptivity when irradiated with collimated solar flux in the direction (θ = 45° and φ = 0°) and (b) Hemispherical emissivity of the material. SCHEMATIC:

ASSUMPTIONS: (1) Gray surface behavior. ANALYSIS: (a) The surface has the directional absorptivity given as

α (θ ,φ ) = α λ ,φ = 0.5[1 − cos φ ] . When irradiated in the direction θ = 45° and φ = 0°, the directional absorptivity for this condition is

α ( 45°, 0° ) = 0.5 1 − cos ( 0° )  = 0.

<

That is, the surface is completely reflecting (or transmitting) for irradiation in this direction. (b) From Kirchhoff’s law,

αθ ,φ = εθ ,φ so that

εθ ,φ = αθ ,φ = 0.5 (1 − cos φ ) . Using Eq. 12.35 find 2π π / 2

∫ ∫ εθ ,φ ,λ cos θ sin θ d θ d φ ε= 0 0 2π π / 2 ∫0 ∫0 cos θ sin θ dθ dφ 2π 2π 0.5 (1 − cos φ ) dφ 0.5 ( φ − sin φ ) ε= 0 = = 0.5. 2π 2 π dφ 0 0

∫

∫

<

PROBLEM 12.63 KNOWN: Area and temperature of an opaque surface. Rate of incident radiation, absorbed radiation and heat transfer by convection. FIND: Surface irradiation, emissive power, radiosity, absorptivity, reflectivity and emissivity. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Adiabatic sides and bottom. ANALYSIS: The irradiation, emissive power and radiosity are G = 1300 W 3m 2 = 433 W m 2

E = G abs − q′′conv = (1000 − 300 ) W 3m 2 = 233 W m 2 J = E + G ref = E + ( G − G abs ) =  233 + ( 433 − 333) W m 2 = 333 W m 2 The absorptivity, reflectivity and emissivity are 433 W m 2 = 0.769 α = G abs G = 333 W m 2

(

)(

)

< < < <

ρ = 1 − α = 0.231 ε = E E b = E σ Ts4 = 233 W m 2 5.67 × 10−8 W m 2⋅ K 4 (500 K ) = 0.066 4

COMMENTS: The expression for E follows from a surface energy balance for which the absorbed irradiation is balanced by emission and convection.

< <

PROBLEM 12.64 KNOWN: Isothermal enclosure at a uniform temperature provides a known irradiation on two small surfaces whose absorption rates have been measured. FIND: (a) Net heat transfer rates and temperatures of the two surfaces, (b) Absorptivity of the surfaces, (c) Emissive power of the surfaces, (d) Emissivity of the surfaces. SCHEMATIC:

ASSUMPTIONS: (1) Enclosure is at a uniform temperature and large compared to surfaces A and B, (2) Surfaces A and B have been in the enclosure a long time, (3) Irradiation to both surfaces is the same. ANALYSIS: (a) Since the surfaces A and B have been within the enclosure a long time, thermal equilibrium conditions exist. That is,

q A,net = q B,net = 0. Furthermore, the surface temperatures are the same as the enclosure, Ts,A = Ts,B = Tenc. Since the enclosure is at a uniform temperature, it follows that blackbody radiation exists within the enclosure (see Fig. 12.12) and 4 G = Eb ( Tenc ) = σ Tenc

(

Tenc = ( G / σ )1/4 = 6300W/m 2 /5.67 ×10 −8 W / m 2 ⋅ K 4 (b) From Eq. 12.45, the absorptivity is Gabs/G, 5600 W / m 2 αA = = 0.89 2

6300 W / m

αB =

630 W / m 2 6300 W / m 2

)

1/4

= 577.4K.

< <

= 0.10.

(c) Since the surfaces experience zero net heat transfer, the energy balance is Gabs = E. That is, the absorbed irradiation is equal to the emissive power, E A = 5600 W / m 2 E B = 630 W / m 2 .

<

(d) The emissive power, E(T), is written as E = ε E b (T ) = ε σ T 4 or

ε = E / σ T4 .

Since the temperature of the surfaces and the emissive powers are known,

 W 4 ε A = 5600 W / m2 /  5.67 × 10−8 577.4K )  = 0.89 (  m2 ⋅ K4  COMMENTS: Note for this equilibrium condition, ε = α.

ε B = 0.10.

<

PROBLEM 12.65 KNOWN: Opaque, horizontal plate, well insulated on backside, is subjected to a prescribed irradiation. Also known are the reflected irradiation, emissive power, plate temperature and convection coefficient for known air temperature. FIND: (a) Emissivity, absorptivity and radiosity and (b) Net heat transfer per unit area of the plate. SCHEMATIC:

ASSUMPTIONS: (1) Plate is insulated on backside, (2) Plate is opaque. ANALYSIS: (a) The total, hemispherical emissivity of the plate according to Eq. 12.37 is E E 1200 W / m 2

ε=

E b ( Ts )

=

=

= 0.34.

<

σ Ts4 5.67 ×10 −8 W / m 2 ⋅ K 4 × ( 227 + 273) 4 K 4 The total, hemispherical absorptivity is related to the reflectivity by Eq. 12.57 for an opaque surface. That is, α = 1 - ρ. By definition, the reflectivity is the fraction of irradiation reflected, Eq. 12.51, such that α = 1 − Gref / G = 1 − 500 W / m 2 / 2500 W / m 2 = 1 − 0.20 = 0.80.

(

)

<

The radiosity, J, is defined as the radiant flux leaving the surface by emission and reflection per unit area of the surface (see Section 12.24). J = ρ G + ε Eb = Gref + E = 500 W / m 2 +1200 W / m 2 = 1700 W / m 2. (b) The net heat transfer is determined from an energy balance,

<

q′′net = q′′in − q′′out = G − G ref − E − q′′conv q′′net = ( 2500 − 500 −1200 ) W / m 2 − 15 W / m 2 ⋅ K ( 227 −127 ) K = −700 W / m 2 .

<

An alternate approach to the energy balance using the radiosity,

q′′net = G − J − q′′conv q′′net = ( 2500 −1700 − 1500 ) W / m 2 q′′net = −700W/m 2 . COMMENTS: (1) Since the net heat rate per unit area is negative, energy must be added to the plate in order to maintain it at Ts = 227°C. (2) Note that α ≠ ε. Hence, the plate is not a gray body. (3) Note the use of radiosity in performing energy balances. That is, considering only the radiation processes, q ′′net = G − J.

PROBLEM 12.66 KNOWN: Horizontal, opaque surface at steady-state temperature of 77°C is exposed to a convection process; emissive power, irradiation and reflectivity are prescribed. FIND: (a) Absorptivity of the surface, (b) Net radiation heat transfer rate for the surface; indicate direction, (c) Total heat transfer rate for the surface; indicate direction. SCHEMATIC:

ASSUMPTIONS: (1) Surface is opaque, (2) Effect of surroundings included in the specified irradiation, (3) Steady-state conditions. ANALYSIS: (a) From the definition of the thermal radiative properties and a radiation balance for an opaque surface on a total wavelength basis, according to Eq. 12.59,

<

α = 1 − ρ = 1 − 0.4 = 0.6. (b) The net radiation heat transfer rate to the surface follows from a surface energy balance considering only radiation processes. From the schematic,

(

)

′′ − E& ′′out q′′net,rad = E& in rad q′′

net,rad = G− ρ G − E= (1380− 0.4×1380− 628 )W / m2 = 200W/m2 .

<

Since q ′′net,rad is positive, the net radiation heat transfer rate is to the surface. (c) Performing a surface energy balance considering all heat transfer processes, the local heat transfer rate is

q′′tot = ( E& in ′′ − E& ′′out )

′′ q′′tot = q′′net,rad − qconv q ′′tot = 200 W / m2 − 28 W / m2 ⋅ K ( 77 − 27 ) K = −1200 W / m2 . The total heat flux is shown as a negative value indicating the heat flux is from the surface. COMMENTS: (1) Note that the surface radiation balance could also be expresses as

q′′net,rad = G − J

α G − E.

or

Note the use of radiosity to express the radiation flux leaving the surface. (2) From knowledge of the surface emissive power and Ts, find the emissivity as ε ≡ E / σ Ts4 = 628 W / m 2 / 5.67 × 10−8 W / m 2 ⋅ K 4 ( 77 + 273)4 K4 = 0.74.

(

Since ε ≠ α, we know the surface is not gray.

)

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PROBLEM 12.67 KNOWN: Temperature and spectral characteristics of a diffuse surface at Ts = 500 K situated in a large enclosure with uniform temperature, Tsur = 1500 K. FIND: (a) Sketch of spectral distribution of E λ and E λ,b for the surface, (b) Net heat flux to the surface, q″rad,in (c) Compute and plot q″rad,in as a function of Ts for the range 500 ≤ Ts ≤ 1000 K; also plot the heat flux for a diffuse, gray surface with total emissivities of 0.4 and 0.8; and (d) Compute and plot ε and α as a function of the surface temperature for the range 500 ≤ Ts ≤ 1000 K. SCHEMATIC:

ASSUMPTIONS: (1) Surface is diffuse, (2) Convective effects are negligible, (3) Surface irradiation corresponds to blackbody emission at 1500 K. ANALYSIS: (a) From Wien’s law, Eq. 12.27, λmax T = 2897.6 µm⋅K. Hence, for blackbody emission from the surface at Ts = 500 K, 2897.6 µ m ⋅ K

λmax =

500 K

<

= 5.80 µ m .

(b) From an energy balance on the surface, the net heat flux to the surface is q″rad,in = αG − E = αEb (1500 K) − εEb (500 K).

(1)

From Eq. 12.46, 4 E λ ,b (1500) ∞ E λ ,b (1500) dλ + 0.8 dλ = 0.4F(0 − 4) − 0.8[1 − F(0 − 4) ]. 0 4 Eb Eb

α = 0.4 ∫

∫

From Table 12.1 with λT = 4µm × 1500 K = 6000 µm⋅K, F(0-4) = 0.738, find α = 0.4 × 0.738 + 0.8 (1 − 0.738) = 0.505. From Eq. 12.38 4 E λ ,b (500) ∞ E λ ,b (500) dλ + 0.8 dλ = 0.4F(0 − 4) + 0.8[1 − F(0 − 4) ] . 0 4 Eb Eb

ε = 0.4 ∫

∫

From Table 12.1 with λT = 4µm × 500 K = 2000 µm⋅K, F(0-4) = 0.0667, find ε = 0.4 × 0.0667 + 0.8 (1 − 0.0667) = 0.773. Hence, the net heat flux to the surface is q′′rad,in = 5.67 × 10−8 W m 2 ⋅ K 4 [0.505 × (1500 K)4 − 0.773 × (500 K)4 ] = 1.422 × 105 W m 2 .

<

Continued...

PROBLEM 12.67 (Cont.) (c) Using the foregoing equations in the IHT workspace along with the IHT Radiation Tool, Band Emission Factor, q′′rad,in was computed and plotted as a function of Ts.

q''radin (W/m^2)

250000 200000 150000 100000 50000 500

600

700

800

900

1000

Surface temperature, Ts (K) Grey surface, eps = 0.4 Selective surface, eps Grey surface, eps = 0.8

The net radiation heat rate, q′′rad,in decreases with increasing surface temperature since E increases with Ts and the absorbed irradiation remains constant according to Eq. (1). The heat flux is largest for the gray surface with ε = 0.4 and the smallest for the gray surface with ε = 0.8. As expected, the heat flux for the selective surface is between the limits of the two gray surfaces. (d) Using the IHT model of part (c), the emissivity and absorptivity of the surface are computed and plotted below. 0.8

eps or alpha

0.7

0.6

0.5

0.4 500

600

700

800

900

1000

Surface temperature, Ts (K) Emissivity, eps Absorptivity, alpha

The absorptivity, α = α (α λ , Tsur ) , remains constant as Ts changes since it is a function of α λ (or ε λ ) and Tsur only. The emissivity, ε = ε (ε λ , Ts ) is a function of Ts and decreases as Ts increases. Could you have surmised as much by looking at the spectral emissivity distribution? Under what condition would you expect α = ε?

PROBLEM 12.68 KNOWN: Opaque, diffuse surface with prescribed spectral reflectivity and at a temperature of 750K is subjected to a prescribed spectral irradiation, Gλ. FIND: (a) Total absorptivity, α, (b) Total emissivity, ε, (c) Net radiative heat flux to the surface. SCHEMATIC:

ASSUMPTIONS: (1) Opaque and diffuse surface, (2) Backside insulated. ANALYSIS: (a) The total absorptivity is determined from Eq. 12.46 and 12.56, ∞ αλ = 1− ρ λ and α= α G dλ /G. 0 λ λ Evaluating by separate integrals over various wavelength intervals.

∫

α=

(1 − ρ λ ,1 ) ∫13 Gλ dλ + (1 − ρ λ ,2 ) ∫36 G λ dλ + (1 − ρ λ ,2 ) ∫68 Gλ dλ 3

∫1

6

8

G λ d λ + ∫ G λ dλ + ∫ G λ dλ 3 6

(1,2)

=

Gabs G

G abs = (1 − 0.6 ) 0.5 × 500W/m2 ⋅ µ m ( 3 − 1) µ m + (1 − 0.2 ) 500W/m2 ⋅ µ m ( 6 − 3 ) µ m









+ (1 − 0.2 )  0.5 × 500W/m2 ⋅ µ m ( 8 − 6) µ m





G = 0.5 × 5 0 0 W / m ⋅ µ m × ( 3 − 1) µ m + 5 0 0 W / m ⋅ µ m ( 6 − 3 ) µm + 0.5 × 5 0 0 W / m ⋅ µ m ( 8 − 6 ) µ m 2

α=

2

2

[ 200 + 1200 + 400]W / m 2 1800W/m2 = [ 500 + 1500 + 500] W / m2 2 5 0 0 W / m2

= 0.720.

<

(b) The total emissivity of the surface is determined from Eq. 12.38 and 12.65,

ελ = αλ

ε λ = 1 − ρλ .

and, hence

(3,4)

The total emissivity can then be expressed as ∞ ∞ ε E ( λ , Ts ) dλ / E b ( Ts ) = 0 (1 − ρ λ ) E λ ,b ( λ , Ts ) dλ / E b ( Ts ) 0 λ λ ,b 3 ∞ ε = 1 − ρ λ ,1 E ( λ , Ts ) dλ / Eb ( Ts ) + 1 − ρλ ,2 3 Eλ ,b ( λ ,Ts ) d λ / E b ( Ts ) 0 λ ,b

ε =∫

∫

( )∫ ( )∫ ε = (1 − ρ λ ,1 ) F( 0 →3 µ m ) + (1 − ρ λ ,2 ) 1 − F( 0 →3 µ m )    ε = (1 − 0.6 ) × 0.111 + (1 − 0.2 )[1 −0.111] = 0.756

<

where Table 12.1 is used to find F(0 - λ) = 0.111 for λ1 Ts = 3 × 750 = 2250 µm⋅K. (c) The net radiative heat flux to the surface is q′′rad = α G − ε Eb ( Ts ) = αG − ε σ Ts4 q′′rad = 0.720 × 2500W/m2 4 −0.756 × 5.67 ×10−8 W / m 2 ⋅ K 4 750K = −11,763W/m 2 .

(

)

<

PROBLEM 12.69 KNOWN: Opaque, gray surface at 27°C with prescribed irradiation, reflected flux and convection process. FIND: Net heat flux from the surface. SCHEMATIC:

ASSUMPTIONS: (1) Surface is opaque and gray, (2) Surface is diffuse, (3) Effects of surroundings are included in specified irradiation. ANALYSIS: From an energy balance on the surface, the net heat flux from the surface is

q′′net = E& ′′out − E& ′′in q′′net = q′′conv + E + G ref − G = h ( Ts − T∞ ) + ε σ Ts4 + G ref − G.

(1)

To determine ε, from Eq. 12.59 and Kirchoff’s law for a diffuse-gray surface, Eq. 12.62,

ε = α = 1 − ρ = 1 − ( Gref / G ) = 1 − ( 800/1000 ) = 1 − 0.8 = 0.2

(2)

where from Eq. 12.51, ρ = Gref/G. The net heat flux from the surface, Eq. (1), is

q′′net = 1 5 W / m 2 ⋅ K ( 27 − 17 ) K + 0.2 × 5.67 ×10−8 W / m 2 ⋅ K 4 ( 27 + 273) K 4 4

+8 0 0 W / m 2 − 1000W/m2 q′′net = (150 + 91.9 + 800 − 1000 ) W / m 2 = 4 2 W / m 2 .

<

COMMENTS: (1) For this situation, the radiosity is

J = Gref + E = ( 800 + 91.9 ) W / m 2 = 8 9 2 W / m 2. The energy balance can be written involving the radiosity (radiation leaving the surface) and the irradiation (radiation to the surface).

q′′net = J −G + q′′conv = ( 892 −1000 + 150 ) W / m 2 = 4 2 W / m2 . (2) Note the need to assume the surface is diffuse, gray and opaque in order that Eq. (2) is applicable.

PROBLEM 12.70 KNOWN: Diffuse glass at Tg = 750 K with prescribed spectral radiative properties being heated in a large oven having walls with emissivity of 0.75 and 1800 K. FIND: (a) Total transmissivity r, total reflectivity ρ, and total emissivity ε of the glass; Net radiative heat flux to the glass, (b) q′′rad,in ; and (c) Compute and plot q′′rad,in as a function of glass temperatures for the range 500 ≤ Tg ≤ 800 K for oven wall temperatures of Tw = 1500, 1800 and 2000 K. SCHEMATIC:

ASSUMPTIONS: (1) Glass is of uniform temperature, (2) Glass is diffuse, (3) Furnace walls large compared to the glass; εw plays no role, (4) Negligible convection. ANALYSIS: (a) From knowledge of the spectral transmittance,τw, and spectral reflectivity, ρ λ , the following radiation properties are evaluated: Total transmissivity, r: For the irradiation from the furnace walls, Gλ = Eλ,b (λ, Tw ). Hence ∞

4 τ = ∫ τ λ E λ ,b ( λ , Tw ) dλ σ Tw ≈ τ λ1F(0 − λ T ) = 0.9 × 0.25 = 0.225 . 0

<

where λT = 1.6 µm × 1800 K = 2880 µm⋅K ≈ 2898 µm⋅K giving F(0-λT) ≈ 0.25. Total reflectivity, ρ: With Gλ = Eλ,b (λ,Tw), Tw = 1800 K, and F0 − λT = 0.25,

(

)

ρ ≈ ρλ1F(0 − λ T ) + ρλ 2 1 − F( 0 − λ T ) = 0.05 × 0.25 + 0.5 (1 − 0.25 ) = 0.388

<

Total absorptivity, α : To perform the energy balance later, we’ll need α. Employ the conservation expression, α = 1 − ρ − τ = 1 − 0.388 − 0.225 = 0.387 . Emissivity, ε: Based upon surface temperature Tg = 750 K, for F0 − λ T ≈ 0.002 . λ T = 1.6 µ m × 750K = 1200 µ m ⋅ K, Hence for λ > 1.6 µm,

ε ≈ ελ ≈ 0.5.

<

(b) Performing an energy balance on the glass, the net radiative heat flux by radiation into the glass is, Continued...

PROBLEM 12.70 (Cont.) ′′ − E′′out q′′net,in = Ein

(

( ))

q′′net,in = 2 α G − ε E b Tg 4 where G = σ Tw

4 4 q′′net,in = 2 0.387σ (1800K ) − 0.5σ ( 750K ) 





q′′net,in = 442.8 kW m 2 . (b) Using the foregoing equations in the IHT Workspace along with the IHT Radiation Tool, Band Emission Factor, the net radiative heat flux, q′′rad,in , was computed and plotted as a function of Tg for selected wall temperatures Tw . 700

q''radin (kW/m^2)

600

500

400

300

200 500

600

700

800

Glass temperature, Tg (K) Tw = 1500 K Tw = 1800 K Tw = 2000 K

As the glass temperature increases, the rate of emission increases so we’d expect the net radiative heat rate into the glass to decrease. Note that the decrease is not very significant. The effect of increased wall temperature is to increase the irradiation and, hence the absorbed irradiation to the surface and the net radiative flux increase.

PROBLEM 12.71 KNOWN: Temperature, absorptivity, transmissivity, radiosity and convection conditions for a semitransparent plate. FIND: Plate irradiation and total hemispherical emissivity. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform surface conditions. ANALYSIS: From an energy balance on the plate

E& in = E& out 2G = 2q ′′conv + 2J. Solving for the irradiation and substituting numerical values,

G = 40 W / m 2 ⋅ K ( 350 − 300 ) K + 5000 W / m 2 = 7000 W / m 2 .

<

From the definition of J,

J = E + ρG + τ G = E + (1 −α ) G. Solving for the emissivity and substituting numerical values,

ε=

J − (1 − α ) G σT

4

5000 W / m 2 ) − 0.6 ( 7000 W / m 2 ) ( = = 0.94. 4 5.67 ×10 −8 W / m 2 ⋅ K 4 ( 350K)

<

Hence,

α ≠ε and the surface is not gray for the prescribed conditions. COMMENTS: The emissivity may also be determined by expressing the plate energy balance as

2α G = 2q ′′conv + 2E. Hence

ε σ T 4 = αG − h ( T − T∞ ) ε=

(

)

0.4 7000 W / m 2 − 40 W / m 2 ⋅ K ( 50 K) 4 5.67 ×10 −8 W / m 2 ⋅ K 4 ( 350 K )

= 0.94.

PROBLEM 12.72 KNOWN: Material with prescribed radiative properties covering the peep hole of a furnace and exposed to surroundings on the outer surface. FIND: Steady-state temperature of the cover, Ts; heat loss from furnace. SCHEMATIC:

ASSUMPTIONS: (1) Cover is isothermal, no gradient, (2) Surroundings of the outer surface are large compared to cover, (3) Cover is insulated from its mount on furnace wall, (4) Negligible convection on interior surface. PROPERTIES: Cover material (given): For irradiation from the furnace interior: τf = 0.8, ρ f = 0; For room temperature emission: τ = 0, ε = 0.8. ANALYSIS: Perform an energy balance identifying the modes of heat transfer,

E& in − E& out = 0 Recognize that

αf Gf + αsur Gsur − 2 ε E b ( Ts ) − h ( Ts − T∞ ) = 0.

(1)

Gf = σ Tf4

(2,3)

4 . Gsur = σ Tsur

α f = 1 − τ f − ρ f = 1 − 0.8 − 0.0 = 0.2.

From Eq. 12.57, it follows that

(4)

Since the irradiation Gsur will have nearly the same spectral distribution as the emissive power of the cover, Eb (Ts), and since Gsur is diffuse irradiation,

αsur = ε = 0.8.

(5)

This reasoning follows from Eqs. 12.65 and 12.66. Substituting Eqs. (2-5) into Eq. (1) and using numerical values,

0.2 × 5.67 ×10−8 ( 450 + 273 ) W / m 2 + 0.8 × 5.67 ×10−8 ×300 4 W / m 2 4

−2 × 0.8 × 5.67 ×10 −8 Ts4 W / m 2 − 50 W / m 2 ⋅ K ( Ts − 300 ) K = 0 9.072 × 10−8 Ts4 + 50Ts = 18,466

or

(2-5)

<

Ts = 344K.

The heat loss from the furnace (see energy balance schematic) is

q f,loss = As αf G f +τ fG f − ε E b ( Ts )  =

π D2 (α f + τf ) G f − ε E b ( Ts )  4 

q f,loss = π ( 0.050m )2 / 4 ( 0.8 + 0.2 )( 723K ) 4  −0.8 ( 344K )4  5.67 × 10−8 W / m 2 ⋅ K4 = 29.2 W. 

<

PROBLEM 12.73 KNOWN: Window with prescribed τλ and ρ λ mounted on cooled vacuum chamber passing radiation from a solar simulator. FIND: (a) Solar transmissivity of the window material, (b) State-state temperature reached by window with simulator operating, (c) Net radiation heat transfer to chamber. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Diffuse behavior of window material, (3) Chamber and room surroundings large compared to window, (4) Solar simulator flux has spectral distribution of 5800K blackbody, (5) Window insulated from its mount, (6) Window is isothermal at Tw. ANALYSIS: (a) Using Eq. 12.55 and recognizing that Gλ,S ~ Eb,λ (λ, 5800K), 1.9 τ S = τ1 E ( λ,5800K ) dλ / E b ( 5800K ) = τ 1  F(0→1.9µm ) − F( 0→0.38µm )  . 0.38 λ ,b

∫

From Table 12.1 at λT = 1.9 × 5800 = 11,020 µm⋅K, F(0 → λ) = 0.932; at λT = 0.38 × 5800 µm⋅K = 2,204 µm⋅K, F(0 → λ) = 0.101; hence

τ S = 0.90 [ 0.932 − 0.101] = 0.748.

<

Recognizing that later we’ll need α S, use Eq. 12.52 to find ρ S

ρS = ρ1 F( 0→0.38µ m) + ρ 2 F( 0→1.9 µ m) − F( 0→0.38µ m)  + ρ 3 1 − F( 0→1.9µ m)      ρS = 0.15 × 0.101 + 0.05[ 0.932 − 0.101] + 0.15[1 −0.932 ] = 0.067 αS = 1 − ρS − τS = 1 − 0.067 − 0.748 = 0.185. (b) Perform an energy balance on the window.

αSGS − q ′′w − c − q ′′w −sur − q ′′conv = 0

(

) (

)

4 − T 4 − εσ T4 − T4 −h T − T = 0. αSGS − εσ Tw ( w ∞) c w sur Recognize that ρ λ (λ > 1.9) = 0.15 and that ε ≈ 1 – 0.15 = 0.85 since Tw will be near 300K. Substituting numerical values, find by trial and error, 4 − 2984 − 774  k 4 − 28 W / m2 ⋅ K T − 298 K = 0 0.185 × 3000 W / m 2 − 0.85 × σ 2Tw ( w )  

Tw = 302.6K = 29.6°C.

<

(c) The net radiation transfer per unit area of the window to the vacuum chamber, excluding the transmitted simulated solar flux is

(

)

4 q ′′w − c = εσ Tw − Tc4 = 0.85 × 5.67 ×10−8 W / m2 ⋅ K4 302.64 − 77 4  K4 = 402 W / m2 .  

<

PROBLEM 12.74 KNOWN: Reading and emissivity of a thermocouple (TC) located in a large duct to measure gas stream temperature. Duct wall temperature and emissivity; convection coefficient. FIND: (a) Gas temperature, T∞ , (b) Effect of convection coefficient on measurement error. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss from TC sensing junction to support, (3) Duct wall much larger than TC, (4) TC surface is diffuse-gray. ANALYSIS: (a) Performing an energy balance on the thermocouple, it follows that q′′w − s − q′′conv = 0 . where radiation exchange between the duct wall and the TC is given by Eq. 1.7. Hence, 4 ε sσ (Tw − Ts4 ) − h (Ts − T∞ ) = 0 . Solving for T∞ with Ts = 180oC, εσ 4 T∞ = Ts − s (Tw − Ts4 ) h

T∞ = (180 + 273)K −

0.6(5.67 × 10−8 W m 2⋅ K 4 ) 2

125 W m ⋅ K

([450 + 273]4 − [180 + 273]4 ) K4 <

T∞ = 453 K − 62.9 K = 390 K = 117$ C . (b) Using the IHT First Law model for an Isothermal Solid Sphere to solve the foregoing energy balance for Ts, with T∞ = 125oC, the measurement error, defined as ∆T = Ts − T∞ , was determined and is plotted as a function of h . 300

Measurement error, delta(C)

250

200

150

100

50

0 0

200

400

600

800

1000

Convection coefficient, hbar(W/m^2.K)

The measurement error is enormous (∆T ≈ 270oC) for h = 10 W/m2⋅K, but decreases with increasing h . However, even for h = 1000 W/m2⋅K, the error (∆T ≈ 8°C) is not negligible. Such errors must always be considered when measuring a gas temperature in surroundings whose temperature differs significantly from that of the gas. Continued...

PROBLEM 12.74 (Cont.) COMMENTS: (1) Because the duct wall surface area is much larger than that of the thermocouple, its emissivity is not a factor. (2) For such a situation, a shield about the thermocouple would reduce the influence of the hot duct wall on the indicated TC temperature. A low emissivity thermocouple coating would also help.

PROBLEM 12.75 KNOWN: Diameter and emissivity of a horizontal thermocouple (TC) sheath located in a large room. Air and wall temperatures. FIND: (a) Temperature indicated by the TC, (b) Effect of emissivity on measurement error. SCHEMATIC:

ASSUMPTIONS: (1) Room walls approximate isothermal, large surroundings, (2) Room air is quiescent, (3) TC approximates horizontal cylinder, (4) No conduction losses, (5) TC surface is opaque, diffuse and gray. PROPERTIES: Table A-4, Air (assume Ts = 25 oC, Tf = (Ts + T∞)/2 ≈ 296 K, 1 atm):

ν = 15.53 × 10 −6 m 2 s, k = 0.026 W m ⋅ K, α = 22.0 × 10−6 m 2 s, Pr = 0.708, β = 1 Tf .

ANALYSIS: (a) Perform an energy balance on the thermocouple considering convection and radiation processes. On a unit area basis, with q′′conv = h(Ts − T∞ ), E in − E out = 0 (1) α G − ε E b (Ts ) − h(Ts − T∞ ) = 0 . Since the surroundings are isothermal and large compared to the thermocouple, G = Eb(Tsur). For the gray-diffuse surface, α = ε. Using the Stefan-Boltzman law, Eb = σT4, Eq. (1) becomes 4 εσ (Tsur − Ts4 ) − h(Ts − T∞ ) = 0 . Using the Churchill-Chu correlation for a horizontal cylinder, estimate h due to free convection.

(2)

2

  6  hD  0.387Ra1/ gβ∆TD3  D = 0.60 + = . Nu D = , Ra D 8 / 27  k να 9 /16   1 + ( 0.559 Pr )     

(3,4)

To evaluate RaD and Nu D , assume Ts = 25oC, giving Ra D =

9.8 m s2 (1 296 K)(25 − 20)K(0.004m)3 15.53 × 10−6 m 2 s × 22.0 × 10−6 m 2 s

= 31.0 2

  1/ 6  0.026 W m ⋅ K  0.387(31.0)  2 h= 0.60 +  = 8.89 W m ⋅ K . 8 / 27 0.004m   1 + ( 0.559 0.708 )9 /16     

(5)

0.4 × 5.67 × 10−8 W m 2 ⋅ K 4 [(30 + 273)4 − Ts4 ]K 4 − 8.89 W m 2 ⋅ K[Ts − (20 + 273)]K = 0 where all temperatures are in kelvin units. By trial-and-error, find

(6)

With ε = 0.4, the energy balance, Eq. (2), becomes

Ts ≈ 22.2oC

< Continued...

PROBLEM 12.75 (Cont.) (b) The thermocouple measurement error is defined as ∆T =Ts − T∞ and is a consequence of radiation exchange with the surroundings. Using the IHT First Law Model for an Isothermal Solid Cylinder with the appropriate Correlations and Properties Toolpads to solve the foregoing energy balance for Ts, the measurement error was determined as a function of the emissivity.

Measurement error, delta(C)

5

4

3

2

1

0 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Emissivity, eps

The measurement error decreases with decreasing ε, and hence a reduction in net radiation transfer from the surroundings. However, even for ε = 0.1, the error (∆T ≈ 1oC) is not negligible.

PROBLEM 12.76 KNOWN: Temperature sensor imbedded in a diffuse, gray tube of emissivity 0.8 positioned within a room with walls and ambient air at 30 and 20 oC, respectively. Convection coefficient is 5 W m 2⋅ K . FIND: (a) Temperature of sensor for prescribed conditions, (b) Effect of surface emissivity and using a fan to induce air flow over the tube. SCHEMATIC:

ASSUMPTIONS: (1) Room walls (surroundings) much larger than tube, (2) Tube is diffuse, gray surface, (3) No losses from tube by conduction, (4) Steady-state conditions, (5) Sensor measures temperature of tube surface. ANALYSIS: (a) Performing an energy balance on the tube, E in − E out = 0 . Hence, q′′rad − q′′conv = 0 , 4 or ε tσ (Tw − Tt4 ) − h(Tt − Tw ) = 0 . With h = 5 W m 2⋅ K and εt = 0.8, the energy balance becomes

0.8 × 5.67 × 10−8 W m 2 ⋅ K 4 (30 + 273) − Tt4  K 4 = 5 W m 2 ⋅ K [Tt − (20 + 273)] K   4

4.5360 × 10−8 3034 − Tt4  = 5 [Tt − 293]





<

which yields Tt = 298 K = 25oC. (b) Using the IHT First Law Model, the following results were determined.

Sensor temperature, Tt(C)

30

28

26

24

22

20 0

5

10

15

20

25

Convection coefficient, h(W/m^2.K) epst = 0.8 epst = 0.5 epst = 0.2

The sensor temperature exceeds the air temperature due to radiation absorption, which must be balanced by convection heat transfer. Hence, the excess temperature Tt − T∞ , may be reduced by increasing h or by decreasing αt, which equals εt for a diffuse-gray surface, and hence the absorbed radiation.

COMMENTS: A fan will increase the air velocity over the sensor and thereby increase the convection heat transfer coefficient. Hence, the sensor will indicate a temperature closer to T∞

PROBLEM 12.77 KNOWN: Diffuse-gray sphere is placed in large oven with known wall temperature and experiences convection process. FIND: (a) Net heat transfer rate to the sphere when its temperature is 300 K, (b) Steady-state temperature of the sphere, (c) Time required for the sphere, initially at 300 K, to come within 20 K of the steady-state temperature, and (d) Elapsed time of part (c) as a function of the convection coefficient for 10 ≤ h ≤ 25 W/ m2⋅K for emissivities 0.2, 0.4 and 0.8. SCHEMATIC:

ASSUMPTIONS: (1) Sphere surface is diffuse-gray, (2) Sphere area is much smaller than the oven wall area, (3) Sphere surface is isothermal. PROPERTIES: Sphere (Given) : α = 7.25 × 10-5 m2 /s, k = 185 W/m⋅K. ANALYSIS: (a) From an energy balance on the sphere find q net = qin − q out q net = α GAs + q conv − EAs q net = ασ To4 As + hAs ( T∞ − Ts ) − εσ Ts4 As .

(1)

Note that the irradiation to the sphere is the emissive power of a blackbody at the temperature of the oven walls. This follows since the oven walls are isothermal and have a much larger area than the sphere area. Substituting numerical values, noting that α = ε since the surface is diffuse-gray and that As = πD2 , find



q net = 0.8 × 5.67 × 10−8 W m 2 ⋅ K



4

(600K )4

+ 15 W m 2⋅ K × ( 400 − 300 ) K



− 0.8 × 5.67 × 10 −8 W m 2 ⋅ K 300 K 4

#

4

π 30 × 10 −3 m

q net = [16.6 + 4.2 − 1.0] W = 19.8 W .

%

2

(1)

<

(b) For steady-state conditions, qnet in the energy balance of Eq. (1) will be zero, 4 0 = ασ To4 As + hAs ( T∞ − Tss ) − εσ Tss As

(2)

Substitute numerical values and find the steady-state temperature as Tss = 538.2K

< Continued...

PROBLEM 12.77 (Cont.) (c) Using the IHT Lumped Capacitance Model considering convection and radiation processes, the temperature- time history of the sphere, initially at Ts (0) = Ti = 300 K, can be determined. The elapsed time required to reach Ts ( t o ) = (538.2 − 20 ) K = 518.2K was found as

<

t o = 855s = 14.3 min

(d) Using the IHT model of part (c), the elapsed time for the sphere to reach within 20 K of its steadystate temperature, tf , as a function of the convection coefficient for selected emissivities is plotted below. Time-to-reach within 20 K of steady-state temperature 2500

Time, tf (s)

2000

1500

1000

500 10

15

20

25

Convection coefficient, h (W/m^2.K) eps = 0.2 eps = 0.4 eps = 0.8

For a fixed convection coefficient, tf increases with decreasing ε since the radiant heat transfer into the sphere decreases with decreasing emissivity. For a given emissivity, the tf decreases with increasing h since the convection heat rate increases with increasing h. However, the effect is much more significant with lower values of emissivity. COMMENTS: (1) Why is tf more strongly dependent on h for a lower sphere emissivity? Hint: Compare the relative heat rates by convection and radiation processes. (2) The steady-state temperature, Tss , as a function of the convection coefficient for selected emmissivities calculated using (2) is plotted below. Are these results consistent with the above plot of tf vs h ? Steady-state temperature, Tss (K)

600

550

500

450

400 10

15

20

Convection coefficient, h (W/m^2.K) eps = 0.2 eps = 0.4 eps = 0.8

25

PROBLEM 12.78 2

KNOWN: Thermograph with spectral response in 9 to 12 µm region views a target of area 200mm with solid angle 0.001 sr in a normal direction. FIND: (a) For a black surface at 60°C, the emissive power in 9 – 12 µm spectral band, (b) Radiant power (W), received by thermograph when viewing black target at 60°C, (c) Radiant power (W) received by thermograph when viewing a gray, diffuse target having ε = 0.7 and considering the surroundings at Tsur = 23°C. SCHEMATIC:

ASSUMPTIONS: (1) Wall is diffuse, (2) Surroundings are black with Tsur = 23°C. ANALYSIS: (a) Emissive power in spectral range 9 to 12 µm for a 60°C black surface is

E t ≡ E b (9 − 12 µ m ) = E b  F ( 0 → 12 µ m ) − F ( 0 − 9 µ m ) where E b ( Ts ) = σ Ts4 . From Table 12.1:

λ2 Ts = 12 × ( 60 + 273 ) ≈ 4000 µ m K,

F ( 0− 12 µ m ) = 0.491

λ1 Ts = 9 × ( 60 + 273) ≈ 3000 µ m K,

F ( 0 − 9 µ m ) = 0.273.

Hence

E t = 5.667 ×10−8 W / m 2 ⋅ K4 × ( 60 + 273 ) K 4 [ 0.491 − 0.273 ] = 144.9 W / m 2 . 4

<

(b) The radiant power, qb (w), received by the thermograph from a black target is determined as

E q b = t ⋅ A s cosθ1 ⋅ω π Et = emissive power in 9 – 12 µm spectral region, part (a) result 2 -4 2 As = target area viewed by thermograph, 200mm (2 × 10 m ) ω = solid angle thermograph aperture subtends when viewed from the target, 0.001 sr θ = angle between target area normal and view direction, 0°.

where

Hence,

qb =

(

)

144.9 W / m2 × 2 ×10−4 m 2 × cos0 °× 0.001sr = 9.23 µ W. π sr

< Continued …..

PROBLEM 12.78 (Cont.) (c) When the target is a gray, diffuse emitter, ε = 0.7, the thermograph will receive emitted power from the target and reflected irradiation resulting from the surroundings at Tsur = 23°C. Schematically:

The power is expressed as

q = q e + q r = ε q b + I r ⋅ As cosθ1 ⋅ω  F( 0→12 µ m) − F( 0→9 µm )    where qb = radiant power from black surface, part (b) result F(0 - λ) = band emission fraction for Tsur = 23°C; using Table 12.1 λ2 Tsur = 12 × (23 + 273) = 3552 µm⋅K, F( 0 − λ ) = 0.394 2

λ1 Tsur = 9 × (23 + 273) = 2664 µm⋅K, F( 0 − λ ) = 0.197 1 Ir = reflected intensity, which because of diffuse nature of surface E (T ) G I r = ρ = (1 − ε ) b sur . π π Hence

5.667 ×10−8 W / m2 ⋅ K4 × ( 273 + 23) 4 K q = 0.7 × 9.23 µ W + (1 − 0.7 ) π sr

(

)

× 2 × 10−4 m 2 × cos0°× 0.001sr [ 0.394 − 0.197] q = 6.46 µ W + 1.64 µW = 8.10 µW.

<

COMMENTS: (1) Comparing the results of parts (a) and (b), note that the power to the thermograph is slightly less for the gray surface with ε = 0.7. From part (b) see that the effect of the irradiation is substantial; that is, 1.64/8.10 ≈ 20% of the power received by the thermograph is due to reflected irradiation. Ignoring such effects leads to misinterpretation of temperature measurements using thermography. (2) Many thermography devices have a spectral response in the 3 to 5 µm wavelength region as well as 9 – 12 µm.

PROBLEM 12.79 KNOWN: Radiation thermometer (RT) viewing a steel billet being heated in a furnace. FIND: Temperature of the billet when the RT indicates 1160K. SCHEMATIC:

ASSUMPTIONS: (1) Billet is diffuse-gray, (2) Billet is small object in large enclosure, (3) Furnace behaves as isothermal, large enclosure, (4) RT is a radiometer sensitive to total (rather than a prescribed spectral band) radiation and is calibrated to correctly indicate the temperature of a black body, (5) RT receives radiant power originating from the target area on the billet. ANALYSIS: The radiant power reaching the radiation thermometer (RT) is proportional to the radiosity of the billet. For the diffuse-gray billet within the large enclosure (furnace), the radiosity is

J = ε E b ( T ) + ρ G = ε E b ( T ) + ( 1− ε ) E b ( Tw ) 4 J = ε σ T 4 + (1 − ε ) σ Tw

(1) 4

where α = ε, G = Eb (Tw) and Eb = σ T . When viewing the billet, the RT indicates Ta = 1100K, referred to as the apparent temperature of the billet. That is, the RT indicates the billet is a blackbody at Ta for which the radiosity will be

E b ( Ta ) = J a = σ Ta4.

(2)

Recognizing that Ja = J, set Eqs. (1) and (2) equal to one another and solve for T, the billet true temperature. 1/4

1−ε 4  1 T =  Ta4 − Tw  ε ε 

.

Substituting numerical values, find 1/4 1 − 0.8  1 4 T =  (1160K ) 4 − 1500K ( )  = 999K. 0.8  0.8 

<

COMMENTS: (1) The effect of the reflected wall irradiation from the billet is to cause the RT to indicate a temperature higher than the true temperature. (2) What temperature would the RT indicate when viewing the furnace wall assuming the wall emissivity were 0.85? (3) What temperature would the RT indicate if the RT were sensitive to spectral radiation at 0.65 µm instead of total radiation? Hint: in Eqs. (1) and (2) replace the emissive power terms with spectral intensity. Answer: 1365K.

PROBLEM 12.80 KNOWN: Irradiation and temperature of a small surface. FIND: Rate at which radiation is received by a detector due to emission and reflection from the surface. SCHEMATIC:

ASSUMPTIONS: (1) Opaque, diffuse-gray surface behavior, (2) As and Ad may be approximated as differential areas. ANALYSIS: Radiation intercepted by the detector is due to emission and reflection from the surface, and from the definition of the intensity, it may be expressed as

qs −d = Ie + rA s cosθ ∆ω . The solid angle intercepted by Ad with respect to a point on As is

∆ω =

Ad r2

= 10− 6 sr.

Since the surface is diffuse it follows from Eq. 12.24 that

I e+ r =

J π

where, since the surface is opaque and gray (ε = α = 1 - ρ),

J = E + ρ G = ε Eb + (1 − ε ) G. Substituting for Eb from Eq. 12.28

J = ε σ Ts4 + (1 − ε ) G = 0.7 × 5.67 ×10 −8

W m2 ⋅ K

500K )4 + 0.3 ×1500 ( 4

W m2

or

J = ( 2481 + 450 ) W / m 2 = 2931W/m 2. Hence

I e+ r = and

2 9 3 1 W / m2 = 933W/m 2 ⋅sr π sr

(

)

q s− d = 933 W / m 2 ⋅ sr 10 −4 m2 × 0.866 10 −6 sr = 8.08 × 10 −8 W.

<

PROBLEM 12.81 KNOWN: Small, diffuse, gray block with ε = 0.92 at 35°C is located within a large oven whose walls are at 175°C with ε = 0.85. FIND: Radiant power reaching detector when viewing (a) a deep hole in the block and (b) an area on the block’s surface. SCHEMATIC:

ASSUMPTIONS: (1) Block is isothermal, diffuse, gray and small compared to the enclosure, (2) Oven is isothermal enclosure. ANALYSIS: (a) The small, deep hole in the isothermal block approximates a blackbody at Ts. The radiant power to the detector can be determined from Eq. 12.54 written in the form: σ Ts4 q = Ie ⋅At ⋅ωt = ⋅ At ⋅ ω t

π

1  4 W q= 5.67 ×10 −8 × (35 + 273 )  × π sr   m 2

(

π 3 ×10−3

)

2

m2

4

× 0.001sr = 1.15 µ W

<

where A t = π D 2t /4. Note that the hole diameter must be greater than 3mm diameter. (b) When the detector views an area on the surface of the block, the radiant power reaching the detector will be due to emission and reflected irradiation originating from the enclosure walls. In terms of the radiosity, Section 12.24, we can write using Eq. 12.24,

q = Ie+ r ⋅ At ⋅ ωt =

J ⋅ A t ⋅ ω t. π

Since the surface is diffuse and gray, the radiosity can be expressed as

J = ε E b ( Ts ) + ρ G = ε E b ( Ts ) + (1 − ε ) E b ( Tsur ) recognizing that ρ = 1 - ε and G = Eb (Tsur). The radiant power is

q= q=

1 ε E b ( Ts ) + (1 − ε ) E b ( Tsur )  ⋅ At ⋅ ωt π

1  0.92 × 5.67 ×10−8 ( 35 + 273 )4 + (1 − 0.92 ) × 5.67 ×10 −8 (175 + 273)4  W / m 2 ×   π sr 

(

π 3 ×10 −3 4

) m 2 × 0.001sr = 1.47 µ W. 2

<

COMMENTS: The effect of reflected irradiation when ε < 1 is important for objects in enclosures. The practical application is one of measuring temperature by radiation from objects within furnaces.

PROBLEM 12.82 KNOWN: Diffuse, gray opaque disk (1) coaxial with a ring-shaped disk (2), both with prescribed temperatures and emissivities. Cooled detector disk (3), also coaxially positioned at a prescribed location. FIND: Rate at which radiation is incident on the detector due to emission and reflection from A1. SCHEMATIC:

2

ASSUMPTIONS: (1) A1 is diffuse-gray, (2) A2 is black, (3) A1 and A3 As, (3) Detector field of view is limited to sample surface. ANALYSIS: (a) The irradiation can be evaluated as Gd = qs-d/Ad and qs-d = Is(e+r) As ωd-s. 2 2 2 -5 Evaluating parameters: ωd-s ≈ Ad/L = 2 mm /(300 mm) = 2.22 × 10 sr, find 4 −8 2 4 Es ε sσ Ts4 0.1 5.67 ×10 W / m ⋅ K ( 400 K ) I s( e ) = = = = 46.2 W / m 2 ⋅ sr

π

π

(

)

π sr

(

)

4 −8 2 4 4 ρs Gs (1 − ε s ) σ Th 0.9 5.67 × 10 W / m ⋅ K ( 273 K ) I s( r ) = = = = 90.2 W / m 2 ⋅ sr π π π sr

(

)

qs −d = ( 46.2 + 90.2 ) W / m 2 ⋅ sr 5× 10−6 m 2 × 2.22 ×10 −5 sr = 1.51× 10−8 W Gd = 1.51×10 −8 W / 2 ×10 −6 m 2 = 7.57 ×10 −3 W / m 2.

<

(b) Since λmax T = 2898 µm⋅K, it follows that λmax(e) = 2898 µm⋅K/400 K = 7.25 µm and λmax(r) = 2898 µm⋅K/273 K = 10.62 µm.

< -4

-8

5

2

λ = 7.25 µm: Table 12.1 → Iλ,b(400 K) = 0.722 × 10 (5.67 × 10 )(400) = 41.9 W/m ⋅µm⋅sr -4

-8

5

2

Iλ,b(273 K) = 0.48 × 10 (5.67 × 10 )(273) = 4.1 W/m ⋅µm⋅sr 2

Iλ = Iλ,e + Iλ,r = ε sIλ,b(400 K) + ρIλ,b(273K) = 0.1 × 41.9 + 0.9 × 4.1 = 7.90 W/m ⋅µm⋅sr -4

-8

5

<

2

λ = 10.62 µm: Table 12.1 → Iλ,b(400 K) = 0.53 × 10 (5.67 × 10 )(400) = 30.9 W/m ⋅µm⋅sr -4

-8

5

2

Iλ,b(273 K) = 0.722 × 10 (5.67 × 10 )(273) = 6.2 W/m ⋅µm⋅sr 2

Iλ = 0.1 × 30.9 + 0.9 × 6.2 = 8.68 W/m ⋅µm⋅sr.

<

COMMENTS: Although Th is substantially smaller than Ts, the high sample reflectivity renders the reflected component of Js comparable to the emitted component.

PROBLEM 12.84 KNOWN: Sample at Ts = 700 K with ring-shaped cold shield viewed normally by a radiation detector. FIND: (a) Shield temperature, Tsh, required so that its emitted radiation is 1% of the total radiant power received by the detector, and (b) Compute and plot Tsh as a function of the sample emissivity for the range 0.05 ≤ ε ≤ 0.35 subject to the parametric constraint that the radiation emitted from the cold shield is 0.05, 1 or 1.5% of the total radiation received by the detector. SCHEMATIC:

ASSUMPTIONS: (1) Sample is diffuse and gray, (2) Cold shield is black, and (3) A d , Ds2 , D 2t A2,a. (c) Using the foregoing equations in the IHT workspace, the variation of the view factors F21 and F22 with L were calculated and are graphed below. Right-circular cone and disk

1

1

0 .8

0.6

0 .6

Fij

Fij

0.8

R ig h t-circu la r cylin d e r a n d d is k , L o = D = 5 0 m m

0.4

0 .4

0.2

0 .2 0

0 0

40

80

120

160

200

0

40

Cone height, L(mm) F21 F22

80

120

160

C o n e h e ig h t, L (m m ) F2 1 F2 2

Note that for both configurations, when L = 0, find that F21 = F13 = 0.1716, the value obtained for coaxial parallel disks. As L increases, find that F22 → 1; that is, the interior of both the cone and cylinder see mostly each other. Notice that the changes in both F21 and F22 with increasing L are greater for the disk-cylinder; F21 decreases while F22 increases. COMMENTS: From the results of part (b), why isn’t the sum of F21 and F22 equal to unity?

200

PROBLEM 13.5 KNOWN: Two parallel, coaxial, ring-shaped disks. FIND: Show that the view factor F12 can be expressed as

F12 =

J 1 6 1 61 6

1 6

4 1 6

1 A 1,3 F 1,3 2,4 − A 3 F3 2,4 − A 4 F4 1,3 − F43 A1

9L

where all the Fig on the right-hand side of the equation can be evaluated from Figure 13.5 (see Table 13.2) for coaxial parallel disks. SCHEMATIC:

ASSUMPTIONS: Diffuse surfaces with uniform radiosities. ANALYSIS: Using the additive rule, Eq. 13.5, where the parenthesis denote a composite surface,

1 6

F1 2,4 = F12 + F14

1 6

F12 = F1 2,4 − F14

(1)

Relation for F1(2,4): Using the additive rule

1 6 1 61 6

1 6

1 6

A 1,3 F 1,3 2,4 = A1 F1 2,4 + A 3 F3 2,4

(2)

where the check mark denotes a Fij that can be evaluated using Fig. 13.5 for coaxial parallel disks. Relation for F14: Apply reciprocity

A1 F14 = A 4 F41

(3)

and using the additive rule involving F41,

1 6

A1 F14 = A 4 F4 1,3 − F43

(4)

Relation for F12: Substituting Eqs. (2) and (4) into Eq. (1),

F12 =

J 1 6 1 61 6

1 6

4 1 6

1 A 1,3 F 1,3 2,4 − A 3 F3 2,4 − A 4 F4 1,3 − F43 A1

9L

COMMENTS: (1) The Fij on the right-hand side can be evaluated using Fig. 13.5. (2) To check the validity of the result, substitute numerical values and test the behavior at special limits. For example, as A3, A4 → 0, the expression reduces to the identity F12 ≡ F12.

<

PROBLEM 13.6 KNOWN: Long concentric cylinders with diameters D1 and D2 and surface areas A1 and A2. FIND: (a) The view factor F12 and (b) Expressions for the view factors F22 and F21 in terms of the cylinder diameters. SCHEMATIC:

ASSUMPTIONS: (1) Diffuse surfaces with uniform radiosities and (2) Cylinders are infinitely long such that A1 and A2 form an enclosure. ANALYSIS: (a) View factor F12. Since the infinitely long cylinders form an enclosure with surfaces A1 and A2, from the summation rule on A1, Eq. 13.4,

F11 + F12 = 1

(1)

and since A1 doesn’t see itself, F11 = 0, giving

< (2)

F12 = 1 That is, the inner surface views only the outer surface.

(b) View factors F22 and F21. Applying reciprocity between A1 and A2, Eq. 13.3, and substituting from Eq. (2),

A1 F12 = A 2 F21 F21 =

(3)

π D1L A1 D F12 = ×1= 1 π D2 L A2 D2

< (4)

From the summation rule on A2, and substituting from Eq. (4),

F21 + F22 = 1 F22 = 1 − F21 = 1 −

D1 D2

<

PROBLEM 13.7 KNOWN: Right-circular cylinder of diameter D, length L and the areas A1, A2, and A3 representing the base, inner lateral and top surfaces, respectively. FIND: (a) Show that the view factor between the base of the cylinder and the inner lateral surface has the form

4 !

F12 = 2 H 1 + H 2

9

1/ 2

−H

"# $

where H = L/D, and (b) Show that the view factor for the inner lateral surface to itself has the form

4

F22 = 1 + H − 1 + H 2

9

1/ 2

SCHEMATIC:

ASSUMPTIONS: Diffuse surfaces with uniform radiosities. ANALYSIS: (a) Relation for F12, base-to-inner lateral surface. Apply the summation rule to A1, noting that F11 = 0

F11 + F12 + F13 = 1 F12 = 1 − F13

(1)

From Table 13.2, Fig. 13.5, with i = 1, j = 3,

F13 =

%K&  K' !

1

6 "#$

1/ 2 1 S − S2 − 4 D3 / D1 2 2

S = 1+

1 + R 23 R12

=

1 R2

(K) K*

(2)

+ 2 = 4 H2 + 2

(3)

where R1 = R3 = R = D/2L and H = L/D. Combining Eqs. (2) and (3) with Eq. (1), find after some manipulation Continued …..

PROBLEM 13.7 (Cont.)

%K 4 &K ! ' 1/ 2  " F12 = 2 H 41 + H 2 9 − H # ! $

9 "#$

1/ 2 2 1 2 2 F12 = 1 − 4 H + 2 − 4 H + 2 − 4 2

(K )K * (4)

(b) Relation for F22, inner lateral surface. Apply summation rule on A2, recognizing that F23 = F21,

F21 + F22 + F23 = 1

F22 = 1 − 2 F21

(5)

Apply reciprocity between A1 and A2,

1

6

F21 = A1 / A 2 F12

(6)

and substituting into Eq. (5), and using area expressions

F22 = 1 − 2

A1 D 1 F12 = 1 − 2 F12 = 1 − F12 A2 4L 2H

(7)

2

where A1 = πD /4 and A2 = πDL. Substituting from Eq. (4) for F12, find

F22 = 1 −

4 !

9

"# $

4

9

1/ 2 1/ 2 1 2 H 1 + H2 − H = 1+ H − 1 + H2 2H

<

PROBLEM 13.8 KNOWN: Arrangement of plane parallel rectangles. FIND: Show that the view factor between A1 and A2 can be expressed as

F12 =

 $  $ $

1 A 1,4 F 1,4 2,3 − A1 F13 − A 4 F42 2 A1

where all Fij on the right-hand side of the equation can be evaluated from Fig. 13.4 (see Table 13.2) for aligned parallel rectangles. SCHEMATIC:

ASSUMPTIONS: Diffuse surfaces with uniform radiosity. ANALYSIS: Using the additive rule where the parenthesis denote a composite surface, * + A F + A F + A F* A(1,4 ) F(*1,4 )( 2,3) = A1 F13 1 12 4 43 4 42

(1)

where the asterisk (*) denotes that the Fij can be evaluated using the relation of Figure 13.4. Now, find suitable relation for F43. By symmetry,

F43 = F21

(2)

and from reciprocity between A1 and A2,

F21 =

A1 F12 A2

(3)

Multiply Eq. (2) by A4 and substitute Eq. (3), with A4 = A2,

A 4 F43 = A 4 F21 = A 4

A1 F12 = A1 F12 A2

(4)

Substituting for A4 F43 from Eq. (4) into Eq. (1), and rearranging,

F12 =

1  * − A F*  A F* − A1 F13 4 42  2 A1  (1,4 ) (1,4 )( 2,3)

<

PROBLEM 13.9 KNOWN: Two perpendicular rectangles not having a common edge. FIND: (a) Shape factor, F12, and (b) Compute and plot F12 as a function of Zb for 0.05 ≤ Zb ≤ 0.4 m; compare results with the view factor obtained from the two-dimensional relation for perpendicular plates with a common edge, Table 13.1. SCHEMATIC:

ASSUMPTIONS: (1) All surfaces are diffuse, (2) Plane formed by A1 + A3 is perpendicular to plane of A2. ANALYSIS: (a) Introducing the hypothetical surface A3, we can write

F2(3,1) = F23 + F21.

(1)

Using Fig. 13.6, applicable to perpendicular rectangles with a common edge, find F23 = 0.19 :

with Y = 0.3, X = 0.5,

F2(3,1) = 0.25 :

Z = Za − Z b = 0.2, and

with Y = 0.3, X = 0.5, Za = 0.4, and

Y X

=

0.3 0.5

Y X

=

0.3 0.5

= 0.6,

= 0.6,

Z

=

0.2

X 0.5 Z 0.4 = = 0.8 X 0.5

= 0.4

Hence from Eq. (1)

F21 = F2(3.1) − F23 = 0.25 − 0.19 = 0.06 By reciprocity,

F12 =

A2 0.5 × 0.3m 2 F21 = × 0.06 = 0.09 2 A1 0.5 × 0.2 m

(2)

<

(b) Using the IHT Tool – View Factors for Perpendicular Rectangles with a Common Edge and Eqs. (1,2) above, F12 was computed as a function of Zb. Also shown on the plot below is the view factor F(3,1)2 for the limiting case Zb → Za.

PROBLEM 13.10 KNOWN: Arrangement of perpendicular surfaces without a common edge. FIND: (a) A relation for the view factor F14 and (b) The value of F14 for prescribed dimensions. SCHEMATIC:

ASSUMPTIONS: (1) Diffuse surfaces. ANALYSIS: (a) To determine F14, it is convenient to define the hypothetical surfaces A2 and A3. From Eq. 13.6, ( A1 + A2 ) F(1,2 )(3,4 ) = A1 F1(3,4 ) + A2 F2(3,4 ) where F(1,2)(3,4) and F2(3,4) may be obtained from Fig. 13.6. Substituting for A1 F1(3,4) from Eq. 13.5 and combining expressions, find A1 F1(3,4 ) = A1 F13 + A1 F14 F14 =

1  ( A1 + A 2 ) F(1,2 )(3,4 ) − A1 F13 − A 2 F2(3,4 )  . A1 

Substituting for A1 F13 from Eq. 13.6, which may be expressed as ( A1 + A2 ) F(1,2 )3 = A1 F13 + A2 F23 . The desired relation is then 1  F14 = ( A1 + A 2 ) F(1,2 )(3,4 ) + A 2 F23 − ( A1 + A 2 ) F(1,2 )3 − A 2 F2(3,4 )  . A1  (b) For the prescribed dimensions and using Fig. 13.6, find these view factors: L +L L +L Surfaces (1,2)(3,4) F(1,2 )(3,4 ) = 0.22 ( Y / X ) = 1 2 = 1, ( Z / X ) = 3 4 = 1.45, W W L L Surfaces 23 F23 = 0.28 ( Y / X ) = 2 = 0.5, ( Z / X ) = 3 = 1, W W L L +L Surfaces (1,2)3 F(1,2 )3 = 0.20 ( Y / X ) = 1 2 = 1, ( Z / X ) = 3 = 1, W W L +L L Surfaces 2(3,4) F2(3,4 ) = 0.31 ( Y / X ) = 2 = 0.5, ( Z / X ) = 3 4 = 1.5, W W Using the relation above, find 1 F14 = [( WL1 + WL2 ) 0.22 + ( WL2 ) 0.28 − ( WL1 + WL2 ) 0.20 − ( WL2 ) 0.31] ( WL1 )

<

F14 = [2 ( 0.22 ) + 1 ( 0.28 ) − 2 (0.20 ) − 1( 0.31)] = 0.01.

<

PROBLEM 13.11 KNOWN: Arrangements of rectangles. FIND: The shape factors, F12. SCHEMATIC:

ASSUMPTIONS: (1) Diffuse surface behavior. ANALYSIS: (a) Define the hypothetical surfaces shown in the sketch as A3 and A4. From the additive view factor rule, Eq. 13.6, we can write √

√

√

A(1,3 ) F(1,3 )( 2,4 ) = A1F12 + A1 F14 + A3 F32 + A3F34

(1)

Note carefully which factors can be evaluated from Fig. 13.6 for perpendicular rectangles with a common edge. (See √). It follows from symmetry that A1F12 = A 4 F43 . (2) Using reciprocity, A 4 F43 = A3F34, (3) then A F =A F . 1 12

3 34

Solving Eq. (1) for F12 and substituting Eq. (3) for A3F34, find that 1  A 1,3 F 1,3 2,4 − A1F14 − A3F32  . F12 =  2A1  ( ) ( )( )

(4)

Evaluate the view factors from Fig. 13.6: Y/X

Fij (1,3) (2,4)

6 9 6

14

6 6

32

3

Z/X

= 0.67

6

=1

6

=2

6

9

6

3

Fij

= 0.67

0.23

=1

0.20

=2

0.14

Substituting numerical values into Eq. (4) yields F12 =

( 6 × 9 ) m 2 × 0.23 − ( 6 × 6 ) m 2 × 0.20 − ( 6 × 3 ) m 2 × 0.14   2 × (6 × 6 ) m  1

2

<

F12 = 0.038. Continued …..

PROBLEM 13.11 (Cont.) (b) Define the hypothetical surface A3 and divide A2 into two sections, A2A and A2B. From the additive view factor rule, Eq. 13.6, we can write √

√

A1,3 F(1,3 )2 = A1F12 + A3F3( 2A ) + A3 F3( 2B ) .

(5)

Note that the view factors checked can be evaluated from Fig. 13.4 for aligned, parallel rectangles. To evaluate F3(2A), we first recognize a relationship involving F(24)1 will eventually be required. Using the additive rule again, √

A 2A F( 2A )(1,3 ) = A 2A F( 2A )1 + A 2A F( 2A )3 .

(6)

Note that from symmetry considerations, A 2A F( 2A )(1,3 ) = A1F12

(7)

and using reciprocity, Eq. 13.3, note that A 2A F2A3 = A3F3( 2A ).

(8)

Substituting for A3F3(2A) from Eq. (8), Eq. (5) becomes √

√

A(1,3 ) F(1,3 )2 = A1F12 + A 2A F( 2A )3 + A3 F3( 2B ) . Substituting for A2A F(2A)3 from Eq. (6) using also Eq. (7) for A2A F(2A)(1,3) find that √ √ √   A(1,3 ) F(1,3 )2 = A1F12 +  A1F12 − A 2A F( 2A )1  + A3 F3( 2B )  



(9)



and solving for F12, noting that A1 = A2A and A(1,3) = A2 F12 =

1 

√ √ √   A 2 F (1,3)2 + A 2A F ( 2A )1 − A3 F 3( 2B )  . 2A1  

(10)

Evaluate the view factors from Fig. 13.4: Fij (1,3) 2

X/L

Y/L

1

=1

1.5

= 1.5

0.25

=1

= 0.5 1 1 =1 1

0.11

1 1

(2A)1

1 1

3(2B)

1

=1

1 0.5

Fij

0.20

Substituting numerical values into Eq. (10) yields F12 =

(1.5 × 1.0 ) m 2 × 0.25 + ( 0.5 × 1) m 2 × 0.11 − (1 × 1) m 2 × 0.20   2 ( 0.5 × 1) m 

F12 = 0.23.

1

2

<

PROBLEM 13.12 KNOWN: Two geometrical arrangements: (a) parallel plates and (b) perpendicular plates with a common edge. FIND: View factors using “crossed-strings” method; compare with appropriate graphs and analytical expressions. SCHEMATIC:

(a) Parallel plates (b) Perpendicular plates with common edge ASSUMPTIONS: Plates infinite extent in direction normal to page. ANALYSIS: The “crossed-strings” method is applicable to surfaces of infinite extent in one direction having an obstructed view of one another. F12 = (1/ 2w1 )[( ac + bd ) − (ad + bc )]. (a) Parallel plates: From the schematic, the edge and diagonal distances are

(

2

2

ac = bd = w1 + L

)

1/ 2

bc = ad = L.

With w1 as the width of the plate, find F12 =

 2 2w1  1

(

2

2

w1 + L

)

1/ 2

 

1

− 2 (L) =

2×4

 2 m

(

2

2

4 +1

)

1/ 2

 

m − 2 (1 m ) = 0.781.

<

Using Fig. 13.4 with X/L = 4/1 = 4 and Y/L = ∞, find F12 ≈ 0.80. Also, using the first relation of Table 13.1,



(

Fij =   Wi + Wj



1/ 2

)2 + 4 

(

−  Wi − Wj

1/ 2 

)2 + 4 

 / 2 Wi 

where wi = wj = w1 and W = w/L = 4/1 = 4, find F12 =

1/ 2 1/ 2   2 2  ( 4 + 4 ) + 4  − ( 4 − 4 ) + 4   / 2 × 4 = 0.781.  

(b) Perpendicular plates with a common edge: From the schematic, the edge and diagonal distances are ac = w1

(

bd = L

2

2

ad = w1 + L

)

bc = 0.

With w1 as the width of the horizontal plates, find

 

 

(

2

)

2

)

F12 = (1 / 2w1 ) 2 ( w1 + L ) −  w1 + L F12 = (1 / 2 × 4 m )

  ( 4 + 1) m − 

2

(

2

1/ 2

1/ 2

4 +1

 

+ 0

 

m + 0  = 0.110.

From the third relation of Table 13.1, with wi = w1 = 4 m and wj = L = 1 m, find



(

)

(

Fij = 1 + w j / w i − 1 + w j / w i 

1/ 2 

)2 

/ 2     2 1/ 2  F12 = 1 + (1 / 4 ) − 1 + (1 / 4 )  / 2 = 0.110.    

<

PROBLEM 13.13 KNOWN: Parallel plates of infinite extent (1,2) having aligned opposite edges. FIND: View factor F12 by using (a) appropriate view factor relations and results for opposing parallel plates and (b) Hottel’s string method described in Problem 13.12 SCHEMATIC:

ASSUMPTIONS: (1) Parallel planes of infinite extent normal to page and (2) Diffuse surfaces with uniform radiosity. ANALYSIS: From symmetry consideration (F12 = F14) and Eq. 13.5, it follows that F12 = (1/ 2 )  F1( 2,3,4 ) − F13 





where A3 and A4 have been defined for convenience in the analysis. Each of these view factors can be evaluated by the first relation of Table 13.1 for parallel plates with midlines connected perpendicularly. W1 = w1 / L = 2

F13:

W2 = w 2 / L = 2

1/ 2

( W1 + W2 )2 + 4    F13 = F1(2,3,4):

1/ 2

− ( W2 − W1 ) + 4  2





2W1

W1 = w1 / L = 2

Hence, find

1/ 2

− ( 2 − 2 ) + 4 



2× 2

2



= 0.618

W( 2,3,4 ) = 3w 2 / L = 6 1/ 2

F1( 2,3,4 ) =

1/ 2

( 2 + 2 )2 + 4    =

( 2 + 6 )2 + 4   

1/ 2

− ( 6 − 2 ) + 4 



2× 2

2



= 0.944.

F12 = (1/ 2 ) [0.944 − 0.618] = 0.163.

<

(b) Using Hottel’s string method, F12 = (1/ 2w1 )[( ac + bd ) − (ad + bc )]

(

ac = 1 + 42 bd = 1

(

)

1/ 2

ad = 12 + 22

)

= 4.123

1/ 2

= 2.236

bc = ad = 2.236 and substituting numerical values find

F12 = (1/ 2 × 2 ) [( 4.123 + 1) − ( 2.236 + 2.236 )] = 0.163. COMMENTS: Remember that Hottel’s string method is applicable only to surfaces that are of infinite extent in one direction and have unobstructed views of one another.

<

PROBLEM 13.14 KNOWN: Two small diffuse surfaces, A1 and A2, on the inside of a spherical enclosure of radius R. FIND: Expression for the view factor F12 in terms of A2 and R by two methods: (a) Beginning with the expression Fij = qij/Ai Ji and (b) Using the view factor integral, Eq. 13.1. SCHEMATIC:

2

ASSUMPTIONS: (1) Surfaces A1 and A2 are diffuse and (2) A1 and A2 100 and heat transfer across the airspace is by free convection, instead of conduction. In this case, convection was evaluated by entering Eqs. 9.58 – 9.60 into the workspace. The results are plotted as follows.

The first graph corresponds to the evacuated space, and the surface temperature decreases with increasing ε1 = ε2, as well as with D2. The increased emissivities enhance the effectiveness of emission at surface 1 and absorption at surface 2, both which have the effect of reducing T1s. Similarly, with increasing D2, more of the radiation emitted from surface 1 is ultimately absorbed at 2 (less of the radiation reflected by surface 2 is intercepted by 1). The second graph reveals the expected effect of a reduction in T1s with inclusion of heat transfer across the air. For small emissivities (ε1 = ε2 < 0.2), conduction across the air is significant relative to radiation, and the small conduction resistance corresponding to D2 = 0.06 m yields the smallest value of T1s. However, with increasing ε, conduction/convection effects diminish relative to radiation and the trend reverts to one of decreasing T1s with increasing D2. COMMENTS: For this situation, the temperature variation within the rod is small and independent of surface conditions.

PROBLEM 13.100 KNOWN: Side wall and gas temperatures for adjoining semi-cylindrical ducts. Gas flow convection coefficients. FIND: (a) Temperature of intervening wall, (b) Verification of gas temperature on one side. SCHEMATIC:

ASSUMPTIONS: (1) All duct surfaces may be approximated as blackbodies, (2) Fully developed conditions, (3) Negligible temperature difference across intervening wall, (4) Gases are nonparticipating media. ANALYSIS: (a) Applying an energy balance to a control surface about the wall yields E in = E out . Assuming Tg,1 > Tw > Tg,2, it follows that q rad (1→ w ) + q conv( g1→ w ) = q rad ( w → 2 ) + q conv( w → g2 )

(

)

(

(

)

)

(

4 4 A1F1wσ T14 − Tw + hA w Tg,1 − Tw = A w Fw2σ Tw − T24 + hA w Tw − Tg,2

and with A1F1w = A w Fw1 = A w Fw2 = A w and substituting numerical values,

(

)

(

4 2σ Tw + 2hTw = σ T14 + T24 + h Tg,1 + Tg,2

)

)

4 11.34 × 10−8 Tw + 10Tw = 13, 900. Trial-and-error solution yields

<

Tw ≈ 526 K. (b) Applying an energy balance to a control surface about the hot gas (g,1) yields E in = E out

(

)

(

hA1 T1 − Tg,1 = hA w Tg,1 − Tw or

(

)

T1 − Tg,1 = [D / (π D / 2 )] Tg,1 − Tw 29°C = 29°C.

) <

COMMENTS: Since there is no change in any of the temperatures in the axial direction, this scheme simply provides for energy transfer from side wall 1 to side wall 2.

PROBLEM 13.101 KNOWN: Temperature, dimensions and arrangement of heating elements between two large parallel plates, one insulated and the other of prescribed temperature. Convection coefficients associated with elements and bottom surface. FIND: (a) Temperature of gas enclosed by plates, (b) Element electric power requirement, (c) Rate of heat transfer to 1 m × 1m section of panel. SCHEMATIC:

ASSUMPTIONS: (1) Diffuse-gray surfaces, (2) Negligible end effects since the surfaces form an enclosure, (3) Gas is nonparticipating, (4) Surface 3 is reradiating with negligible conduction and convection. ANALYSIS: (a) Performing an energy balance for a unit control surface about the gas space, E in − E out = 0. h1π D ( T1 − Tm ) − h 2s ( Tm − T2 ) = 0

Tm =

hπ DT1 + h 2sT2 h1π D + h 2s

=

10 W / m 2 ⋅ Kπ ( 0.025 m ) 600 K + 2 W / m 2 ⋅ K ( 0.05 m ) 400 K 10 W / m 2 ⋅ K π ( 0.025 m ) + 2 W / m 2 ⋅ K ( 0.05 m )

<

Tm = 577 K. (b) The equivalent thermal circuit is

The energy balance on surface 1 is ′ ′ ′ = q1,conv + q1,rad q1,elec ′ where q1,rad can be evaluated by considering a unit cell of the form

A1′ = π D = π ( 0.025 m ) = 0.0785 m A′2 = A′3 = s = 0.05 m

Continued …..

PROBLEM 13.101 (Cont.) The view factors are: 1/ 2

2 F21 = 1 − 1 − ( D / s ) 





F21 = 1 − [1 − 0.25]

1/ 2

(

)

1/ 2 + ( D / s ) tan −1  s 2 − D 2 / D2 



+ 0.5 tan −1 ( 4 − 1)

1/ 2



= 0.658 = F31

F23 = 1 − F21 = 0.342 = F32 . For the unit cell, A′2 F21 = sF21 = 0.05 m × 0.658 = 0.0329 m = A1′ F12 = A′3F31 = A1′ F13 A′2 F23 = sF23 = 0.05 m × 0.342 = 0.0171 m = A′3F32 . Hence, ′ q1,rad =

E b1 − E b2

R ′equiv + (1 − ε 2 ) / ε 2 A′2

−1 R ′equiv = A1′ F12 +



1 1/ A1′ F13 + 1/ A′2 F23

=  0.0329 +

 

 m (0.0329 )−1 + (0.0171)−1  1

R ′equiv = 22.6 m −1. Hence ′ q1,rad =

(

)

5.67 × 10−8 W / m 2 ⋅ K 4 6004 − 4004 K 4

[22.6 + (1 − 0.5) / 0.5 × 0.05] m−1

= 138.3 W / m

′ q1,conv = h1π D ( T1 − Tm ) = 10 W / m 2 ⋅ Kπ ( 0.025 m )( 600 − 577 ) K = 17.8 W / m ′ = (138.3 + 17.8 ) W / m = 156 W / m. q1,elec

<

(c) Since all energy added via the heating elements must be transferred to surface 2, q′2 = q1′ . Hence, since there are 20 elements in a 1 m wide strip, ′ q 2(1m×1m ) = 20 × q1,elec = 3120 W.

<

COMMENTS: The bottom panel would have to be cooled (from below) by a heat sink which could 2 dissipate 3120 W/m .

PROBLEM 13.102 KNOWN: Flat plate solar collector configuration. FIND: Relevant heat transfer processes. SCHEMATIC:

The incident solar radiation will experience transmission, reflection and absorption at each of the cover plates. However, it is desirable to have plates for which absorption and reflection are minimized and transmission is maximized. Glass of low iron content is a suitable material. Solar radiation incident on the absorber plate may be absorbed and reflected, but it is desirable to have a coating which maximizes absorption at short wavelengths. Energy losses from the absorber plate are associated with radiation, convection and conduction. Thermal radiation exchange occurs between the absorber and the adjoining cover plate, between the two cover plates, and between the top cover plate and the surroundings. To minimize this loss, it is desirable that the emissivity of the absorber plate be small at long wavelengths. Energy is also transferred by free convection from the absorber plate to the first cover plate and between cover plates. It is transferred by free or forced convection to the atmosphere. Energy is also transferred by conduction from the absorber through the insulation. The foregoing processes provide for heat loss from the absorber, and it is desirable to minimize these losses. The difference between the solar radiation absorbed by the absorber and the energy loss by radiation, convection and conduction is the energy which is transferred to the working fluid. This transfer occurs by conduction through the absorber and the tube wall and by forced convection from the tube wall to the fluid.

PROBLEM 13.103 KNOWN: Operating conditions of a flat plate solar collector. FIND: Expressions for determining the rate at which useful energy is collected per unit area. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform surface heat fluxes and temperatures, (3) Opaque, diffuse-gray surface behavior for long-wave thermal radiation, (4) Complete absorption of solar radiation by absorber plate (αap,S = 1). ANALYSIS: From an energy balance on the absorber plate, E ′′in = E ′′out ,

(

)

α ap,S τ cp,S GS = q′′u + q′′conv,i + q′′rad,ap − cp . Hence with complete absorption of solar radiation by the absorber plate,

(

)

q′′u = τ cp,SGS − h i Tap − Tcp −

(

4 4 σ Tap − Tcp

)

(1)

1/ ε ap + 1/ ε cp − 1

<

where Fap-cp ≈ 1 and Eq. 13.24 is used to obtain q′′rad,ap − cp . To determine q′′u from Eq. (1), however, Tcp must be known. From an energy balance on the cover plate,

α cp,SGS + q′′conv,i + q′′rad,ap − cp = q′′conv,o + q ′′rad,cp −sky or

(

)

α cp,SGS + h i Tap − Tcp +

(

(

4 4 σ Tap − Tcp

)

1/ ε ap + 1/ ε c − 1

)

(

)

4 4 = h o Tcp − T∞ + ε cpσ Tcp − Tsky .

(2)

<

Eq. (2) may be used to obtain Tcp. COMMENTS: With Tap presumed to be known, Tcp may be evaluated from Eq. (2) and q′′u from Eq. (1).

PROBLEM 13.104 KNOWN: Ceiling temperature of furnace. Thickness, thermal conductivity, and/or emissivities of alternative thermal insulation systems. Convection coefficient at outer surface and temperature of surroundings. FIND: (a) Mathematical model for each system, (b) Temperature of outer surface Ts,o and heat loss q′′ for each system and prescribed conditions, (c) Effect of emissivity on Ts,o and q′′. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Diffuse/gray surfaces, (3) Surroundings form a large enclosure about the furnace, (4) Radiation in air space corresponds to a two-surface enclosure of large parallel plates. -4

2

PROPERTIES: Table A-4, air (Tf = 730 K): k = 0.055 W/m⋅K, α = 1.09 × 10 m /s, ν = 7.62 × -5 2 -1 10 m /s, β = 0.001335 K , Pr = 0.702. ANALYSIS: (a) To obtain Ts,o and q′′, an energy balance must be performed at the outer surface of the shield. q′′cond = q′′conv,o + q′′rad,o = q′′

Insulation:

k=

(Ts,i − Ts,o ) = h

(

)

L ′′ q conv,i + q′′rad,i = q′′conv,o + q′′rad,o = q′′

Air Space:

(

)

h i Ts,i − Ts,o +

(

4 4 o Ts,o − T∞ + ε oσ Ts,o − Tsur

(

4 4 σ Ts,i − Ts,o

1

εi

+

1

εo

)=h

−1

(

) (

)

4 4 o Ts,o − T∞ + ε oσ Ts,o − Tsur

)

where Eq. 13.24 has been used to evaluate q′′rad,i and hi is given by Eq. 9.49 hL 3 0.074 Nu L = i = 0.069Ra1/ L Pr k (b) For the prescribed conditions (εi = εo = 0.5), the following results were obtained. Insulation:

The energy equation becomes

(

)

0.09 W / m ⋅ K 900 − Ts,o K 0.025 m

2

(

)

= 25 W / m ⋅ K Ts,o − 300 K + 0.5 × 5.67 × 10

−8

2

W/m ⋅K

4

(

4

Ts,o − 300

4

)

K

4

Continued …..

PROBLEM 13.104 (Cont.) and a trial-and-error solution yields q′′ = 1920 W / m 2

Ts,o = 366 K Air-Space:

<

The energy equation becomes

(

)

h i 900 − Ts,o K +

5.67 × 10

−8

2

W / m ⋅K

4

(900

4

4

)

− Ts,o K

4

3 2

(

)

= 25 W / m ⋅ K Ts,o − 300 K + 0.5 × 5.67 × 10

−8

2

W /m ⋅K

4

(T

)

4 4 4 s,o − 300 K

where hi =

0.055 W / m ⋅ K 0.025 m

3 0.074 0.069 Ra1/ L Pr

(1)

3

and RaL = gβ(Ts,i – Ts,o)L /αν. A trial-and-error solution, which includes reevaluation of the air properties, yields Ts,o = 598 K

q′′ = 10,849 W / m 2

<

The inner and outer heat fluxes are q′′conv,i = 867 W / m 2 , q′′rad,i = 9982 W / m 2 , q′′conv,o = 7452 2 W/m , and q′′rad,o = 3397 W / m 2 .

(c) Entering the foregoing models into the IHT workspace, the following results were generated. Insulation:

Continued …..

PROBLEM 13.104 (Cont.) As expected, the outer surface temperature decreases with increasing εo. However, the reduction in Ts,o is not large since heat transfer from the outer surface is dominated by convection.

In this case Ts,o increases with increasing εo = εi and the effect is significant. The effect is due to an increase in radiative transfer from the inner surface, with q′′rad,i = q′′conv,i = 1750 W / m 2 for εo = εi = 0.1 and q′′rad,i = 20,100 W / m 2 >> q′′conv,i = 523 W / m 2 for εo = εi = 0.9. With the increase in Ts,o,

(

)

the total heat flux increases, along with the relative contribution of radiation q′′rad,o to heat transfer from the outer surface. COMMENTS: (1) With no insulation or radiation shield and εi = 0.5, radiative and convective heat 2 fluxes from the ceiling are 18,370 and 15,000 W/m , respectively. Hence, a significant reduction in the heat loss results from use of the insulation or the shield, although the insulation is clearly more effective. (2) Rayleigh numbers associated with free convection in the air space are well below the lower limit of applicability of Eq. (1). Hence, the correlation was used outside its designated range, and the error associated with evaluating hi may be large. (3) The IHT solver had difficulty achieving convergence in the first calculation performed for the radiation shield, since the energy balance involves two nonlinear terms due to radiation and one due to convection. To obtain a solution, a fixed value of RaL was prescribed for Eq. (1), while a second 3 value of RaL,2 ≡ gβ(Ts,i – Ts,o)L /αν was computed from the solution. The prescribed value of RaL was replaced by the value of RaL,2 and the calculations were repeated until RaL,2 = RaL.

PROBLEM 13.105 KNOWN: Dimensions of a composite insulation consisting of honeycomb core sandwiched between solid slabs. FIND: Total thermal resistance. SCHEMATIC: Because of the repetitive nature of the honeycomb core, the cell sidewalls will be adiabatic. That is, there is no lateral heat transfer from cell to cell, and it suffices to consider the heat transfer across a single cell.

ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Equivalent conditions for each cell, (3) Constant properties, (4) Diffuse, gray surface behavior. PROPERTIES: Table A-3, Particle board (low density): k1 = 0.078 W/m⋅K; Particle board (high density): k2 = 0.170 W/m⋅K; For both board materials, ε = 0.85; Table A-4, Air ( T ≈ 7.5°C, 1 atm): -6 2 -6 2 -3 -1 ν = 14.15 × 10 m /s, k = 0.0247 W/m⋅K, α = 19.9 × 10 m /s, Pr = 0.71, β = 3.57 × 10 K . ANALYSIS: The total resistance of the composite is determined by conduction, convection and radiation processes occurring within the honeycomb and by conduction across the inner and outer slabs. The corresponding thermal circuit is shown.

The total resistance of the composite and equivalent resistance for the honeycomb are R = R cond,i + R eq + R cond,o

(

−1 −1 −1 −1 R eq = R cond + R conv + R rad

)

.

hc

The component resistances may be evaluated as follows. The inner and outer slabs are plane walls, for which the thermal resistance is given by Eq. 3.6. Hence, since L1 = L3 and the slabs are constructed from low-density particle board. R cond,i = R cond,o =

L1 k1W

2

=

0.0125 m 0.078 W / m ⋅ K ( 0.01 m )

2

= 1603 K / W.

Continued …..

PROBLEM 13.105 (Cont.) Similarly, applying Eq. 3.6 to the side walls of the cell R cond,hc =

L2 k 2 W − (W − t )



=

=

2

2



L2

(

k 2 2Wt − t

2

)

0.050 m 0.170 W / m ⋅ K  2 × 0.01 m × 0.002 m − ( 0.002 m )

2



= 8170 K / W.



From Eq. 3.9 the convection resistance associated with the cellular airspace may be expressed as R conv,hc = 1/ h ( W − t ) . 2

The cell forms an enclosure that may be classified as a horizontal cavity heated from below, and the appropriate form of the Rayleigh number is Ra L = gβ ( T1 − T2 ) L32 / αν . To evaluate this parameter, however, it is necessary to assume a value of the cell temperature difference. As a first approximation, T1 − T2 = 15°C − ( −5°C ) = 20°C, Ra L =

(

9.8 m / s 2 3.57 × 10−3 K −1

)(20 K )(0.05 m )3 = 3.11×105.

19.9 × 10−6 m 2 / s × 14.15 × 10−6 m 2 / s Applying Eq. 9.49 as a first approximation, it follows that h = ( k / L 2 )  0.069Ra L



1/ 3

Pr

0.074 



0.0247 W / m ⋅ K

=

0.05 m

  0.069

(

3.11 × 10

5

)

1/ 3

 ( 0.71)0.074  = 2.25 

2

W / m ⋅ K.

The convection resistance is then R conv,hc =

1 2.25 W / m ⋅ K ( 0.01 m − 0.002 m )

2

2

= 6944 K / W.

The resistance to heat transfer by radiation may be obtained by first noting that the cell forms a threesurface enclosure for which the sidewalls are reradiating. The net radiation heat transfer between the 2 end surfaces of the cell is then given by Eq. 13.30. With ε1 = ε2 = ε and A1 = A2 = (W – t) , the equation reduces to q rad =

( W − t )2 σ

(T14 − T24 )

2 (1/ ε − 1) +  F12 + [( F1R + F2R ) / F1R F2R 

−1

.

However, with F1R = F2R = (1 – F12), it follows that q rad =

( W − t )2 σ

(T14 − T24 )

− 2 1

(1 − F12 )  1   2  − 1  +  F12 + 2 (1 − F12 )   ε   

=

( W − t )2 σ

(T14 − T24 ) .

 1 − 1 + 2   ε  1 + F12

2

The view factor F12 may be obtained from Fig. 13.4, where X Y W − t 10 mm − 2 mm = = = = 0.16. L L L2 50 mm Hence, F12 ≈ 0.01. Defining the radiation resistance as T −T R rad,hc = 1 2 q rad it follows that Continued …..

PROBLEM 13.105 (Cont.) R rad,hc =

(

2 (1/ ε − 1) + 2 / (1 + F12 )

( W − t )2 σ

) (

where T14 − T24 = T12 + T22

(T12 + T22 )(T1 + T2 )

)(T1 + T2 )(T1 − T2 ). Accordingly,

2    1   2  0.85 − 1 + 1 + 0.01  R rad,hc = (0.01 m − 0.002 m )2 × 5.67 × 10−8 W / m2 ⋅ K 4 ( 288 K )2 + ( 268 K )2  ( 288 + 268 ) K where, again, it is assumed that T1 = 15°C and T2 = -5°C. From the above expression, it follows that 0.353 + 1.980 = 7471 K / W. R rad,hc = 3.123 × 10−4 In summary the component resistances are R cond,i = R cond,o = 1603 K / W R cond,hc = 8170 K / W

R conv,hc = 6944 K / W

R rad,hc = 7471 K / W.

The equivalent resistance is then 1 1   1 + +   8170 6944 7471 

R eq = 

−1

= 2498 K / W

and the total resistance is R = 1603 + 2498 + 1603 = 5704 K / W.

<

COMMENTS: (1) The problem is iterative, since values of T1 and T2 were assumed to calculate Rconv,hc and Rrad,hc. To check the validity of the assumed values, we first obtain the heat transfer rate q from the expression Ts,1 − Ts,2 25°C − ( −10°C ) = = 6.14 × 10−3 W. q= R 5704 K / W Hence T1 = Ts,i − qR cond,i = 25°C − 6.14 × 10−3 W × 1603 K / W = 15.2°C T2 = Ts,o + qR cond,o = −10°C + 6.14 × 10−3 W × 1603 K / W = −0.2°C. Using these values of T1 and T2, Rconv,hc and Rrad,hc should be recomputed and the process repeated until satisfactory agreement is obtained between the initial and computed values of T1 and T2. 2

(2) The resistance of a section of low density particle board 75 mm thick (L1 + L2 + L3) of area W is 9615 K/W, which exceeds the total resistance of the composite by approximately 70%. Accordingly, use of the honeycomb structure offers no advantages as an insulating material. Its effectiveness as an insulator could be improved (Req increased) by reducing the wall thickness t to increase Rcond, evacuating the cell to increase Rconv, and/or decreasing ε to increase Rrad. A significant increase in Rrad,hc could be achieved by aluminizing the top and bottom surfaces of the cell.

PROBLEM 13.106 KNOWN: Dimensions and surface conditions of a cylindrical thermos bottle filled with hot coffee and lying horizontally. FIND: Heat loss. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss from ends (long infinite cylinders), (3) Diffuse-gray surface behavior. PROPERTIES: Table A-4, Air (Tf = (T1 + T2)/2 = 328 K, 1 atm): k = 0.0284 W/m⋅K, ν = 23.74 × -6 2 -6 2 -3 -1 10 m /s, α = 26.6 × 10 m /s, Pr = 0.0703, β = 3.05 × 10 K . ANALYSIS: The heat transfer across the air space is q = q rad + q conv . From Eq. 13.25 for concentric cylinders q rad =

(

σ (π D1L ) T14 − T24 1 1 − ε 2  r1  +   ε1 ε 2  r2 

) = 5.67 ×10−8 W / m2 ⋅ K4π (0.07 × 0.3) m2 (3484 − 3084 ) K4 4 + 3 ( 0.035 / 0.04 )

q rad = 3.20 W. From Eq. 9.25, Ra L =

gβ ( T1 − T2 ) L3

αν

=

(

9.8 m / s 2 3.05 × 10−3 K −1

)(40 K )(0.005 m )3 = 236.7.

26.6 × 10−6 m 2 / s × 23.74 × 10−6 m2 / s

Hence from Eq. 9.60

[

]

[

]

4 4 ln ( D2 / D1 ) Ra L ln ( 0.08 / 0.07 ) 236.7 ∗ = = 7.85. Ra c = 5 5 3 −0.6 −0.6 −0.6 −0.6 −3 3 + D2 L D1 m (0.005 m ) 0.07 + 0.08

(

)

(

)

However, the implication of such a small value of Ra ∗c is that free convection effects are negligible. Heat transfer across the airspace is therefore by conduction (keff = k). From Eq. 3.27 2π Lk ( T1 − T2 ) 2π × 0.3 m × 0.0284 W / m ⋅ K ( 75 − 35 ) K = = 16.04 W. q cond = ln ( r2 / r1 ) ln ( 0.04 / 0.035 ) Hence the total heat loss is q = q rad + q cond = 19.24 W. COMMENTS: (1) End effects could be considered in a more detailed analysis, (2) Conduction losses could be eliminated by evacuating the annulus.

<

PROBLEM 13.107 KNOWN: Thickness and height of a vertical air space. Emissivity and temperature of adjoining surfaces. FIND: (a) Heat loss per unit area across the space, (b) Heat loss per unit area if space is filled with urethane foam. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Diffuse-gray surface behavior, (3) Air space is a vertical cavity, (4) Constant properties, (5) One-dimensional conduction across foam. -6

2

PROPERTIES: Table A-4, Air (Tf = 4°C, 1 atm): ν = 13.84 × 10 m /s, k = 0.0245 W/m⋅K, α = -6 2 -3 -1 19.5 × 10 m /s, Pr = 0.71, β = 3.61 × 10 K ; Table A-3, Urethane foam: k = 0.026 W/m⋅K. ANALYSIS: (a) With the air space, heat loss is by radiation and free convection or conduction. From Eq. 13.24,

(

σ T14 − T24

q′′rad =

)

1/ ε1 + 1/ ε 2 − 1

=

1.222

With Ra L =

gβ ( T1 − T2 ) L3

)

(

5.67 × 10−8 W / m 2 ⋅ K 4 2914 − 2634 K 4

=

(

= 110.7 W / m 2 .

)

9.8 m 2 / s 3.61× 10−3 K −1 (18 + 10 ) K ( 0.1 m )

13.84 × 10−6 m 2 / s × 19.5 × 10−6 m 2 / s and H/L = 30, Eq. 9.53 may be used as a first approximation to obtain

να

(

3 6 Nu L = 0.046Ra1/ L = 0.046 3.67 × 10

h=

k

Nu L =

0.0245 W / m ⋅ K

L 0.1 m The convection heat flux is

)

1/ 3

3

= 3.67 × 106

= 7.10

7.10 = 1.74 W / m 2 ⋅ K.

q′′conv = h ( T1 − T2 ) = 1.74 W / m 2 ⋅ K (18 + 10 ) K = 48.7 W / m 2 . The heat loss is then q′′ = q′′rad + q′′conv = 110.7 + 48.7 = 159 W / m 2 . (b) With the foam, heat loss is by conduction and k 0.026 W / m ⋅ K q′′ = q′′cond = ( T1 − T2 ) = (18 + 10 ) K = 7.3 W / m2 . L 0.1 m

<

COMMENTS: Use of the foam insulation reduces the heat loss considerably. Note the significant effect of radiation.

PROBLEM 13.108 KNOWN: Temperatures and emissivity of window panes and critical Rayleigh number for onset of convection in air space. FIND: (a) The conduction heat flux across the air gap for the optimal spacing, (b) The total heat flux for uncoated panes, (c) The total heat flux if one or both of the panes has a low-emissivity coating. SCHEMATIC:

ASSUMPTIONS: (1) Critical Rayleigh number is RaL,c = 2000, (2) Constant properties, (3) Radiation exchange between large (infinite), parallel, diffuse-gray surfaces. -6

2

PROPERTIES: Table A-4, air [T = (T1 + T2)/2 = 1°C = 274 K]: ν = 13.6 × 10 m /s, k = 0.0242 -6 2 -1 W/m⋅K, α = 19.1 × 10 m /s, β = 0.00365 K . ANALYSIS: (a) With Ra L,c = g β ( T1 − T2 ) L3op / αν 1/ 3

 αν Ra L,c  Lop =    g β (T1 − T2 ) 

1/ 3

 −12 m 4 / s 2 × 2000  19.1×13.6 × 10  =  2 1 − 42°C   9.8 m / s 0.00365 K 

)

(

= 0.0070m

The conduction heat flux is then

q′′cond = k ( T1 − T2 ) / Lop = 0.0242 W / m ⋅ K ( 42°C ) / 0.0070m = 145.2 W / m 2

<

(b) For conventional glass (εg = 0.90), Eq. (13.24) yields, σ T14 − T24 5.67 × 10−8 W / m 2 ⋅ K 4 2954 − 2534 K 4 q′′rad = = = 161.3 W / m 2 2

)

(

εg

(

)

1.222

−1

and the total heat flux is

q′′tot = q′′cond + q′′rad = 306.5 W / m 2

<

(c) With only one surface coated, 5.67 × 10−8 W / m 2 ⋅ K 4 2954 − 2534 q′′rad = = 19.5 W / m 2 1 1

(

0.90

+

0.10

)

−1

Continued …..

PROBLEM 13.108 (Cont.) q′′tot = 164.7 W / m 2

<

With both surfaces coated, 5.67 × 10−8 W / m 2 ⋅ K 4 2954 − 2534 q′′rad = = 10.4 W / m 2 1 1

(

0.10 q′′tot = 155.6 W / m 2

+

0.10

)

−1

<

COMMENTS: Without any coating, radiation makes a large contribution (53%) to the total heat loss. With one coated pane, there is a significant reduction (46%) in the total heat loss. However, the benefit of coating both panes is marginal, with only an additional 3% reduction in the total heat loss.

PROBLEM 13.109 KNOWN: Dimensions and emissivity of double pane window. Thickness of air gap. Temperatures of room and ambient air and the related surroundings. FIND: (a) Temperatures of glass panes and rate of heat transfer through window, (b) Heat rate if gap is evacuated. Heat rate if special coating is applied to window. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Negligible glass pane thermal resistance, (3) Constant properties, (4) Diffuse-gray surface behavior, (5) Radiation exchange between interior window surfaces may be approximated as exchange between infinite parallel plates, (6) Interior and exterior surroundings are very large. PROPERTIES: Table A-4, Air (p = 1 atm). Obtained from using IHT to solve for conditions of Part -6 2 -6 2 (a): Tf,i = 287.4 K: νi = 14.8 × 10 m /s, ki = 0.0253 W/m⋅K, αi = 20.8 × 10 m /s, Pri = 0.71, βi = -1 -6 2 0.00348 K . T = (Ts,i + Ts,o)/2 = 273.7 K: ν = 13.6 × 10 m /s, k = 0.0242 W/m⋅K, α = 19.0 × 10 -1 -6 2 6 2 m /s, Pr = 0.71, β = 0.00365 K . Tf,o = 259.3 K: νo = 12.3 × 10 m /s, ko = 0.023 W/m⋅K, αo = -6 2 -1 17.1 × 10 m /s, Pro = 0.72, βo = 0.00386 K . ANALYSIS: (a) The heat flux through the window may be expressed as

)

(

(

4 − T4 + h T − T q′′ = q′′rad,i + q′′conv,i = ε g σ Tsur,i i ∞,i s,i s,i

q′′ = q′′rad,gap + q′′conv,gap =

(

4 − T4 σ Ts,i s,o

1 1 + −1 εg εg

(

)+h )

)

(

gap Ts,i − Ts,o

(

4 − T4 q′′ = q′′rad,o + q′′conv,o = ε g σ Ts,o sur,o + h o Ts,o − T∞,o

(1)

)

(2)

)

(3)

where radiation exchange between the window panes is determined from Eq. (13.24) and radiation exchange with the surroundings is determined from Eq. (13.27). The inner and outer convection coefficients, h i and h o , are determined from Eq. (9.26), and h gap is obtained from Eq. (9.52).

(

)

The foregoing equations may be solved for the three unknowns q′′, Ts,i , Ts,o . Using the IHT software to effect the solution, we obtain

<

Ts,i = 281.8 K = 8.8°C Continued …..

PROBLEM 13.109 (Cont.) Ts,o = 265.6 K = −7.4°C

<

q = 91.3 W

<

(

)

(b) If the air space is evacuated h g = 0 , we obtain

Ts,i = 283.6 K = 10.6°C

<

Ts,o = 263.8 K = 9.2°C

<

q = 75.5 W

<

If the space is not evacuated but the coating is applied to inner surfaces of the window panes,

Ts,i = 285.9 K = 12.9°C

<

Ts,o = 261.3 K = −11.7°C

<

q = 55.9 W

<

If the space is evacuated and the coating is applied,

Ts,i = 291.7 K = 18.7°C

<

Ts,o = 254.7 K = −18.3°C

<

q = 9.0 W

<

COMMENTS: (1) For the conditions of part (a), the convection and radiation heat fluxes are comparable at the inner and outer surfaces of the window, but because of the comparatively small convection coefficient, the radiation flux is approximately twice the convection flux across the air gap. (2) As the resistance across the air gap is progressively increased (evacuated, coated, evacuated and coated), the temperatures of the inner and outer panes increase and decrease, respectively, and the heat loss decreases. (3) Clearly, there are significant energy savings associated with evacuation of the gap and application of the coating. (4) In all cases, solutions were obtained using the temperaturedependent properties of air provided by the software. The property values listed in the PROPERTIES section of this solution pertain to the conditions of part (a).

PROBLEM 13.110 KNOWN: Absorber and cover plate temperatures and spectral absorptivities for a flat plate solar collector. Collector orientation and solar flux. FIND: (a) Rate of solar radiation absorption per unit area, (b) Heat loss per unit area. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Adiabatic sides and bottom, (3) Cover is transparent to solar radiation, (4) Sun emits as a blackbody at 5800 K, (5) Cover and absorber plates are diffuse-gray to long wave radiation, (6) Negligible end effects, (7) L 12, τ < τ ∗ , cos τ = 0.5, Ra L cos τ = 11, 302 Nu L

1708   = 1 + 1.44 1 −  11, 302 

h = Nu L

k L

= 2.30 ×

 1708 (sin 108° )1.6   11, 302 1/ 3  1 −  +   − 1 11, 302    5830  

0.0279 W / m ⋅ K 0.02 m

= 3.21 W / m 2 ⋅ K.

Hence, the convective heat flux is q′′conv = 3.21 W / m 2 ⋅ K (343 − 300 ) K = 138.0 W / m 2 . The radiative exchange can be determined from Eq. 13.24 treating the cover and absorber plates as a two-surface enclosure, q′′rad =

(

σ Ta4 − Tc4

)

1/ ε a + 1/ ε c − 1

=

4 4 5.67 × 10−8 W / m 2 ⋅ K 4 (343 K ) − (300 K ) 





1/ 0.2 + 1/ 0.75 − 1

q′′rad = 61.1 W / m 2 . Hence, the total heat loss per unit area from the collector q′′loss = (138.0 + 61.1) = 199 W / m 2 .

<

COMMENTS: (1) Non-solar components of radiation transfer are concentrated at long wavelength for which αa = εa = 0.2 and αc = εc = 0.75. (2) The collector efficiency is

η=

669.3 − 199.1 669.3

× 100 = 70%.

This value is uncharacteristically high due to specification of nearly optimum αa(λ) for absorber.

PROBLEM 13.111 KNOWN: Diameters and temperatures of a heated tube and a radiation shield. FIND: (a) Total heat loss per unit length of tube, (b) Effect of shield diameter on heat rate. SCHEMATIC:

ASSUMPTIONS: (1) Opaque, diffuse-gray surfaces, (2) Negligible end effects. PROPERTIES: Table A-4, Air (Tf = 77.5°C ≈ 350 K): k = 0.030 W/m⋅K, Pr = 0.70, ν = 20.92 × 10 -6 2 -1 6 2 m /s, α = 29.9 × 10 m /s, β = 0.00286 K . ANALYSIS: (a) Heat loss from the tube is by radiation and free convection

q′ = q′rad + q′conv q′rad =

From Eq. (13.25)

(

σ (π Di ) Ti4 − To4 1 1 − ε o  ri  +   εi ε o  ro 

)

or

5.67 × 10−8 q′rad =

W m ⋅ K4

(π × 0.1m )

(3934 − 3084 ) K 4

1 0.9  0.05  + 0.8 0.1  0.06 

= 30.2

W m

g β (Ti − To ) L3 9.8 m / s 2 × 0.00286 K −1 (85 K )( 0.01m ) = = 3809 6 2 6 2 − − να 20.92 × 10 m / s 29.9 × 10 m / s 3

Ra L =

)(

(

)

Hence from Eq. (9.60) 4 4  n ( Do / Di ) Ra L  n (0.12 / 0.10 ) 3809 ∗ = = 171.6 Ra c = 5 5 L3 Di−3/ 5 + Do−3/ 5 (0.01m )3 (0.1m )−0.6 + (0.12m )−0.6 

(

)

and from Eq. (9.59) 1/ 4

Pr   k eff = 0.386 k    0.861 + Pr 

( ) Ra ∗c

1/ 4

Continued …..

-

PROBLEM 13.111 (Cont.) 1/ 4

W  0.7   k eff = 0.386  0.03   m ⋅ K   0.861 + 0.7  

(171.6 )1/ 4 = 0.0343

W m⋅K

Hence from Eq. (9.58)

W   2π  0.0343 2π k eff W m ⋅ K  q′conv = (Ti − To ) =  (120 − 35) K = 100.5 m n ( Do / Di ) n (0.12 / 0.10 ) q′ = (30.2 + 100.5 )

W W = 130.7 m m

(b) As shown below, both convection and radiation, and hence the total heat rate, increase with increasing shield diameter. In the limit as Do → ∞, the radiation rate approaches that corresponding to net transfer between a small surface and large surroundings at To. The rate is independent of ε.

Heat rates (W/m)

200 150 100 50 0 0.1

0.15

0.2

0.25

Shield diameter, Do(m) Radiation heat rate (W/m) Convection heat rate (W/m) Total heat rate (W/m)

COMMENTS: Designation of a shield temperature is arbitrary. The temperature depends on the nature of the environment external to the shield.

PROBLEM 13.112 KNOWN: Diameters of heated tube and radiation shield. Tube surface temperature and temperature of ambient air and surroundings. FIND: Temperature of radiation shield and heat loss per unit length of tube. SCHEMATIC:

ASSUMPTIONS: (1) Opaque, diffuse-gray surfaces, (2) Negligible end effects, (3) Large surroundings, (4) Quiescent air, (5) Steady-state. PROPERTIES: Determined from use of IHT software for iterative solution. Air, (Ti + To)/2 = -5 2 -5 2 -1 362.7 K: νi = 2.23 × 10 m /s, ki = 0.031 W/m⋅K, αi = 3.20 × 10 m /s, β i = 0.00276 K , Pri = -5 2 -5 2 0.698. Air, Tf = 312.7 K: νo = 1.72 × 10 m /s, ko = 0.027 W/m⋅K, αo = 2.44 × 10 m /s, β o = -1 0.0032 K , Pro = 0.705. ANALYSIS: From an energy balance on the radiation shield, q′i = q′o or q′rad,i + q′conv,i = q′rad,o + q′conv,o . Evaluating the inner and outer radiation rates from Eqs. (13.25) and (13.27), respectively, and the convection heat rate in the air gap from Eq. (9.58),

(

σ π Di Ti4 − To4

) + 2π keff (Ti − To ) = σ π D n ( Do / Di )

1 1 − ε o  Di  +   εi ε o  Do 

(

)

4 4 o ε o To − Tsur + π Do h o ( To − T∞ )

From Eqs. (9.59) and (9.60) 1/ 4

  Pri k eff = 0.386 k i    0.861 + Pri 

( ) Ra ∗c

1/ 4

4  n ( Do / Di ) Ra L ∗ Ra c = 5 L3 Di−3/ 5 + Do−3/ 5

)

(

3

where RaL = g β i (Ti – To)L /νi αi and L = (Do – Di)/2. From Eq. (9.34), the convection coefficient on the outer surface of the shield is 2

  1/ 6   0.387 Ra D k  h o = o 0.60 + 8 / 27  Do   1 + ( 0.559 / Pr )9 /16     

Continued …..

PROBLEM 13.112 (Cont.) The solution to the energy balance is obtained using the IHT software, and the result is

To = 332.5 K = 59.5°C

<

The corresponding value of the heat loss is

q′i = 88.7 W / m

<

COMMENTS: (1) The radiation and convection heat rates are q ′rad,i = 23.7 W / m, q ′rad,o = 10.4 W / m, q ′conv,i = 65.0 W / m, and q ′conv,o = 78.3 W / m. Convection is clearly the dominant mode of heat transfer. (2)With a value of To = 59.5°C > 35°C, the heat loss is reduced (88.7 W/m compared to 130.7 W/m if the shield is at 35°C).

PROBLEM 13.113 KNOWN: Dimensions and inclination angle of a flat-plate solar collector. Absorber and cover plate temperatures and emissivities. FIND: (a) Rate of heat transfer by free convection and radiation, (b) Effect of the absorber plate temperature on the heat rates. SCHEMATIC:

ASSUMPTIONS: (1) Diffuse-gray, opaque surface behavior. PROPERTIES: Table A-4, air ( T = ( T1 + T2 ) / 2 = 323 K ) : ν = 18.2 × 10 m /s, k = 0.028 -6

-6

2

2

-1

W/m⋅K, α = 25.9 × 10 m /s, Pr = 704, β = 0.0031 K . ANALYSIS: (a) The convection heat rate is q conv = hA ( T1 − T2 ) 2

where A = wH=4 m and, with H/L > 12 and τ < τ* = 70 deg, h is given by Eq. 9.54. With a Rayleigh number of Ra L =

gβ ( T1 − T2 ) L3

αν



Nu L = 1 + 1.44 1 −



=

(

9.8 m / s 2 0.0031 K −1

)(40°C)(0.03 m )3 = 69, 600

25.9 × 10−6 m 2 / s × 18.2 × 10−6 m 2 / s

  1708 ( 0.923)   0.5 × 69, 600 1/ 3   − 1  1 −  +  0.5 ( 69, 600 )   0.5 ( 69, 600 )   5830    1708

Nu L = 1 + 1.44 [0.951][0.955] + 0.814 = 3.12 h = ( k / L ) Nu L = ( 0.028 W / m ⋅ K / 0.03 m ) 3.12 = 2.91 W / m 2 ⋅ K

(

q conv = 2.91 W / m 2 ⋅ K 4 m 2

)(70 − 30)°C = 466 W

<

The net rate of radiation exchange is given by Eq. 13.24. q=

(

Aσ T14 − T24 1 1 + −1 ε1 ε 2

) = (4 m2 )5.67 ×10−8 W / m2 ⋅ K4 (3434 − 3034 ) K4 = 1088 W 1 1 + −1 0.96 0.92

<

(b) The effect of the absorber plate temperature was determined by entering Eq. 9.54 into the IHT workspace and using the Properties and Radiation Toolpads. Continued …..

PROBLEM 13.113 (Cont.)

4

As expected, the convection and radiation losses increase with increasing Ti, with the T dependence providing a more pronounced increase for the radiation. COMMENTS: To minimize heat losses, it is obviously desirable to operate the absorber plate at the lowest possible temperature. However, requirements for the outlet temperature of the working fluid may dictate operation at a low flow rate and hence an elevated plate temperature.

PROBLEM 13.114 KNOWN: Disk heated by an electric furnace on its lower surface and exposed to an environment on its upper surface. FIND: (a) Net heat transfer to (or from) the disk qnet,d when Td = 400 K and (b) Compute and plot qnet,d as a function of disk temperature for the range 300 ≤ Td ≤ 500 K; determine steady-state temperature of the disk. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Disk is isothermal; negligible thermal resistance, (3) Surroundings are isothermal and large compared to the disk, (4) Non-black surfaces are graydiffuse, (5) Furnace-disk forms a 3-surface enclosure, (6) Negligible convection in furnace, (7) Ambient air is quiescent. -6

2

PROPERTIES: Table A-4, Air (Tf = (Td + T∞)/2 = 350 K,1 atm): ν = 20.92 × 10 m /s, k = 0.30 -6 2 W/m⋅K, α = 29.9 × 10 m /s. ANALYSIS: (a) Perform an energy balance on the disk identifying: qrad as the net radiation exchange between the disk and surroundings; qconv as the convection heat transfer; and q3 as the net radiation leaving the disk within the 3-surface enclosure. (1) q net,d = E in − E out = −q rad − q conv − q3 Radiation exchange with surroundings: The rate equation is of the form

(

4 q rad = ε d,2 Adσ Td4 − Tsur

)

(2)

(

)

q rad = 0.8 (π / 4 )( 0.400 m ) 5.67 × 10−8 W / m 2 ⋅ K 4 4004 − 3004 K 4 = 99.8 W. 2

Free convection: The rate equation is of the form q conv = hAd ( Td − T∞ )

(3)

where h can be estimated by an appropriate convection correlation. Find first, Ra L = gβ∆TL3 / να

(4)

Ra L = 9.8 m / s 2 (1/ 350 K )( 400 − 300 ) K ( 0.400 m / 4 ) / 20.92 × 10−6 m / s 2 × 29.9 × 10−6 m 2 / s 3

Ra L = 4.476 × 106 4

7

where L = Ac/P = D/4. For the upper surface of a heated plate for which 10 ≤ RaL ≤ 10 , Eq. 9.30 is the appropriate correlation, Continued …..

PROBLEM 13.114 (Cont.) 1/ 4

Nu L = hL / k = 0.54 Ra L

(5)

(

h = 0.030 W / m ⋅ K / ( 0.400 m / 4 ) × 0.54 4.476 × 106

)

1/ 4

= 7.45 W / m 2 ⋅ K

Hence, from Eq. (3), q conv = 7.45 W / m 2 ⋅ K (π / 4 )( 0.400 m ) ( 400 − 300 ) K = 93.6 W. Furnace-disk enclosure: From Eq. 13.20, the net radiation leaving the disk is 2

q3 =

J 3 − J1

+

J3 − J 2

( A3F31 )−1 ( A3F32 )−1

= A 3 [F31 ( J 3 − J1 ) + F32 ( J 3 − J 2 )].

(6)

The view factor F32 can be evaluated from the coaxial parallel disks relation of Table 13.1 or from Fig. 13.5. R i = ri / L = 200 mm / 200 mm = 1, R j = rj / L = 1,

(

)

( )

S = 1 + 1 + R 2j / R 2j = 1 + 1 + 12 12 = 3



 

(

F31 = 1/ 2 S − S2 − 4 rj / ri 



1/ 2 

)2 

  2 1/ 2  2  = 1/ 2 3 − 3 − 4 (1)   = 0.382.   

(7)

From summation rule, F32 = 1 – F33 – F31 = 0.618 with F33 = 0. Since surfaces A2 and A3 are black, J 2 = E b2 = σ T24 = σ (500 K ) = 3544 W / m 2 4

J 3 = E b3 = σ T34 = σ ( 400 K ) = 1452 W / m 2 . 4

To determine J1, use Eq. 13.21, the radiation balance equation for A1, noting that F12 = F32 and F13 = F31, E b1 − J1 J −J J −J = 1 2 + 1 3 (1 − ε1 ) / ε1A1 ( A1F12 )−1 ( A1F13 )−1 3544 − J1

J − 3544 J − 1452 = 1 + 1 − 1 (1 − 0.6 ) / 0.6 (0.618 ) (0.382 )−1

J1 = 3226 W / m 2 .

(8)

Substituting numerical values in Eq. (6), find q 3 = (π / 4 )( 0.400 m )

2

 0.382 (1452 − 3226 ) W / m 2 + 0.618 (1452 − 3544 ) W / m 2  = −248 W.  

Returning to the overall energy balance, Eq. (1), the net heat transfer to the disk is q net,d = −99.8 W − 93.6 W − ( −248 W ) = +54.6 W

<

That is, there is a net heat transfer rate into the disk. (b) Using the energy balance, Eq. (1), and the rate equation, Eqs. (2) and (3) with the IHT Radiation Tool, Radiation, Exchange Analysis, Radiation surface energy balances and the Correlation Tool, Free Convection, Horizontal Plate (Hot surface up), the analysis was performed to obtain qnet,d as a function of Td. The results are plotted below. The steady-state condition occurs when qnet,d = 0 for which

<

Td = 413 K Continued …..

PROBLEM 13.114 (Cont.)

COMMENTS: The IHT workspace for the foregoing analysis is shown below.

Continued …..

PROBLEM 13.114 (Cont.)

PROBLEM 13.115 KNOWN: Radiation shield facing hot wall at Tw = 400 K is backed by an insulating material of known thermal conductivity and thickness which is exposed to ambient air and surroundings at 300 K. FIND: (a) Heat loss per unit area from the hot wall, (b) Radiosity of the shield, and (c) Perform a parameter sensitivity analysis on the insulation system considering effects of shield reflectivity ρs, insulation thermal conductivity k, overall coefficient h, on the heat loss from the hot wall. SCHEMATIC:

ASSUMPTIONS: (1) Wall is black surface of uniform temperature, (2) Shield and wall behave as parallel infinite plates, (3) Negligible convection in region between shield and wall, (4) Shield is 2 diffuse-gray and very thin, (5) Prescribed coefficient h = 10 W/m ⋅K is for convection and radiation. ANALYSIS: (a) Perform an energy balance on the shield to obtain q′′w − s = q′′cond But the insulating material and the convection process at the exposed surface can be represented by a thermal circuit.

In equation form, using Eq.13.24 for the wall and shield, q′′w − s =

(

(

4 σ Tw − Ts4

)

1/ ε w + 1/ ε s − 1

σ 4004 − Ts4

)=

1 + 1/ 0.05 − 1

=

Ts − T∞ L / k + 1/ h

(1,2)

(Ts − 300 ) K (0.025 / 0.016 + 1/10 ) m 2 ⋅ K / W

Ts = 350 K. where εs = 1 - ρs. Hence, q′′w − s =

(350 − 300 ) K = 30 W / m 2 . 2 (0.025 / 0.016 + 1/10 ) m ⋅ K / W

<

(b) The radiosity of the shield follows form the definition,

( )

( )

4 J s = ρs G s + ε s E b (Ts ) = ρs σ Tw + (1 − ρs ) σ Ts4 .

(3)

J s = 0.95σ ( 400 K ) + (1 − 0.95 )σ (350 K ) = 1421 W / m 2 .

<

4

-8

2

4

4

with σ = 5.67 × 10 W/m ⋅K . Continued …..

PROBLEM 13.115 (Cont.) (c) Using the Eqs. (1) and (2) in the IHT workspace, q′′w − s can be computed and plotted for selected ranges of the insulation system variables, ρs, k, and h. Intuitively we know that q′′w − s will decrease with increasing ρs, decreasing k and decreasing h. We chose to generate the following family of curves plotting q′′w − s vs. k for selected values of ρs and h.

Considering the base condition with variable k, reducing k by a factor of 3, the heat loss is reduced by 2 a factor of 2. The effect of changing h (4 to 24 W/m ⋅K) has little influence on the heat loss. However, the effect of shield reflectivity change is very significant. With ρs = 0.98, probably the upper limit of a practical reflector-type shield, the heat loss is reduced by a factor of two. To improve the performance of the insulation system, it is most advantageous to increase ρs and decrease k.

PROBLEM 13.116 KNOWN: Diameter and surface temperature of a fire tube. Gas low rate and temperature. Emissivity of tube and partition. FIND: (a) Heat transfer per unit tube length, q′, without the partition, (b) Partition temperature, Tp, and heat rate with the partition, (c) Effect of flow rate and emissivity on q′ and Tp. Effect of emissivity on radiative and convective contributions to q′. SCHEMATIC:

ASSUMPTIONS: (1) Fully-developed flow in duct, (2) Diffuse/gray surface behavior, (3) Negligible gas radiation. -7

2

PROPERTIES: Table A-4, air (Tm,g = 900 K): µ = 398 × 10 N⋅s/m , k = 0.062 W/m⋅K, Pr = 0.72; -7 2 air (Ts = 385 K): µ = 224 × 10 N⋅s/m .  ANALYSIS: (a) Without the partition, heat transfer to the tube wall is only by convection. With m

−7  / π Dµ = 4 ( 0.05 kg / s ) / π ( 0.07 m ) 398 × 10 = 0.05 kg/s and ReD = 4 m N ⋅ s / m 2 = 22,850, the g flow is turbulent. From Eq. (8.61),

Nu D = 0.027 Re 4D/ 5 Pr1/ 3 ( µ / µs )

0.14

h=

k D

Nu D =

0.062 W / m ⋅ K 0.07 m

(

= 0.027 ( 22,850 )

4/5

(0.72 )1/ 3 (398 / 224 )0.14 = 80.5

80.5 = 71.3 W / m 2 ⋅ K

)

q′ = hπ D Tm,g − Ts = 71.3 W / m 2 ⋅ K (π ) 0.07 m (900 − 385 ) = 8075 W / m (b) The temperature of the partition is determined from an energy balance which equates net radiation exchange with the tube wall to convection from the gas. Hence, q′′rad = q′′conv , where from Eq. 13.23, q′′rad =

<

(

σ Tp4 − Ts4 1 − εp

εp

+

)

1 1 − εs Ap + ε s As Fps

where F12 = 1 and Ap/As = D/(πD/2) = 2/π = 0.637. The flow is now in a noncircular duct for which 2   / 2 = 0.025 Dh = 4Ac/P = 4(πD /8)/(πD/2+D) = πD/(π + 2) = 0.611 D = 0.0428 m and m g 1/ 2 = m 2

2

  kg/s. Hence, ReD = m 1/ 2 Dh/Acµ = m 1/ 2 Dh/(πD /8)µ = 8(0.025 kg/s) (0.0428 m)/π(0.07 m) 398 ×

-7

2

10 N⋅s/m = 13,970 and Nu D = 0.027 (13, 970 )

4/5

h=

k Dh

Nu D =

(0.72 )1/ 3 (398 / 224 )0.14 = 54.3

0.062 W / m ⋅ K 0.0428 m

54.3 = 78.7 W / m 2 ⋅ K Continued …..

PROBLEM 13.116 (Cont.)

(

)

Hence, with εs = εp = 0.5 and q′′conv = h Tm,g − Tp ,

(

)

5.67 × 10−8 W / m 2 ⋅ K 4 Tp4 − 3854 K 4 1 + 1 + 0.637

(

)

= 78.7 W / m 2 ⋅ K 900 − Tp K

21.5 × 10−8 Tp4 + 78.7Tp − 71, 302 = 0 A trial-and-error solution yields

<

Tp = 796 K The heat rate to one-half of the tube is then

(

Dσ Tp4 − Ts4

′ 2 = q′ps + qconv ′ q1/ = 1 − εp

εp ′ 2= q1/

(

0.07 m 5.67 × 10

−8

2

W /m ⋅K

4

)

1 1 − εs Ap + + ε s As Fps

)(796.4

4

− 385

4

)K

2.637

(

+ h (π D / 2 ) Tm,g − Ts

)

4

+ 78.7 W / m ⋅ K ( 0.110 m )(900 − 385 ) K 2

′ 2 = 572 W / m + 4458 W / m = 5030 W / m q1/ The heat rate for the entire tube is ′ 2 = 10, 060 W / m q′ = 2q1/

<

(c) The foregoing model was entered into the IHT workspace, and parametric calculations were performed to obtain the following results.

Radiation transfer from the partition increases with increasing εp = εs, thereby reducing Tp while  T and q ′ also increase with m.  increasing q′. Since h increases with increasing m, p Continued …..

PROBLEM 13.116 (Cont.)

Although the radiative contribution to the heat rate increases with increasing εp = εs, it still remains small relative to convection. COMMENTS: Contrasting the heat rate predicted for part (b) with that for part (a), it is clear that use of the partition enhances heat transfer to the tube. However, the effect is due primarily to an increase in h and secondarily to the addition of radiation.

PROBLEM 13.117 KNOWN: Height and width of a two-dimensional cavity filled with helium. Temperatures and emissivities of opposing vertical plates. FIND: (a) Heat rate per unit length, (b) Effect of L on heat rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Isothermal plates, (3) Diffuse-gray surfaces, (4) Reradiating cavity sidewalls. -6

2

PROPERTIES: Table A-4, Helium (T = 318 K, 1 atm): ν = 136 × 10 m /s, k = 0.158 W/m⋅K, α = -6 2 -1 201 × 10 m /s, Pr = 0.679, β = 0.00314 K . ANALYSIS: (a) The power generated by the electronics leaving the surface A1 is q′ = q′conv + q′rad , or q′ = hH ( T1 − T2 ) +

(

Hσ T14 − T24

)

1 − ε1 1 1 − ε2 + ε1 F + [(1/ F ) + (1/ F )]−1 ε2 12 1R 2R

The free convection coefficient can be estimated using Eq. 9.50 with Ra L =

gβ ( T1 − T2 ) L3

αν

=

(

)

9.8 m / s 2 0.00314 K −1 60 K ( 0.02 m )

3

201 × 10−6 × 136 × 10−6 m 4 / s 2

= 540

However, since RaL < 1000, free convection effects can be neglected, in which case heat transfer is by conduction and Nu L = 1. Hence, h = Nu L ( k / L ) = 1.0 ( 0.158 W / m ⋅ K / 0.02m ) = 7.9 W / m 2 ⋅ K. The view factor can be found from Fig. 13.4 with X/L = 8 and Y/L = ∞. Hence, F12 = 0.9 and F1R = F2R = 0.1. It follows that q′ = 7.9 W / m 2 ⋅ K ( 0.16 m )( 60°C ) +

)

(

0.16 m × 5.67 × 10−8 W / m 2 ⋅ K 3484 − 2884 K 4 0.25 +

1 0.9 + 10 + 10−1 



+ 0.25



<

q′ = 75.8 W / m + 45.5 W / m = 121 W / m.

Continued …..

PROBLEM 13.117 (Cont.) (b) To assess the effect of plate spacing on convection heat transfer, q′conv = hH ( T1 − T2 ) was

(

)

computed by using both Eq. 9.50 and the conduction limit Nu L = 1 to determine h. These expressions were entered into the IHT workspace, and the Radiation Toolpad was used to obtain the appropriate radiation rate equation and view factor.

The cross-over at L = 28 mm marks the plate spacing below and above which, respectively, the conduction limit and Eq. 9.50 are applicable. Although there is a slight increase in q′conv with increasing L for L > 28 mm, the increase pales by comparison with that corresponding to a reduction in L for L < 28 mm. As the two plates are brought closer to each other in the conduction limit, the reduction in the corresponding thermal resistance significantly increases the heat rate. The total heat rate and the conduction and radiative components are also plotted for 5 ≤ L ≤ 25 mm. There is an increase in q′rad with decreasing L, due to an increase in F12 (F12 = 0.97 for L = 5 mm). However, the increase is small, and conduction is the dominant heat transfer mode. COMMENTS: Even for small values of L, the total heat rate is small and the scheme is poorly suited for electronic cooling. Note that helium is preferred over air on the basis of its larger thermal conductivity.

PROBLEM 13.118 KNOWN: Diameters, temperatures, and emissivities of concentric spheres. FIND: Rate at which nitrogen is vented from the inner sphere. Effect of radiative properties on evaporation rate. SCHEMATIC:

ASSUMPTIONS: Diffuse-gray surfaces. 5

PROPERTIES: Liquid nitrogen (given): hfg = 2 × 10 J/kg; Table A-4, Helium ( T = (Ti + To)/2 = -6

2

-6

2

180 K, 1 atm): ν = 51.3 × 10 m /s, k = 0.107 W/m⋅K, α = 76.2 × 10 m /s, Pr = 0.673, β = -1 0.00556 K . ANALYSIS: (a) Performing an energy balance for a control surface about the liquid nitrogen, it  follows that q = qconv + qrad = mh fg . From the Raithby-Hollands expressions for free convection between concentric spheres, qconv = keffπ(Di Do/L)(To – Ti), where 1/ 4

  1/ 4  Pr L    k eff = 0.74 k      0.861 + Pr   4 −7 / 5 −7 / 5 5  ( D D ) Di + D o   o i Ra L =

gβ ( To − Ti ) L3

να

=

)

(

)(206 K )(0.05m )3 = 3.589 ×105. (51.3 ×10−6 m2 / s )(76.2 ×10−6 m2 / s ) (

9.8 m / s 2 0.00556 K −1

Hence, 1/ 4 

0.673   k eff = 0.74 ( 0.107 W / m ⋅ K )    0.861 + 0.673  The heat rate by convection is

1/ 4

0.05 3.589 × 105 

   (1.1)4 (1 + 0.875 )5 

(

= 0.309 W / m ⋅ K.

)

q conv = ( 0.309 W / m ⋅ K )π 1.10 m 2 / 0.05 m 206 K = 4399 W. From Table 13.3, q rad = q oi =

(

σπ D12 To4 − Ti 4

)

1/ ε i + ((1 − ε o ) / ε o ) ( Di / Do )

2

2 5.67 × 10−8 W / m 2 ⋅ K 4 )π (1 m ) ( 2834 − 77 4 ) K 4 ( = = 216 W.

1/ 0.3 + ( 0.7 / 0.3)(1/1.1)

2

Continued …..

PROBLEM 13.118 (Cont.) Hence,

5  =q/h m fg = ( 4399 + 216 ) W / 2 × 10 J / kg = 0.023 kg / s.

<

With the cavity evacuated, IHT was used to compute the radiation heat rate as a function of εi = εo.

Clearly, significant advantage is associated with reducing the emissivities and qrad = 31.8 W for εi = εo = 0.05. COMMENTS: The convection heat rate is too large. It could be reduced by replacing He with a gas of smaller k, a cryogenic insulator (Table A.3), or a vacuum. Radiation effects are second order for small values of the emissivity.

PROBLEM 13.119 KNOWN: Dimensions, emissivity and upper temperature limit of coated panel. Arrangement and power dissipation of a radiant heater. Temperature of surroundings. FIND: (a) Minimum panel-heater separation, neglecting convection, (b) Minimum panel-heater separation, including convection. SCHEMATIC:

ASSUMPTIONS: (1) Top and bottom surfaces of heater and panel, respectively, are adiabatic, (2) Bottom and top surfaces of heater and panel, respectively are diffuse-gray, (3) Surroundings form a large enclosure about the heater-panel arrangement, (4) Steady-state conditions, (5) Heater power is dissipated entirely as radiation (negligible convection), (6) Air is quiescent and convection from panel may be approximated as free convection from a horizontal surface, (7) Air is at atmospheric pressure. -6

2

PROPERTIES: Table A-4, Air (Tf = (400 + 298)/2 ≈ 350 K, 1 atm): ν = 20.9 × 10 m /s, k = 0.03 -6 2 -3 -1 W/m⋅K, Pr = 0.700, α = 29.9 × 10 m /s, β = 2.86 × 10 K . ANALYSIS: (a) Neglecting convection effects, the panel constitutes a floating potential for which the net radiative transfer must be zero. That is, the panel behaves as a re-radiating surface for which Eb2 = J2. Hence J − E b2 J1 − E b3 + q1 = 1 (1) 1/ A1F12 1/ A1F13 and evaluating terms E b2 = σ T24 = 5.67 × 10−8 W / m 2 ⋅ K 4 ( 400 K ) = 1452 W / m 2 4

E b3 = σ T34 = 5.67 × 10−8 W / m 2 ⋅ K 4 ( 298 K ) = 447 W / m 2 4

F13 = 1 − F12

A1 = 25 m 2

find that 75, 000 W 25 m

2

J − 1452 J − 447 = 1 + 1 1/ F12 1/ (1 − F12 )

3000 W / m 2 = F12 ( J1 − 1452 ) + ( J1 − 447 ) − F12 ( J1 − 447 ) J1 = 3447 + 1005F12 .

(2)

Performing a radiation balance on the panel yields J1 − E b2 E b2 − E b3 = . 1/ A1F12 1/ A 2 F23 Continued …..

PROBLEM 13.119 (Cont.) With A1 = A2 and F23 = 1 – F12 F1 ( J1 − 1452 ) = (1 − F12 )(1452 − 447 ) or 447F12 = F12 J1 − 1005.

(3)

Substituting for J1 from Eq. (2), find 447F12 = F12 (3447 + 1005F12 ) − 1005 2 1005F12 + 3000F12 − 1005 = 0

F12 = 0.30. Hence from Fig. 13.4, with X/L = Y/L and Fij = 0.3, X / L ≈ 1.45

<

L ≈ 5 m /1.45 = 3.45 m. (b) Accounting for convection from the panel, the net radiation transfer is no longer zero at this surface and Eb2 ≠ J2. It then follows that q1 =

J1 − J 2

J − E b3 + 1 1/ A1F12 1/ A1F13

(4)

where, from an energy balance on the panel, J 2 − E b2

(1 − ε 2 ) / ε 2 A 2

= q conv,2 = hA 2 ( T2 − T∞ ) .

(5)

2

With L ≡ As/P = 25 m /20 m = 1.25 m, Ra L =

gβ ( Ts − T∞ ) L3

να

Hence

=

( 20.9 × 29.9 )10

(

3 9 Nu L = 0.15Ra1/ L = 0.15 8.94 × 10

h = 311 k / L = 311

)

(

9.8 m / s 2 2.86 × 10−3 K −1 (102 K )(1.25 m )

0.03 W / m ⋅ K 1.25 m

)

1/ 3

−12

3

4

m /s

2

= 8.94 × 109.

= 311

= 7.46 W / m 2 ⋅ K

q′′conv,2 = 7.46 W / m 2 ⋅ K (102 K ) = 761 W / m 2 . From Eq. (5) J 2 = E b2 +

1 − ε2

ε2

q′′conv,2 = 1452 +

0.7 0.3

761 = 3228 W / m 2 . Continued …..

PROBLEM 13.119 (Cont.) From Eq. (4), 75, 000 25

J − 3228 J − 447 = 1 + 1 1/ F12 1/ (1 − F12 )

3000 = F12 ( J1 − 3228 ) + J1 − 447 − F12 ( J1 − 447 ) J1 = 3447 + 2781F12 .

(6)

From an energy balance on the panel, E −J J 2 − E b2 + b3 2 = = q conv,2 1/ A1F12 1/ A 2 F23 (1 − ε 2 ) / ε 2 A 2 J1 − J 2

F12 ( J1 − 3228 ) + (1 − F12 )( 447 − 3228 ) = 761 F12 J1 − 447F12 = 3542. Substituting from Eq. (6), F12 (3447 + 2781F12 ) − 447F12 = 3542 2 + 3000F12 − 3542 = 0 2781F12

F12 = 0.71. Hence from Fig. 13.4, with X/L = Y/L and Fij = 0.71, X / L = 5.7

L ≈ 5 m / 5.7 = 0.88 m.

<

COMMENTS: (1) The results are independent of the heater surface radiative properties. (2) Convection at the heater surface would reduce the heat rate q1 available for radiation exchange and hence reduce the value of L.

PROBLEM 13.120 KNOWN: Diameter and emissivity of rod heater. Diameter and position of reflector. Width, emissivity, temperature and position of coated panel. Temperature of air and large surroundings. FIND: (a) Equivalent thermal circuit, (b) System of equations for determining heater and reflector temperatures. Values of temperatures for prescribed conditions, (c) Electrical power needed to operate heater. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Diffuse-gray surfaces, (3) Large surroundings act as blackbody, (4) Surfaces are infinitely long (negligible end effects), (5) Air is quiescent, (6) Negligible convection at reflector, (7) Reflector and panel are perfectly insulated. -6

2

PROPERTIES: Table A-4, Air (Tf = 350 K, 1 atm): k = 0.03 W/m⋅K, ν = 20.9 × 10 m /s, α = 29.9 -6 2 -6 2 × 10 m /s, Pr = 0.70; (Tf = (1295 + 300)/2 = 800 K): k = 0.0573 W/m⋅K, ν = 84.9 × 10 m /s, α = -6 2 120 × 10 m /s. ANALYSIS: (a) We have assumed blackbody behavior for A1 and A4; hence, J = Eb. Also, A2 is insulated and has negligible convection; hence q = 0 and J2 = Eb2. The equivalent thermal circuit is:

(b) Performing surface energy balances at 1, 2 and 3: E − E b2 E b1 − J3 E b1 − E b4 + + q1 − q conv,1 = b1 1/ A1F12 1/ A1F13 1/ A1F14 E b1 − E b2

(2)

E b1 − J3

(3a)

= q conv,3 .

(3b)

0= J3 − E b3

(1 − ε 3 ) / ε 3A3

(1)

=

J − E b2 E b4 − E b2 + 3 + 1/ A 2 F21 1/ A 2 F23 1/ A 2 F24

E −J E −J + b2 3 + b4 3 1/ A3F31 1/ A3F32 1/ A3F34

where J3 − E b3

(1 − ε 3 ) / ε 3A3

Continued …..

PROBLEM 13.120 (Cont.) Solution procedure with Eb3 and Eb4 known: Evaluate qconv,3 and use Eq. (3b) to obtain J3; Solve Eqs. (2) and (3a) simultaneously for Eb1 and Eb2 and hence T1 and T2; Evaluate qconv,1 and use Eq. (1) to obtain q1. For free convection from a heated, horizontal plate: Lc =

As P

Ra L =

=

(W × L)

( 2L + 2W )

≈

gβ ( T3 − T∞ ) L3c

αν

W 2

=

= 0.5 m −1

9.8 m / s 2 (350 K )

20.9 × 29.9 × 10

(

3 8 Nu L = 0.15Ra1/ L = 0.15 5.6 × 10

h3 =

k Lc

Nu L =

(100 K )(0.5 m )3

)

1/ 3

0.03 W / m ⋅ K × 123.6 0.5 m

−12

4

m /s

2

= 5.6 × 108

= 123.6

= 7.42 W / m 2 ⋅ K.

q′′conv,3 = h3 ( T3 − T∞ ) = 742 W / m 2 . Hence, with E b3 = σ T34 = 5.67 × 10 −8 W / m 2 ⋅ K 4 ( 400 K ) = 1451 W / m 2 4

using Eq. (3b) find J 3 = E b3 +

1 − ε3

ε 3A3

q conv,3 = (1451 + [0.3 / 0.7 ] 742 ) = 1769 W / m 2 . -1

-1

View Factors: From symmetry, it follows that F12 = 0.5. With θ = tan (W/2)/H = tan (0.5) = 26.57°, it follows that F13 = 2θ / 360 = 0.148. From summation and reciprocity relations, F14 = 1 − F12 − F13 = 0.352 F21 = ( A1 / A 2 ) F12 = ( 2D1 / D 2 ) F12 = 0.02 × 0.5 = 0.01 F23 = ( A3 / A 2 ) F32 = ( 2 / π )( F32′ − F31 ). For X/L = 1, Y/L ≈ ∞, find from Fig. 13.4 that F32′ ≈ 0.42. Also find, F31 = ( A1 / A3 ) F13 = (π × 0.01/1) 0.148 = 0.00465 ≈ 0.005 F23 = ( 2 / π )(0.42 − 0.005 ) = 0.264 F22 ≈ 1 − F22′ = 1 − ( A′2 / A 2 ) F2′2 = 1 − ( 2 / π ) = 0.363 F24 = 1 − F21 − F22 − F23 = 0.363 Continued …..

PROBLEM 13.120 (Cont.) F31 = 0.005,

F32 = 0.415

F34 = 1 − F32′ = 1 − 0.42 = 0.58. With E b4 = σ T44 = 5.67 × 10−8 W / m 2 ⋅ K 4 (300 K ) = 459 W / m 2 , Eq. (3a) → 0.005(Eb1 – 1769) + 0.415(Eb2 – 1769) + 0.58(459 – 1769) = 742 0.005Eb1 + 0.415Eb2 = 2245 4

Eq. (2) → 0.01(Eb1 – Eb2) + 0.264(1769 – Eb2) + 0.363(459 – Eb2) = 0 0.01Eb1 – 0.637Eb2 + 633.6 = 0.

(4)

(5)

Hence, manipulating Eqs. (4) and (5), find E b2 = 0.0157E b1 + 994.7 0.005E b1 + ( 0.415 )( 0.0157E b1 + 994.7 ) = 2245. T1 = ( E b1 / σ )

1/ 4

E b1 = 159, 322 W / m 2

E b2 = 0.0157 (159, 322 ) + 994.7 = 3496 W / m 2

<

= 1295 K T2 = ( E b2 / σ )

1/ 4

= 498 K.

<

(c) With T1 = 1295 K, then Tf = (1295 + 300)/2 ≈ 800 K, and using Ra D =

gβ ( T1 − T∞ ) D13

αν

9.8 m / s 2 (1/ 800 K )(1295 − 300 ) K ( 0.01 m )

3

=

120 × 84.9 × 10−12 m 4 / s 2

= 0.85 (1196 ) Nu D = 0.85Ra 0.188 D

0.188

= 1196

= 3.22

h1 = ( k / D1 ) Nu D = (0.0573 / 0.01) × 3.22 = 18.5 W / m 2 ⋅ K. The convection heat flux is q′′conv,1 = h1 ( T1 − T∞ ) = 18.5 (1295 − 300 ) = 18, 407 W / m 2 , Using Eq. (1), find q1′′ = q′′conv,1 + F12 ( E b1 − E b2 ) + F13 ( E b1 − J3 ) + F14 ( E b1 − E b4 ) q1′′ = 18, 407 + 0.5 (159, 322 − 3496 )

+0.148 (159, 322 − 1769 ) + 0.352 (159, 322 − 459 )

q1′′ = 18, 407 + ( 77, 913 + 23, 314 + 55, 920 ) q1′′ = 18, 407 + 236, 381 = 254, 788 W / m 2 q1′ = π D1q1′′ = π ( 0.01) 254, 788 = 8000 W / m.

<

COMMENTS: Although convection represents less than 8% of the net radiant transfer from the heater, it is equal to the net radiant transfer to the panel. Since the reflector is a re-radiating surface, results are independent of its emissivity.

PROBLEM 13.121 KNOWN: Temperature, power dissipation and emissivity of a cylindrical heat source. Surface emissivities of a parabolic reflector. Temperature of air and surroundings. FIND: (a) Radiation circuit, (b) Net radiation transfer from the heater, (c) Net radiation transfer from the heater to the surroundings, (d) Temperature of reflector. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Heater and reflector are in quiescent and infinite air, (3) Surroundings are infinitely large, (4) Reflector is thin (isothermal), (5) Diffuse-gray surfaces. -6

2

PROPERTIES: Table A-4, Air (Tf = 750 K, 1 atm): ν = 76.37 × 10 m /s, k = 0.0549 W/m⋅K, α = -6 2 109 × 10 m /s, Pr = 0.702. ANALYSIS: (a) The thermal circuit is

(b) Energy transfer from the heater is by radiation and free convection. Hence, ′ P1′ = q1′ + q1,conv where

′ = hπ D1 ( T1 − T∞ ) q1,conv

and Ra D =

gβ ( T1 − T∞ ) D3

=

−1

9.8 m / s 2 ( 750 K )

(900 K )(0.005 m )3

να 76.37 × 109 × 10−12 m 4 / s 2 Using the Churchill and Chu correlation, find

= 176.6.

2

Nu D

2

    1/ 6 6     0.387 176.6 0.387Ra1/ ( )   D = 0.6 + = 0.6 + = 1.85  8 / 27 8 / 27  9 /16 9 /16     1 + ( 0.559 / Pr )  1 + ( 0.559 / 0.702 )          h = Nu D ( k / D ) = 1.85 (0.0549 W / m ⋅ K / 0.005 m ) = 20.3 W / m 2 ⋅ K. Continued …..

PROBLEM 13.121 (Cont.) Hence, ′ = 20.3 W / m 2 ⋅ Kπ ( 0.005 m )(1200 − 300 ) K = 287 W / m q1,conv q1′ = 1500 W / m − 287 W / m = 1213 W / m.

<

(c) The net radiative heat transfer from the heater to the surroundings is q1′ (sur ) = A1′ F1sur ( J1 − Jsur ) . The view factor is F1sur = (135 / 360 ) = 0.375 and the radiosities are 4 J sur = σ Tsur = 5.67 × 10−8 W / m 2 ⋅ K 4 (300 K ) = 459 W / m 2 4

J1 = E b1 − q1′ (1 − ε1 ) ε1A1′ = 5.67 × 10−8 W / m 2 ⋅ K 4 (1200 K )

4

−1213 W / m [0.2 / 0.8π ( 0.005 m )]

J1 = 98, 268 W / m 2 . Hence q1′ (sur ) = π ( 0.005 m ) 0.375 (98, 268 − 459 ) W / m 2 = 576 W / m.

<

(d) Perform an energy balance on the reflector, q′2i = q′2o + q′2,conv J 2i − E b2

(1 − ε 2i ) / ε 2i A′2

=

E b2 − Jsur + 2h 2 A′2 ( T2 − T∞ ) . (1 − ε 2o ) / ε 2o A′2 + 1/ A′2 F2o(sur )

The radiosity of the reflector is q1′ ( 2i ) (1213 − 576 ) W / m J 2i = J1 − = 98, 268 W / m 2 − A1′ F1( 2i ) π ( 0.005 m )( 225 / 360 ) J 2i = 33, 384 W / m 2 . Hence

( )=

( )

33, 384 − 5.67 × 10−8 T24

5.67 × 10−8 T24 − 459

(0.9 / 0.1 × 0.2 m )

(0.2 / 0.8 × 0.2 m ) + (1/ 0.2m ×1)

+ 2 × 0.4 ( T2 − 300 )

741.9 − 0.126 × 10−8 T24 = 0.907 × 10−8 T24 − 73.4 + 0.8T2 − 240 1.033 × 10−8 T24 + 0.8T2 = 1005 and from a trial and error solution, find T2 = 502 K.

<

COMMENTS: Choice of small ε2i and large ε2o insures that most of the radiation from heater is reflected to surroundings and reflector temperature remains low.

PROBLEM 13.122 KNOWN: Geometrical conditions associated with tube array. Tube wall temperature and pressure of water flowing through tubes. Gas inlet velocity and temperature when heat is transferred from products of combustion in cross-flow, or temperature of electrically heated plates when heat is transferred by radiation from the plates. FIND: (a) Steam production rate for gas flow without heated plates, (b) Steam production rate with heated plates and no gas flow, (c) Effects of inserting unheated plates with gas flow. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible gas radiation, (3) Tube and plate surfaces may be approximated as blackbodies, (4) Gas outlet temperature is 600 K. 3

PROPERTIES: Table A-4, Air ( T = 900 K, 1 atm): ρ = 0.387 kg/m , cp = 1121 J/kg⋅K, ν = 102.9 × -6 2 3 10 m /s, k = 0.062 W/m⋅K, Pr = 0.720; (T = 400 K): Pr = 0.686; (T = 1200 K): ρ = 0.29 kg/m ; 6 Table A-6, Sat. water (2.5 bars): hfg = 2.18 × 10 J/kg. ANALYSIS: (a) With Vmax = [ST / (ST − D )] V = 20 m / s V D 20 m / s (0.01 m ) Re D = max = = 1944 ν 102.9 × 10−6 m 2 / s and from the Zhukauskas correlation with C = 0.27 and m = 0.63, Nu D = 0.27 (1944 )

0.63

(0.720 )0.36 (0.720 / 0.686 )1/ 4 = 28.7

h = 0.062 W / m ⋅ K × 28.7 / 0.01 m = 178 W / m 2 ⋅ K. The outlet temperature may be evaluated from

 hA Ts − Tm,o = exp  −  mc Ts − Tm,i   p

  hNπ DL   = exp  −    ρ VN TST Lc p 

  178 W / m 2 ⋅ K × 100 × π × 0.01 m = exp  −   0.29 kg / m3 × 10 m / s × 5 × 0.02 m × 1121 J / kg ⋅ K  400 − 1200  

400 − Tm,o

Tm,o = 543 K.

Continued …..

PROBLEM 13.122 (Cont.) With ∆T"m =

(Ts − Tm,i ) − (Ts − Tm,o ) = −800 − ( −143) = −382 K ln (800 /143) ln ( Ts − Tm,i ) / ( Ts − Tm,o )

find q = hA∆T"m = 178 W / m 2 ⋅ K (100 )π ( 0.01 m )1 m ( −382 K ) q = −214 kW.  hfg, If the water enters and leaves as saturated liquid and vapor, respectively, it follows that –q = m hence 214, 000 W  = = 0.098 kg / s. m 2.18 × 106 J / kg

<

(b) The radiation exchange between the plates and tube walls is

(

)

q =  A p Fpsσ Tp4 − Ts4  ⋅ 2 ⋅ NT



where the factor of 2 is due to radiation transfer from two plates. The view factor and area are 1/ 2

2 Fps = 1 − 1 − ( D / S ) 





(

)

1/ 2 + ( D / S ) tan −1  S2 − D 2 / D2 





Fps = 1 − 0.866 + 0.5 tan −1 1.732 = 1 − 0.866 + 0.524 Fps = 0.658 A p = N L ⋅ SL ⋅ 1 m = 20 × 0.02 m × 1 m = 0.40 m 2 . Hence,

(

)

q = 5 ×  0.80 m 2 × 0.658 × 5.67 × 10−8 W / m 2 ⋅ K 4 12004 − 4004 K 4 





q = 305, 440 W and the steam production rate is  = m

305, 440 W 2.18 × 106 J / kg

= 0.140 kg / s.

<

(c) The plate temperature is determined by an energy balance for which convection to the plate from the gas is equal to net radiation transfer from the plate to the tube. Conditions are complicated by the fact that the gas transfers energy to both the plate and the tubes, and its decay is not governed by a simple exponential. Insertion of the plates enhances heat transfer to the tubes and thereby increases the steam generation rate. However, for the prescribed conditions, the effect would be small, since in case (a), the heat transfer is already ≈ 80% of the maximum possible transfer.

PROBLEM 13.123 KNOWN: Gas-fired radiant tube located within a furnace having quiescent gas at 950 K. At a particular axial location, inner wall and gas temperature measured by thermocouples. FIND: Temperature of the outer tube wall at the axial location where the thermocouple measurements are being made. SCHEMATIC:

ASSUMPTIONS: (1) Silicon carbide tube walls have negligible thermal resistance and are diffusegray, (2) Tubes are positioned horizontally, (3) Gas is radiatively non-participating and quiescent, (4) Furnace gas behaves as ideal gas, β = 1/T. 3

-6

2

PROPERTIES: Gas (given): ρ = 0.32 kg/m , ν = 130 × 10 m /s, k = 0.070 W/m⋅K, Pr = 0.72, α = -4 2 ν/Pr = 1.806 × 10 m /s. ANALYSIS: Consider a segment of the outer tube at the Prescribed axial location and perform an energy balance, E ′in − E ′out = 0 q′rad,i + q′conv,i − q′rad,o − q′conv,o = 0

(1)

The heat rates by radiative transfer are: Inside: For long concentric cylinders, Eq. 13.25, q′rad,i =

q′rad,i =

(

4 4 σπ Di Ts,i − Ts,o

)

1/ ε1 + (1 − ε 2 ) / ε 2 ( Di / Do )

(

)

4 5.67 × 10−8 W / m 2 ⋅ K 4π ( 0.10 m ) 12004 − Ts,o K4

1/ 0.6 + (1 − 0.6 ) / 0.6 ( 0.10 / 0.20 )

(

)

4 q′rad,i = 8.906 × 10−9 12004 − Ts,o .

Outside: For the outer tube surface to large surroundings,

(

)

(2)

(

)

4 4 4 − Tsur = 0.6π ( 0.20 m ) 5.67 × 10−8 W / m 2 ⋅ K 4 Ts,o − 9504 K 4 q′rad,o = επ Doσ Ts,o

(

)

4 − 9504 . q′rad,o = 2.138 × 10−8 Ts,o

(3)

The heat rates by convection processes are: Continued …..

PROBLEM 13.123 (Cont.) Inside: The rate equation is q′conv,i = h iπ Do Tm,g − Ts,o .

(

)

(4)

(

)

Find the Reynolds number with A c = π Do2 − Di2 / 4 and D h = 4 A c / P,

Dh =

(

 / ρ A = 0.13 kg / s /  0.32 kg / m × π / 4 0.2 − 0.1 um = m c

Re D = u m D h / ν

(

4 (π / 4 ) D o − Di 4

2

π ( D o + Di )

) = π (0.2

2

2

− 0.1

)m

3



2

( 0.2 + 0.1) m

= 0.100 m

Re D =

2

2

) m  = 17.2 m / s 2

17.2 m / s × 0.100 m 130 × 10

−6

2

= 13, 231.

m /s

The flow is turbulent and assumed to be fully developed; from the Dittus-Boelter correlation, 0.3 Nu D = hDh / k = 0.023 Re0.8 D Pr 0.070 W / m ⋅ K 0.8 0.3 × 0.023 (13, 231) ( 0.720 ) = 28.9 W / m 2 ⋅ K hi = 0.100 m Substituting into Eq. (4),

(

)

(

)

q′conv,i = 28.9 W / m 2 ⋅ K × π ( 0.20 m ) 1050 − Ts,o K = 18.16 1050 − Ts,o .

(5)

Outside: The rate equation is q′conv,o = h oπ Do Ts,o − T∞ .

(

)

Evaluate the Rayleigh number assuming Ts,o = 1025 K so that Tf = 987 K, Ra D =

gβ∆TD3o

9.8 m 2 / s 2 (1/ 987 K )(1025 − 950 ) K ( 0.20 m )

3

=

−6

να 130 × 10 m / s × 1.806 × 10 For a horizontal tube, using Eq. 9.33 and Table 9.1, 2

(

n = 0.48 2.537 × 105 Nu D = h o Do / k = CRa D

)

1/ 4

−4

2

= 2.537 × 105.

m /s

= 10.77

h o = ( 0.070 W / m ⋅ K ) / 0.20 m × 10.77 = 3.77 W / m 2 ⋅ K. Substituting into Eq. (6)

(

)

(

)

q′conv,o = 3.77 W / m 2 ⋅ K × π ( 0.20 m ) Ts,o − 950 K = 2.369 Ts,o − 950 .

(7)

Returning to the energy balance relation on the outer tube, Eq. (1), substitute for the individual rates from Eqs. (2, 5, 3, 7), 8.906 × 10

−9

(1200

4

4

)

(

)

− Ts,o + 18.16 1050 − Ts,o − 2.138 × 10

By trial-and-error, find

−8

(T

)

(

)

4 4 s,o − 950 − 2.369 Ts,o − 950 = 0

(8)

<

Ts,o = 1040 K.

COMMENTS: (1) Recall that in estimating ho we assumed Ts,o = 1025 K, such that ∆T = 75 K (vs. 92 K using Ts,o = 1042 K) for use in evaluating the Rayleigh number. For an improved estimate of Ts,o, it would be advisable to recalculate ho. (2) Note from Eq. (8) that the radiation processes dominate the heat transfer rate: q′rad ( W / m ) Inside Outside

7948 7839

q′conv ( W / m ) 136 219

PROBLEM 13.124 KNOWN: Temperature and emissivity of ceramic plate which is separated from a glass plate of equivalent height and width by an air space. Temperature of air and surroundings on opposite side of glass. Spectral radiative properties of glass. FIND: (a) Transmissivity of glass, (b) Glass temperature Tg and total heat rate qh, (c) Effect of external forced convection on Tg and qh. SCHEMATIC:

ASSUMPTIONS: (1) Spectral distribution of emission from ceramic approximates that of a blackbody, (2) Glass surface is diffuse, (3) Atmospheric air in cavity and ambient, (4) Cavity may be approximated as a two-surface enclosure with infinite parallel plates, (5) Glass is isothermal. PROPERTIES: Table A-4, air (p = 1 atm): Evaluated at T = (Tc + Tg)/2 and Tf = (Tg +T∞)/2 using IHT Properties Toolpad. ANALYSIS: (a) The total transmissivity of the glass is ∞

τ λ E λ b dλ λ2 =1.6 µ m ∫ o τ= = ∫ (Eλ ,b / E b ) dλ = F(0→λ2 ) − F(0→λ1 ) E b

λ1 = 0.4 µ m

With λ2T = 1600 µm⋅K and λ1T = 400 µm⋅K, respectively, Table 12.1 yields F( 0→ λ ) = 0.0197 and 2 F( 0→ λ

1)

= 0.0. Hence,

<

τ = 0.0197 With so little transmission of radiation from the ceramic, the glass plate may be assumed to be opaque to a good approximation. Since more than 98% of the incident radiation is at wavelengths exceeding 1.6 µm, for which αλ = 0.9, αg ≈ 0.9. Also, since Tg < Tc, nearly 100% of emission from the glass is at λ > 1.6 µm, for which ελ = αλ = 0.9, εg = 0.9 and the glass may be approximated as a gray surface. (b) The glass temperature may be obtained from an energy balance of the form q′′conv,i + q′′rad,i = q′′conv,o + q′′rad,o . Using Eqs. 13.24 and 13.27 to evaluate q′′rad,i and q′′rad,o , respectively, it follows that

(

)

hi Tc − Tg +

(

σ Tc4 − Tg4 1

εc

+

1

εg

)=h

−1

(

)

(

4 4 o Tg − T∞ + ε gσ Tg − Tsur

) Continued …..

PROBLEM 13.124 (Cont.) where, assuming 104 ≤ Ra L ≤ 107 , hi and h o are given by Eqs. 9.52 and 9.26, respectively, k 4 0.012 hi = i Ra1/ ( H / L )−0.3 L Pri L

  1/ 6  ko  0.387Ra  H ho = 0.825 + 8 / 27  H   1 + ( 0.492 / Pr )9 /16  o     3

2

3

with RaL = gβ i (Tc – Tg)L /νiαi and RaH = gβ o (Tg - T∞) H /νoαo. Entering the energy balance into the IHT workspace and using the Correlations, Properties and Radiation Toolpads to evaluate the convection and radiation terms, the following result is obtained. Tg = 825 K

<

The corresponding value of qh is q h = 108 kW

<

where qconv,i = 3216 W, qrad,i = 104.7 kW, qconvo,o = 15,190 W and qrad,o = 92.8 kW. The convection coefficients are hi = 4.6 W / m 2 ⋅ K and ho = 7.2 W / m 2 ⋅ K. (c) For the prescribed range of ho , IHT was used to generate the following results.

With increasing ho , the glass is cooled more effectively and Tg must decrease. With decreasing Tg, qconv,i, qrad,i and hence qh must increase. Note that radiation makes the dominant contribution to heat transfer across the airspace. Although qrad,o decreases with decreasing Tg, the increase in qconv,o exceeds the reduction in qrad,o.

PROBLEM 13.125 KNOWN: Conditions associated with a spherical furnace cavity. FIND: Cooling rate needed to maintain furnace wall at a prescribed temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Blackbody behavior for furnace wall, (3) N2 is non-radiating. ANALYSIS: From an energy balance on a unit surface area of the furnace wall, the cooling rate per unit area must equal the absorbed irradiation from the gas (Eg) minus the portion of the wall’s emissive power absorbed by the gas q′′c = E g − α g E b ( Ts ) q′′c = ε gσ Tg4 − α gσ Ts4 . Hence, for the entire furnace wall,

(

)

q c = Asσ ε g Tg4 − α g Ts4 . The gas emissivity, εg, follows from Table 13.4 with Le = 0.65D = 0.65 × 0.5 m = 0.325 m = 1.066 ft. p c L e = 0.25 atm × 1.066 ft = 0.267 ft − atm and from Fig. 13.18, find εg = εc = 0.09. From Eq. 13.42,

 Tg  α g = α c = Cc    Ts 

0.45

(

)

× ε c Ts , p c L e Ts / Tg  .

With Cc = 1 from Fig. 13.19,

α g = 1 (1400 / 50 )

0.45

× ε c (500K, 0.095 ft − atm )

where, from Fig. 13.18,

ε c (500K, 0.095 ft − atm ) = 0.067. Hence

α g = 1(1400 / 500 )

0.45

× 0.067 = 0.106

and the heat rate is 2 4 4 q c = π ( 0.5 m ) 5.67 × 10−8 W / m 2 ⋅ K 4 0.09 (1400 K ) − 0.106 (500 K ) 



q c = 15.1 kW.



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PROBLEM 13.126 KNOWN: Diameter and gas pressure, temperature and composition associated with a gas turbine combustion chamber. FIND: Net radiative heat flux between the gas and the chamber surface. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Blackbody behavior for chamber surface, (3) Remaining species are non-radiating, (4) Chamber may be approximated as an infinitely long tube. ANALYSIS: From Eq. 13.40 the net rate of radiation transfer to the surface is

(

q net = Asσ ε g Tg4 − α g Ts4

)

or

(

q′net = π Dσ ε g Tg4 − α g Ts4

)

with As = πDL. From Table 13.4, Le = 0.95D = 0.95 × 0.4 m = 0.380 m = 1.25 ft. Hence, pwLe = pcLe = 0.152 atm × 1.25 ft = 0.187 atm-ft.

( ) Fig.13.18 ( Tg = 1273 K ) , → ε c ≈ 0.085. Fig.13.20 ( p w / ( p c + p w ) = 0.5, Lc ( p w + pc ) = 0.375 ft − atm, Tg ≥ 930°C ) , → ∆ε ≥ 0.01. Fig.13.16 Tg = 1273 K , → ε w ≈ 0.069.

From Eq. 13.38, ε g = ε w + ε c − ∆ε = 0.069 + 0.085 − 0.01 ≈ 0.144. From Eq. 13.41 for the water vapor,

(

α w = C w Tg / Ts

)0.45 × ε w (Ts , p w Lc Ts / Tg  )

where from Fig. 13.16 (773 K, 0.114 ft-atm), → εw ≈ 0.083,

α w = 1(1273 / 773)

0.45

× 0.083 = 0.104.

From Eq. 13.42, using Fig. 13.18 (773 K, 0.114 ft-atm), → εc ≈ 0.08,

α c = 1(1273 / 773)

0.45

× 0.08 = 0.100.

From Fig. 13.20, the correction factor for water vapor at carbon dioxide mixture,

(p w / ( pc + p w ) = 0.1, Le (p w + pc ) = 0.375, Tg ≈ 540°C ) , → ∆α ≈ 0.004 and using Eq. 13.43

α g = α w + α c − ∆α = 0.104 + 0.100 − 0.004 ≈ 0.200. Hence, the heat rate is 4 4 q′net = π ( 0.4 m ) 5.67 × 10−8 W / m 2 ⋅ K 4  0.144 (1273) − 0.200 ( 773 )  = 21.9 kW / m.





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PROBLEM 13.127 KNOWN: Pressure, temperature and composition of flue gas in a long duct of prescribed diameter. FIND: Net radiative flux to the duct surface. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Duct surface behaves as a blackbody, (3) Other gases are non-radiating, (4) Flue may be approximated as an infinitely long tube. ANALYSIS: With As = πDL, it follows from Eq. 13.40 that

(

q′net = π Dσ ε g Tg4 − α g Ts4

)

From Table 13.4, Le = 0.95D = 0.95 × 1 m = 0.95 m = 3.12 ft. Hence p w Le = 0.12 atm × 3.12 m = 0.312 atm − ft p c L e = 0.05 atm × 3.12 m = 0.156 atm − ft. With Tg = 1400 K, Fig. 13.16 → εw = 0.083; Fig. 13.18 → εc = 0.072. With pw/(pc + pw) = 0.67, Le(pw +pc) = 0.468 atm-ft, Tg ≥ 930°C, Fig. 13.20 → ∆ε = 0.01. Hence from Eq. 13.38,

ε g = ε w + ε c − ∆ε = 0.083 + 0.072 − 0.01 = 0.145. From Eq. 13.41,

(

α w = C w Tg / Ts

)0.45 × ε w (Ts , pw Le Ts / Tg  )

α w = 1(1400 / 400 )

0.45

× ε w Fig. 13.16 → ε w ( 400 K, 0.0891 atm − ft ) = 0.1

α w = 0.176. From Eq. 13.42,

(

α c = Cc Tg / Ts

)0.45 × ε c (Ts , pcLeTs / Tg )

α c = 1(1400 / 400 )

0.45

× ε c Fig. 13.18 → ε c ( 400 K, 0.0891 atm − ft ) = 0.053

α c = 0.093. With pw/(pc + pw) = 0.67, Le(pw + pc) = 0.468 atm-ft, Tg ≈ 125°C, Fig. 13.20 gives ∆α ≈ 0.003. Hence from Eq. 13.43, α g = α w + α c − ∆α = 0.176 + 0.093 − 0.003 = 0.266. The heat rate per unit length is 4 4 q′net = π (1 m ) 5.67 × 10−8 W / m 2 ⋅ K 4 0.145 (1400 K ) − 0.266 ( 400 K ) 



q′net = 98 kW / m.



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PROBLEM 13.128 KNOWN: Gas mixture of prescribed temperature, pressure and composition between large parallel plates of prescribed separation. FIND: Net radiation flux to the plates. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Furnace wall behaves as a blackbody, (3) O2 and N2 are non-radiating, (4) Negligible end effects. ANALYSIS: The net radiative flux to a plate is

(

)

′′ = Gs − Es = ε gσ Tg4 − 1 − τ g σ Ts4 qs,1 where G s = ε gσ Tg4 + τ g Es , E s = σ Ts4 and τ g = 1 − α g ( Ts ) . From Table 13.4, Le = 1.8L = 1.8 × 0.75 m = 1.35 m = 4.43 ft. Hence pwLe = pcLe = 1.33 atm-ft. From Figs. 3.16 and 3.18 find εw ≈ 0.22 and εc ≈ 0.16 for p = 1 atm. With (pw + p)/2 = 1.15 atm, Fig. 13.17 yields Cw ≈ 1.40 and from Fig. 13.19, Cc ≈ 1.08. Hence, the gas emissivities are

ε w = C w ε w (1 atm ) ≈ 1.40 × 0.22 = 0.31

ε c = Ccε c (1 atm ) ≈ 1.08 × 0.16 = 0.17.

From Fig. 13.20 with pw/(pc + pw) = 0.5, Le(pc + pw) = 2.66 atm-ft and Tg > 930°C, ∆ε ≈ 0.047. Hence, from Eq. 13.38, ε g = ε w + ε c − ∆ε ≈ 0.31 + 0.17 − 0.047 ≈ 0.43. To evaluate αg at Ts, use Eq. 13.43 with

(

α w = C w Tg / Ts

)0.45 ε w (Ts , pw L2Ts / Tg ) = Cw (1300 / 500 )0.45 ε w (500, 0.51)

α w ≈ 1.40 (1300 / 500 )

0.45

α c = Cc (1300 / 500 )

0.45

0.22 = 0.47

ε c (500, 0.51) ≈ 1.08 (1300 / 500 )

0.45

0.11 = 0.18.

From Fig. 13.20, with Tg ≈ 125°C and Le(pw + pc) = 2.66 atm-ft, ∆α = ∆ε ≈ 0.024. Hence α g = α w + α c − ∆α ≈ 0.47 + 0.18 − 0.024 ≈ 0.63 and τ g = 1 − α g ≈ 0.37. Hence, the heat flux from Eq. (1) is ′′ = 0.43 × 5.67 × 10−8 W / m 2 ⋅ K 4 (1300 K ) − 0.63 × 5.67 × 10−8 W / m 2 ⋅ K 4 (500 K ) qs,1 4

4

q′′s,1 ≈ 67.4 kW / m 2 . The net radiative flux to both plates is then q′′s,2 ≈ 134.8 kW / m 2 .

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PROBLEM 13.129 KNOWN: Flow rate, temperature, pressure and composition of exhaust gas in pipe of prescribed diameter. Velocity and temperature of external coolant. FIND: Pipe wall temperature and heat flux. SCHEMATIC:

ASSUMPTIONS: (1) L/D >> 1 (infinitely long pipe), (2) Negligible axial gradient for gas temperature, (3) Gas is in fully developed flow, (4) Gas thermophysical properties are those of air, (5) Negligible pipe wall thermal resistance, (6) Negligible pipe wall emission. 3

-7

PROPERTIES: Table A-4: Air (Tm = 2000 K, 1 atm): ρ = 0.174 kg/m , µ = 689 × 10 kg/m⋅s, k = 3

-6

0.137 W/m⋅K, Pr = 0.672; Table A-6: Water (T∞ = 300 K): ρ = 997 kg/m , µ = 855 × 10 kg/s⋅m, k = 0.613 W/m⋅K, Pr = 5.83. ANALYSIS: Performing an energy balance for a control surface about the pipe wall, q′′r + q′′c,i = q′′c,o

ε gσ Tg4 + h i ( Tm − Ts ) = ho ( Ts − T∞ ) The gas emissivity is ε g = ε w + ε c = ∆ε where Le = 0.95D = 0.238 m = 0.799 ft p c L e = p w L e = 0.1 atm × 0.238 m = 0.0238 atm − m = 0.0779 atm − ft and from Fig. 13.16 → εw ≈ 0.017; Fig. 13.18 → εc ≈ 0.031; Fig. 13.20 → ∆ε ≈ 0.001. Hence εg = 0.047. Estimating the internal flow convection coefficient, find Re D =

 4m

π Dµ

=

4 × 0.25 kg / s

π ( 0.25 m ) 689 × 10−7 kg / m ⋅ s

= 18, 480

and for turbulent flow, 4 / 5 0.3 Nu D = 0.023 ReD Pr = 0.023 (18, 480 )

4/5

h i = Nu D

k D

= 52.9

0.137 W / m ⋅ K 0.25 m

(0.672 )0.3 = 52.9

= 29.0 W / m 2 ⋅ K. Continued …..

PROBLEM 13.129 (Cont.) Estimating the external convection coefficient, find Re D =

ρ VD 997 kg / m3 × 0.3 m / s × 0.25 m = = 87, 456. µ 855 × 10−6 kg / s ⋅ m

Hence 0.37 Nu D = 0.26 Re0.6 ( Pr/ Prs ) D Pr

1/ 4

.

Assuming Pr/Prs ≈ 1, Nu D = 0.26 (87, 456 )

0.6

(5.83)0.37 = 461

ho = Nu D ( k / D ) = 461 (0.613 W / m ⋅ K / 0.25 m ) = 1129 W / m 2 ⋅ K. Substituting numerical values in the energy balance, find 0.047 × 5.67 × 10−8 W / m 2 ⋅ K 4 ( 2000 K ) + 29 W / m 2 ⋅ K ( 2000 − Ts ) K 4

= 1129 W / m 2 ⋅ K ( Ts − 300 ) K Ts = 380 K.

<

The heat flux due to convection is q′′c,i = h i ( Tm − Ts ) = 29 W / m 2 ⋅ K ( 2000 − 379.4 ) K = 46, 997 W / m2 and the total heat flux is q′′s = q′′r + q′′c,i = 42, 638 + 46, 997 = 89, 640 W / m 2 .

<

COMMENTS: Contributions of gas radiation and convection to the wall heat flux are approximately the same. Small value of Ts justifies neglecting emission from the pipe wall to the gas. Prs = 1.62 for Ts = 380 → (Pr/Prs)1/4 = 1.38. Hence the value of ho should be corrected. The value would ↑, and Ts would ↓.

PROBLEM 13.130 KNOWN: Flowrate, composition and temperature of flue gas passing through inner tube of an annular waste heat boiler. Boiler dimensions. Steam pressure.  . FIND: Rate at which saturated liquid can be converted to saturated vapor, m s

SCHEMATIC:

ASSUMPTIONS: (1) Inner wall is thin and steam side convection coefficient is very large; hence Ts = Tsat(2.455 bar), (2) For calculation of gas radiation, inner tube is assumed infinitely long and gas is approximated as isothermal at Tg. -7

PROPERTIES: Flue gas (given): µ = 530 × 10 kg/s⋅m, k = 0.091 W/m⋅K, Pr = 0.70; Table A-6, Saturated water (2.455 bar): Ts = 400 K, hfg = 2183 kJ/kg. ANALYSIS: The steam generation rate is  =q/h m s fg = ( q conv + q rad ) / h fg where

(

q rad = Asσ ε g Tg4 − α g Ts4

)

with

ε g = ε w + ε c − ∆ε

α g = α w + α c − ∆α .

From Table 13.4, find Le = 0.95D = 0.95 m = 3.117 ft. Hence p w Le = 0.2 atm × 3.117 ft = 0.623 ft − atm p c L e = 0.1 atm × 3.117 ft = 0.312 ft − atm. From Fig. 13.16, find εw ≈ 0.13 and Fig. 13.18 find εc ≈ 0.095. With pw/(pc + pw) = 0.67 and Le(pw + pc) = 0.935 ft-atm, from Fig. 13.20 find ∆ε ≈ 0.036 ≈ ∆α. Hence εg ≈ 0.13 + 0.095 – 0.036 = 0.189. Also, with pwLe(Ts/Tg) = 0.2 atm × 0.95 m(400/1400) = 0.178 ft-atm and Ts = 400 K, Fig. 13.16 yields εw ≈ 0.14. With pcLe(Ts/Tg) = 0.1 atm × 0.95 m(400/1400) = 0.089 ft-atm and Ts = 400 K, Fig. 13.18 yields εc ≈ 0.067. Hence

(

α w = Tg / Ts

)0.45 ε w (Ts , p w LeTs / Tg )

α w = (1400 / 400 )

0.45

0.14 = 0.246

and

(

α c = Tg / Ts

)0.65 ε c (Ts , pcLeTs / Tg ) Continued …..

PROBLEM 13.130 (Cont.) α c = (1400 / 400 )

0.65

0.067 = 0.151

α g = 0.246 + 0.151 − 0.036 = 0.361. Hence 4 4 q rad = π (1 m ) 7 m × 5.67 × 10−8 W / m 2 ⋅ K 4  0.189 (1400 K ) − 0.361 ( 400 K ) 





q rad = (905.3 − 11.5 ) kW = 893.8 kW. For convection,

(

q conv = hπ DL Tg − Ts

)

with Re D =

 4m

π Dµ

=

4 × 2 kg / s

= 48, 047

π × 1 m × 530 × 10−7 kg / s ⋅ m

and assuming fully developed turbulent flow throughout the tube, the Dittus-Boelter correlation gives Nu D = 0.023 Re 4D/ 5 Pr 0.3 = 0.023 ( 48, 047 )

4/5

(0.70 )0.3 = 115

h = ( k / D ) Nu D = ( 0.091 W / m ⋅ K /1 m )115 = 10.5 W / m 2 ⋅ K. Hence q conv = 10.5 W / m 2 ⋅ Kπ (1 m ) 7 m (1400 − 400 ) K = 230.1 kW and the vapor production rate is  = m s

q h fg

=

(893.8 + 230.1) kW 2183 kJ / kg

 = 0.515 kg / s. m s

=

1123.9 kW 2183 kJ / kg

<

COMMENTS: (1) Heat transfer is dominated by radiation, which is typical of heat recovery devices having a large gas volume. (2) A more detailed analysis would account for radiation exchange involving the ends (upstream and downstream) of the inner tube. (3) Using a representative specific heat of cp = 1.2 kJ/kg⋅K, the temperature drop of the gas passing through the tube would be ∆Tg = 1123.9 kW/(2 kg/s × 1.2 kJ/kg⋅K) = 468 K.

PROBLEM 13.131 KNOWN: Wet newsprint moving through a drying oven. FIND: Required evaporation rate, air velocity and oven temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible freestream turbulence, (3) Heat and mass transfer analogy applicable, (4) Oven and newsprint surfaces are diffuse gray, (5) Oven end effects negligible. 3

PROPERTIES: Table A-6, Water vapor (300 K, 1 atm): ρsat = 1/vg = 0.0256 kg/m , hfg = 2438 -6 2 kJ/kg; Table A-4, Air (300 K, 1 atm): η = 15.89 × 10 m /s; Table A-8, Water vapor-air (300 K, 1 -4 2 atm): DAB = 0.26 × 10 m /s, Sc = η/DAB = 0.611. ANALYSIS: The evaporation rate required to completely dry the newsprint having a water content of m′′A = 0.02 kg / m 2 as it enters the oven (x = L) follows from a species balance on the newsprint.    M A,in − M A,out = M st    M L − M 0 − M A,s = 0.

The rate at which moisture enters in the newsprint is  M L = m′′A VW hence, 2 −3  M A,s = m′′A VW = 0.02 kg / m × 0.2 m / s × 1 m = 4 × 10 kg / s.

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The required velocity of the airstream through the oven, u∞, can be determined from a convection analysis. From the rate equation,

(

)

 M A,s = h m WL ρ A,s − ρ A,∞ = h m WLρ A,sat (1 − φ∞ )  hm = M A,s / WLρ A,sat (1 − φ∞ )

h m = 4 × 10−3 kg / s /1 m × 20 m × 0.0256 kg / m3 (1 − 0.2 ) = 9.77 × 10−3 m / s. Now determine what flow velocity is required to produce such a coefficient. Assume flow over a flat plate with Sh L = h m L / DAB = 9.77 × 10−3 m / s × 20 m / 0.26 × 10−4 m 2 / s = 7515 Continued …..

PROBLEM 13.131 (Cont.) and 1/ 3  2

2

Re L = Sh L / 0.664Sc1/ 3  =  7515 / 0.664 ( 0.611)







8  = 1.78 × 10 .

5

Since ReL > ReLc = 5 × 10 , the flow must be turbulent. Using the correlation for mixed laminar and turbulent flow conditions, find Re 4L/ 5 = Sh L / Sc1/ 3 + 871 / 0.037





Re 4L/ 5 =  7515 / (0.611)

1/ 3



+ 871 / 0.037



Re L = 5.95 × 106 noting ReL > ReLc. Recognize that u ∗∞ is the velocity relative to the newsprint, u ∗∞ = Re L ν / L = 5.95 × 106 × 15.89 × 10 −6 m 2 / s / 20 m = 4.73 m / s. The air velocity relative to the oven will be, u ∞ = u ∗∞ − V = ( 4.73 − 0.2 ) m / s = 4.53 m / s.

<

The temperature required of the oven surface follows from an energy balance on the newsprint. Find E in − E out = 0 q rad − q evap = 0 where −3 3  q evap = M A,s h fg = 4.0 × 10 kg / s × 2438 × 10 J / kg = 9752 W

and the radiation exchange is that for a two surface enclosure, Eq. 13.23, q rad =

(

σ T14 − T24

)

(1 − ε1 ) / ε1A1 + 1/ A1F12 + (1 − ε 2 ) / ε 2A 2

Evaluate, A1 = π / 2 WL, hence, with ε1 = 0.8,

(

A 2 = WL,

.

F21 = 1, and A1F12 = A 2 F21 = WL

)

q rad = σ WL T14 − T24 / [(1/ 2π ) + 1] T14 = T24 + q rad [(1/ 2π ) + 1] / σ WL T14 = (300 K ) + 9752 W [(1/ 2π + 1)] / 5.67 × 10−8 W / m 2 ⋅ K 4 × 1 m × 20 m 4

T1 = 367 K. COMMENTS: Note that there is no convection heat transfer since T∞ = Ts = 300 K.

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PROBLEM 13.132 KNOWN: Configuration of grain dryer. Emissivities of grain bed and heater surface. Temperature of grain. FIND: (a)Temperature of heater required for specified drying rate, (b) Convection mass transfer coefficient required to sustain evaporation, (c) Validity of assuming negligible convection heat transfer. SCHEMATIC:

ASSUMPTIONS: (1) Diffuse/gray surfaces, (2) Oven wall is a reradiating surface, (3) Negligible convection heat transfer, (4) Applicability of heat/mass transfer analogy, (5) Air is dry. 3

6

PROPERTIES: Table A-6, saturated water (T = 330 K): vg = 8.82 m /kg, hfg = 2.366 × 10 J/kg. 3 -6 2 Table A-4, air (assume T ≈ 300 K): ρ = 1.614 kg/m , cp = 1007 J/kg⋅K, α = 22.5 × 10 m /s. Table -4 2 A-8, H2O(v) – air (T = 298 K): DAB = 0.26 × 10 m /s. ANALYSIS: (a) Neglecting convection, the energy required for evaporation must be supplied by net radiation transfer from the heater plate to the grain bed. Hence,

(

)

6 ′ q′rad = m evap h fg = ( 2.5 kg / h ⋅ m ) 2.366 × 10 J / kg / 3600 s / h = 1643 W / m

where q′rad is given by Eq. 13.30. With A′p = A′g ≡ A′, q′rad =

(

A′ E bp − E bg 1 − εp

εp

+

)

+

1

(

) (

)

Fpg +  1/ FpR + 1/ FgR 

−1

1 − εg

εg

where A′ = R = 1 m, Fpg = 0 and FpR = FgR = 1. Hence, q′rad =

(

) = 2.40 ×10−8 T4 − 3204 = 1643 W / m (p ) 0.25 + 2 + 0.111 σ Tp4 − 3204

2.40 × 10 −8 Tp4 − 2518 = 1643

<

Tp = 530 K  ′evap , and ρA,∞ = 0, (b) The evaporation rate is given by Eq. 6.12, and with A′s = 1 m, n ′A = m

Continued …..

PROBLEM 13.132 (Cont.) hm =

n ′A vg 2.5 kg / h ⋅ m n ′A 1 m3 = = × × 8.82 = 6.13 × 10−3 m / s ′ ′ ρ As A,s As 1m 3600 s kg

(c) From the heat and mass transfer analogy, Eq. 6.92, h = h m ρ c p Le 2 / 3 where Le = α/DAB = 22.5/26.0 = 0.865. Hence

(

)

h = 6.13 × 10 −3 m / s 1.161kg / m3 1007 J / kg ⋅ K ( 0.865 )

2/3

= 6.5 W / m 2 ⋅ K.

The corresponding convection heat transfer rate is

(

)

q′conv = hA′ Tg − T∞ = 6.5 W / m 2 ⋅ K (1 m )(330 − 300 ) K = 195 W / m Since q′conv T2, the axial temperature gradient (dT/dx) will result in an axial density gradient. However, since dρ/dx < 0 there will be no buoyancy driven, convective motion of the mixture. There will also be axial species density gradients, d ρO /dx and d ρN /dx. However, there is no 2 2

(

)

gradient associated with the mass fractions dmO /dx = 0, d mN /dx = 0 . Hence, from Fick’s 2

2

law, Eq. 14.1, there is no mass transfer by diffusion. (b) If T1 < T2, dρ /dx > 0 and there will be a buoyancy driven, convective motion of the mixture. However, dmO /dx = 0 and dmN /dx = 0, and there is still no mass transfer. Hence, although 2 2 there is motion of each species with the convective motion of the mixture, there is no relative motion between species. COMMENTS: The commonly used special case of Fick’s law,

jA = −D AB

dρ A dx

would be inappropriate for this problem since ρ is not uniform. If applied, this special case indicates that mass transfer would occur, thereby providing an incorrect result.

PROBLEM 14.6 KNOWN: Pressure and temperature of hydrogen stored in a spherical steel tank of prescribed diameter and thickness. FIND: (a) Initial rate of hydrogen mass loss from the tank, (b) Initial rate of pressure drop in the tank. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional species diffusion in a stationary medium, (2) Uniform total molar concentration, C, (3) No chemical reactions. ANALYSIS: (a) From Table 14.1

NA,r =

NA,r =

CA,o − CA,L R m,dif

(

=

CA,o

(1 / 4π D AB ) (1 / ri − 1 / ro )

)

4π 0.3 ×10−12 m2 / s 1.5 kmol/m3

(1/0.05 m − 1/0.052 m )

= 7.35 ×10−12 kmol/s

or

n A,r = M A N A,r = 2 kg/kmol × 7.35 ×10 −12 kmol/s = 14.7 ×10 −12 kg/s.

<

(b) Applying a species balance to a control volume about the hydrogen,

& & M A,st = −M A,out = −n A,r 3 3 3 & A,st = d ( ρAV ) = π D d ρA = π D dp A = π D M A dp A M dt 6 dt 6R A T dt 6ℜT dt Hence

(

)

6 0.08314 m 3 ⋅ bar/kmol ⋅ K ( 300 K ) dpA 6ℜ T =− n A,r = − ×14.7 ×10 −12 kg/s 3 dt πD M A π 0.1m 3 2 kg/kmol

(

)

dpA = −3.50 × 10−7 bar/s. dt

< 2

COMMENTS: If the spherical shell is appoximated as a plane wall, Na,x = DAB(CA,o) πD /L = 7.07 -12 × 10 kmol/s. This result is 4% lower than that associated with the spherical shell calculation.

PROBLEM 14.7 KNOWN: Molar concentrations of helium at the inner and outer surfaces of a plastic membrane. Diffusion coefficient and membrane thickness. FIND: Molar diffusion flux. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3) Stationary medium, (4) Uniform C = CA + CB. ANALYSIS: The molar flux may be obtained from Eq. 14.50,

N′′A,x =

DAB 10 −9 m 2 / s CA,1 − CA,2 = ( 0.02 − 0.005 ) kmol/m3 L 0.001m

(

)

N′′A,x = 1.5 ×10−8 kmol/s ⋅ m2 . COMMENTS: The mass flux is

n ′′A,x = M A N′′A,x = 4 kg/kmol ×1.5 ×10 −8 kmol/s ⋅ m 2 = 6 ×10 −8 k g / s ⋅ m2 .

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PROBLEM 14.8 KNOWN: Mass diffusion coefficients of two binary mixtures at a given temperature, 298 K. FIND: Mass diffusion coefficients at a different temperature, T = 350 K. ASSUMPTIONS: (a) Ideal gas behavior, (b) Mixtures at 1 atm total pressure. -4

2

PROPERTIES: Table A-8, Ammonia-air binary mixture (298 K), DAB = 0.28 × 10 m /s; -4 2 Hydrogen-air binary mixture (298 K), DAB = 0.41 × 10 m /s. ANALYSIS: According to treatment of Section 14.1.5, assuming ideal gas behavior,

D AB ~ T3 / 2 where T is in kelvin units. It follows then, that for

NH3 − Air:

DAB ( 350 K ) = 0.28 ×10 −4 m 2 / s ( 350 K / 2 9 8 K)

3/2

DAB ( 350 K ) = 0.36 ×10 −4 m 2 / s H 2 − Air:

<

D AB (350 K ) = 0.41 ×10−4 m 2 / s ( 350/298 )

3/2

DAB ( 350 K ) = 0.52 ×10 −4 m 2 / s COMMENTS: Since the H2 molecule is smaller than the NH3 molecule, it follows that

DH 2−Air > DNH3−Air as indeed the numerical data indicate.

<

PROBLEM 14.9 KNOWN: The inner and outer surfaces of an iron cylinder of 100-mm length are exposed to a carburizing gas (mixtures of CO and CO2). Observed experimental data on the variation of the carbon composition (weight carbon, %) in the iron at 1000°C as a function of radius. Carbon flow rate under steady-state conditions.

 $% is a constant if the diffusion

FIND: (a) Beginning with Fick’s law, show that dρ c / d n r

coefficient, DC-Fe, is a constant; sketch of the carbon mass density, ρc(r), as function of ln(r) for such a diffusion process; (b) Create a graph for the experimental data and determine whether DC-Fe for this diffusion process is constant, increases or decreases with increasing mass density; and (c) Using the experimental data, calculate and tabulate DC-Fe for selected carbon compositions over the range of the experiment. SCHEMATIC:

3

PROPERTIES: Iron (1000°C). ρ = 7730 kg/m .Experimental observations of carbon composition r (mm) Wt. C (%)

4.49 1.42

4.66 1.32

4.79 1.20

4.91 1.09

5.16 0.82

5.27 0.65

5.40 0.46

5.53 0.28

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial diffusion in a stationary medium, and (3) Uniform total concentration. ANALYSIS: (a) For the one-dimensional, radial (cylindrical) coordinate system, Fick’s law is

jA = − D AB A r

dρ A dr

(1)

where Ar = 2πrL. For steady-state conditions, jA is constant, and if DAB is constant, the product

r

dρ A = C1 dr

(2)

must be a constant. Using the differential relation dr/r = d (ln r), it follows that

dρ A = C1 d ln r

 $

(3)

so that on a ln(r) plot, ρA is a straight line. See the graph below for this behavior. Continued …..

PROBLEM 14.9 (Cont.) (b) To determine whether DC-Fe is a constant for the experimental diffusion process, the data are represented on a ln(r) coordinate.

Wt. carbon distribution - experimental observations 1.6

1.4 Exp data 1.2

Wt Carbon (%)

1

0.8

0.6

0.4

0.2

0 1.45

1.5

1.55

1.6

1.65

1.7

1.75

ln (r, mm)

Since the plot is not linear, DC-Fe is not a constant. From the treatment of part (a), if DAB is not a constant, then

DAB

dρ A = C2 d ln r

 $

must be constant. We conclude that DC-Fe will be lower at the radial position where the gradient is higher. Hence, we expect DC-Fe to increase with increasing carbon content. (c) From a plot of Wt - %C vs. r (not shown), the mass fraction gradient is determined at three locations and Fick’s law is used to calculate the diffusion coefficient,

jc = − ρ ⋅ A r ⋅ DC − Fe



∆ Wt − % C ∆r

$

where the mass flow rate is



$

jc = 3.6 × 10−3 kg / 100 h 3600 s / h = 1 × 10−8 kg / s 3

and ρ = 7730 kg/m , density of iron. The results of this analysis yield, Wt - C (%) 1.32 0.955 0.37

r (mm) 4.66 5.04 5.47

∆ Wt-C/∆r (%/mm) -0.679 -1.08 -1.385

DC-Fe × 10

11

6.51 3.79 2.72

2

(m /s)

PROBLEM 14.10 KNOWN: Three-dimensional diffusion of species A in a stationary medium with chemical reactions. FIND: Derive molar form of diffusion equation. SCHEMATIC:

ASSUMPTIONS: (1) Uniform total molar concentration, (2) Stationary medium. ANALYSIS: The derivation parallels that of Section 14.2.2, except that Eq. 14.33 is applied on a molar basis. That is,

& A,g − NA,x + dx − N A,y+ dy − N A,z+ dz = N & A,st . NA,x + N A,y + N A,z + N With

NA,x + dx = NA,x +

∂NA,x

NA,x = −DAB ( dydz ) & A,g = N & A ( dxdydz ) , N

∂x

dx,

∂C A , ∂x

NA,y + dy = .... NA,y = .... & A,st = ∂CA dxdydz N ∂t

It follows that

∂  ∂CA  ∂  ∂C A  ∂  ∂C A  & ∂C A .  DAB  +  D AB  +  DAB  + NA = ∂x  ∂x  ∂y  ∂y  ∂z  ∂z  ∂t COMMENTS: If DAB is constant, the foregoing result reduces to Eq. 14.38b.

<

PROBLEM 14.11 KNOWN: Gas (A) diffuses through a cylindrical tube wall (B) and experiences chemical reactions at & . a volumetric rate, N A FIND: Differential equation which governs molar concentration of gas in plastic. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial diffusion, (2) Uniform total molar concentration, (3) Stationary medium. ANALYSIS: Dividing the species conservation requirement, Eq. 14.33, by the molecular weight, and applying it to a differential control volume of unit length normal to the page,

A,

& A,g − NA,r +dr = N & A,st NA,r + N where

′′ = −2π rDAB NA,r = ( 2πr ⋅1) NA,r NA,r + dr = NA,r +

∂NA,r ∂r

∂C A ∂r

dr

& A,g = − N & A ( 2π r ⋅ dr ⋅ 1) N

∂ C 2π rdr ⋅1)  & A,st =  A ( N . ∂t

Hence

& A ( 2π rdr ) + 2π DAB −N

∂  ∂C A  ∂C r dr = 2π rdr A   ∂r  ∂r  ∂t

or

DAB ∂  ∂ CA  & ∂ CA r − NA = .   r ∂r  ∂r  ∂t COMMENTS: (1) The minus sign in the generation term is necessitated by the fact that the reactions deplete the concentration of species A. & ( r , t ) , the foregoing equation could be solved for CA (r,t). (2) From knowledge of N A (3) Note the agreement between the above result and the one-dimensional form of Eq. 14.39 for uniform C.

<

PROBLEM 14.12 KNOWN: One-dimensional, radial diffusion of species A in a stationary, spherical medium with chemical reactions. FIND: Derive appropriate form of diffusion equation. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, radial diffusion, (2) Uniform total molar concentration, (3) Stationary medium. ANALYSIS: Dividing the species conservation requirement, Eq. 14.33, by the molecular weight, and applying it to the differential control volume, it follows that

A,

& A,g − NA,r +dr = N & A,st NA,r + N where

NA,r = −DAB 4π r 2 NA,r + dr = NA,r +

(

∂CA ∂r

∂NA,r ∂r

dr

(

)

∂ CA 4π r 2 dr   & A,st =  N . ∂t

)

& A,g = N & A 4π r 2 dr , N Hence

(

)

& A 4π r 2 dr + 4π ∂  DABr 2 ∂CA  dr = 4π r 2 ∂CA dr N ∂r  ∂r  ∂t or

1

∂

 DAB r r 2 ∂r 

2 ∂CA  + N&

∂r 

A =

∂C A . ∂t

<

COMMENTS: Equation 14.40 reduces to the foregoing result if C is independent of r and variations in φ and θ are negligible.

PROBLEM 14.13 KNOWN: Oxygen pressures on opposite sides of a rubber membrane. FIND: (a) Molar diffusion flux of O2, (b) Molar concentrations of O2 outside the rubber. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Stationary medium of uniform total molar concentration, C = CA + CB, (3) Perfect gas behavior. -9

2

PROPERTIES: Table A-8, Oxygen-rubber (298 K): DAB = 0.21 × 10 m /s; Table A-10, Oxygen-3 3 rubber (298 K): S = 3.12 × 10 kmol/m ⋅bar. ANALYSIS: (a) For the assumed conditions

N′′A,x = J ∗A,x = −DAB

C ( 0 ) − CA ( L) dCA = DAB A . dx L

From Eq. 14.33,

CA ( 0) = SpA,1 = 6.24 ×10−3 kmol/m3 CA ( L ) = Sp A,2 = 3.12 × 10−3 lmol/m 3 . Hence

N′′A,x

= 0.21 ×10−9 m 2 / s

(6.24 ×10−3 − 3.12× 10−3 ) kmol/m3 0.0005 m

N′′A,x = 1.31 ×10−9 kmol/s ⋅ m2 .

<

(b) From the perfect gas law

p 2 bar CA,1 = A,1 = = 0.0807 kmol/m3 3 ℜT 0.08314 m ⋅ bar/kmol ⋅ K 298 K

<

CA,2 = 0.5CA,1 = 0.0404 kmol/m3 .

<

(

)

COMMENTS: Recognize that the molar concentrations outside the membrane differ from those within the membrane; that is, CA,1 ≠ CA(0) and CA,2 ≠ CA(L).

PROBLEM 14.14 KNOWN: Water vapor is transferred through dry wall by diffusion. FIND: The mass diffusion rate through a 0.01 × 3 × 5 m wall. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional species diffusion, (3) Homogeneous medium, (4) Constant properties, (5) Uniform total molar concentration, (6) Stationary medium with xA 1 and N′′A,s =

C D AB x A,L 0.0187 kmol / m3 × 10−4 m 2 / s × 0.0012 = = 2.24 × 10−7 kmol / s ⋅ m 2 L 0.01m

COMMENTS: If the process is reaction limited, N′′A,s → 0 as k1′′ → 0.

<

PROBLEM 14.25 KNOWN: Partial pressures and temperatures of CO2 at opposite ends of a circular tube which also contains nitrogen. FIND: Mass transfer rate of CO2 through the tube. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional diffusion, (3) Uniform temperature and total pressure. -4

2

PROPERTIES: Table A-8, CO2 – N2 (T ≈ 298 K, 1 atm): DAB = 0.16 × 10 m /s. ANALYSIS: From Eq. 14.70 the CO2 molar transfer rate is

NA = NA =

(

)

DAB π D 2 / 4 p A,0 − pA,L ℜT

L

0.16 × 10−4 m 2 / s (π / 4)( 0.05 m )2

(100 − 50) mmHg

0.08205 m3 ⋅ atm/kmol ⋅ K× 298 K 1 m × 760 mmHg/atm

NA = 8.45 × 10−11 kmol/s. The mass transfer rate is then

n A = M A NA = 44kg/kmol × 8.45 ×10 −11 kmol/s n A = 3.72 × 10−9 kg/s.

< -11

COMMENTS: Although the molar transfer rate of N2 in the opposite direction is NB = 8.45 × 10 kmol/s, the mass transfer rate is

n B = M B NB = 28 kg/kmol × 8.45 ×10 −11 kmol/s = 2.37 ×10 −9 kg/s.

PROBLEM 14.26 KNOWN: Conditions associated with evaporation from a liquid in a column, with vapor (A) transfer occurring in a gas (B). In one case B has unlimited solubility in the liquid; in the other case it is insoluble. FIND: Case characterized by the largest evaporation rate and ratio of evaporation rates if pA = 0 at the top of the column and pA = p/10 at the liquid interface. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional species transfer, (3) Uniform temperature and total pressure in the column, (4) Constant properties. ANALYSIS: If gas B has unlimited solubility in the liquid, the solution corresponds to equimolar counter diffusion of A and B. From Eqs. 14.63 and 14.68, it follows that

N′′A,x = −CDAB

− x A,L x dx A = CDAB A,0 . dx L

(1)

If gas B is completely insoluble in the liquid, the diffusion of A is augmented by convection and from Eqs. 14.73 and 14.77

N′′A,x = −CDAB

dx A CDAB 1 − x A,L + CA v∗x = ln . dx L 1 − xA,0

(2)

Comparing Eqs. (1) and (2), it is obvious that the evaporation rate for the second case exceeds that for the first case. Also

( CDAB / L ) ( x A,0 − x A,L ) 0.1 − 0 = N′′A,x ( insol) ( CDAB / L) ln (1 − x A,L ) / (1 − x A,0 ) ln  (1 − 0 ) / (1 − 0.1)  N′′A,x (sol ) N′′A,x (sol ) N′′A,x ( insol)

=

= 0.949.

<

COMMENTS: The above result suggests that, since the mole fraction of the saturated vapor is typically small, the rate of evaporation in a column is well approximated by the result corresponding to equimolar counter diffusion.

PROBLEM 14.27 KNOWN: Water in an open pan exposed to prescribed ambient conditions. FIND: Evaporation rate considering (a) diffusion only and (b) convective effects. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional diffusion, (3) Constant properties, (4) Uniform T and p, (5) Perfect gas behavior. -4

2

PROPERTIES: Table A-8, Water vapor-air (T = 300 K, 1 atm), DAB = 0.26 × 10 m /s; Table A-6, 3 Water vapor (T = 300 K, 1 atm), psat = 0.03513 bar, vg = 39.13 m /kg. ANALYSIS: (a) The evaporation rate considering only diffusion follows from Eq. 14.63 simplified for a stationary medium. That is,

NA,x = N′′A,x ⋅ A = −D AB A

dCA . dx

Recognizing that φ ≡ pA/pA,sat = CA/CA,sat , the rate is expressed as

− CA,s DAB A C NA,x = −DABA A,∞ = CA,sat (1 −φ ∞ ) L L

N A,x =

0.26 ×10

−4

(

m / s ( π / 4 )( 0 . 2 m ) 2

2

80 ×10

where CA,s = 1/ v g M A

−3

1 3

39.13 m / k g ×18 kg/kmol

m

(1 − 0.25 ) = 1.087 × 10−8 kmol/s

) with M A = 18 kg/kmol.

(b)The evaporation rate considering convective effects using Eq. 14.77 is

NA,x = N′′A,x ⋅ A =

CDAB A 1 − xAL ln . L 1 − x A,0

Using the perfect gas law, the total concentration of the mixture is C = p / ℜT = 1.0133 bar/ 8.314 ×10 −2 m 3 ⋅ bar/kmol ⋅ K× 300K = 0.04063 kmol/m3

(

)

where p = 1 atm = 1.0133 bar. The mole fractions at x = 0 and x = L are

p 0.03531bar x A,0 = A,s = = 0.0348 p 1.0133bar

x A,L = φ∞ x A,0 = 0.0087.

Hence −4

0.04063 k m o l / m × 0.26 × 10 m / s ( π / 4 )( 0.2 m ) 3

N A,x =

80 × 10

−3

2

m

2

ln

1 − 0.0087 1 − 0.0348

= 1.107 × 10

−8

kmol/s.

<

COMMENTS: For this situation, the convective effect is very small but does tend to increase (by 1.5%) the evaporation rate as expected.

PROBLEM 14.28 KNOWN: Vapor concentrations at ends of a tube used to grow crystals. Presence of an inert gas. Ends are impermeable to the gas. Constant temperature. FIND: Vapor molar flux and spatial distribution of vapor molar concentration. Location of maximum concentration gradient. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Constant properties, (3) Constant pressure, hence C is constant. ANALYSIS: Physical conditions are analogous to those of the evaporation problem considered in Section 14.4.4 with CA,0 > CA,L → diffusion of vapor from source to crystal, CB,L > CB,0 → diffusion of inert gas from crystal to source,

(

)

Impermeable ends → absolute flux of species B is zero N′′B,x = 0 ; hence v B,x = 0. Diffusion of B from crystal to source must be balanced by advection from source to crystal. The ∗ advective velocity is v x = N′′A,x /C. The vapor molar flux is therefore determined by Eq. 14.77,

N′′A,x =

CDAB  1 − x A,L ln   1− xA,0 L 

  

and the vapor molar concentration is given by Eq. 14.75, x/L  1 − x A,L  CA xA = = 1 − 1 − x A,0  .  1 − x A,0  C   From Eq. 14.72,

(

)

< <

dx A = −N′′A,x (1 − x A ) /CD AB dx N′′ dCA = − A,x (1 − x A ) . dx DAB Hence maximum concentration gradient corresponds to minimum xA and occurs at

x = L. COMMENTS: Vapor transfer is enhanced by the advection, which is induced by presence of the inert gas.

PROBLEM 14.29 KNOWN: Spherical droplet of liquid A and radius ro evaporating into stagnant gas B. FIND: Evaporation rate of species A in terms of pA,sat , partial pressure pA(r), the total pressure p and other pertinent parameters. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial, species diffusion, (3) Constant properties, including total concentration, (4) Droplet and mixter air at uniform pressure and temperature, (5) Perfect gas behavior. ANALYSIS: From Eq. 14.31 for a radial spherical coordinate system, the evaporation rate of liquid A into a binary gas mixture A + B is

NA,r = − DABAr

dCA CA + NA,r dr C

2

where Ar = 4πr and NA,r = NA, a constant,

C dC NA  1 − A  = − DAB ⋅ 4π r 2 ⋅ A . C  dr  From perfect gas behavior, CA = p A / ℜT and C = p / ℜT, p dpA NA ( p − pA ) = −DAB ⋅ 4π r 2 ⋅

ℜT dr

Separating variables, setting definite limits, and integrating

− NA

r dr p ℜT 1 dpA = ∫ A,r ∫ pA,r p − p A p 4π DAB ro r 2 o

find that

NA = 4π ro DAB

p − p A (r ) p 1 ln ℜT 1 − ro / r p − pA,o

where p A,o = p A ( ro ) = pA,sat , the saturation pressure of liquid A at temperature T. COMMENTS: Compare the method of solution and result with the content of Section 14.4.4, Evaporation in a Column.

<

PROBLEM 14.30 KNOWN: Vent pipe on a methanol distillation system condenser discharges to atmosphere at 1 bar. 3 Cooler and vent at 21°C. Vapor volume of cooler is 0.005 m . FIND: (a) Weekly loss of methanol vapor due to diffusion out the vent pipe and (b) Weekly loss due to expulsion of methanol vapor in the cooler once per hour caused by process heat rate change. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional species transport, (3) Uniform temperature and total pressure in vent pipe, (4) Constant properties, (5) Perfect gas behavior. -4

2

PROPERTIES: Methanol-air mixture (given, 273 K): DAB = 0.13 × 10 m /s. ANALYSIS: (a) The methanol transfer rate through the vent follows from Eq. 14.77 π D2 CDAB 1 − p A,2 / p NA,x = N′′A,x ⋅ Ac = ln 4 L 1 − pA,1 / p where pA,2 = 0 and pA,1 = pA = 100 mmHg = 0.1333 bar = 13.3 kPa,

C=

p 1bar = = 4.093× 10−2 kmol/m3 ℜT 8.314 × 10−2 m 3 ⋅ bar/kmol ⋅ K ( 21 + 273) K

D AB ( 294K ) = D AB ( 273 )( 294/273 )

3/2

= 0.13 × 10

−4

m / s ( 294/273 ) 2

3/2

= 0.145 × 10

Substituting numerical values, find the rate on a weekly basis as π ( 0.035 m ) 2 0.145 × 10−4 m 2 / s NA = × 4.093× 10−2 kmol/m 3 × ln

4

0.5 m

−4

2

m /s.

1− 0 1 − 0.1333/1

×3600 s / h × 24 h / d a y × 7day/week = 9.883 ×10 −5 kmol/week m A = NA M A = 9.883 ×10 −5 kmol/week ×32 kg/kmol = 0.00316 kg/week.

<

3

(b) The methanol vapor in the cooler of volume 0.005 m is expelled once per hour, so that the additional mass loss is m A = n A M A , where nA is

p V 0.1333bar × 0.005m3 nA = A = = 2.728 ×10 −5 kmol − 2 3 ℜT 8.314 ×10 m ⋅ bar/kmol ⋅ K × 294 K from which it follows that

m A = 2.728 ×10 −5 kmol/ × 24 × 7 ×32 kg/kmol = 0.1467 kg/week.

<

COMMENTS: Note that the loss through the vent is approximately 2% that lost by expulsion when the process heat rate is varied.

PROBLEM 14.31 2

KNOWN: Clean surface with pure steam has condensate rate of 0.020 kg/m ⋅s for the prescribed conditions. With the presence of stagnant air in the steam, the condensate surface drops from 28°C to 24°C and the condensate rate is halved. FIND: Partial pressure of air in the air-steam mixture as a function of distance from the condensate film. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties including pressure in air-steam mixture, (3) Perfect gas behavior. PROPERTIES: Table A-6, Water vapor: psat (28°C = 301 K) = 0.03767 bar; psat (24°C = 297 K) = -4 2 0.02983 bar; Table A-8, Water-air (298 K, 1 bar): DAB = 0.26 × 10 m /s. ANALYSIS: The partial pressure distribution of the air as a function of distance y can be found from the species (A) rate expression, Eq. 14.77,

(

)(

)

N′′A,y = ( CDAB / y ) ln 1 − x A,y / 1 − x A,0 .

With C = p / ℜT, x B,y = 1 − x A,y and x B,0 = 1 − x A,0, recognizing that xB = pB/p, find

 ℜT  p B ( y ) = p B,0 ⋅ exp  N ′′A,y y pDAB   p B,0 = p − p A,0 = p sat ( 28 °C ) − p sat ( 24°C ) = ( 0.03767 − 0.02983) bar = 0.00784 bar. With N′′A,y = − ( 0.020/2 ) k g / m 2 ⋅ s/28kg/kmol = 3.57 ×10−4 kmol/m 2 ⋅ s, −2 3   −4 2 8.314 ×10 m ⋅ bar/kmol ⋅ K × 299 K   p B ( y ) = 0.0784 bar × exp 3.57 ×10 kmol/m ⋅ s  0.03767 bar × 6.902× 10−4 m2 / s  

p B ( y ) = 784 kPa × exp ( −0.3415y ) with pB in [kPa] and y in [mm], where T = 26°C = 299 K, the average temperature of the air-steam -1 3/2 -4 2 3/2 -4 2 mixture, and DAB ≈ p T = 0.26 × 10 m /s (1/0.03767) (299/298) = 6.902 × 10 m /s. Selected values for the pressure are shown below and the distribution is shown above: y (mm) pB(y) (kPa)

0 784

5 142

10 25.8

15 4.7

COMMENTS: To minimize inert gas effects, the usual practice is to pass vapor over the surfaces so that the inerts are eventually collected near the outlet region of the condenser. Our estimate shows that the effective region to be swept is approximately 10 mm thick.

PROBLEM 14.32 KNOWN: Column containing liquid phase of water (A) evaporates into the air (B) flowing over the mouth of the column. 2

FIND: Evaporation rate of water (kg/h⋅m ) using the known value of the binary diffusion coefficient for the water vapor - air mixture. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, one-dimensional diffusion in the column, (2) Constant properties, (3) Uniform temperature and pressure throughout the column, (4) Water vapor exhibits ideal gas behavior, and (5) Negligible water vapor in the chamber air. PROPERTIES: Table A-6, water (T = 320 K): psat = 0.1053 bar; Table A-8, water vapor-air (0.25 -1 3/2 atm, 320 K): Since DAB ~ p T find



$

$

DAB = 0.26 × 10−4 m2 / s 1.00 / 0.25 320 / 298 3/ 2 = 1157 . × 10−4 m2 / s ANALYSIS: Equimolar counter diffusion occurs in the vertical column as water vapor, evaporating at the liquid-vapor interface (x = 0), diffuses up the column through air out into the chamber. From Eq. 14.7, the molar flow rate per unit area is

N ′′A,x =

C DAB 1 − xA,L ln L 1 − xA,0

where C is the mixture concentration determined from the ideal gas law as

C=

p 0.25 atm = = 0.009397 kmol/m3 Ru T 8.205 × 10−2 m3 ⋅ atm/kmol ⋅ K × 320 K

where Ru = 8.205 × 10−2 m3 ⋅ atm/kmol ⋅ K. The mole fractions at x = 0 and x = L are xA,L = 0 (no water vapor in air above column)

xA,0 = p A / p = 01053 . / 0.25 = 0.4212 where pA is the saturation pressure for water at T = 320 K. Substituting numerical values 1− 0 0.009397 kmol / m3 × 1157 . × 10−4 m2 / s N ′′A,x = ln

 $

0.150 m

N ′′A,x = 3.964 × 10−6 kmol / m2 ⋅ s

1 − 0.4212$

or, on a mass basis,

m′′A,x = N ′′A,x M A mA,x ′′ = 3.964 × 10−6 kmol / m2 ⋅ s × 3600 s / h × 18 kg / kmol m′′A,x = 0.257 kg / m2 ⋅ h

<

PROBLEM 14.33 KNOWN: Ground level flux of NO2 in a stagnant urban atmosphere. FIND: (a) Vertical distribution of NO2 molar concentration, (b) Critical ground level flux of NO2, N ′′A,0,crit .

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional diffusion in a stationary medium, (3) Total molar concentration C is uniform, (4) Perfect gas behavior. ANALYSIS: (a) For the prescribed conditions the molar concentration of NO2 is given by Eq. 14.80, subject to the following boundary conditions.

N′′A,0 dCA  =− .  dx  x = 0 DAB

CA ( ∞ ) = 0,

From the first condition, C1 = 0. From the second condition,

− mC2 = − N′′A,0 / DAB .

Hence

CA ( x ) =

N′′A,0 mD AB

where m = (k1/DAB)

e− mx

<

1/2

.

(b) At ground level, CA ( 0 ) =

N′′A,0 mD AB

p A ( 0 ) = C A ( 0 ) ℜT =

ℜTN′′A,0 mD AB

-4 1/2

Hence, with m = (0.03/0.15 × 10 )

N′′A,0,crit =

. Hence, from the perfect gas law,

-1

-1

m = 44.7 m .

mD ABp A ( 0 )crit ℜT

.

=

44.7m −1 × 0.15 × 10−4 m 2 / s × 2 × 10−6 bar 8.314 × 10−2 m 3 ⋅ bar/kmol ⋅ K× 300 K

N′′A,0,crit = 5.38 × 10−11 kmol/s ⋅ m2 .

<

COMMENTS: Because the dispersion of pollutants in the atmosphere is governed strongly by convection effects, the above model should be viewed as a first approximation which describes a worst case condition.

PROBLEM 14.34 KNOWN: Radius of a spherical organism and molar concentration of oxygen at surface. Diffusion and reaction rate coefficients. FIND: (a) Radial distribution of O2 concentration, (b) Rate of O2 consumption, (c) Molar concentration at r = 0. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, one-dimensional diffusion, (2) Stationary medium, (3) Uniform total molar concentration, (4) Constant properties (k0, DAB). ANALYSIS: (a) For the prescribed conditions and assumptions, Eq. (14.40) reduces to

D AB d  2 d C A  r  − k0 = 0 2 dr dr   r r2

k r3 d CA = 0 + C1 dr 3D AB

k 0 r 2 C1 CA = − + C2 6 DAB r With the requirement that CA(r) remain finite at r = 0, C1 = 0. With CA(ro) = CA,o

k 0 ro2 C2 = CA,o − 6 DAB

(

CA = CA,o − ( k 0 / 6 DAB ) ro2 − r 2

)

<

Because CA cannot be less than zero at any location within the organism, the right-hand side of the foregoing equation must always exceed zero, thereby placing limits on the value of CA,o. The smallest possible value of CA,o is determined from the requirement that CA(0) ≥ 0, in which case CA,o ≥ k 0 ro2 / 6 DAB

(

)

<

(b) Since oxygen consumption occurs at a uniform volumetric rate of k0, the total respiration rate is R = ∀ k 0 , or

R = ( 4 / 3)π ro3 k 0

< Continued …..

PROBLEM 14.34 (Cont.) (c) With r = 0,

CA ( 0 ) = CA,o − k 0 ro2 / 6 DAB

(

CA ( 0 ) = 5 × 10−5 kmol / m3 − 1.2 × 10−4 kmol / s ⋅ m3 10−4 m

)

2

/ 6 × 10−8 m 2 / s

CA ( 0 ) = 3 × 10−5 kmol / m3

<

COMMENTS: (1) The minimum value of CA,o for which a physically realistic solution is possible is CA,o = k 0 ro2 / 6 DAB = 2 × 10−5 kmol / m3. (2) The total respiration rate may also be obtained by applying Fick’s law at r = ro, in which case

(

)

(

R = − N A ( ro ) = + DAB 4π ro2 d CA / dr r = r = D AB 4π R o2 o The result agrees with that of part (b).

)(ko / 6 DAB ) 2ro = (4 / 3)π ro3k 0 .

PROBLEM 14.35 KNOWN: Radius of a spherical organism and molar concentration of oxygen at its surface. Diffusion and reaction rate coefficients. FIND: (a) Radial distribution of O2 concentration, (b) Expression for rate of O2 consumption, (c) Molar concentration at r = 0 and rate of oxygen consumption for prescribed conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, one-dimensional diffusion, (2) Stationary medium, (3) Uniform total molar concentration, (4) Constant properties (k1, DAB). ANALYSIS: (a) For the prescribed conditions and assumptions, Eq. (14.40) reduces to

1 d 2 d CA  DAB r 2 dr r dr 

  − k1 CA = 0  2

With y ≡ r CA, d CA/dr = (1/r) dy/dr – y/r and 2  DAB d  dy  DAB  d y    r y r = − =      r 2 dr  dr r 2  dr 2 

1 d d CA DAB r 2  dr r 2 dr  The species equation is then

d2y dr 2

−

k1 y=0 DAB

The general solution is of the form

y = C1 sinh ( k1 / D AB )

1/ 2

r + C2 cosh ( k1 / DAB )

1/ 2

r

or

C C 1/ 2 1/ 2 CA = 1 sinh ( k1 / DAB ) r + 2 cosh ( k1 / DAB ) r r r Because CA must remain finite at r = 0, C2 = 0. Hence, with CA (ro) = CA,o,

C1 =

CA,o ro sinh ( k1 / D AB )

1/ 2

ro

and Continued …..

PROBLEM 14.35 (Cont.) r CA = CA,o  o  r 

 sinh ( k1 / DAB )1/ 2 r  1/ 2  sinh ( k1 / DAB ) ro

<

(b) The total O2 consumption rate corresponds to the rate of diffusion at the surface of the organism.

(

)

R = − N A ( ro ) = + DAB 4π ro2 d CA / dr ro  1 1 1/ 2 1/ 2  R = 4π ro2 D AB CA,o ro  − + ( k1 / DAB ) cot ( k1 / DAB ) ro   ro2 ro  R = 4π ro DAB CA,o (α coth α − 1)

<

)

(

1/ 2 . where α ≡ k1 ro2 / D AB

(c) For the prescribed conditions, (k1/DAB)

CA =

5 × 10

−5

3

kmol / m × 10

sinh ( 4.472 )

−4

m

1/2

-1

sinh ( k1 / D AB )

1/ 2

×

-8

2

= (20 s ÷ 10 m /s)

r

r

1/2

-1

= 44,720 m and α = 4.472.

1/ 2 r −10 kmol sinh ( k1 / D AB ) = 1.136 × 10 × 3 r

m

In the limit of r → 0, the foregoing expression yields

CA ( r → 0 ) = 5.11× 10−6 kmol / m3

<

R = 4π × 10−4 m × 10−8 m 2 / s × 5 × 10−5 kmol / m3 ( 4.472 coth 4.472 − 1) = 2.18 × 10−15 kmol / s COMMENTS: The total respiration rate may also be obtained by integrating the volumetric rate of  =− N  d∀ = ro k C ( r ) 4π r 2dr. consumption over the volume of the organism. That is, R A 0 1 A

∫

∫

PROBLEM 14.36 KNOWN: Radius and catalytic reaction rate of a porous spherical pellet. Surface mole fraction of reactant and effective diffusion coefficient. FIND: (a) Radial distribution of reactant concentration in pellet, total reactant consumption rate, and pellet effectiveness, (b) CO consumption rate and effectiveness for prescribed conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial diffusion, (3) Constant properties, (4) Homogeneous chemical reactions, (5) Isothermal, constant pressure conditions within pellet, (6) Stationary medium. ANALYSIS: (a) In spherical coordinates, the mass diffusion equation is given by

1 ∂ ∂x  & CDABr 2 A  + N  A =0 2 ∂r  r ∂r  & A = −k1′ Av CA . Hence where C, DAB are constant and N 1 d  2 dx A  k1′ Av xA = 0.  r dr  − D  r 2 dr  eff The boundary conditions are xA(ro) = xA,s and xA(0) is finite. Transform the dependent variable, y ≡ rxA, with 2 dx A 1 d y y 1 d  2 dx A  1 d  dy  1  d y  = − or r = r − y = r .   dr r dr r 2 dr  r 2 dr  dr  r 2  dr 2  r 2 dr 





Hence

d2 y

k′ A − 1 v y = 0. dr 2 Deff

The general solution is of the form

y = C1 sinh ( ar ) + C2 cosh ( ar )

1/2 where a ≡ ( k1′ A v / Deff ) giving

C C x A = 1 sinh ( ar ) + 2 cosh ( ar ) r r and using the boundary conditions,

x A ( 0 ) finite → C2 = 0

x A ( ro ) = x A,s → C1 = x A , sro /sinh ( aro ) . Continued …..

PROBLEM 14.36 (Cont.) Hence

x A ( r) = x A,s ( ro / r )

sinh ( ar ) . sinh ( aro )

<

& A,in + N& A,g = 0, the total Applying conservation of species to a control volume about the pellet, N rate of consumption of A in the pellet is ∗ & A,g = N & A,in = NA,r ( ro ) = 4π ro2 J A,r −N ( ro ) . Hence

 sinh ( ar ) cosh ( ar )  dx A  4π ro3  NA,r ( ro ) = 4π ro2  −CDeff = CDeff x A,s  −   dr r =ro sinh ( aro )  r2  r2  r = ro

)

(

  aro NA,r ( ro ) = 4π ro CDeff xA,s 1 − .  tanh ( aro ) 

<

The pellet effectiveness ε is defined as ε ≡ NA,r(ro)/[NA,r(ro)]max and the maximum consumption occurs if xA(r) = xA,s for all 0 ≤ r ≤ ro. Hence

& A Vp = − k1′ Av Cx A,s 4 π ro3  NA,r ( ro )  =N max 3 ε=−

 3  aro 1−  . a 2ro2  tanh ( aro ) 

<

(b) To evaluate the rate, first determine values for these parameters:

C=

p 1.2atm = = 0.0178 kmol/m3 3 ℜT 0.08205 m ⋅ atm/kmol ⋅ K ×823 K 1/2

 k′A  a = 1 v   Deff 

 10−3 m / s × 108 m2 / m 3   =   2 ×10−5 m 2 / s  

aro = 176.8

tanh ( aro ) = 1.

1/2

= 7.07 ×10 4m −1

Hence the consumption rate is

NA,r ( ro ) = 4π ( 0.0025 m) 0.0178 kmol/m 3 × 2 ×10− 5 m 2 / s × 0.04 (1 − 176.8 ) NA,r ( ro ) = −7.86 ×10−8 kmol/s

<

and the effectiveness is

ε=−

(

3

)

2 2 7.07 ×104 m −1 ( 0.0025 m )

[1 − 176.8] = 0.0169

COMMENTS: For the range of conditions of interest, ε ≈ 3/aro. Hence ε may be increased by

<

↓ ro , ↓ k′1, ↓ A v and ↑ Deff . However, NA,r ( ro ) would decrease with ↓ ro , ↓ k 1′ and ↓ A v .

PROBLEM 14.37 KNOWN: Molar concentrations of oxygen at inner and outer surfaces of lung tissue. Volumetric rate of oxygen consumption within the tissue. FIND: (a) Variation of oxygen molar concentration with position in the tissue, (b) Rate of oxygen transfer to the blood per unit tissue surface area. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional species transfer by diffusion through a plane wall, (3) Homogeneous, stationary medium with uniform total molar concentration and constant diffusion coefficient. ANALYSIS: (a) From Eq. 14.78 the appropriate form of the species diffusion equation is d 2CA DAB − ko = 0. 2

dx

Integrating,

dCA /dx = ( ko / DAB ) x + C1

CA =

ko x 2 + C1x + C2. 2DAB

With CA = CA ( 0 ) at x = 0 and CA = CA ( L ) at x = L,

C2 = CA ( 0 )

C ( L ) − C A (0 ) k o L C1 = A − . L 2DAB

Hence

CA ( x ) =

ko x x ( x − L ) + CA (L ) − C A ( 0 )  + CA ( 0 ) . 2D AB L

<

(b) The oxygen assimilation rate per unit area is

N′′A,x (L ) = −D AB ( dCA /dx ) x =L  k L k L N′′A,x (L ) = −DAB  o − o  DAB 2DAB

 DAB  C ( L ) − CA ( 0)  − L  A 

k L D N′′A,x = − o + AB  CA ( 0 ) − CA ( L)  . 2 L COMMENTS: The above model provides a highly approximate and simplified treatment of a complicated problem. The lung tissue is actually heterogeneous and conditions are transient.

<

PROBLEM 14.38 KNOWN: Combustion at constant temperature and pressure of a hydrogen-oxygen mixture adjacent to a metal wall according to the reaction 2H 2 + O2 → 2H2O. Molar concentrations of hydrogen, 3 oxygen, and water vapor are 0.10, 0.10 and 0.20 kmol/m , respectively. Generation rate of water -2 3 vapor is 0.96 × 10 kmol/m ⋅s. FIND: (a) Expression for CH as function of distance from wall, plot qualitatively, (b) CH at the 2

2

wall, (c) Sketch also curves for CO ( x ) and CH O ( x ) , and (d) Molar flux of water at x = 10mm. 2 2 SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional diffusion, (3) Stationary mixture, (4) Constant properties including pressure and temperature. -5

PROPERTIES: Species binary diffusion coefficient (given, for H2, O2 and H2O): DAB = 0.6 × 10 2 m /s.

ANALYSIS: (a) The species conservation equation, Eq. 14.38b, and its general solution are & &A d2 CA N N + A =0 C A (x ) = − x + C1x + C2. (1,2) D AB 2D AB dx 2 The boundary condition at the wall must be dCA(0)/dx = 0, such that C1 = 0. For the species hydrogen, & H = −N & H O , according evaluate C2 from knowledge of CH (10 mm ) = 0.10 kmol/m3 and N 2 2 2 to the chemical reaction. Hence,

0.10 kmol/m 3 = −

( −0.96× 10−2 kmol/m3 ⋅ s) (0.010 m )2 + 0 + C2 2 × 0.6 ×10−5 m 2 / s

C2 = 0.02 kmol/m 3. Hence, the hydrogen species concentration distribution is

C H2 ( x ) = −

&H N 2 2DAB

x 2 + 0.02 = 800x 2 + 0.02

<

which is parabolic with zero slope at the wall; see sketch above. (b) The value of CH at the wall is, 2

CH2 (0 ) = ( 0 + 0.02 ) kmol/m 3 = 0.02 kmol/m 3.

< Continued …..

PROBLEM 14.38 (Cont.) (c) The concentration distribution for water vapor species will be of the same form,

C H2 O ( x ) = −

&H O N 2 x2 +C x +C 1 2 2D AB

(3)

With C1 = 0 for the wall condition, find C2 from CH O (10 mm ) , 2

0.20 kmol/m

3

( 0.96 ×10 =−

−2

2 × 0.6 × 10

3

kmol/m −5

2

) ( 0.010 m )

2

+ C2

3

C2 = 0.28 k m o l / m .

m /s

Hence, CH O at the wall is, 2

CH2 O ( 0 ) = 0 + 0 + C 2 = 0.28 kmol/m3 & & and its distribution appears as above. Recognizing that N O2 = −0.5N H2O , by the same analysis, find

CO2 ( 0 ) = 0.06 kmol/m3 and its shape, also parabolic with zero slope at the wall is shown above. (d) The molar flux of water vapor at x = 10 mm is given by Fick’s law

N′′H2 O,x = −DAB

dCH2 O dx

and using the concentration distribution of Eq. (3), find

N′′H2 O,x = −DAB

&H O  d  N 2 x2  = +N & H Ox − 2  dx  2DAB 

and evaluation at the location x = 10 mm, the species flux is

)

(

N′′H O, (10 mm) = + 0.96 × 10− 2 kmol/m3 ⋅ s × 0.010 m = 9.60 × 10− 5 kmol/m 2 ⋅ s. 2 x

<

COMMENTS: Note that the generation rate of water vapor is a positive quantity. Whereas for H2 & H and N & O are negative. According to the chemical and O2, species are consumed and hence N 2

2

reaction one mole of H2 and 0.5 mole of O2 are consumed to generate one mole of H2O. Therefore,

& H = −N & H O and N & O = −0.5 N & H O. N 2 2 2 2

PROBLEM 14.39 KNOWN: Ground level flux of NO2 in a stagnant urban atmosphere. FIND: (a) Governing differential equation and boundary conditions for the molar concentration of NO2, (b) Concentration of NO2 at ground level three hours after the beginning of emissions. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional diffusion in a stationary medium, (2) Uniform total molar concentration, (3) Constant properties. ANALYSIS: (a) Applying the species conservation requirement, Eq. 14.33, on a molar basis to a unit area of the control volume,

N′′A,x − ( k1CA ) dx− N′′A,x + dx =

(

∂C A dx. ∂t

)

With N′′A,x + dx = N′′A,x + ∂ N′′A,x / ∂x dxand N′′A,x = − DAB ( ∂ CA / ∂x ) , it follows that

DAB

∂ 2CA ∂x 2

− k1CA =

∂CA . ∂t

<

Initial Condition: CA ( x,0 ) = 0. Boundary Conditions:

− DAB

< ∂C A  = N′′A,0 , ∂x  x =0

CA ( ∞ , t ) = 0.

<

(b) The present problem is analogous to Case (2) of Fig. 5.7 for heat conduction in a semi-infinite medium. Hence by analogy to Eq. 5.59, with k ↔ DAB and α ↔ DAB , 1/2

 t  CA ( x,t ) = 2N′′A,0    π DAB 

  x 2  N′′A,0x x exp  − − erfc   4DAB t  D AB  2 ( D t )1/2   AB 

   

At ground level (x = 0) and 3h, 1/2

 t  CA ( 0,3h ) = 2N′′A,0    π DAB 

(

C A ( 0,3h ) = 2 3 × 10

−11

kmol/s ⋅ m

2

)(1 0 , 8 0 0 s / π × 0.15 ×10

−4

2

m /s

)

1/2

= 9.08 ×10

−7

3

km o l / m .

<

COMMENTS: The concentration decays rapidly to zero with increasing x, and at x = 100 m it is, for all practical purposes, equal to zero.

PROBLEM 14.40 KNOWN: Carbon dioxide concentration at water surface and reaction rate constant. FIND: (a) Differential equation which governs variation with position and time of CO2 concentration in water, (b) Appropriate boundary conditions and solution for a deep body of water with negligible chemical reactions. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional diffusion in x, (2) Constant properties, including total density ρ, (3) Water is stagnant. ANALYSIS: (a) From Eq. 14.37b, it follows that, for the prescribed conditions, ∂ 2ρ A ∂ρ DAB − k 1ρA = A . 2

∂t

∂x

<

The first term on the left-hand side represents the net transport of CO2 into a differential control volume by diffusion. The second term represents the rate of CO2 consumption due to chemical reactions. The term on the right-hand side represents the rate of increase of CO2 storage within the control volume. (b) For a deep body of water, appropriate boundary conditions are

ρ A ( 0,t ) = ρ A,0 ρ A ( ∞, t ) = 0 and, with negligible chemical reactions, the species diffusion equation reduces to

∂ 2 ρA ∂x 2

=

∂ρ A . DAB ∂t 1

With an initial condition, ρ A(x,0) ≡ ρ A,i = 0, the problem is analogous to that involving heat transfer in a semi-infinite medium with constant surface temperature. By analogy to Eq. 5.57, the species concentration is then

ρA ( x,t ) − ρA,0 − ρA,0

 x = erf   2 ( D t )1/2 AB 

 x ρ A ( x,t ) = ρ A,0erfc   2 ( D t )1/2 AB 

 .  

   

<

PROBLEM 14.41 KNOWN: Initial concentration of hydrogen in a sheet of prescribed thickness. Surface concentrations for time t > 0. FIND: Time required for density of hydrogen to reach prescribed value at midplane of sheet. SCHEMATIC:

3

CA(x,0) = 3 kmol/m = CA,i 3 CA(0,tf) = 1.2 kg/m /2 kg/kmol 3 CA(0,tf) = 0.6 kmol/m = CA CA(20 mm,t) = 0 = CA,s

ASSUMPTIONS: (1) One-dimensional diffusion in x, (2) Constant DAB, (3) No internal chemical reactions, (4) Uniform total molar concentration. ANALYSIS: Using Heisler chart with heat and mass transfer analogy

γ∗ =

CA − CA,s CA,i − CA,s

=

0.6 − 0 = 0.2 = γ o∗ 3.0 − 0

With Bim = ∞, Fig. D.1 may be used with θ o∗ = 0.2, Bi −1 = 0

Fo ≈ 0.75.

Hence

Fo m =

DAB t f L2

= 0.75

t f = 0.75 ( 0.02 m ) /9 ×10−7 m2 / s 2

<

t f = 333s. COMMENTS: If the one-term approximation to the infinite series solution

θ∗ =

∞

∑ Cn exp ( −ς n2Fo ) cos ( ς n x∗ )

n =1

is used, it follows that

(

)

γ o∗ ≈ C1 exp −ς12 Fom = 0.2 Using values of ς1 = 1.56 and C1 = 1.27, it follows that

exp  − (1.56 )2 Fom  = 0.157   Fo m = 0.76 which is in excellent agreement with the result from the chart.

PROBLEM 14.42 KNOWN: Sheet material has high, uniform concentration of hydrogen at the end of a process, and is then subjected to an air stream with a specified, low concentration of hydrogen. Mass transfer parameters specified include: convection mass transfer coefficient, hm, and the mass diffusivity and solubility of hydrogen (A) in the sheet material (B), DAB and SAB, respectively. FIND: (a) The final mass density of hydrogen in the material if the sheet is exposed to the air stream for a very long time, ρA,f, (b) Identify and evaluate the parameter that can be used to determine whether the transient mass diffusion process in the sheet can be characterized by a uniform concentration at any time; Hint: this situation is analogous to the lumped capacitance method for a transient heat transfer process; (c) Determine the time required to reduce the hydrogen concentration to twice the limiting value calculated in part (a). SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional diffusion, (2) Material B is stationary medium, (3) Constant properties, (4) Uniform temperature in air stream and material, and (5) Ideal gas behavior. ANALYSIS: (a) The final content of H2 in the material will depend upon the solubility of H2 (A) in the material (B) and its partial pressure in the free stream. From Eq. 14.44,

CA,f = SAB p A,∞ = 160 kmol / m3 ⋅ atm × 0.1 atm = 16 kmol / m3

ρ f = M A CA,f = 2 kg / kmol × 16 kmol / m3 = 32 kg / m3

<

(b) The parameters associated with transient diffusion in the material follow from the analogous treatment of Section 5.2 (Fig. 5.3) and are represented in the schematic.

In the material, from Fick’s law, the diffusive flux is

3

8

N ′′A,dif = DAB CA,1 − CA,2 / L

(1)

At the surface, x = L, the rate equation, Eq. 6.8, convective flux of species A is

3

N ′′A,conv = h m CA,s − CA,∞

8

Continued …..

PROBLEM 14.42 (Cont.) and substituting the ideal gas law, Eq. 14.9, and introducing the solubility relation, Eq. 14.44,

3

N ′′A,conv =

hm SAB p A,s − SAB p A,∞ SAB Ru T∞

N ′′A,conv =

hm C2,s − CA,∞ SAB Ru T∞

3

8

8

(2)

where CA,∞ = CA,f, the final concentration in the material after exposure to the air stream a long time. Considering a surface species flux balance, as shown in the schematic above, with the rate equations (1) and (2),

3

8

DAB CA,1 − CA,2 hm CA,s − CA,f = L SAB Ru T∞

3

8

CA,1 − CA,2

Rm,dif ′′ h /S R T = m AB u ∞ = = Bi m CA,s − CA,f DAB / L Rm,conv ′′

(3)

and introducing resistances to species transfer by diffusion, Eq. 14.51, and convection. Recognize from the analogy to heat transfer, Eq. 5.10 and Table 14.2, that when Bim < 0.1, the concentration can be characterized as uniform during the transient process. That is, the diffusion resistance is negligible compared to the convection resistance,

Bi m = Bi m =

h mL . < 01 SAB Ru T∞ D AB

(4)

115. m / h × 3600 s / h6 × 0.003 m

160 kmol / m3 ⋅ atm × 8.205 × 10-2 m3 ⋅ atm / kmol ⋅ K × 555 K × 2.68 × 10-8 m2 / s Bi m = 6.60 × 10 −3 < 0.1

Hence, the mass transfer process can be treated as a nearly uniform concentration situation. From conservation of species on the material with uniform concentration,

 ′′ − N A,conv =N ′′ A,st −

3

8

hm d CA CA − CA,f = L SAB Ru T∞ dt

Integrating, with the initial condition CA (0) = CA,i, find

CA − CA,f CA,i − CA,f

 

= exp −

hm t L SAB Ru T∞

 

(5)

<

Continued …..

PROBLEM 14.42 (Cont.) which is similar to the analogous heat transfer relation for the lumped capacitance analysis, Eq. 5.6.

(c) The time, to, required for the material to reach a concentration twice that of the limiting value, CA (To) = 2 CA,f, can be calculated from Eq. (5).

 1.5 m / h × t o 12 − 16 × 16 kmol / m3 = exp − − 3 3 3 2  0.003 m × 160 kmol / m ⋅ atm × 8.205 × 10 m ⋅ atm / kmol ⋅ K × 555 K  1320 -166 kmol / m t o = 42.9 hour

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PROBLEM 14.43 KNOWN: Hydrogen-removal process described in Problem 14.3 (S), but under conditions for which -11 2 the mass diffusivity of hydrogen gas (A) in the sheet (B) is DAB = 1.8 × 10 m /s (instead of -8 2 2.6 × 10 m /s). With a smaller DAB, a uniform concentration condition may no longer be assumed to exist in the material during the removal process. FIND: (a) The final mass density of hydrogen in the material if the sheet is exposed to the air stream for a very long time, ρA,f, (b) Identify and evaluate the parameters that describe the transient mass transfer process in the sheet; Hint: this situation is analogous to that of transient heat conduction in a plane wall; (c) Assuming a uniform concentration in the sheet at any time during the removal process, determine the time required to reach twice the limiting mass density calculated in part (a); (d) Using the analogy developed in part (b), determine the time required to reduce the hydrogen concentration to twice the limiting value calculated in part (a); Compare the result with that from part (c). SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional diffusion, (2) Material B is a stationary medium, (3) Constant properties, (4) Uniform temperature in air stream and material, and (5) Ideal gas behavior. ANALYSIS: (a) The final content of H2 in the material will depend upon the solubility of H2 (A) in the material (B) at its partial pressure in the free stream. From Eq. 14.44,

CA,f = SAB p A,∞ = 160 kmol / m3 ⋅ atm × 0.1 atm = 16 kmol / m3

ρ f = M A CA,f = 2 kg / kmol × 16 kmol / m3 = 32 kg / m3

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(b) For the plane wall shown in the schematic below, the heat and mass transfer conservation equations and their initial and boundary conditions are Heat transfer ∂T ∂ 2T

∂t

=α

Mass (Species A) transfer ∂ CA ∂ 2 CA = D AB 2

∂t

∂ x2

1 6 ∂ CA 10, t6 = 0 ∂x

1 6 ∂T 10, t6 = 0 ∂x

T x,0 = Ti

−k

1 6

∂x

CA x,0 = CA,i

1 6

∂T L, t = h T L, t − T∞ ∂x

− DAB

1 6

1 6

∂ CA hm L, t = CA x, t − C f ∂x SAB Ru T Continued …..

PROBLEM 14.43 (Cont.)

The derivation for the species transport surface boundary condition is developed in the solution for Problem 14.3 (S). The solution to the mass transfer problem is identical to the analogous heat transfer problem provided the transport coefficients are represented as

h h /S R T m AB u k DAB

(1)

(c) The uniform concentration transient diffusion process is analogous to the heat transfer lumpedcapacitance process. From the solution of Problem 14.3 (S), the time to reach twice the limiting concentration, CA (to) = 2 CA,f, can be calculated as

1 6

 

CA t o − CA,f hm to = exp − CA,i − CA,f L SAB Ru T

 

(2)

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t o = 42.9 hour For the present situation, the mass transfer Biot number is

Bi m =

Bi m =

hm L SAB Ru T DAB

115. m / h / 3600 s / h6 × 0.003 m

160 kmol / m3 ⋅ atm × 8.205 × 10-2 m3 ⋅ atm / kmol ⋅ K × 555 K × 1.8 × 10-11 m2 / s

Bi m = 9.5 >> 0.1 and hence the concentration of A within B is not uniform (d) Invoking the analogy with the heat transfer situation, we can use the one-term series solution, Eq. 5.40, with Bi m Bi and

Fo m Fo

Fo m =

D AB t

(3)

L2 Continued …..

PROBLEM 14.43 (Cont.) With Bim = 9.5, find ζ1 = 1.4219 rad and C1 = 1.2609 from Table 5.1, so that Eq. 5.41 becomes

1 6

CA t o − CA,f = C1 exp −ς 12 Fo m CA,i − CA,f

4

9

12 − 16 × 16 kmol / m3 = 12609 . exp4 −1.4219 2 Fo m 9 3 1320 − 166kmol / m Fo m =

18 . × 10−11 m2 / s × t o

10.003 m62

= 1571 .

t o = 218 hour

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COMMENTS: (1) Since Bim = 9.5, the uniform concentration assumption is not valid, and we expect the analysis to provide a longer time estimate to reach CA(to) = 2 CA,f. (2) Note that the uniform concentration analysis model of part (c) does not include DAB. Why is this so?

PROBLEM 14.44 KNOWN: Radius and temperature of air bubble in water. FIND: Time to reach 99% of saturated vapor concentration at center. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional radial diffusion of vapor in air, (2) Constant properties, (3) Air is initially dry. -4

2

PROPERTIES: Table A-8, Water vapor-air (300 K): DAB = 0.26 × 10 m /s. ANALYSIS: Use Heisler charts with heat and mass transfer analogy,

γ∗ ≡

CA − CA,s CA,i − CA,s

=1 −

CA . CA,s

−1 = 0, from Fig. D.7 find Fo ≈ 0.52. Hence with For γ o∗ = 1 − 0.99 = 0.01and Bim m

Fo m =

DAB t ro2

= 0.52

(

)

t = 0.52 10−6 m 2 /0.26× 10−4 m 2 /s = 0.02s.

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COMMENTS: (1) This estimate is likely to be conservative, since shear driven motion of air within the bubble would enhance vapor transport from the surface to the center. (2) If the one-term approximation to the infinite series solution,

θ∗ =

∞

(

∑

Cn exp −ςn2Fo n =1

) ς(n r∗ ) sin ς n r∗

is used, it follows that with sin 0/0 = 1,

(

)

γ o∗ ≈ C1 exp −ζ12 Fo m = 0.1. Using values of C1 = 2.0 and ς1 = 3.11for Bi m = 100, it follows that

0.01 = 2.0 exp  − (3.11) 2 Fo m   

or

Fo m = 0.55

which is in reasonable agreement with the Heisler chart result.

PROBLEM 14.45 KNOWN: Initial carbon content and prescribed surface content for heated steel. FIND: Time required for carbon mole fraction to reach 0.01 at a distance of 1 mm from the surface. SCHEMATIC:

ASSUMPTIONS: (1) Steel may be approximated as a semi-infinite medium, (2) One-dimensional diffusion in x, (3) Isothermal conditions, (4) No internal chemical reactions, (5) Uniform total molar concentration. ANALYSIS: Conditions within the steel are governed by the species diffusion equation of the form ∂2 C A 1 ∂CA = 2 DAB ∂t ∂x or, in molar form,

∂2 x A ∂x 2

=

∂x A . DAB ∂t 1

The initial and boundary conditions are of the form

x A ( x,0 ) = 0.001

x A ( 0,t ) = x A,s = 0.02

x A ( ∞ , t ) = 0.001.

The problem is analogous to that of heat transfer in a semi-infinite medium with constant surface temperature, and by analogy to Eq. 5.57, the solution is

x A ( x,t ) − x A,s x A,i − x A,s where

  x   = erf  2 ( D t )1/2  AB  

DAB = 2 ×10 −5 exp [−17,000/1273] = 3.17 ×10 −11 m 2 /s.

Hence

    0.01 − 0.02 0.001m = 0.526 = erf   0.001 − 0.02  2 3.17 × 10−11t 1/2     

(

)

where erf w = 0.526 → w ≈ 0.51,

(

0.51 = 0.001/2 3.17 ×10 −11 t

)

1/2

or

t = 30,321s = 8.42 h.

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PROBLEM 14.46 KNOWN: Thick plate of pure iron at 1000°C subjected to a carburizing process with sudden exposure to a carbon concentration CC,s at the surface. FIND: (a) Consider the heat transfer analog to the carburization process; sketch the mass and heat transfer systems; explain correspondence between variables; provide analytical solutions to the mass and heat transfer situation; (b) Determine the carbon concentration ratio, CC (x, t)/CC,s, at a depth of 1 mm after 1 hour of carburization; and (c) From the analogy, show that the time dependence of the 1/ 2 mass flux of carbon into the plate can be expressed as n ′′C = ρ C,s D C − Fe / π t ; also, obtain an expression for the mass of carbon per unit area entering the iron plate over the time period t.

1

6

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional transient diffusion, (2) Thick plate approximates a semiinfinite medium for the transient mass and heat transfer processes, and (3) Constant properties. ANALYSIS: (a) The analogy between the carburizing mass transfer process in the plate and the heat transfer process is illustrated in the schematic above. The basis for the mass - heat transfer analogy stems from the similarity of the conservation of species and energy equations, the general solution to the equations, and their initial and boundary conditions. For both processes, the plate is a semiinfinite medium with initial distributions, CC (x, t ≤ 0) = CC,i = 0 and T (x, t ≤ 0) = Ti, suddenly subjected to a surface potential, CC (0, t > 0) = CC,s and T (0, t > 0) = Ts. The heat transfer situation corresponds to Case 1, Section 5.7, from which the following relations were obtained. Mass transfer Rate equation

jC ′′ = − DAB

Heat transfer

∂ Cc ∂x

q ′′x = − k

Diffusion equation

∂ ∂x

 ∂ CC  = 1  ∂ x  DAB

∂ CC ∂t

14.84

Polential distribution

1 6

CC x, t − CC,s = 0 − CC,s

1 6

  1

∂ ∂x

 ∂ T = 1  ∂ x α

1 6

  6 

∂T ∂x ∂T ∂t

2.15

  1 6 

T x, t − Ts x = erf Ti − Ts 2 αt 1/ 2

  

5.58

CC x, t x = erfc 1/ 2 CC,s 2 D AB t

Continued …..

PROBLEM 14.46 (Cont.) Flux

1Ts − Ti 6 1 6 k πα 1 t61/ 2

q s′′ t =

See Part (c)

5.58

(b) Using the concentration distribution expression above, with L = 1 mm, t = 1 h and -11 2 DAB = 3 × 10 m /s, find the concentration ratio,

  CC 11 mm, 1 h6   = 0.0314 0.001 m = erfc CC,s  243 × 10-11 m2 / s × 3600 s91/ 2 

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(c) From the heat flux expression above, the mass flux of carbon can be written as

n′′C,s =

3

8 = ρc,s 1DC− Fe / π t61/ 2

DC − Fe ρ C,s − 0

1

6

π D C − Fe t 1/ 2

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The mass per unit area entering the plate over the time period follows from the integration of the rate expression t

1/ 2 m′′C ( t ) = ∫ n′′C,s dt = ρC,s ( DAB / π ) 0

t

∫0 t

-1/2 dt = 2 ρ C,s

( DC− Fe t/π )1/ 2

PROBLEM 14.47 KNOWN: Thickness, initial condition and bottom surface condition of a water layer. FIND: (a) Time to reach 25% of saturation at top, (b) Amount of salt transfer in that time, (c) Final concentration of salt solution at top and bottom. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional diffusion, (2) Uniform total mass density, (3) Constant DAB. ANALYSIS: (a) With constant ρ and DAB and no homogeneous chemical reactions, Eq. 14.37b reduces to ∂ 2 ρA 1 ∂ρ A = . DAB ∂t ∂x 2 with the origin of coordinates placed at the top of the layer, the dimensionless mass density is

(

)

(

)

γ ∗ x∗ ,Fom =

ρ − ρA,s γ ρ = A = 1− A γ i ρ A,i − ρ A,s ρ A,s

( )

Hence, γ ∗ 0, Fo m,1 = 1 − 0.25 = 0.75. The initial condition is γ ∗ x∗ ,0 = 1, and the boundary conditions are ∂γ ∗

∂x∗

x∗ = 0

=0

γ ∗ (1, Fom ) = 0

where the condition at x∗ = 1 corresponds to Bim = ∞. Hence, the mass transfer problem is analogous to the heat transfer problem governed by Eq. 5.34 to 5.37. Assuming applicability of a oneterm approximation (Fom > 0.2), the solution is analogous to Eq. 5.40.

(

) ( )

γ ∗ = C1 exp −ς12 Fom cos ς1x ∗ . With Bi m = ∞, ς1 = π / 2 = 1.571 rad and, from Table 5.1, C1 ≈ 1.274. Hence, for x∗ = 0,

0.75 = 1.274exp − (1.571) 2 Fo m,1    Fo m,1 = − ln ( 0.75/1.274 ) / (1.571) = 0.215. 2

Hence,

(1 m )2 L2 t1 = Fom,1 = 0.215 = 1.79 ×108 s = 2071days. − 9 2 DAB 1.2 ×10 m / s

< Continued …..

PROBLEM 14.47 (Cont.) (b) The change in the salt mass within the water is

(

)

L

∆M A = M A ( t1 ) − M A,i = ∫ ρA − ρA,i dV = A∫ ρ Adx 0 Hence, L

(

)

∆M ′′A = ρ A,s ∫ ρ A / ρ A,s dx 0

(

)

∆M ′′A = ρ A,s L∫ 1 − γ ∗ dx ∗ 0 1

(

) ( ) dx∗

1 ∆M ′′A = ρA,s L ∫ 1 −C 1 exp −ς12Fo m cos ς1 x∗ 0 

(

)

∆M ′′A = ρA,s L 1 − C1 exp −ς12 Fom sin ς1 / ς1  .   Substituting numerical values,

 1.274exp  − (1.571)2 0.215 1     ∆M ′′A = 380 k g / m 3 (1 m ) 1 −  1.571rad     ∆M ′′A = 198.7 k g / m 2 .

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(c) Steady-state conditions correspond to a uniform mass density in the water. Hence,

ρ A ( 0, ∞ ) = ρ A ( L, ∞ ) = ∆M′′A / L = 198.7 k g / m3 .

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COMMENTS: (1) The assumption of constant ρ is weak, since the density of salt water depends strongly on the salt composition. (2) The requirement of Fom > 0.2 for the one-term approximation to be valid is barely satisfied.

PROBLEM 14.48 KNOWN: Temperature distribution expression for a semi-infinite medium, initially at a uniform temperature, that is suddenly exposed to an instantaneous amount of energy, Q o′′ J / m2 .

"

'

Analogous situation of a silicon (Si) wafer with a 1-µm layer of phosphorous (P) that is placed in a furnace suddenly initiating diffusion of P into Si. FIND: (a) Explain the correspondence between the variables in the analogous temperature and concentration distribution expressions, and (b) Determine the mole fraction of P at a depth of 0.1 mm in the Si after 30 s. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, transient diffusion, (2) Wafer approximates a semi-infinite medium, (3) Uniform properties, and (4) Diffusion process for Si and P is initiated when the wafer reaches the elevated temperature as a consequence of the large temperature dependence of the diffusion coefficient. -17

2

PROPERTIES: Given in statement: DP-Si = 1.2 × 10 m /s; Mass densities of Si and P: 2000 and 3 2300 kg/m ; Molecular weights of Si and P: 30.97 and 28.09 kg/kmol. ANALYSIS: (a) For the thermal process illustrated in the schematic, the temperature distribution is

 $

T x, t − Ti =

Qo′′

 $

ρc παt

1/ 2

"

exp − x 2 / 4αt

'

(HT)

where Ti is the initial, uniform temperature of the medium. For the mass transfer process, the P concentration has the form

 $

C P x, t =



M ′′P,o

π D P −Si t

$

1/ 2

"

exp − x2 / 4 D P −Si t

'

(MT)

2

where M ′′P,o is the molar area density (kmol/m ) of P represented by the film of concentration CP and thickness do. The correspondence between mass and heat transfer variables in the equations HT and MT involves the following conditions. The LHS represents the increase with time of the temperature or concentration above the initial uniform distribution. The initial concentration is zero, so only the CP (x, t) appears. On the RHS note the correspondence of the terms in the exponential parenthesis and in the denominator. The thermal diffusivity and diffusion coefficient are directly analogous; this can be seen by comparing the MT and HT diffusion equations, Eq. 2.15 and 14.84. The terms Q ′′o / ρc and M ′′P,o for HT and MT represent the energy and mass instantaneously appearing at the surface. The product ρc is the thermal capacity per unit area and appears in the storage term of the HT diffusion equation. For MT, the “capacity” term is the volume itself. Continued …..

PROBLEM 14.48 (Cont.) 2

(b) The molar area density (kmol/m ) of P associated with the film of thickness do = 1 µm and concentration CP,o is



$

M ′′P,o = C P,o ⋅ d o = ρ P / M P d o

"

'

M ′′P,o = 2000 kg / m3 / 30.97 kmol / kg × 1 × 10−6 m M ′′P,o = 6.458 × 10−5 kmol / m2 Substituting numerical values into the MT equation, find



$

C p 0.1 mm, 30 s =

"

6.458 × 10−5 kmol / m2

'

π × 1.2 × 10-17 m2 / s × 30 s



'

$ "

exp − 0.0001 m 2 / 4 × 12 . × 10 −7 m2 / s × 30 s

C p = 0.08188 kmol / m3 The mole fraction of P in the Si wafer is



x P = C P / CSi = C P / ρ Si / MSi

"

$

x P = 0.08188 kmol / m3 / 2300 kg / m3 / 28.09 kmol / kg x P = 2.435 × 10−5

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