CHAPTER TWO 2.1 (a) (b) (c)
2.2 (a) (b) (c)
3 wk
7d
24 h 3600 s 1000 ms
1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s 3.2808 554 m 4
1d
ft
h
1
1000 g
m
1 h
0.0006214 mi 3600 s 1
kg
35.3145 ft 3
5.37 × 10 3 kJ 1 min 1000 J min
60 s
1
m
4
= 3.85 × 10 4 cm 4 / min⋅ g
= 340 m / s
1 m3
921 kg 2.20462 lb m m3
= 25.98 mi / h ⇒ 26.0 mi / h
1 kg 10 8 cm 4
1h
d ⋅ kg 24 h 60 min
760 mi
1 h
= 1.8144 × 10 9 ms
= 57.5 lb m / ft 3
1.34 × 10 -3 hp
1 kJ
1
J/s
= 119.93 hp ⇒ 120 hp
2.3 Assume that a golf ball occupies the space equivalent to a 2 in × 2 in × 2 in cube. For a classroom with dimensions 40 ft × 40 ft × 15 ft : 40 × 40 × 15 ft 3 (12)3 in 3 1 ball n balls = = 5.18 × 10 6 ≈ 5 million balls ft 3 2 3 in 3 The estimate could vary by an order of magnitude or more, depending on the assumptions made. 2.4 4.3 light yr 365 d 24 h 1 yr 1 d
3600 s 1.86 × 10 5 mi 1 h
1
s
3.2808 ft 0.0006214 mi
1 step = 7 × 1016 steps 2 ft
2.5 Distance from the earth to the moon = 238857 miles 238857 mi
1
m
1 report
0.0006214 mi
0.001 m
= 4 × 10 11 reports
2.6 19 km 1000 m 0.0006214 mi 1000 L = 44.7 mi / gal 1 L 1 km 1 m 264.17 gal Calculate the total cost to travel x miles. Total Cost
Total Cost
American
European
= $14,500 +
= $21,700 +
$1.25 1 gal gal 28 mi
$1.25 1 gal gal 44.7 mi
x (mi)
Equate the two costs ⇒ x = 4.3 × 10 5 miles
2-1
= 14,500 + 0.04464 x
x (mi)
= 21,700 + 0.02796x
2.7
106 cm 3 220.83 imp. gal
5320 imp. gal 14 h 365 d plane ⋅ h 1 d 1 yr
0.965 g 1 cm 3
1 kg 1 tonne 1000 g 1000 kg
tonne kerosene plane ⋅ yr 9 4.02 ×10 tonne crude oil 1 tonne kerosene plane ⋅ yr 5 yr 7 tonne crude oil 1.188 ×10 tonne kerosene = 1.188 ×105
= 4834 planes ⇒ 5000 planes
2.8 (a) (b) (c)
2.9
2.10 2.11
32.1714 ft / s2
25.0 lb m
1
lb f
32.1714 lb m ⋅ ft / s 2 25 N
1 kg ⋅ m / s2
1 9.8066 m / s 2
10 ton
1N
1 lb m 5 × 10
50 × 15 × 2 m 3
-4
980.66 cm / s2
35.3145 ft 3
85.3 lb m
3
3
1
kg
2.20462 lb m
m
1 ft 1 m3 11.5 kg
1 dyne 1 g ⋅ cm / s2
2.20462 lb m
1 500 lbm
= 2.55 kg ⇒ 2.6 kg
1000 g ton
= 25.0 lb f
32.174 ft 1 s
≈ 5 × 10 2
2
= 9 × 10 9 dynes
1 lb f 32.174 lb m / ft ⋅ s
FG 1IJ FG 1 IJ ≈ 25 m H 2K H 10K
2
= 4.5 × 10 6 lb f
3
(a) mdisplaced fluid = mcylinder ⇒ ρ f V f = ρ cVc ⇒ ρ f hπr 2 = ρ c Hπr 2 ρfh
(30 cm − 14.1 cm)(100 . g / cm 3 ) ρc = = = 0.53 g / cm 3 H 30 cm ρ H (30 cm )( 0.53 g / cm 3 ) (b) ρ f = c = = 1.71 g / cm 3 h (30 cm - 20.7 cm)
2-2
ρc H ρf h
2.12
Vs =
πR 2 H πR 2 H πr 2 h R r R ; Vf = − ; = ⇒r= h 3 3 3 H h H
FG IJ = πR FG H − h IJ H K 3H HK πR F h I πR H H− =ρ G J 3 H 3 H K 2
πR 2 H πh Rh ⇒ Vf = − 3 3 H ρ f V f = ρ s Vs ⇒ ρ f H
⇒ ρ f = ρs
H−
2.13
h3
2
r
2
2
3
H
2
ρs
s
2
= ρs
h
3
H3 H 3 − h3
1−
H2
R
1
= ρs
FG h IJ H HK
3
Say h( m ) = depth of liquid
y y= 1 dA y=y=1––1+h h x
⇒ A(m 2 )
h
1m
x = 1– y 2 y= –1
dA
z
1− y 2
dA = dy ⋅
dx = 2 1 − y 2 dy
− 1− y
d i=2
⇒Am
2
E
2
z
−1+ h
1 − y 2 dy
−1
Table of integrals or trigonometric substitution
d i
A m 2 = y 1 − y 2 + sin−1 y
b g
W N =
h −1 −1
b g b g
= h −1 1− h −1
4 m × A( m 2 ) 0.879 g 10 6 cm 2 cm
3
1m
E Substitute for A L W bN g = 3.45 × 10 Mbh − 1g 1 − bh − 1g N 4
3
2
b g
+ sin −1 h − 1 +
1 kg
9.81 N
3
kg {
10 g
= 3.45 × 10 4 A
g g0
2
b g π2 OPQ
+ sin −1 h − 1 +
2.14 1 lb f = 1 slug ⋅ ft / s2 = 32 .174 lb m ⋅ ft / s2 ⇒ 1 slug = 32.174 lb m 1 1 poundal = 1 lb m ⋅ ft / s2 = lb f 32.174
2-3
π 2
ρf
2.14(cont’d) (a) (i) On the earth: 175 lb m M= W=
175 lb m
(ii) On the moon 175 lb m M= W=
175 lb m
1 slug = 5.44 slugs 32.174 lbm 32.174 ft 1 poundal 3 2 2 = 5.63 × 10 poundals s 1 lb m ⋅ ft / s 1 slug = 5.44 slugs 32.174 lb m 32.174 ft 1 poundal = 938 poundals 2 6 s 1 lb m ⋅ ft / s2
(b ) F = ma ⇒ a = F / m =
1 lb m ⋅ ft / s 2 1 poundal
355 poundals 25.0 slugs
1 slug 32.174 lb m
1m 3.2808 ft
= 0.135 m / s2 2.15 (a) F = ma ⇒ 1 fern = (1 bung)(32.174 ft / s 2 ) ⇒
FG 1IJ = 5.3623 bung ⋅ ft / s H 6K
2
1 fern 5.3623 bung ⋅ ft / s2
3 bung 32.174 ft 1 fern = 3 fern 2 6 s 5.3623 bung ⋅ ft / s2 On the earth: W = (3)(32.174 ) / 5.3623 = 18 fern
(b) On the moon: W =
2.16 (a) ≈ (3)(9 ) = 27
≈
(d)
≈ 50 × 10 3 − 1 × 10 3 ≈ 49 × 10 3 ≈ 5 × 10 4
( 2.7)(8.632 ) = 23
(c) ≈ 2 + 125 = 127 2.365 + 125.2 = 127 .5
2.17 R ≈
4.0 ×10−4 ≈ 1× 10−5 40 (3.600 ×10−4 ) / 4 5 = 8.0 × 10−6
(b)
4.753 × 10 4 − 9 × 10 2 = 5 × 10 4
(7 × 10−1 )(3 × 105 )(6)(5 ×104 ) ≈ 42 ×10 2 ≈ 4 ×103 (Any digit in range 2-6 is acceptable) 6 (3)(5 × 10 )
Rexact = 3812.5 ⇒ 3810 ⇒ 3.81 ×103
2-4
2.18 (a) A: R = 731 . − 72.4 = 0.7 o C 72.4 + 731 . + 72.6 + 72.8 + 73.0 = 72.8 o C 5
X=
( 72.4 − 72.8) 2 + ( 731 . − 72.8) 2 + (72.6 − 72.8 ) 2 + (72.8 − 72.8 ) 2 + (73.0 − 72 .8) 2 5 −1 o = 0.3 C
s=
B: R = 1031 . − 97.3 = 58 . oC X=
97.3 + 1014 . + 98.7 + 1031 . + 100.4 = 100.2 o C 5
s=
(97.3 − 1002 . ) 2 + (1014 . − 1002 . ) 2 + (98.7 − 1002 . ) 2 + (1031 . − 100.2) 2 + (100.4 − 100.2)2 5 −1
= 2.3o C
(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate. 2.19 (a)
12
X =
∑
C min=
12
Xi
i =1
= 735 . s= 12 = X − 2 s = 73.5 − 2 (1.2) = 711 .
∑ ( X − 73.5) i =1
12 − 1
2
= 12 .
C max= = X + 2 s = 735 . + 2(12 . ) = 75.9
(b) Joanne is more likely to be the statistician, because she wants to make the control limits stricter. (c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor temperature (failure of reactor control system), problems with the color measurement system, operator carelessness
2-5
2.20 (a),(b) (a) Run 1 2 X 134 131 Mean(X) 131.9 Stdev(X) 2.2 Min 127.5 Max 136.4 (b) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14
X 128 131 133 130 133 129 133 135 137 133 136 138 135 139
Min 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5
3 129
4 5 6 7 8 9 10 11 12 13 14 15 133 135 131 134 130 131 136 129 130 133 130 133
Mean 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9
Max 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4
140 138 136 134 132 130 128 126 0
5
10
15
(c) Beginning with Run 11, the process has been near or well over the upper quality assurance limit. An overhaul would have been reasonable after Run 12.
2.21 (a) Q' =
2.36 × 10 −4 kg ⋅ m 2 h
2.22 N Pr = N Pr ≈
Cp µ k
1 h
2
3600 s
kg
m
−4
(2 × 10 )(2 )(9 ) ≈ 12 × 10 ( −4− 3) ≈ 12 . × 10 −6 lb ⋅ ft 2 / s 3 × 10 3 = 1.48 × 10 −6 lb ⋅ ft 2 / s = 0.00000148 lb ⋅ ft 2 / s
(b) Q' approximate ≈ Q' exact
2.10462 lb 3.2808 2 ft 2
=
0.583 J / g ⋅ o C 1936 lb m 1 h 3.2808 ft 1000 g o 0.286 W / m ⋅ C ft ⋅ h 3600 s m 2.20462 lbm
(6 × 10 −1 )(2 × 10 3 )(3 × 10 3 ) 3 × 10 3 ≈ ≈ 15 . × 10 3 . The calculator solution is 1.63 × 10 3 (3 × 10 −1 )(4 × 10 3 )(2 ) 2
2.23 Duρ 0.48 ft 1 m 2.067 in 1 m = −3 µ s 3.2808 ft 0.43 × 10 kg / m ⋅ s 39.37 in
Re ≈
(5 × 10 −1 )(2 )(8 × 10 − 1 )(10 6 ) 5 × 101−( −3) ≈ ≈ 2 × 10 4 ⇒ the flow is turbulent (3)(4 × 10)(10 3 )(4 × 10 −4 ) 3
2-6
0.805 g cm 3
1 kg 10 6 cm 3 1000 g 1 m3
Re =
2.24 (a)
kgdp y D
FG µ IJ FG d uρ IJ H ρD K H µ K 1/3
= 2.00 + 0.600
1/2
p
L 1.00 × 10 N ⋅ s / m = 2.00 + 0.600M . kg / m )(100 . × 10 m N (100 −5
−5
3
= 44.426 ⇒
OP LM (0.00500 m)(10.0 m / s)(100 . kg / m ) O PQ / s) Q N (100 . × 10 N ⋅ s / m ) 1/ 3
2
k g (0.00500 m)(0.100) 1.00 × 10 −5 m 2 / s
3
−5
2
1/2
2
= 44.426 ⇒ k g = 0.888 m / s
(b) The diameter of the particles is not uniform, the conditions of the system used to model the equation may differ significantly from the conditions in the reactor (out of the range of empirical data), all of the other variables are subject to measurement or estimation error. (c) dp (m) y D (m2/s) µ (N-s/m 2) ρ (kg/m 3) u (m/s) kg 0.005 0.1 1.00E-05 1.00E-05 1 10 0.889 0.010 0.1 1.00E-05 1.00E-05 1 10 0.620 0.005 0.1 2.00E-05 1.00E-05 1 10 1.427 0.005 0.1 1.00E-05 2.00E-05 1 10 0.796 0.005 0.1 1.00E-05 1.00E-05 1 20 1.240 2.25 (a) 200 crystals / min ⋅ mm; 10 crystals / min ⋅ mm 2 (b) r = 200 crystals
0.050 in 25.4 mm
min ⋅ mm
in
= 238 crysta ls / min ⇒
b g
(c) D mm =
bg
D ′ in
10 crystals
−
(25.4) 2 mm 2
min ⋅ mm2
238 crystals 1 min min
0.050 2 in 2
= 4.0 crystals / s
60 s
FG H
in 2
IJ K
crystals 60 s 25.4 mm crystals = r′ = 60r ′ = 25.4 D ′ ; r min s 1 min 1 in
b
g b
⇒ 60r ′ = 200 25.4 D ′ − 10 25.4 D ′
g
2
b g
⇒ r ′ = 84 .7 D ′ − 108 D ′
2
2.26 (a) 70.5 lb m / ft 3 ; 8.27 × 10 -7 in 2 / lb f −7 2 9 × 10 6 N 14.696 lb f /in 2 (b) ρ = (70.5 lb / f t 3) exp 8.27 × 10 in m lbf m 2 1.01325 × 10 5 N/m 2 70.57 lb m 35.3145 ft 3 1 m 3 1000 g = = 1.13 g /cm 3 3 3 6 3 ft m 10 cm 2.20462 lb m
FG lb IJ = ρ ′ g H ft K cm F lb IJ = P ' N PG H in K m
(c) ρ
m 3
f 2
3
1 lb m
28,317 cm 3
453.593 g
1 ft 3
0.2248 lb f
12
2
d
⇒ 62.43 ρ ′ = 70.5 exp 8.27 × 10
m2 2
1N
39.37 in −7
= 62.43ρ ′
2
= 1.45 × 10 −4 P'
. × 10 P 'i ⇒ ρ ′ = 113 . expd1.20 × 10 id145 −4
−10
P' = 9.00 × 10 6 N / m 2 ⇒ ρ ' = 113 . exp[(1.20 × 10 −10 )(9.00 × 10 6 )] = 113 . g / cm 3
2-7
P'
i
cm = 16.39V ' ; t bsg = 3600t ′b hr g d i V ' din i 28,317 1728 in ⇒ 16.39V ' = expb3600t ′g ⇒ V ' = 0.06102 expb3600t ′ g
2.27 (a) V cm 3 =
3
3
3
(b) The t in the exponent has a coefficient of s-1 . 2.28 (a) 3.00 mol / L, 2.00 min -1 (b) t = 0 ⇒ C = 3.00 exp[(-2.00)(0)] = 3.00 mol / L t = 1 ⇒ C = 3.00 exp[(-2.00)(1)] = 0.406 mol / L 0.406 − 3.00 For t=0.6 min: Cint = ( 0.6 − 0) + 3.00 = 1.4 mol / L 1− 0 Cexact = 3.00 exp[(-2.00)(0.6)] = 0.9 mol / L For C=0.10 mol/L:
t int = t exact
1− 0 (0.10 − 3.00) + 0 = 112 . min 0.406 − 3 1 C 1 0.10 =ln = - ln = 1.70 min 2.00 3.00 2 3.00
(c) 3.5 C exact vs. t
3 C (mol/L)
2.5 2
(t=0.6, C=1.4)
1.5 1
(t=1.12, C=0.10)
0.5 0 0
1
2
t (min)
p* =
2.29 (a) (b)
c
1 902 *
903 2
60 − 20 (185 − 166.2) + 20 = 42 mm Hg 1998 . − 1662 .
MAIN PROGRAM FOR PROBLEM 2.29 IMPLICIT REAL*4(A–H, 0–Z) DIMENSION TD(6), PD(6) DO 1 I = 1, 6 READ (5, *) TD(I), PD(I) CONTINUE WRITE (5, 902) FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X, ‘ (C) (MM HG)’/) DO 2 I = 0, 115, 5 T = 100 + I CALL VAP (T, P, TD, PD) WRITE (6, 903) T, P FORMAT (10X, F5.1, 10X, F5.1) CONTINUE END
2-8
2.29 (cont’d) SUBROUTINE VAP (T, P, TD, PD) DIMENSION TD(6), PD(6) I=1 1 IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2 I=I+1 IF (I.EQ.6) STOP GO TO 1 2 P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I)) RETURN END DATA OUTPUT 98.5 1.0 TEMPERATURE VAPOR PRESSURE 131.8 5.0 (C) (MM HG) 100.0 1.2 M M 215.5 100.0 105.0 1.8 M M 215.0 98.7 2.30 (b) ln y = ln a + bx ⇒ y = ae bx b = (ln y 2 − ln y 1 ) / ( x 2 − x 1 ) = (ln 2 − ln 1) / (1 − 2 ) = −0.693 ln a = ln y − bx = ln 2 + 0.63(1) ⇒ a = 4.00 ⇒ y = 4.00e −0.693 x
(c) ln y = ln a + b ln x ⇒ y = ax b b = (ln y 2 − ln y 1 ) / (ln x 2 − ln x 1 ) = (ln 2 − ln 1) / (ln 1 − ln 2) = −1 ln a = ln y − b ln x = ln 2 − ( −1) ln(1) ⇒ a = 2 ⇒ y = 2 / x (d) ln( xy ) = ln a + b ( y / x ) ⇒ xy = ae by / x ⇒ y = ( a / x )e by / x [can' t get y = f ( x )] b = [ln( xy ) 2 − ln( xy )1 ] / [( y / x ) 2 − ( y / x ) 1 ] = (ln 807.0 − ln 40.2 ) / (2.0 − 1.0) = 3 ln a = ln( xy ) − b( y / x ) = ln 807.0 − 3 ln(2.0) ⇒ a = 2 ⇒ xy = 2e 3 y / x ⇒ y = (2 / x )e 3y / x
(e) ln( y 2 / x ) = ln a + b ln( x − 2) ⇒ y 2 / x = a ( x − 2) b ⇒ y = [ax ( x − 2 ) b ]1/2 b = [ln( y 2 / x ) 2 − ln( y 2 / x )1 ] / [ln( x − 2 ) 2 − ln( x − 2 )1 ] = (ln 807.0 − ln 40.2 ) / (ln 2.0 − ln 1.0) = 4.33 ln a = ln( y 2 / x ) − b ( x − 2 ) = ln 807 .0 − 4.33 ln(2.0) ⇒ a = 40.2 ⇒ y 2 / x = 40.2 ( x − 2) 4.33 ⇒ y = 6.34 x 1/2 ( x − 2 ) 2.165
2.31 (b) Plot y 2 vs. x 3 on rectangular axes. Slope = m, Intcpt = − n
(c)
1 1 a 1 = + x ⇒ Plot vs. ln( y − 3) b b ln( y − 3)
(d)
1 1 = a ( x − 3) 3 ⇒ Plot vs. ( x − 3) 3 [rect. axes] , slope = a , intercept = 0 2 2 ( y + 1) ( y + 1)
OR
2-9
x [rect. axes], slope =
a 1 , intercept = b b
2.31 (cont’d) 2 ln( y + 1) = − ln a − 3 ln( x − 3) Plot ln( y + 1) vs. ln( x − 3) [rect.] or (y + 1) vs. (x - 3) [log] 3 ln a ⇒ slope = − , intercept = − 2 2
(e) ln y = a x + b Plot ln y vs.
x [rect.] or y vs.
x [semilog ], slope = a, intercept = b
(f) log10 ( xy ) = a ( x 2 + y 2 ) + b Plot log10 ( xy ) vs. ( x 2 + y 2 ) [rect.] ⇒ slope = a, intercept = b
(g)
1 b x x = ax + ⇒ = ax 2 + b ⇒ Plot vs. x 2 [rect.], slope = a , intercept = b y x y y OR
1 b 1 b 1 1 = ax + ⇒ = a + 2 ⇒ Plot vs. 2 [rect.] , slope = b, intercept = a y x xy x xy x
2.32 (a) A plot of y vs. R is a line through ( R = 5 , y = 0.011 ) and ( R = 80 , y = 0.169 ).
0.18 0.16 0.14
y
0.12 0.1 0.08 0.06 0.04 0.02 0 0
20
40
60
80
100
R
y=aR +b
U| V| W
0.169 − 0.011 = 2.11 × 10 −3 80 − 5 ⇒ y = 2 .11 × 10 −3 R + 4.50 × 10 −4 −3 −4 b = 0.011 − 2.11 × 10 5 = 4.50 × 10
a=
d
d
ib g
ib g
(b) R = 43 ⇒ y = 2.11 × 10 −3 43 + 4.50 × 10 −4 = 0.092 kg H 2O kg
b1200 kg hgb0.092 kg H O kgg = 110 kg H O h 2
2
2-10
2.33 (a) ln T = ln a + b ln φ ⇒ T = aφ b b = (ln T2 − ln T1 ) / (ln φ 2 − ln φ 1) = (ln 120 − ln 210) / (ln 40 − ln 25) = −119 . ln a = ln T − b ln φ = ln 210 − (−119 . ) ln(25) ⇒ a = 9677 .6 ⇒ T = 9677.6φ −1.19
b
(b) T = 9677.6φ −1.19 ⇒ φ = 9677.6 / T
b
g
T = 85 o C ⇒ φ = 9677.6 / 85
0.8403
b g T = 290 C ⇒ φ = b9677.6 / 290g T = 175 o C ⇒ φ = 9677.6 / 175 o
g
0.8403
= 53.5 L / s
0.8403 0.8403
= 29.1 L / s = 19.0 L / s
(c) The estimate for T=175°C is probably closest to the real value, because the value of temperature is in the range of the data originally taken to fit the line. The value of T=290°C is probably the least likely to be correct, because it is farthest away from the date range.
2-11
ln ((C A-CAe )/(CA0-C Ae))
2.34 (a) Yes, because when ln[(CA − C Ae ) / (C A0 − CAe )] is plotted vs. t in rectangular coordinates, the plot is a straight line. 0
50
100
150
200
0 -0.5 -1 -1.5 -2 t (min)
Slope = -0.0093 ⇒ k = 9.3 × 10 -3 min −1
(b) ln[(CA − C Ae ) / (C A0 − C Ae )] = −kt ⇒ C A = (C A 0 − CAe )e − kt + CAe −3
CA = (0.1823 − 0.0495)e− (9.3×1 0
)(120)
+ 0.0495 =9.300 ×10-2 g/L
9.300 ×10-2 g 30.5 gal 28.317 L C =m/ V ⇒ m=CV = = 10.7 g L 7.4805 gal 2.35 (a) ft 3 and h-2 , respectively (b) ln(V) vs. t2 in rectangular coordinates, slope=2 and intercept= ln(3.53 × 10 −2 ) ; or V(logarithmic axis) vs. t2 in semilog coordinates, slope=2, intercept= 3.53 × 10−2 (c) V ( m 3 ) = 100 . × 10 −3 exp(15 . × 10− 7 t 2 ) 2.36 PV k = C ⇒ P = C / V k ⇒ ln P = ln C − k ln V 8.5
lnP
8 7.5 7 6.5 6 2.5
3
lnP = -1.573(lnV ) + 12.736
3.5
4
lnV
k = − slope = −( −1573 . ) = 1573 . (dimensionless) Intercept = ln C = 12.736 ⇒ C = e 12.736 = 3.40 × 10 5 mm Hg ⋅ cm 4.719
G − GL 1 G −G G −G = ⇒ 0 = K L Cm ⇒ ln 0 = ln K L + mln C m G0 − G K L C G − GL G − GL ln(G 0 -G)/(G-G L )= 2.4835lnC - 10.045
3 ln(G 0-G)/(G-G L)
2.37 (a)
2 1 0 -1 3.5
4
4.5
5
lnC
2- 12
5.5
2.37 (cont’d) m = slope = 2.483 (dimensionless) Intercept = ln KL = −10.045 ⇒ K L = 4.340 × 10 −5 ppm-2.483 G − 180 . × 10 −3 = 4.340 × 10 −5 (475) 2.483 ⇒ G = 1806 . × 10 −3 3.00 × 10 −3 − G C=475 ppm is well beyond the range of the data.
(b) C = 475 ⇒
2.38 (a) For runs 2, 3 and 4: Z = aV& b p c ⇒ ln Z = ln a + b ln V& + c ln p ln(3.5) = ln a + b ln(1.02 ) + c ln(9.1) ln(2.58) = ln a + b ln(102 . ) + c ln(11.2 ) ln(3.72 ) = ln a + b ln(175 . ) + c ln(11.2 )
b = 0.68 ⇒ c = −1.46 a = 86.7 volts ⋅ kPa 1.46 / (L / s) 0.678
& . Slope=b, Intercept= ln a + c ln p (b) When P is constant (runs 1 to 4), plot ln Z vs. lnV 2
lnZ
1.5 1 0.5 0 -1
-0.5
0
0.5
1
1.5
lnV
lnZ = 0.5199lnV + 1.0035
b = slope = 0.52
Intercept = lna + c ln P = 1.0035 & When V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept= ln a + c ln V& 2
lnZ
1.5 1 0.5 0 1.5
1.7
1.9
2.1
2.3
lnP
lnZ = -0.9972lnP + 3.4551
c = slope = −0.997 ⇒ 1.0 Intercept = lna + b ln V& = 3.4551
Z
Plot Z vs V& b P c . Slope=a (no intercept) 7 6 5 4 3 2 1 0.05
0.1
0.15 b
b c
Z = 31.096V P
VP
0.2
c
a = slope = 311 . volt ⋅ kPa / (L / s) .52
The results in part (b) are more reliable, because more data were used to obtain them.
2- 13
2.39 (a) sxy = sxx =
n
1 n
∑x y
1 n
∑x
i
i
= [( 0.4 )(0.3) + (2.1)(1.9) + (3.1)(3.2 )] / 3 = 4.677
i =1 n
2 i
= (0.32 + 1.9 2 + 3.2 2 ) / 3 = 4 .647
i =1
1 n 1 xi = (0.3 + 1.9 + 3.2 ) / 3 = 1.8; s y = n i =1 n sxy − sx sy 4.677 − (1.8)(1.867) a= = = 0.936 2 4.647 − (18 . )2 sxx − sx
∑
sx =
n
∑y
i
= (0.4 + 2.1 + 3.1) / 3 = 1.867
i =1
b g
b=
sxx s y − sxy sx
b g
sxx − sx
2
(4.647 )(1867 . ) − (4.677 )(18 .) = 0182 . 2 4.647 − (18 .)
=
y = 0.936x + 0.182
(b) a =
sxy sxx
=
4.677 = 10065 . ⇒ y = 1.0065 x 4.647
4
y
3
y = 0.936x + 0.182
2
y = 1.0065x
1 0 0
1
2
3
4
x
2.40 (a) 1/C vs. t. Slope= b, intercept=a a = Intercept = 0.082 L / g
3 2.5 2 1.5 1 0.5 0
2 1.5 C
1/C
(b) b = slope = 0.477 L / g ⋅ h;
1 0.5 0
0
1
1/C = 0.4771t + 0.0823
2
3
4
5
6
t
1 C
2 C-fitted
3
4
5
t
(c) C = 1 / ( a + bt ) ⇒ 1 / [0.082 + 0.477 (0)] = 12.2 g / L t = (1 / C − a ) / b = (1 / 0.01 − 0.082 ) / 0.477 = 209.5 h (d) t=0 and C=0.01 are out of the range of the experimental data.
(e) The concentration of the hazardous substance could be enough to cause damage to the biotic resources in the river; the treatment requires an extremely large period of time; some of the hazardous substances might remain in the tank instead of being converted; the decomposition products might not be harmless.
2- 14
2.41 (a) and (c)
y
10
1 0.1
1
10
100
x
(b) y = ax b ⇒ ln y = ln a + b ln x; Slope = b, Intercept = ln a ln y = 0.1684ln x + 1.1258 2 ln y
1.5 1 0.5
b = slope = 0168 .
0 -1
0
1
2 ln x
3
4
5
Intercept = ln a = 11258 . ⇒ a = 3.08
2.42 (a) ln(1-Cp /CA0 ) vs. t in rectangular coordinates. Slope=-k, intercept=0 (b)
600
0
800
ln(1-Cp/Cao)
400
ln(1-Cp/Cao)
0 200 0 -1 -2 -3 -4 ln(1-Cp/Cao) = -0.0062t
100
400
500
t
Lab 1
600
400
600
-4
-6 ln(1-Cp/Cao) = -0.0111t
t
Lab 2
k = 0.0111 s-1
800
0
0
ln(1-Cp/Cao)
ln(1-Cp/Cao)
200
300
-2
k = 0.0062 s-1
0
200
0
-2 -4
200
400
600
800
0 -2 -4
-6 ln(1-Cp/Cao)= -0.0064t
-6 ln(1-Cp/Cao) = -0.0063t t
Lab 3
k = 0.0063 s-1
t
Lab 4
k = 0.0064 s-1
(c) Disregarding the value of k that is very different from the other three, k is estimated with the average of the calculated k’s. k = 0.0063 s-1 (d) Errors in measurement of concentration, poor temperature control, errors in time measurements, delays in taking the samples, impure reactants, impurities acting as catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty reactor.
2- 15
2.43 y i = ax i ⇒ φ (a ) =
n
∑
d i2 =
i =1
n
⇒a =
i
i =1
i
2
⇒
dφ = 0= da
∑b n
i =1
g
2 yi − axi xi ⇒
n
∑ i =1
yi xi − a
n
∑x
2 i
i =1
n
∑y x /∑x
2 i
i i
i =1
2.44
∑ b y − ax g n
i =1
DIMENSION X(100), Y(100) READ (5, 1) N C N = NUMBER OF DATA POINTS 1FORMAT (I10) READ (5, 2) (X(J), Y(J), J = 1, N 2FORMAT (8F 10.2) SX = 0.0 SY = 0.0 SXX = 0.0 SXY = 0.0 DO 100J = 1, N SX = SX + X(J) SY = SY + Y(J) SXX = SXX + X(J) ** 2 100SXY = SXY + X(J) * Y(J) AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN CALCULATE SLOPE AND INTERCEPT A = (SXY - SX * SY)/(SXX - SX ** 2) B = SY - A * SX WRITE (6, 3) 3FORMAT (1H1, 20X 'PROBLEM 2-39'/) WRITE (6, 4) A, B 4FORMAT (1H0, 'SLOPEb -- bAb =', F6.3, 3X 'INTERCEPTb -- b8b =', F7.3/) C CALCULATE FITTED VALUES OF Y, AND SUM OF SQUARES OF RESIDUALS SSQ = 0.0 DO 200J = 1, N YC = A * X(J) + B RES = Y(J) - YC WRITE (6, 5) X(J), Y(J), YC, RES 5FORMAT (3X 'Xb =', F5.2, 5X /Yb =', F7.2, 5X 'Y(FITTED)b =', F7.2, 5X * 'RESIDUALb =', F6.3) 200SSQ = SSQ + RES ** 2 WRITE (6, 6) SSQ 6FORMAT (IH0, 'SUM OF SQUARES OF RESIDUALSb =', E10.3) STOP END $DATA 5 1.0 2.35 1.5 5.53 2.0 8.92 2.5 12.15 3.0 15.38 SOLUTION: a = 6.536, b = −4 .206
2- 16
=0
2.45 (a) E(cal/mol), D0 (cm2 /s) (b) ln D vs. 1/T, Slope=-E/R, intercept=ln D0 . (c) Intercept = ln D0 = -3.0151 ⇒ D0 = 0.05 cm 2 / s .
3.0E-03
2.9E-03
2.8E-03
2.7E-03
2.6E-03
2.5E-03
2.4E-03
2.3E-03
2.2E-03
2.1E-03
2.0E-03
Slope = − E / R = -3666 K ⇒ E = (3666 K)(1.987 cal / mol ⋅ K) = 7284 cal / mol
-10.0 ln D
-11.0 -12.0 -13.0 -14.0
ln D = -3666(1/T) - 3.0151
1/T
(d) Spreadsheet T 347 374.2 396.2 420.7 447.7 471.2
D 1.34E-06 2.50E-06 4.55E-06 8.52E-06 1.41E-05 2.00E-05
1/T 2.88E-03 2.67E-03 2.52E-03 2.38E-03 2.23E-03 2.12E-03 Sx Sy Syx Sxx -E/R ln D0
lnD (1/T)*(lnD) -13.5 -0.03897 -12.9 -0.03447 -12.3 -0.03105 -11.7 -0.02775 -11.2 -0.02495 -10.8 -0.02296 2.47E-03 -12.1 -3.00E-02 6.16E-06 -3666 -3.0151
D0
7284
E
0.05
2- 17
(1/T)**2 8.31E-06 7.14E-06 6.37E-06 5.65E-06 4.99E-06 4.50E-06
CHAPTER THREE 3.1
16 × 6 × 2 m 3 1000 kg
(a) m =
m3
(b) m& =
b
gb gb gd i
≈ 2 × 10 5 2 103 ≈ 2 × 105 kg
8 oz 1 qt 10 6 cm3 1g 4 × 106 ≈ ≈ 1 × 10 2 g / s 3 3 2 s 32 oz 1056.68 qt cm 3 × 10 10
b
gd i
(c) Weight of a boxer ≈ 220 lb m 12 × 220 lb m 1 stone Wmax ≥ ≈ 220 stones 14 lb m (d)
dictionary V= ≈
πD L 314 . 4.5 ft = 4 4 2
2
d
2
800 miles 5880 ft 7.4805 gal 1 barrel 1 mile 1 ft 3 42 gal
i d
i
3 × 4 × 5 × 8 × 10 2 × 5 × 10 3 × 7 4 × 4 × 10
(e) (i) V ≈
6 ft × 1 ft × 0.5 ft 28,317 cm 3
(ii) V ≈
1 ft 3
≈ 1 × 10 7 barrels
≈ 3 × 3 × 10 4 ≈ 1 × 105 cm 3
1 ft 3
28,317 cm 3
62.4 lb m
1 ft 3
150 lb m
150 × 3 × 104 ≈ 1 × 105 cm 3 60
≈
(f) SG ≈ 105 . 3.2
995 kg
(a) (i)
m
(ii)
3
0.028317 m 3
1 lb m 0.45359 kg
1 ft
995 kg / m 3 62.43 lb m / ft 3 1000 kg / m 3
3
= 62.12 lb m / ft 3
= 62.12 lb m / ft 3
(b) ρ = ρ H2 O × SG = 62.43 lb m / ft 3 × 5.7 = 360 lb m / ft 3 3.3
(a)
(b)
(c)
50 L
0.70 × 10 3 kg
1 m3
m3 10 3 L
m 3 1000 L 1 min
1150 kg min 10 gal
= 35 kg
0.7 × 1000 kg 1 ft 3
2 min 7.481 gal
1 m3
0.70 × 62.43 lb m 1 ft 3
60 s
= 27 L s
≅ 29 lb m / min
3- 1
3.3 (cont’d) (d) Assuming that 1 cm 3 kerosene was mixed with Vg (cm 3 ) gasoline
d
i d i 1dcm kerosene i ⇒ 0.82 dg kerosene i d0.70V + 0.82idg blendi = 0.78 ⇒ V SG = V + 1dcm blend i Vg cm 3gasoline ⇒ 0.70Vg g gasoline 3
g
3
g
g
Volumetric ratio =
3.4
In France: In U.S.:
3.5
50.0 kg
=
0.82 − 0.78 3 = 0.5 0 cm 0.78 − 0.70
Vgasoline 0.50 cm 3 = = 0.50 cm 3 gasoline / cm3 kerosene Vkerosene 1 cm3 L
5 Fr
$1
= $68.42 1L 5.22 Fr 1 gal $1.20 = $22.64 0.70 × 1.0 kg 3.7854 L 1 gal
0.7 × 1.0 kg 50.0 kg L
V&B ( ft 3 / h ), m& B ( lb m / h )
V& ( ft 3 / h ), SG = 0.850
V& H ( ft 3 / h ), m& H ( lb m / h )
700 lb m / h
700 lb m ft 3 (a) V& = = 13.19 ft 3 / h h 0.850 × 62.43 lb m 3 & V ft 0.879 × 62.43 lb m & B kg / h m& B = B = 54.88 V 3 h ft m& H = V&H 0.659 × 62.43 = 4114 . V&H kg / h
d i bg d hb
b
g
b
g
g
V&B + V&H = 1319 . ft 3 / h m& B + m& H = 54.88V&B + 4114 . V&H = 700 lb m ⇒ V&B = 114 . ft 3 / h ⇒ m& B = 628 lb m / h benzene
& H = 71.6 lb m / h hexane V&H = 1.74 ft 3 / h ⇒ m
(b) – No buildup of mass in unit. – ρ B and ρ H at inlet stream condit ions are equal to their tabulated values (which are o strictly valid at 20 C and 1 atm.) – Volumes of benzene and hexane are additive. – Densitometer gives correct reading.
3- 2
3.6
(a) V =
195.5 kg H 2SO 4
1 kg solution
L
0.35kg H 2SO 4 1.2563 × 1.000 kg
= 445 L
(b) L 18255 . × 1.00 kg 195.5 kg H 2SO 4 0.65 kg H 2O + 0.35 kg H 2SO 4 470 − 445 % error = × 100% = 5.6% 445
V ideal =
3.7
195.5 kg H 2SO 4
b gE
b
L = 470 L 1.000 kg
g
Buoyant force up = Weight of block down
Mass of oil displaced + Mass of water displaced = Mass of block
b
g
ρ oil 0.542 V + ρ H
2O
b1 − 0.542gV = ρ
V
c
From Table B.1: ρ c = 2.26 g / cm3 , ρ w = 100 . g / cm 3 ⇒ ρ oil = 3.325 g / cm3 moil = ρ oil × V = 3.325 g / cm3 × 35.3 cm 3 = 117.4 g moil + flask = 117.4 g + 124.8 g = 242 g
3.8
b g
b
g
Buoyant force up = Weight of block down
⇒ Wdisplaced liquid = Wblock ⇒ ( ρVg ) disp. Liq = ( ρVg ) block
b g
b g
Expt. 1: ρ w 15 . A g = ρB 2 A g ⇒ ρ B = ρ w × ρ w =1.00 g/cm 3
bg
15 . 2
b g
ρ B = 0.75 g / cm 3 ⇒ SG
B
= 0.75
b g
b g
Expt. 2: ρ soln A g = ρ B 2 A g ⇒ ρ soln = 2 ρ B = 15 . g / cm3 ⇒ SG
= 15 .
soln
3.9 Let ρ w = density of water. Note: ρ A > ρ w (object sinks)
d
Volume displaced: Vd 1 = Abhsi = Ab h p1 − hb1
hs 1
WA + WB
Archimedes ⇒
h ρ1
hb 1
h p1 − hb1 =
d
WA + WB p w gAb
i
(2)
b ig
d
Volume of pond water: Vw = Ap h p1 − Vd1 ⇒ Vw = Ap h p1 − Ab h p1 − hb1 subst. ( 2 ) for b p1 − hb 1
W A + WB V W + WB ⇒ hp1 = w + A ρw g Ap ρ w gAp
(3)
Vw (W A + WB ) 1 1 + − Ap ρ wg Ap Ab
(4)
Vw = Ap h p1 −
subst. ( 3for ) hp1 in
(2, )solve for
hb1
hb1 =
(1)
ρ wVd 1g = WA + WB 123 weight of displaced water
Subst. (1) for Vd 1 , solve for h p1 − hb1
Before object is jettisoned
i
3- 3
i
3.9 (cont’d) hs 2
WB WA
Let V A = volume of jettisoned object = hρ 2
h b2
After object is jettisoned
d
Volume displaced by boat: Vd 2 = Ab h p2 − hb 2
E , solve for dh
p2
− hb 2
(7)
Volume of pond water: Vw = Ap h p2 − Vd 2 − VA ⇒ h p2 = h p 21
subst. ( 8 )
⇒
for h p 2 in (7, )solve for hb 2
i
i
WB ρ w gAb
solve for
(5)
Archimedes ⇒ ρ WVd 2 g = WB Subst. for Vd 2
h p 2 − hb2 =
WA ρ Ag
(5), (6 ) & ( 7 )
Vw = Ap h p 2 −
Vw WB WA + + Ap ρw gAp ρ A gAp
hb2 =
WB W − A ρwg ρ A g
(8)
Vw WB WA WB + + − Ap ρ w gAp ρ A gAp ρ w gAb
(9)
(a) Change in pond level ( 8)− (3)
h p2 − hp1 =
WA Ap g
1 1 W A ( ρw − ρ A ) ρ − ρ = ρ ρ gA < 0 (since ρw < ρ A ) W A w p A
⇒ the pond level falls
(b)
Change in boat level >0 >0 } 644744 8 5) ( 9)− (4 ) W 1 ( A 1 1 V ρ p A h p 2 − hp1 = − + − 1 > 0 = A 1 + A Ap g ρ A Ap ρW A p ρW A b Ap ρW Ab ⇒ the boat rises
3.10 (a) ρ bulk =
2.93 kg CaCO 3 0.70 L CaCO3 L CaCO3
(b) Wbag = ρ bulkVg =
L total
= 2.05 kg / L
2.05 kg 50 L 9.807 m / s2
1N
= 1.00 × 10 3 N L 1 kg ⋅ m / s2 Neglected the weight of the bag itself and of the air in the filled bag.
(c) The limestone would fall short of filling three bags, because – the powder would pack tighter than the original particles. – you could never recover 100% of what you fed to the mill.
3- 4
(6)
3.11 (a) Wb = mb g =
122.5 kg 9.807 m / s2
1N 1 kg ⋅ m / s2
= 1202 N
Wb − WI (1202 N - 44.0 N) 1 kg ⋅ m / s2 = = 119 L ρwg 0.996 kg / L × 9.807 m / s2 1N m 122 .5 kg ρb = b = = 1.03 kg / L Vb 119 L
Vb =
m f + mnf = mb
(b)
xf =
mf mb
(1)
⇒ m f = mb x f
(2)
d
(1),( 2) ⇒ mnf = mb 1 − x f mf
V f + Vnf = Vb ⇒
b 2g ,b 3g
⇒ mb
(c) x f =
Fx GH ρ
+
f
b
ρ nf
1 / ρ b − 1 / ρ nf 1 / ρ f − 1 / ρ nf
ρf
I =m JK ρ
1− xf
f
+
i
mnf ρ nf
(3) =
⇒ xf
b
mb ρb
F1 GH ρ
−
f
1 ρ nf
I= 1 − 1 JK ρ ρ b
⇒ xf =
nf
1 / ρ b − 1 / ρ nf 1 / ρ f − 1 / ρ nf
1 / 1.03 − 1 / 1.1 = 0.31 1 / 0.9 − 1 / 1.1
=
(d) V f + Vnf + Vlungs + Vother = Vb mf ρf
+
mnf ρ nf
m =m x f b f mnf = mb (1 − x f )
⇒ xf
mb ρb
+ Vlungs + Vother = mb
Fx GH ρ
f f
−
1− x f ρ nf
lungs
+ Vother ) = mb
F1− 1I GH ρ ρ JK b
nf
F 1 − 1 I = 1 − 1 − V +V GH ρ ρ JK ρ ρ m F 1 − 1 I − F V + V I F 1 1 I F 12. + 0.1I GH ρ ρ JK GH m JK GH 1.03 − 11. JK − GH 122.5 JK = = = 0.25 F1− 1I FG 1 − 1 IJ GH ρ ρ JK H 0.9 11. K lungs
f
nf
b
nf
b
nf
other
b
lungs
⇒ xf
I + (V JK
other
b
f
nf
3- 5
Conc. (g Ile/100 g H2O)
3.12 (a) 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0.987
y = 545.5x - 539.03 R2 = 0.9992
0.989
0.991
0.993
0.995
0.997
Density (g/cm3)
From the plot above, r = 5455 . ρ − 539.03 (b) For ρ = 0.9940 g / cm 3 , m& Ile =
150 L
0.994 g
h
3
cm
r = 3.197 g Ile / 100g H 2O 1000 cm 3
3.197 g Ile
L
1 kg
103.197 g sol 1000 g
= 4.6 kg Ile / h
(c) The density of H2 O increases as T decreases, therefore the density was higher than it should have been to use the calibration formula. The valve of r and hence the Ile mass flow rate calculated in part (b) would be too high.
3.13 (a)
Mass Flow Rate (kg/min)
1.20 1.00
y = 0.0743x + 0.1523 R2 = 0.9989
0.80 0.60 0.40 0.20 0.00 0.0
2.0
4.0
6.0
8.0
10.0
12.0
Rotameter Reading
b g
& = 0.0743 5.3 + 0.1523 = 0.55 kg / min From the plot, R = 5.3 ⇒ m
3- 6
3.13 (cont’d) (b) Rotameter Collection Collected Reading Time Volume (min) (cm3) 2 1 297 2 1 301 4 1 454 4 1 448 6 0.5 300 6 0.5 298 8 0.5 371 8 0.5 377 10 0.5 440 10 0.5 453
Mass Flow Rate (kg/min) 0.297 0.301 0.454 0.448 0.600 0.596 0.742 0.754 0.880 0.906
b
Difference Duplicate (Di)
Mean Di
0.004 0.006 0.004
0.0104
0.012 0.026
g
1 0.004 + 0.006 + 0.004 + 0.012 + 0.026 = 0.0104 kg / min 5 95% confidence limits: ( 0.610 ± 174 . Di ) kg / min = 0.610 ± 0.018 kg / min Di =
There is roughly a 95% probability that the true flow rate is between 0.592 kg / min and 0.628 kg / min . 3.14 (a) (b)
(c) (d) (e) (f) (g) (h)
15.0 kmol C6 H 6 15.0 kmol C6 H 6
15,000 mol C 6H 6
78.114 kg C 6H 6
= 117 . × 10 3 kg C6 H 6 kmol C6 H 6 1000 mol = 15 . × 10 4 mol C6 H 6 kmol lb - mole 453.6 mol
15,000 mol C6 H 6
= 33.07 lb - mole C 6H 6
6 mol C 1 mol C 6H 6
15,000 mol C 6H 6
6 mol H 1 mol C6 H6
90,000 mol C 12.011 g C mol C 90,000 mol H 1.008 g H mol H 15,000 mol C 6H 6
= 90,000 mol C = 90,000 mol H
= 1.08 × 10 6 g C
= 9.07 × 10 4 g H
6.022 × 10 23 mol
= 9.03 × 1027 molecules of C6 H 6
3- 7
3.15 (a) m& = (b) n& =
175 m3
1000 L
0.866 kg 3
h
m
2526 kg min
L
1000 mol 1 min 92.13 kg
60 s
1h 60 min
= 2526 kg / min
= 457 mol / s
(c) Assumed density (SG) at T, P of stream is the same as the density at 20o C and 1 atm 3.16 (a)
200.0 kg mix 0150 . kg CH 3OH kg mix
(b) m& mix =
3.17
M=
100.0 lb - mole MA h
0.25 mol N 2
m& N 2 =
3000 kg h
kmol CH3OH 1000 mol 32.04 kg CH3OH
74.08 lb m MA
1 kmol
1 lb m mix
1 lb - mole MA 0.850 lb m MA 28.02 g N 2
+
0.75 mol H 2
mol N 2 kmol 0.25 kmol N 2 8.52 kg
kmol feed
= 936 mol CH3OH
= 8715 lb m / h
2.02 g H2
= 8.52 g mol mol H 2 28.02 kg N 2 = 2470 kg N 2 h kmol N 2
3.18 M suspension = 565 g − 65 g = 500 g , M CaCO3 = 215 g − 65 g = 150 g (a) V& = 455 mL min , m& = 500 g min (b) ρ = m& / V& = 500 g / 455 mL = 110 . g mL (c) 150 g CaCO3 / 500 g suspension = 0.300 g CaCO 3 g suspension 3.19
Assume 100 mol mix. mC 2 H5 OH =
10.0 mol C2 H5OH
46.07 g C 2 H5OH
= 461 g C2 H5OH mol C 2H 5OH 75.0 mol C4 H8 O2 88.1 g C 4 H8O2 mC 4 H8 O2 = = 6608 g C 4 H8O2 mol C 4 H8O2 15.0 mol CH 3COOH 60.05 g CH 3COOH mCH 3COOH = = 901 g CH 3COOH mol CH3COOH 461 g x C 2 H5OH = = 0.0578 g C 2 H 5OH / g mix 461 g + 6608 g + 901 g 6608 g x C4 H8 O2 = = 0.8291 g C4 H 8O 2 / g mix 461 g + 6608 g + 901 g 901 g x CH 3COOH = = 0.113 g CH3COOH / g mix 461 g + 6608 g + 901 g 461 g + 6608 g + 901 g MW = = 79.7 g / mol 100 mol 25 kmol EA 100 kmol mix 79.7 kg mix m= = 2660 kg mix 75 kmol EA 1 kmol mix
3- 8
3.20 (a) Unit Crystallizer Filter Dryer
Function Form solid gypsum particles from a solution Separate particles from solution Remove water from filter cake 0.35 kg CaSO 4 ⋅ 2 H 2 O = 0. 35 kg CaSO 4 ⋅ 2 H 2 O L slurry
(b) m gypsum = 1 L slurry
L CaSO4 ⋅ 2H2 O = 0151 . L CaSO 4 ⋅2 H2O 2.32 kg CaSO4 ⋅ 2 H2O 0.35 kg gypsum 136.15 kg CaSO 4 CaSO 4 in gypsum: m = = 0.277 kg CaSO 4 172.18 kg gypsum 1 − 0151 . L sol 1.05 kg 0.209 kg CaSO 4 CaSO 4 in soln.: m = = 0.00186 kg CaSO4 L 100.209 kg sol
Vgypsum =
0.35 kg CaSO4 ⋅ 2H2 O
b
(c) m =
0.35 kg gypsum
g
0.05 kg sol 0.209 g CaSO 4 = 384 . × 10 -5 kg CaSO4 0.95 kg gypsum 100.209 g sol
0.277 g + 3.84 × 10 -5 g % recovery = × 100% = 99.3% 0.277 g + 0.00186 g
3.21 CSA: FB:
45.8 L
0.90 kg
min L 55.2 L 0.75 kg min L
kmol
U| |V || W
kmol 0.5496 mol CSA 75 kg min ⇒ = 12 . kmol kmol 0.4600 mol FB = 0.4600 90 kg min = 0.5496
She was wrong. The mixer would come to a grinding halt and the motor would overheat. 3.22 (a)
150 mol EtOH
46.07 g EtOH
= 6910 g EtOH mol EtOH 6910 g EtO H 0.600 g H 2 O = 10365 g H 2 O 0.400 g Et OH 6910 g EtOH L 10365 g H 2 O L V= + = 19.123 L ⇒ 19.1 L 789 g EtOH 1000 g H 2 O (6910 +10365) g L SG = = 0903 . 191 . L 1000 g
(b) V ′ =
(6910 + 10365) g mix
% error =
L = 18.472 L ⇒ 18.5 L 935.18 g
(19.123 − 18.472 ) L × 100% = 3.5% 18.472 L
3-9
3.23
0.09 mol CH 4
16.04 g 0.91 mol Air 29.0 g Air + = 27 .83 g mol mol mol 700 kg kmol 0.090 kmol CH 4 = 2.264 kmol CH 4 h h 27.83 kg 1.00 kmol mix 2.264 kmol CH 4 0.91 kmol air = 22.89 kmol air h h 0.09 kmol CH 4 M =
5% CH 4 ⇒
2.264 kmol CH 4
0.95 kmol air
h
0.05 kmol CH 4
b
= 43.01 kmol air h
g
Dilution air required: 43.01 - 22.89 kmol air 1000 mol = 20200 mol air h h
1 kmol
Product gas: 700 kg + 20.20 kmol Air 29 kg Air = 1286 kg h h
h
kmol Air
43.01 kmol Air 0.21 kmol O2 32.00 kg O2 h kg O2 = 0.225 h 1.00 kmol Air 1 kmol O2 1286 kg total kg
3.24
xi =
mi m M , ρi = i , ρ = M Vi V
m m 1 A: ∑ x i ρ i = ∑ i i = M Vi M B:
xi
∑
Not helpful.
mi Vi 1 V 1 = Vi = = Correct. ∑ mi M M ρ i xi 0.60 0.25 0.15 = + + = 1.091 ⇒ ρ = 0.917 g / cm 3 ρ i 0.791 1.049 1.595
∑ρ
1 = ρ
mi2 ∑ Vi ≠ ρ
=
∑M
3.25 (a) Basis: 100 mol N 2 ⇒ 20 mol CH 4
R|20 × 80 = 64 mol CO 25 ⇒S |T 20 × 40 = 32 mol CO 25
2
N total = 100 + 20 + 64 + 32 = 216 mol
32 64 = 0.15 mol CO / mol , x CO 2 = = 0.30 mol CO 2 / mol 216 216 20 100 = = 0 .09 mol CH 4 / mol , x N 2 = = 0.46 mol N 2 / mol 216 216
xC O =
xC H 4
(b) M = ∑ yi M i = 015 . × 28 + 0.30 × 44 + 0.09 × 16 + 0.46 × 28 = 32 g / mol
3-10
3.26 (a) Samples Species
MW
k
1
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
Peak Area 3.6 2.8 2.4 1.7
Mole Fraction 0.156 0.233 0.324 0.287
Mass Fraction 0.062 0.173 0.353 0.412
moles
mass
0.540 0.804 1.121 0.991
8.662 24.164 49.416 57.603
2
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
7.8 2.4 5.6 0.4
0.249 0.146 0.556 0.050
0.111 0.123 0.685 0.081
1.170 0.689 2.615 0.233
18.767 20.712 115.304 13.554
3
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
3.4 4.5 2.6 0.8
0.146 0.371 0.349 0.134
0.064 0.304 0.419 0.212
0.510 1.292 1.214 0.466
8.180 38.835 53.534 27.107
4
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
4.8 2.5 1.3 0.2
0.333 0.332 0.281 0.054
0.173 0.324 0.401 0.102
0.720 0.718 0.607 0.117
11.549 21.575 26.767 6.777
5
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
6.4 7.9 4.8 2.3
0.141 0.333 0.329 0.197
0.059 0.262 0.380 0.299
0.960 2.267 2.242 1.341
15.398 68.178 98.832 77.933
(b) REAL A(10), MW(10), K(10), MOL(10), MASS(10), MOLT, MASST INTEGER N, ND, ID, J READ (5, *) N CN-NUMBER OF SPECIES READ (5, *) (MW(J), K(J), J = 1, N) READ (5, *) ND DO 20 ID = 1, ND READ (5, *)(A(J), J = 1, N) MOLT = 0. 0 MASST = 0. 0 DO 10 J = 1, N MOL(J) = MASS(J) = MOL(J) * MW(J) MOLT = MOLT + MOL(J) MASST = MASST + MASS(J) 10 CONTINUE DO 15 J = 1, N MOL(J) = MOL(J)/MOLT MASS(J) = MASS(J)/MASST 15 CONTINUE WRITE (6, 1) ID, (J, MOL(J), MASS (J), J = 1, N) 20 CONTINUE 1 FORMAT (' SAMPLE: `, I3, /,
3-11
3.26 (cont’d)
∗ ' SPECIES MOLE FR. MASS FR.', /, ∗ 10(3X, I3, 2(5X, F5.3), /), /) END $DATA ∗ 4 16. 04 0. 150
30 . 07
0. 287
44 . 09
0. 467
58 . 12
0. 583
5 3. 6 2. 8 2. 4 1. 7 7. 8 2. 4
5. 6 0. 4
3. 4
4. 5
2. 6
0. 8
4 .8
2. 5
1. 3
0. 2
6. 4 7. 9 4. 8 2. 3 [OUTPUT] SAMPLE : 1 SPECIES MOLE FR
1
0.156
0.062
2
0.233
0.173
3
0. 324
0. 353
0. 287
0.412
4 SAMPLE: 2 (ETC.)
3.27 (a)
MASS FR
(8.7 × 10 6 × 0.40) kg C
44 kg CO 2 = 128 . × 10 7 kg CO 2 ⇒ 2 .9 × 10 5 kmol CO 2 12 kg C
(11 . × 10 6 × 0.26) kg C 28 kg CO = 6.67 × 10 5 kg CO ⇒ 2.38 × 10 4 kmol CO 12 kg C ( 3.8 × 10 5 × 0.10) kg C
16 kg CH 4 12 kg C
= 5.07 × 10 4 kg CH 4 ⇒ 3.17 × 10 3 kmol CH 4
(1.28 × 10 + 6 .67 × 10 + 5.07 × 10 4 ) kg 1 metric ton metric tons m= = 13,500 yr 1000 kg 7
M =
∑y M i
i
5
= 0.915 × 44 + 0.075 × 28 + 0.01 × 16 = 42.5 g / mol
3.28 (a) Basis: 1 liter of solution 1000 mL
1.03 g 5 g H 2 SO 4 mL 100 g
mol H 2 SO 4 = 0.525 mol / L ⇒ 0.525 molar solution 98.08 g H 2 SO 4
3-12
3.28 (cont’d) (b) t = V = V& 55 gal
&
55 gal
3.7854 L
min
60 s
gal
87 L
min
3.7854 L gal
(c) u = V =
87 L
10 3 mL 1.03 g 1L mL
m3
A
= 144 s
0.0500 g H 2 SO 4 g
1 min
min 1000 L 60 s L 45 m t= = = 88 s u 0.513 m / s
(π × 0.06 2 / 4) m 2
1 lbm = 23.6 lb m H 2 SO 4 453.59 g
= 0.513 m / s
3.29 (a) n&1 (mol/min)
n& 2 (mol/min)
0.180 mol C6H14/mol 0.820 mol N2/mol
0.050 mol C6H14/mol 0.950 mol N2/mol
1.50 L C6H14(l)/min n& 3 (mol C6H14(l)/min)
n& 3 =
150 . L 0.659 kg 1000 mol min
L
86.17 kg
= 1147 . mol / min
UV W
R| S|T
Hexane balance: 0.180n&1 = 0050 . n&2 + 1147 . (mol C6 H14 / min) solve n&1 = 838 . mol / min ⇒ & n2 = 72.3 mol / min Nitrogen balance: 0.820n&1 = 0950 . n&2 (mol N2 / min) (b) Hexane recovery =
30 mL 3.30
n&3 1147 . × 100% = × 100% = 76% n&1 0180 . 838 .
b g
1 L 0.030 mol 172 g = 0155 . g Nauseum 103 mL lL 1 mol
3-13
ln(CA)
3.31 (a) kt is dimensionless ⇒ k (min-1 ) (b) A semilog plot of CA vs. t is a straight line ⇒ ln CA = ln CAO − kt 1 0 -1 -2 -3 -4 -5
y = -0.4137x + 0.2512 R2 = 0.9996
0.0
5.0 t (min)
10.0
k = 0.414 min −1
ln CAO = 02512 . ⇒ CAO = 1286 . lb - moles ft 3
FG 1b - molesIJ = C ′ mol 28.317 liter H ft K liter 1 ft t ′b sg 1 min t b ming = = t ′ 60 60 s
(c) C A
A
3
2.26462 lb- moles = 0.06243C′A 1000 mol
3
CA = C A 0 exp( − kt )
b
g
0.06243C ′A = 1334 . exp −0.419 t ′ 60
drop primes
⇒
b
g
b
C A mol / L = 214 . exp −0.00693t
g
t = 200 s ⇒ C A = 5.30 mol / L
3.32 (a)
(b)
2600 mm Hg
14.696 psi = 50.3 psi 760 mm Hg
275 ft H 2O 101.325 kPa = 822.0 kPa 33.9 ft H 2 O
3.00 atm 101325 . × 105 N m 2 12 m2 (c) = 30.4 N cm 2 2 2 1 atm 100 cm
(d)
280 cm Hg 10 mm 101325 . × 10 6 dynes cm 2 1002 cm 2
(e) 1 atm −
1 cm
760 mm Hg
2
1 m
20 cm Hg 10 mm 1 atm = 0.737 atm 1 cm 760 mm Hg
3-14
2
= 3.733 × 1010
dynes m2
3.32 (cont’d) (f)
(g)
b
g
b
25.0 psig 760 mm Hg gauge = 1293 mm Hg gauge 14.696 psig
b25.0 + 14.696gpsi
g
b g
760 mm Hg = 2053 mm Hg abs 14.696 psi
b
g
(h) 325 mm Hg − 760 mm Hg = −435 mm Hg gauge (i) Eq. (3.4-2) ⇒ h = =
P ρg 35.0 lbf in
2
144 in 2 1 ft
s2 32.174 lbm ⋅ ft
ft3 1.595 × 62.43 lbm
2
32.174 ft
lbf ⋅ s
2
100 cm 3.2808 ft
= 1540 cm CCl4
0.92 × 1000 kg 9.81 m / s2 m3 ⇒ h (m) = 0.111Pg (kPa)
3.33 (a) Pg = ρgh =
h (m)
1N 1 kg ⋅ m / s2
1 kPa 103 N / m 2
h
Pg Pg = 68 kPa ⇒ h = 0.111 × 68 = 7.55 m
FG H
moil = ρV = 0.92 × 1000
IJ FG K H
IJ K
kg 16 2 3 × 7 . 55 × π × m = 14 . × 106 kg m3 4
(b) Pg + Patm = Ptop + ρgh
b
g b g
68 + 101 = 115 + 0.92 × 1000 × 9.81 / 103 h ⇒ h = 5.98 m
3.34 (a) Weight of block = Sum of weights of displaced liquids ρ h + ρ 2 h2 ( h1 + h2 ) Aρb g = h1 Aρ1 g + h2 Aρ 2 g ⇒ ρb = 1 1 h1 + h2
3-15
3.34 (cont’d) (b)
Ptop = Patm + ρ1gh0 , Pbottom = Patm + ρ1g(h0 + h1) + ρ2gh2 , Wb = ρb (h1 + h2 ) A ⇒ Fdown = ( Patm + ρ1gh0 ) A + ρb (h1 + h2 ) A , Fup = [Patm + ρ1g(h0 + h1 ) + ρ2gh2 ]A Fdown = Fup ⇒ ρb (h1 + h2 ) A = ρ1gh1 A + ρ2 gh2 A ⇒ Wblock =Wliquid displaced 3.35
b
g
∆ P = Patm + ρgh − Pinside
. g1000 kg b105 = 1 atm − 1 atm + m3
F=
3.36
154 N 65 cm2 2
cm
= 100 . × 104 N ×
9.8066 m 150 m 12 m 2 1N 2 2 2 s 100 cm 1 kg ⋅ m / s2
FG 0.22481 lb IJ = 2250 lb H 1N K f
f
1.4 × 62.43 lb m 1 ft 3 2.3 × 106 gal = 2.69 × 107 lb m 3 ft 7.481 gal P = P0 + ρgh . × 62.43 lb m 32.174 ft 30 ft 1 lb f 12 ft 2 lb 14 = 14.7 f2 + in ft 3 s2 32.174 lb m ⋅ ft / s2 12 2 in2 = 32.9 psi
m = ρV =
— Structural flaw in the tank. — Tank strength inadequate for that much force. — Molasses corroded tank wall π × 24 2 × 3 in 3 1 ft 3 8.0 × 62.43 lb m = 392 lb m 3 3 4 12 in ft 3 392 lb m 32.174 ft / s 2 1 lb f W = mhead g = = 392 lb f 32.174 lb m ⋅ ft / s2
3.37 (a) mhead =
( 30 + 14.7 ) lb f π × 202 in 2 Fnet = Fgas − Fatm − W = 2 in 4 −
14.7 lbf in
2
π × 24 2 in 2 4
− 392 lb f = 7.00 ×10 3 lb f
The head would blow off.
3-16
3.37 (cont’d) 7.000 × 10 lbf Fnet = 392 lb m mhead 3
Initial acceleration: a =
32.174 lbm ⋅ ft/s 2 1 lbf
= 576 ft/s
2
(b) Vent the reactor through a valve to the outside or a hood before removing the head. 3.38 (a)
a
b
2m 1m
Pa = ρgh + Patm , Pb = Patm If the inside pressure on the door equaled Pa , the force on the door would be F = Adoor ( Pa − Pb ) = ρghAdoor Since the pressure at every point on the door is greater than Pa , Since the pressure at every point on the door is greater than Pa , F >ρghAdoor
(b) Assume an average bathtub 5 ft long, 2.5 ft wide, and 2 ft high takes about 10 min to fill. V 5 × 25 . × 2 ft 3 & Vtub = ≈ = 2.5 ft 3 / min ⇒ V& = 5 × 2.5 = 125 . ft 3 / min t 10 min For a full room, h = 10 m
(i)
1000 kg 9.81 m 1N 10 m 2 m 2 ⇒ F > 2.0 × 105 N 3 2 2 m s 1 kg ⋅ m / s The door will break before the room fills ⇒F>
(ii)
If the door holds, it will take 5 × 15 × 10 m 3 35.3145 ft 3 1h V t fill = room = = 31 h V& 12.5 ft 3 / min 1 m3 60 min He will not have enough time.
b
d i dP i
3.39 (a) Pg g
tap
=
g
25 m H2 O
101.3 kPa = 245 kPa 10.33 m H 2O 25 + 5 m H 2O 101.3 kPa = = 294 kPa 10.33 m H 2O
b
junction
g
(b) Air in the line. (lowers average density of the water.) (c) The line could be clogged, or there could be a leak between the junction and the tap.
3-17
3.40
Pabs = 800 mm Hg Pgauge = 25 mm Hg Patm = 800 − 25 = 775 mm Hg
b
g
3.41 (a) P1 + ρ A g h1 + h2 = P2 + ρ B gh1 + ρ C gh2
b
g
b
g
⇒ P1 − P2 = ρ B − ρ A gh1 + ρ C − ρ A gh2
LMb1.0 − 0.792 g g 981 cm 30.0 cm + b137 . − 0.792 g g cm s cm N F 1 dyne I F I = 123.0 kPa 101.325 kPa × G J G H 1 g ⋅ cm / s K H 1.01325 × 10 dynes / cm JK
(b) P1 = 121 kPa +
3
2
2
3.42
3
6
OP Q
981 cm 24.0 cm s2
2
(a) Say ρt (g/cm3) = density of toluene, ρm (g/cm3) = density of manometer fluid ρ t g (500 − h + R ) = ρ m gR ⇒ R =
500 − h ρm −1 ρt
(i) Hg: ρ t = 0.866, ρ m = 13.6, h = 150 cm ⇒ R = 238 . cm (ii) H 2 O: ρ t = 0.866, ρ m = 1.00, h = 150 cm ⇒ R = 2260 cm
Use mercury, because the water manometer would have to be too tall. (b) If the manometer were simply filled with toluene, the level in the glass tube would be at the level in the tank. Advantages of using mercury: smaller manometer; less evaporation. (c) The nitrogen blanket is used to avoid contact between toluene and atmospheric oxygen, minimizing the risk of combustion. P 3.43 Patm = ρ f g 7.23 m ⇒ ρ f = atm 7.23 g Patm Pa − Pb = ρ f − ρ w g 26 cm = − ρ w g 26 cm 7.23 m
b
d
g ib
g FGH
IJ b K
F756mmHg 1 m −1000 kg 9.81 m/s =G H 7.23 m 100 cm m
2
3
⇒ Pa − Pb = 81 . mm Hg
3-18
g
Ib g JK
1N 760mmHg 1m 26 cm 2 5 2 1 kg⋅ m/s 1.01325×10 N m 100 cm
3.44 (a) ∆h = 900 − hl =
75 . psi 760 mm Hg = 388 mm Hg ⇒ hl =900 − 388 =512 mm 14.696 psi
(b) ∆h = 388 − 25 × 2 = 338 mm ⇒ Pg =
338 mm Hg
14.696 psi = 6.54 psig 760 mm Hg
3.45 (a) h = L sin θ
b
g b g
(b) h = 8.7 cm sin 15° = 2.3 cm H 2 O = 23 mm H 2 O 3.46 (a) P = Patm − Poil − PHg = 765 − 365 −
9.81 m / s2 0.10 m
920 kg m3
1N 760 mm Hg 2 1 kg ⋅ m/ s 1.01325 × 105 N / m2
= 393 mm Hg (b) — Nonreactive with the vapor in the apparatus. — Lighter than and immiscible with mercury. — Low rate of evaporation (low volatility).
c
h
c
3.47 (a) Let ρ f = manometer fluid density 110 . g cm 3 , ρ ac = acetone density 0.791 g cm 3
d
i
Differential manometer formula: ∆P = ρ f − ρ ac gh ∆P ( mm Hg ) =
(1.10 − 0.791) g
981 cm h (mm)
3
cm
s
1 cm
1 dyne
10 mm 1 g ⋅ cm/s
2
= 0.02274 h ( mm )
V& ( mL s )
62
87
107
123
138
151
5
10
15
20
25
30
h ( mm )
∆P ( mm Hg )
0.114 0.227 0.341 0.455 0.568 0.682
b g
(b) lnV& = n ln ∆P + ln K 6
ln(V)
5.5
y = 0.4979x + 5.2068
5 4.5 4 -2.5
-2
-1.5
-1
-0.5
0
ln( P)
3-19
760 mm Hg 2
1.01325×106 dyne/cm2
h
3.47 (cont’d)
b g
. ln ∆P + 52068 . From the plot above, ln V& = 04979
⇒n = 04979 . ≈ 05 . , ln K = 5.2068 ⇒ K = 183
b
ml s
bmm Hgg
gb g
0 .5
b
g
(c) h = 23 ⇒ ∆P = 0.02274 23 = 0.523 mm Hg ⇒ V& = 183 0.523
0. 5
= 132 mL s
132 mL 0.791 g 104 g 1 mol = 104 g s = 180 . mol s s mL s 58.08 g
. = 544° R / 18 . = 303 K − 273 = 30°C 3.48 (a) T = 85° F + 4597 . = 474° R − 460 = 14° F (b) T = −10°C + 273 = 263 K × 18 (c) ∆T =
(d)
85° C 10 . °K 85° C 18 . °F 85° C 1.8° R = 85° K; = 153° F; = 153° R 10 . °C 1° C 1.0° C
150° R 1° F 1° R
= 150° F;
150° R 1.0o K 1.8° R
= 83.3° K;
150° R 1.0° C 1.8° R
= 83.3° C
3.49 (a) T = 0.0940 × 1000o FB + 4.00 = 98.0o C ⇒ T = 98.0 × 1.8 + 32 = 208o F (b) ∆T (o C) = 0.0940 ∆T (o FB) = 0.94o C ⇒ ∆T (K) = 0.94 K 0.94o C 1.8o F o ∆T ( F) = = 1.69o F ⇒ ∆T (o R) = 1.69o R o 1.0 C
o o (c) T1 = 15o C ⇒ 100o L ; T2 = 43 C⇒1000 L T ( o C) = aT (o L) + b o F oCI ; 43 − 15g C b a= = 0.0311G H o L JK b1000 - 100go L
b = 15 − 0.0311 × 100 = 119 . oC
⇒ T ( o C) = 0.0311T ( o L) + 11.9 and T ( o L) = 32.15T ( o C) − 382.6 (d) Tbp = −88.6o C ⇒ 184.6 K ⇒ 332.3o R ⇒ -127.4o F ⇒ −9851 . o FB ⇒ −3232o L (e) ∆T = 50.0o L ⇒ 1.56o C ⇒ 16.6o FB ⇒ 156 . K ⇒ 2.8o F ⇒ 2.8o R
3-20
3.50
bT g = 100° C bT g (a) V b mV g = aT b° Cg + b b H O 2
m AgCl
= 455° C
5.27 = 100a + b a = 0.05524 mV ° C ⇒ 24.88 = 455a + b b = −0.2539 mV V mV = 0.05524T ° C − 0.2539
b g
b g
⇓ T ° C = 1810 . V mV + 4.596
b g
b g
. mV→136 . mV ⇒1856 . ° C→2508 . °C ⇒ (b) 100 3.51 (a) ln T = ln K + n ln R
n=
b
g
b
g
. − 1856 . °C dT 2508 = = 326 . °C / s dt 20 s
T = KR n
ln 250.0 110.0 = 1184 . ln 40.0 20.0
b
g
. ln K = ln 1100 . −1184 . (ln 200 . ) = 1154 . ⇒ K = 3169 . ⇒T = 3169 . R1184
320 I (b) R = FG JK H 3169 .
1/ 1.184
= 49.3
(c) Extrapolation error, thermocouple reading wrong.
3.52 (a) PV = 0.08206nT
b g Pbatmg = 14696 .
P′ psig + 14696 .
b g b
g
n mol = n′ lb - moles ×
bg d i
28317 . ft 3 , V L = V ′ ft × L 3
453.59 mol T ′( o F) − 32 , T(o K) = + 27315 . lb − moles 1.8
b P′ + 14.696g × V ′ × 28.317 = 008206 453.59 L (T ′ − 32) O . × n′ × ×M + 27315 . P 14.696 1 N 1.8 Q
⇒
b
g
⇒ P ′ + 14.696 × V ′ =
b
g
0.08206 × 14.696 × 453.59 × n′ × T ′ + 459.7 28.317 × 18 .
b
b
g
⇒ P′ + 14.696 V ′ = 1073 . n′ T ′ + 459.7
3-21
g
3.52 (cont’d)
b500 + 14.696g × 3.5 = 0.308 lb - mole 10.73 × b85 + 459.7g
(b) ntot ′ =
mCO =
(c) T ′ =
0308 . lb- mole 0.30 lb- mole CO 28 lbm CO = 2.6 lb m CO lb- mole lb - mole CO
b3000 + 14.696g × 35. − 459.7 = 2733o F 10.73 × 0.308
b g
b
g
3.53 (a) T ° C = a × r ohms + b
UV ⇒ W
0 = 23624 . a +b 100 = 33028 . a +b
a = 10634 . ⇒ T °C = 10634 . r ohms − 25122 . b = −25122 .
b g
FG kmol IJ = n&′ (kmol) 1 min = n& ′ H s K min 60 s 60 P ′bmm Hgg 1 atm P′ Pbatmg = = 760 mm Hg 760 F m I m 1 min = V& ′ V&G J = V& ′ H s K min 60 s 60
b g
(b) n&
3
,
bg b g
T K = T ′ °C + 27316 .
3
b g d b g
i
0.016034P′ mm Hg V& ′ m3 min V& ′ n&′ 12.186 P ′ = ⇒ n& ′ = 60 T ′ ° C + 27316 . 760 T ′ + 27316 . 60 (c) T = 10.634 r − 251.22 r1 = 26.159 ⇒ T1 = 26.95° C ⇒ r2 = 26157 . ⇒ T2 = 26.93° C r3 = 44.789 ⇒ T3 = 2251 . °C P (mm Hg) = h + Patm = h + ( 29.76 in Hg)
FG 760 mm Hg IJ = h + 755.9 H 29.92 in Hg K
h1 = 232 mm ⇒ P1 = 987.9 mm Hg . mm Hg ⇒ h2 = 156 mm ⇒ P2 = 9119 h3 = 74 mm ⇒ P3 = 829.9 mm Hg
3-22
3.53 (cont’d)
b0.016034 gb987.9gb947 60g = 0.8331 kmol CH 26.95 + 27316 . b0.016034gb9119. gb195g = 9.501 kmol air min =
(d) n&1 = n& 2
4
min
26.93 + 27316 . n& 3 = n&1 + n& 2 = 10.33 kmol min (e) V3 =
b
. g g = b10.33gb2251. + 27316 0.016034 P b0.016034gb829.9g = 387 m
n&3 T2 + 27316 .
3
min
3
(f)
0.8331 kmol CH 4 16.04 kg CH 4 kg CH 4 = 13.36 min kmol min 0.21× 9.501 kmol O2 32.0 kg O2 0.79 × 9.501 kmol N2 + min kmol O2 min xC H4 =
3.54
13.36 kg CH 4 min = 0.0465 kg CH4 kg (13.36 + 274) kg / min
REAL, MW, T, SLOPE, INTCPT, KO, E REAL TIME (100), CA (100), TK (100), X (100), Y(100) INTEGER IT, N, NT, J READ 5, ∗ MW, NT DO 10 IT=1, NT READ 5, ∗ TC, N TK(IT) = TC + 273.15 READ 5, ∗ (TIME (J), CA (J), J = 1 , N) DO 1 J=1, N CA J = CA J / MW
b g b g b g
bg bg XbJ g = TIMEbJ g YbJg = 1./CAbJg
1
CONTINUE CALL LS (X, Y, N, SLOPE, INTCPT)
b g
K IT = SLOPE
WRITE (E, 2) TK (IT), (TIME (J), CA (J), J = 1 , N) WRITE (6, 3) K (IT) 10 CONTINUE DO 4 J=1, NT X J = 1./TK J
bg bg YbJg = LOGcKbJgh
3-23
28.0 kg N 2 kg air = 274 kmol N2 min
3.54 (cont’d) 4
CONTINUE CALL LS (X, Y, NT, SLOPE, INTCPT) KO = EXP INTCPT
b
2
3 5
10
g
E = −8.314 = SLOPE WRITE (6, 5) KO, E FORMAT (' TEMPERATURE (K): ', F6.2, / * ' TIME CA', /, * ' (MIN) (MOLES)', / * 100 (IX, F5.2, 3X, F7.4, /)) FORMAT (' K (L/MOL – MIN): ', F5.3, //) FORMAT (/, ' KO (L/MOL – MIN) : ', E 12.4, /, ' E (J/MOL): ', E 12.4) END SUBROUTINE LS (X, Y, N, SLOPE, INTCPT) REAL X(100), Y(100), SLOPE, INTCPT, SX, SY, SXX, SXY, AN INTEGER N, J SX=0 SY=0 SXX=0 SXY=0 DO 10 J=1,N SX = SX + X(J) SY = SY + Y(J) SXX = SXX + X(J)**2 SXY = SXY + X(J)*Y(J) CONTINUE AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN SLOPE = (SXY – SX*SY)/(SXX – SX**2) INTCPT = SY – SLOPE*SX RETURN END $ DATA 65.0 94.0 10.0 20.0
4 6 8.1 4.3
[OUTPUT] TEMPERATURE (K): 367.15 TIME CA (MIN) (MOLS/L) 10.00 0.1246
3-24
3.54 (cont’d) 30.0 40.0 50.0 60.0
3.0 2.2 1.8 1.5
20.00 30.00 40.00 50.00 60.00
b
0.0662 0.0462 0.0338 0.0277 0.0231
g
K L / MOL ⋅ MIN : 0.707 110. 10.0 20.0 30.0 40.0 50.0 60.0
6 3.5 1.8 1.2 0.92 0.73 0.61
127.
6
M M ETC
bat 94° Cg
TEMPERATURE (K): 383.15 M
b
g
K L / MOL ⋅ MIN : 1.758
M
b g E bJ / MOLg: 0.6690E + 05
K0 L / MOL − MIN : 0.2329E + 10
3-25
CHAPTER FOUR 4.1
a.
Continuous, Transient
b.
Input – Output = Accumulation No reactions ⇒ Generation = 0, Consumption = 0
6.00 c.
4.2
t=
kg kg dn dn kg − 3.00 = ⇒ = 3.00 s s dt dt s
1.00 m3 1000 kg 1s = 333 s 1 m3 3.00 kg
a.
Continuous, Steady State
b.
k = 0 ⇒ CA = CA0
c.
Input – Output – Consumption = 0 Steady state ⇒ Accumulation = 0 A is a reactant ⇒ Generation = 0
k = ∞ ⇒ CA = 0
FG IJ FG IJ FG IJ FG IJ H K H K H K H K
FG IJ H K
m3 mol m3 mol mol C A0 & & V CA 0 =V CA + kVC A ⇒ CA = 3 3 kV s m s m s 1+ V& 4.3
b
m& v kg / h
a.
g
100 kg / h
0.850kg B / kg
0.550kg B / kg
0.150kg T / kg
0.450kg T / kg
m& l kg / h
b
g
Input – Output = 0 Steady state ⇒ Accumulation = 0 No reaction ⇒ Generation = 0, Consumption = 0
0.106kg B / kg 0.894kg T / kg
(1) Total Mass Balance: 100.0 kg / h = m& v + m& l
& v + 0106 &l (2) Benzene Balance: 0.550 × 100.0 kg B / h = 0.850 m . m
& v = 59.7 kg h, m& l = 40.3 kg h Solve (1) & (2) simultaneously ⇒ m b.
The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by masses (kg). The balance equations are also identical (initial input = final output).
c.
Possible explanations ⇒ a chemical reaction is taking place, the process is not at steady state, the feed composition is incorrect, the flow rates are not what they are supposed to be, other species are in the feed stream, measurement errors.
4- 1
4.4
b.
c.
n(mol) 0 .500 mol N 2 mol 0 .500 mol CH 4 mol 100.0 g / s
b g bg C H g g bg C H g g
xE g C 2 H6 g xP xB
d.
3
8
4
10
b
g
b
0.500n mol N 2 28 g N 2 1 kg = 0.014 n kg N 2 mol N 2 1000 g n& E =
b
100x E g C2H 6 s
b
g
1 lb m lb - mole C2H 6 3600 s 453.593 g 30 lb m C2H 6 h
= 26.45 x E lb - mole C2H 6 / h
b g R|n& blb - mole DA sg U| . lb - moleO lb - mole DA V S| 021 . lb- moleN lb - mole DA|W T 079 n& 1 lb - mole H2 O s
b
n& O2 = 0.21n& 2 lb - mole O2 / s
g
g
2
2
2
x H2 O =
x O2 =
e.
n ( mol)
FG H
n&1 lb - mole H2O &n1 + n& 2 lb - mole
FG H
0.21n&2 lb - mole O 2 n&1 + n&2 lb - mole
IJ K
IJ K
nN 2O4 = n 0.600 − y NO2 ( mol N 2O 4 )
0.400mol NO mol yNO 2 ( mol NO 2 mol) 0.600 − yNO 2 ( mol N 2 O4 mol )
4- 2
g
4.5
a.
Basis: 1000 lbm C3H8 / h fresh feed (Could also take 1 h operation as basis flow chart would be as below except that all / h would be deleted.)
b
n& 6 lb m / h
b
g
n& 7 lb m / h
1000 lb m C 3H 8 / h
g
002 . lb m C3H 8 / lb m 098 . lb m C3 H 6 / lbm
097 . lb m C3H 8 / lb m
Still
0.03 lb m C3 H 6 / lb m
Compressor
b n& b lb n& b lb n& b lb
g C H / hg CH / h g H / hg
b b
n& 1 lb m C3H 8 / h
n&1 lb m C3 H8 / h
Reactor
2
m
3
m
4
m
3
6
4
2
b b
n& 3 lb m CH 4 / h n& 4 lb m H 2 / h
Note: the compressor and the off gas from the absorber are not mentioned explicitly in the process description, but their presence should be inferred.
g
g
b
n& 5 lbm / h
g
Stripper
Absorber
b b n& b lb
n&1 lb m C3 H8 / h
g g
n& 2 lb m C3 H 6 / h 5
4.6
g g
n& 2 lb m C3H 6 / h
m
oil / h
g
b.
Overall objective : To produce C3 H6 from C3 H8 . Preheater function: Raise temperature of the reactants to raise the reaction rate. Reactor function: Convert C3 H8 to C3H6 . Absorption tower function: Separate the C3H8 and C3 H6 in the reactor effluent from the other components. Stripping tower function: Recover the C3H8 and C3H6 from the solvent. Distillation column function: Separate the C3 H5 from the C3 H8.
a.
3 independent balances (one for each species)
b.
7 unknowns ( m& 1 , m & 3 , m& 5 , x2 , y2 , y 4 , z4 ) – 3 balances – 2 mole fraction summations 2 unknowns must be specified
c.
y2 = 1 − x2
FG kg A IJ = m& + b1200gb0.70g FG kg A IJ H h K H hK F kg I F kg I Overall Balance: m& + 5300 G J = m& + 1200 + m& G J HhK H hK F kg B IJ = 1200 y + 0.60m& FG kg BIJ B Balance: 0.03m & + 5300 x G H hK H hK A Balance: 5300 x2
3
1
1
3
2
5
4
z4 = 1 − 0.70 − y 4
4- 3
5
4.7
a.
3 independent balances (one for each species)
b. Water Balance:
bg b g
400 g 0.885 g H2O m& R g 0.995 g H 2O & R = 356 g min = ⇒m min g min g
b gb
Acetic Acid Balance: 400 0115 .
g FGH g CHminOOH IJK = 0.005m& 3
R
&E + 0.096m
& E = 461g min ⇒m
3
FG g IJ = m& + m& FG g IJ ⇒ m& = 417 g min H minK H minK F g IJ = b0.096gb461g FG g IJ ⇒ 44 g min = 44 g min . gb 400g − b0.005gb356g G b0115 H min K H min K
Overall Balance: m& C + 400 c.
FG g CH OOH IJ H min K
R
E
C
d. CH3 COOH
H 2O someCH 3COOH CH3COOH H 2O
Extractor
C4 H 9 OH
C 4H 9OH CH 3COOH
Distillation Column
C 4 H9 OH
4.8
a.
X-large: 25 broken eggs/min 35 unbroken eggs/min
120 eggs/min 0.30 broken egg/egg 0.70 unbroken egg/egg
b.
Large: n1 broken eggs/min n2 unbroken eggs/min
b
U| n = 11 V| ⇒ n = 39 W
g
120 = 25 + 35 + n1 + n2 eggs min ⇒ n1 + n2 = 50
b0.30gb120g = 25 + n
2
1
c.
n1 + n2 = 50 large eggs min
1
b
g
n1 large eggs broken/50 large eggs = 11 50 = 0.22 d.
b
g
22% of the large eggs (right hand) and 25 70 ⇒ 36% of the extra-large eggs (left hand) are broken. Since it does not require much strength to break an egg, the left hand is probably poorly controlled (rather than strong) relative to the right. Therefore, Fred is right-handed.
4- 4
4.9
a.
b
m1 lb m strawberries 015 . lb m S / lb m 0.85 lb m W / lb m
c
m2 lb m S sugar
g
h
b
m3 lb m W evaporated
g
1.00 lb m jam 0.667 lb m S / lb m 0.333 lb m W / lb m
b.
3 unknowns ( m1 , m2 , m3 ) – 2 balances – 1 feed ratio 0 DF
c.
Feed ratio: m1 / m2 = 45/55
(1)
S balance: 0.15m1 + m2 = 0.667 (2) Solve simultaneously ⇒ m1 = 0.49 lb m strawberries, m 2 = 0.59 lb m sugar 4.10
a.
300 gal
b g
m1 lb m
b g
0.750 lb m C 2 H 5OH / lb m 0.250 lb m H 2O / lb m
m3 lb m
0.600 lb m C 2 H 5OH / lb m 0.400 lb m H 2O / lb m
b g m b lb g
4 unknowns ( m1, m2 ,V40 , m3 ) – 2 balances – 2 specific gravities 0 DF
V40 gal 2
m
0.400 lb m C 2 H 5 OH / lb m 0.600 lb m H 2O / lb m
b.
m1 =
300gal
1 ft3 0.877 × 62.4 lb m = 2195 lb m 7.4805 gal ft 3
Overall balance: m1 + m2 = m3 C2 H5OH balance: 0.750m1 + 0.400m2 = 0.600m3 Solve (1) & (2) simultaneously ⇒ m2 = 1646 lb m, , m3 = 3841lb m V40 =
1646 lb m
ft 3 7.4805gal = 207 gal 0.952 × 62.4lb m 1ft 3
4- 5
(1) (2)
4.11
a.
b
3 unknowns ( n&1 , n&2 , n&3 ) – 2 balances 1 DF
g
n&1 mol / s
0.0403 mol C3H 8 / mol 0.9597 mol air / mol
b
b
n&3 mol / s
g
0.0205 mol C 3H 8 / mol 0.9795 mol air / mol
g
n&2 mol air / s
0.21 mol O2 / mol 0.79 mol N 2 / mol
b.
b
g
Propane feed rate: 0.0403n&1 = 150 ⇒ n&1 = 3722 mol / s
b
g
Propane balance: 0.0403n&1 = 0.0205n&3 ⇒ n&3 = 7317 mol / s
b
g
Overall balance: 3722 + n&2 = 7317 ⇒ n&2 = 3600 mol / s c.
> . The dilution rate should be greater than the value calculated to ensure that ignition is not possible even if the fuel feed rate increases slightly.
4.12
a.
b
m & kg / h
g
0.960 kg CH3OH / kg 0.040 kg H2O / kg
1000 kg / h 0.500 kg CH3OH / kg 0.500 kg H 2O / kg
2 unknowns ( m & ,x ) – 2 balances 0 DF
673 kg / h
b
g 1 − x bkg H O / kg g x kg CH3OH / kg 2
b.
& + 673 ⇒ m & = 327 kg / h Overall balance: 1000 = m
b g
b g b g
Methanol balance: 0.500 1000 = 0.960 327 + x 673 ⇒ x = 0.276 kg CH3OH / kg Molar flow rates of methanol and water: 673 kg 0.276 kg CH3OH 1000 g molCH3OH = 5.80 × 103 mol CH3OH / h h kg kg 32.0 g CH3OH 673 kg 0.724 kg H 2O 1000 g mol H 2O = 2.71 × 10 4 mol H 2O / h h kg kg 18 g H 2O Mole fraction of Methanol: 5.80 × 10 3 = 0176 . mol CH3OH / mol 5.80 × 103 + 2.71 × 104
c.
Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the system is not at steady state.
4- 6
4.13
a.
Product Feed
Reactor effluent
Reactor
2 2 5 3 kg
Purifier
2253k g R = 388
1239 k g R = 583
Waste
b g
mw k g R = 140
Analyzer Calibration Data 1
xp
x p = 0.000145R
1.364546
0.1
0.01 100
b.
1000
R
b g = 0.494 kg P / kg Product: x = 0.000145b583g = 0.861 kg P / kg Waste: x = 0.000145b140g = 0123 . kg P / kg 0.861b1239g Efficiency = × 100% = 95.8% 0.494b2253g Effluent: x p = 0.000145 388
1.3645
1.3645
p
1.3645
p
c.
Mass balance on purifier: 2253 = 1239 + mw ⇒ mw = 1014 kg P balance on purifier: Input: 0.494 kg P / kg 2253 kg = 1113 kg P
b
gb g Output: b0.861 kg P / kg gb1239 kg g + b 0123 . kg P / kg gb1014 kgg = 1192 kg P
The P balance does not close . Analyzer readings are wrong; impure feed; extrapolation beyond analyzer calibration data is risky -- recalibrate; get data for R > 583; not at steady state; additional reaction occurs in purifier; normal data scatter.
4- 7
4.14
a.
b
g
n1 lb - mole/ h & 00100 . lb -mole H2O/ lb -mole 09900 . lb- mole DA / lb -mole
b d i
b
g
n3 lb- mole/ h & 0100 . lb -mole H2O/ lb- mole 0900 . lb- mole DA/ lb - mole
g
n2 lb- mole HO/ h 2 & v2 ft 3 / h &
4 unknowns ( n&1 , n&2 , n& 3 , v& ) – 2 balances – 1 density – 1 meter reading = 0 DF Assume linear relationship: v& = aR + b v& − v& 96.9 − 40.0 Slope : a = 2 1 = = 1626 . R 2 − R1 50 − 15 Intercept: b = v&a − aR1 = 40.0 − 1.626 15 = 15.61
b g
c
v&2 = 1.626 95 + 15.61 = 170 ft / h n& 2 =
3
h
b g b
3
170 ft 62 .4 lb m lb - mol = 589 lb - moles H 2 O / h h ft 3 18.0 lb m
g
DA balance: 0.9900n&1 = 0.900n& 3 (1) Overall balance: n&1 + n&2 = n& 3 (2) & & Solve (1) & (2) simultaneously ⇒ n1 = 5890 lb - moles / h , n 3 = 6480 lb - moles / h b. 4.15
Bad calibration data, not at steady state, leaks, 7% value is wrong, v& − R relationship is not linear, extrapolation of analyzer correlation leads to error.
a.
b
m& kg / s
100 kg / s
g
0.900 kg E / kg 0100 . kg H 2 O / kg
0.600 kg E / kg 0.050 kg S / kg 0.350 kg H 2 O / kg
b
& kg / s m
g
b g x b kgS / kg g 1 − x − x b kg H O / kgg x E kg E / kg S
E
b.
S
2
b
g = 0100 . b kg S / kg g
Overall balance: 100 = 2m & ⇒ m& = 50.0 kg / s
b g b g
S balance: 0.050 100 = xS 50 ⇒ xS
b g
b g b g 0.300b50 g kg Ein bottom stream kg Ein bottom stream = = 0.25 kg E in feed 0.600b100g kg E in feed
E balance: 0.600 100 = 0.900 50 + x E 50 ⇒ x E = 0.300 kg E / kg
4- 8
3 unknowns ( m & , xE , xS ) – 3 balances 0 DF
4.15 (cont’d) c. x = aR b ⇒ ln x = ln a + b ln R
bg bg b g lnb x / x g lnb0.400 / 0100 . g b= = = 1491 . lnb R / R g lnb38 / 15g lnbag = lnb x g − b ln b R g = lnb 0100 . g − 1491 . lnb15g = −6340 . ⇒ a = 1764 . × 10 2
1
2
1
1
−3
1
x = 1764 . × 10−3 R1.491 . F x I F 0900 IJ R =G J =G H a K H 1764 . × 10 K 1 b
−3
d.
1 1.491
= 655 .
Device not calibrated – recalibrate. Calibration curve deviates from linearity at high mass fractions – measure against known standard. Impurities in the stream – analyze a sample. Mixture is not all liquid – check sample. Calibration data are temperature dependent – check calibration at various temperatures. System is not at steady state – take more measurements. Scatter in data – take more measurements.
4- 9
4.16
a.
b.
b
4.00 mol H 2SO 4 0.098 kg H2SO 4 L of solution = 0.323 kg H2SO 4 / kg solution L of solution molH2SO 4 1.213kg solution
bg
5 unknowns ( v1 , v2 , v3 , m2 , m3 ) – 2 balances – 3 specific gravities 0 DF
v1 L
bg m b kg g
100 kg
v3 L
0.200 kg H 2 SO 4 / kg
g
3
0.800 kg H 2 O / kg
0.323 kg H 2SO 4 / kg 0.677 kg H 2 O / kg SG = 1213 .
SG = 1139 .
bg m b kg g v2 L 2
0.600 kg H 2SO 4 / kg 0.400 kg H 2 O / k g SG = 1.498
UV ⇒ m = 44.4 kg Water balance: 0.800b100 g + 0.400 m = 0.677 m W m = 144 kg Overall mass balance: 100 + m2 = m3
2
2
v1 =
100 kg
v2 =
44.4 kg
3
3
L = 87.80 L20% solution 1139 . kg L = 29.64 L 60%solution 1498 . kg
v1 87.80 L 20% solution = = 2.96 v 2 29.64 L 60% solution
c.
4.17
1250 kg P 44.4 kg 60% solution L = 257 L / h h 144 kg P 1.498 kg solution
b g
m1 kg @$18 / kg 0.25 kg P / kg 0.75 kg H2O / kg
100 . kg 017 . kg P/ kg 0.83 kg H2O / kg
b g
m2 kg @$10 / kg 012 . kg P / kg 0.88 kg H2O / kg
Overall balance: m1 + m2 = 100 .
b g Cost of blend: 0.385b$18.00g + 0.615b$10.00g = $13.08 per kg Selling price: 110 . b$13.08g = $14.39 per kg
(1)
Pigment balance: 0.25m1 + 0.12m2 = 0.17 1.00 (2) Solve (1) and (2) simultaneously ⇒ m1 = 0.385 kg 25% paint , m2 = 0.615kg12% paint
4- 10
4.18
b
a.
gb
m1 kg H 2 O 85%of enteringwater
g
100 kg 0.800 kgS / kg 0.200 kg H 2O / kg
b g m b kg H Og m2 kgS 3
2
b
gb g Sugar balance: m = 0.800b100g = 80.0 kg S
85% drying: m1 = 0.850 0.200 100 = 17.0 kg H2O 2
Overall balance: 100 = 17 + 80 + m3 ⇒ m3 = 3 kg H 2O xw =
3 kg H2O = 0.0361 kg H 2O / kg 3 + 80 kg
b
g
m1 17 kg H2O = = 0.205 kg H2O / kg wet sugar m2 + m3 80 + 3 kg
b
b.
g
1000 tonswet sugar 3 tonsH 2 O = 30 tons H 2 O / day day 100 tonswet sugar
1000 tons WS 0.800 tons DS 2000 lb m $0.15 365days = $8.8 × 107 per year day ton WS ton lb m year
c.
b
g
1 x w1 + x w 2 +...+ x w10 = 0.0504 kg H 2 O / kg 10 1 2 2 SD = x w1 − x w +...+ x w10 − x w = 0.00181 kg H 2 O / kg 9 Endpoints = 0.0504 ± 3 0.00181 xw =
b
g
b
b
g
g
Lower limit = 0.0450, Upper limit = 0.0558
4.19
d.
The evaporator is probably not working according to design specifications since x w = 0.0361 < 0.0450 .
a.
v1 m 3
c h m b kg H O g 1
2
SG = 1.00
d i m b kg suspension g v3 m 3 3
d i
v2 m
3
SG = 1.48
5 unknowns ( v1 , v2 , v3 , m1 , m3 ) – 1 mass balance – 1 volume balance – 3 specific gravities 0 DF
400 kg galena S G = 7 .44
Total mass balance: m1 + 400 = m3
(1)
4- 11
4.19 (cont’d) Assume volume additivity:
b g
m1 kg
b g
m3 400 kg m 3 m kg m3 + = 3 (2) 1000 kg 7440 kg 1480 kg
Solve (1) and (2) simultaneously ⇒ m1 = 668 kg H 2O, m3 = 1068 kg suspension v1 =
4.20
668 kg
m3 = 0.668 m 3 water fed to tank 1000 kg
b.
Specific gravity of coal < 1.48 < Specific gravity of slate
c.
The suspension begins to settle. Stir the suspension. 1.00 < Specific gravity of coal < 1.48
a.
b
n&1 mol / h
g
b
n&2 mol/ h
b
0.040 mol H2O / mol 0.960 mol DA / mol
g
x mol H 2O / mol
b
g
g
1 − x mol DA / mol
b
n&3 mol H 2O adsorbed / h
g
97% of H 2O in feed
Adsorption rate: n& 3 =
b3.54 − 3.40g kg b
5h
g
molH2O = 1556 . molH2O / h 0.0180 kg H2O
97% adsorbed: 156 . = 0.97 0.04n&1 ⇒ n&1 = 401 . mol/ h Total mole balance: n&1 = n& 2 + n& 3 ⇒ n&2 = 401 . − 1556 . = 38.54 mol / h
Water balance: 0.040 ( 40.1) = 1.556 + x ( 38.54 ) ⇒ x = 1.2 × 10−3 ( molH 2O/mol )
4.21
b.
The calcium chloride pellets have reached their saturation limit. Eventually the mole fraction will reach that of the inlet stream, i.e. 4%.
a.
300lb m / h 0.55 lb m H 2SO 4 / lb m
b
0.45 lb m H 2O / lb m
b
& B lb m / h m
& C lb m / h m
g
g
0.75 lb m H2SO 4 / lb m 0.25 lb m H2O / lb m
0.90 lb m H2SO 4 / lb m 0.10 lb m H2O / lb m
Overall balance: 300 + m& B = m& C
(1)
b g
H2 SO4 balance: 0.55 300 + 0.90m & B = 0.75m& C & B = 400lb m / h , m & C = 700 lb m / h Solve (1) and (2) simultaneously ⇒ m
4- 12
(2)
4.21 (cont’d) b.
500 − 150 & A = 7.78 R A − 44.4 RA − 25 ⇒ m 70 − 25 800 − 200 & B − 200 = & B = 15.0 RB − 100 m RB − 20 ⇒ m 60 − 20 ln 100 − ln 20 ln x − ln 20 = Rx − 4 ⇒ ln x = 0.2682 Rx + 1.923 ⇒ x = 6.841e0.2682 Rx 10 − 4 300 + 44.4 400 + 100 mA = 300 ⇒ RA = = 44.3, m B = 400 ⇒ RB = = 33.3, 7.78 15.0 1 55 x = 55% ⇒ Rx = ln = 7.78 0.268 6.841
b b
& A − 150 = m
g g
b
g
FG H
c.
IJ K
Overall balance: m& A + m& B = m& C
b
g
H2 SO4 balance: 0.01xm & A + 0.90m& B = 0.75m& C = 0.75 m & A + m& B ⇒ m& B =
d
0.75 − 0.01 6.841e
⇒ 15.0 RB − 100 =
d
⇒ RB = 2.59 − 0.236 e
0.2682 Rx
i
0.2682 Rx
i b7.78 R
A
− 44.4
b0.75 − 0.01xgm&
g
A
0.15
015 . RA + 135 . e 0.2682 Rx − 813 .
Check: RA = 44.3, Rx = 7.78 ⇒ RB = 2.59 − 0.236e 0.2682b7.78g 44.3 + 135 . e 0.2682b 7.78 g − 813 . = 33.3
e
4.22
a.
b
n& A kmol / h
g 100 kg / h
0.10 kmolH2 / kmol 0.90 kmolN 2 / kmol
b
n& B kmol / h
j
b
n& P kmol / h
g
0.20 kmolH 2 / kmol
g
0.80 kmol N 2 / kmol
0.50 kmolH2 / kmol 0.50 kmolN 2 / kmol
b
g
b
g
MW = 0.20 2.016 + 0.80 28.012 = 22.813 kg / kmol 100 kg kmol = 4.38 kmol / h h 22.813 kg Overall balance: n& A + n& B = 4.38 H2 balance: 0.10 n& A + 0.50n& B = 0.20 4.38 ⇒ n& P =
b g
Solve (1) and (2) simultaneously ⇒ n& A = 3.29 kmol / h , n& B = 110 . kmol / h
4- 13
(1) (2)
4.22 (cont’d) b.
n& P =
m& P 22.813
m &P 22.813 x m & H2 balance: x An& A + xB n& B = P P 22.813 m& P xB − x P ⇒ n& A = 22.813 xB − x A Overall balance: n& A + n& B =
b b
c.
Trial 1 2 3 4 5 6 7 8 9 10 11 12
XA 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10
g g
XB 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50
bx 22.813 b x m& P
n& B =
XP 0.10 0.20 0.30 0.40 0.50 0.60 0.10 0.20 0.30 0.40 0.50 0.60
mP 100 100 100 100 100 100 250 250 250 250 250 250
P
− xA
B
− xA
g g
nA 4.38 3.29 2.19 1.10 0.00 -1.10 10.96 8.22 5.48 2.74 0.00 -2.74
nB 0.00 1.10 2.19 3.29 4.38 5.48 0.00 2.74 5.48 8.22 10.96 13.70
The results of trials 6 and 12 are impossible since the flow rates are negative. You cannot blend a 10% H2 mixture with a 50% H2 mixture and obtain a 60% H2 mixture. d. 4.23
Results are the same as in part c. Venous blood 1950 . ml / min 1.75 mg urea / ml
Arterialblood 200.0 ml / min 190 . mg urea / ml
Dialysate
b b
Dialyzing fluid 1500 ml / min
a.
g
v ml / min & c mg urea / ml Water removal rate: 200.0 − 195.0 = 5.0 ml / min
b
g
b
g
g
Urea removal rate: 1.90 200.0 − 1.75 1950 . = 38.8 mg urea / min b.
v& = 1500 + 5.0 = 1505 ml / min
c=
38.8mgurea/min = 0.0258mgurea/ml 1505ml/min
4- 14
4.23 (cont’d) c. 2.7 − 11 . mg removed 1 min 10 3 ml 5.0 L = 206 min (3.4 h) ml 38.8 mg removed 1L
b
4.24
a.
g
b
n&1 kmol / min
g b
20.0 kg CO2 / min
b
n& 2 kmol / min
n&3 kmol / min
g
g
0.023 kmol CO2 / kmol
0.015kmol CO2 / kmol
20.0 kg CO2 kmol = 0.455 kmolCO2 / min min 44 .0 kg CO2 Overall balance: 0.455 + n&2 = n& 3 CO2 balance: 0.455 + 0.015 n& 2 = 0.023n&3 Solve (1) and (2) simultaneously ⇒ n&2 = 55.6 kmol / min , n&3 = 561 . kmol / min n&1 =
b.
u=
150 m = 8.33 m / s 18 s
1 561 . kmol m3 1 min s A = πD 2 = ⇒ D = 108 . m 4 min 0.123 kmol 60 s 8.33 m
b
g
Spectrophotometer calibration: C = kA ====> C µg / L = 3.333 A
4.25
A = 0.9 C =3
b
gb g
Dye concentration: A = 018 . ⇒ C = 3333 . 018 . = 0.600 µg / L Dye injected =
b
0.60 cm 3
1L 10 3 cm 3
5.0 mg 10 3 µ g 1L
1 mg
= 3.0 µ g
g bg
⇒ 3.0 µ g V L = 0.600 µ g / L ⇒ V = 5.0 L
4.26
a.
1000 LB / min
b g V& d m / min i n& b kmol / min g y b kmol SO / kmolg 1 − y b kmol A / kmolg & 2 kg B / min m 3
1 1
1
2
1
b g y b kmol SO / kmol g 1 − y b kmol A / kmol g & bkg / min g m x b kg SO / kg g 1 − x b kg B / kg g n&3 kmol / min 3
2
3
4
4
2
4
4- 15
(1) (2)
4.26 (cont’d)
– – – – –
b.
8 unknowns ( n&1 , n&3 , v&1, m& 2 , m & 4 , x4 , y1 , y3 ) 3 material balances 2 analyzer readings 1 meter reading 1 gas density formula 1 specific gravity 0 DF
Orifice meter calibration: A log plot of V& vs. h is a line through the points h1 = 100, V&1 = 142 and h2 = 400, V&2 = 290 .
d
ln V& = b ln h + ln a ⇒ V& = ah b
d h = lnb290 142g = 0.515 lnb h h g lnb400 100g ln a = ln V& − b ln h = lnb142g − 0.515ln 100 = 2.58 ⇒ a = e b=
i d
ln V&2 V&1 2
1
1
1
2 .58
= 13.2 ⇒ V& = 13.2 h 0.515
Analyzer calibration: ln y = bR + ln a ⇒ y = ae bR b=
b
ln y 2 y 1 R2 − R1
ln a = ln y 1 − bR1
E
a = 5.00 × 10 −4
c.
U| 90 − 20 || = lnb0.00166g − 0.0600b20 g = −7.60V ⇒ y = 5.00 × 10 || |W
g = lnb0.1107 0.00166g = 0.0600
n&1 =
e 0.0600R
b g = 207.3 m min b12.2g b150 + 14.7g 14.7 batm g = 0.460 mol / L = 0.460 kmol / m = b75 + 460g 18. b Kg E
h1 = 210 mm ⇒ V&1 = 13.2 210 ρ feed gas
−4
0.515
3
207.3 m 3 0.460 kmol = 95.34 kmol min min m3
b g exp b00600 . ×116 . g = 0.00100 kmol SO
R1 = 82.4 ⇒ y1 = 5.00 ×10−4 exp 0.0600 × 82.4 = 0.0702 kmol SO 2 kmol R3 = 116 . ⇒ y3 = 500 . × 10 &2 = m
−4
2
1000 L B 130 . kg = 1300 kg / min min LB
4- 16
kmol
3
i
4.26 (cont’d) A balance: 1 − 0.0702 95.34 = 1 − 0.00100 n3 ⇒ n3 = 88.7 kmol min
b
SO2
gb g b g & x balance: b0.0702gb9534 . g(64.0 kg / kmol) = b0.00100 gb88.7g(64 ) + m
4 4
& 4 (1 − x4 ) B balance: 1300 = m
(1) (2)
& 4 = 1723 kg / min, x4 = 0.245 kg SO2 absorbed / kg Solve (1) and (2) si multaneously ⇒ m & 4 x4 = 422 kg SO 2 / min SO2 removed = m
4.27
d.
Decreasing the bubble size increases the bubble surface-to-volume ratio, which results in a higher rate of transfer of SO 2 from the gas to the liquid phase.
a.
V&2 m 3 / min
b g y b kmolSO / kmolg 1 − y b kmol A / kmolg
d i & b kg B / min g m
n& 3 kmol / min 3
2
3
d i n& b kmol / min g y b kmolSO / kmolg 1 − y b kmol A / kmolg V&1 m 3 / min
R3
b g x bkgSO kg g 1 − x bkg B / kg g & 4 kg / min m
1
1
2
2
4
1
2
4
P1 , T1 , R1 , h1
b.
& 2, n&3, y3, R3 , m & 4 , x4 ) 14 unknowns ( n&1,V&1, y1, P1, T1, R1, h1,V&2, m – 3 material balances – 3 analyzer and orifice meter readings – 1 gas density formula (relates n&1 andV&1 ) & and V& ) – 1 specific gravity (relates m 2
2
6 DF
b
g b
g
A balance: 1 − y1 n&1 = 1 − y3 n&3 &4 x4 m SO2 balance: y1n&1 = y3n& 3 + 64 kgSO 2 / kmol & 2 = 1 − x4 m &4 B balance: m
b
Calibration formulas:
(1) (2)
g
(3)
−4 0.060 R1
y1 = 5.00 × 10 e
(4)
y 3 = 5.00 × 10 −4 e 0.060 R3 V& = 13.2 h 0.515
(5)
1
Gas density formula : n&1 =
b
1
g
12.2 P1 + 14.7 / 14.7 & V1 T1 + 460 / 18 .
b
g
b g
& kg m m3 Liquid specific gravity: SG = 130 . ⇒ V&2 = 2 h 1300 kg
4- 17
(6) (7) (8)
4.27 (cont’d) c.
T1
75 °F
y1
P1
150 psig
V1
207 m3/h
h1
210 torr
n1
95.26 kmol/h
R1
0.07 kmol SO2 /kmol
82.4 x4 (kg SO 2/kg) y3 (kmol SO 2/kmol) V2 (m3/h) n3 (kmol/h) 0.10 0.050 0.89 93.25 0.10 0.025 1.95 90.86 0.10 0.010 2.56 89.48 0.10 0.005 2.76 89.03 0.10 0.001 2.92 88.68 0.20 0.050 0.39 93.25 0.20 0.025 0.87 90.86 0.20 0.010 1.14 89.48 0.20 0.005 1.23 89.03 0.20 0.001 1.30 88.68
Trial 1 2 3 4 5 6 7 8 9 10
m4 (kg/h) m2 (kg/h) 1283.45 1155.11 2813.72 2532.35 3694.78 3325.31 3982.57 3584.31 4210.72 3789.65 641.73 513.38 1406.86 1125.49 1847.39 1477.91 1991.28 1593.03 2105.36 1684.29
3
V 2 (m /h)
V2 vs. y 3 3.50 3.00 2.50 2.00 1.50 1.00 0.50 0.00 0.000
0.020
0.040
0.060
y 3 (kmol SO 2 /kmol) x4 = 0.10
x4 = 0.20
For a given SO 2 feed rate removing more SO 2 (lower y3 ) requires a higher solvent feed rate (V&2 ). For a given SO 2 removal rate (y3 ), a higher solvent feed rate (V& ) tends to a more dilute 2
SO2 solution at the outlet (lower x4 ). d. 4.28
Answers are the same as in part c. Maximum balances: Overall - 3, Unit 1 - 2; Unit 2 - 3; Mixing point - 3 Overall mass balance ⇒ m& 3 Mass balance - Unit 1 ⇒ m& 1 A balance - Unit 1 ⇒ x1 Mass balance - mixing point ⇒ m& 2 A balance - mixing point ⇒ x 2 C balance - mixing point ⇒ y2
4- 18
4.29
a.
100 mol / h 0.300 mol B / mol 0.250 mol T / mol 0.450 mol X / mol
b
n&2 mol / h
b
g
b g Column 1 x b mol T / molg 1 − x − x bmol X / molg n& bmol / hg
n&4 mol / h
0.940 mol B / mol
x B 2 mol B / mol T2
B2
Column 2 0.060 molT / mol
T2
3
0.020 molT / mol 0980 . molX / mol
b
n&5 mol / h
Column 1 4 unknowns ( n& 2 , n&3, x B 2, x T 2 ) –3 balances – 1 recovery of X in bot. (96%) 0 DF
b g x bmol T / molg 1 − x − x bmol X / molg T5
gb g
(1)
Total mole balance: 100 = n& 2 + n&3
(2)
B balance: 0.300 100 = x B 2n&2
(3)
& + 0.020 n& 3
T 2 n2
Column 2 97% B recovery: 0.97 xB 2n&2 = 0.940 n&4
b.
(4)
(5)
Total mole balance: n& 2 = n&4 + n&5
(6)
B balance: xB 2n&2 = 0.940 n&4 + xB5n&5
(7)
T balance: xT 2 n&2 = 0.060n& 4 + xT 5n&5
(8)
(1) ⇒ n& 3 = 44 .1 mol / h (2) ⇒ n& 2 = 55.9 mol / h ( 3) ⇒ x B 2 = 0.536 molB / mol ( 4) ⇒ x T 2 = 0.431 molT / mol (5) ⇒ n& 4 = 30.95 mol / h ( 6) ⇒ n& 5 = 24.96 mol / h ( 7) ⇒ x B5 = 0.036 mol B / mol (8) ⇒ x T 5 = 0.892 mol T / mol 0.940 30.95 Overall benzene recovery: × 100% = 97% 0.300 100
b
Overall toluene recovery:
T5
Column 2: 4 unknowns ( n& 3 , n&4 , n&5 , y x ) – 3 balances – 1 recovery of B in top (97%) 0 DF
Column 1 96% X recovery: 0.96 0.450 100 = 0.98 n&3
b g T balance: 0.250b100g = x
g
xB 5 molB / mol
B5
b
g
b g 0.892b24.96g × 100 = 89% 0.250b100g
4- 19
g
4.30
a.
100 kg / h 0.035 kg S / kg 0.965 kg H 2 O / kg
b
m & w kg H 2O / h
b.
b b
& 3 kg / h m
g
1
g
b
1 − x3 kg H2 O / kg
b
0100 . m& w kg H 2O / h
g
b b
m& 4 kg / h
x3 kg S / kg
4
g
g
b
b
& w kg H 2 O / h 0100 . m
m& 10 (kg / h) 0.050 kg S/kg 0.950 kg H2 O/kg m& w ( kg H 2 O / h )
b g
Salt balance: 0.035 100 = 0.050m& 10
& w + m& 10 Overall balance: 100 = m H2 O yield: Yw =
b b
g
& w kg H2O recovered m 96.5 kg H2Oin freshfeed
g
First 4 evaporators
b b
g
m4 kg/ h & x 4 kg S/ kg 1 − x4 kg H2 O / kg
100 kg/ h 0.035 kg S/ kg 0965 . kg H2 O / kg
b
g
&4 Overall balance: 100 = 4 0100 . m& w + m
b g
&4 Salt balance: 0.035 100 = x4 m c.
g
b g 4 × 0100 . m b kg H O / hg & w
Yw = 0 .31 x4 = 0.0398
4- 20
b
& 10 kg / h m
g
1 − x4 kg H2 O / kg
Overall process 100 kg/h 0.035 kg S/kg 0.965 kg H2 O/kg
g
x4 kg S / kg
2
g
g
10
g
0.050 kg S / kg 0.950 kg H2 O / kg
b
& w kg H 2 O / h 0.100 m
g
4.31
b g
a.
2 n&1 mol
Condenser
0.97 mol B / mol 0.03 mol T / mol
b g
b g
100 mol 0.50 mol B / mol 0.50 mol T / mol
Still
n&1 mol
n&1 mol ( 89.2% of Bin feed )
0.97 mol B / mol 0.03 mol T / mol
0.97 mol B / mol 0.03 mol T / mol
b gb y bmol B / mol g 1 − y bmol T / mol g
n&4 mol 45% of feed to reboiler
g
B
B
b g z bmolB / molg 1 − z bmolT / molg n& 2 mol B
b g x b mol B / mol g 1 − x b mol T / molg n&3 mol
Reboiler
B
B
B
Overall process:
Condenser:
3 unknowns ( n&1 , n&3 , xB ) Still: 5 unknowns ( n&1 , n&2 , n&4 , y B , zB ) – 2 balances – 2 balances – 1 relationship (89.2% recovery) 3 DF 0 DF
1 unknown ( n&1 ) – 0 balances 1 DF
Reboiler:
6 unknowns ( n& 2 , n& 3 , n& 4 , x B , y B , z B ) – 2 balances – 2 relationships (2.25 ratio & 45% vapor) 3 DF
Begin with overall process. b.
Overall process 89.2% recovery: 0.892 0.50 100 = 0.97 n&1
b gb g
Overall balance: 100 = n&1 + n& 3
b g
B balance: 0.50 100 = 0.97n&1 + x B n& 3 Reboiler
e j = 2.25 / b1 − x g
yB / 1− yB Composition relationship:
xB
Percent vaporized: n& 4 = 0.45 n& 2 Mole balance: n& 2 = n&3 + n& 4
B
(1) (2)
(Solve (1) and (2) simultaneously.) B balance: z Bn&2 = x Bn&3 + yB n&4
4- 21
4.31 (cont’d) c. B fraction in bottoms : xB = 0100 . mol B / mol Moles of overhead: n&1 = 46.0 mol Recovery of toluene: 4.32
Moles of bottoms : n& 3 = 54.0 mol
b1 − x gn& × 100% = b1 − 0.10gb54.02g × 100% = 97% 0.50b100g 0.50b100 g B
3
a.
b
m3 kg H2O
g
Bypass Basis: 100 kg 100 kg 0.12 kg S / kg 0.88 kg H2O / kg
Mixing point
b g
Evaporator
m1 kg
0.12 kg S / kg 0.88 kg H2 O / kg
b g
b g
m4 kg
m5 kg
0.58 kg S / kg 0.42 kg H2O / kg
0.42 kg S / kg 0.58 kg H2O / kg
b g
m2 kg
0.12 kg S / kg 0.88 kg H2 O / kg
Overall process:
Evaporator:
2 unknowns ( m3 , m5 ) – 2 balances 0 DF
3 unknowns ( m1 , m3 , m4 ) – 2 balances 1 DF
Bypass:
Mixing point:
2 unknowns ( m1 , m2 ) – 1 independent balance 1 DF
3 unknowns ( m2 , m4 , m5 ) – 2 balances 1 DF
b g
Overall S balance: 0.12 100 = 0.42m5 Overall mass balance: 100 = m3 + m5 Mixing point mass balance: m4 + m2 = m5
(1)
Mixing point S balance: 0.58m4 + 012 . m2 = 0.42m5
(2)
Solve (1) and (2) simultaneously Bypass mass balance: 100 = m1 + m2 b.
m1 = 90.05 kg , m2 = 9 .95 kg , m3 = 71.4 kg, m4 = 18.65 kg , m5 = 28.6 kg product
Bypass fraction: c.
m2 = 0.095 100
Over-evaporating could degrade the juice; additional evaporation could be uneconomical; a stream consisting of 90% solids could be hard to transport.
4- 22
4.33
a.
b
m& 4 kg Cr / h
b
g
b
g
m & 1 kg / h
m& 2 kg / h
0.0515 kgCr / kg 09485 . kg W / kg
0.0515 kgCr / kg 0.9485 kg W / kg
g
b g x b kg Cr / kg g 1 − x bkg W / kgg & 5 kg / h m
Treatment Unit
b
5
5
b g x b kg Cr / kg g 1 − x b kg W / kgg m& 6 kg / h 6
6
g
m& 3 kg / h
0.0515 kgCr / kg 0.9485 kg W / kg
b.
b
&1 = 6000 kg / h ⇒ m & 2 = 4500 kg / h maximum allowed value m Bypass point mass balance: m& 3 = 6000 − 4500 = 1500 kg / h & 4 = 0.95 0.0515 4500 = 2202 95% Cr removal: m . kg Cr / h
b
g
gb g
Mass balance on treatment unit : m& 5 = 4500 − 220.2 = 4279.8 kg / h
b
g
0.0515 4500 − 220.2 = 0.002707 kg Cr / kg 47798 . & 6 = 1500 + 42798 Mixing point mass balance: m . = 5779.8 kg / h
Cr balance on treatment unit : x 5 =
Mixing point Cr balance: x 6 =
c.
b g
b
g
0.0515 1500 + 0.0002707 4279.8 = 0.0154 kg Cr / kg 5779.8
m 1 (kg/h) m 2 (kg/h) m 3 (kg/h) m 4 (kg/h) m 5 (kg/h) 1000 1000 0 48.9 951 2000 2000 0 97.9 1902 3000 3000 0 147 2853 4000 4000 0 196 3804 5000 4500 500 220 4280 6000 4500 1500 220 4280 7000 4500 2500 220 4280 8000 4500 3500 220 4280 9000 4500 4500 220 4280 10000 4500 5500 220 4280
4- 23
x5 m 6 (kg/h) 0.00271 951 0.00271 1902 0.00271 2853 0.00271 3804 0.00271 4780 0.00271 5780 0.00271 6780 0.00271 7780 0.00271 8780 0.00271 9780
x6 0.00271 0.00271 0.00271 0.00271 0.00781 0.0154 0.0207 0.0247 0.0277 0.0301
4.33 (cont’d)
x 6 (kg Cr/kg)
m 1 vs. x 6 0.03500 0.03000 0.02500 0.02000 0.01500 0.01000 0.00500 0.00000 0
2000
4000
6000
8000 10000 12000
m 1 (kg/h)
d.
4.34
Cost of additional capacity – installation and maintenance, revenue from additional recovered Cr, anticipated wastewater productio n in coming years, capacity of waste lagoon, regulatory limits on Cr emissions.
a.
b
175 kg H 2 O / s 45% of water fed to evaporator
b
& 1 kg / s m
g
b b
m & 4 kg K2SO 4 / s
0196 . kg K 2SO4 / kg 0.804 kg H 2 O / kg
& 5 kg H2O / s m
g
g
b b
m & 6 kg K2SO 4 / s
Evaporator
& 7 kg H2 O / s m
g
g
g Crystallizer Filter
Filter cake
b
& 2 kg K2SO4 / s 10m
b
g
RS0.400 kg K SO / kgUV T0.600 kg H O / kg W 2
Filtrate & 3 kg / s m
b
g
m & 2 kg soln / s
4
2
g
0.400 kg K2SO 4 / kg 0.600 kg H2 O / kg
Let K = K2 SO4 , W = H2 Basis: 175 kg W evaporated/s Overall process: 2 unknowns ( m& 1, m& 2 ) - 2 balances 0 DF Evaporator:
4 unknowns ( m& 4 , m& 5 , m& 6 , m& 7 ) – 2 balances – 1 percent evaporation 1 DF
Mixing point:
Crystallizer:
&1, m &2 Strategy: Overall balances ⇒ m % evaporation ⇒ m &5 Balances around mixing point ⇒ m & 3, m &4 &7 Balances around evaporator ⇒ m& 6 , m
4- 24
4 unknowns ( m&1 , m& 3, m& 4 , m& 5 ) - 2 balances 2 DF
4 unknowns ( m& 2 , m& 3 , m& 6 , m& 7 ) – 2 balances 2 DF
U| verify that each |Vchosen subsystem involves | no more than two |W unknown variables
4.34 (cont’d) & 1 = 175 + 10m &2 + m &2 Overall mass balance: m Overall K balance: 0196 . m& 1 = 10m& 2 + 0.400m &2 &2 Production rate of crystals = 10 m
U| V| W
45% evaporation: 175 kg evaporated min = 0.450m& 5 & 1 + 0.600m & 3 = m& 5 W balance around mixing point: 0.804 m & 3 = m& 4 + m& 5 Mass balance around mixing point: m& 1 + m &4 K balance around evaporator: m& 6 = m
W balance around evaporator: m& 5 = 175 + m& 7 Mole fraction of K in stream entering evaporator =
b.
&1 = 221 kg / s Fresh feed rate: m
m& 4 m &4 + m &5
bg
& 2 = 416 Production rate of crystals = 10 m . kg K s s
Recycle ratio:
c.
b
g
& 3 kg recycle s m 352.3 kg recycle = = 160 . & m1 kg fresh feed s 220.8 kg fresh feed
b
g
Scale to 75% of capacity. Flow rate of stream entering evaporator = 0.75(398 kg / s) = 299 kg / s 46.3% K, 53.7% W
d.
Drying . Principal costs are likely to be the heating cost for the evaporator and the dryer and
the cooling cost for the crystallizer.
4- 25
4.35
a.
Overall objective : Separate components of a CH 4 -CO2 mixture, recover CH4 , and discharge CO2 to the atmosphere. Absorber function: Separates CO 2 from CH4. Stripper function: Removes dissolved CO2 from CH3 OH so that the latter can be reused.
b.
The top streams are liquids while the bottom streams are gases. The liquids are heavier than the gases so the liquids fall through the columns and the gases rise.
c.
b
n& 1 mol / h
g
b g n& b molCO / h g n& 5 mol N 2 / h
0.010 molCO 2 / mol 0.990 mol CH 4 / mol
b
n& 2 mol / h
Absorber
100 mol / h 0.300 molCO 2 / mol 0.700 molCH 4 / mol
6
g
Stripper
0005 . molCO 2 / mol 0995 . molCH3 OH / mol
2
b
n& 5 mol N 2 / h
g
b g n& b mol CH OH / h g n& 3 mol CO 2 / h 4
Overall:
Stripper:
3
3 unknowns ( n&1, n&5 , n& 6 ) – 2 balances 1 DF
Absorber:
4 unknowns ( n&1, n& 2 , n&3 , n&4 ) – 3 balances 1 DF
4 unknowns ( n& 2 , n& 3 , n&4 , n& 5 ) – 2 balances – 1 percent removal (90%) 1 DF
b
gb g b mol CH / hg = 0.990n& Overall mole balance: 100bmol / h g = n& + n& Overall CH4 balance: 0.700 100
4
1
1
6
Percent CO2 stripped: 0.90 n&3 = n& 6 Stripper CO2 balance: n& 3 = n& 6 + 0.005 n& 2 Stripper CH3 OH balance: n& 4 = 0.995n&2 d.
n&1 = 70.71 mol / h , n&2 = 651.0 mol / h , n&3 = 32.55 mol CO2 / h, n&4 = 6477 . mol CH 3OH / h , n& 6 = 29.29 mol CO 2 / h 30.0 − 0.010n&1 Fractional CO2 absorption: f CO2 = = 0.976 molCO 2 absorbed / mol fed 30.0
4- 26
4.35 (cont’d) Total molar flow rate of liquid feed to stripper and mole fraction of CO2 : n&3 n& 3 + n&4 = 680 mol / h , x3 = = 0.0478 molCO 2 / mol n&3 + n&4 e.
Scale up to 1000 kg/h (=106 g/h) of product gas:
b
g
b
g
MW1 = 0.01 44 g CO2 / mol + 0.99 16 g CH 4 / mol = 16.28 g / mol
bn& g = d1.0 × 10 g / hib16.28 g / molg = 6.142 × 10 mol / h . × 10 mol / h) / (70.71 mol / h) = 8.69 × 10 bn& g = b100 mol / hg (6142 6
4
1 new
4
4
feed new
4.36
mol / h
f.
Ta < Ts The higher temperature in the stripper will help drive off the gas. Pa > Ps The higher pressure in the absorber will help dissolve the gas in the liquid.
g.
The methanol must have a high solubility for CO2 , a low solubility for CH4 , and a low volatility at the stripper temperature.
a.
Basis: 100 kg beans fed
e
m kg C H 5 6 14
e
m kg C H 1 6 14
j
300 kg C6H14
Ex
b g x b kg S / kgg y bkg oil / kgg 1 − x − y b kg C H
j
Condenser
m2 kg
F
2
2
2
2
6
14
/ kg
g
13.0 kg oil 87.0 kg S
b g y b kg oil / kgg 1 − y b kg C H 4
4
6
14
/ kg
Ev
g
g
b g
m3 kg
0.75 kg S / kg
b
y3 kg oil / kg
b
g
0.25 − y3 kg C6 H14 / kg
Overall:
b
m6 kg oil
m4 kg
4 unknowns ( m1 , m3 , m6 , y3 ) – 3 balances 1 DF
Extractor:
g
3 unknowns ( m2 , x2 , y2 ) – 3 balances 0 DF
2 unknowns ( m1 , m5 ) Evaporator: 4 unknowns ( m4 , m5 , m6 , y4 ) – 1 balance – 2 balances 1 DF 2 DF Filter: 7 unknowns ( m2 , m3 , m4 , x2 , y2 , y3 , y4 ) – 3 balances – 1 oil/hexane ratio 3 DF Mixing Pt:
Start with extractor (0 degrees of freedom) Extractor mass balance: 300 + 87.0 + 13.0 kg = m2
4- 27
4.36
(cont’d) Extractor S balance: 87.0 kg S = x2m2 Extractor oil balance: 13.0 kg oil = y2 m2 Filter S balance: 87.0 kg S = 0.75m3
b g
Filter mass balance: m2 kg = m3 + m4 Oil / hexane ratio in filter cake: y3 0.25 − y3
=
y2 1 − x2 − y2
Filter oil balance: 13.0 kg oil = y3m3 + y4 m4
b
g
Evaporator hexane balance: 1 − y 4 m4 = m5 Mixing pt. Hexane balance: m1 + m5 = 300 kg C 6H14 Evaporator oil balance: y4m4 = m6 b.
b
g
m6 11.8 kg oil = = 0118 . kg oil / kg beans fed 100 100 kg beans fed m 28 kg C6H 14 Fresh hexanefeed = 1 = = 0.28 kg C 6 H14 / kg beans fed 100 100 kg beans fed m 272 kg C 6H14 recycled Recycle ratio = 5 = = 9.71 kg C6H14 recycled / kg C6H14 fed m1 28 kg C6H14 fed
Yield =
b b
c.
g
g
Lower heating cost for the evaporator and lower cooling cost for the condenser.
4.37
b
g
m lb m dirt 1 98 lb m dry shirts 3 lb m Whizzo
100 lbm 2 lbm dirt 98 lb m dry shirts
b
m lb m Whizzo 2
g
Tub
b g
Filter
b g
m lb m 3 0.03 lb m dirt / lb m
m lb m 4 0 .13 lb m dirt / lb m
0.97 lb m Whizzo / lb m
0 .87 lb m Whizzo / lb m
m
b g
lb 5 m 0 .92 lb m dirt / lb m 0 .08 lb m Whizzo / lb m
b g b g b lb Whizzo/ lb g
m6 lb m 1− x lbm dirt / lb m x
m
m
Strategy 95% dirt removal ⇒ m1 ( = 5% of the dirt entering) Overall balances: 2 allowed (we have implicitly used a clean shirt balance in labeling the chart) ⇒ m2, m5 (solves Part (a))
4- 28
4.37
(cont’d)
b g around the filter bm , m , x g , but the tub only involves 2 bm , m g and 2 balances are Balances around the mixing point involve 3 unknowns m3 , m6 , x , as do balances 4
6
3
4
allowed for each subsystem. Balances around tub ⇒ m3, m4 Balances around mixing point ⇒ m6 , x (solves Part (b)) a.
b gb g Overall dirt balance: 2.0 = 010 . + b0.92g m ⇒ m = 2.065 lb dirt Overall Whizzo balance: m = 3 + b0.08 gb2.065g blb Whizzog = 317 . lb 95% dirt removal: m1 = 0.05 2.0 = 010 . lb m dirt 5
5
2
b.
m
m
Tub dirt balance: 2 + 0.03m3 = 010 . + 013 . m4 Tub Whizzo balance: 0.97 m3 = 3 + 0.87 m4 Solve (1) & (2) simultaneously ⇒ m3 = 20.4 lb m , m4 = 19.3 lb m Mixing pt. mass balance: 317 . + m6 = 20.4 lb m ⇒ m6 = 17.3 lb m Mixing pt. Whizzo balance:
m
Whizzo
(1) (2)
3.17 + x (17.3) = ( 0.97 )( 20.4) ⇒ x = 0.961 lbm Whizzo/lb m ⇒ 96%Whizzo, 4% dirt
4.38
a.
2720 kg S mixer 3 Discarded C 3L kg L C 3S kg S 3300 kg S
Filter 3
C 2L kg L C 2S kg S
F 3L kg L F 3S kg S
620 kg L mixer 1 Filter 1 F 1L F 1S
mixer 2 C 1L kg L C 1S kg S kg L kg S
Filter 2 F 2L kg L F 2S kg S To holding tank
b g g U| V| W
mixer filter 1: 0.01 620 = F1L ⇒ F1L = 6.2 kg L balance: 620 = 6.2 + C1L ⇒ C1L = 6138 . kg L 0 . 01 6138 . + F = F mixer filter 2 : F2 L = 6.2 kg L 3L 2L balance: 6138 . + F3L = F2 L + C3L ⇒ C2 L = 613.7 kg L mixer filter 3: 0.01C2 L = F3L F3L = 61 . kg L balance: 613.7 = 6.1+ C3L ⇒ C3L = 6076 . kg L
b
4- 29
4.38 (cont’d) Solvent m f 1: balance: m f 2: balance: m f 3: balance:
b g U| b g 495 + F = C + F | V⇒ 015 . b2720 + C g = C | 2720 + C = F + C |W 015 . 3300 = C1S ⇒ 3300 = 495 + F1S ⇒ 015 . 495 + F3S = C2 S 3S
2S
2S
2S
2S
C1S = 495 kg S F1S = 2805 kg S C2S = 482.6 kg S F2 S = 2734.6 kg S C3S = 480.4 kg S F3S = 2722.2 kg S
3S
3S
3S
Holding Tank Contents 6.2 + 6.2 = 12.4 kg leaf 2805 + 27346 . = 5540 kg solvent
b.
b g
5540 kg S
b g
Q R kg
0165 . kg E / kg 0.835 kg W / kg
b g Q b kg Fg
Q0 kg
. kg E / kg Extraction 013 0.15 kg F / kg Unit
Steam Stripper
0.855 kg W / kg
QD kg D
0.026 kg F / kg 0.774 kg W / kg QB kg
b g
b g Q bkg D g Q b kg Fg QE kg E
F
0.200 kg E / kg
0.013 kg E / kg 0.987 kg W / kg
D F
b
Q3 kg steam
1 kg D
g
620 kg leaf
= 0.62 kg D = QD 1000 kg leaf Water balance around extraction unit: 0.835 5540 = 0.855QR ⇒ QR = 5410 kg Ethanol balance around extraction unit: 0.165 5540 = 013 . 5410 + QE ⇒ QE = 211 kg ethanol in extract
Mass of D in Product:
b g
b g
c.
b
g
b
g
F balance around stripper 0.015 5410 = 0.026 Q0 ⇒ Q0 = 3121 kg mass of stripper overhead product
b g
b
g
E balance around stripper 0.13 5410 = 0.200 3121 + 0.013QB ⇒ QB = 6085 kg mass of stripper bottom product
b g b g b W balance around stripper 0.855b5410g + Q = 0.774b 3121g + 0.987b 6085g ⇒ Q S
4.39
a.
S
C 2 H 2 + 2 H 2 → C2 H 6 2 mol H 2 react / mol C 2 H 2 react 0.5 kmol C 2 H 6 formed / kmol H 2 react
4- 30
g
= 3796 kg steam fed to stripper
4.39 (cont’d) b. nH 2
= 1.5 < 2.0 ⇒ H2 is limiting reactant
nC 2 H2
15 . molH2 fed ⇒ 1.0 mol C2 H2 fed ⇒ 0.75 molC 2H2 required (theoretical) 1.0 mol fed − 0.75 mol required % excess C2H 2 = × 100% = 333% . 0.75 mol required
c.
4 × 10 6 tonnes C2 H 6 1 yr 1 day 1 h 1000 kg 1 kmol C2 H 6 2 kmolH 2 2.00 kg H 2 yr 300 days 24 h 3600 s tonne 30.0 kg C2 H 6 1 kmolC 2H 6 1 kmol H 2 = 20.6 kg H2 / s
4.40
d.
The extra cost will be involved in separating the product from the excess reactant.
a.
4 NH3 + 5 O 2 → 4 NO + 6 H2O 5 lb - mole O2 react = 125 . lb - mole O 2 react / lb - mole NO fo rmed 4 lb - mole NO formed
b.
dn i O2
=
theoretical
100 kmol NH3 5 kmol O2 = 125 kmol O2 h 4 kmol NH3
d i
40% excess O 2 ⇒ n O2
c.
fed
b
g
= 1.40 125 kmol O2 = 175 kmol O2
b50.0 kg NH gb1 kmol NH / 17 kg NH g = 2.94 kmol NH . kmol O b100.0 kg O gb1 kmol O / 32 kg O g = 3125 F n I = 3125 F n I = 5 = 1.25 . = 1 . 06 < GH n JK 2.94 GH n JK 4 3
3
2
2
2
O2
NH3
3
3
2
O2
NH3
fed
stoich
⇒ O2 is the limiting reactant
Required NH3 :
3125 . kmol O2 4 kmol NH3 = 2.50 kmolNH3 5 kmol O2
2.94 − 2.50 × 100% = 17.6% excess NH3 2.50 Extent of reaction: nO2 = nO2 − vO2 ξ ⇒ 0 = 3125 . − −5 ξ ⇒ ξ = 0.625kmol = 625 mol % excess NH3 =
d i
Mass of NO:
4.41
a.
b g
0
3125 . kmol O2 4 kmol NO 30.0 kg NO = 75.0 kg NO 5 kmol O2 1 kmol NO
By adding the feeds in stoichometric proportion, all of the H2 S and SO 2 would be consumed. Automation provides for faster and more accurate response to fluctuations in the feed stream, reducing the risk of release of H2 S and SO 2 . It also may reduce labor costs.
4- 31
4.41 (cont’d) b.
n& c =
3.00 × 10 2 kmol 0.85 kmol H 2S 1 kmol SO 2 = 1275 . kmol SO 2 / h h kmol 2 kmol H 2S
c.
Calibration Curve 1.20
X (mol H 2S/mol)
1.00 0.80 0.60 0.40 0.20 0.00 0.0
20.0
40.0
60.0
80.0
100.0
Ra (mV)
X = 0.0199Ra − 0.0605
b
d.
n& c kmol SO2 / h
b
n& f kmol / h
b
g
x kmol H 2S / kmol
g
g
Blender
Flowmeter calibration:
UV W
n& f = aR f 20 n& = R &n f = 100 kmol / h , R f = 15 mV f 3 f
Control valve calibration:
UV W
n&c = 25.0 kmol / h, R c = 10.0 mV 7 5 n& = R + n& c = 60.0 kmol / h , Rc = 25.0 mV c 3 c 3
FG H
IJ b K
1 7 5 1 20 n& f x ⇒ Rc + = R f 0.0119 Ra − 0.0605 2 3 3 2 3 10 5 ⇒ Rc = R f 0.0119 Ra − 0.0605 − 7 7
Stoichiometric feed: n& c =
b
n& f = 3.00 × 10 2 kmol / h ⇒ R f =
3 n& f = 45 mV 20
4- 32
g
g
4.41 (cont’d)
b gb b g
e.
gb g
10 5 45 0.0119 76.5 − 0.0605 − = 53.9 mV 7 7 7 5 ⇒ n&c = 53.9 + = 127.4 kmol / h 3 3 Faulty sensors, computer problems, analyzer calibration not linear, extrapolation beyond range of calibration data, system had not reached steady state yet. Rc =
4.42 165 mol / s
b
g
x mol C2H 4 / mol
b
b
n& mol / s
g
0.310 mol C 2 H 4 / mol
g
1 − x mol HBr / mol
0173 . mol HBr / mol 0.517 mol C 2H 5Br / mol
C 2 H 4 + HBr → C 2H 5Br C balance:
b
g
b
gb g b
gb g
165 mol x mol C 2H4 2 mol C = n& 0.310 2 + n& 0.517 2 s mol mol C2 H4
Br balance: 165 (1 − x )( 1) = n& ( 0.173 )( 1) + n& ( 0.517 )(1)
(1) (2)
Solve (1) and (2) simultaneously ⇒ n& = 108.77 mol / s, x = 0.545mol C2H 4 / mol
b g
⇒ 1 − x = 0.455mol HBr / mol
Since the C2 H4 /HBr feed ratio (0.545/0.455) is greater than the stoichiometric ration (=1), HBr is the limiting reactant .
bn& g = b165mol / sgb0.455mol HBr / molg = 75.08 mol HBr 75.08 − b 0173 . gb108.8 g Fractional conversion of HBr = × 100% = 0.749 mol HBr react / molfed HBr fed
dn& i = 75.08 molC H dn& i = b165 mol / sgb 0.545mol C H C2 H4 stoich C2 H4
2
75.08
4
2
fed
4
g
/ mol = 89.93 mol C2 H4
89.93 − 75.08 = 19.8% 75.08 Extentof reaction: n&C 2H5 Br = n& C2 H5 Br + v C2 H5 Brξ ⇒ 1088 . 0.517 = 0 + 1 ξ ⇒ ξ = 56.2 mol / s
% excess of C2H 4 =
d
i
0
4- 33
b
gb
g
bg
4.43
a.
2HCl +
1 O 2 → Cl 2 + H 2 O 2
Basis: 100 mol HCl fed to reactor
b g n bmolO g n b mol N g n bmol Cl g n b mol H Og
100 mol HCl
b
n1 mol air
n2 mol HCl
g
0.21 mol O 2 / mol
3
2
4
2
5
0.79 mol N2 / mol
2
6
35% excess
2
mol O = 25 mol O bO g stoic = 100 mol HCl 0.5 2 mol HCl 35% excess air: 0.21n b mol O fedg = 1.35 × 25 ⇒ n = 160.7 mol air fed 2
2
2
1
2
1
85% conversion ⇒ 85 mol HCl react ⇒ n2 = 15 mol HCl n5 =
85 mol HCl react
1 mol Cl 2 2 mol HCl
b gb g
= 42.5 mol Cl 2
n6 = 85 1 2 = 42.5 mol H2O N 2 balance:
b160.7gb0.79g = n
4
⇒ n4 = 127 mol N2
O balance:
b160.7gb0.21g mol O
2
2 mol O 1 mol O 2
= 2 n3 +
42.5 mol H 2 O
1 mol O 1 mol H2O
⇒ n3 = 12.5 mol O 2
Total moles: 5
∑ n j = 239.5 mol ⇒ j =2
15 mol HCl mol HCl molO 2 mol N 2 = 0.063 , 0.052 , 0.530 , 239.5 mol mol mol mol 0177 .
b.
molCl 2 mol H 2 O , 0177 . mol mol
As before, n1 = 160.7 mol air fed, n2 = 15 mol HCl 1 2HCl + O2 → Cl2 + H2O 2
b g
ni = ni
E
0
+ vi ξ
HCl: 15 = 100 − 2ξ ⇒ ξ = 42.5 mol
4- 34
4.43 (cont’d)
b g 12 ξ = 12.5 mol O = 0.79b160.7g = 127 mol N
O 2 : n 3 = 0.21 160.7 − N 2 : n4
2
2
Cl 2 : n 5 = ξ = 42.5 mol Cl 2 H 2 O: n6 = ξ = 42.5 mol H 2 O
c.
4.44
These molar quantities are the same as in part (a), so the mole fractions would also be the same. Use of pure O2 would eliminate the need for an extra process to remove the N2 from the product gas, but O2 costs much more than air. The cheaper process will be the process of choice.
b g Fe O + 3H SO → Fe bSO g + 3H O bTiOgSO + 2H O → H TiO bsg + H SO H TiO bsg → TiO bsg + H O
FeTiO3 + 2H 2SO 4 → TiO SO 4 + FeSO 4 + 2H 2O 2
3
2
4
2
4
2
2
2
3
2
4 3
2
3
2
4
2
Basis: 1000 kg TiO 2 produced 1000 kg TiO 2
kmol TiO 2
1 kmol FeTiO 3
79.90 kg TiO2
1 kmol TiO 2
12.52 kmol FeTiO3 dec.
1 kmol FeTiO3 feed 0.89 kmol FeTiO3 dec.
14.06 kmol FeTiO3
b
= 12.52 kmol FeTiO 3 decomposes = 14.06 kmol FeTiO3 fed
1 kmol Ti
47.90 kg Ti
1 kmol FeTiO3
kmol Ti
g
= 6735 . kg Ti fed
673.5 kg Ti / M kg ore = 0.243 ⇒ M = 2772 kg ore fed
b
g
Ore is made up entirely of 14.06 kmol FeTiO 3 + n kmol Fe 2O3 (Assumption!) n = 2772 kg ore − 638.1 kg Fe2 O3
14.06 kmol FeTiO3 151.74 kg FeTiO3 kmol FeTiO3 kmol Fe 2O3
159.69 kg Fe 2O3
= 6381 . kg Fe 2O3
= 4.00 kmol Fe2 O3
14.06 kmol FeTiO3 2 kmol H2SO4 4.00 kmol FeTiO3 3 kmol H2SO4 + = 4012 . kmol H2SO4 1 kmol FeTiO3 1 kmol Fe2O3
b
g
50% excess: 15 . 4012 . kmol H2SO 4 = 6018 . kmol H2SO 4 fed Mass of 80% solution:
60.18 kmol H 2SO 4 98.08 kg H2SO 4
b
g
1 kmol H2SO 4
= 59024 . kg H2SO 4
5902.4 kg H 2SO 4 / M a kg soln = 0.80 ⇒ M a = 7380 kg 80% H 2SO 4 feed
4- 35
4.45
a.
Plot C (log scale) vs. R (linear scale) on semilog paper, get straight line through
d R = 10, C = 0.30 g m i and FH R 3
1
1
2
= 48, C2 = 2.67 g m3
IK
ln C = bR + ln a ⇔ C = ae br b=
b
g = 0.0575 , ln a = lnb2.67g − 0.0575b48g = −178 . ⇒a=e
ln 2.67 0.30
−1.78
48 − 10 ⇒ C = 0169 . e 0.0575 R
i C ′(ftlb E
d
C g m3 =
m)
4536 . g 35.31 ft 3
3
1 m3
1 lb m
d
= 0.169
= 16,020C ′
i
16 ,020C' = 0.169e 0.0575R ⇒ C ′ lb m SO 2 ft 3 = 1.055 × 10 −5 e 0.0575 R
b.
d2867 ft sib 60 s min g = 138 ft 3
1250 lb m min
d
3
lb m coal
i
R = 37 ⇒ C ′ lb m SO 2 ft 3 = 1055 . × 10 −5 e b0.0575 gb37g = 8.86 × 10 8.86 × 10 −5 lb m SO 2 ft
c.
3
138 ft 3 1 lb m coal
= 0.012 < 0.018
−5
lb m SO 2 ft 3
lb m SO 2 compliance achieved lb m coal
S + O 2 → SO 2
1250 lb m coal 0.05 lb m S 64.06 lb m SO 2 min
1 lb m coal
32.06 lb m S
= 1249 . lb m SO 2 generated min
2867 ft3 60 s 8.86 ×10−5 lb m SO 2 = 15.2 lb m SO 2 min in scrubbed gas s 1 min ft3 air 1250 lbm coal/min 62.5 lb m S/min
% removal =
d.
scrubbing fluid
furnace stack gas 124.9 lbm SO2 /min ash
b124.9 − 15.2g lb
scrubber scrubbed gas 15.2 lb m SO2 /min liquid effluent (124.9 – 15.2)lbm SO2 (absorbed)/min
SO 2 scrubbed min × 100% = 88% 1249 . lb m SO 2 fed to scrubber min m
The regulation was avoided by diluting the stack gas with fresh air before it exited from the stack. The new regulation prevents this since the mass of SO 2 emitted per mass of coal burned is independent of the flow rate of air in the stack.
4- 36
4.46
a.
A + B ===== C + D nA = nA − ξ 0
e = en = en = en
j − ξj n ξj n + ξj n
n B = nB − ξ
y A = n A − ξ nT
nC = nC + ξ
yB
n D = nD + ξ
yC
0 0
0
nI = nI Total nT = ∑ ni At equilibrium:
yD
0
b b
0
B0
T
C0 + D0
gb gb
T T
g g
nC0 + ξ c nD0 + ξ c yC y D = = 487 . (nT ’s cancel) y A yB n A0 − ξ c n B0 − ξ c
b
c
gh b
g
387 . ξ 2c − nC0 + nD0 + 487 . nA0 + nB0 ξ c − nC0 nD0 − 487 . nA0 nB0 = 0 [aξ 2c + bξ c + c = 0] a = 387 . 1 2 ∴ ξc = −b ± b − 4ac where b = − nC0 + nD 0 + 4.87 nA0 + nB0 2a c = − nC0nD0 − 4.87nA0nB0
e
b.
b
j
Basis: 1 mol A feed
nA0 = 1 nB0 = 1 nC0 = nD0 = nI0 = 0
Constants: a = 3.87 b = −9.74
ξe =
(
1 9.74 ± 2 ( 3.87 )
g
c = 4.87
( 9.74) 2 − 4 ( 3.87 )( 4.87 )
)⇒ξ
e1
= 0.688
(ξ e 2 = 1.83 is also a solution but leads to a negative conversion ) Fractional conversion: X A ( = X B ) = c.
nA0 − nA ξ e1 = = 0.688 nA 0 nA 0
nA0 = 80, nC0 = nD 0 = nJ 0 = 0 nC 0 = 0 nC = 70 = nC 0 + ξ c =======> ξ c = 70 mol n A = nA 0 − ξ c = n A0 − 70 mol nB = n B0 − ξ c = 80 − 70 = 10 mol nC = nC 0 + ξ c = 70 mol nD = nD 0 + ξ c = 70 mol 4.87 =
b gb g b gb g
70 70 yC y D nC nD = ⇒ = 4.87 ⇒ n A0 = 170.6 mol methanol fed y A y B n An B n A0 − 70 10
4- 37
4.46 (cont’d) y A = 0.401 mol CH3OH mol Product gas n A = 170.6 − 70 = 100.6 mol y B = 0.040 mol CH3COOH mol n B = 10 mol ⇒ y C = 0.279 mol CH3COOCH3 mol nC = 70 mol y D = 0.279 mol H 2O mol n D = 70 mol
U || V| |W
ntotal = 250.6 mol
4.47
d.
Cost of reactants, selling price for product, market for product, rate of reaction, need for heating or cooling, and many other items.
a.
CO + H2O ← → CO 2 + H2 (A)
(B)
(C)
(D)
b g n bmol H O g n b mol CO g n b mol H g n b mol Ig
1.00 mol
n A mol CO
0.20 mol CO / mol
B
010 . mol CO 2 / mol 0.40 mol H 2O / mol
2
C
D
0.30 mol I / mol
2
2
I
Degree of freedom analysis :
6 unknowns ( n A , nB , nC , n D , n I ,ξ )
bg
– 4 expressions for ni ξ – 1 balance on I – 1 equilibrium relationship 0 DF b.
Since two moles are prodcued for every two moles that react, n total out = ntotal in = 1.00 mol
b g b g
b g
n A = 0.20 − ξ nB = 0.40 − ξ nC = 010 . +ξ nD = ξ
(1) (2) (3) (4)
n I = 0.30
(5)
ntot = 1.00 mol
At equilibrium:
b
y D = n D = ξ = 0110 . mol H 2
c.
b
gb g
FG H
IJ K
010 . +ξ ξ yC yD nC n D 4020 = = = 0.0247exp ⇒ ξ = 0110 . mol y A yB n A nB 0.20 − ξ 0.40 − ξ 1123
b / molg
gb
The reaction has not reached equilibrium yet.
4- 38
g
4.47
(cont’d) d.
T (K) 1223 1123 1023 923 823 723 623 673 698 688
x (CO) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
x (H2O) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
1123 1123 1123 1123
0.2 0.4 0.3 0.5
0.4 0.2 0.3 0.4
x (CO2)
Keq Keq (Goal Seek) Extent of Reaction 0.6610 0.6610 0.2242 0.8858 0.8856 0.2424 1.2569 1.2569 0.2643 1.9240 1.9242 0.2905 3.2662 3.2661 0.3219 6.4187 6.4188 0.3585 15.6692 15.6692 0.3992 9.7017 9.7011 0.3785 7.8331 7.8331 0.3684 8.5171 8.5177 0.3724
0 0 0 0 0 0 0 0 0 0 0.1 0.1 0 0
0.8858 0.8858 0.8858 0.8858
0.8863 0.8857 0.8856 0.8867
0.1101 0.1100 0.1454 0.2156
y (H2) 0.224 0.242 0.264 0.291 0.322 0.358 0.399 0.378 0.368 0.372 0.110 0.110 0.145 0.216
The lower the temperature, the higher the extent of reaction. An equimolar feed ratio of carbon monoxide and water also maximizes the extent of reaction. 4.48
a.
A + 2B → C
b g g = lnd10.5 / 2.316 × 10 i = 11458
ln Ke = ln A0 + E T K E=
b
ln Ke1 / K e2 1 T1 − 1 T2
−4
1 373 − 1 573
ln A0 = ln K e1 − 11458 T1 = ln 10.5 − 11458 373 = −28.37 ⇒ A0 = 4.79 × 10 −13
b gh
c
K e = 4.79 × 10 −13 exp 11458 T K atm −2 ⇒ K e (450 K ) = 00548 . atm −1 b.
U − 2ξ || V +ξ | − 2ξ |W
n A = n A0 − ξ nB = n B0 nC = nC 0 nT = nT 0
b = bn = bn
g bn − 2ξg − 2ξ g b n − 2ξ g + ξ g bn − 2ξ g +n +n g
y A = n A0 − ξ yB B0 ⇒ yC C0 nT 0 = n A0
b
T0
T0
T0
B0
C0
At equilibrium,
b b
gb gb
n + ξ n − 2ξ e yC 1 = C0 e T 0 2 2 y A yB P n A0 − ξ e nB 0 − 2ξ e
c.
g g
2 2
bg
bg
1 = Ke T (substitute for K T from Part a.) e P2
Basis: 1 mol A (CO) n A0 = 1 n B0 = 1 nC 0 = 0 ⇒ nT 0 = 2 , P = 2 atm , T = 423K
b
ξ e 2 − 2ξ e
g
2
b1 − ξ gb1 − 2ξ g e
e
2
b g
1 -2 2 . =0 2 = Ke 423 = 0.278 atm ⇒ ξ e − ξ e + 01317 4 atm
4- 39
4.48 (cont’d) (For this particular set of initial conditions, we get a quadratic equation. In general, the equation will be cubic.) ξ e = 0156 . , 0.844
Reject the second solution, since it leads to a negative n B .
. gh ⇒ y = 0.500 b g c2 − 2b0156 y = c1 − 2b 0156 . gh c2 − 2b0156 . gh ⇒ y = 0.408 y = b0 + 0.156 g c2 − 2b 0156 . gh ⇒ y = 0.092 n −n ξ Fractional Conversion of CO b Ag = = n n y A = 1 − 0156 .
A
B
B
C
C
A0
A
A0
d.
= 0.156 mol A reacted / mol A feed
A0
Use the equations from part b. i) ii) iii) iv)
Fractional conversion decreases with increasing fraction of CO. Fractional conversion decreases with increasing fraction of CH3 OH. Fractional conversion decreases with increasing temperature. Fractional conversion increases with increasing pressure.
*
1
2
REAL TRU, A, E, YA0, YC0, T, P, KE, P2KE, C0, C1, C2, C3, EK, EKPI, FN, FDN, NT, CON, YA, YB, YC INTEGER NIT, INMAX TAU = 0.0001 INMAX = 10 A = 4.79E–13 E = 11458. READ (5, *) YA0, YB0, YC0, T, P KE = A * EXP(E/T) P2KE = P*P*KE C0 = YC0 – P2KE * YA0 * YB0 * YB0 C1 = 1. – 4. * YC0 + P2KE * YB0 * (YB0 + 4. * YA0) C2 = 4. * (YC0 –1. – P2KE * (YA0 + YB0)) C3 = 4. * (1. + P2KE) EK = 0.0 (Assume an initial value ξ e = 0 . 0 ) NIT = 0 FN = C0 + EK * (C1 + EK * (C2 + EK * C3)) FDN = C1 + EK * (2. * C2 + EK * 3. * C3) EKPI = EK - FN/FDN NIT = NIT + 1 IF (NIT.EQ.INMAX) GOTO 4 IF (ABS((EKPI – EK)/EKPI).LT.TAU) GOTO 2 EK = EKPI GO TO 1 NT = 1. – 2. * EKPI YA = (YA0 – EKPI)/NT YB = (YB0 – 2. + EKPI)/NT YC = (YC0 + EKPI)/NT
4- 40
4.48 (cont’d) CON = EKPI/YA0 WRITE (6, 3) YA, YB, YC, CON STOP 4 WRITE (6, 5) INMAX, EKPI 3 FORMAT (' YA YB YC CON', 1, 4(F6.3, 1X)) FORMAT ('DID NOT * CONVERGE IN', I3, 'ITERATIONS',/, * 'CURRENT VALUE = ', F6.3) END $ DATA 0.5 0.5 0.0 423. 2. RESULTS: YA = 0.500, YB = 0.408, YC = 0.092, CON = 0.156 Note: This will only find one root — there are two others that can only be found by choosing different initial values of ξ a 4.49
a.
CH4 + O 2 → HCHO + H 2O
(1)
CH4 + 2O 2 → CO2 + 2H 2O
(2)
100 mol / s
b g n& b mol O / sg n& b mol HCHO / sg n& b mol H O / sg n&n& bmol (molCO CO g /s) n&1 mol CH4 / s
0.50 mol CH4 / mol 0.50 mol O 2 / mol
2
2
3 4 5
5
7 unknowns ( n&1, n& 2 , n&3, n&4 , n&5 , ξ& 1,ξ& 2 ) – 5 equations for n& ξ& , ξ& i
e
1
2
2
2
2
j
2 DF b.
n&1 = 50 − ξ& 1 − ξ& 2 n& = 50 − ξ& − 2ξ& 2
1
n& 3 = ξ& 1 n& = ξ& + 2ξ& 4
1
n& 5 = ξ& 2
c.
(1) (2)
2
(3) (4)
2
(5)
Fractional conversion: Fractional yield :
( 50 − n&1 ) = 0.900 ⇒ n&
1
50
= 5.00 mol CH /s 4
n&3 = 0.855 ⇒ n&3 = 42.75 mol HCHO/s 50
4- 41
4.49 (cont’d)
y CH = 0.0500 mol CH 4 /mol 4 Equation 3 ⇒ ξ1 = 42.75 y O = 0.0275 mol O 2 /mol Equation 1 ⇒ ξ2 = 2.25 2 y Equation 2 ⇒ n&2 = 2.75 ⇒ HCHO = 0.4275 mol HCHO/mol Equation 4 ⇒ n&4 = 47.25 y H O = 0.4725 mol H 2O/mol 2 Equation 5 ⇒ n&5 = 2.25 y CO = 0.0225 mol CO 2 /mol 2
Selectivity: [(42.75mol HCHO/s)/(2.25molCO 2 /s) = 19.0 mol HCHO/mol CO2
4- 42
4.50
a.
Design for low conversion and feed ethane in excess. Low conversion and excess ethane make the second reaction unlikely.
b.
C2 H6 + Cl2 → C2 H5 Cl + HCl, C2 H5 Cl + Cl2 → C2H4 Cl2 + HCl Basis: 100 mol C2 H5 Cl produced
c.
n 1 (mol C2 H6 )
100 mol C2 H5 Cl
n 2 (mol Cl2 )
n 3 (mol C2 H6 ) n 4 (mol HCl) n 5 (mol C2 H5 Cl2 )
5 unknowns –3 atomic balances 2 D.F.
Selectivity: 100 mol C 2 H 5Cl = 14 n5 (mol C 2 H 4 Cl 2 ) ⇒ n5 = 7.143 mol C 2 H 4 Cl 2
U| ⇒ n = 714.3 mol C H in V 2 n = 2b100g + 2 n + 2b7.143g|W n = 114.3 mol C H out 6b714.3g = 5b100g + 6b114.3g + n + 4b7 .143g ⇒ n = 607 .1 mol HCl 2 n = 100 + 607.1 + 2b7 .143g ⇒ n = 114 .3 mol Cl b
g
15% conversion: 1 − 0.15 n1 = n 3 C balance:
H balance: Cl balance:
1
3
1
2
6
3
2
6
4
2
4
2
2
. mol Cl 2 / mol C 2 H 6 Feed Ratio : 114.3 mol Cl 2 / 714.3 mol C 2 H 6 = 016 Maximum possible amount of C2 H5 Cl: 114.3 mol Cl 2 1 mol C 2 H 5Cl n max = = 114.3 mol C 2 H 5Cl 1 mol Cl 2 Fractional yield of C2 H5 Cl:
4.51
nC2 H5 Cl n max
=
100 mol = 0.875 114.3 mol
d.
Some of the C2 H4 Cl2 is further chlorinated in an undesired side reaction: C2 H5 Cl2 + Cl2 → C2H4 Cl3 + HCl
a.
C2 H4 + H2O → C2 H5OH, 2 C2H5 OH → (C2 H5 )2 O + H2O Basis: 100 mol effluent gas 100 mol 0.433 mol C 2 H 4 / mol 0.025 mol C H OH / mol 2 5 0.0014 mol (C H ) O / mol 2 5 2 0.093 mol I / mol
n1 (mol C 2 H 4 ) n [mol H O (v)] 2 2 n 3 (mol I)
3 unknowns -2 independent atomic balances -1 I balance 0 D. F.
0.4476 mol H O (v) / mol 2
b
(1) C balance: 2 n1 = 100 2∗0.433 + 2∗0.025 + 4∗0.0014
b
g
(2) H balance: 4 n1 + 2n 2 = 100 4∗0.433 + 6∗0.025 + 10∗0.0014 + 2∗0.4476
b
g
g
(3) O balance: n 2 = 100 0.025 + 0.0014 + 0.4476 Note; Eq. (1)∗2 + Eq. (3)∗2 = Eq. (2 ) ⇒2 independent atomic balances (4) I balance: n3 = 9.3
4-43
4.51 (cont'd) b. (1) ⇒ n1 = 46.08 mol C 2 H 6 (3) ⇒ n 2 = 47.4 mol H 2 O ⇒ Reactor feed contains 44.8% C 2 H 6 , 46.1% H 2 O, 9.1% I (4) ⇒ n 3 = 9.3 mol I
U| V| W
46.08 − 43.3 × 100% = 6.0% 46.08 If all C2 H4 were converted and the second reaction did not occur, n C2 H5OH
% conversion of C2 H4 :
d
⇒ Fractional Yield of C2 H5 OH: n C2 H5OH / nC2 H5 OH
i
max
b
d
g
i
max
= 46.08 mol
= 2.5 / 46.08 = 0.054
Selectivity of C2 H5 OH to (C2 H5)2 O: 2.5 mol C 2 H 5OH = 17.9 mol C 2 H 5OH / mol (C2 H 5 ) 2 O 0.14 mol (C 2 H 5 ) 2 O c. 4.52
Keep conversion low to prevent C2 H5OH from being in reactor long enough to form significant amounts of (C2 H5)2 O. Separate and recycle unreacted C2H4 .
bg
bg
bg
bg
CaF2 s + H 2SO 4 l → CaSO 4 s + 2HF g 1 metric ton acid 1000 kg acid 0.60 kg HF
1 metric ton acid
1 kg acid
= 600 kg HF
Basis: 100 kg Ore dissolved (not fed) 100 kg Ore dissolved 0.96 kg CaF 2/kg 0.04 kg SiO 2/kg nA (kg 93% H2 SO4 ) 0.93 H2 SO4 kg/kg 0.07 H2 O kg/kg
n1 n2 n3 n4
(kg CaSO4) (kg HF) (kg H 4SiF6 ) (kg H 2 SO 4)
n5 (kg H2 O)
Atomic balance - Si:
b g
0.04 100 kg SiO2
28.1 kg Si 60.1 kg SiO 2
=
n 3 (kg H 4 SiF6 )
28.1 kg Si 146.1 kg H 4 SiF6
Atomic balance - F:
b g
n (kg HF) 19.0 kg F 38.0 kg F = 2 20.0 kg HF 78.1 kg CaF2 9.72 kg H 4SiF6 114.0 kg F + ⇒ n 2 = 41.2 kg HF 146.1 kg H 4 SiF6
0.96 100 kg CaF2
600 kg HF 100 kg ore diss. 41.2 kg HF
1 kg ore feed 0.95 kg ore diss.
4-44
= 1533 kg ore
⇒ n 3 = 9.72 kg H 4 SiF6
4.53
a.
C 6 H 6 + Cl 2 → C 6 H 5 Cl + HCl C 6 H 5 Cl + Cl 2 → C 6 H 4 Cl 2 + HCl C 6 H 4 Cl 2 + Cl 2 → C 6 H 3Cl 3 + HCl Convert output wt% to mol%: Basis 100 g output
species C6H 6 C 6 H 5Cl C 6 H 4 Cl 2 C 6 H 3Cl 3
g 65.0 32.0 2.5 0.5
Mol. Wt. 78.11 112.56 147.01 181.46
mol 0.832 0.284 0.017 0.003 total 1.136
mol % 73.2 25.0 1.5 0.3
Basis: 100 mol output
n1 (mol C6 H6 ) n2 (mol Cl 2) n3 (mol I)
n 4 (mol HCl(g)) n 3 (mol I) 65.0 mol C6 H6 73.2 mol C6 H6 32.0 mol C 6 H 5 Cl 25.0 mol C6 H5 Cl 2.5 Cl 2 1.5mol molCC66H H44 Cl 2 0.5 mol C H Cl 3 0.3 mol C6 H 3Cl 6
b.
3
4 -3 -1 0
unknowns atomic balances wt% Cl 2 in feed D.F.
3
C balance: 6n1 = 6 ( 73.2 + 25.0 + 1.5 + 0.3 ) ⇒ n1 = 100 mol C 6 H 6
H balance: 6 (100) = 6 ( 73.2 ) + 5 ( 25.0 ) + 4 (1.5 ) + 3 ( 0.3) + n4 ⇒ n4 = 28.9 mol HCl Cl balance: 2 n2 = 28.9 + 25.0 + 2 (1.5 ) + 3 ( 0.3) ⇒ n2 = 28.9 mol Cl 2
Theoretical C 6 H 6 = 28.9 mol Cl 2 (1 mol C6H 6 1 mol Cl 2 ) = 28.9 mol C6 H6
(100 − 28.9 ) 28.9 × 100% = 246% excess C6H6 Fractional Conversion: (100 − 73.2) 100 = 0.268 mol C6 H6 react/mol fed Excess C 6 H 6 :
Yield: (25.0 mol C 6 H 5Cl) (28.9 mol C 6 H 5Cl maximum)=0.865
g gas ⇒ 0.268 g liquid 78.11 g C6 H6 Liquid feed: (100 mol C6 H6 ) = 7811 g liquid mol C6H6
Gas feed:
28.9 mol Cl2 70.91 g Cl2 1 g gas = 2091 g gas mole Cl 2 0.98 g Cl2
c.
Low conversion ⇒ low residence time in reactor ⇒ lower chance of 2nd and 3rd reactions occurring. Large excess of C 6 H 6 ⇒ Cl 2 much more likely to encounter C 6 H 6 than substituted C 6 H 6 ⇒ higher selectivity.
d. e.
Dissolve in water to produce hydrochloric acid. Reagent grade costs much more. Use only if impurities in technical grade mixture affect the reaction rate or desired product yield.
4-45
4.54
a.
2CO 2 ⇔ 2CO + O 2
2A ⇔ 2B + C
O 2 + N 2 ⇔ 2NO
C + D ⇔ 2E
n A = n A 0 − 2ξ e1
b = bn = bn = bn = bn
y A = n A0 − 2ξ e1
n B = n B0 + 2ξ e2
yB
n C = n C 0 + ξ e1 − ξ e2 ⇒ y C n D = n D 0 − ξ e2 yD n E = n E 0 + 2ξ e2 yE
bn
ntotal = n T 0 + ξ e1
T0
g bn g bn
B0
+ 2ξ e1
C0
+ ξ e1 − ξ e 2
D0
− 1ξ e 2
E0
+ 2ξ e 2
g bn g bn
T0
+ ξ e1
T0
+ ξ e1
g bn
T0
g g
+ ξ e1
g g
T0
+ ξ e1
T0
+ ξ e1
g
= n A0 + n B0 + n C 0 + n D0 + n E 0
g
Equilibrium at 3000K and 1 atm y 2B y C y 2A
bn =
g bn + ξ − ξ g = 0.1071 bn − 2ξ g bn + ξ g
B0
+ 2ξ e 1
2
A0
e1
C0 2
e1
T0
e2
e1
yE2 ( nE 0 + 2ξe 2 ) = = 0.01493 yC y D ( nA0 + ξe 1 − ξ e 2)( nD0 − ξe 2 ) 2
E
b
f 1 = 0.1071 n A0 − 2ξ e1
b
g bn 2
T0
g b gbn
g bn + ξ − ξ g = 0U| Defines functions g − bn + 2ξ g = 0 V|W f bfξ bξ, ξ, ξg and g
+ ξ e 1 − n B 0 + 2ξ e1
f 2 = 0.01493 nC 0 + ξ e1 − ξ e 2
D0
− ξ e2
2
C0
e1
e2
2
E0
1
1
2
e2
2
1
2
b.
Given all n io ’s, solve above equations for ξ e1 and ξ e2 ⇒ n A , n B, n C, n D , n E ⇒ yA , yB, yC, yD , yE
c.
n A0 = n C0 = nD0 = 0.333, n B0 = n E0 = 0 ⇒ ξ e1 =0.0593, ξ e2 = 0.0208 ⇒ yA = 0.2027, yB = 0.1120, yC = 0.3510, yD = 0.2950, yE = 0.0393
d.
a 11d 1 + a 12 d 2 = − f 1 a f − a 22 f 1 d 1 = 12 2 a 11a 22 − a 12 a 21
bξ g
e1 new
= ξ e1 + d 1
a 21d 1 + a 22 d 2 = − f 2 a f − a 11 f 2 d 2 = 21 1 a11a 22 − a12 a 21
bξ g
e 2 new
= ξ e1 + d 2
(Solution given following program listing.) . 1 30
IMPLICIT REAL * 4(N) WRITE (6, 1) FORMAT('1', 30X, 'SOLUTION TO PROBLEM 4.57'///) READ (5, *) NA0, NB0, NC0, ND0, NE0 IF (NA0.LT.0.0)STOP WRITE (6, 2) NA0, NB0, NC0, ND0, NE0
4-46
4.54 (cont’d) 2
FORMAT('0', 15X, 'NA0, NB0, NC0, ND0, NE0 *', 5F6.2/) NTO = NA0 + NB0 + NC0 + ND0 + NE0 NMAX = 10 X1 = 0.1 X2 = 0.1 DO 100 J = 1, NMAX NA = NA0 – X1 – X1 NB = NB0 + X1 + X1 NC = NC0 + X1 – X2 ND = ND0 – X2 NE = NE0 + X2 + X2 NAS = NA ** 2 NBS = NB ** 2 NES = NE ** 2 NT = NT0 + X1 F1 = 0.1071 * NAS * NT – NBS * NC F2 = 0.01493 * NC * ND – NES A11 = – 0.4284 * NA * NT * 0.1071 * NAS – 4.0 * NB * NC – NBS A12 = NBS A21 = 0.01493 * ND A22 = –0.01493 * (NC + ND) – 4.0 * NE DEN = A11 * A22 – A12 * A21 D1 = (A12 * F2 – A22 * F1)/DEN D2 = (A21 * F1 – A11 * F2)/DEN X1C = X1 + D1 X2C = X2 + D2 WRITE (6, 3) J, X1, X2, X1C, X2C 3 FORMAT(20X, 'ITER *', I3, 3X, 'X1A, X2A =', 2F10.5, 6X, 'X1C, X2C =', * 2F10.5) IF (ABS(D1/X1C).LT.1.0E– 5.AND.ABS(D2/X2C).LT.1.0E– 5) GOTO 120 X1 = X1C X2 = X2C 100 CONTINUE WRITE (6, 4) NMAX 4 FORMAT('0', 10X, 'PROGRAM DID NOT CONVERGE IN', I2, 'ITERATIONS'/) STOP 120 YA = NA/NT YB = NB/NT YC = NC/NT YD = ND/NT YE = NE/NT WRITE (6, 5) YA, YB, YC, YD, YE 5 FORMAT ('0', 15X, 'YA, YB, YC, YD, YE =', 1P5E14.4///) GOTO 30 END $DATA 0.3333 0.00 0.3333 0.3333 0.0 0.50 0.0 0.0 0.50 0.0 0.20 0.20 0.20 0.20 0.20
SOLUTION TO PROBLEM 4.54 NA0, NB0, NC0, ND0, NE0 = 0.33 0.00 0.33 ITER = 1 X1A, X2A = 0.10000 0.10000 ITER = 2 X1A, X2A = 0.06418 0.05181 ITER = 3 X1A, X2A = 0.05969 0.02486
4-47
0.33 X1C, X1C, X1C,
0.00 X2C = 0.06418 X2C = 0.05969 X2C = 0.05937
0.05181 0.02986 0.02213
4.54 (cont’d)
ITER = 4 X1A, X2A = 0.05437 ITER = 5 X1A, X2A = 0.05931 ITER = 6 X1A, X2A = 0.05930
0.02213 0.02086 0.02083
X1C, X2C = 0.05931 X1C, X2C = 0.05930 X1C, X2C = 0.05930
YA, YB, YC, YD, YE =
2.0270E − 01
1.1197 E − 01
2 .9501E − 01
3.9319 E − 02
NA0, NB0, NC0, ND0, NE0 = 0.20 ITER = 1 X1A, X2A = 0.10000 ↓ ITER = 7 X1A, X2A = –0.02244
YA, YB, YC, YD, YE=
4.55
0.20
0.20
0.20 0.10000
0.02086 0.02083 0.02083
3.5100E − 01
0.20 X1C, X2C = 0.00012
–0.08339 X1C, X2C = – 0.02244
0.00037 –0.08339
2.5051E − 01 1.5868E − 01 2.6693E − 01 2.8989E − 01 3.3991E − 02
a.
(B) m & B 0 (kg A/h) 1 kg B/kg A fed to reactor
( P)
( A) & A0 (kg A/h) m
m& B 0 (kg A/h)
xRA (kg R/kg A)
x RA (kg R/kg A)
R → S
& 3 (kg A/h) m
m & P (kg P/h)
xR 3 (kg R/kg)
0.0075 kg R/kg P
99% conv. f m& A0 (kg A/h)
xRA (kg R/kg A)
Splitting point: 1 allowed material balance Reactor: 1 mass balance + 99% conversion of R (=> 2 equations) Mixing point: 2 allowed material balances (1 mass, 1 on R) & A0 , f , x RA ,m& B0 , m & 3 , x R 3 , m& P ) − 5 equations = 2 degrees of freedom ⇒ 7 unknowns ( m
b.
Mass balance on splitting point: mA0 = mB0 + f mA0
(1)
Mass balance on reactor: 2 mB0 = m3 99% conversion of R: xR3 m3 = 0.01 xRA mB0 Mass balance on mixing point: m3 + f mA0 = mP R balance on mixing point: xR3 m3 + xRA f mA0 = 0.0075 mP Given xRA and mP , solve simultaneously for mA0 , mB0 , f, m3 , xR3
(2) (3) (4) (5)
4-48
4.55(cont’d) c. mA0 = 2778 kg A/h mB0 = 2072 kg B/h fA = 0.255 kg bypass/kg fresh feed mP 4850 4850 4850 4850 4850 4850 4850 4850 4850
xRA 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
mA0 3327 3022 2870 2778 2717 2674 2641 2616 2596
mB0 1523 1828 1980 2072 2133 2176 2209 2234 2254
f 0.54 0.40 0.31 0.25 0.21 0.19 0.16 0.15 0.13
mP 2450 2450 2450 2450 2450 2450 2450 2450 2450
xRA 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
mA0 1663 1511 1435 1389 1359 1337 1321 1308 1298
mB0 762 914 990 1036 1066 1088 1104 1117 1127
f 0.54 0.40 0.31 0.25 0.22 0.19 0.16 0.15 0.13
f vs. x RA f (kg bypass/kg fresh feed)
d.
0.60 0.50 0.40 0.30 0.20 0.10 0.00 0.00
0.02
0.04
0.06 x R A (kg R/kg A)
4-49
0.08
0.10
0.12
4.56
a.
900 kg HCHO 1 kmol HCHO = 30.0 kmol HCHO / h h 30.03 kg HCHO
n (kmol CH OH / h) 1 3
30.0 kmol HCHO / h n2 (kmol H 2 / h) n3 (kmol CH 3OH /h)
% conversion:
30.0 = 0.60 ⇒ n1 = 50.0 kmol CH 3 OH / h n1
b.
n (kmol CH OH / h) 1 3
30.0 kmol HCHO / h
30.0 kmol HCHO / h
n2 (kmol H 2 /h)
n2 (kmol H 2 /h)
n3 (kmol CH 3OH / h)
n (kmol CH OH / h) 3 3
Overall C balance: n1 (1) = 30.0 (1) ⇒ n1 = 30.0 kmol CH3 OH/h (fresh feed) Single pass conversion:
c.
4.57
a.
30.0 = 0.60 ⇒ n 3 = 20.0 kmol CH 3 OH / h n1 + n3
n 1 + n 3 = 50.0 kmol CH3 OH fed to reactor/h Increased xsp will (1) require a larger reactor and so will increase the cost of the reactor and (2) lower the quantities of unreacted methanol and so will decrease the cost of the separation. The plot would resemble a concave upward parabola with a minimum around xsp = 60%. Convert effluent composition to molar basis. Basis: 100 g effluent: 10.6 g H2
1 mol H2 2.01 g H 2
64.0 g CO
= 5.25 mol H 2
1 mol CO 28.01 g CO
25.4 g CH3 OH
= 2.28 mol CO
⇒
H : 0.631 mol H / mol 2 2 CO: 0.274 mol CO / mol CH 3 OH: 0.0953 mol CH 3 OH / mol
1 mol CH 3OH 32.04 g CH3 OH
= 0.793 mol CH3O H
4-50
4.57 (cont’d) n4 (mol / min) 0.004 mol CH 3OH(v)/ mol x (mol CO /mol) (0.896 - x) (mol H 2 / mol) Cond. Reactor
350 mol/ min 0.631 mol CH 3OH(v)/ mol
n1 (mol CO/ min) n 2 (mol H 2 / min)
n 3 (mol CH 3OH(l) / min)
0.274 mol CO/ mol CO+ H 2 → CH 3OH 0.0953 mol H / mol 2
Condenser 3 unknowns (n3 , n4 , x) -3 balances 0 degrees of freedom Balances around condenser
Overall process 2 unknowns (n1 , n2 ) -2 independent atomic balances 0 degrees of freedom
U| V| W
n = 32.1 mol CH 3OH(l) / min CO: 350∗0.274 = n ∗ x 3 4 H : 350∗0.631 = n ∗ (0.996 − x ) ⇒ n = 318.7 mol recyc le / min 2 4 4 CH OH: 350∗0.0953 = n + 0.004∗ n x =.301 molCO / mol 3 3 4
Overall balances
U|V |W
n = 32.08 mol / min CO in feed C: n = n 1 3 ⇒ 1 H: 2n = 4n n = 64.16 mol / min H 2 in feed 2 2 2
Single pass conversion of CO:
(32.08 + 318 .72 ∗ 0.3009 ) − 350 ∗ 0.274 × 100 % = 25 .07 % (32 .08 + 318 .72 ∗ 0.3009 )
32 .08 − 0 × 100 % = 100 % 32 .08 Reactor conditions or feed rates drifting. (Recalibrate measurement instruments.) Impurities in feed. (Re-analyze feed.) Leak in methanol outlet pipe before flowmeter. (Check for it.)
Overall conversion of CO: b.
– – –
4-51
4.58
a.
Basis: 100 kmol reactor feed/hr n3 (kmol CH4 /h) 100 kmol /h Reactor n1 (kmol CH4 /h) 80 kmol CH4 /h n2 (kmol Cl2 /h) 20 kmol Cl 2 /h
n3 (kmol CH4 /h) n4 (kmol HCl /h) 5n5 (kmol CH3 Cl /h) n5 (kmol CH2 Cl 2 /h)
Cond.
Solvent
Absorb n3 (kmol CH4 /h) n4 (kmol HCl/h)
n4 (kmol HCl/h)
5n5 (kmol CH3Cl /h) Still 5n5 (kmol CH3Cl /h) n5 (kmol CH2Cl 2 /h)
n5 (kmol CH2Cl 2 /h)
Overall process: 4 unknowns (n1 , n2 , n4 , n5 ) -3 balances = 1 D.F. Mixing Point: 3 unknowns (n1 , n2 , n3 ) -2 balances = 1 D.F. Reactor: 3 unknowns (n3 , n4 , n5 ) -3 balances = 0 D.F. Condenser: 3 unknowns (n3 , n4 , n5 ) -0 balances = 3 D.F. Absorption column: 2 unknowns (n3 , n4 ) -0 balances = 2 D.F. Distillation Column: 2 unknowns (n4 , n5 ) -0 balances = 2 D.F. Atomic balances around reactor: 1) C balance : 80 = n 3 + 5n 5 + n 5 2) H balance : 320 = 4n 3 + n 4 + 15n 5 + 2n 5 ⇒ Solve for n 3 , n 4 , n 5 3) Cl balance : 40 = n 4 + 5n 5 + 2n 5
CH4 balance around mixing point: n1 = (80 – n3 ) Cl2 balance: n2 = 20 b.
c.
Solve for n1
For a basis of 100 kmol/h into reactor n1 = 17.1 kmol CH4 /h n2 = 20.0 kmol Cl2 /h n3 = 62.9 kmol CH4 /h
n4 = 20.0 kmol HCl/h 5n5 = 14.5 kmol CH3 Cl/h
(1000 kg CH3 Cl/h)(1 kmol/50.49 kg) = 19.81 kmol CH3 Cl/h Scale factor =
19 .81 kmol CH 3 Cl/h 14 .5 kmol CH 3Cl/h
= 1.366
n tot = 50 .6 kmol/h n = (17 .1)(1.366 ) = 23.3 kmol CH 4 /h Fresh feed: 1 ⇒ n 2 = (20 .0 )(1 .366 ) = 27.3 kmol Cl 2 /h 46.0 mol% CH 4 , 54.0 mole% Cl 2
Recycle: n3 = (62.9)(1.366) = 85.9 kmol CH4 recycled/h
4-52
4.59
a.
Basis: 100 mol fed to reactor/h ⇒ 25 mol O2 /h, 75 mol C2 H4 /h n1 (mol C2H 4 //h) n2 (mol O2 /h)
Seperator
reactor nC2H4 ( mol C2 H 4 /h) nO2 (mol O2 /h)
75 mol C2H 4 //h 25 mol O2 /h
n1 (mol C 2H 4 //h) n2 (mol O2 /h) n3 (mol C2H 4O /h) n4 (mol CO2 /h) n5 (mol H2O /h)
n3 (mol C2 H 4O /h)
n4 (mol CO2 /h) n5 (mol H2O /h)
Reactor 5 unknowns (n1 - n5 ) -3 atomic balances -1 - % yield -1 - % conversion 0 D.F. Strategy: 1. Solve balances around reactor to find n1 - n5 2. Solve balances around mixing point to find nO2 , nC2H4 (1) % Conversion ⇒ n1 = .800 * 75 (2) % yield: (.200 )(75) mol C 2 H 4 ×
90 mol C 2 H 4 O = n 3 (productio n rate of C 2 H 4 O) 100 mol C 2 H 4
(3) C balance (reactor): 150 = 2 n1 + 2 n3 + n4 (4) H balance (reactor): 300 = 4 n1 + 4 n3 + 2 n5 (5) O balance (reactor): 50 = 2 n2 + n3 + 2 n4 + n5 (6) O2 balance (mix pt): nO2 = 25 – n2 (7) C2 H4 balance (mix pt): nC2H4 = 75 – n1 Overall conversion of C2 H4 : 100% b.
n1 = 60.0 mol C2 H4 /h n2 = 13.75 mol O2 /h n3 = 13.5 mol C2 H4 O/h n4 = 3.00 mol CO2 /h
c. Scale factor =
n5 = 3.00 mol H2 O/h nO2 = 11.25 mol O2 /h nC2H4 = 15.0 mol C2 H4 /h 100% conversion of C2 H4
2000 lbm C 2 H 4 O 1 lb - mole C 2 H 4 O h lb − mol / h = 3.363 h 44 .05 lbm C 2 H 4 O 13 .5 mol C 2 H 4 O mol / h
nC2H4 = (3.363)(15.0) = 50.4 lb-mol C2 H4 /h nO2 = (3.363)(11.25) = 37.8 lb-mol O2 /h
4-53
4.60
a.
Basis: 100 mol feed/h 100 mol/h 32 mol CO/h 64 mol H 2 / h 4 mol N 2 / h
&11 (mol /h) nn
reactor
n&n23 (mol CH 3 OH / h)
cond.
.13 mol N 2 /mol
500 mol / h x1 (mol N 2 /mol) x2 (mol CO / mol) 1-x1-x 2 (mol H 2 / h)
n3 (mol / h) (mol/h) x1n&(mol N 2 /mol) 3 x2x(mol CO mol) (mol N2/ /mol) 1 1-x1-x 2 (mol H 2 / h) Purge
x 2 (mol CO/mol) 1 − x1 − x2 (mol H2 /mol)
Mixing point balances: total: (100) + 500 = n&1 ⇒ n&1 = 600 mol/h N2 : 4 + x1 * 500 = .13 * 600 ⇒ x1 = 0.148 mol N2 /mol Overall system balances: N2 : 4 = n&3 (0.148)
⇒ n&3 = 27 mol/h
Atomic C:32(1) = n& 2 (1) + 27 x 2 (1)
Atomic H:64(2) = n& 2 (4) + 27(1 − 0.148 − x2 )(2)
=>
n& 2 = 24.3 mol CH3 OH/h x 2 = 0.284 mol CO/mol
Overall CO conversion: 100*[32-0.284(27)]/32 = 76% Single pass CO conversion: 24.3/ (32+.284*500) = 14% b.
Recycle: To recover unconsumed CO and H2 and get a better overall conversion. Purge: to prevent buildup of N2 .
4.61
a.
3H2NH -> NH 3 2 N2 + 2N 3H22 +→ 3 (1-yp) (1-fsp) n1 (mol N 2) (1-yp) (1-fsp) 3n1 (mol H 2) (1-yp) n2 (mol I)
1 mol (1-XI0)/4 (mol N2 / mol) 3/4 (1-XI0) (mol H2 / mol) XI0 (mol I / mol)
nr (mol) n1 (mol N 2) 3n1 (mol H2) n2 (mol I)
Reactor
4-54
(1-fsp) n1 (mol N 2) (1-fsp) 3n1 (mol H 2) n2 (mol I)
nr (mol) (1-fsp) n1 (mol N2) (1-fsp) 3n1 (mol H 2) n2 (mol I) 2 fsp n1 (mol NH 3)
yp (1-fsp) n1 (mol N 2) yp (1-fsp) 3n1 (mol H 2) yp n2 (mol I)
Condenser
np (mol) 2 fsp n1 (mol NH 3)
4.61 (cont’d) At mixing point: N2 : (1-XI0)/4 + (1-yp )(1-fsp ) n1 = n1 I: XI0 + (1-yp ) n2 = n2 Total moles fed to reactor: nr = 4n1 + n2 Moles of NH3 produced: np = 2fsp n1 Overall N2 conversion:
b.
(1 − X I 0 ) / 4 − y p (1 − f sp ) n 1 (1 − X I 0 ) / 4
XI0 = 0.01 fsp = 0.20 yp = 0.10 n1 = 0.884 mol N2 n2 = 0.1 mol I
× 100 %
nr = 3.636 mol fed np = 0.3536 mol NH3 produced N2 conversion = 71.4%
c.
Recycle: recover and reuse unconsumed reactants. Purge: avoid accumulation of I in the system.
d.
Increasing XI0 results in increasing nr , decreasing np , and has no effect on fov . Increasing fsp results in decreasing nr , increasing np , and increasing fov . Increasing yp results in decreasing nr, decreasing np , and decreasing fov . Optimal values would result in a low value of nr and fsp , and a high value of np , this would give the highest profit. XI0 0.01 0.05 0.10 0.01 0.01 0.01 0.10 0.10 0.10
fsp 0.20 0.20 0.20 0.30 0.40 0.50 0.20 0.20 0.20
yp 0.10 0.10 0.10 0.10 0.10 0.10 0.20 0.30 0.40
nr 3.636 3.893 4.214 2.776 2.252 1.900 3.000 2.379 1.981
4-55
np 0.354 0.339 0.321 0.401 0.430 0.450 0.250 0.205 0.173
fov 71.4% 71.4% 71.4% 81.1% 87.0% 90.9% 55.6% 45.5% 38.5%
4.62
a.
i - C 4 H 10 + C 4 H 8 = C 8 H 18
D
Basis : 1-hour operation
n 2 (n-C 4H10) n 3 (i-C 4H 10) n 1 (C 8H 18) m4 (91% H 2SO 4)
F decanter
E
Units of n: kmol Units of m: kg
still
n 1 (C 8H 18) n 2 (n-C 4H 10) n 3 (i-C 4H 10)
n 5 (n-C 4H 10) n 6 (i-C 4H 10) n 7 (C 8H 18) m8 (91% H 2SO 4)
reactor
C B
n 1 (C 8H 18) n 2 (n-C4 H10)
P
m 4 (kg 91% H 2SO 4)
40000 kg A n 0 kmol 0.25 i-C4 H 10 0.50 n-C4 H10 0.25 C4 H 8
n 3 (i-C 4H 10)
Calculate moles of feed
b gb
g b gb
g
M = 0.25 M L− C 4 H10 + 0.50 M n− C4 H10 + 0.25 M C 4 H8 = 0.75 58.12 + 0.25 56.10 = 57 .6 kg kmol
b
gb
g
n 0 = 40000 kg 1 kmol 57.6 kg = 694 kmol
b gb g
Overall n - C4 H10 balance: n 2 = 0.50 694 = 347 kmol n - C4 H 10 in product C 8 H 18 balance: n1 =
b0.25gb 694g kmol C H 4
8
react 1 mol C8 H 18 1 mol C 4 H 8
= 1735 . kmol C 8 H 8 in product
b
At (A), 5 mol i - C 4 H 10 1 mole C 4 H 8 ⇒ n mol i - C 4 H 10
b
694g = 867.5 kmol g = b15gb404.252gb44 3 A
i -C 4 H10 at
moles C 4 H8 at A =173.5
g
Note: n mol C 4 H 8 = 173.5 at (A), (B) and (C) and in feed
b gb g
i - C 4 H 10 balance around first mixing point ⇒ 0.25 694 + n3 = 867 .5 ⇒ n3 = 694 kmol i - C 4 H 10 recycled from still
At C, 200 mol i - C4 H 10 mol C4 H 8
b
⇒ n mol i - C4 H 10
g = b200gb1735. g = 34,700 kmol i - C H 4
C
4-56
10
bA g and b B g
4.62 (cont’d) i - C 4 H 10 balance around second mixing point ⇒ 8675 . + n6 = 34 ,700 ⇒ n6 = 33,800 kmol C 4 H 10 in recycle E
Recycle E: Since Streams (D) and (E) have the same composition,
b g = n bmoles i - C H g ⇒ n n bmoles n - C H g n b moles i - C H g n bmoles C H g n = ⇒ n = 8460 kmol C H n b moles C H g n n 5 moles n - C 4 H 10
E
6
4
10 E
10 D
3
4
10 D
5
2
4
7
8
18 E
6
1
8
18 D
3
7
4
= 16,900 kmol n - C 4 H10
18
Hydrocarbons entering reactor:
kg I b347 + 16900gb kmol n - C H g FGH 58.12 kmol JK kg I kg I F F + b867 .5 + 33800gb kmol i - C H g G 58.12 H kmol JK + 173.5 kmol C H GH 56.10 kmolJK kg I F + 8460 kmol C H G114.22 J = 4.00 × 10 kg . H kmol K H SO solution entering reactor 4.00 × 10 kg HC 2 kg H SO baq g = band leaving reactor g 1 kg HC = 8.00 × 10 kg H SO b aq g m bH SO in recycle g n b n - C H in recycle g = 8.00 × 10 bH SO leaving reactor g n + n bn - C H leaving reactorg ⇒ m = 7 .84 × 10 kg H SO b aq g in recycle E 4
10
4
10
4
8
6
8
2
18
6
4
2
4
6
2
8
2
4
4
5
4
10
5
4
10
6
2
4
2
6
8
2
4
m4 = H 2 SO 4 entering reactor − H 2 SO 4 in E
b g
= 1.6 × 10 5 kg H 2 SO 4 aq recycled from decanter
b g d ib g d1.6 × 10 ib0.09gkg H O b1 kmol 18.02 kg g = 799 kmol H
⇒ 1.6 × 10 5 0.91 kg H 2SO 4 1 kmol 98.08 kg = 1480 kmol H 2SO 4 in recycle 5
2
2O
from decanter
Summary: (Change amounts to flow rates) Product: 173.5 kmol C8 H 18 h , 347 kmol n - C 4 H10 h Recycle from still: 694 kmol i - C 4 H 10 h Acid recycle: 1480 kmol H 2 SO 4 h , 799 kmol H 2 O h Recycle E: 16,900 kmol n - C 4 H 10 h , 33,800 kmol L - C4 H10 h , 8460 kmol C 8 H18 h, 7.84 × 10 6 kg h 91% H 2 SO 4 ⇒ 72,740 kmol H 2SO 4 h , 39,150 kmol H 2 O h
4-57
4.63
a.
A balance on ith tank (input = output + consumption) v& L min C A , i−1 mol L = vC & Ai + kC Ai C Bi mol liter ⋅ min V L
b
g
b
g
b
gbg
E ÷ v&, note V / v& = τ
C A, i− 1 = C Ai + kτ C Ai C Bi
B balance. By analogy, C B, i −1 = CBi + k τ C Ai CBi Subtract equations ⇒ CBi − C Ai = CB, i−1 − C A, i−1
=
A from balances on bi −1g tank
CB , i −2 − C A, i − 2 =K = CB 0 − C A0
st
b.
C Bi − C Ai = CB 0 − C A0 ⇒ C Bi = C Ai + C B0 − C A 0 . Substitute in A balance from part (a).
b
g
C A, i−1 = C Ai + kτ C Ai C Ai + CB0 − C A0 . Collect terms in C 2Ai , C 1Ai , C 0Ai . 2 C Ai
b
g
k τ + C AL 1 + kτ C B0 − C A0 − C A, i−1 = 0
b
g
⇒ α C 2AL + β C AL + γ = 0 where α = kτ , β = 1 + k τ CB 0 − C A0 , γ = −C A, i−1 − β + β 2 − 4αγ
(Only + rather than ±: since αγ is negative and the 2α negative solution would yield a negative concentration.)
Solution: C Ai =
c. k= v= V= CA0 = CB0 = alpha = beta =
36.2 5000 2000 0.0567 0.1000 14.48 1.6270
N 1 2 3 4 5 6 7 8 9 10 11 12 13 14
gamma -5.670E-02 -2.791E-02 -1.512E-02 -8.631E-03 -5.076E-03 -3.038E-03 -1.837E-03 -1.118E-03 -6.830E-04 -4.182E-04 -2.565E-04 -1.574E-04 -9.667E-05 -5.939E-05
CA(N) 2.791E-02 1.512E-02 8.631E-03 5.076E-03 3.038E-03 1.837E-03 1.118E-03 6.830E-04 4.182E-04 2.565E-04 1.574E-04 9.667E-05 5.939E-05 3.649E-05
xA(N) 0.5077 0.7333 0.8478 0.9105 0.9464 0.9676 0.9803 0.9880 0.9926 0.9955 0.9972 0.9983 0.9990 0.9994
(xmin = 0.50, N = 1), (xmin = 0.80, N = 3), (x min = 0.90, N = 4), (xmin = 0.95, N = 6), (xmin = 0.99, N = 9), (xmin = 0.999, N = 13). As xmin → 1, the required number of tanks and hence the process cost becomes infinite. d.
(i) k increases ⇒ N decreases (faster reaction ⇒ fewer tanks) ( ii) v& increases ⇒ N increases (faster throughput ⇒ less time spent in reactor ⇒ lower conversion per reactor) (iii) V increases ⇒ N decreases (larger reactor ⇒ more time spent in reactor ⇒ higher conversion per reactor)
4-58
4.64
a.
Basis: 1000 g gas Species
m (g)
MW
n (mol)
mole % (wet)
mole % (dry)
C3 H8
800
44.09
18.145
77.2%
87.5%
C4 H10
150
58.12
2.581
11.0%
12.5%
H2 O
50
18.02
2.775
11.8%
Total
1000
23.501
100%
100%
Total moles = 23.50 mol, Total moles (dry) = 20.74 mol Ratio: 2.775 / 20.726 = 0.134 mol H 2 O / mol dry gas
b.
C3 H8 + 5 O2 → 3 CO2 + 4 H2 O, C4 H10 + 13/2 O2 → 4 CO2 + 5 H2O Theoretical O2 : C3H8: C 4 H 10 :
100 kg gas 80 kg C 3 H 8 1 kmol C 3 H 8 5 kmol O 2 = 9.07 kmol O 2 / h h 100 kg gas 44.09 kg C3 H 8 1 kmol C 3 H8 100 kg gas 15 kg C 4 H10 1 kmol C 4 H 10 6.5 kmol O 2 = 1.68 kmol O 2 / h h 100 kg gas 58.12 kg C4 H 10 1 kmol C4 H10
Total: (9.07 + 1.68) kmol O2 /h = 10.75 kmol O2 /h Air feed rate :
10.75 kmol O 2 1 kmol Air 1.3 kmol air fed = 66.5 kmol air / h h .21 kmol O 2 1 kmol air required
The answer does not change for incomplete combustion 4.65
5 L C 6 H 14 0.659 kg C 6 H 14 1000 mol C 6 H14 = 38.3 mol C6 H14 L C 6 H 14 86 kg C 6 H 14 4 L C 7 H 16 0.684 kg C 7 H 16 1000 mol C7 H16 = 27.36 mol C 7 H 16 L C 7 H 16 100 kg C 7 H16
C6 H14 +19/2 O2 → 6 CO2 + 7 H2 O
C6 H14 +13/2 O2 → 6 CO + 7 H2 O
C7 H16 + 11 O2 → 7 CO2 + 8 H2 O
C7 H16 + 15/2 O2 → 7 CO + 8 H 2 O
Theoretical oxygen: 38.3 mol C6 H14
9.5 mol O 2 27.36 mol C7 H 16 + mol C 6 H14
11 mol O 2 = 665 mol O2 required mol C 7 H16
O2 fed: (4000 mol air )(.21 mol O2 / mol air) = 840 mol O2 fed Percent excess air:
840 − 665 × 100% = 26.3% excess air 665
4-59
4.66
CO +
1 O 2 → CO 2 2
H2 +
1 O2 → H2O 2
175 kmol/h 0.500 kmol N2/kmol x (kmol CO/mol) (0.500–x) (kmol H2/kmol) 20% excess air
Note: Since CO and H 2 each require 0 .5 mol O 2 / mol fuel for complete combustion, we can calculate the air feed rate without determining x C O . We include its calculation for illustrative purposes.
b
g
A plot of x vs. R on log paper is a straight line through the points R1 = 10.0 , x1 = 0.05 and
bR
2
g
= 99.7 , x 2 = 1.0 .
b
g b
g ln a = ln b1.0g − 1.303lnb99.7 g = −6.00 a = expb−6.00 g = 2.49 × 10 −3
b = ln 1.0 0.05 ln 99.7 10.0 = 1.303
ln x = b ln R + ln a
@
x = a Rb R = 38.3 ⇒ x = 0.288
⇒ x = 2.49 × 10 −3 R1.303
moles CO mol
Theoretical O 2 : 175 kmol 0.288 kmol CO 0.5 kmol O2 h kmol kmol CO +
Air fed:
4.67 a.
175 kmol h
0.212 kmol H 2 kmol
kmol O2 0.5 kmol O2 = 43.75 kmol H 2 h
43.75 kmol O 2 required
1 kmol air
1.2 kmol air fed
h
0.21 kmol O 2
1 kmol air required
CH 4 + 2O 2 → CO 2 + 2H 2 O 7 O → 2CO 2 + 3H 2 O 2 2 C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O 13 O → 4CO 2 + 5H 2 O 2 2
Theoretical O2 :
b g
0.944 100 kmol CH 4 h
b g
kmol air h
100 kmol/h 0.944 CH4 0.0340 C2H6 0.0060 C3H8 0.0050 C4H10
C2H6 +
C 4 H 10 +
= 250
17% excess air n a (kmol air/h) 0.21 O2 0.79 N2
b g
2 kmol O 2 0.0340 100 kmol C 2 H 6 3.5 kmol O 2 + 1 kmol CH 4 h 1 kmol C 2 H 6
b g
0.0060 100 kmol C 3 H 8 5 kmol O 2 0.0050 100 kmol C 4 H 10 6.5 kmol O 2 + h 1 kmol C 3 H 3 h 1 kmol C 4 H 10 = 207.0 kmol O 2 h +
4- 60
4.67 (cont’d) Air feed rate: n f =
207.0 kmol O 2
1 kmol air
h
0.21 kmol O2
b
gb
1.17 kmol air fed
gb
kmol air req.
= 1153 kmol air h
g
b.
n a = n f 2 x1 + 3.5 x2 + 5 x3 + 6.5 x4 1 + Pxs 100 1 0.21
c.
n& f = aR f , (n& f = 75.0 kmol / h, R f = 60) ⇒ n& f = 1.25 R f n& a = bRa , ( n&a = 550 kmol / h, Ra = 25) ⇒ n&a = 22 .0 Ra
x i = kAi ⇒
∑x
i
=k
i
∑A
i
=1 ⇒ k =
i
1
∑A
i
i
⇒ xi =
Ai , i = CH 4 , C 2 H 4 , C 3 H 8 , C 4 H 10 Ai
∑ i
4.68
Run 1 2 3
Pxs 15% 15% 15%
Rf 62 83 108
A1 248.7 305.3 294.2
A2 19.74 14.57 16.61
A3 6.35 2.56 4.78
A4 1.48 0.70 2.11
Run 1 2 3
nf 77.5 103.8 135.0
x1 0.900 0.945 0.926
x2 0.0715 0.0451 0.0523
x3 0.0230 0.0079 0.0150
x4 0.0054 0.0022 0.0066
na 934 1194 1592
d.
Either of the flowmeters could be in error, the fuel gas analyzer could be in error, the flowmeter calibration formulas might not be linear, or the stack gas analysis could be incorrect.
a.
C4H10 + 13/2 O2 → 4 CO2 + 5 H2O Basis:
100 mol C4H10
n CO2 (mol CO2) n H2O (mol H2O) n C4H10 (mol C4H10) n O2 (mol O2) n N2 (mol N2)
Pxs (% excess air) n air (mol air) 0.21 O2 0.79 N2 D.F. analysis 6 unknowns (n, n1, n2, n3, n4, n5) -3 atomic balances (C, H, O) -1 N2 balance -1 % excess air -1 % conversion 0 D.F.
4- 61
Ra 42.4 54.3 72.4
4.68 (cont’d) b. i) Theoretical oxygen = (100 mol C4H10)(6.5 mol O2/mol C4H10) = 650 mol O2 n air = ( 650 mol O 2 )(1 mol air / 0.21 mol O 2 ) = 3095 mol air
100% conversion ⇒ n C4H10 = 0 , n O2 = 0
b gb b b
g
U| V| W
n N2 = 0.79 3095 mol = 2445 mol 73.1% N 2 n CO2 = 100 mol C 4 H 10 react 4 mol CO 2 mol C 4 H 10 = 400 mol CO 2 12.0% CO2 n H2O = 100 mol C 4 H10 react 5 mol H 2O mol C 4 H 10 = 500 mol H 2 O 14.9% H 2 O
gb gb
g g
ii) 100% conversion ⇒ n C4H10 = 0 20% excess ⇒ nair = 1.2(3095) = 3714 mol (780 mol O2, 2934 mol N2) Exit gas: 400 mol CO2
10.1% CO2
500 mol H2O
12.6% H2O
130 mol O2
3.3% O2
2934 mol N2
74.0% N 2
iii) 90% conversion ⇒ n C4H10 = 10 mol C4H10 (90 mol C4H10 react, 585 mol O2 consumed) 20% excess: nair = 1.2(3095) = 3714 mol (780 mol O2, 2483 mol N2) Exit gas: 10 mol C4H10
0.3% C4H10
360 mol CO2
9.1% CO2
450 mol H2O (v)
4.69
a.
11.4% H2O
195 mol O2
4.9% O2
2934 mol N2
74.3% N 2
C3H8 + 5 O2 → 3 CO2 + 4 H2O
H2 +1/2 O2 → H2O
C3H8 + 7/2 O2 → 3 CO + 4 H2O Basis: 100 mol feed gas 100 mol 0.75 mol C3H8 0.25 mol H2
n 1 (mol C3H8) n 2 (mol H2) n 3 (mol CO2) n 4 (mol CO) n 5 (mol H2O) n 6 (mol O2) n 7 (mol N2)
n 0 (mol air) 0.21 mol O2/mol 0.79 mol N2/mol Theoretical oxygen:
75 mol C 3H 8
5 mol O 2 25 mol H 2 0.50 mol O 2 + = 387.5 mol O 2 mol C 3 H 8 mol H 2
4- 62
4.69 (cont’d) Air feed rate: n 0 =
387.5 mol O 2 1 kmol air 1.25 kmol air fed = 2306.5 mol air h 0.21 kmol O2 1 kmol air req'd.
90% propane conversion ⇒ n1 = 0.100(75 mol C3 H 8 ) = 7.5 mol C 3 H 8 (67.5 mol C 3 H 8 reacts) 85% hydrogen conversion ⇒ n 2 = 0150 . (25 mol C 3 H 8 ) = 3.75 mol H 2 95% CO 2 selectivity ⇒ n 3 =
0.95(67.5 mol C 3 H 8 react) 3 mol CO 2 generated mol C3 H 8 react
= 192.4 mol CO 2 5% CO selectivity ⇒ n3 =
0.05( 67.5 mol C 3 H 8 react) 3 mol CO generated = 10.1 mol CO mol C 3 H 8 react
FG H
H balance: (75 mol C 3H 8 ) 8
IJ K
mol H + ( 25 mol H 2 )(2) mol C 3 H 8
= (7.5 mol C 3 H 8 )(8 ) + ( 3.75 mol H 2 )(2 ) + n 5 ( mol H 2 O)(2) ⇒ n5 = 291.2 mol H 2 O
mol O ) = (192.4 mol CO2 )( 2) mol O 2 + (10.1 mol CO)(1) + (2912 . mol H 2 O)(1) + 2 n6 ( mol O 2 ) ⇒ n6 = 1413 . mol O2
O balance: (0.21 × 2306.5 mol O 2 )(2
N 2 balance: n 7 = 0.79 (2306.5) mol N 2 = 1822 mol N 2
Total moles of exit gas = (7.5 + 3.75 + 192.4 + 10.1 + 291.2 + 141.3 + 1822) mol = 2468 mol CO concentration in exit gas =
b.
10.1 mol CO × 10 6 = 4090 ppm 2468 mol
If more air is fed to the furnace, (i)
more gas must be compressed (pumped), leading to a higher cost (possibly a larger pump, and greater utility costs)
(ii) The heat released by the combustion is absorbed by a greater quantity of gas, and so the product gas temperature decreases and less steam is produced.
4- 63
4.70
a.
C5H12 + 8 O2 → 5 CO2 + 6 H2O Basis: 100 moles dry product gas n 1 (mol C5H12)
100 mol dry product gas (DPG) 0.0027 mol C5H12/mol DPG 0.053 mol O2/mol DPG 0.091 mol CO2/mol DPG 0.853 mol N2/mol DPG n 3 (mol H2O)
Excess air n 2 (mol O2) 3.76n 2 (mol N2)
3 unknowns (n1, n2, n3) -3 atomic balances (O, C, H) -1 N2 balance -1 D.F. ⇒ Problem is overspecified b.
N2 balance: 3.76 n2 = 0.8533 (100) ⇒ n2 = 22.69 mol O2 C balance: 5 n1 = 5(0.0027)(100) + (0.091)(100) ⇒ n1 = 2.09 mol C5H12 H balance: 12 n1 = 12(0.0027)(100) + 2n3 ⇒ n3 = 10.92 mol H2O O balance: 2n2 = 100[(0.053)(2) + (0.091)(2)] + n3 ⇒ 45.38 mol O = 39.72 mol O Since the 4th balance does not close, the given data cannot be correct.
c. n 1 (mol C5H12)
100 mol dry product gas (DPG) 0.00304 mol C5H12/mol DPG 0.059 mol O2/mol DPG 0.102 mol CO2/mol DPG 0.836 mol N2/mol DPG n 3 (mol H2O)
Excess air n 2 (mol O2) 3.76n 2 (mol N2)
N2 balance: 3.76 n2 = 0.836 (100) ⇒ n2 = 22.2 mol O2 C balance: 5 n1 = 100 (5*0.00304 + 0.102) ⇒ n1 = 2.34 mol C5H12 H balance: 12 n1 = 12(0.00304)(100) + 2n3 ⇒ n3 = 12.2 mol H2O O balance: 2n2 = 100[(0.0590)(2) + (0.102)(2)] + n3 ⇒ 44.4 mol O = 44.4 mol O √ 2.344 − 100 × 0 .00304 = 0.870 mol react/mol fed 2 .344 Theoretical O2 required: 2.344 mol C5H12 (8 mol O2/mol C5H12) = 18.75 mol O2
Fractional conversion of C5H12:
% excess air:
22.23 mol O 2 fed - 18.75 mol O 2 required × 100 % = 18 .6% excess air 18.75 mol O2 required
4- 64
4.71
a.
12 L CH 3 OH 1000 ml 0 .792 g mol = 296 .6 mol CH 3 OH / h h L ml 32 .04 g
CH3OH + 3/2 O2 → CO2 +2 H2O, CH3OH + O2 → CO +2 H2O n& 2 ( mol dry gas / h) 0.0045 mol CH3OH(v)/mol DG 0.0903 mol CO2/mol DG 0.0181 mol CO/mol DG x (mol N2/mol DG) (0.8871–x) (mol O2/mol DG) n& 3 ( mol H 2 O(v) / h)
296.6 mol CH3OH(l)/h n&1 (mol O 2 / h) 3.76n&1 (mol N 2 / h)
4 unknowns ( n&1 , n& 2 , n& 3 , x ) – 4 balances (C, H, O, N2) = 0 D.F. b.
Theoretical O2: 296.6 (1.5) = 444.9 mol O2 / h C balance: 296.6 = n& 2 (0.0045 + 0.0903 + 0.0181) ⇒ n& 2 = 2627 mol/h
H balance: 4 (296.6) = n& 2 (4*0.0045) + 2 n& 3 ⇒ n& 3 = 569.6 mol H2O / h
O balance : 296.6 + 2n 1 = 2627[0.0045 + 2(0.0903) + 0.0181 + 2(0.8871 - x)] + 569.6 N2 balance: 3.76 n& 1 = x ( 2627)
Solving simultaneously ⇒ n&1 = 574.3 mol O 2 / h, x = 0.822 mol N 2 / mol DG Fractional conversion:
296 .6 − 2627 (0.0045 ) = 0.960 mol CH3 OH react/mol fed 296 .6
574 .3 − 444 .9 × 100 % = 29.1% 444 .9 569 .6 mol H 2 O Mole fraction of water: = 0.178 mol H 2 O/mol (2627 + 569 .6 ) mol
% excess air:
4.72
c.
Fire, CO toxicity. Vent gas to outside, install CO or hydrocarbon detector in room, trigger alarm if concentrations are too high
a.
G.C. Say ns mols fuel gas constitute the sample injected into the G.C. If xCH4 and xC 2 H 6 are the mole fractions of methane and ethane in the fuel, then
b g b n b mol gx bmol CH
gb g = 20 mol gb1 mol C 1 mol CH g 85
ns mol x C 2 H6 mol C 2 H 2 mol 2 mol C 1 mol C2 H 6
E
s
CH 4
b b mol CH
4
4
x C 2 H6 mol C 2 H 6 mol fuel xC H4
4
mol fuel
g
g = 0.1176 mole C H 2
4- 65
6
mole CH 4 in fuel gas
4.72 (cont’d)
b1.134 g H Ogb1 mol 18.02 gg = 0.126
mole H 2O 0.50 mol product gas mole product gas Basis: 100 mol product gas. Since we have the most information about the product stream composition, we choose this basis now, and would subsequently scale to the given fuel and air flow rates if it were necessary (which it is not). 2
Condensation measurement:
CH 4 + 2O 2 → CO2 + 2H 2 O 7 C 2 H6 + O2 → 2CO2 + 3H 2O 2 100 mol dry gas / h
n1 (mol CH4 ) 0.1176 n1 (mol C2H6) n2 (mol CO2)
0.126 mol H2O / mol 0..874 mol dry gas / mol 0.119 mol CO2 / mol D.G. x (mol N2 / mol) (0.881-x) (mol O2 / mol D.G.)
n3 (mol O2 / h) 376 n3 (mol N2 / h)
N 2 balance
C balance ⇒ n 2 ;
Strategy: H balance ⇒ n ; 1
b gb
g b gb
gb g
O balance
UV ⇒ n , x W 3
H balance: 4 n1 + 6 0.1176n1 = 100 0.126 2 ⇒ n1 = 5.356 mol CH 4 in fuel
⇒ 0.1176(5.356) = 0.630 mol C2H6 in fuel
b gb
g
b gb
gb
g
C balance: 5.356 + 2 0.630 + n2 = 100 0.874 0.119 ⇒ n2 = 3.784 mol CO2 in fuel Composition of fuel: 5.356 mol CH 4 , 0.630 mol C 2 H 6 , 3.784 mols CO2 ⇒ 0.548 CH 4 , 0.064 C 2 H 6 , 0.388 CO 2
b gb
g
N 2 balance: 3.76n3 = 100 0.874 x
b gb
g
b gb
g b gb
gb g
b
O balance: 2 3.784 + 2 n3 = 100 0126 . + 100 0.874 2 0119 . + 0.881 − x
g
Solve simultaneously: n3 = 18.86 mols O2 fed , x = 0.813 5.356 mol CH 4 2 mol O2 0.630 mol C2 H 6 3.5 mol O2 Theoretical O2 : + 1 mol CH 4 1 mol CH 4 = 12 .92 mol O2 required Desired O2 fed:
(5.356 + 0.630 + 3.784) mol fuel 7 mol air 0.21 mol O 2 = 14.36 mol O2 1 mol fuel mol air
Desired % excess air:
b.
Actual % excess air:
14 .36 − 12 .92 × 100 % = 11 % 12 .92
18 .86 − 12 .92 × 100 % = 46% 12 .92
Actual molar feed ratio of air to fuel:
(18 .86 / 0.21) mol air = 9 :1 9 .77 mol feed
4- 66
4.73
a.
C3H8 +5 O2 → 3 CO2 + 4 H2O, C4H10 + 13/2 O2 → 4 CO2 + 5 H2O Basis 100: mol product gas n 1 (mol C3H8) n 2 (mol C4H10)
100 mol 0.474 mol H2O/mol x (mol CO2/mol) (0.526–x) (mol O2/mol)
n 3 (mol O2) Dry product gas contains 69.4% CO2 ⇒
x 69 .4 = ⇒ x = 0.365 mol CO2 /mol 0.526 − x 30 .6
3 unknowns (n 1, n 2, n 3) – 3 balances (C, H, O) = 0 D.F. O balance: 2 n 3 = 152.6 ⇒ n 3 = 76.3 mol O2 C balance : 3 n 1 + 4 n 2 = 36.5 n = 7.1 mol C 3 H8 ⇒ 1 ⇒ 65 .1 % C 3 H8 , 34.9% C 4 H10 H balance : 8 n1 + 10 n 2 = 94.8 n 2 = 3.8 mol C 4 H10
b.
n c=100 mol (0.365 mol CO2/mol)(1mol C/mol CO2) = 365 mol C n h = 100 mol (0.474 mol H2O/mol)(2mol H/mol H2O)=94.8 mol H ⇒ 27.8%C, 72.2% H From a: 7.10 mol C 3 H 8
3.80 mol C 4 H10 4 mol C 3 mol C + mol C 3 H 8 mol C 4 H10
7.10 mol C3 H 8 11 mol (C + H) 3.80 mol C 4 H10 14 mol (C + H) + mol C 3 H 8 mol C 4 H10
4.74
Basis: 100 kg fuel oil Moles of C in fuel:
100 kg 0.85 kg C 1 kmol C = 7.08 kmol C kg 12.01 kg C
Moles of H in fuel:
100 kg 0.12 kg H 1 kmol H = 12.0 kmol H kg 1 kg H
Moles of S in fuel:
100 kg 0.017 kg S 1 kmol S = 0.053 kmol S kg 32.064 kg S
1.3 kg non-combustible materials (NC)
4- 67
× 100 % = 27.8% C
4.74 (cont’d)
100 kg fuel oil 7.08 kmol C 12.0 kmol H 0.053 kmol S 1.3 kg NC (s) 20% excess air n 1 (kmol O2) 3.76 n 1 (kmol N2)
n 2 (kmol N2) n 3 (kmol O2) n 4 (kmol CO2) (8/92) n 4 (kmol CO) n 5 (kmol SO2) n 6 (kmol H2O)
C + O2 → CO2 C + 1/2 O2 → CO 2H + 1/2 O2 → H2O S + O2 → SO2
Theoretical O2: 7.08 kmol C 1 kmol O 2 12 kmol H .5 kmol O 2 0.053 kmol S 1 kmol O 2 + + = 10 .133 kmol O 2 1 kmol C 2 kmol H 1 kmol S
20 % excess air: n1 = 1.2(10.133) = 12.16 kmol O2 fed O balance: 2 (12.16) = 2 (6.5136) + 0.5664 + 2 (0.053) + 6 + 2 n3 ⇒ n3 = 2.3102 kmol O2 C balance: 7.08 = n4+8n4/92 ⇒ n4 = 6.514 mol CO2 ⇒ 8 (6.514)/92 = 0.566 mol CO S balance: n5 = 0.53 kmol SO 2 H balance: 12 = 2n6 ⇒ n6 = 6.00 kmol H2O N2 balance: n2 = 3.76(12.16) = 45.72 kmol N2 Total moles of stack gas = (6.514 + 0.566 + 0.053 + 6.00 + 2.310 + 45.72) kmol = 61.16 kmol ⇒ 10.7% CO, 0.92% CO, 0.087% SO 2 , 9.8% H 2 O, 3.8% O 2 , 74.8% N 2
4.75 a. Basis: 5000 kg coal/h; 50 kmol air min = 3000 kmol air h 5000 kg coal / h 0.75 kg C / kg 0.17 kg H / kg 0.02 kg S / kg 0.06 kg ash / kg
C + 02 --> CO2 2H + 1/2 O2 -->H2O S + O 2 --> SO2 C + 1/2 O2 --> CO
3000 kmol air / h 0.21 kmol O2 / kmol 0.79 kmol N2 / kmol
Theoretical O 2:
C:
b
g
n1 (kmol O2 / h) n2 (kmol N2 / h) n3 (kmol CO 2 / h) 0.1 n3 (kmol CO / h) n4 (kmol SO2 / h) n5 (kmol H2O / h)
mo kg slag / h
0.75 5000 kg C
1 kmol C
1 kmol O2
h
12.01 kg C
1 kmol C
4- 68
= 312.2 kmol O 2 h
4.75 (cont’d) H: S:
b
g
0.17 5000 kg H 1 kmol H 1 kmol H 2O h
1.01 kg H
2 kmol H
0.02 5000 kg S
1 kmol S
1 kmol O2
h
32.06 kg S
1 kmol S
b
g
1 kmol O 2 2 kmol H2 O
= 210.4 kmol O 2 h
= 3.1 kmol O2/h
Total = (312.2+210.4 + 3.1) kmol O2/h = 525.7 kmol O 2 h
b g
O 2 fed = 0.21 3000 = 630 kmol O2 h
Excess air:
630 − 525.7 × 100% = 19.8% excess air 525.7
b. Balances: 0.94 0.75 5000 kg C react 1 kmol C C: = n& 3 + 0.1n&3 h 12 .01 kg C
b gb gb
g
⇒ n&3 = 266.8 kmol CO 2 h , 0.1n& 3 = 26.7 kmol CO h
H:
b0.17gb5000g kg H
S:
(from part a)
N2 :
O:
h
1 kmol H 1 kmol H 2 O 1.01 kg H
b
2 kmol H
3.1 kmol O 2 for SO2
g
= n5 ⇒ n5 = 420.8 kmol H 2O h
1 kmol SO2
h
1 kmol O 2
. kmol SO 2 h = n& 4 ⇒ n&4 = 31
b0.79gb3000g kmol N h = n& ⇒ n& = 2370 kmol N h b0.21g(3000)b2g = 2n& + 2b2668. g + 1b 26.68g + 2b31. g + b1gb420.8g 2
2
2
2
1
⇒ n&1 = 136.4 kmol O 2 / h Stack gas total = 3223 kmol h Mole fractions: xC O = 26.7 3224 = 8.3 × 10 − 3 mol CO mol xSO 2 = 31 . 3224 = 9.6 × 10 −4 mol SO 2 mol
c.
1 SO 2 + O2 → SO 3 2 SO 3 + H 2O → H2SO 4
3.1 kmol SO 2 1 kmol SO 3 1 kmol H2SO 4 h
1 kmol SO 2
1 kmol SO 3
4- 69
98.08 kg H 2SO 4 kmol H2SO 4
= 304 kg H2SO 4 h
4.76 a.
Basis: 100 g coal as received (c.a.r.). Let a.d.c. denote air-dried coal; v.m. denote volatile matter 100 g c. a.r. 1.147 g a. d.c. 1.207 g c. a. r. 95.03 g a. d. c
= 95.03 g air - dried coal; 4.97 g H 2 O lost by air drying
b1.234 − 1.204g g H O = 2.31 g H O lost in second drying step 2
2
1.234 g a.d. c.
Total H 2 O = 4.97 g + 2 .31 g = 7.28 g moisture 95.03 g a. d. c
b1.347 − 0.811g g b v. m.+ H Og − 2.31 g H O = 35.50 g volatile matter 2
2
1.347 g a. d.c.
95.03 g a. d.c
0.111 g ash 1.175 g a. d.c.
= 8.98 g ash
b
g
Fixed carbon = 100 − 7.28 − 3550 . − 8.98 g = 48.24 g fixed carbon 7.28 g moisture 48.24 g fi xed carbon 35.50 g volatile matter ⇒ 8.98 g ash 100 g coal as received
7.3% moisture 48.2% fixed carbon 35.5% volatile matter 9.0% ash
b. Assume volatile matter is all carbon and hydrogen. C + CO 2 → CO 2 :
2H +
1 mol O 2
1 mol C
10 3 g
1 mol air
1 mol C
12.01 g C
1 kg
0.21 mol O 2
= 396.5 mol air kg C
1 0.5 mol O 2 1 mol H 10 3 g 1 mol air O2 → H 2O : = 1179 mol air kg H 2 2 mol H 1.01 g H 1 kg 0.21 mol O 2
Air required:
1000 kg coal 0.482 kg C 396.5 mol air kg coal + +
kg C
1000 kg 0.355 kg v.m.
6 kg C
396.5 mol air
kg 7 kg v. m. kg C 1000 kg 0.355 kg v.m. 1 kg H 1179 mol air kg
7 kg v. m.
4- 70
kg H
= 3.72 × 105 mol air
4.77
a.
Basis 100 mol dry fuel gas. Assume no solid or liquid products!
n1 (mol C) n2 (mol H) n3 (mol S)
100 mol dry gas C + 02 --> CO2 C + 1/2 O2 --> CO 2H + 1/2 O2 -->H2O S + O 2 --> SO2
0.720 mol CO2 / mol 0.0257 mol CO / mol 0.000592 mol SO2 / mol 0.254 mol O2 / mol
n4 (mol O2) (20% excess)
n5 (mol H2O (v))
O balance : 2 n 4 = 100 [ 2(0.720) + 0.0257 + 2 (0.000592) + 2 (0.254)] + n 5 20 % excess O 2 : (1.20) (74.57 + 0.0592 + 0.25 n 2 ] = n 4
H balance : n 2 = 2 n 5
⇒ n 2 = 183.6 mol H, n 4 = 144.6 mol O2, n 5 = 91.8 mol H2O Total moles in feed: 258.4 mol (C+H+S) ⇒ 28.9% C, 71.1% H, 0.023% S
4.78
Basis: 100 g oil Stack SO 2, N2, O 2, CO ,2H O2 (612.5 ppm SO )2
x n3 mol SO 2 (N2 , O 2, CO 2, H O) 2
0.10 (1 – x) n 5 mol SO 2 (N2 , O 2, CO 2, H O) 2
100 g oil 0.87 g C/g 0.10 g H/g 0.03 g S/g n1 mol O2 3.76 n1 mol N2 (25% excess)
furnace Alkaline solution (1 – x) n5 mol SO 2 (N2 , O 2, CO 2, H O) 2 n2 n3 n4 n5 n6
CO 2: H2 O:
b g
0.87 100 g C
mol N 2 mol O 2 mol CO 2 mol SO 2 mol H 2O
0.90 (1 – x) n5 mol SO 2
1 mol C
1 mol CO 2
12.01 g C
1 mol C
b g
0.10 100 g H 1 mol H 1 mol H 2 O 1.01 g H
scrubber
2 mol H
4- 71
FG 7.244 mol O IJ H consumed K F 2.475 mol O IJ = 4.95 mol H OG H consumed K
⇒ n 4 = 7.244 mol CO 2
⇒ n6
2
2
2
4.78 (cont’d) SO 2 :
b g
0.03 100 g S
1 mol S
1 mol SO 2
32.06 g S
1 mol S
b
⇒ n5 = 0.0936 mol SO2
g
FG 0.0956 mol O IJ H consumed K 2
25% excess O 2 : n1 = 1.25 7.244 + 2.475 + 0.0936 ⇒ 12.27 mol O 2
b
g
O 2 balance: n 3 = 12.27 mol O 2 fed − 7.244 + 2.475 + 0.0936 mol O 2 consumed = 2.46 mol O2
b
g
N 2 balance: n 2 = 3.76 12.27 mol = 46.14 mol N 2
b
g
SO 2 in stack SO 2 balance around mixing point :
F H
I K
b gb
g
b
x 0.0936 + 010 . 1 − x 0.0936 = 0.00936 + 0.0842x mol SO 2 n5
g
Total dry gas in stack (Assume no CO2 , O 2 , or N 2 is absorbed in the scrubber)
b
g
b
g
7.244 + 2.46 + 46.14 + 0.00936 + 0.0842 x = 55.85 + 0.0842 x mol dry gas
bC O g bO g 2
bN g
2
b SO g
2
b
2
g
612.5 ppm SO 2 dry basis in stack gas 0.00936 + 0.0842 x 612 .5 = ⇒ x = 0.295 ⇒ 30% bypassed 5585 . + 0.0842x 1.0 × 10 6
Basis: 100 mol stack gas
4.79
n 1 (mol C) n 2 (mol H) n 3 (mol S) n 4 (mol O 2) 3.76 n 4 (mol O 2)
a.
C + O 2 → CO 2 1 2H + O 2 → H 2O 2 S + O 2 → SO 2
b gb b gb
100 mol 0.7566 N 2 0.1024 CO 2 0.0827 H O 2 0.0575 O 2 0.000825 SO
2
g gb g
C balance: n1 = 100 0.1024 = 10.24 mol C 10.24 mol C mol C ⇒ = 0.62 H balance: n 2 = 100 0.0827 2 = 16.54 mol H 16.54 mol H mol H
The C/H mole ratio of CH 4 is 0.25, and that of C2H 6 is 0.333; no mixture of the two could have a C/H ratio of 0.62, so the fuel could not be the natural gas. b.
b gb
g
S balance: n 3 = 100 0.000825 = 0.0825 mol S 122.88 10.24 mol C 12.0 g 1 mol = 122.88 g C = 7.35 g C g H 16.54 mol H 1.01 g 1 mol = 16.71 g H ⇒ 16.71 ⇒ No. 4 fuel oil 2.65 × 100% = 1 .9% S 0.0825 mol S 32.07 g 1 mol = 2.65 g S 142.24
b b b
gb gb gb
g
g
U| V |W
g
4- 72
4.80
a.
Basis: 1 mol CpHqOr 1 mol CpHqO r no (mol S) Xs (kg s/ kg fuel)
C + 02 --> CO2 2H + 1/2 O2 -->H2O S + O2 --> SO2
P (% excess air) n1 (mol O2) 3.76 n1 (mol N2)
n2 (mol CO 2) n3 (mol SO2) n4 (mol O2) 3.76 n1 (mol N2) n5 (mol H2O (v))
Hydrocarbon mass: p (mol C) ( 12 g / mol) = 12 p (g C) q (mol H) (1 g / mol) =
q (g H)
⇒ (12 p + q + 16 r) g fuel
r (mol O) (16 g / mol) = 16 r (g O) S in feed: n o=
(12 p + q + 16r) g fuel
Theoretical O2:
X s (g S) X (12 p + q + 16 r) 1 mol S = s (mol S) (1) (1 - Xs ) (g fuel) 32.07 g S 32 .07 (1 - X s )
p (mol C) 1 mol O 2 q (mol H) 0.5 mol O 2 ( r mol O) 1 mol O 2 + − 1 mol C 2 mol H 2 mol O = (p + 1/4 q − 1/2 r) mol O 2 required
% excess ⇒ n 1 = (1 + P/100) (p +1/4 q – ½ r) mol O2 fed
(2)
C balance: n 2 = p
(3)
H balance: n 5 = q/2
(4)
S balance: n 3 = n 0
(5)
O balance: r + 2n 1 = 2n 2 + 2n 3 + 2n 4 + n 5 ⇒ n 4 = ½ (r+2n 1-2n 2-2n 3-n 5)
(6)
Given: p = 0.71, q= 1.1, r = 0.003, Xs = 0.02 P = 18% excess air (1) ⇒ n0 = 0.00616 mol S
(5) ⇒ n3 = 0.00616 mol SO 2
(2) ⇒ n1 = 1.16 mol O2 fed
(6) ⇒ n4 = 0.170 mol O2
(3) ⇒ n2 = 0.71 mol CO2
(4) ⇒ n5 = 0.55 mol H2O
(3.76*1.16) mol N2 = 4.36 mol N2 Total moles of dry product gas = n2 + n3 + n4 + 3.76 n1=5.246 mol dry product gas Dry basis composition yCO2 = (0.710 mol CO2/ 5.246 mol dry gas) * 100% = 13.5% CO2 yO2 = (0.170 / 5.246) * 100% = 3.2% O2 yN2 = (4.36 / 5.246) * 100% = 83.1% N2 ySO2 = (0.00616 / 5.246) * 106 = 1174 ppm SO 2
4- 73
CHAPTER FIVE 5.1
Assume volume additivity
1 0.400 0.600 = + ⇒ ρ = 0.719 kg L ρ 0.703 kg L 0.730 kg L
Av. density (Eq. 5.1-1):
a.
m
A
& + m0 ⇒ m & = = mt
A
mass of tank at time t
mass of empty tank
& / min) = ⇒ V(L
b.
5.2
A
A
ρO
ρD
b250 − 150gkg = 14.28 kg min b m& = mass flow rate of liquid g b10 − 3g min
14.28 kg 1L & m(kg / min) & = ⇒ V = 19.9 L min min 0.719 kg ρ (kg / L)
bg
& = 150 − 14.28 3 = 107 kg m0 = m(t) - mt
b
g
void volume of bed: 100 cm3 − 2335 . − 184 cm 3 = 50.5 cm 3
porosity: 50.5 cm3 void 184 cm3 total = 0.274 cm 3 void cm3 total bulk density: 600 g 184 cm 3 = 3.26 g cm 3
b
g
absolute density: 600 g 184 − 50.5 cm 3 = 4 .49 g cm 3
5.3 C 6 H 6 ( l)
m & B (kg / min) & = 20.0 L / min V B
& (kg / min) m & (L / min) V
C 7 H 8 (l )
m & T (kg / min) & (L / min) V T
2 2 & = ∆ V = πD ∆h = π (5.5 m) 0.15 m = 0.0594 m3 / min V ∆t 4 ∆t 4 60 min Assume additive volumes &T = V & -V & B = 59.4 − 20.0 L / min = 39.4 L / min V
b
g
& B + ρT ⋅ V & T = 0.879 kg 20.0 L + 0.866 kg 39.4 L = 51.7 kg / min & = ρB ⋅V m L min L min xB =
m & B ( 0.879 kg / L)(20.0 L / min) = = 0.34 kg B / kg m & (51.7 kg / min)
5-1
a.
F IF I e jge j hbmgGH 11 N JK GH 11 Pa JK = ρ gh
b
kg m3
m s2
kg ⋅m s2
sl
N m2
g
1 − xc 1 x = c + ⇒ check units! ρ sl ρ c ρl 1 kg crystals / kg slurry kg liquid / kg slurry = + kg slurry / L slurry kg crystals / L crystals kg liquid / L liquid L slurry L crystals L liquid L slurry = + = kg slurry kg slurry kg slurry kg slurry ∆P 2775 c. i.) ρ sl = = = 1415 kg / m 3 gh 9.8066 0.200
b.
ii.)
b gb g 1 x b1 - x g ⇒ x FG 1 − 1 IJ = FG 1 − 1 IJ = + ρ ρ ρ Hρ ρ K Hρ ρ K F 1 I GG 1415 kg / m − 12. d10001kg / m i JJ H K = 0.316 kg crystals / kg slurry x = F I GG 2.3d10001kg / m i − 1.2d10001kg / m i JJ H K c
c
c
sl
c
l
c
l
sl
3
l
3
c
3
3
m sl 175 kg 1000 L = = 123.8 L ρ sl 1415 kg / m 3 m 3
iii.) Vsl =
b
gb
g
iv.) mc = x c m sl = 0.316 kg crystals / kg slurry 175 kg slurry = 55.3 kg crystals v.) mCuSO4 =
55.3 kg CuSO 4 ⋅ 5H 2O 1 kmol 1 kmol CuSO4 159 .6 kg = 35.4 kg CuSO 4 249 kg 1 kmol CuSO4 ⋅ 5 H 2 O 1 kmol
b
g
b
gb
g
vi.) ml = 1 − x c m sl = 0.684 kg liquid / kg slurry 175 kg slurry = 120 kg liquid solution vii.)
Vl =
h(m) ρl(kg/m^3) ρc(kg/m^3) ∆P(Pa) xc ρsl(kg/m^3)
0.2 1200 2300 2353.58 0 1200.00
ml 120 kg 1000 L = = 100 L 3 ρl m3 1.2 1000 kg / m
b gd
i
d. 2411.24 0.05 1229.40
2471.80 0.1 1260.27
2602.52 0.2 1326.92
2747.84 0.3 1401.02
2772.61 0.316 1413.64
2910.35 0.4 1483.87
3093.28 0.5 1577.14
Effect of Slurry Density on Pressure Measurement
0.6 Solids Fraction
5.4
U| V| W
P1 = P0 + ρ slgh 1 P2 = P0 + ρ slgh 2 ⇒ ∆ P = P1 − P2 = ρ sl h = h1 − h 2
0.5 0.4 0.3
∆P = 2775, ρ = 0.316
0.2 0.1 0 2300.00
2500.00
2700.00
2900.00
Pressure Difference (Pa)
5-2
3100.00
5.4 (cont’d) e.
b
g d
i
Basis: 1 kg slurry ⇒ x c kg crystals , Vc m 3 crystals =
b1 - x gb kg liquid g, V d m c
l
3
b
g
x c kg crystals
d
ρ c kg / m
3
i
g i b1- ρx dgbkgkg/ mliquid i
liquid =
c
3
l
ρ sl =
5.5
b
1 kg 1 = 3 xc 1 − xc Vc + Vl m + ρc ρl
b
gd i
g
Assume Patm = 1 atm 3 $ = RT ⇒ V $ = 0.08206 m ⋅ atm 313.2 K 1 kmol = 0.0064 m 3 mol PV kmol ⋅ K 4.0 atm 103 mol
ρ=
5.6
a.
1 mol 0.0064 m air V=
1 kg
mol 10 3 g
= 4.5 kg m3
nRT 1.00 mol 0.08206 L ⋅ atm 373.2 K = = 3.06 L P mol ⋅ K 10 atm
b. % error = 5.7
29.0 g
3
b3.06L - 2.8Lg × 100% = 9.3% 2.8L
Assume Patm = 1.013 bar a.
b10 + 1.013gbar PV = nRT ⇒ n =
b.
g
PV nRT T P n = ⇒ n = V⋅ s ⋅ ⋅ s Ps Vs n s RTs T Ps Vs n=
5.8
b
kmol ⋅ K 28.02 kg N2 = 249 kg N 2 kmol 25 + 273.2 K 0.08314 m3 ⋅ bar 20.0 m 3
20.0 m 3
273K 298.2K
b10 + 1.013gbar 1.013 bar
1 kmol
28.02 kg N 2 = 249 kg N 2 kmol 22.415 m STP 3
b g
a.
R=
Ps Vs 1 atm 22.415 m 3 atm ⋅ m 3 = = 8.21 × 10 −2 n sTs 1 kmol 273 K kmol ⋅ K
b.
R=
Ps Vs 1 atm 760 torr 359.05 ft 3 torr ⋅ ft 3 = = 555 n sTs 1 lb - mole 1 atm 492 o R lb - mole ⋅o R
5-3
5.9
P = 1 atm +
10 cm H 2O
1m
1 atm = 1.01 atm 10 cm 10.333 m H 2O 2
3
& = 2.0 m = 0.40 m 3 min = 400 L min T = 25o C = 298.2 K , V 5 min & = n& mol / min ⋅ MW g / mol m
b
a.
& = m
b.
m & =
g
b
g
L & 28.02 PV 1.01 atm 400 min ⋅ MW = L⋅atm RT 0.08206 mol⋅K 298.2 K
400
L min
28.02 273 K 1 mol 298.2 K 22.4 L STP
b g
g mol
g mol
= 458 g min
= 458 g min
& P u FG mIJ = V& d m si = nRT H s K Ad m i π D 4 ⇒ u 3
2
5.10 Assume ideal gas behavior: u
1
& nR T P D2 ⋅ 2 ⋅ 1 ⋅ 12 nR & T1 P2 D2
(1.80 + 1.013 ) bar ( 7.50 cm ) (1.53 + 1.013) bar ( 5.00 cm )2
2
T P D 2 60.0 m 333.2K u 2 = u1 2 1 12 = T1 P2D 2 sec 300.2K 5.11 Assume ideal gas behavior: n =
2
2
=
b
= 165 m sec
g
PV 1.00 + 1.00 atm 5 L = = 0.406 mol L⋅ atm 0.08206 mol 300 K RT ⋅K
MW = 13.0 g 0.406 mol = 32.0 g mol ⇒ Oxygen
5.12 Assume ideal gas behavior: Say m t = mass of tank, n g = mol of gas in tank
gU|V ⇒ n = 0.009391 mol g |W m = 37.0256 g b37.062 − 37.0256gg = 3.9 g mol ⇒ Helium MW = b b
N2 : 37.289 g = m t + n g 28.02 g mol CO2 : 37.440 g = m t + n g 44.1 g mol
unknown:
5.13 a.
b.
g t
0.009391 mol
3 3 ∆V &Vstd cm3 STP min = ∆ V liters 273K 763 mm Hg 10 cm = 9253 . ∆t ∆ t min 296.2K 760 mm Hg 1L
b g
b g
U| |V straight line plot ||φ = 0.031EV& + 0.93 W 22.4 litersbSTP g 10 cm = 224 cm / min
φ
& cm3 STP min V std
5.0 9 .0 12.0
139 268 370
& std = 0.010 mol N 2 V min
d
std
3
3
3
1 mole
1L
i
φ = 0.031 224 cm 3 / min + 0.93 = 7 .9 5-4
b Fρ I si ⋅ G J Hρ K
g b
g
n kmol M(kg / kmol) 5.14 Assume ideal gas behavior ρ kg L = VL
d
i d
V2 cm 3 s = V1 cm 3
1
bg
n P = V RT
====> RT
PM
12
= V1 P1M 1T2 P2 M 2 T1
12
2
LM N
cm3 758 mm Hg 28.02 g mol 323.2K = 350 s 1800 mm Hg 2.02 g mol 298.2K
OP Q
12
= 881 cm3 s
a.
VH 2
b.
M = 0.25MCH4 + 0.75MC3 H8 = 0.25 16.05 + 0.75 44.11 = 37 .10 g mol
cm 3 Vg = 350 s
b gb
g b gb
LM b758gb28.02gb 323.2g OP N b1800gb 37.10gb2982. g Q
= 205 cm3 s
g
12
5.15 a. Reactor
∆h soap
b.
n& CO2
n& C O2 = 5.16
d
2 & PV πR 2∆ h π 0.012 m & = ⇒ V= = RT ∆t 4
i
2
1.2 m 60 s = 11 . × 10− 3 m 3 / min 7.4 s min
755 mm Hg 1 atm 1.1 ×10-3 m 3 /min 1000 mol = 0.044 mol/min 3 ⋅atm 0.08206 mkmolK 760 mm Hg 300 K 1 kmol ⋅
& air = 10.0 kg / h m
n& air (kmol / h)
n& (kmol / h) yCO 2 (kmol CO 2 / kmol)
3 & V CO2 = 20.0 m / h
n& CO (kmol / h) o 150 C, 1.5 bar 2
Assume ideal gas behavior 10.0 kg 1 kmol n& air = = 0.345 kmol air / h h 29.0 kg air n& CO2 =
& PV 1.5 bar 100 kPa 20.0 m 3 / h = = 0.853 kmol CO 2 / h 3 RT 8.314 mkmol⋅kPa 1 bar 423.2 K ⋅K
y CO2 × 100% =
b
0.853 kmol CO 2 / h × 100% = 71.2% 0.853 kmol CO 2 / h + 0.345 kmol air / h
g
5-5
5.17 Basis: Given flow rates of outlet gas. Assume ideal gas behavior & 1 (kg / min) m 0.70 kg H 2 O / kg 0.30 kg S / kg
311 m 3 / min, 83o C, 1 atm n& 3 (kmol / min) 0.12 kmol H 2 O / kmol 0.88 kmol dry air / kmol
n& 2 (kmol air / min) & (m 3 / min) V 2 o 167 C, - 40 cm H 2O gauge
m & 4 (kg S / min)
a.
n& 3 =
1 atm
311 m3
356.2K
min
kmol ⋅ K 0.08206 m 3 ⋅ atm
H 2O balance: 0.70 m1 =
10.64 kmol 0.12 kmol H 2O 18.02 kg kmol kmol min
& 1 = 32.9 kg min milk ⇒m
b
g
= 10.64 kmol min
b
g
&4⇒m & 4 = 9.6 kg S min S olids balance: 0.30 32.2 kg min = m
Dry air balance: n& 2 = 0.88 (10.64 kmol min ) ⇒ n& 2 = 9.36 kmol min air V& 2 =
9.36 kmol 0.08206 m 3 ⋅ atm kmol ⋅ K
min
440K
(1033 − 40 ) cm H2O
1033 cm H 2O 1 atm
= 352 m air min 3
u air (m/min)=
& air (m3 /s) 352 m3 1 min V = A (m2 ) min 60 s
π
4
⋅ (6 m) 2
= 0.21 m/s
b. If the velocity of the air is too high, the powdered milk would be blown out of the reactor by the air instead of falling to the conveyor belt.
5.18 SG CO2 =
5.19 a.
ρ CO2 ρ air
=
x CO2 = 0.75
PM CO 2 RT PM air RT
=
M CO2 M air
=
44 kg / kmol = 152 . 29 kg / kmol
x air = 1 − 0.75 = 0.25
Since air is 21% O 2 , x O2 = (0.25)(0.21) = 0.0525 = 5.25 mole% O2
b.
mC O2 = n ⋅ x CO2 ⋅ M CO2 =
b
g
2 × 1.5 × 3 m 3 0.75 kmol CO2 44.01 kg CO 2 1 atm =12 kg m3 ⋅atm 298.2 K kmol kmol CO 2 0.08206 kmol ⋅K
More needs to escape from the cylinder since the room is not sealed.
5-6
5.19 (cont’d) c. With the room closed off all weekend and the valve to the liquid cylinder leaking, if a person entered the room and closed the door, over a period of time the person could die of asphyxiation. Measures that would reduce hazards are: 1. Change the lock so the door can always be opened from the inside without a key. 2. Provide ventilation that keeps air flowing through the room. 3. Install a gas monitor that sets off an alarm once the mole% reaches a certain amount. 4. Install safety valves on the cylinder in case of leaks.
5.20 n CO2 =
15.7 kg
1 kmol
= 0.357 kmol CO2 44.01 kg Assume ideal gas behavior, negligible temperature change T = 19° C = 292.2 K
b
a.
g
P1V n 1RT n1 P 102kPa = ⇒ = 1 = P2 V n 1 + 0.357 RT n 1 + 0.357 P2 3.27 × 103 kPa
b
g
⇒ n1 = 0.0115 kmol air in tank
b.
Vtank =
n1RT 0.0115 kmol 292.2 K 8.314 m3 ⋅ kPa 10 3L = = 274 L P1 102kPa kmol ⋅ K m3
15700 g CO2 + 11.5 mol air ⋅ (29.0 g air / mol) = 58.5 g / L 274 L c. CO 2 sublimates ⇒ large volume change due to phase change ⇒ rapid pressure rise. Sublimation causes temperature drop; afterwards, T gradually rises back to room temperature, increase in T at constant V ⇒ slow pressure rise. ρf =
b
gb
g
5.21 At point of entry, P1 = 10 ft H 2 O 29.9 in. Hg 33.9 ft H 2O + 28.3 in. Hg = 37.1 in. Hg . At surface, P2 = 28.3 in. Hg, V2 = bubble volume at entry 1 x x 0.20 0.80 Mean Slurry Density: = solid + solution = + 3 ρ sl ρ solid ρ solution (1.2 )(1.00 g / cm ) (100 . g / cm 3 ) = 0.967
cm 3 1.03 g 2.20 lb 5 × 10 −4 ton 106 cm 3 ⇒ ρ sl = = 4.3 × 10 −3 ton / gal g 1 lb 264.17 gal cm 3 1000 g
a.
300 ton gal 40.0 ft 3 (STP) 534.7 o R 29.9 in Hg = 2440 ft 3 / hr −3 o hr 4.3 × 10 ton 1000 gal 492 R 37 .1 in Hg
b.
4 π P2 V2 nRT V2 P1 3 = ⇒ = ⇒ P1V1 nRT V1 P2 4 π 3
% change =
e j e j D2 2 D1 2
3 3
=
37.1 ⇒ D32 = 1.31D13 28.3
b2.2 - 2.0g mm × 100 = 10% 2.0 mm
5-7
==> D D1 = 2 mm
2
= 2.2 mm
5.22 Let B = benzene n1 , n 2 , n 3 = moles in the container when the sample is collected, after the helium is added, and after the gas is fed to the GC. n inj = moles of gas injected n B , n air , n He = moles of benzene and air in the container and moles of helium added n BGC , m BGC = moles, g of benzene in the GC y B = mole fraction of benzene in room air
a.
P1V1 = n1RT1 (1 ≡ condition when sample was taken): P1 = 99 kPa, T1 = 306 K
n1 =
99 kPa 2 L mol ⋅ K = 0.078 mol = n air + n B kPa 101.3 atm 306 K .08206 L ⋅ atm
P2 V2 = n 2 RT2 (2 ≡ condition when charged with He): P2 = 500 kPa, T2 = 306 K
n2 =
500 kPa 2 L mol ⋅ K = 0.393 mol = n air + n B + n He kPa 101.3 atm 306 K .08206 L ⋅ atm
P3V3 = n 3RT3 (3 ≡ final condition in lab): P3 = 400 kPa, T3 = 296K
n3 =
400 kPa 2 L mol ⋅ K = 0.325 mol = (n air + n B + n He ) − n inj kPa 101.3 atm 296 K .08206 L ⋅ atm
n inj = n 2 − n 3 = 0.068 mol n B = n BGC ×
y B (ppm) =
n2 0.393 mol m BGC (g B) 1 mol = = 0.0741⋅ m BGC n inj 0.068 mol 78.0 g
nB 0.0741⋅ m BGC × 10 6 = × 106 = 0.950 × 10 6 ⋅ m BGC n1 0.078
U| | ) = 0.749 ppm VThe avg. is below the PEL | ) = 0.864 ppm| W
9 am: y B = (0.950 × 10 6 )(0.656 × 10− 6 ) = 0.623 ppm 1 pm: y B = ( 0.950 × 10 6 )(0.788 × 10 −6 5 pm: y B = ( 0.950 × 10 6 )(0.910 × 10 −6
b. Helium is used as a carrier gas for the gas chromatograph, and to pressurize the container so gas will flow into the GC sample chamber. Waiting a day allows the gases to mix sufficiently and to reach thermal equilibrium. c. (i) It is very difficult to have a completely evacuated sample cylinder; the sample may be dilute to begin with. (ii) The sample was taken on Monday after 2 days of inactivity at the plant. A reading should be taken on Friday. (iii) Helium used for the carrier gas is less dense than the benzene and air; therefore, the sample injected in the GC may be Herich depending on where the sample was taken from the cylinder. (iv) The benzene may not be uniformly distributed in the laboratory. In some areas the benzene concentration could be well above the PEL.
5- 8
b g
4 3 5.23 Volume of balloon = π 10 m = 4189 m 3 3 Moles of gas in balloon
b
g
n kmol =
a.
4189 m 3 492° R 3 atm
1 kmol
b g = 5159. kmol
535° R 1 atm 22.4 m 3 STP
He in balloon:
b
gb
g
m = 5159 . kmol ⋅ 4.003 kg kmol = 2065 kg He
mg =
b.
dP dP
2065 kg 9.807 m 1N = 20,250 N 2 s 1 kg ⋅ m / s2
gas in balloon air displaced
iV = n
iV = n
Fbuoyant
gasRT
air RT
⇒ n air =
Pair 1 atm ⋅ n gas = ⋅ 515.9 kmol = 172.0 kmol Pgas 3 atm
Fbuoyant = Wair displaced =
172.0 kmol 29.0 kg 9.807 m 1 N = 48,920 N 1 kmol s2 1 kg2⋅m s
Since balloon is stationary, Wtotal
Fcable
∑F = 0 1
Fcable = Fbuoyant − Wtotal = 48920 N −
b2065 + 150gkg
9.807 m 1 N = 27,200 s2 1 kg⋅2m s
c. When cable is released, Fnet
dAi = 27200 N = M ⇒a=
27200 N
tota
b2065 + 150g kg
1 kg ⋅ m / s2 N
= 12 .3 m s2
d. When mass of displaced air equals mass of balloon + helium the balloon stops rising. Need to know how density of air varies with altitude. e. The balloon expands, displacing more air ⇒ buoyant force increases ⇒ balloon rises until decrease in air density at higher altitudes compensates for added volume. 5.24 Assume ideal gas behavior, Patm = 1 atm 3 PN VN 5.7 atm 400 m / h 3 a. PN VN = Pc Vc ⇒ Vc = = = 240 m h 9.5 atm Pc
b.
Mass flow rate before diversion:
400 m 3 273 K 5.7 atm 1 kmol 44.09 kg kg C 3 H 6 = 4043 3 h 303 K 1 atm 22.4 m ( STP ) kmol h
5- 9
5.24 (cont’d) Monthly revenue:
( 4043 c.
kg h )( 24 h day )( 30 days month )( $0.60 kg ) = $1,747,000 month
Mass flow rate at Noxious plant after diversion: 400 m 3 hr
273 K 303 K
2.8 atm 1 atm
1 kmol 22.4 m
3
44.09 kg kmol
= 1986 kg hr
Propane diverted = ( 4043 − 1986 ) kg h = 2057 kg h
5.25 a. PHe = y He ⋅ P = 0.35 ⋅ (2.00 atm) = 0.70 atm PCH 4 = y CH 4 ⋅ P = 0.20 ⋅ (2.00 atm) = 0.40 atm PN 2 = y N 2 ⋅ P = 0.45 ⋅ (2.00 atm) = 0.90 atm
b. Assume 1.00 mole gas 4.004 g 0.35 mol He = 1.40 g He mol
U| || 16.05 gI F 0.20 mol CH G H mol JK = 3.21 g CH V|17.22 g ⇒ mass fraction CH F 28.02 gIJ = 12.61 g N || 0.45 mol N G H mol K W FG H
IJ K
4
2
4
4
=
3.21 g = 0.186 17.22 g
2
g of gas = 17.2 g / mol mol P MW 2.00 atm 17.2 kg / kmol m n MW = = = = = 115 . kg / m 3 3 V V RT 0.08206 mkmol⋅atm 363 . 2 K ⋅K
c.
MW =
d.
ρ gas
d i d i b e
gb
jb
g g
5.26 a. It is safer to release a mixture that is too lean to ignite. If a mixture that is rich is released in the atmosphere, it can diffuse in the air and the C3 H8 mole fraction can drop below the UFL, thereby producing a fire hazard. b. fuel-air mixture n& 1 ( mol / s) y C3 H8 = 0.0403 mol C3 H8 / mol n& C3 H8 = 150 mol C3H 8 / s
n& 3 (mol / s) 0.0205 mol C3 H 8 / mol
diluting air n& 2 ( mol / s)
n& 1 =
150 mol C 3H8 mol = 3722 mol / s s 0.0403 mol C 3H8
Propane balance: 150 = 0.0205⋅ n& 3 ⇒ n& 3 = 7317 mol / s
5- 10
5.26 (cont’d) Total mole balance: n& 1 + n& 2 = n& 3 ⇒ n& 2 = 7317 − 3722 = 3595 mol air / s
c.
b g
n& 2 = 1.3 n& 2
min
= 4674 mol / s
U| |V | |W
3 & = 4674 mol / s 8.314 m ⋅ Pa 398.2 K = 118 m 3 / s V 2 & mol ⋅ K 131,000 Pa V m 3 diluting air 2 = 141 . & 3 V m 3 fuel gas 1 &V = 3722 mol 8.314 m ⋅ Pa 298.2 K = 83.9 m 3 / s 1 s mol ⋅ K 110000 Pa
y2 =
150 mol / s 150 mol / s = × 100% = 1.8% n& 1 + &n 2 3722 mol / s + 4674 mol / s
b
g
d. The incoming propane mixture could be higher than 4.03%. If n& 2 = n& 2 min , fluctuations in the air flow rate would lead to temporary explosive conditions.
b g
5.27
Basis: (12 breaths min )( 500 mL air inhaled breath ) = 6000 mL inhaled min o
24 C, 1 atm 6000 mL / min
lungs
n& in (mol / min) 0.206 O 2 0.774 N 2 0.020 H 2O
a.
n& in =
blood
37o C, 1 atm n& out (mol / min) 0.151 O 2 0.037 CO2 0.750 N 2 0.062 H 2 O
6000 mL 1L 273K 1 mol = 0.246 mol min 3 min 10 mL 297K 22.4 L STP
b g
b
gb
g
N 2 balance: 0.774 0.246 = 0.750 &n out ⇒ n& out = 0.254 mol exhaled min
O 2 transferred to blood:
b0.246gb0.206g − b0.254 gb0.151g bmol O
2
g
min 32.0 g mol
= 0.394 g O2 min CO2 transferred from blood:
b0.254gb0.037g b mol CO
2
g
min 44.01 g mol
= 0.414 g CO2 min H2 O transferred from blood:
b0.254 gb0.062g − b0.246gb0.020g b mol H O ming 18.02 g mol 2
= 0.195 g H 2 O min
5- 11
5.27 (cont’d) PVin n RT = in in PVout n out RTout
FG IJ FG T IJ = FG 0.254 mol min IJ FG 310KIJ = 1.078 mL exhaled ml inhaled H K H T K H 0.246 mol min K H 297KK b0.414 g CO lost ming + b0.195 g H O lost ming − b 0.394 g O gained ming = 0.215 g min ⇒
b.
Vout n = out Vin n in
out in
2
5.28
2
2
STACK Ts (K) M s (g/mol) Ps (Pa)
LL(m) ( M)
PM RT
Ideal gas: ρ =
a.
b g
D = ρgL
Ta (K) Ma (g/mol) Pc (Pa)
combust.
b g
− ρgL
stack
=
L MN
Pa M a PM P gL M a M a gL − a s gL = a − RTa RTs R Ta Ts
O PQ
b gb g b gb g b gb g
b. M s = 0.18 44.1 + 0.02 32 .0 + 0.80 28.0 = 31.0 g mol , Ts = 655K , Pa = 755 mm Hg M a = 29.0 g mol , Ta = 294 K , L = 53 m
D=
755 mm Hg
×
1 atm 53.0 m 9.807 m kmol - K 2 760 mm Hg s 0.08206 m3 − atm
LM 29.0 kg kmol − 310. kg kmolOP × FG 1 N IJ = 323 N 1033 cm H O 655K N 294K Q H 1 kg ⋅ m / s K m 1.013 × 10 N m 2
2
2
5
2
= 3.3 cm H 2 O
5.29 a.
b.
b g =======>
MWCCl 2 O = 98.91 g /mol ρ CCl 2 O 98.91 P MW = = 3.41 RT ρ air 29.0 Phosgene, which is 3.41 times more dense than air, will displace air near the ground.
ρ=
Vtube =
b g
2
π Din L π = 0.635 cm - 2 0.0559 cm 4 4
m CCl2 O = Vtube ⋅ ρ CCl2 O =
c.
n CCl 2 O(l) =
b
3.22 cm 3
g b15.0 cmg = 3.22 cm 2
3
1L 1 atm 98.91 g / mol = 0.0131 g 3 3 L⋅atm 10 cm 0.08206 mol⋅K 296.2 K
3.22 cm 3 1.37 × 1.000 g mol = 0.0446 mol CCl 2 O cm 3 98.91 g
5- 12
5.29 (cont’d) PV 1 atm 2200 ft 3 28.317 L mol ⋅ K = = 2563 mol air 3 RT 296.2K .08206 L ⋅ atm ft
n air = n CCl2 O n air
=
0.0446 = 17.4 × 10 −6 = 17.4 ppm 2563
The level of phosgene in the room exceeded the safe level by a factor of more than 100. Even if the phosgene were below the safe level, there would be an unsafe level near the floor since phosgene is denser than air, and the concentration would be much higher in the vicinity of the leak. d. Pete’s biggest mistake was working with a highly toxic substance with no supervision or guidance from an experienced safety officer. He also should have been working under a hood and should have worn a gas mask. 5.30 CH 4 + 2O 2 → CO 2 + 2 H 2 O 7 C 2 H6 + O2 → 2 CO 2 + 3H 2O 2 C 3H 8 + 5O 2 → 3CO 2 + 4 H 2 O 3
o
1450 m / h @ 15 C, 150 kPa n& 1 (kmol / h) 0.86 CH 4 , 0.08 C2 H 6 , 0.06 C 3 H 8 n& 2 (kmol air / h) 8% excess, 0.21 O 2 , 0.79 N2
n& 1 =
1450 m 3 273.2K h
b101.3 + 150gkPa
1 kmol
101.3 kPa
22.4 m 3 STP
288.2K
b g = 152 kmol h
Theoretical O 2:
LM F MN GH
I JK
F GH
I JK
F GH
152 kmol 2 kmol O 2 3.5 kmol O 2 5 kmol O2 0.86 + 0.08 + 0.06 h kmol CH 4 kmol C 2 H 6 kmol C3 H 8
b
g
& air = 1.08 349.6 kmol O2 Air flow: V h
b g = 4.0 × 10
1 kmol Air
22.4 m 3 STP
0.21 kmol O 2
kmol
5- 13
I OP = 349.6 kmol h O JK PQ 4
b g
2
m 3 STP h
5.31 Calibration formulas
bT = 25.0; R = 14g , bT = 35.0, R = 27g ⇒ Tb° Cg = 0.77R + 14.2 d P = 0; R = 0i , dP = 20.0, R = 6i ⇒ P bkPag = 3.33R d V& = 0; R = 0i , dV& = 2.0 × 10 , R = 10i ⇒ V& d m hi = 200R dV& = 0; R = 0h , dV& = 1.0 × 10 , R = 25i ⇒ V& d m hi = 4000R T
g
T
p
g
T
r
gauge
p
3
F
p
F
3
F
F
5
A
A
F
3
A
A
A
A
& (m 3 / h), T, P V F g n& F (kmol / h) xA (mol CH 4 / mol) xB (mol C 2 H 6 / mol) xC (mol C 3 H8 / mol) xD (mol n - C 4 H 10 / mol) xE (mol i - C 4 H10 / mol)
CH 4 + 2O 2 → CO2 + 2 H 2 O 7 C 2 H 6 + O 2 → 2CO2 + 3H 2 O 2 C 3 H 8 + 5O 2 → 3CO2 + 4H 2 O 13 C 4 H 10 + O 2 → 4CO 2 + 5H 2 O 2
& ( m3 / h) (STP) V A & A (kmol / h) n 0.21 mol O2 / mol 0.79 mol N2 / mol
n& F =
d
& F m3 h V
i
273.2K
d P + 101.3i kPa
1 kmol
g
bT + 273.2gK 101.3 kPa & d P + 101.3i kmol 0.12031V FG IJ = b T + 273g H h K F
b g
22.4 m 3 STP
g
Theoretical O 2:
dn& i
o2 Th
c
b
gh
= n& F 2x A + 3.5x B + 5x C + 6.5 x D + x E kmol O 2 req. h
Air feed: n& A =
dn& i o2
Th
kmol O 2 req. 1 kmol air h 0.21 kmol O 2
FG P IJ d n i H 100 K b kmol air hg d22.4 m bSTPg kmol i = 22.4n& = 4.762 1 +
& A = n& a V
b1 + P
g
100 kmol feed 1 kmol req. x
x
o2
Th
3
A
b g
m3 STP h
RT T(C) Rp Pg(kPa) Rf xa xb xc xd xe PX(%) nF nO2, th nA Vf(m3/h) Va (m3/h) Ra 23.1 32.0 7.5 25.0 7.25 0.81 0.08 0.05 0.04 0.02 15 72.2 183.47 1004.74 1450 22506.2 5.63 7.5 20.0 19.3 64.3 5.8 0.58 0.31 0.06 0.05 0.00 23 78.9 226.4 1325.8 1160 29697.8 7.42 46.5 50.0 15.8 52.6 2.45 0.00 0.00 0.65 0.25 0.10 33 28.1 155.2 983.1 490 22022.3 5.51 21 30.4 3 10.0 6 0.02 0.4 0.35 0.1 0.13 15 53.0 248.1 1358.9 1200 30439.2 7.6 23 31.9 4 13.3 7 0.45 0.12 0.23 0.16 0.04 15 63.3 238.7 1307.3 1400 29283.4 7.3 25 33.5 5 16.7 9 0.5 0.3 0.1 0.04 0.06 15 83.4 266.7 1460.8 1800 32721.2 8.2 27 35.0 6 20.0 10 0.5 0.3 0.1 0.04 0.06 15 94.8 303.2 1660.6 2000 37196.7 9.3
5- 14
5.32 NO + 12 O2 ⇔ NO2 1 mol 0.20 mol NO / mol 0.80 mol air / mol 0.21 O 2 0.79 N 2 P0 = 380 kPa
R| S| T
a.
n1 (mol NO) n2 (mol O 2 ) n3 (mol N2 ) n4 (mol NO2 ) Pf (kPa)
U| V| W
Basis: 1.0 mol feed 90% NO conversion: n1 = 0.10 (0.20) = 0.020 mol NO ⇒ NO reacted = 0.18 mol
0.18 mol NO 0.5 mol O 2 = 0.0780 mol O2 mol NO N2 balance: n 3 = 0.80(0.79 ) = 0.632 mol N 2
O 2 balance: n 2 = 0.80(0.21) −
n4 = y NO
0.18 mol NO 1 mol NO2 = 018 . mol NO 2 ⇒ n f = n1 + n 2 + n 3 + n 4 = 0.91 mol 1 mol NO 0.020 mol NO mol NO = = 0.022 0.91 mol mol
y O2 = 0.086
mol O2 mol N 2 mol NO2 y N2 = 0.695 y NO2 = 0.198 mol mol mol
FG H
IJ K
Pf V n f RT n 0.91 mol = ⇒ Pf = P0 f = 380 kPa = 346 kPa P0 V n 0RT n0 1 mol
b.
Pf 360 kPa = (1 mol) = 0.95 mol P0 380 kPa n i = n i0 + υ iξ nf = n0
E
n1 (mol NO) = 0.20 − ξ n 2 ( mol O2 ) = ( 0.21)(0.80 ) − 0.5ξ n 3 (mol N 2 ) = (0.79 )(0.80) n 4 ( mol NO2 ) = ξ n f = 1 − 0.5ξ = 0.95 ⇒ ξ = 010 . ⇒ n 1 = 0.10 mol NO , n 2 = 0.118 mol O 2 , n 3 = 0.632 mol N 2 , n 4 = 0.10 mol NO 2 ⇒ y NO = 0.105, y O2 = 0.124, y N 2 = 0.665, y NO2 = 0.105 NO conversion =
b0.20- n g × 100% = 50% 1
0.20 360 kPa P (atm) = = 3.55 atm 101.3 kPa atm
Kp =
(y NO 2 P ) ( y NO P)( y O2 P )
0.5
=
(y NO2 ) 0.5
( y NO )( y O2 ) P
0.5
5- 15
=
b
0.105
(0.105) 0.124
g b3.55g 0.5
1
0.5
= 151 . atm 2
5.33 Liquid composition: 49.2 kg M 1 kmol = 0.437 kmol M 112.6 kg
100 kg liquid ⇒
0.481 kmol M / kmol
29.6 kg D 1 kmol = 0.201 kmol D 147.0 kg
⇒ 0.221 kmol D / kmol
21.2 kg B 1 kmol = 0.271 kmol B 78.12 kg
0.298 kmol B / kmol
0.909 kmol
a.
Basis: 1 kmol C6 H 6 fed 3
o
V1 (m ) @ 40 C, 120 kPa n1 (kmol) 0.920 HCl 0.080 Cl2 1 kmol C6 H 6 ( 78.12 kg) n 0 (kmol Cl2 )
n 2 (kmol)
0.298 C6 H 6 0.481 C 6 H5Cl 0.221 C 6 H4 Cl 2
C 6H 6 + Cl2 → C 6H 5Cl + HCl
C balance:
1 kmol C6 H 6
C 6H 5Cl + Cl 2 → C6 H 5Cl 2 + HCl
6 kmol C = n 2 0.298 × 6 + 0.481 × 6 + 0.221 × 6 1 kmol C6 H 6
⇒ n 2 = 100 . kmol
H balance:
1 kmol C6 H6
b
g
6 kmol H = n 1 0.920 (1) 1 kmol C6 H6
+ n 2 0.298 × 6 + 0.481 × 5 + 0.221 × 4 ⇒ n1 = 1.00 kmol V1 =
n1RT 1.00 kmol 101.3 kPa 0.08206 m3 ⋅ atm 313.2 K = = 21.7 m3 P 120 kPa 1 atm kmol ⋅ K
⇒
b.
V1 217 . m3 = = 0.278 m 3 / kg B m B 78.12 kg B
2 & & ( m3 / s) = u(m / s) ⋅ A(m 2 ) = u ⋅ πd ⇒ d 2 = 4 ⋅ Vgas V gas 4 π ⋅u
d2 =
& B0 (kg B) 0.278 m3 4m s 1 min 10 4 cm 2 = 5.90 m & B0 (cm2 ) min kg B π (10) m 60 s m2
b g
& B0 ⇒ d(cm) = 2.43⋅ m
1 2
c. Decreased use of chlorinated products, especially solvents.
5- 16
5.34 Vb ( m 3 / min) @900o C, 604 mtorr 60% DCS conversion n& 1 (mol DCS / min) n& 2 (mol N 2 O / min) n& b (mol / min) n& 3 (mol N 2 / min) n& 4 (mol HCl(g) / min)
3.74 SCMM
U| V| W
& a ( mol / min) n 0.220 DCS 0.780 N 2 O
SiH 2Cl 2(g) + 2 N 2O(g) → SiO2(s) + 2 N 2(g) + 2 HCl (g)
a.
n& a =
3.74 m 3 (STP) 10 3 mol =167 mol / min min 22.4 m3 (STP)
mol DCS I gFGH 0.220 mol JK b167 mol / ming = 14.7 mol DCS / min mol DCS DCS reacted: b0.60gb 0.220gb167 g = 22.04 mol DCS reacted / min min
b
60% conversion: n& 1 = 1 - 0.60
b g molminN O
N2 O balance: &n 2 = 0.780 167 −
N2 balance: n& 3 =
2
22.04 mol DCS 2 mol N 2O = 86.18 mol N 2O / min min mol DCS
22.04 mol DCS 2 mol N 2 = 44.08 mol N 2 / min min mol DCS
HCl balance: n& 4 =
22.04 mol DCS 2 mol HCl = 44.08 mol HCl / min min mol DCS
n& B = n& 1 + n& 2 + n& 3 + n& 4 = 189 mol / min & = ⇒V B
n& B RT 189 mol 62.36 L ⋅ torr 0.001 m3 1173 K = = 2.29 × 104 m 3 / min P min mol ⋅ K L 0.604 torr
n& 14.7 mol DCS/min ⋅ P= 1 P= ⋅ 604 mtorr=47.0 mtorr DCS DCS n& 189 mol/min B n& 86.2 mol N 2O/min p N2 O = x N 2O ⋅ P= 2 P= ⋅ 604 mtorr=275.5 mtorr &n B 189 mol/min
b. p
=x
r = 3.16 × 10−8 p DCS ⋅ p N2 O0.65 = 3.16 ×10-8 ( 47.0 )( 275.5
)
0.65
= 5.7 × 10 −5
mol SiO 2 m2 ⋅ s
& MW 5.7 × 10 −5 mol SiO2 60 s 120 min 60.09 g/mol 1010 A & c. h(A)=r ⋅t ⋅ = ρSiO2 min m2 ⋅ s 2.25 × 10 6 g/m 3 1 m (Table B.1)
& =1.1 ×105 A
The films will be thicker closer to the entrance where the lower conversion yields higher pDCS and p N 2O values, which in turn yields a higher deposition rate.
5- 17
5.35 Basis: 100 kmol dry product gas n1 (kmol C x H y ) m1 (kg C x H y ) kmol dry gasU R|100 |V 0.105 CO S|0.053 O |W T0.842 N
& (m3 ) V 2 n2 (kmol air) 0.21 O 2 0.79 N 2
2
2
2
o
n 3 (kmol H2 O)
30 C, 98 kPa
a.
N2 balance: 0.79n 2 = 0.842 (100) ⇒ n 2 = 106.6 kmol air
b
g
b
g b
g
O balance: 2 0.21n 2 = 100 2 0.105 + 2 0.053 + n 3 ⇒ n 3 = 1317 . kmol H 2O C balance:
d
n 1 kmol C x H y
i x bkmol Cg = 100b0105 . g ⇒ n x = 10.5 dkmol C H i
b1g
1
x
y
b2g
n 3 =13.17
H balance: n 1y = 2n 3 ====> n1y = 26.34
b g b g yx = 2610.34.5 = 2.51 mol H / mol C fed: 0.21b106.6 kmol air g = 22.4 kmol in excess = 5.3 kmol ⇒ Theoretical O = b22.4 - 5.3g kmol = 17.1 kmol
Divide 2 by 1 ⇒ O2
O2
2
% excess =
b.
V2 = m1 =
5.3 kmol O2 × 100% = 31% excess air 17.1 kmol O 2
106.6 kmol N 2 22.4 m 3 (STP) 101.3 kPa 303 K = 2740 m 3 kmol 98 kPa 273 K
b
g
b
g
n1x kmol C 12.0 kg n1y kmol H 101 . kg n1x=10.5 + =====> m 1 = 152.6 kg kmol kmol n1y =26.34
V2 2740 m3 air m3 air = = 18.0 m1 152 .6 kg fuel kg fuel
5.36 3N 2 H 4 → 6 xH 2 + (1 + 2 x)N 2 + (4 − 4 x)NH3 a. 0 ≤ x ≤ 1 b.
n N 2 H4 =
50 L 0.82 kg 1 kmol = 1.28 kmol L 32.06 kg
LM 6x kmol H + b1 + 2xg kmol N + b4 − 4xg kmol NH OP 3 kmol N H N 3 kmol N H 3 kmol N H Q 1.28 = b6x + 1 + 2x + 4 − 4xg = 1.707 x + 2.13 kmol 3 n product = 1.28 kmol N2 H 4
2
2
2
4
5- 18
2
4
3
2
4
5.36 (cont’d) nproduct 2.13 2.30 2.47 2.64 2.81 2.98 3.15 3.32 3.50 3.67 3.84
Vp (L) 15447.92 16685.93 17923.94 19161.95 20399.96 21637.97 22875.98 24113.99 25352.00 26590.01 27828.02
Volume of Product Gas 30000.00
25000.00
20000.00 V (L)
x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
15000.00
10000.00
5000.00
0.00 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
c.
Hydrazine is a good propellant because as it decomposes generates a large number of moles and hence a large volume of gas.
5.37
& A (g A / h) m
c
h
& m3 / h V air
3
C A (g A / m )
a. (i) Cap left off container of liquid A and it evaporates into room, (ii) valve leak in cylinder with A in it, (iii) pill of liquid A which evaporates into room, (iv) waste containing A poured into sink, A used as cleaning solvent. b.
c. d.
m &A
yA
FG kg A IJ H hK
FG kg A IJ H h K C e j⋅ V mol A = = mol air M e j ⋅ n =m &A
in
gA A m3 gA A mol
yA = 50 × 10−6
d h & V air
out
air
F I FG GH JK H
3 & m C kg A =V air A h m3
===================> m & PV CA = A ; nair = k⋅Vair RT
IJ K
yA =
& A RT m k ⋅ Vair M A P
m & A = 90 . g/h
& 8.314 mmol⋅Pa m RT 9.0 g / h 293 K ⋅K = A = = 83 m3 / h 3 −6 kyA MA P 0.5 50 × 10 101.3 × 10 Pa 104.14 g / mol 3
min
d
i
Concentration of styrene could be higher in some areas due to incomplete mixing (high concentrations of A near source); 9.0 g/h may be an underestimate; some individuals might be sensitive to concentrations < PEL. e. Increase in the room temperature could increase the volatility of A and hence the rate of & : At higher T, need evaporation from the tank. T in the numerator of expression for V air a greater air volume throughput for y to be < PEL.
5- 19
C 3H 6 + H 2 ⇔ C3H 8
5.38 Basis: 2 mol feed gas
U| V| W
np (mol C3 H8 ) (1- np )(mol C 3 H 6 ) n 2 = n p + 2(1 − n p ) = 2 − n p (1- np )(mol H 2 )
1 mol C3 H 6 1 mol H 2 25o C, 32 atm
235o C, P2
a. At completion, n p = 1 mol , n 2 = 2 − 1 = 1 mol 1 mol 508K P2 V n 2 RT2 n T = ⇒ P2 = 2 2 P1 = 2 mol 298K P1V n 1RT1 n 1 T1
b.
32.0 atm
= 27.3 atm
P2 = 35.1 atm
n2 =
35.1 atm 298K P2 T1 n1 = 32.0 atm 508K P1 T2
2 mol
= 1.29 mol
1.29 = 2 − n p ⇒ n p = 0.71 mol C3H 8 produced
b
g
⇒ 1 - 0.71 = 0.29 mol C 3H 6 unreacted ⇒ 71% conversion of propylene
c. n2 1.009 1.028 1.083 1.101 1.156 1.174 1.211 1.229 1.248 1.266 1.285 1.358 1.431 1.468
C3H8 prod. 0.99075 0.9724 0.91735 0.899 0.84395 0.8256 0.7889 0.77055 0.7522 0.73385 0.7155 0.6421 0.5687 0.532
%conv. 99.075 97.24 91.735 89.9 84.395 82.56 78.89 77.055 75.22 73.385 71.55 64.21 56.87 53.2
Pressure vs Fraction Conversion 120
100
80 % conversion
P2 (atm) 27.5 28.0 29.5 30.0 31.5 32.0 33.0 33.5 34.0 34.5 35.0 37.0 39.0 40.0
60
%conv.
40
20
0 25.0
27.0
29.0
31.0
33.0 Pressure (atm)
5- 20
35.0
37.0
39.0
41.0
5.39
Convert fuel composition to molar basis Basis: 100 g ⇒
b
UV W
g g
95 g CH4 1 mol 16.04 g = 5.92 mol CH 4 97.2 mol % CH 4 ⇒ 5 g C 2 H6 1 mol 30.07 g = 017 . mol C2 H 6 2.8 mol % C 2 H6
b
500 m3 / h
n& 2 (kmol CO2 / h) n& 3 (kmol H2 O / h)
n& 1 (mol / h) 0.972 CH 4
n& 4 (kmol O 2 / h) n& 5 (kmol N 2 / h)
0.028 C 2 H 6 40 o C, 1.1 bar
& (SCMH) V air 25% excess air
n& 1 =
& P1V 11 . bar 500 m3 1 = RT1 313K h
kmol ⋅ K 0.08314 m3 ⋅ bar
7 C2 H 6 + O 2 → 2 CO2 + 3H 2O 2
CH 4 + 2O 2 → CO 2 + 2 H2 O Theoretical O 2 =
= 211 . kmol h
LM N
21.1 kmol 0.972 kmol CH 4 kmol h
2 kmol O 2 1 kmol CH 4
OP = 43.1 kmol O kmol 1 kmol C H Q h 125 . b431 . kmol O g 1 kmol Air 22.4 m bSTPg Air Feed: = 5700 SCMH +
0.028 kmol C 2 H 6
3.5 kmol O 2 2
2
6
3
2
h
0.21 kmol O 2
1 kmol
5.40 Basis: 1 m3 gas fed @ 205° C, 1.1 bars Ac = acetone 1 m 3 @205 o C, 1.1 bar
n3 (kmol), 10o C, 40 bar
condenser
n1 (kmol) y1 (kmol Ac / kmol) (1 - y1 )(kmol air / kmol) pAC = 0100 . bar
y 3 (kmol Ac / kmol) (1- y 3 )(kmol air / kmol) p AC = 0.379 bar
n 2 (kmol Ac(l))
a.
n1 =
y1 =
100 . m 3 273K
110 . bars
1 kmol
b g = 0.0277 kmol
478K 10132 . bars 22.4 m3 STP
0.100 bar 0.379 bar = 0.0909 kmol Ac kmol , y 3 = = 9.47 × 10 −3 kmol Ac kmol 1.1 bars 40.0 bars
b
gb
g
Air balance: 0.0277 0.910 = (1 − 9.47 × 10 −3 )n 3 ⇒ n 3 = 0.0254 kmol Mole balance: 0.0277 = 0.0254 + n 2 ⇒ n 2 = 0.0023 kmol Ac condensed
Acetone condensed =
0.0023 kmol Ac 58.08 kg Ac 1 kmol Ac
5- 21
= 0.133 kg acetone condensed
5.40 (cont’d) Product gas volume =
b.
20.0 m 3 effluent
b g
0.0254 kmol 22.4 m 3 STP
273K
0.0277 kmol feed 0.0149 m effluent
5.41 Basis: 1.00 × 10 6 gal. wastewater day.
kmol feed
kmol Ac
= 0.0149 m 3
= 196 kg Ac h
Neglect evaporation of water. Effluent gas: 68o F, 21.3 psia(assume)
1.00 × 106 gal / day
n1 (lb - moles H2 O / day) 0.03n1 (lb - moles NH3 / day)
a.
40.0 bars
0.0909 kmol Ac 58.08 kg Ac
3
h
283K 1.0132 bars
n2 (lb - moles air / day) n3 (lb - moles NH3 / day)
300 × 106 ft 3 air / day
Effluent liquid
68o F, 21.3 psia n 2 (lb-moles air/day)
n1 (lb - moles H2 O / day) n4 (lb - moles NH 3 / day)
Density of wastewater: Assume ρ = 62.4 lb m ft 3
LMn lb - moles H O N day 1
2
= 1.00 × 10 6
OP Q
18.02 lbm 0.03 n1 lb m NH3 17.03 lbm 1 ft 3 + × 1 lb - mole day 1 lb - mole 62.4 lb m
7.4805 gal 1 ft 3
gal day
⇒ n 1 = 450000 lb - moles H 2 O fed day , 0.03n 1 = 13500 lb - moles NH 3 fed day n2 =
300 × 106 ft 3
492 o R
21.3 psi
1 lb - mole
527.7 R 14.7 psi
359 ft 3 STP
o
day
b g
= 113 . × 106 lb - moles air day
93% stripping: n 3 = 0.93 × 13500 lb - moles NH 3 fed day = 12555 lb - moles NH 3 day Volumetric flow rate of effluent gas
PVout n out RT n 300 × 10 6 ft 3 = ⇒ Vout = Vin out = PVin n in RT n in day
d1.13 × 10
6
i
+ 12555 lb - moles day
1.13 × 10 lb - moles day 6
= 303 × 10 ft day 6
3
Partial pressure of NH 3 = y NH 3 P =
12555 lb - moles NH 3 day
d1.129 × 10
= 0.234 psi
5- 22
6
i
+ 12555 lb - moles day
× 21.3 psi
5.42 Basis: 2 liters fed / min Cl ads.=
2.0 L soln 1130 g 0.12 g NaOH 1 mol 0.23 NaOH ads. 1 mol Cl 2 mol = 0.013 60 min L g soln 40.0 g mol NaOH 2 mol NaOH min
2 L / min @ 23o C, 510 mm H 2O n& 1 (mol / min) y (mol Cl2 / mol) (1 - y)(mol air / mol)
n 2 (mol air / mol)
0.013 mol Cl 2 / min
b g = b10.33 + 0.510g m H O = 10.84 m H O
Assume Patm = 10.33 m H 2O ⇒ Pabs
n& 1 =
2L min
2
273K 10.84 m H 2O 1 mol = 0.0864 mol min 296K 10.33 m H 2O 22.4 L STP
b g
Cl balance: 0.0864y = 0.013 ⇒ y = 0150 .
5.43
2
in
mol Cl 2 ,∴ specification is wrong mol
& (L / min) @ 65o C, 1 atm V 3
125 L / min @ 25o C, 105 kPa n& 1 ( mol / min) y1 (mol H2 O / mol) 1- y1 ( mol dry gas / mol) 0.235 mol C2 H 6 / mol DG 0.765 mol C2 H 4 / mol DG
R|b g S| T
n& C 2 H6 (mol C 2 H 6 / min) n& C 2 H4 (mol C2 H 4 / min) n& air (mol air / min) n& H 2 O (mol H 2 O / min)
U| V| W
355 L / min air @ 75o C, 115 kPa n& 2 (mol / min) y2 ( mol H2 O / mol) (1- y2 )( mol dry air / mol)
a.
Hygrometer Calibration ln y = bR + ln a
b=
b
ln y 1 y 2 R2 − R1
d y = ae i bR
g = lnd0.2 10 i = 0.08942 −4
90 − 5
bg
ln a = ln y1 − bR 1 = ln 10 −4 − 0.08942 5 ⇒ a = 6.395 × 10 −5 ⇒ y = 6.395 × 10 −5 e 0.08942R
b.
n& 1 =
125 L min
273K 105 kPa 298K 101 kPa
1 mol = 5.315 mol min wet gas 22.4 L STP
n& 2 =
355 L 273K 115 kPa 1 mol = 14.156 mol min wet air min 348K 101 kPa 22.4 L STP
b g
b g
R1 = 86.0 → y 1 = 0.140 , R 2 = 12.8 → y 2 = 2.00 × 10 −4 mol H 2O mol
5- 23
5.43 (cont’d)
b
C 2 H6 balance: n& C 2 H6 = 5.315 mol min
C H I gFGH b1 − 0.140g molmolDG IJK FGH 0.235 mol J mol DG K 2
6
= 1.07 mol C2 H 6 min C 2 H 4 balance: n& C 2 H 4 = 5.315 0.860 0.765 = 3.50 mol C2 H 4 min
b
gb gb g Dry air balance: n& = b14.156gd1 − 2.00 × 10 i = 14.15 mol DA min Water balance: n& = b5.315gb0.140g + b14.156gd1.00 × 10 i = 0.746 mol H O min n& = b1.07 + 3.50 + 14.15g mol min = 18.72 mol min , n& = b18.72 + 0.746g = 19.47 mol min & = 19.47 mol min 22.4 L bSTPg 338K = 540 liters min V −4
air
−4
H 2O
2
dry product gas
total
3
mol
Dry basis composition:
c.
p H 2O = y H 2 Ol ⋅ P =
273K
FG 1.07 IJ × 100% = 5.7% C H , 18.7% C H , 75% dry air H 18.72K 2
6
2
4
0.746 mol H 2O × 1 atm = 0.03832 atm 19.47 mol
y H 2O = 0.03832 ⇒ R =
FG H
IJ K
1 0.03832 ln = 71.5 0.08942 6.395 × 10 −5
5.44 CaCO 3 → CaO + CO 2 n& CO2 =
1350 m3
273K
1 kmol
b g = 12.92 kmol CO
1273K 22.4 m 3 STP
h
12.92 kmol CO2 1 kmol CaCO3 100.09 kg CaCO3 h
1 kmol CO2 0.17 kg clay
h
0.83 kg limestone
0.95 kg CaCO 3
= 279 kg clay h
Weight % F e 2O3
b0.07g
kg Fe2 O 3 kg clay
279
. Fe O b g × 100% = 18%
1362 + 279 − 12.92 44.1 kg clay 14243
kg limestone
2
kg CO2 evolved
5- 24
h
1 kg limestone
1 kmol CaCO 3
1362 kg limestone
2
3
= 1362 kg limestone h
5.45
Basis:
R|864.7 g C b1 mol 12.01 g g = 72.0 mol C |116.5 g H b1 mol 1.01 gg = 1153. mol H 1 kg Oil ⇒ S |13.5 g S b1 mol 32.06 gg = 0.4211 mol S |T5.3 g I 53 . gI n1 (mol CO 2 ) n 2 (mol CO) n 3 (mol H 2 O) n 4 (mol SO 2 ) n5 (mol O 2 ) n 6 (mol N 2 )
72.0 mol C 115.3 mol H 0.4211 mol S 5.3 g I
C + O 2 → CO2 1 C + O 2 → CO 2 S + O 2 → SO 2 1 2H + O 2 → H 2O 2
na (mol), 0.21 O 2 , 0.79 N 2 15% excess air o 175 C, 180 mm Hg (gauge)
a.
Theoretical O 2:
72.0 mol C 1 mol O
2 + 1 mol C
+
115.3 mol H
0.25 mol O 2 1 mol H
0.4211 mol S 1 mol O 2 = 101.2 mol O 2 1 mol S
(
1.15 101.2 mol O Air Fed:
)
1 mol Air 0.21 mol O
554 mol Air 1 kg oil
2
22.4 liter ( STP ) mol
1 m3 3
10 liter
= 554 mol Air = n 2
448K
760 mm Hg
273K
940 mm Hg
a
= 16.5 m air kg oil 3
b. S balance: n 4 = 0.4211 mol SO 2 H balance: 115.3 = 2n 3 ⇒ n 3 = 57.6 mol H 2O
b g
C balance: 0.95 72.0 = n 1 ⇒ n1 = 68.4 mol CO 2 ⇒ 0.05(72.0) = n 2 = 3.6mol CO
N2 balance: 0.79 ( 554 ) =n 6 ⇒ n 6 = 437.7 mol N 2
O balance: 0.21 ( 554 ) 2=57.6+3.6+2(68.4)+2 ( 0.4211 ) +2n 5 ⇒ n 5 = 16.9 mol O 2
Total moles ( excluding inerts ) wet: 585 mols dry: 527 mols dry basis:
wet basis:
3.6 mol CO 527 mol 3.6 mol CO 585 mol
= 6.8 ×10 −3
mol CO mol
,
× 10 6 = 6150 ppm CO ,
5- 25
0.4211 mol SO2 527 mol
= 7.2 × 10 −4
0.4211 mol SO 2 585 mol
mol SO 2 mol
× 10 6 = 720 ppm SO 2
bg
5.46 Basis: 50.4 liters C 5H 12 l min 50.4 L C5 H12 (l) / min n& 1 (kmol C5 H12 /min)
heater
n& 1, n& 2
Combustion chamber
15% excess air, V& air (L / min) &n 2 kmol air 0.21 O 2 0.79 N 2 336 K, 208.6 kPa (gauge)
& 3 (kmol C5 H 12 / min) n & 4 (kmol O 2 / min) n & 5 (kmol N 2 / min) n & 6 (kmol CO 2 / min) n & 7 (kmol H 2 O / min) n
Condenser
C5 H 12 + 80 2 → 5CO 2 + 6 H 2O
a.
& (L/min) V liq & m=3.175 kg C5 O12 /min n& 3 (kmol C5O12 /min) n& 7 (kmol H 2O(l) / m i n )
n& 1 =
50.4 L
0.630 kg
1 kmol
min
L
72.15 kg
n& 3 =
3175 . kg 1 kmol = 0.044 kmol / min min 72.15 kg
frac. convert = n& 2 =
V& gas (L/min), 275 K, 1 atm &n 4 (kmol O2 / min) & 5 (kmol N 2 / min) n &n 6 (kmol CO 2 / min)
= 0.440 kmol min
0.440 - 0.044 kmol × 100 = 90% C5 H12 converted 0.440
0.440 kmol C5 H12 1.15 (8 kmol O2 ) 1 mol air = 19.28 kmol air min min kmol C5 H12 0.21 mol O2
b g
& air = 19.28 kmol 22.4 L STP V min mol
336K
101 kPa 1000 mol = 173000 L min 273K 309.6 kPa kmol
n& 4 = [(0.21)(19.28) − (0.90)(0.440)(8)]
kmol O 2 = 0.882 kmol O 2 /min min
19.28 kmol air 0.79 kmol N2 = 15.23 kmol N2 /min min kmol air 0.90(0.440 kmol C5H12 ) 5 kmol CO2 n& 6 = = 1.98 kmol CO2 / min min kmol C 5 H12
n& 5 =
& = 0.882+15.23+1.98 kmol 22.4 L(STP) 275 K 1000 mol = 4.08 × 10 5 L/min V gas min mol 273 K kmol
5- 26
5.46 (cont’d) n& 7 =
0.9( 0.440 kmol C5H 12 ) 6 kmol H 2 O = 2.38 kmol H2 O( l ) / min min kmol C5H 12
Condensate:
&C H = V 5 12 &H O = V 2
0.044 kmol 72.15 kg min
kmol
2.38 kmol 18.02 kg min
L
kmol
0.630 kg L 1 kg
= 5.04 L min
= 42 .89 L min
Assume volume additivity (liquids are immiscible) & liq = 5.04 + 42.89 = 47.9 L min V b. C5 H12 ( l)
bg
C5 H12 l
bg
H2 O l
bg
H2 O l
5.47 o n& air (kmol / min), 25 C, 1 atm 0.21 O 2 0.79 N 2
n& 0 (kmol / min)
n& 1 (kmol H 2 S / min) Furnace
H 2 S + 32 O 2 → SO 2 + H 2 O
0.20 kmol H 2S / mol 0.80 kmol CO 2 / mol
Reactor
&n 2 (kmol H 2S / min)
2H 2 S +SO 2 → 3S(g) + 2 H 2 O
10.0 m 3 / min @ 380o C, 205 kPa n& exit (kmol / min) n& 3 (kmol N 2 / min) n& 4 (kmol H 2 O / min) n& 5 (kmol CO 2 / min) n& 6 (kmol S / min)
n& exit =
& PV 205 kPa 10.0 m 3 / min = = 0.377 kmol / min m3 ⋅ kPa RT 8.314 kmol 653 K ⋅K
b g
n& 1 = 0.20 n& 0 / 3 = 0.0667n& 0 ; n& 2 = 2 n& 1 = 0.133n& 0
5- 27
5.47 (cont’d) Air feed to furnace: n& air =
0.0667 &n 0 (kmol H 2S fed) 15 . kmol O 2 1 kmol air (min) 1 kmol H 2 S 0.21 kmol O 2
= 0.4764 n& 0 kmol air / min
0.4764 n& 0 (kmol air) 0.79 kmol N 2 = 0.3764 n& 0 (kmol N 2 / min) (min) min 0.200n& 0 (kmol H 2S) 1 kmol S Overall S balance: n& 6 = = 0.200 n& 0 (kmol S / min) (min) 1 kmol H 2S Overall N2 balance: n& 3 =
Overall CO 2 balance: n& 5 = 0.800n& 0 (kmol CO2 / min) Overall H balance:
0.200n& 0 (kmol H 2 S) 2 kmol H n& kmol H 2 O 2 kmol H = 4 (min) 1 kmol H 2S min 1 kmol H 2 O
⇒ n& 4 = 0.200n& 0 (kmol H 2 O / min)
b
g
n& exit = n& 0 0.376 + 0.200 + 0.200 + 0.800 = 0.377 kmol / min ⇒ n& 0 = 0.24 kmol / min
n& air = 0.4764(0.24 kmol air / min) = 0114 . kmol air / min
5.48 Basis: 100 kg ore fed ⇒ 82.0 kg FeS 2 (s), 18.0 kg I.
b
gb
g
n FeS 2 fed = 82.0 kg FeS2 1 kmol / 120.0 kg = 0.6833 kmol FeS 2
100 kg ore
Vout m3 (STP)
0.6833 kmol FeS2 18 kg I
n2 n3 n4 n5
40% excess air n 1 (kmol) 0.21 O 2 0.79 N 2 V1 m 3 (STP)
(kmol SO2 ) (kmol SO 3 ) (kmol O 2 ) (kmol N2 )
m6 (kg FeS2 ) m7 (kg Fe 2O 3 ) 18 kg I
2 FeS2(s) + 11 O2(g) → Fe 2 O3(s) + 4SO 2(g) 2 2 FeS2(s) + 152 O2(g) → Fe 2O3(s) + 4SO 3(g)
a.
n1 =
0.6833 kmol FeS2 7 .5 kmol O2 1 kmol air req' d 1.40 kmol air fed = 17.08 kmol air 2 kmol FeS 2 0.21 kmol O2 kmol air req'd
b
gb
g
V1 = 17.08 kmol 22.4 SCM / kmol = 382 SCM / 100 kg ore
n2 =
( 0.85)(0.40)0.6833 kmol FeS 2 4 kmol SO 2 = 0.4646 kmol SO2 2 kmol FeS2
5- 28
5.48 (cont’d) n3 =
(0.85)(0.60) 0.6833 kmol FeS2 4 kmol SO 2 = 0.6970 kmol SO 3 2 kmol FeS 2
n = (0.21 ×17.08 ) kmol O fed − 4 2 −
b
.4646 kmol SO
2
5.5 kmol O 4 kmol SO
2
2
.697 kmol SO 7.5 kmol O 3 2 = 1.641 kmol O 2 4 kmol SO 3
g
n5 = 0.79 × 17.08 kmol N2 = 13.49 kmol N 2 Vout = (0.4646+0.6970+1.641+13.49 ) kmol [ 22.4 SCM (STP)/kmol] = 365 SCM/100 kg ore fed
ySO 2 =
0.4646 kmol SO 2 × 100% = 2.9%; y SO 3 = 4.3%; y O2 = 10.1%; y N 2 = 82.8% 16.285 kmol
b.
e j
Product gas, T o C Converter
0.4646 kmol SO2 0.697 kmol SO 3 1633 . kmol O 2 1349 . kmol N 2
nSO2 (kmol) nSO3 (kmol) nO2 (kmol) nN 2 (kmol)
Let ξ (kmol) = extent of reaction
n SO 2 = 0.4646 − ξ n SO3 = 0.697 + ξ n O2 = 1.641 − 12 ξ n N 2 = 13.49 n=16.29- 12 ξ K p (T)=
0.4646 − ξ 0.697 + ξ ySO = , y SO3 = 2 1 16.29- 2 ξ 16.29- 12 ξ ⇒ 1.641 − 12 ξ 13.49 yO 2 = , y N2 = 1 16.29- 2 ξ 16.29- 12 ξ (0.697 + ξ ) (16.29 − 12 ξ ) 2 1
P ⋅ y SO3 1 2
P ⋅ y SO2 (P ⋅ yO2 )
⇒
(0.4646 − ξ ) (1.641 − 12 ξ )
1 2
-1
⋅ P 2 = K p (T)
P=1 atm, T=600 o C, K p = 9.53 atm - 2 ⇒ ξ = 0.1707 kmol 1
⇒ n SO2 = 0.2939 kmol ⇒ fSO2 =
( 0.4646 − 0.2939 ) kmol SO2 0.4646 kmol SO 2 fed
reacted
= 0.367
P=1 atm, T=400 o C, Kp = 397 atm 2 ⇒ ξ = 0.4548 kmol -1
⇒ n SO2 = 0.0098 kmol ⇒ fSO2 = 0.979
The gases are initially heated in order to get the reaction going at a reasonable rate. Once the reaction approaches equilibrium the gases are cooled to produce a higher equilibrium conversion of SO 2 .
5- 29
5.48 (cont’d) c.
SO 3 leaving converter: (0.6970 + 0.4687) kmol = 1.156 kmol
1.156 kmol SO 3 1 kmol H 2SO 4 98 kg H 2 SO 4 = 1133 . kg H 2SO 4 min 1 kmol SO 3 kmol
⇒
Sulfur in ore:
0.683 kmol FeS 2 2 kmol S 32.1 kg S = 438 . kg S kmol FeS 2 kmol
113.3 kg H 2SO 4 kg H 2SO 4 = 2.59 43.8 kg S kg S
100% conv.of S: ⇒
0.683 kmol FeS2 2 kmol S 1 kmol H2SO 4 98 kg = 1339 . kg H2SO 4 kmol FeS 2 1 kmol S kmol
133.9 kg H 2SO 4 kg H2SO 4 = 3.06 43.8 kg S kg S
The sulfur is not completely converted to H2 SO4 because of (i) incomplete oxidation of FeS2 in the roasting furnace, (ii) incomplete conversion of SO 2 to SO 3 in the converter. 5.49 N 2 O 4 ⇔ 2 NO 2
dP
gauge
i
+ 1.00 V
n0 =
b.
n1 = mol NO 2 , n 2 = mol N 2 O 4
RT0
=
b2.00 atmgb2.00 Lg b473Kgb0.08206 L ⋅ atm mol - Kg = 0.103 mol NO
a.
F n IP⇒ K = n P GH n + n JK n bn + n g Ideal gas equation of state ⇒ PV = b n + n gRT ⇒ n + n = PV / RT b1g p NO2 = y NO 2 P =
F n IP, p GH n + n JK
2
1
1
N 2O 4
=
2 1
2
p
2
1
1
2
2
2
1
1
2
2
Stoichiometric equation ⇒ each mole of N 2O 4 present at equilibrium represents a loss of two moles of NO 2 from that initially present ⇒ n1 + 2n 2 = 0.103
b3g ,
Solve (1) and (2) ⇒ n1 = 2(PV / RT) − 0.103
b2 g
b4g
n 2 = 0.103 − (PV / RT)
Substitute (3) and (4) in the expression for Kp , and replace P with Pgauge + 1
b2n − 0.103g d P = n b0.103 − n g 2
gauge
t
T(K) 350 335 315 300
Pgauge(atm) 0.272 0.111 -0.097 -0.224
i
+ 1 where n t =
dP
gauge
nt Kp(atm) 0.088568 5.46915 0.080821 2.131425 0.069861 0.525954 0.063037 0.164006
(1/T) ln(Kp) 0.002857 1.699123 0.002985 0.756791 0.003175 -0.64254 0.003333 -1.80785
i
+1V
RT
t
⇒ nt =
V =2 L
2 1
i
T
y = -7367x + 22.747 R2 = 1
0 -1 -2 0.0028
0.003
0.0032 1/T
5- 30
d
24.37 Pg + 1
Variation of Kp with Temperature
ln Kp
Kp
t
0.0034
5.49 (cont’d) c.
A semilog plot of Kp vs.
1 is a straight line. Fitting the line to the exponential law T
yields ln K p = −
FG H
IJ K
7367 −7367 a = 7.567 × 109 atm + 22.747 ⇒ K p = 7.567 × 10 9 exp ⇒ T T b = 7367K 10.00 atm
5.50 & 1 (kmol A / h) n &n 2 (kmol H 2 / h)
5.00 kmol S / h &n 4 (kmol A / h) n& 5 (kmol H 2 / h)
3n& 3 (kmol A / h) &n 3 (kmol H 2 / h)
5.00 kmol S / h
n& 4 (kmol A / h) n& 5 (kmol H 2 / h) V&rcy (SCMH)
A + H2
S
5.00 kmol S 1 kmol A react = 5.00 kmol A / h h 1 kmol S form 5.00 kmol S 1 kmol H 2 react Overall H 2 balance: n 2 = = 5.00 kmol H 2 / h h 1 kmol S form Overall A balance: n1 =
Extent of reaction equations: &n i = n& i0 + ν iξ& A + H2 ↔ S
n& 4 = 3 &n 3 − ξ& H 2 : n& 5 = n& 3 − ξ& S: 5.00 = ξ& =====> n& 4 = 3n& 3 − 5.00 n& 5 = n& 3 − 5.00 n& 3n& - 5.00 ⇒ pA = yA P = 4 P = 3 10.0 &n S = 5.00 n& tot 4 n& 3 − 5.00 A:
n& tot = 4 n& 3
U| |V | − 5.00 |W
n& 5 n& - 5.00 P= 3 10.0 &n tot 4 n& 3 − 5.00 5.00 p S = yS P = 10.0 4 n& 3 − 5.00 p H 2 = yH 2 P =
Kp =
b
g
5.00 4n& 3 − 5.00 pS = = 0.100 ⇒ n& 3 = 1194 . kmol H 2 / h p A p H 2 10.0 3n& 3 − 5.00 n& 3 − 5.00
b
gb
g
n& 4 = 3(11.94 ) - 5.00 = 30.82 kmol A / h n& 5 = 11.94 − 5.00 = 6.94 kmol H 2 / h & = 30.82 + 6.94 kmol / h 22.4 m 3 (STP ) / kmol = 846 SCMH V rcy
b
g
d
i
5- 31
5.51 n& 4 (kmol CO / h) n& 5 (kmol H 2 / h)
Reactor & 1 (kmol CO / h) n &n 2 (kmol H 2 / h)
n& 4 (kmol CO/h) n& 5 (kmol H 2 /h) n& 6 (kmol CH 3OH/h) T, P
100 kmol CO / h &n 3 (kmol H2 / h) T (K), P (kPa)
H xs (% H 2 excess)
a.
Separator
n& 6 (kmol CH 3 OH/h)
%XS H 2 , 2 atomic balances, Eq. relation ⇒ four equations in n& 3 , n& 4 , n& 5 , and n& 6
5% excess H2 in reactor feed: n&3 =
100 mol CO 2 mol H2 req'd 1.05 mol H2 fed mol H 2 = 210 h mol CO 1 mol H2 req'd h
C balance: 100(1) = n&4 (1) + n&6 (1) ⇒ n&4 = 1 − n&6 H balance: 210(2) = n&5 (2) + n&6 (4) ⇒ n&5 = 210 − 2n&6
(1) (2)
n& T = n& 4 + &n5 + &n6 = (100 − n& 6 ) + ( 210 − 2n& 6 ) + n& 6 = 310 − 2n& 6
b
g
K p T = 500K = 1390 . × 10
−4
F 21.225 + 9143.6 − 7.492lnb500K g I G JJ K exp GH +4.076 × 10500b500K g - 1.161 × 10 b500Kg K -3
= 9.11 × 10 −7 kPa-2
Kp =
yM P
(
y CO P y H2 P
K p P = 9.11 × 10 2
)
−7
2
(1) − (3)
yM
⇒ Kp P 2 =
( )
y CO y H2
kPa
-2
( 5000 kPa )
2
2
====>
= 22.775 =
2
-8
n& 6 ( 310 − 2n& 6 )
(100 − n& 6 ) ( 210 − 2n& 6 ) ( 310 − 2n& 6 ) ( 310 − 2n& 6 )2
2
n& 6 (310 − 2n& 6 )
2
(100 − n& 6 )( 210 − 2n& 6 )
2
(3)
Solving for n& 6 ⇒ n& 6 = 75.7 kmol M/h ⇒ n& 4 = 100 − n& 6 = 24.3 kmol CO/h , n& 5 = 210 − 2n& 6 = 58.6 kmol H2 / h 1 kmol CO 2 kmol H 2 n& 6 = 75.7 kmol CO/h , n& 2 = n& 6 = 151 kmol H2 /h 1 kmol M 1 kmol M 3 & = ( n& + n& ) 22.4 m (STP) = 1860 SCMH V rec 4 5 kmol n& 1 =
5-32
5.51 (cont’d) b. P(kPa) 1000 5000 10000 5000 5000 5000 5000 5000 5000
`
T(K) Hxs(%) 500 5 500 5 500 5 400 5 500 5 600 5 500 0 500 5 500 10
n6(kmol ntot M/h) (kmol/h) KpcE8 25.55 258.90 9.1E-01 9.00 292.00 2.3E-01 86.72 136.56 9.1E+01 98.93 112.15 7.8E+03 75.68 158.64 2.3E+01 14.58 280.84 4.1E-01 73.35 153.30 2.3E+01 75.68 158.64 2.3E+01 77.77 164.45 2.3E+01
Kp(T)E8 9.1E+01 9.1E+01 9.1E+01 3.1E+04 9.1E+01 1.6E+00 9.1E+01 9.1E+01 9.1E+01
KpP^2 0.91 22.78 91.11 7849.77 22.78 0.41 22.78 22.78 22.78
KpP^2- n1(kmol KpcP^2 CO/h) 1.3E-05 25.55 2.3E+01 9.00 4.9E-03 86.72 3.2E-08 98.93 3.4E-03 75.68 -2.9E-04 14.58 9.8E-03 73.35 3.4E-03 75.68 -3.1E-03 77.77
n3(kmol n4(kmol n5(kmol H2/h) CO/h) H2/h) 210 74.45 158.90 210 91.00 192.00 210 13.28 36.56 210 1.07 12.15 210 24.32 58.64 210 85.42 180.84 200 26.65 53.30 210 24.32 58.64 220 22.23 64.45 n2(kmol Vrec H2/h) (SCMH) 51.10 5227 18.00 6339 173.44 1116 197.85 296 151.36 1858 29.16 5964 146.70 1791 151.36 1858 155.55 1942
c. Increase yield by raising pressure, lowering temperature, increasing Hxs . Increasing the pressure raises costs because more compression is needed. d. If the temperature is too low, a low reaction rate may keep the reaction from reaching equilibrium in a reasonable time period. e. Assumed that reaction reached equilibrium, ideal gas behavior, complete condensation of methanol, not steady-state measurement errors.
5.52
CO 2 ⇔ CO + 12 O 2
1.0 mol CO 2 1.0 mol O 2 1.0 mol N 2 T = 3000 K, P = 5.0 atm
K1
dp =
1/ 2
i = 0.3272 atm
1/ 2
p CO 2 1 1 O2 + 2 N 2 ⇔ NO 2 p NO K2 = = 01222 . 1/ 2 p N 2 p O2
d
1 A ⇔ B+ C 2 1 1 C+ D = E 2 2
CO p O2
i
A − CO 2 , B − CO , C − O 2 , D − N 2 , E − NO
ξ 1 - extent of rxn 1
n A0 = n C0 = n D0 = 1 , n B0 = n E0 = 0
ξ 2 - extent of rxn 2
5-33
5.52 (cont’d) nA = 1 − ξ1 nB = ξ1 1 1 nC = 1 + ξ1 − ξ2 2 2 1 nD = 1 − ξ 2 2 nE = ξ 2 1 6 + ξ1 n tot = 3 + ξ 1 = 2 2
U| || y = n n = 2 b1 − ξ g b6 + ξ g | yy == 22ξ + ξb6 +− ξξ g 6 + ξ gb g V b || y = b2 − ξ g b 6 + ξ g || y = 2ξ b6 + ξ g |W A
B
A
tot
1
1
1
C
1
D
2
1
2
1
g b5g p b gb g ⇒ 0.3272b1 − ξ gb6 + ξ g = 2.236ξ b2 + ξ − ξ g
K1 =
p CO p1O22
=
CO2
b
y B y 1C 2 b1+ 12 −1g 2ξ 1 2 + ξ 1 − ξ 2 p = yA 2 1 − ξ1 6 + ξ1
12
K2 =
d
p NO p O2 p N 2
b
i
12
=
⇒ 0.1222 2 + ξ 1 − ξ 2
= 0.3272
12
1
yE 12 12 yC yD
12
12
12
1
p i = y iP
1
2
E
1
1
p 1−1 2 −1 2 =
1
2ξ 2
b2 + ξ − ξ g b2 − ξ g 12
1
g b2 − ξ g 12
2
12
(1)
2
2
12
= 01222 .
2
= 2ξ 2
(2)
Solve (1) and (2) simultaneously with E-Z Solve ⇒ ξ 1 = 0.20167 , ξ 2 = 012081 . ,
b
gb
g
y A = 2 1 − ξ 1 6 + ξ 1 = 0.2574 mol CO2 mol y D = 0.3030 mol N 2 mol y B = 0.0650 mol CO mol y E = 0.0390 mol NO mol y C = 0.3355 mol O2 mol
5.53 a.
n& 4 (kmol / h)
0.04 O2 0.96 N2
PX=C8 H10 , TPA=C6H 10O 4 , S=Solvent & (m 3 / h) @105o C, 5.5 atm V 3 n& 3O (kmol O2 / h) n& 3N (kmol N 2 / h) n& 3W (kmol H 2O(v) / h) & (m 3 / h) at 25o C, 6.0 atm V 2 n& 2 (kmol / h) 0.21 O 2 0.79 N 2
condenser
n& 3W (kmol H2 O(v) / h) & (m 3 / h) V 3W
reactor
n& 1 (kmol PX / h) ( n& 1 + n& 3p ) kmol PX / h & s (kg S / h) m 3 kg S / kg PX
n& 3p (kmol PX / h) & s (kg S / h) m 5-34
n& 3p ( kmolPX / h ) 100 kmol TPA / h m & s (kg S / h)
separator
100 mol TPA / s
5.53 (cont’d) b. Overall C balance: n& 1
c.
FG kmol PXIJ 8 kmol C = 100 kmol TPA H h K kmol PX h
8 kmol C ⇒ n& 1 = 100 kmol PX / h kmol TPA
100 kmol TPA 3.0 kmol O 2 = 300 kmol O2 /h h 1 kmol TPA kmol O 2 Overall O2 balance: 0.21n& 2 = 300 +0.04n& 4 n& 2 = 1694 kmol air/h h ⇒ n& 4 = 1394 kmol/h Overall N2 balance: 0.79n& 2 = 0.96n& 4 100 kmol TPA 2 kmol H2 O Overall H2 O balance: n& 3W = = 200 kmol H 2O / h h 1 kmol TPA O2 consumed =
3 & = n& 2RT = 1694 kmol 0.08206 m ⋅ atm 298 K = 6.90 × 103 m3 air/h V 2 P h kmol ⋅ K 6.0 atm 3 & & & = ( n 3W + n 4 ) RT = ( 200+1394 ) kmol 0.08206 m ⋅ atm 378 K = 8990 m3 /h V 3 P h kmol ⋅ K 5.5 atm
3 &V3W = 200 kmol H 2O (l) 18.0 kg 1 m = 3.60 m 3 H 2O(l) / h leave condenser h kmol 1000 kg
d
i
n& 1 =100
d. 90% single pass conversion ⇒ n& 3p = 0.10 n& 1 + n& 3p ====> n& 3p = 111 . kmol PX / h m& recycle = m& S + m& 3 P =
(100+11.1) kmol PX h
3 kg S 11.1 kmol PX + h 1 kmol PX kg PX
106 kg PX
106 kg
kmol PX
= 3.65 × 104
kg h
e. O2 is used to react with the PX. N2 does not react with anything but enters with O2 in the air. The catalyst is used to accelerate the reaction and the solvent is used to disperse the PX. f. The stream can be allowed to settle and separate into water and PX layers, which may then be separated. 5.54
n& 1 (kmol CO / h), n& 3 (kmol H 2 / h), 0.10n & 2 (kmol H 2 / h)
Separator &n 6 (kmol CO / h) &n 7 (kmol H 2 / h) n& 8 (kmol CO 2 / h) 2 kmol N 2 / h
0.90n& 2 2 kmol N2 / h
n& 1 , n& 2 , n& 3 2 kmol N2 / h 0.300 kmol CO / kmol 0.630 kmol H 2 / kmol 0.020 kmol N 2 / kmol 0.050 kmol CO 2 / kmol
Reactor
n& 1 (kmol CO / h) n& 2 (kmol H 2 / h) n& 3 (kmol CO2 / h) n& 4 (kmol M / h) n& 5 (kmol H 2 O / h) 2 kmol N 2 / h
5-35
Separator
n& 4 (kmol M / h) &n5 (kmol H 2 O / h)
5.54 (cont’d) CO + 2H 2 ⇔ CH3OH(M) CO2 + 3H2 ⇔ CH 3OH + H 2O
a.
Let ξ 1 ( kmol / h) = extent of rxn 1, ξ 2 ( kmol / h) = extent of rxn 2
CO: n& 1 = 30 - ξ 1 H2 : n& 2 = 63 - 2ξ 1 − 3ξ 2
U| M: n& = ξ + ξ || P ⋅y P ⋅ y H O: n& = ξ V ⇒ dK i = P ⋅ y P ⋅ Py ⋅ y , dK i = bP ⋅ y gd P ⋅ y i || d i d id i N : n& = 2 n& = 100 - 2ξ − 2ξ |W n& − 2ξ − 2 ξ g &n = 84 .65 (1) dK i ⋅ P = n& F n& I = bbξ30+−ξξ gbgb100 63 − 2ξ − 3ξ g n& GH n& JK FG n& IJ FG n& IJ (2) dK i ⋅ P = FH nn&& IKFH nn&& IK = ξ bbξ5 −+ξξ gbgb63100− 2−ξ2ξ− 3−ξ2ξg g = 1.259 GH n& JK GH n& JK CO2 : n& 3 = 5 - ξ 2
2 2
4
1
5
2
2
M
p 1
CO
N2
tot
1
CO2
H2
2
2
1
tot
p 1
2
1
2
2
2
H2 O
3
H2
4
2
M
p 2
2
1
2
tot
tot
2
1
4
5
tot
tot
3
2
tot
tot
p 2
1
2
2
2
1
2
1
2 2
3
2
1
2
Solve (1) and (2) fo r ξ 1 , ξ 2 ⇒ ξ 1 = 25.27 kmol / h ξ 2 = 0.0157 kmol / h
n& 1 = 30.0 − 25.27 = 4.73 kmol CO / h
⇒
9.98% CO
n& 2 = 63.0 − 2 (25.27 ) − 3(0.0157 ) = 12.4 kmol H 2 / h
26.2% H 2
n& 3 = 5.0 − 0.0157 = 4.98 kmol CO2 / h
10.5% CO 2
n& 4 = 25.27 + 0.0157 = 25.3 kmol M / h
⇒
n& 5 = 0.0157 = 0.0157 kmol H 2 O / h n total = 49 .4 kmol / h
53.4% M 0.03% H 2 O
UV W
n& 6 = 25.4 kmol CO / h C balance: n& 4 = 25.3 kmol / h ⇒& n 8 = 0.02 kmol CO 2 / h O balance: n& 6 + 2 n& 8 = n& 4 + n& 5 = 25.44 mol / s
H balance: 2n& 7 = 2 (0.9 n& 2 ) + 4 n& 4 + 2 n& 5 = 1237 . ⇒ n& 7 = 618 . mol H 2 / s
b. (n& 4 ) process = 237 kmol M / h ⇒ Scale Factor =
237 kmol M / h 25.3 kmol / h
5-36
5.54 (cont’d)
237 kmol / h I F 22.4 m (STP)I gFGH 25.3 J kmol JK = 18,700 SCMH kmol / h K GH F 237 kmol / hIJ = 444 kmol / h Reactor effluent flow rate: b 49.4 kmol / h gG H 25.3 kmol / sK F kmol IJ FG 22.4 m (STP) IJ = 9946 SCMH ⇒ V& G 444 H h K H kmol K
b
3
Process feed: 25.4 + 618 . + 0.02 + 2.0
3
std
c.
9950 m 3 (STP ) ⇒ V&actual = h 3 & $ = V = 354 m / h 1000 L V n& 444 kmol / h m 3
4732 . K 1013 . kPa = 354 m 3 / h 273.2 K 4925 kPa 1 kmol = 0.8 L / mol 1000 mol
$ < 20 L / mol====> ideal gas approximation is poor V (5.2-36)
& from n& using the ideal gas equation of state is likely Most obviously, the calculation of V to lead to error. In addition, the reaction equilibrium expressions are probably strictly valid only for ideal gases, so that every calculated quantity is likely to be in error.
5.55 a.
$ PV B RT = 1 + $ ⇒ B = c Bo + ωB1 RT Pc V From Table B.1 for ethane: Tc = 305.4 K, Pc = 48.2 atm
b
g
From Table 5.3-1 ω = 0.098 0.422 0.422 Bo = 0.083 − 1.6 = 0.083 − = −0.333 1.6 Tr 3082 . K 305.4 K 0.172 0.172 B1 = 0.139 − 4.2 = 0.139 − = −0.0270 4 .2 Tr 308.2 K 305.4K RTc 0.08206 L ⋅ atm 305.4 K B(T) = Bo + ωB1 = −0.333 − 0.098 0.0270 Pc mol ⋅ K 48.2 atm = −01745 . L / mol
b
e
j
e
j
g
FG H
b
IJ K
g
$2 PV mol ⋅ K $ 2 $ $ - B = 10.0 atm −V V − V + 0.1745 = 0 RT 308.2K 0.08206 L ⋅ atm
$ = ⇒V
b
gb
g = 2.343 L / mol, 0.188 L / mol
1 ± 1 - 4 0.395 mol / L 0.1745 L / mol
b
2 0.395 mol / L
g
V$ideal = RT / P = 0.08206 × 3082 . / 10.0 = 2.53, so the second solution is
b. c.
likely to be a mathematical artifact. $ PV 10.0 atm 2.343 L / mol z= = = 0.926 L ⋅atm RT 0.08206 mol 308.2K ⋅K
& = m
& V 1000 L mol 30.0 g 1 kg MW = = 12.8 kg / h $ h 2.343 L mol 1000 g V
5-37
5.56
$ PV B RT = 1 + $ ⇒ B = c Bo + ωB1 RT Pc V
b
g
b g T bC H g = 369.9 K, P = 42 .0 atm From Table 5.3-1 ω bCH OHg = 0.559, ω bC H g = 0.152 From Table B.1 Tc CH3OH = 513.2 K, Pc = 78.50 atm c
3
8
c
3
0.422
Bo (CH 3OH) = 0.083 − Bo (C 3H 8 ) = 0.083 −
B1 (C3 H8 ) = 0.139 −
B(CH 3OH) =
Tr
1.6
0.422
= 0.083 −
= 0.083 −
Tr 1.6
B1 (CH 3OH) = 0.139 −
3
e
8
0.422
e
513.2K 0.422
373.2K
369.9K 0172 .
j
0.172
= 0.139 −
Tr 4.2
b
RTc Bo + ω B1 Pc
e
513.2K 0.172
373.2K
369.9K
j
j
4 .2
= −0.619
= −0.333
1.6
0.172 = 0.139 − Tr 4.2 373.2K
e
j
1.6
373.2K
4 .2
= −0.516
= −0.0270
g
0.08206 L ⋅ atm 513.2K −0.619 − 0.559 0.516 = −0.4868 mol ⋅ K 78.5 atm RT B(C3 H8 ) = c Bo + ωB1 Pc
c
=
b
=
Bmix =
h
0.08206 L ⋅ atm 369.9 K −0.333 − 0.152 0.0270 = −0.2436 mol ⋅ K 42.0 atm
b
c
i
b
g
L mol
g
∑∑ y y B i
b
j
ij
d
⇒B ij = 0.5 Bii + B jj
j
g
g
h
L mol
i
Bij = 0.5 −0.4868 − 0.2436 L / mol = -0.3652 L / mol
b gb gb
g b gb gb
g b gb gb
Bmix = 0.30 0.30 −0.4868 + 2 0.30 0.70 −0.3652 + 0.70 0.70 −0.2436 = −0.3166 L / mol
FG H
g
IJ K
$2 PV mol ⋅ K $ 2 $ $ - B = 10.0 atm −V V − V + 0.3166 = 0 mix RT 373.2K 0.08206 L ⋅ atm
$ $ = Solve for V:V
b
gb
g = 2.70 L / mol, 0.359 L / mol
1 ± 1 - 4 0.326 mol / L 0.3166 L / mol
b
g
2 0.326 mol / L
RT 0.08206 L ⋅ atm 3732 . K $ $ V = = 3.06 L / mol ⇒ V ideal = virial = 2.70 L / mol P mol ⋅ K 10.0 atm $ & = 2.70 L / mol & = Vn V
15.0 kmol CH 3OH / h 1000 mol 1 m 3 =135 m3 / h 0.30 kmol CH3OH / kmol 1 kmol 1000 L
5-38
5.57 a.
van der Waals equation: P =
d
i
d
RT a2 − 2 $ $ -b V V
i
$2 V $ - b ⇒ PV $ 3 − PV $ 2 b = RTV $ 2 − aV $ + ab Multiply both sides by V
b
g
$ 3 + -Pb - RT V $ 2 + aV $ - ab = 0 PV c 3 = P = 50.0 atm
b
g b
gb
g c
c 2 = -Pb - RT = −50.0 atm 0.0366 L / mol − 0.08206 c 1 = − a = 1.33 atm ⋅ L2 / mol 2
d
ib
L⋅ atm mol⋅ K
hb223 Kg = −20.1 L ⋅ atm / mol
g
c 0 = −ab = - 1.33 atm ⋅ L2 / mol2 0.0366 L / mol = −0.0487
b.
atm ⋅ L3 mol 3
RT 0.08206 L ⋅ atm 223 K $ V = = 0.366 L / mol ideal = P mol ⋅ K 50.0 atm
c. T(K) 223 223 223 223 223
P(atm) 1.0 10.0 50.0 100.0 200.0
c3
c2
1.0 10.0 50.0 100.0 200.0
c1
-18.336 -18.6654 -20.1294 -21.9594 -25.6194
c0
1.33 1.33 1.33 1.33 1.33
-0.0487 -0.0487 -0.0487 -0.0487 -0.0487
V(ideal) V f(V) % error (L/mol) (L/mol) 18.2994 18.2633 0.0000 0.2 1.8299 1.7939 0.0000 2.0 0.3660 0.3313 0.0008 10.5 0.1830 0.1532 -0.0007 19.4 0.0915 0.0835 0.0002 9.6
d. 1 eq. in 1 unknown - use Newton-Raphson. $ =0 b1g ⇒ gd V$ i = 50.0V$ + b-20.1294 gV$ + b1.33gV-.0487 3
2
∂g $ 2 − 40.259 V $ +1.33 = 150 V $ ∂V solve −g Eq. (A.2-14) ⇒ ad = −g ⇒ d = a
Eq. (A.2-13) ⇒ a =
$ (k +1) = V $ (k) + d Guess V $ (1) = V $ Then V ideal = 0.3660 L / mol .
1 2 3 4
$ (k) V 0.3660 0.33714 0.33137 0.33114
5-39
$ (k +1) V 0.33714 0.33137 0.33114 0.33114 converged
b
d
5.58 C 3H 8 : TC = 369 .9 K Specific Volume
PC = 42.0 atm 4.26 × 10 6 Pa
5.0 m3
44.09 kg
1 kmol
75 kg
1 kmol
10 3 mol
i
ω = 0.152
= 2.93 × 10 −3 m 3 mol
Calculate constants a=
0.42747
b=
0.08664
d8.314 m ⋅ Pa mol ⋅ Ki b369.9 Kg 2
3
4.26 × 10 Pa 6
2
= 0.949 m6 ⋅ Pa mol 2
d8.314 m ⋅ Pa mol ⋅ Ki b369.9 Kg = 6.25 × 10 3
4.26 × 10 6 Pa
b
g
b
m = 0.48508 + 1.55171 0.152 − 0.15613 0.152
e
α = 1 + 0.717 1 − 298.2 369.9
j
2
g
2
−5
m 3 mol
= 0.717
= 115 .
SRK Equation:
d8.314 m ⋅ Pa mol ⋅ Kib298.2 Kg − d2.93 × 10 − 6.25 × 10 i m mol 2.93 × 10
d
115 . 0.949 m 6 ⋅ Pa mol 2
3
P=
−3
−5
3
−3
d
i
i
m3 mol 2.93 × 10 −3 + 6.25 × 10 −5 m 3 mol
⇒ P = 7.40 × 10 6 Pa ⇒ 7.30 atm
P=
Ideal:
ib
d
(8.35 − 7.30) atm × 100% = 14.4% 7.30 atm
Percent Error:
TC = 304.2 K
5.59 CO2 :
g
3 . K RT 8.314 m ⋅ Pa mol ⋅ K 2982 = = 8.46 × 10 6 Pa ⇒ 8.35 atm − 3 3 $V 2.93 × 10 m mol
PC = 72.9 atm ω = 0.225
TC = 151.2 K PC = 48.0 atm ω = −0.004 $ = 35.0 L / 50.0 mol = 0.70 L mol P = 510 . atm , V
Ar:
Calculate constants (use R = 0.08206 L ⋅ atm mol ⋅ K ) CO 2 : a = 3.65 Ar:
bg
a = 1.37
f T =
L2 ⋅ atm mol2 L ⋅ atm 2
mol2
, m = 0.826 , b = 0.0297 , m = 0.479 , b = 0.0224
e
RT a − 1 + m 1 − T TC $ $ V $ +b V−b V
d
i
j
2
e L , α = 1 + 0.479e1 − mol
−P =0
Use E-Z Solve. Initial value (ideal gas): L L ⋅ atm Tideal = 51.0 atm 0.70 0.08206 = 4350 . K mol mol ⋅ K
gFGH E - Z Solve ⇒ bT g b
max CO2
IJ FG K H
= 455.4 K ,
bT g
max Ar
IJ K
= 431.2 K
5-40
j T 1512 . j
L , α = 1 + 0.826 1 − T 304.2 mol
2
2
b
g
5.60 O2 : TC = 154.4 K ; PC = 49.7 atm ; ω = 0.021 ; T = 208.2 K 65° C ; P = 8.3 atm ; m & = 250 kg h ; R = 0.08206 L ⋅ atm mol ⋅ K
SRK constants: a = 1.38 L2 ⋅ atm mol2 ; b = 0.0221 L mol ; m = 0.517 ; α = 0.840
d i dV$RT− bi − V$ dVa$α+ bi − P = 0 =====> V$ = 2.01 L / mol
$ = f V
SRK equation:
E-Z Solve
& = ⇒V
5.61
W
∑F
y
250 kg
kmol
h
32.00 kg
= PCO 2 ⋅ A - W = 0
103 mol 2.01 L 1 kmol
mol
e
= 15,700 L h
where W = mg = 5500 kg 9.81
m s2
j = 53900 N
PCO2 ⋅ A
a.
PCO2 =
W A piston
=
53900 N π 4
b0.15 mg
2
b. SRK equation of state: P =
1 atm = 30.1 atm 1.013 × 10 5 N / m2 RT αa − $ -b V $ V $ +b V
d i d
i
For CO 2: Tc = 304 .2, Pc = 72.9 atm , ω = 0.225
a = 3.654 m 6 ⋅ atm / kmol2 , b = 0.02967 m 3 / kmol, m = 0.8263, α (25o C) = 1.016
e0.08206 jb298.2 Kg − b1.016ge3.654 j 301 . atm = e V$ - 0.02967 j V$ dV$ + 0.02967 i m 3 ⋅atm kmol⋅ K
m 6 ⋅atm kmol2
m6 2 kmol
m3 kmol
$ = 0.675 m3 / kmol =====> V E-Z Solve
b g Vbafter expansiong = 0.030 m
V before expansion = 0.030 m 3
mC O2 =
3
b
g b15. mg = 0.0565 m
+ π4 0.15 m
2
3
V 0.0565 m3 44.01 kg MW = = 3.68 kg 3 $V 0.675 m / kmol kmol
mC O2 ( initially) =
PV 1 atm 0.030 m 3 44.01 kg MW = = 0.0540 kg 3 RT 298.2 K kmol 0.08206 mkmol⋅atm ⋅K
mCO2 (added) = 3.68 - 0.0540 kg = 3.63 kg
5-41
5.61 (cont’d) c.
W = 53,900 N
V
h
add 3.63 kg CO2
n o (kmol) Vo (m3 ) 1 atm, 25o C
ho
========> n (kmol) P (atm), 25o C
ho
d(m)
d(m)
Given T, Vo , h, find d Initial: n o =
Vo RT
Final: V = Vo + Vo +
bP = 1g o
πd 2 h 3.63 (kg) V , n = no + = o + 0.0825 4 44 (kg / kmol) RT
πd 2 h 4
$ =V= V n Vo + 0.0825 RT W RT αa 53,900 RT αa P= = − ⇒ 2 = − $ $ $ V $ +b $ V $ +b A piston V - b V πd / 4 V - b V
b1g i d i $ in b1g ⇒ one equation in one unknown. Solve for d . Substitute expression for V d
5.62 a. Using ideal gas assumption: Pg =
nRT 35.3 lb m O 2 1 lb - mole 10.73 ft 3 ⋅ psia 509.7 o R − Patm = − 14.7 psia = 2400 psig V 32.0 lb m lb - mole ⋅ o R 2.5 ft 3
b. SRK Equation of state: P =
RT αa − $ -b V $ V $ +b V
d i d
i
3 ft 3 $ = 2.5 ft 32.0 lb m / lb - mole = 2.27 V 35.3 lb m lb - mole
For O2: Tc = 277.9 o R, Pc = 7304 . psi, ω = 0.021 a = 5203.8
ft 6 ⋅ psi ft 3 , b = 0 . 3537 , m = 0.518, α 50o F = 0.667 lb - mole 2 lb - mole
d
i
e10.73 jd509.7 R i − b 0.667 ge52038. b2400 + 14.7g psi = V$ - 0.3537 $ dV $ + 0.3537 i V d i 3
ft ⋅ psi lb-mole⋅o R
o
ft 3 lb -mole
$ = 2.139 ft 3 / lb - mole E - Z Solve ⇒ V V 2.5 ft 3 32.0 lb m mO 2 = MW = = 37.4 lb m 3 $ 2.139 ft / lb - mole lb - mole V
5-42
6
ft ⋅ psi lb-mole2
j
ft 6 2 lb-mole
5.62 (cont’d) Ideal gas gives a conservative estimate. It calls for charging less O2 than the tank can safely hold. c. 1. 2. 3. 4.
Pressure gauge is faulty The room temperature is higher than 50°F Crack or weakness in the tank Tank was not completely evacuated before charging and O2 reacted with something in the tank 5. Faulty scale used to measure O2 6. The tank was mislabeled and did not contain pure oxygen.
5.63 a.
SRK Equation of State: P =
RT αa − $ $ $ +b V-b V V
d i d
i
d id i $ i = PV $ dV $ - bidV $ + b i − RTV $ dV $ + b i + α ad V $ - bi = 0 fdV $ i = PV $ − RTV $ + dαa - b P - bRTiV $ - α ab = 0 fdV
$ V $ -b V $ +b : ⇒ multiply both sides of the equation by V
3
2
2
b. Problem 5.63-SRK Equation Spreadsheet Species Tc(K) Pc(atm) ω a b m
CO2 304.2 R=0.08206 m^3 atm/kmol K 72.9 0.225 3.653924 m^6 atm/kmol^2 0.029668 m^3/kmol 0.826312
f(V)=B14*E14^3-0.08206*A14*E14^2+($B$7*C14-$B$8^2*B14-$B$8*0.08206*A14)*E14-C14*$B$7*$B$8 T(K) 200 250 300 300 300
P(atm) 6.8 12.3 6.8 21.5 50.0
alpha 1.3370 1.1604 1.0115 1.0115 1.0115
V(ideal) 2.4135 1.6679 3.6203 1.1450 0.4924
V(SRK) 2.1125 1.4727 3.4972 1.0149 0.3392
f(V) 0.0003 0.0001 0.0001 0.0000 0.0001
c. E-Z Solve solves the equation f(V)=0 in one step. Answers identical to VSRK values in part b. d.
REAL T, P, TC, PC, W, R, A, B, M, ALP, Y, VP, F, FP INTEGER I CHARACTER A20 GAS DATA R 10.08206/ READ (5, *) GAS WRITE (6, *) GAS 10 READ (5, *) TC, PC, W READ (5, *) T, P IF (T.LT.Q.) STOP
5-43
5.63 (cont’d) R = 0.42747 *R*R/PC*TC*TC B = 0.08664 *R*TC/PC M = 0.48508 + W = 155171 . − W∗015613 .
d
c b
b
g
g
hi
ALP = 1.+ M∗ 1 − T / TC ∗∗0.5 ∗∗2 .
VP = R∗ T / P DO 20 I = 7 , 15 V = VP F = R * T/(V – B) – ALP * A/V/(V + B) – P FP = ALP * A * (2. * V + B)/V/V/(V + B) ** 2 – R * T/(V – B) ** 2. VP = V – F/FP IF (ABS(VP – V)/VP.LT.0.0001) GOTO 30 20 CONTINUE WRITE (6, 2) 2 FORMAT ('DID NOT CONVERGE') STOP 30 WRITE (6, 3) T, P, VP 3 FORMAT (F6.1, 'K', 3X, F5.1, 'ATM', 3X, F5.2, 'LITER/MOL') GOTO 10 END
$ DATA CARBON 304.2 200.0 250.0 300.0 –1
72.9 6.8 12.3 21.5 0.
DIOXIDE 0.225
RESULTS CARBON DIOXIDE 200.0 K 6.8 ATM 250.0 K 12.3 ATM 300.0 K 6.8 ATM 300.0 K 21.5 ATM 300.0 K 50.0 ATM
5.64 a.
b.
2.11 LITER/MOL 1.47 LITER/MOL 3.50 LITER/MOL 1.01 LITER/MOL 0.34 LITER/MOL
b
g
U| V |W
Tr = 40 + 273.2 126.2 = 2.48 N2 : TC = 126.2 K ⇒ 40 MPa 10 atm PC = 33.5 atm Pr = = 11.78 33.5 atm 1.013 MPa
He: TC = 5.26 K PC = 2.26 atm
⇒
Fig. 5.4-4
. U| g b5.26 + 8g = 552 V| P = 350 b2.26 + 8 g = 34.11 W
b
Tr = −200 + 2732 . r
↑ Newton’s correction
5-44
⇒ z = 12 .
Fig. 5.4-4
⇒ z = 1.6
d
i
5.65 a. ρ kg / m3 = =
b.
m (kg) 3
V (m )
=
(MW)P RT
30 kg kmol 9.0 MPa 10 atm = 69 .8 kg m3 3 m ⋅ atm 465 K 0.08206 kmol ⋅ K 1.013 MPa
UV P = 9.0 4.5 = 2.0 W
Tr = 465 310 = 1.5
Fig. 5.4-3
⇒ z = 0.84
r
ρ=
(MW)P 69.8 kg m3 = = 83.1 kg m 3 zRT 0.84
100 lb m CO2
1 lb - mole CO 2
= 2.27 lb - moles 44.01 lb m CO2 TC = 304.2 K (1600 + 14.7 ) psi 1 atm = 1.507 ⇒ Pr = P PC = 72.9 atm 14.7 psi PC = 72.9 atm
5.66 Moles of CO 2:
ˆ 10.0 ft 3 72.9 atm lb-mole ⋅°R 1k ˆ = VPC = V = 0.80 r 3 RTC 2.27 lb-moles 304.2 K 0.7302 ft ⋅ atm 1.8 °R
Fig. 5.4-3: Pr = 1.507 , Vr = 0.80 ⇒ z = 0.85 PV 1614.7 psi 10.0 ft 3 lb - mole⋅° R 1 atm T= = = 779° R = 320 ° F 3 znR 0.85 2.27 lb - moles 0.7302 ft ⋅ atm 14.7 psi
Pr1 Tr2 Pr2
V2 = V1 V2 =
|UVz = 1.00 (Fig. 5.4 - 2) = 1 49.7 = 0.02 |W = 358 154.4 = 2.23 U| Vz = 1.61 b Fig. 5.4 - 4g = 1000 49.7 = 20.12|W
Tr1 = 298 154.4 = 1.93
5.67 O : T = 154.4 K 2 C PC = 49.7 atm
1
2
z 2 T2 P1 z 1 T1 P2
127 m3 1.61 358 K 1 atm = 0.246 m3 h h 1.00 298 K 1000 atm
b
g
Tr = 27 + 273.2 154.4 = 1.94 Pr1 = 175 49.7 = 3.52 ⇒ z1 = 0.95
5.68 O 2: TC = 154.4 K PC = 49.7 atm
(Fig. 5.3-2)
Pr2 = 1.1 49.7 = 0.02 ⇒ z 2 = 1.00
n1 − n 2 =
F GH
I JK
10.0 L V P1 P2 − = RT z1 z2 300.2 K
FG H
IJ K
mol ⋅ K 175 atm 11 . atm − = 74.3 mol O2 0.08206 L ⋅ atm 0.95 1.00
5- 45
5.69 a.
b.
$ = V = 50.0 mL 44.01 g = 440.1 mL / mol V n 5.00 g mol RT 82.06 mL ⋅ atm 1000 K P= $ = = 186 atm mol ⋅ K 440.1 mL / mol V For CO 2: Tc = 304.2 K, Pc = 72.9 atm
T 1000 K = = 3.2873 Tc 304.2 K $ VP 4401 . mL 72.9 atm mol ⋅ K c Vr ideal = = = 1.28 RTc mol 304.2 K 82.06 mL ⋅ atm
Tr =
Figure 5.4-3: Vr ideal = 1.28 and Tr = 3.29 ⇒ z=1.02 P=
c.
zRT 1.02 82.06 mL ⋅ atm mol 1000 K = = 190 atm ˆ mol ⋅ K 440.1 mL V
a = 3.654 × 10 6 mL2 ⋅ atm / mol2 , b = 29 .67 mL / mol, m = 0.8263, α (1000 K) = 0.1077
. ge3.654 × 10 j = 198 atm c82.06 hb1000 Kg − b01077 P= b440.1 - 29.67g 440.1b440.1+ 29.67 g 6 mL2 ⋅atm mol 2
mL ⋅atm mol ⋅ K
2
mL mol
mL mol2
5.70 a. The tank is being purged in case it is later filled with a gas that could ignite in the presence of O 2 . b. Enough N2 needs to be added to make x O2 = 10 × 10 −6 . Since the O2 is so dilute at this condition, the properties of the gas will be that of N2 . Tc = 126.2 K, Pc = 33.5 atm, Tr = 2.36 n initial = n 1 =
PV 1 atm 5000 L = = 204.3 mol L⋅ atm RT 0.08206 mol⋅ K 298.2 K
FG 0.21 mol O IJ = 42.9 mol O H mol air K
n O2 = 204.3 mol air n O2 n2
2
2
= 10 × 10 −6 ⇒ n 2 = 4.29 × 10 −6 mol
$ = V
5000 L = 116 . × 10 -3 L / mol 4.29 × 10 6 mol $ . × 10 −3 L mol ⋅ K 33.5 atm $ ideal = VPc = 116 V = 3.8 × 10 −3 r RTc mol 0.08206 L ⋅ atm 126.2 K ⇒ not found on compressibility charts Ideal gas: P =
RT 0.08206 L ⋅ atm 2982 . K = = 2.1 × 10 4 atm −3 $ mol ⋅ K 116 V . × 10 L / mol
The pressure required will be higher than 2.1 × 10 4 atm if z ≥ 1, which fro m Fig. 5.3- 3 is very likely.
ib
d
g
n added = 4.29 × 106 − 204.3 ≅ 4.29 × 106 mol N 2 0.028 kg N 2 / mol = 1.20 × 105 kg N2
5- 46
5.70 (cont’d) c.
143 . kmol N 2
143 . kmol N 2
n initial = 0.204 kmol y O2 = 0.21 kmol O 2 / kmol
1.43 kmol N 2 y1
1.43 kmol N 2 y2
Fig 5.4-2
N 2 at 700 kPa gauge = 7.91 atm abs. ⇒ Pr = 0.236, Tr = 2.36 =======> z = 0.99 n2 =
P2 V 7.91 atm 5000 L = = 1.633 kmol L ⋅atm zRT 0.99 0.08206 mol 298.2 K ⋅K
y1 =
0.21 0.204 y init n init = = 0.026 1.634 1634 .
y2 =
y 1n init n init = y init 1.634 1.634
b g FG H
IJ K
= 0.0033
Fy I FG n IJ ⇒ n = GH y JK = 4.8 ⇒ Need at least 5 stages H 1.634 K F n IJ ln G H 1.634K = 5b143 . kmol N gb28.0 kg / kmolg = 200 kg N ln
n
y n = y init
2
n
init
init
init
Total N 2
2
2
d. Multiple cycles use less N2 and require lower operating pressures. The disadvantage is that it takes longer.
5.71
& = MW a. m
b.
UV W
Tc = 369.9 K = 665.8o R ⇒ Tr = 0.85 Pc = 42.0 atm ⇒ Pr = 0.16 & = 60.4 m
F GH
I JK
& & & & PV SPV 44.09 lb m / lb - mol SPV SPV & = MW ⇒ Cost ($ / h) = mS = = 60 . 4 3 ft ⋅ atm RT RT T T 0.7302 lb-mol ⋅o R Fig. 5.4-2
⇒ z = 0.91
& m & PV & ideal = ideal = 110 . m zT z
⇒ Delivering 10% more than they are charging for (undercharging their customer)
5- 47
5.72 a.
For N2 : Tc = 12620 . K = 227.16 o R, Pc = 335 . atm
U| |V | |W
609.7 o R = 2.68 22716 . oR ⇒ z = 1.02 600 psia 1 atm Pr = = 1.2 33.5 atm 14.7 psia
After heater: Tr =
n& =
150 SCFM = 0.418 lb - mole / min 359 SCF / lb - mole
. 0.418 lb - mole 10.73 ft 3 ⋅ psia 609.7 o R & = zRTn& = 102 V = 4 .65 ft 3 / min o P min 600 psia lb - mole ⋅ R
b.
tank =
0.418 lb - mole 28 lb m / lb - mole 60 min 24 h 7 days 2 weeks min 0.81 62.4 lb m / ft 3 h day week
b g
= 4668 ft 3 = 34,900 gal
5.73 a.
For CO: Tc = 133.0 K, Pc = 34 .5 atm
U| |V || W
300 K = 2.26 133.0 K 2514.7 psia 1 atm Pr1 = = 5.0 34.5 atm 14.7 psia
Initially: Tr1 =
n1 =
Fig. 5.4-3
⇒ z = 1.02
2514.7 psia 150 L 1 atm mol ⋅ K = 1022 mol 1.02 300 K 14.7 psia 0.08206 L ⋅ atm
U| |V || W
300 K = 2 .26 133.0 K 2258.7 psia 1 atm Pr1 = = 4 .5 34.5 atm 14.7 psia
After 60h: Tr1 =
n2 =
n1 − n2 = 1.73 mol / h 60 h
n 2 = y 2 n air = y 2
t min =
⇒ z = 1.02
2259.7 psia 150 L 1 atm mol ⋅ K = 918 mol 1.02 300 K 14.7 psia 0.08206 L ⋅ atm
n& leak =
b.
Fig. 5.4-3
PV 200 × 10 −6 mol CO 1 atm 30.7 m 3 1000 L = = 0.25 mol L ⋅atm RT mol air 0.08206 mol 300 K m3 ⋅K
n2 0.25 mol = = 014 . h n& leak 1.73 mol / h
⇒ t min would be greater because the room is not perfectly sealed
c. (i) CO may not be evenly dispersed in the room air; (ii) you could walk into a high concentration area; (iii) there may be residual CO left from another tank; (iv) the tank temperature could be higher than the room temperature, and the estimate of gas escaping could be low.
5- 48
5.74 CH 4 : Tc = 190.7 K , Pc = 45.8 atm C 2 H6 : Tc = 305.4 K , Pc = 48.2 atm C 2H 4 : Tc = 2831 . K , Pc = 50.5 atm
b gb g b gb g b gb g Pseudocritical pressure: P ′ = b0.20gb45.8g + b0.30gb48.2g + b 0.50gb50.5g = 48.9 atm U| b90 + 273.2gK = 1.34 Reduced temperature: T = |V ⇒ z = 0.71 2713 . K 200 bars 1 atm Reduced pressure: P = = 4.04 | |W 48.9 atm 1.01325 bars Mean molecular weight of mixture: M = b0.20g M + b0.30g M + b0.50gM = b0.20gb16.04g + b0.30gb30.07 g + b0.50gb28.05g Pseudocritical temperature: Tc′ = 0.20 1907 . + 0.30 305.4 + 0.50 2831 . = 2713 . K c
r
Figure 5.4 -3
r
CH4
C2 H 6
C 2 H4
= 26.25 kg kmol V=
znRT 0.71 10 kg 1 kmol = P 26.25 kg
0.08314 m3 ⋅ bar kmol ⋅ K
UV W
b90 + 273g K = 0.041 m (41 L) 3
200 bars
b g b g
b
g
5.75 N 2 : Tc = 126.2 K, PC = 33.5 atm T ′ = 0.10 309.5 + 0.90 126.2 = 144.5 K c N 2 O: Tc = 309.5 K, PC = 71.7 atm Pc′ = 0.10 71.7 + 0.90 33.5 = 37.3 atm
b
g
b
b g
g
M = 0.10 44.02 + 0.90 28.02 = 29.62
b
g
n = 5.0 kg 1 kmol 29.62 kg = 0.169 kmol = 169 mol
a.
b
g
37.3 atm $ = 30 L V r 169 mol 144.5 K
P=
b.
U| mol ⋅ K V ⇒ z = 0.97b Fig. 5.4 - 3g = 0.56| 0.08206 L ⋅ atm W
Tr = 24 + 273.2 144.5 = 2.06
0.97 169 mol 297.2 K 0.08206 L ⋅ atm mol ⋅ K
30 L
|UV g |W
= 133 atm ⇒ 132 atm gauge
Pr = 273 37.3 = 7.32 ⇒ z = 1.14 Fig. 5.4 - 3 $ = 0.56 from a. V r
b
T=
b
g
273 atm 30 L mol ⋅ K = 518 K ⇒ 245° C 1.14 169 mol 0.08206 L ⋅ atm
5- 49
UV W
b g b g
b
g g
5.76 CO: Tc = 133.0 K, Pc = 34.5 atm Tc′ = 0.60 133.0 + 0.40 33 + 8 = 96.2 K H 2 : Tc = 33 K, Pc = 12 .8 atm Pc′ = 0.60 34.5 + 0.40 12.8 + 8 = 29.0 atm
b
U| 1 atm V → z ≈ 1.01 = 4.69 | 14.7 psi W
b
g
Tr = 150 + 273.2 96.2 = 4.4
Turbine inlet:
Pr =
Fig. 5.4-1
2000 psi 29.0 atm
Turbine exit: Tr = 373.2 96.2 = 3.88
⇒ z=1.0
Pr = 1 29.0 = 0.03
& Pin V z nRTin P z T ft 3 14.7 psia 1.01 423.2K in = in ⇒ Vin = Vout × out in in = 15,000 & Pout V z out n RTout Pin z out Tout min 2000 psia 1.00 373.2 out = 126 ft 3 / min
If the ideal gas equation of state were used, the factor 1.01 would instead be 1.00 ⇒ −1% error
UV W
b g b g
b
g
5.77 CO: Tc = 133.0 K, Pc = 34.5 atm Tc′ = 0.97 133.0 + 0.03 304.2 = 138.1 K CO2 : Tc = 304.2 K, Pc = 72.9 atm Pc′ = 0.97 34.5 + 0.03 72.9 = 35.7 atm = 524.8 psi
UV → z = 2014.7 524.8 = 3.8 W
Initial: Tr = 303.2 138.1 = 2.2 Pr
Fig. 5.4-3
1
b g
= 0.97
Final: Pr = 1889.7 524.8 = 3.6 ⇒ z 1 = 0.97 Total moles leaked: n1 − n 2 =
FG P − P IJ V = b2000 − 1875gpsi 0.97 H z z K RT 1
2
30.0 L
1
2
303 K 14.7 psi
mol ⋅ K
1 atm
0.08206 L ⋅ atm
= 10.6 mol leaked
b g
Moles CO leaked: 0.97 10.6 = 10.3 mol CO
Total moles in room: Mole% CO in room =
24.2 m 3 10 3 L 273 K 1m
3
1 mol
b g = 9734. mol
303 K 22.4 L STP
10.3 mol CO × 100% = 1.0% CO 9734 . mol
5- 50
CO + 2H 2 → CH 3OH
5.78 Basis: 54.5 kmol CH 3OH h n& 1 (kmol CO / h) 2n& 1 (kmol H 2 / h) 644 K 34.5 MPa
Catalyst Bed
Condenser
CO, H 2
545 . kmol CH 3OH( l) / h
a.
n& 1 =
54.5 kmol CH 3OH 1 kmol CO react h
1 kmol CO fed
1 kmol CH 3OH
b g
0.25 kmol CO react
b
= 218 kmol h CO
g
2n& 1 = 2 218 = 436 kmol H 2 h ⇒ 218 + 436 = 654 kmol h (total feed)
CO: Tc = 133.0 K
Pc = 34.5 atm
H 2 : Tc = 33 K
Pc = 12.8 atm
⇓ Newton’s corrections
b g b g 1 2 P ′ = b34.5g + b12.8 + 8g = 25.4 atm 3 3 T = 644 71.7 = 8.98 U| 34.5 MPa 10 atm V → z P = = 13.45| 24.5 atm 1.013 MPa W Tc′ =
1 2 1330 . + 33 + 8 = 71.7 K 3 3
c
r
Fig. 5.4 -4
1
= 1.18
r
644 K & feed = 1.18 654 kmol V h 34.5 MPa Vcat =
120 m 3 h
0.08206 m 3 ⋅ atm 1.013 MPa kmol ⋅ K
1 m3 cat 25,000 m3 / h
10 atm
= 120 m 3 h
= 0.0048 m 3 catalyst (4.8 L)
b. CO, H 2 n& 4 kmol CO / h 2n& 4 kmol H 2 / h
54.5 kmol CH3OH (l) / h
Overall C balance ⇒ n& 4 = 54.5 mol CO h Fresh feed:
54.5 kmol CO h 109.0 kmol H 2 h 163.5 kmol feed gas h
644 K & feed = 1.18 163.5 kmol V h 34.5 MPa
0.08206 m 3 ⋅ atm 1.013 MPa kmol ⋅ K
5- 51
10 atm
= 29.9 m 3 h
5.79
H 2 : Tc = ( 33.3 + 8) K = 41.3 K
1 - butene: Tc = 419.6 K
Pc = (12.8 + 8 ) atm = 20.8 atm
Pc = 39.7 atm
Tc ' = 0.15(41.3 K) + 0.85(419.6 K) = 362.8 K
UV P ' = 0.27 W
Tr ' = 0.89
Pc ' = 0.15(20.8 atm) + 0.85(39.7 atm) = 36.9 atm
Fig. 5.4-2
⇒
z = 0.86
r
& 0.86 35 kmol 0.08206 m3 ⋅ atm 323 K 1 h & = znRT V = = 133 . m3 / min P h kmol ⋅ K 10 atm 60 min
F I FG IJ d i GH JK H K
3 2 & & m = u m A m 2 = u × πd ⇒ d = 4V = V min min 4 πu
5.80
CH4 :
d
i FG 100 cmIJ = 10.6 cm π b150 m / ming H m K
4 1.33 m 3 / min
Tc = 190.7 K Pc = 45.8 atm
C2 H4 : Tc = 283.1 K Pc = 50.5 atm C2 H6 : Tc = 305.4 K Pc = 48.2 atm
U| V P ' = 0.15(45.8 atm) + 0.60(50.5 atm) + 0.25(48.2 atm) = 49.2 atm =====> P ' = 35 . |W T=90 o C
Tc ' = 0.15(1907 . K) + 0.60(283.1 K) + 0.25(305.4 K) = 274.8 K ====>
Tr ' = 1.32
P=175 bar
c
r
5.4-3 Fig. → z = 0.67
F I FG IJ d i FG GH JK H K H
3 & m = u m A m 2 = 10 m V s s s
n& =
5.81
IJ FG 60 s IJ π b0.02 mg K H min K 4
2
= 0.188
m3 min
& PV 175 bar 1 atm kmol ⋅ K 0.188 m 3 / min = = 1.63 kmol / min zRT 0.67 1.013 bar .08206 m3 ⋅ atm 363 K
N2 :
Tc = 126.2 K = 227.16o R Pc = 33.5 atm
acetonitrile: Tc = 548 K = 986.4o R
Pc = 47.7 atm Fig. 5.4-3
Tank 1 (acetonitrile): T1 = 550o F, P1 = 4500 psia ⇒ Tr1 = 1.02 Pr1 = 6.4 ⇒ z 1 = 0.80 ⇒ n& 1 =
P1V1 306 atm 0.200 ft 3 = z 1RT1 0.80 1009 .7 o R
lb - mole ⋅o R = 0.104 lb - mole 0.7302 ft 3 ⋅ atm Fig. 5.4-3
Tank 2 (N 2 ): T2 = 550 o F, P2 = 10 atm ⇒ Tr2 = 4.4 , Pr2 = 6.4 ⇒ z 2 = 1.00 ⇒ n& 2 =
P2 V2 10.0 atm 2.00 ft 3 = z 2 RT2 1.00 1009.7 o R
5- 52
lb - mole ⋅o R = 0.027 lb - mole .7302 ft 3 ⋅ atm
5.81 (cont’d)
FG 0.104IJ 986.4 R + FG 0.027 IJ 227.16 R = 830 R H 0.131K H 0.131K . F 0104 IJ 47.7 atm + FG 0.027 IJ 33.5 atm = 44.8 atm P '=G H 0.131K H 0.131 K
Final: Tc ' =
o
o
o
o
T=550 F→
Tr ' = 1.22
c
dV$ i r
P=
=
ideal
$ ' Fig. 5.4-2 VP 2.2 ft 3 44.8 atm lb - mole ⋅o R c = = 1.24 ⇒ z = 0.85 RTc ' 0.131 lb - mole 830 o R 0.7302 ft 3 ⋅ atm
znRT 0.85 0.131 lb - mole .7302 ft 3 ⋅ atm 1009.7 o R = = 37.3 atm V lb - mole ⋅o R 2.2 ft 3
5.82 3.48 g Ca H bO c , 26.8o C, 499.9 kPa n c (mol C), n H (mol H), n O (mol O)
1 L @483.4o C, 1950 kPa n p (mol) 0.387 mol CO2 / mol 0.258 mol O 2 / mol 0.355 mol H 2O / mol
n O2 (mol O 2 ) 26.8o C, 499.9 kPa
a.
d
i
Volume of sample: 3.42 g 1 cm3 159 . g = 2.15 cm 3
O 2 in Charge: n O2 =
d
1.000 L − 2.15 cm 3 10 −3 L km 3 L ⋅ atm mol ⋅ K
0.08206
i
499.9 kPa
1 atm
300 K
101.3 kPa
= 0.200 mol O 2
Product
1.000 L 1950 kPa 1 atm L ⋅ atm = 0.310 mol product 0.08206 756.6 K 101.3 kPa mol ⋅ K Balances: np =
b
g
b
g b
g
O: 2 0.200 + n O = 0.310 2 0.387 + 2 0.258 + 0.355 ⇒ n O = 0.110 mol O in sample
b
g
C: n C = 0.387 0.310 = 0.120 mol C in sample
b
gb
g
H: n H = 2 0.355 0.310 = 0.220 mol H in sample
Assume c = 1 ⇒ a = 0.120 0.110 = 1.1 b = 0.220 0.110 = 2 Since a, b, and c must be integers, possible solutions are (a,b,c) = (11,20,10), (22,40,20), etc. b.
b g
b g
MW = 12.01a + 1.01b + 16.0c = 12.01 1.1c + 1.01 2c + 16.0c = 31.23c 300 < MW < 350 ⇒ c = 10 ⇒ C11H 20 O10
5- 53
bg
C5 H10 +
5.83 Basis: 10 mL C5H10 l charged to reactor
15 O 2 → 5CO2 + 5H 2O 2
bg
10 mL C 5H10 l n1 (mol C5H10 )
n 2 (mol air) 0.21 O 2 0.79 N 2 27 o C, 11.2 L, Po (bar)
a.
bg
10.0 mL C5H 10 l
n1 =
Stoichiometric air: n 2 = Po =
n 3 (mol CO 2 ) n 4 mol H 2 O(v) n 5 (mol N 2 ) 75.3 bar (gauge), Tad
b
0.745 g
1 mol
mL
70.13 g
g
d Ci o
= 0.1062 mol C5 H 10
0.1062 mol C5H10
7.5 mol O2
1 mol air
1 mol C 2 H10 0.21 mol O 2
= 3.79 mol air
nRT 3.79 mol 0.08314 L ⋅ bar 300K = = 8.44 bars V 11.2 L mol ⋅ K
(We neglect the C5H 10 that may be present in the gas phase due to evaporation) Initial gauge pressure = 8.44 bar − 1 bar = 7.44 bar
b.
U || |V || |W
5 mol CO2 = 0.531 mol CO 2 1 mol C 5 H10 0.531 mol CO 2 1 mol H 2 O n4 = = 0.531 mol H 2 O ⇒ 4.052 mol product gas 1 mol CO 2 n 5 = 0.79 3.79 = 2.99 mol N 2 n3 =
0.1062 mol C5 H 10
b g
CO 2 : y 3 = 0.531 / 4.052 = 0.131 mol CO 2 / mol, Tc = 304.2 K Pc = 72.9 atm H 2 O: y 4 = 0.531 / 4.052 = 0.131 mol H 2 O / mol, N2:
y 5 = 2.99 / 4.052 = 0.738 mol N 2 / mol,
Tc = 647.4 K Pc = 218.3 atm Tc = 126.2 K Pc = 33.5 atm
Tc ' = 0.131(304.2 K) + 0.131(647.4 K) + 0.738(126.2 K) = 217.8 K Pc ' = 0.131(72.9 atm) + 0.131(218.3 atm) + 0.738(33.5 atm) = 62.9 atm ⇒ Pr ' = 1.21 $ mol ⋅ K $ ideal = VPc ' = 11.2 L 62.9 atm V = 9.7 ⇒ z ≈ 1.04 (Fig. 5.4 - 3) r RTc ' 4.052 mol 217 .8 K .08206 L ⋅ atm T=
b
g
75.3 + 1 bars 11.2 L PV mol ⋅ K = = 2439 K - 273= 2166o C znR 1.04 4.052 mol 0.08314 L ⋅ bar
5- 54
CHAPTER SIX 6.1
a.
AB: Heat liquid - -V ≈ constant BC: Evaporate liquid - -V increases, system remains at point on vapor - liquid equilibrium curve as long as some liquid is present. T = 100 o C. CD: Heat vapor - -T increas es, V increas es .
b.
Point B: Neglect the variation of the density of liquid water with temperature, so ? = 1.00 g/mL and VB = 10 mL Point C: H2 O (v, 100°C) n=
10 mL
1.00 g 1 mol = 0.555 mol mL 18.02 g
PCVC = nRTC ⇒ VC =
6.2
nRTC 0.555 mol 0.08206 L ⋅ atm 373 K = = 17 L 1 atm mol ⋅ K PC
a. Pfinal = 243 mm Hg . Since liquid is still present, the pressure and temperature must lie on the vapor-liquid equilibrium curve, where by definition the pressure is the vapor pressure of the species at the system temperature. b. Assuming ideal gas behavior for the vapor, (3.000 - 0.010) L mol ⋅ K 243 mm Hg 1 atm 119.39 g m(vapor) = = 4.59 g (30 + 273.2) K 0.08206 L ⋅ atm 760 mm Hg mol m(liquid) =
10 mL
1.489 g mL
= 14.89 g
m total = m(vapor) + m(liquid) = 19.5 g x vapor =
6.3
a.
4.59 = 0.235 g vapor / g total 19 .48
log10 p ∗ = 7.09808 −
1238.71 = 2.370 ⇒ p * = 10 2.370 = 2345 . mm Hg 45 + 217
d
i
* * ∆H$ v 1 ∆H$ v ln p2 / p1 b. ln p = − + B⇒− = 1 1 = R T R T2 − T1 ∗
B = ln( p1* ) +
b
g
ln 760 / 118.3
b
1 77 .0 + 273.2 K
g
−b
∆H$ v / R 4151 K = ln 118.3 + = 18.49 T1 29.5 + 273.2 K
b
g b
g
6-1
1 29 .5+ 273.2 K
g
= −4151K
6.3 (cont’d) ln p ∗ (45 o C) = −
b
4151 + 18.49 ⇒ p ∗ = 231.0 mm Hg 45 + 273.2
g
231.0 − 234.5 × 100% = −1.5% error 2345 .
c.
p∗ =
FG 1183. − 760 IJ b45 − 29.5g + 118.3 = 327.7 mm Hg H 29.5 − 77 K
327.7 − 234.5 × 100% = 39.7% error 2345 .
b
g
1 (rect. scale) on semilog paper T + 273.2 ⇒ straight line: slope = −7076 K , intercept = 21.67
Plot p ∗ log scale vs
b
g
ln p∗ mm Hg =
b
7076 K ∆ Hv = 7076K ⇒ ∆ H$ v = R
g
8.314 J
1 kJ
mol ⋅ K 10 3 J
OP Q
= 58.8 kJ mol
ln p* = A/T(K) + B p*(mm Hg) 5 20 40 100 400 760
1/T(K) 0.002834 0.002639 0.002543 0.002410 0.002214 0.002125
ln(p*) p*(fitted) 1.609 5.03 2.996 20.01 3.689 39.26 4.605 101.05 5.991 403.81 6.633 755.13
7 6 5 4 3 2 1
1/T
6-2
0.003
0.0026
0.0024
0.0022
0
y = -7075.9x + 21.666 0.002
T( oC) 79.7 105.8 120.0 141.8 178.5 197.3
ln(p*)
6.5
LM N
−7076 −7076 + 2167 . ⇒ p∗ mm Hg = exp + 2167 . o T ( C) + 2732 . T ( C) + 2732 . o
0.0028
6.4
T( oC) p*(fitted) 50 0.80 80 5.12 110 24.55 198 760.00 230 2000.00 Least confidence (Extrapolated)
6.6
a. T(°C)
1/T(K)
42.7 58.9 68.3 77.9 88.6 98.3 105.8
3.17×10 -3 3.01×10 -3 2.93×10 -3 2.85×10 -3 2.76×10 -3 2.69×10 -3 2.64×10 -3
p * (mm Hg) =758.9 + hright -hleft 34.9 78.9 122.9 184.9 282.9 404.9 524.9
b. Plot is linear, ln p∗ = −
∆H$ v −51438 . K + B ⇒ ln p∗ = + 19 .855 RT T
At the normal boiling point, p∗ = 760 mmHg ⇒ Tb = 116° C 8.314 J 5143.8 K 1 kJ ∆ H$ v = = 42.8 kJ mol mol ⋅ K 10 3 J
c. Yes — linearity of the ln p∗ vs 1 / T plot over the full range implies validity.
6.7
a.
b
g
ln p∗ = a T + 273.2 + b ⇒ y = ax + b
b
g
y = ln p∗ ; x = 1 T + 273.2
Perry' s Handbook, Table 3 - 8: T1 = 39.5° C , p1 ∗ = 400 mm Hg ⇒ x 1 = 31980 . × 10 − 3 , y1 = 5.99146 T2 = 56.5° C , p 2 ∗ = 760 mm Hg ⇒ x 2 = 3.0331 × 10 −3 , y 2 = 6.63332 T = 50° C ⇒ x = 3.0941 × 10 −3 x − x1 y = y1 + y 2 − y1 = 6.39588 ⇒ p∗ 50° C = e6.39588 = 599 mm Hg x 2 − x1
F GH
Ib JK
g
b
g
b. 50° C = 122° F
c.
6.8
Cox Chart ⇒ p∗ =
12 psi 760 mm Hg
log p∗ = 7.02447 −
1161.0 = 2.7872 ⇒ p∗ = 10 2.7872 = 613 mm Hg 50 + 224
b
14.6 psi
g
Estimate p ∗ 35° C : Assume ln p∗ =
= 625 mm Hg
a + b , interpolate given data. T K
b g
U| ln p∗ b35° Cg = − 6577.1 + 25.97 = 4.630 35 + 273.2 − − |V ⇒ a 6577.1 b = ln p ∗− = lnb50g + = 25.97 | p∗ b35° Cg = e = 102.5 mm Hg |W T 25 + 273.2
a=
b
g=
ln p 2 ∗ p1 ∗ 1 T2
1 T1
b
g
ln 200 50
1 45+ 273.2
1 25 + 273.2
= − 65771 .
4 .630
1
1
6-3
6.8 (cont’d) Moles in gas phase: n =
150 mL
b
273 K 102.5 mm Hg 1L 1 mol 3 35 + 273.2 K 760 mm Hg 10 mL 22.4 L STP
g
b g
= 8.0 × 10 −4 mol
6.9
m = 2 π = 2 ⇒ F = 2 + 2 − 2 = 2 . Two intensive variable values (e.g., T & P) must be specified to determine the state of the system. 1209.6 b. log p∗ MEK = 6.97421 − = 2.5107 ⇒ p∗ MEK = 10 2.5107 = 324 mm Hg 55 + 216. Since vapor & liquid are in equilibrium p MEK = p∗ MEK = 324 mm Hg
a.
⇒ y MEK = p MEK / P = 324 1200 = 0.27 > 0115 . The vessel does not constitute an explosion hazard.
6.10 a. The solvent with the lower flash point is easier to ignite and more dangerous. The solvent with a flash point of 15°C should always be prevented from contacting air at room temperature. The other one should be kept from any heating sources when contacted with air. b. At the LFL, y M = 0.06 ⇒ p M = p *M = 0.06 × 760 mm Hg = 45.60 mm Hg 1473.11 Antoine ⇒ log 10 45.60 = 7.87863⇒ T = 6.85° C T + 230 c. The flame may heat up the methanol-air mixture, raising its temperature above the flash point. 6.11 a. At the dew point, p ∗ ( H 2 O) = p( H 2 O) = 500 × 0.1= 50 mm Hg ⇒ T = 38.1° C from Table B.3.
b. VH2 O =
30.0 L
273 K 500 mm Hg 1 mol 0.100 mol H2 O 18.02 g 1 cm3 =134 . cm3 (50 + 273) K 760 mm Hg 22.4 L (STP) mol mol g
c. (iv) (the gauge pressure)
6-4
6.12 a.
b g = 110°C , p ∗ = 755 mm Hg − b577 − 222 gmm Hg = 400 mm Hg
T1 = 58.3° C , p1 ∗ = 755 mm Hg − 747 − 52 mm Hg = 60 mm Hg T2
2
ln p∗ = a=
a +b T K
b g
b
g=
b
ln p 2 ∗ p1 ∗ 1 T2
−
b = ln p1 ∗−
1 T1
g
ln 400 60 1 110+ 273.2
−
1 58.3 + 273 .2
= −46614 .
b g
a 4661.4 = ln 60 + = 18.156 T1 58.3 + 273.2
T=130o C=403.2 K −46614 . + 18.156 T ln p∗ 130° C = 6.595 ⇒ p∗ 130° C = e 6.595 = 7314 . mm Hg
ln p∗ =
b
b.
g
b
g
Basis: 100 mol feed gas CB denotes chlorobenzene. n 1 mol @ 58.3°C, 1atm y1 (mol CB(v)/mol) (sat’d) (1-y1 ) (mol air/mol)
100 mol @ 130°C, 1atm y0 (mol CB(v)/mol) (sat’d) (1-y0 ) (mol air/mol)
n 2 mol CB (l)
b
g
Saturation condition at inlet: y o P = p CB ∗ 130° C ⇒ y o =
731 mm Hg = 0.962 mol CB mol 760 mm Hg
b
60 mm Hg = 0.0789 mol CB mol g 760 mm Hg Air balance: 100b1 − y g = n b1 − y g ⇒ n = b100gb1 − 0.962g b1 − 0.0789g = 4.126 mol Total mole balance: 100 = n + n ⇒ n = 100 − 4.126 = 95.87 mol CBbl g Saturation condition at outlet: y 1 P = p CB ∗ 58.3° C ⇒ y 1 = o
1
1
1
% condensation:
b
2
1
2
95.87 mol CB condensed × 100% = 99 .7% 0.962 × 100 mol CB feed
g
c. Assumptions: (1) Raoult’s law holds at initial and final conditions; (2) CB is the only condensable species (no water condenses); (3) Clausius-Clapeyron estimate is accurate at 130°C. 6.13 T = 78° F = 25.56° C , Pbar = 29.9 in Hg = 759.5 mm Hg , hr = 87%
b
y H 2O P = 0.87 p∗ 25.56° C
g
Table B.3
yH 2O =
0.87 ( 24.559 mm Hg)
( )
759.5 mm Hg
Dew Point: p ∗ Tdp = yH2 O P = 0.0281( 759.5 ) = 21.34 mm Hg
6-5
= 0.0281 mol H 2 O mol air
Table B.3
Tdp = 23.2 °C
6.13 (cont’d) 0.0281 = 0.0289 mol H2 O mol dry air 1 − 0.0281 0.0289 mol H2O 18.02 g H2O mol dry air ha = = 0.0180 g H2 O g dry air mol dry air mol H2O 29.0 g dry air hm =
hp =
hm
p ∗ ( 25.56 °C ) P − p ∗ ( 25.56 °C )
×100% =
0.0289 × 100 = 86.5% 24.559 [759.5 − 24.559 ]
6.14 Basis I : 1 mol humid air @ 70° F (21.1° C), 1 atm, h r = 50%
b
h r = 50% ⇒ y H 2 O P = 0.50 p H 2 O ∗ 21.1° C Table B.3
Mass of air:
y H 2O =
g
0.50 × 18.765 mm Hg mol H 2 O = 0.012 760.0 mm Hg mol
0.012 mol H 2 O 18.02 g 0.988 mol dry air 29.0 g + = 28.87 g 1 mol 1 mol
Volume of air:
1 mol
b g b273.2 + 21.1gK = 24.13 L
22.4 L STP 1 mol
273.2K
28.87 g = 1196 . g L 24.13 L Basis II : 1 mol humid air @ 70° F (21.1° C), 1 atm, hr = 80% Density of air =
b
h r = 80% ⇒ y H 2 O P = 0.80 p H 2 O ∗ 21.1° C Table B.3
Mass of air:
y H 2O =
0.80 × 18.765 mm Hg mol H 2 O = 0.020 760.0 mm Hg mol
0.020 mol H 2 O 18.02 g 1 mol
Volume of air:
Density of air =
1 mol
g
+
0.980 mol dry air 29.0 g 1 mol
b g b273.2 + 21.1gK = 24.13 L
22.4 L STP 1 mol
273.2K
28.78 g = 1193 . g L 24.13 L
Basis III: 1 mol humid air @ 90° F (32.2° C), 1 atm, h r = 80%
b
h r = 80% ⇒ y H2 O P = 0.80 p H2 O ∗ 32.2° C Table B.3
y H 2O =
g
0.80 × 36.068 mm Hg mol H 2 O = 0.038 760.0 mm Hg mol
6-6
= 28.78 g
6.14 (cont’d) Mass of air:
0.038 mol H 2 O 18.02 g
Volume of air:
Density =
1 mol
+
0.962 mol dry air
29.0 g 1 mol
= 28.58 g
b g b 273.2 + 32.2g K = 25.04 L
1 mol
22.4 L STP 1 mol
273.2K
28.58 g = 1141 . g L 25.04 L
Increase in T ⇒ increase in V ⇒ decrease in density Increase in hr ⇒ more water (MW = 18), less dry air (MW = 29) ⇒ decrease i n m ⇒ decrease in density Since the density in hot, humid air is lower than in cooler, dryer air, the buoyancy force on the ball must also be lower. Therefore, the statement is wrong.
b
6.15 a. h r = 50% ⇒ y H2 O P = 0.50 p H2 O ∗ 90° C Table B.3
y H 2O =
g
0.50 × 52576 . mm Hg = 0.346 mol H 2 O / mol 760.0 mm Hg
b g
d i
Dew Point: y H2 O p = p∗ Tdp = 0.346 760 = 262.9 mm Hg
Table B.3
Tdp = 72.7 ° C
Degrees of Superheat = 90 − 72.7 = 17.3° C of superheat
b. Basis:
1 m3 feed gas 10 3 L 273K m
3
363K
mol
b g = 33.6 mol
22.4 L STP
n 1 mol @ 25°C, 1atm y1 (mol H2 O (v)/mol) (sat’d) (1-y1 ) (mol air/mol)
33.6 mol @ 90°C, 1atm 0.346 H2 O mol /mol 0.654 mol air/mol
n 2 mol H2 O (l)
Saturation Condition: y 1 =
b
p *H 2O 25° C
g = 23.756 = 0.0313 mol H O mol
P 760 Dry air balance: 0.654 (33.6 ) = n1 (1 − 0.0313 ) ⇒ n1 = 22.7 mol
2
Total mol balance: 33.6=22.7+n2 ⇒ n2 = 10.9 mol H 2 O condense/m 3
b
g
c. y H 2O P = p∗ 90° C ⇒ P =
p * ( 90° C) 525.76 mmHg = = 1520 mm Hg = 2.00 atm yH 2 O 0.346
6-7
6.16 T = 90° F = 32.2° C , p = 29.7 in Hg = 754.4 mm Hg , hr = 95% Basis: 10 gal water condensed/min n&condensed =
1 ft 3
10 gal H2O min
62.43 lb m
7.4805 gal
ft
18.02 lb m
y2 (lb-mol H2 O (v)/lb-mol) (sat’d) (1-y2 ) (lb-mol DA/lb-mol) 40o F (4.4o C), 754 mm Hg
y1 (lb-mol H2 O (v)/lb-mol) (1-y1 ) (lb-mol DA/lb-mol) h r =95%, 90o F (32.2o C), 29.7 in Hg (754 mm Hg)
4.631 lb-moles H2 O (l)/min
b
g
95% hr at inlet: y H2 O P = 0.95 p ∗ 32.2° C
y H 2O =
= 4.631 lb-mole/min
n& 2 (lb - moles / min)
V&1 (ft 3 / min) n&1 (lb - moles / min)
Table B.3
1 lb-mol
3
b
g
0.95 36.068 mm Hg = 0.045 lb - mol H 2O lb - mol 754.4 mm Hg
Raoult's law: y2 P = p* ( 4.4° C)
Table B.3
y2 =
6.274 = 0.00817 lb-mol H 2O lb-mol 754.4
Mole balance: n&1 = n&2 + 4.631 n&1 = 124.7 lb-moles/min ⇒ Water balance: 0.045 n&1 = 0.00817 n&2 + 4.631 n& 2 = 120.1 lb-moles/min Volume in: V& =
124.7 lb-moles 359 ft 3 (STP) (460+90) oR min
lb-moles
o
492 R
760 mm Hg 754 mm Hg
= 5.04 ×104 ft 3 /min
6.17 a. Assume no water condenses and that the vapor at 15°C can be treated as an ideal gas. p final =
760 mm Hg
(15 + 273) K (200 + 273) K
= 462.7 mm Hg ⇒ ( p H2 O ) final = 0.20 × 462.7 = 92.6 mm Hg
p * (15° C) = 12.79 mm Hg < p H2 O . Impossible ⇒ condensation occurs.
Tfinal 288 K = ( 0.80 × 760) mm Hg × = 370.2 mm Hg Tinitial 473 K = 370.2 + 12.79 = 383 mm Hg
( p air ) final = ( p air ) initial
P = p H 2 O + p air
b. Basis:
1 L 273 K 473 K
mol 22.4 L (STP)
= 0.0258 mol
6-8
6.17 (cont’d)
n 1 mol @ 15°C, 383.1 mm Hg y1 (mol H2 O (v)/mol) (sat’d) (1-y1 ) (mol dry air/mol)
0.0258 mols @ 200°C, 760 mm Hg 0.20 H2 O mol /mol 0.80 mol air/mol
n 2 mol H2 O (l)
Saturation Condition: y1 =
b
b
p *H2 O 15° C P
g = 12.79 mm Hg = 0.03339 mol H O mol 2
383.1 mm Hg
g b
g
Dry air balance: 0.800 0.0258 = n1 1 − 0.03339 ⇒ n1 = 0.02135 mol
c.
Total mole balance: 0.0258 = 0.02135 + n 2 ⇒ n 2 = 0.00445 mol
Mass of water condensed =
0.00445 mol
18.02 g mol
= 0.0802 g
6.18 Basis: 1 mol feed 3
n2 (mol), 15.6°C, 3 atm y 2 (mol H 2O (v)/mol)(sat'd) (1 – y 2) (mol DA/mol)
V1(m ) 1 mol, 90°C, 1 atm 0.10 mol H O 2 (v)/mol 0.90 mol dry air/mol
heat
100°C, 3 atm n2 (mol) 3
V2(m )
n3 (mol) H 2O( l), 15.6°C, 3 atm
Saturation: y2 =
b
p *H2 O 15.6° C
g
Table B.3
P
y2 =
bg b g H O mol balance: 0.10b1g = 0.00583b 0.9053g + n
13.29 mm Hg
atm
3 atm
760 mm Hg
= 0.00583
Dry air balance: 0.90 1 = n 2 1 − 0.00583 ⇒ n 2 = 0.9053 mol 2
Fraction H 2 O condensed: hr =
3
⇒ n3 = 0.0947 mol
0.0947 mol condensed = 0.947 mol condense mol fed 0100 . mol fed
b
g
y2 P × 100% 0.00583 3 atm = × 100% = 1.75% p∗ 100° C 1 atm
b
g
6-9
6.18 (cont’d)
b g
V2 =
0.9053 mol 22.4 L STP
V1 =
1 mol 22.4 L STP
273K 3 atm 10 L 1 m3
363K
= 2 .98 × 10 −2 m 3 feed air @ 90° C
273K 10 3 L
mol
= 9.24 × 10 −3 m 3 outlet air @ 100° C
3
mol
b g
1 m3
373K 1 atm
V2 9.24 × 10 −3 m 3 outlet air = = 0.310 m 3 outlet air m 3 feed air −2 3 V1 2.98 × 10 m feed air
25 L 1.00 kg
6.19 Liquid H 2 O initially present: Saturation at outlet: y H2 O =
⇒
b
P
Evaporation Rate:
b g
15 L STP min
23.76 mm Hg = 0.0208 mol H 2 O mol air 1.5 × 760 mm Hg
1 mol = 0.670 mol dry air min 22.4 L STP
b g
0.670 mol dry air
0.0212 mol H 2 O
min
mol dry air
Complete Evaporation:
1.387 kmol 10 3 mol
1h
0.0142 mol 60 min
= 1628 h
b67.8 daysg
7.069 × 103 ft 3 7.481 gal π ⋅ 302 ⋅ (18 − 8) = = 5.288 × 104 gal / day 4 day ft 3
5.288 × 10 4 gal
1 ft 3
0.703 × 62.43 lb m
day
7.481 gal
ft 3
( SG ) C8 H18 = 0.703 ⇒
0.703 × 62 .43 lb m
= 0.0142 mol H 2 O min
min
kmol
6.20 a. Daily rate of octane use =
∆p =
g=
= 1.387 kmol H 2 O l
18.02 kg
0.0208 = 0.0212 mol H 2 O mol dry air 1 − 0.0208
Flow rate of dry air:
b.
L
p *H2 O 25° C
bg
1 kmol
= 3.10 × 10 5 lb m C 8 H 18 / day
32.174 ft
ft 3
c. Table B.4: p *C8 H18 (90 o F) =
s2
1 lb f lb ⋅ ft 32.174 m2 s
20.74 mm Hg
(18 - 8) ft
14.696 psi
29.921 in Hg 14.696 lb f / in 2
= 6.21 in Hg
= 0.40 lb f / in 2 = p octane = y octane P 760 mm Hg Octane lost to environment = octane vapor contained in the vapor space displaced by liquid during refilling. Volume:
5.288 × 10 4 gal
1 ft 3 7.481 gal
= 7069 ft 3
6-10
6.20 (cont’d) pV (16.0 + 14.7) psi 7069 ft 3 = = 36.77 lb - moles RT 10.73 ft 3 ⋅ psi / (lb - mole ⋅o R) (90 + 460) o R pC H 0.40 psi Mole fraction of C 8 H18 : y = 8 18 = = 0.0130 lb - mole C 8 H18 / lb - mole P (16.0 + 14.7) psi
Total moles: n =
Octane lost = 0.0130( 36.77 ) lb - mole = 0.479 lb - mole (= 55 lb m = 25 kg)
d. A mixture of octane and air could ignite. * * 6.21 a. Antoine equation ⇒ p tol (85o F) = p tol (29.44 o C) = 35.63 mmHg = p tol
Mole fraction of toluene in gas: y =
p tol 35.63 mmHg = = 0.0469 lb - mole toluene / lb - mole P 760 mmHg
yPV RT 0.0469 lb - mole tol
Toluene displaced = yntotal = =
1 atm ft ⋅ atm 3
lb - mole
0.7302
lb - mole ⋅ o R
1 ft 3
900 gal
(85 + 460) o R 7.481 gal
92.13 lb m tol lb - mole
= 1.31 lb m toluene displaced
b. Basis: 1mol 0.0469 mol C7 H8 (v)/mol 0.9531 mol G/mol
n V (mol) y (mol C7 H8 (v)/mol) (1-y) (mol G/mol) T(o F), 5 atm
Assume G is noncondensable
n L [mol C7 H8 (l)] 90% of C7 H8 in feed
90% condensation ⇒ n L = 0.90( 0.0469 )(1) mol C 7 H 8 = 0.0422 mol C 7 H 8 ( l ) Mole balance: 1 = nV + 0.0422 ⇒ nV = 0.9578 mol Toluene balance: 0.0469(1) = y ( 0.9578 ) + 0.0422 ⇒ y = 0.004907 mol C 7 H 8 / mol Raoult’s law:
* p tol = yP = (0.004907 )(5 × 760) = 18.65 mmHg = p tol (T )
Antoine equation: T=
B − C ( A − log 10 p* ) 1346.773 − 219.693(6.95805 − log10 18.65) = = 17.11o C=62.8 o F A − log10 p * 6.95805 − log10 18.65
6-11
6.22 a. Molar flow rate: n& =
& VP 100 m 3 kmol ⋅ K 2 atm = = 6.53 kmol / h -3 3 RT h 82.06 × 10 m ⋅ atm (100 + 273) K
b. Antoine Equation: 1175.817 = 3.26601 100+224.867 ⇒ p * = 1845 mm Hg
log 10 p *Hex (100°C)=6.88555-
p Hex = y Hex ⋅ P =
0.150(2.00) atm 760 mm Hg atm
p *Hex (T ) = 228 mm Hg ⇒ log10 228=6.88555-
c. 6.53 kmol/h 0.15 C6 H14 (v) 0.85 N2
= 228 mm Hg < p *Hex ⇒ not saturated
1175.817 = 2.35793 ⇒ T = 34.8°C T+224.867
n V (kmol/h) y (kmol C6 H14 (v)/kmol), sat’d (1-y) (kmol N2 /kmol) T (o C), 2 atm
n L (kmol C6 H14 (l)/h) 80% of C6 H14 in feed
80% condensation: n L = 0.80(0.15)(6.53 kmol / h) = 0.7836 kmol C 6 H 14 ( l) / h Mole balance: 6.53 = nV + 0.7836 ⇒ nV = 5.746 kmol / h Hexane balance: 0.15(6.53) = y (5.746 ) + 0.7836 ⇒ y = 0.03409 kmol C 6 H 14 / kmol p Hex = yP = ( 0.03409)(2 × 760 mmHg) = 51.82 mmHg = p *Hex (T ) 1171530 . Antoine equation: log10 5182 . = 6.87776 − ⇒ T = 2.52 o C T + 224.366
Raoult’s law:
6.23 Let H=n-hexane a. n& 0 ( kmol / min) y 0 (kmol H(v)/kmol (1-y 0 ) (kmol N2 /kmol) 80o C, 1 atm, 50% rel. sat’n
Condenser
n&1 ( kmol / min) 0.05 kmol H(v)/kmol, sat’d 0.95 kmol N2 /kmol T (o C), 1 atm 1.50 kmol H(l)/min
50% relative saturation at inlet: y o P = 0.500 p *H (80 o C) Table B.4
yo =
(0.500)(1068 mmHg) = 0.703 kmolH / kmol 760 mmHg
Saturation at outlet: 0.05 P = p *H (T1 ) ⇒ p *H ( T1 ) = 0.05(760 mmHg) = 38 mmHg
6-12
6.23 (cont’d) Antoine equation: log 10 38 = 6.88555 −
1175.817 ⇒ T1 = −3.26o C T1 + 224.867
UV ⇒ RSn& = 2.18 kmol / min = 0.95n& W Tn& = 0.682 kmol / min
Mole balance: n&0 = n&1 + 150 . N 2 balance: (1 − 0.703)n&0
0
1
1
(0.95)0.682 kmol 22.4 m 3 (STP ) N2 volume: V&N 2 = = 14.5 SCMM min kmol
b.
Assume no condensation occurs during the compression
2.18 kmol/min 0.703 H(v) 0.297 N2 80o C, 1 atm
Compressor
V&0 ( m 3 / min ) 2.18 kmol/min 0.703 H(v) 0.297 N2 T0 (o C), 10 atm, 50% R.S.
V&1 (m3 / min) 0.682 kmol/min 0.05 H(v), sat’d 0.95 N2 T1 (o C), 10 atm Condenser
1.5 kmol H(l)/min
50% relative saturation at condenser inlet: 0.500 p *H (T0 ) = 0.703(7600 mmHg) ⇒ p *H (T0 ) = 1.068 × 10 4 mmHg
Saturation at outlet: 0.050(7600 mmHg) = 380 mmHg = p *H (T1 ) Volume ratio :
Antoine
Antoine
T0 = 187 o C T1 = 48.2° C
V&1 n1R T1 / P n1 (T1 + 273.2) 0.682 kmol/min 321 K m 3 out = = = × = 0.22 3 V&0 n0 RT0 / P n0 (T0 + 273.2) 2.18 kmol/min 460 K m in
c. The cost of cooling to −3.24 o C (installed cost of condenser + utilities and other operating costs) vs. the cost of compressing to 10 atm and cooling at 10 atm. 6.24 a.
Maximum mole fraction of nonane achieved if all the liquid evaporates and none escapes. (SG)nonane n max =
15 L C 9 H 20 (l ) 0.718 × 1.00 kg
kmol
L C 9 H 20
128.25 kg
Assume T = 25o C, P = 1 atm n gas =
2 × 10 4 L 273 K
1
kmol
298 K 22.4 × 10 L(STP) 3
6-13
= 0.818 kmol
= 0.084 kmol C 9 H 20
6.24 (cont’d) y max =
n max 0.084 kmol C 9 H 20 = = 0.10 kmol C 9 H 20 / kmol (10 mole%) n gas 0.818 kmol
As the nonane evaporates, the mole fraction will pass through the explosive range (0.8% to 2.9%). The answer is therefore yes . The nonane will not spread uniformly—it will be high near the sump as long as liquid is present (and low far from the sump). There will always be a region where the mixture is explosive at some time during the evaporation. b. ln p * = −
A +B T
T1 = 258 . o C = 299 K, p1* = 5.00 mmHg T2 = 66.0 o C = 339 K, p *2 = 40.0 mmHg
ln( 40.0 / 5.00) 5269 5269 ⇒ A = 5269, B = ln(5.00) + = 19.23 ⇒ p * = exp(19.23 − ) 1 1 299 T ( K) − 339 299 At lower explosion limit, y = 0.008 kmol C 9 H 20 / kmol ⇒ p * (T ) = yP = (0.008)(760 mm Hg) −A =
= 6.08 mm Hg
Formula for p *
T = 302 K = 29 o C
c. The purpose of purge is to evaporate and carry out the liquid nonane. Using steam rather than air is to make sure an explosive mixture of nonane and oxygen is never present in the tank. Before anyone goes into the tank, a sample of the contents should be drawn and analyzed for nonane. 6.25 Basis: 24 hours of breathing n0 (mol H 2O) 23°C, 1 atm n1 (mol) @ hr = 10% 0.79 mol N 2/mol y1 (mol H 2O/mol) + O2 , CO2
Lungs
O2
Air inhaled: n 1 =
37°C, 1 atm n2 (mol), saturated 0.75 mol N /mol 2 y 2 (mol H 2O/mol) + O2 , CO 2
CO2
12 breaths 500 ml 1 liter min breath 10 3 ml
b
273K 1 mol 23 + 273 K 22.4 liter STP
g
60 min 24 hr 1 hr 1 day
b g
= 356 mol inhaled day
Inhaled air - -10% r. h.: y1 = Inhaled air - -50% r. h.: y1 =
b
0.10 p∗ H2 O 23° C P
b
g = 0.10b 2107 . mm Hgg = 2.77 × 10
0.50 p∗ H 2O 23° C P
−3
mol H 2 O mol
−2
mol H 2 O mol
760 mm Hg
g = 0.50b2107 . mm Hg g = 1.39 × 10 760 mm Hg
6-14
6.25 (cont’d) H 2 O balance: n 0 = n 2 y 2 − n1 y 1 ⇒ ( n0 ) 10%
FG H
= 356
IJ L K MN
rh
− ( n0 ) 50%
rh
= ( n1 y 1 ) 50% − (n1 y 1 ) 10%
mol H 2 O mol (0.0139 − 0.00277) day mol
OPFG 18.0 gIJ = 71 g / day QH 1 mol K
Although the problem does not call for it, we could also calculate that n 2 = 375 mol exhaled/day, y2 = 0.0619, and the rate of weight loss by breathing at 23o C and 50% relative humidity is n 0 (18) = (n 2y2 - n 1 y1 )18 = 329 g/day. 6.26 a. To increase profits and reduce pollution. b. Assume condensation occurs. A=acetone n 1 mol @ To C, 1 atm 1 mol @ 90o C, 1 atm
y1 mol A(v)/mol (sat’d) (1-y1 ) mol N2 /mol
0.20 mol A(v)/mol 0.80 mol N2 /mol
n 2 mol A(l)
For cooling water at 20o C
(
)
log 10 p *A 20 o C = 7.11714 −
(
)
1210.595 = 2.26824 ⇒ p*A 20 o C = 184.6 mmHg 20.0 + 229.664
d
i
Saturation: y1 ⋅ P = p *A 20 o C ⇒ y1 =
184.6 = 0.243 > 0.2 , so no saturation occurs. 760
For refrigerant at –35o C
(
)
log 10 p *A −35 o C = 7.11714 −
(
)
1210.595 = 0.89824 ⇒ p *A −35 o C = 7.61 mmHg −35.0 + 229.644
Note: –35o C is outside the range of Antoine equation coefficients in Table B.4. If the correct vapor pressure of acetone at that temperature is looked up (e.g., in Perry’s Handbook) and used, the final result is almost identical. 7.61 = 0.0100 d i 760 mole balance: 1b0.8g = n b1 − 0.01g ⇒ n = 0.808 mol
Saturation: y1 ⋅ P = p *A −35 o C ⇒ y1 = N2
1
1
Total mole balance: 1 = 0.808 + n2 ⇒ n 2 = 0192 . mol 0.192 ×100% = 96% 2 c. Costs of acetone, nitrogen, cooling tower, cooling water and refrigerant d. The condenser temperature could never be as low as the initial cooling fluid temperature because heat is transferred between the condenser and the surrounding environment. It will lower the percentage acetone recovery.
Percentage acetone recovery:
6-15
6.27
Basis:
12500 L 1 mol 273 K 103000 Pa = 5285 . mol / h h 22.4 L(STP) 293 K 101325 Pa
n o (mol/h) @ 35o C, 103 KPa y0 [mol H2 O(v)/mol] yo1–mol H2 O(v)/mol y0 (mol DA/mol) (1-y o ) mol DA/mol h r=90% h r =90%
528.5 (mol/h) @ 20o C, 103 KPa y0 [mol H2 O(v)/mol] y11–mol H2 O(v)/mol (sat’d) y0 (mol DA/mol) (1-y1 ) mol DA/mol H2 O(l)/h] 2 [mol n 2nmol H2O(l)/h
Inlet: y o =
d
hr ⋅ p H* 2 O 35 o C P
Outlet: y1 =
p *H 2O
b
i = 0.90 × 42.175 mmHg 103000 Pa
101325 Pa = 0.4913 mol H 2 O / mol 760 mmHg
d20 Ci = 17.535 mmHg 101325 Pa = 0.02270 mol H O / mol o
P
103000 Pa
g b
760 mmHg
gb
2
g
Dry air balance: 1 − 0.04913 n o = 1 − 0.02270 528.5 ⇒ n o = 5432 . mol / h 5432 . mol 22.4 L(STP) 308 K 101325 Pa = 13500 L / h h mol 273 K 103000 Pa Total balance: 543.2 = 528.5 + n 2 ⇒ n 2 = 14.7 mol / h
Inlet air :
Condensatio n rate:
6.28 Basis:
14.7 mol 18.02 g H 2 O 1 kg = 0.265 kg / h h 1 mol H 2 O 1000 g
10000 ft 3 1 lb - mol 492 o R 29.8 in Hg = 24.82 lb - mol / min min 359 ft 3 (STP) 550 o R 29.92 in Hg
n 1 lb -mole/min 40o F, 29.8 in.Hg y1 [lb-mole H2 O(v)/lb-mole] 1- y1 (lb-mole DA/mol)
24.82 lb -mole/min 90o F, 29.8 in.Hg y0 [lb-mole H2 O(v)/mol 1- y0 (lb-mole DA/mol) h r = 88%
n 1 lb-mole/min 65o F, 29.8 in.Hg y1 [lb-mole H2 O(v)/lb-mole] 1- y1 (lb-mole DA/lb-mole)
n 2 [lb-mole H2 O(l)/min]
Inlet: y o =
d
i = 0.88b36.07 mmHgg
hr ⋅ p *H2 O 90 o F
Outlet: y1 =
P
29 .8 in Hg
d
1 in Hg = 0.0419 lb - mol H 2O / lb - mol 25.4 mmHg
i = 6.274 mmHg
p H* 2 O 40 o F P
b
1 in Hg = 0.00829 lb - mol H 2 O / lb - mol 29 .8 in Hg 25.4 mmHg
g b
g
Dry air balance: 24.82 1 − 0.0419 = n1 1 − 0.00829 ⇒ n1 = 23.98 lb - mol / min Total balance: 24.82 = 23.98 + n 2 ⇒ n 2 = 0.84 lb - mole / min
6-16
6.28 (cont’d) 0.84 lb - mol 18.02 lb m 1 ft 3 7.48 gal = 181 . gal / min min lb − mol 62.4 lb m 1 ft 3
Condensation rate:
Air delivered @ 65o F:
23.98 lb - mol 359 ft 3 (STP) 525 o R 29.92 in Hg = 9223 ft 3 / min min 1 lb − mol 492 o R 29.8 in Hg
6.29 Basis: 100 mol product gas
no mol, 32oC, 1 atm yo mol H2O(v)/mol (1-yo) mol DA/mol hr=70%
100 mol, T1, 1 atm
100 mol, 25oC,1 atm
y1 mol H2O(v)/mol, (sat’d) (1-y1) mol DA/mol
y1 mol H2O(v)/mol, (1-y1) mol DA/mol hr=55%
(mol HH22O(l)) nn2 2lb-mol O(l)/min
Outlet: y1 =
d
h r ⋅ p H* 2 O 25o C
i = 0.55b23.756g = 0.0172 mol H O / mol 2
P 760 Saturation at T1 : 0.0172 760 = 13.07 = p *H2 O T1 ⇒ T1 = 15.3 o C
Inlet: yo =
hr ⋅
b g b g d32 Ci = 0.70b35.663g = 0.0328 mol H O / mol
p *H2 O
o
P
2
760
b
g
b
g
Dry air balance: n o 1 − 0.0328 = 100 1 − 0.0172 ⇒ no = 1016 . mol Total balance: 101.6 + n 2 = 100.0 ⇒ n 2 = −1.6 mol (i.e. removed) kg H 2 O removed :
kg dry air : Ratio:
b
16 . mol 18.02 g 1 kg = 0.0288 kg H 2 O 1 mol 1000 g
g
100 1 − 0.0172 mol 29.0 g 1 kg = 2.85 kg dry air 1 mol 1000 g
0.0288 = 0.0101 kg H 2O removed / kg dry air 2.85
6-17
6.30 a.
Room air − T = 22 ° C , P = 1 atm , hr = 40% :
b
g
y1 P = 0.40 P∗ H 2 O 22° C ⇒ y 1 =
b 0.40g19.827 mm Hg = 0.01044 mol H O mol 2
760 mm Hg
Second sample − T = 50° C , P = 839 mm Hg , saturated:
y2 P = p ∗H 2 O (50 °C ) ⇒ y2 =
92.51 mm Hg = 0.1103 mol H2 O mol 839 mm Hg
ln y = bH + ln a ⇔ y = ae bH , y 1 = 0.01044 , H1 = 5 , y 2 = 01103 . , H 2 = 48
b=
b
ln y 2 y 1
g = lnb0.1103 0.01044 g = 0.054827
H2 − H1
48 − 5
b
g b gb g expb 0.054827 H g
b
g
ln a = ln y1 − bH 1 = ln 0.01044 − 0.054827 5 = −4.8362 ⇒ a = exp −4.8362 = 7.937 × 10 −3
⇒ y = 7.937 × 10 −3 b.
Basis:
1 m3 delivered air
273K
b22 + 273gK
1 k mol
10 3 mol
22.4m 3 STP
1 kmol
b g
o 41.31 mol, 1 atm 41.4 mol, 22o22 C,1C,atm
41.31 mol, T, 1 atm n o mol, 35o C, 1 atm
0.0104 mol H22 O(v)/mol, (sat’d) sat’d mol DA/mol 0.09896 0.9896 mol DA/mol
yo mol H2 O(v)/mol (1-yo ) mol DA/mol H=30
= 41.31 mol air delivered
0.0104 0.0104mol molH2HO(v)/mol 2 O(v)/mol 0.09896 0.9896 mol mol DA/mol DA/mol
n 1 mol H2 O(l)
Saturation condition prior to reheat stage: yH 2 O P = p*H2 O (T ) ⇒ ( 0.01044 )( 760 mm Hg ) = 7.93 mm Hg ⇒ T = 7.8°C (from Table B.3)
bg
Part a
Humidity of outside air: H = 30 ⇒ y 0 = 0.0411 mol H 2 O mol
Overall dry air balance: n0 (1 − y0 ) = 41.31 ( 0.9896 ) ⇒ n0 =
( 41.31)(0.9896 ) = 42.63 mol (1 − 0.0411)
Overall water balance: n0 y0 = n1 + ( 41.31)( 0.0104 ) ⇒ n1 = ( 42.63 )( 0.0411) − ( 41.31)( 0.0104 ) = 1.32 mol H2 O condensed Mass of condensed water =
1.32 mol H 2 O 18.02 g H 2 O 1 kg 1 mol H 2 O 10 3 g
= 0.024 kg H 2 O condensed m 3 air delivered
6-18
6.31 a.
Basis: n& 0 mol feed gas . S = solvent , G = solvent - free gas n 1 (mol) @ T f (°C), P4 (mm Hg) y 1 [mol S(v)/mol] (sat’d) (1–y 1 ) (mol G/mol)
n 0 (mol) @ T0 (°C), P0 (mm Hg) y 0 (mol S/mol) (1-y 0 ) (mol G/mol) Td0 (°C) (dew point)
Inlet dew point = T0 ⇒
n 2 (mol S (l))
b g
y o Po = p ∗ Tdo ⇒ y o =
b g
p ∗ Tdo Po
d i
Saturation condition at outlet: y 1 Pf = p∗ T f ⇒ y1 = Fractional condensation of S = f ⇒ n 2 = n0 y 0 f Total mole balance: n& 0 = n1 + n 2 ⇒ n1 = n&0 − n2
(1)
d i
p∗ T f
(2)
Pf
b gP n& fp∗ bT g −
(1) → n 2 = n0 fp∗ T0
bg
Eq. 3 for n1
⇒
n1 = n& 0
b gb g
0
0
do
Po
S balance: n0 y 0 = n1 y 1 + n2 (1) - (4)
b g OPFG p∗dT PQGH P
iIJ + n& fp∗bT g JK p LM fp∗ eTdo j OP p∗ dT i 1 − MM PP P o b1 − f g p∗bT g = LM1 − fp∗ bT g OP p∗ dT i ⇒ P = N Q P P P MN PQ p∗ T b1 − f g eP do j b g LM MN
n& 0 p∗ Tdo n& 0 fp∗ Tdo = n& 0 − Po Po
f
0
f
do
o
f
⇒
do
f
do
f
o
o
f
o
b.
Condensation of ethylbenzene from nitrogen Antoine constants for ethylbenzene A= 6.9565 B= 1423.5 C= 213.09 Run T0 P0 Td0 f 1 2 3 4
50 50 50 50
765 765 765 765
40 40 40 40
0.95 0.95 0.95 0.95
Tf
p* (Td0) p*(Tf)
45 40 35 20
21.472 21.472 21.472 21.472
6-19
27.60 21.47 16.54 7.07
Pf 19139 14892 11471 4902
Crefr Ccomp Ctot 2675 4700 8075 26300
107027 109702 83329 88029 64239 72314 27582 53882
(3)
(4)
6.31 (cont’d) When Tf decreases, Pf decreases. Decreasing temperature and increasing pressure both to c. increase the fractional condensation. When you decrease Tf, less compression is required to achieve a specified fractional condensation. d.
6.32 a.
A lower Tf requires more refrigeration and therefore a greater refrigeration cost (Crefr). However, since less compression is required at the lower temperature, Ccomp is lower at the lower temperature. Similarly, running at a higher Tf lowers the refrigeration cost but raises the compression cost. The sum of the two costs is a minimum at an intermediate temperature. Basis : 120m 3 min feed @ 1000 o C(1273K), 35 atm . Use Kay’s rule. Cmpd. Tc ( K )
Pc ( atm )
(T c )corr . ( Pc )corr ( Apply Newton's corrections for H2 )
H2
33.2
12.8
41.3
20.8
CO
133.0
34.5
−
−
CO 2
304.2
72.9
−
−
CH 4
190.7
45.8
−
−
∑y T P′ = ∑ y P
Tc′ = c
i ci i
ci
b g b g b g b g = 0.40b20.8g + 0.35b 34.5g + 0.20b72.9 g + 0.05b458 . g = 37.3 atm
= 0.40 41.3 + 0.35 133.0 + 0.20 304.2 + 0.05 1907 . = 133.4 K
Feed gas to heater Tr = 1273 K 133.4 K = 9.54 Fig. 5.3-2 ⇒ z = 1.02 Pr = 35.0 atm 37.3 atm = 0.94 Vˆ =
1.02
8.314 N ⋅ m
1273 K
1 atm
mol ⋅ K 35 atm 101325 N m
⇒ n&gas feed =
120 m 3 min
−3
3.04 ×10
mol
1 kmol
3
103 mol
m
2
= 3.04 × 10 −3 m 3 mol
= 39.5 kmol min
Feed gas to absorber Fig. 5.4-1 Tr = 283K/133.4K = 2.12, Pr = 35.0 atm/37.3 atm = 0.94 → z = 0.98 znRT 0.98 39.5 kmol V& = = P min
8.314 N ⋅ m
283 K
1 atm
mol ⋅ K 35.0 atm 101325 N/m 2
103 mol 1 kmol
n 1 (kmol/min), 261 K, 35atm
1.2(39.5) kmol/min MeOH(l)
y NaOH sat’d y H2 y CH4 (2% of feed) y CO
39.5 kmol/min, 283K, 35 atm 0.40 mol H2 /mol 0.35 mol CO/mol 0.20 mol CO2 /mol 0.05 mol CH4 /mol
n 2 (kmol/min), liquid y MeOH y CO2 y CH4 (98% of feed)
6-20
= 25.7
m3 min
6.32 (cont’d)
Saturation at Outlet: y McOH
10 7.87863 −1473 .11 b −12+ 2300g mm Hg p∗ MeOH 261K = = P 35 atm 760 mm Hg atm
b
g
b
g
= 4.97 × 10 −4 mol MeOH mol y McOH =
b.
6.33
nMeOH
n MeOH n + H 2 + n CH4
A = input
A =0.02 of input
+ n CO
=
nMeOH
A
=input
n MeOH + 39.5(0.40 + 0.02 ( 0.05) + 0.35)
E
n MeOH = 0.0148 kmol min MeOH in gas
The gas may be used as a fuel. CO2 has no fuel value, so that the cost of the added energy required to pump it would be wasted.
n& 0 (kmol/min wet air) @ 28°C, 760 mmHg y 1 (mol H2 O/mol) (1-y 1 ) (mol dry air/mol) 50% rel. sat.
y 2 (mol H2 O/mol) (1-y 2 ) (mol dry air/mol) Tdew point = 40.0o C
m& 1 (kg/min wet pulp)
1500 kg/min wet pulp 0.75 /(1 + 0.75) kg H2 O/kg 1/1.75 kg dry pulp/kg
Dry pulp balance: 1500 ×
n& 1 (kmol/min wet air) @ 80°C, 770
0.0015 kg H2 O/kg 0.9985 kg dry pulp/kg
1 =m & 1 (1 − 0.0015) ⇒ m & 1 = 858 kg / min 1 + 0.75
50% rel. sat’n at inlet: y1 P = 0.50 pH* 2 O (28 o C) ⇒ y1 = 0.50(28.349 mm Hg)/(760 mm Hg) = 0.0187 mol H 2 O/mol
40 C dew point at outlet: y 2 P = o
p *H2 O ( 40 o C)
⇒ y 2 = (55.324 mm Hg) / (770 mm Hg) = 0.0718 mol H 2O / mol
Mass balance on dry air: n& 0 (1 − 0.0187 ) = n&1 (1 − 0.0718)
(1)
Mass balance on water: n& 0 ( 0.0187 )(18.0 kg / kmol ) + 1500 ( 0.75 / 1.75) = n&1 ( 0.0718 )(18 ) + 858 (0 .0015) (2 ) Solve (1) and (2) ⇒ n& 0 = 622.8 kmol / min, n&1 = 658.4 kmol / min Mass of water removed from pulp : [1500(0.75/1.75)–858(.0015)]kg H2 O = 642 kg / min 622.8 kmol 22.4 m3 (STP) (273 + 28) K Air feed rate : V&0 = = 1.538 × 10 4 m 3 / min min kmol 273 K
6-21
6.34
Basis: 500 lb m hr dried leather (L) n&1 (lb - moles / h)@130o F, 1 atm
n& 0 ( lb - moles dry air / h)@140o F, 1 atm
y1 (lb - moles H2 O / lb - mole) (1 - y1 )(lb - moles dry air / lb - mole)
m& 0 (lb m / h)
500 lb m / h
0.61 lb m H 2O(l) / lb m 0.39 lb m L / lb m
0.06 lb m H 2 O(l) / lb m 0.94 lb m L/ lb
b gb g
Dry leather balance: 0.39 m0 = 0.94 500 ⇒ m0 = 1205 lb m wet leather hr
b
g
Humidity of outlet air: y 1 P = 0.50 p∗ H 2O 130° F ⇒ y 1 =
b gb
g
a
0.50 (115 mm Hg) mol H 2 O = 0.0756 760 mmHg mol
fb
H 2 O balance: 0.61 1205 lb m hr = ( 0.06) 500 lb m hr +
b
E
g
0.0756n1 lb - moles H 2 O 18.02 lb m hr 1 lb - mole
n1 = 517.5 lb - moles hr
g 359 ft bSTPg b140 + 460 g° R = 2.09 × 10
Dry air balance: n 0 = 1 − 0.0756 (5175 . ) lb - moles hr = 478.4 lb - moles hr
Vinlet =
6.35 a.
478.4 lb - moles hr
3
5
492 ° R
1 lb - mole
ft 3 hr
Basis: 1 kg dry solids n1 (kmol)N 2, 85°C
n 2 (kmol) 80°C, 1 atm y 2 (mol Hex/mol) (1 – y )2 (mols N 2/mol) 70% rel. sat.
dryer 1.00 kg solids 0.78 kg Hex
condenser
n 3 (kmol) 28°C, 5.0 atm y 3 (mol Hex/mol) sat'd (1 – y)3 (mols N 2/mol) n 4 (kmol) Hex(l)
0.05 kg Hex 1.00 kg solids
Mol Hex in gas at 80° C
b0.78 − 0.05gkg
kmol 86.17 kg
= 8.47 × 10 −3 kmol Hex
Antoine eq.
↓
70% rel. sat.: y 2 =
0.70 p ∗hex ( 80°C ) P
6.885551175.817 − (80+ 224.867) 0.70 )10 ( =
760
6-22
= 0.984 mol Hex mol
6.35 (cont’d) n2 =
8.47 × 10 −3 kmol Hex
1 kmol 0.984 kmol Hex
b
= 0.0086 kmol
g
N 2 balance on dryer: n1 = 1 − 0.984 0.0086 = 1376 . × 10 −4 kmol Antoine Eq.
↓
− ( 28+224.867) p ∗hex ( 28 °C ) 106.885551175.817 Saturation at outlet: y3 = = = 0.0452 mol Hex mol P 5 ( 760)
b
g
Overall N 2 balance: 1.376 × 10 -4 = n 3 1 − 0.0452 ⇒ n 3 = 1.44 × 10 −4 kmol Mole balance on condenser: 0.0086 = 1.44 × 10 −4 + n 4 ⇒ n 4 = 0.0085 kmol
Fractional hexane recovery:
b.
0.0085 kmol cond. 86.17 kg 0.78 kg feed
kmol
= 0.939 kg cond. kg feed
Basis: 1 kg dry solids
0.9n
3
heater
0.9n 3 (kmol) @ 28°C, 5.0 atm
y3 (1 – y3)
n 1 (kmol)N 2 85°C dryer 1.00 kg solids 0.78 kg Hex
y 3 (mol Hex/mol) sat'd (1 – y3) (mol N 2/mol)
n 2 (kmol) 80°C, 1 atm y 2 (mol Hex/mol) (1 – y 2) (mols N2 /mol) 70% rel. sat.
condenser
n3 (kmol) y3 (1 – y 3)
0.1n3
n4 (kmol) Hex(l)
0.05 kg Hex 1.00 kg solids
Mol Hex in gas at 80°C: 8.47x10-3 + 0.9n 3 (0.0452) = n 2 (0.984)
(1)
N2 balance on dryer: n1 + 0.9n 3 (1 − 0.0452 ) = n 2 (1 − 0.984 )
(2 )
Overall N2 balance: n1 = 0.1n3 (1 − 0.0452)
(3)
n1 = 1.38 × 10 kmol Equations (1) to (3) ⇒ n2 = 0.00861 kmol −4 n3 = 1.44 × 10 kmol 1.376 × 10 -4 −1.38 × 10−5 Saved fraction of nitrogen= ×100% = 90% 1.376 ×10 −4 −5
Introducing the recycle leads to added costs for pumping (compression) and heating.
6-23
6.36 b. m& 1 (lb m/h)
300 lb m/h wet product 0.2 = 0167 . lb m T(l) / lb m 1 + 0.2 0.833 lb m D / lb m
0.02 / (1.02 ) = 0.0196 lb m T(l) / 0.9804 lb m D / lb m
Dryer
n&3 ( lb-mole/h)
n& 1 (lb-mole/h)
y1 (lb-mole T(v)/lb-mole) (1–y1 ) (lb-mole N2 /lbmole) T=toluene 70% r.s.,150o F, 1.2 atm D=dry solids
@ 200O F, y3 (lb-mole T/lb-mole) (1-y3 )( lb-mole N2 /lb-mole) Heater
n&3 ( lb-mole/h)
y3 (lb-mole T(v)/lb-mole) (1-y3 ) (lb-mole N2 /lb-mole)
Condenser
Eq.@ 90O F, 1atm
n& 2 ( lb-mole T(l)/h )T(l)
Strategy: Overall balance⇒ m& 1 & n& 2 ; Relative saturation⇒y1; , Gas and liquid equilibrium⇒y3 Balance over the condenser⇒ n&1 & n& 3
UV RS W T
Toluene Balance: 300 × 0.167 = m& 1 × 0.0196 + n&2 × 92.13 & = 255 lb m / h m ⇒ 1 & 1 × 0.9804 Dry Solids Balance: 300 × 0.833 = m n&2 = 0.488 lb - mole / h
70% relative saturation of dryer outlet gas: pC*7 H8 (150 O F=65.56 O C)=10
(6.95805 −
y1 P = 0.70 pC* 7 H8 (150 O F) ⇒ y 1 =
1346.773 ) 65.56 + 219.693
0.70 pC*7 H8 P
=
= 172.47 mmHg
(0.70)(172.47) = 01324 . lb - mole T(v) / lb - mole 1.2 × 760
Saturation at condenser outlet: 1346.773 (6.95805) * o o 32.22 + 219.693 pC7 H8 (90 F=32.22 C)=10
y3 =
p*C7 H8 P
=
= 40.90 mmHg
40.90 = 0.0538 mol T(v)/mol 760
Condenser Toluene Balance: n&1 × 0.1324 = 0.488 + n& 3 × 0.0538 Condenser N 2 Balance: n&1 × (1 − 0.1324) = n& 3 × (1 − 0.0538)
6-24
UV ⇒ RSn& = 5.875 lb - mole / h W Tn& = 5.387 lb - mole / h 1 3
6.36 (cont’d) Circulation rate of dry nitrogen = 5.875 × (1 - 0.1324) =
5.097 lb - mole lb - mole h 28.02 lb m
= 0.182 lb m / h
Vinlet =
6.37
b g
5.387 lb - moles 359 ft 3 STP hr
(200 + 460)° R 492° R
1 lb - mole
C 6 H 14 +
Basis: 100 mol C 6 H 14 100 mol C 6H
= 2590 ft 3 h
19 O2 → 6CO 2 + 7H 2 O 2 n1 (mol) dry gas, 1 atm 0.821 mol N /mol D.G. 2 0.069 mol CO 2 /mol D.G. 0.069 mol CO /mol D.G. 2 0.021 mol CO/mol D.G. 0.021 mol CO/mol D.G. 0.00265 mol C6 H14 /mol 0.086 molOO/mol) /mol D.G. 2 x (mol 2 0.00265 mol C H /mol D.G. 6 14/mol) (0.907–x) (mol N 2 n2n (mol H O) 2 O) (mol H
14
n 0 (mol) air 0.21 mol O 2/mol 0.79 mol N 2/mol
L O C balance: 6b100g = n M0.069 + 0.021+ 6b0.00265gP ⇒ n = 5666 mol dry gas MN b g b g b g PQ 100 − 0.00265b5666g mol reacted Conversion: × 100% = 85.0% 100 mol fed H balance: 14b100g = 2 n + 5666b14 gb0.00265g ⇒ n = 595 mol H O p∗ d T i 595 Dew point: y = = ⇒ p∗ dT i = 72.2 mm Hg ⇒ 595 + 5666 760 mm Hg 2
1
C O2
2
1
CO
C6 H14
2
2
2
Table B.3
dp
H 2O
dp
N2 balance: 0.79n0 = 5666(0.907 − x ) O balance: 0.21(n0 )(2) = 5666[(0.069)(2) + 0.021 + 2 x ) + 595
Solve simultaneously to obtain n 0 = 5888 mol air, x = 0.086 mol O2 /mol Theoretical air:
Excess air:
100 mol C 2 H14
19 mol O 2
1 mol air
2 mol C 2 H 14
0.21 mol O 2
5888 − 4524 × 100% = 30.2% excess air 4524
6-25
= 4524 mol air
Tdp = 451 . °C
6.38 Basis: 1 mol outlet gas/min n& 0 ( mol / min) y0 ( mol CH 4 / mol) (1 − y0 ( mol C 2 H 6 / mol)
1 mol / min @ 573K, 105 kPa y1 (mol CO2 / mol)
n&1 (mol O2 / min)
y 2 (mol H 2 O / mol)
3.76n& 1 (mol N 2 / min)
(1 − y1 − y2 ) mol N 2 / mol
CH 4 + 2O 2 → CO 2 + 2H 2 O p CO2 = 80 mmHg ⇒ y1 =
C2 H6 +
7 O 2 → 2CO 2 + 3H 2 O 2
80 mmHg 101325 Pa = 0.1016 mol CO2 / mol 105000 Pa 760 mmHg 7 2
b
g
100% O2 conversion : 2no y o + no 1 − y o = n1
(1)
. C balance: no yo + 2no 1 − yo = 01016
(2)
N2 balance: 376 . n1 = 1 − y1 − y2
(3)
H balance: 4 no y o + 6no 1 − y o = 2 y2
(4)
b
g
b
g
R|n = 0.0770 mol |y = 0.6924 mol CH / mol Solve equations 1 to 4 ⇒ S ||n = 0.1912 mol O . mol H O / mol Ty = 01793 Dew point: 01793 . b105000g Pa 760 mmHg = 141.2 mmHg ⇒ T p dT i = o
o
4
1
2
2
* H2 O
dp
2
dp
101325 Pa
b
6.39 Basis: 100 mol dry stack gas n P (mol C 3H 8) n B (mol C 4 H10 ) n out (mol) 0.21 O2 0.79 N2
P = 780 mm Hg Stack gas: Tdp = 46.5°C 100 mol dry gas 0.000527 mol C3 H 8/mol 0.000527 mol C 4 H 10/mol 0.0148 mol CO/mol 0.0712 mol CO 2/mol + O2, N 2
nw (mol H2 O)
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
C 4 H10 +
6-26
g
= 58.8 o C Table B.3
13 O 2 → 4CO 2 + 5H 2 O 2
6.39 (cont’d)
b
g
Dew point = 46.5° C ⇒ y w P = p ∗ w 46.5° C ⇒ y w =
But yw =
77.6 mm Hg mol H 2 O = 0.0995 780 mm Hg mol
nw = 0.0995 ⇒ nw = 11.05 mol H2 O (Rounding off strongly affects the result) 100 + n w
b gb
gb g b
gb g
C balance: 3n p + 4 n B = 100 0.000527 3 + 0.000527 4 + 0.0148 + 0.0712
⇒
b1g
3n p + 4 n B = 8.969
H balance: 8 n p + 10 nB = (100 ) ( 0.000527 )(8 ) + ( 0.000527 )(10 ) + (11.05 )( 2 )
( 2)
⇒ 8n p + 10 nB = 23.047
49% C3 H8 n p = 1.25 mol C3H8 Solve (1) & ( 2 ) simultaneously: ⇒ ⇒ nB = 1.30 mol C4 H10 51% C 4 H10 ( Answers may vary ± 8% due to loss of precision )
6.40 a.
L&1 (lb - mole C10 H 22 / h)
L& 2 (lb - mole / h) x 2 (lb - mole C 3 H 8 / lb - mole) 1 − x 2 (lb - mole C10 H 22 / lb - mole)
G& 1 (lb - mole / h)
G& 2 = 1 lb - mole / h
y1 (lb - mole C 3 H8 / lb - mole)
0.07 (lb - mole C 3 H 8 / lb - mole)
1 − y 1 (lb - mole N 2 / lb - mole)
0.93 (lb - mole N 2 / lb - mole)
Basis: G& 2 = 1 lb - mole h feed gas
b gb g
b
g
b
g = b1 − 0.985gb1gb0.07 g ⇒ G& y
b1g b2 g
N 2 balance: 1 0.93 = G& 1 1 − y1 ⇒ G& 1 1 − y 1 = 0.93
98.5% propane absorption ⇒ G& 1 y1 . × 10 1 1 = 105 1 & 2 ⇒ G&1 = 0.93105 lb - mol h , y1 = 1128 . × 10 −3 mol C 3H 8 mol Assume G& 2 − L& 2 streams are in equilibrium
bg b g
−3
From Cox Chart (Figure 6.1-4), p * C3 H8 (80 o F ) = 160 lb / in 2 = 10.89 atm
b
g
Raoult's law: x 2 p∗ C3 H8 80° F = 0.07 p ⇒ x 2 =
b0.07gb1.0 atmg = 0.006428 mol H
b gb g
10.89 atm
Propane balance: 0.07 1 = G& 1 y1 + L&2 x 2 ⇒ L& 2 = = 10.726 lb - mole h
b
gd h b d L& / G& h
b
gd
mol
0.07 − 0.93105 1128 . × 10 −3
i
0.006428
gb
g
Decane balance: L&1 = 1 − x 2 L& 2 = 1 − 0.006428 10.726 = 10.66 lb - mole h
⇒
1
2 min
= 10.7 mol liquid feed / mol gas feed
6-27
2O
6.40 (cont’d) b. The flow rate of propane in the exiting liquid must be the same as in Part (a) [same feed rate and fractional absorption], or n& C 3H 8 =
10.726 lb - mole 0.006428 lb - mole C 3H 3 h
lb - mole
= 0.06895 lb - mol C 3 H 8 h
The decane flow rate is 1.2 x 10.66 = 12.8 lb-moles C10H22 /h ⇒ x2 =
b
0.06895 lb - mole C 3 H 8 h = 0.00536 lb - mole C 3 H 8 / lb - mole 0.06895 + 12.8 lb - moles h
g
c. Increasing the liquid/gas feed ratio from the minimum value decreases the size (and hence the cost) of the column, but increases the raw material (decane) and pumping costs. All three costs would have to be determined as a function of the feed ratio. Let B = n - butane , HC = other hydrocarbons
6.41 a. Basis: 100 mol/s liquid feed stream
n& 4 (mol/s) @ 30°C, 1 atm y 4 (mol B/mol) (1-y 4 ) (mol N2 /mol)
100 mol/s @ 30o C, 1 atm xB =12.5 mol B/s 87.5 mol other hydrocarbon/s
n& 3 (mol N2 /s)
88.125 mol/s 0.625 mol B/s (5% of B fed) 87.5 mol HC/s
p *B (30 o C) ≅ 41 lb / in 2 = 2120 mm Hg (from Figure 6.1-4) x B p *B (30 o C) 0.125 × 2120 = = 0.3487 P 760 95% n-butane stripped: n& 4 ⋅ 0.3487 = 12.5 0.95 ⇒ n& 4 = 34.06 mol / s Raoult's law: y 4 P = x B p B* (30 o C) ⇒ y 4 =
b
g b gb g
Total mole balance: 100 + n&3 = 34.06 + 88.125 ⇒ n&3 = 22.18 mol/s
⇒ mol gas fed = 22.18 mol/s = 0.222 mol gas fed/mol liquid fed mol liquid fed
100 mol/s
b. If y 4 = 0.8 × 0.3487 = 0.2790 , following the same steps as in Part (a),
b
g b gb g
95% n-butane is stripped: n& 4 ⋅ 0.2790 = 12.5 0.95 ⇒ n&4 = 42 .56 mol / s Total mole balance: 100 + n&3 = 42.56 + 88.125 ⇒ n&3 = 30.68 mol / s mol gas fed 30.68 mol/s ⇒ = = 0.307 mol gas fed/mol liquid fed mol liquid fed 100 mol/s c. When the N2 feed rate is at the minimum value calculated in (a), the required column length is infinite and hence so is the column cost. As the N2 feed rate increases for a given liquid feed rate, the column size and cost decrease but the cost of purchasing and compressing (pumping) the N2 increases. To determine the optimum gas/liquid feed ratio, you would need to know how the column size and cost and the N2 purchase and compression costs depend on the N2 feed rate and find the rate at which the cost is a minimum.
6-28
6.42 Basis: 100 mol NH 3 Preheated air 100 mol NH 3 780 kPasat’d sat'd 820 kPa, 820 kPa, sat’d
N2 O2
converter
n3 n4 n5 n6
n 1 (mol) O2 3.76 n 1 (mol) N2 n 2 (mol) H2 O 1 atm, 30°C h r= 0.5
a. i)
hydrator absorber
(mol NO) (mol N 2) (mol O2 ) (mol H 2O)
55 wt% HNO 3 (aq ) n 8 (mol HNO 3 ) n 9 (mol H 2O)
n7 (mol H 2O)
b g
NH 3 feed: P = P∗ Tsat = 820 kPa = 6150 mm Hg = 8.09 atm
Antoine : log10 6150 = 7 .55466 − 1002.711 Tsat + 247885 . ⇒ Tsat = 18.4° C = 291.6 K
b g
b
Table B.1 ⇒ V NH 3 =
g
UV ⇒ z = 0.92 W
Pc = 1113 . atm ⇒ Pr = 8.09 / 1113 . = 0.073 Tc = 405.5 K ⇒ Tr = 291.6 / 405.5 = 0.72
b
0.92 100 mol
g
8.314 Pa
291.6 K
mol - K
820 × 10 3 Pa
(Fig. 5.3-1)
= 0.272 m 3 NH 3
Air feed: NH 3 + 2 O 2 → HNO 3 + H 2 O
n1 =
100 mol NH 3
2 mol O 2 mol NH 3
b
= 200 mol O 2
g
hr ⋅ p * 30° C 0.500 × 31824 . Water in Air: y H2 O = = = 0.02094 p 760 n2 ⇒ 0.02094 = ⇒ n 2 = 20.36 mol H 2O n 2 + 4.76( 200)
A
( 4 .76 mol air mol O2 )
Vair =
b g
b g
4.76 200 + 20.36 mol 22.4 L STP 1 mol
303K
1 m3 3
273K 10 L
= 24.2 m 3 air
ii) Reactions: 4 NH 3 + 5O 2 → 4 NO + 6 H 2 O , 4 NH 3 + 3O 2 → 2 N 2 + 6 H 2 O Balances on converter NO: n 3 =
97 mol NH 3
4 mol NO 4 mol NH 3
= 97 mol NO
6- 29
6.42 (cont’d)
b g
N 2 : n 4 = 3.76 2.00 mol + O 2 : n 5 = 200 mol −
3 mol NH 3
2 mol N 2 4 mol NH 3
97 mol NH 3
= 7535 . mol N 2
5 mol O 2 4 mol NH 3
−
3 mol NH 3
3 mol O 2 4 mol NH 3
H 2 O: n 6 = 20.36 mol +
100 mol NH 3
= 76.5 mol O2
6 mol H 2 O 4 mol NH 3
= 1704 . mol H 2 O
⇒ n total = ( 97 + 7535 . + 76.5 + 170.4 )mol = 1097 mol converter effluent 8.8% NO, 68.7% N 2 , 7.0% O 2 , 15.5% H 2 O
iii) Reaction: 4 NO + 3O 2 + 2 H 2 O → 4 HNO 3 HNO 3 bal. in absorb er: n8 = H 2 O in product: n9 =
97 mol NO react 4 mol HNO 3 4 mol NO
97 mol HNO3
63.02 g HNO 3
= 97 mol HNO 3
45 g H 2 O
mol
1 mol H 2 O
55 g HNO 3 18.02 g H 2 O
= 27756 . mol H 2 O
b
gb g
b
gb gb
H balance on absorber: 1704 . 2 + 2n 7 = 97 + 277 .6 2 mol H
g
⇒ n7 = 1557 . mol H 2 O added VH 2 O =
b.
155.7 mol H 2 O 18.02 g H 2 O 1 cm 3 1 mol
1 m3 10 6 cm 3
1g
bg
= 2.81 × 10 −3 m 3 H 2 O l
63.02 g HNO3 277.6 mol H 2 O 18.02 g HNO3 + mol mol = 11115 g = 11115 . kg
M acid in old basis =
Scale factor =
97 mol HNO 3
b1000 metric tonsgb1000 kg metric ton g = 8.997 × 10 11.115 kg
d
id i = d8.997 × 10 id24.2 m air i = 2 .18 × 10 m air = d8.997 × 10 id2.81 × 10 m H Oi = 253 m H Obl g
VNH3 = 8.997 × 10 4 0.272 m 3 NH 3 = 2.45 × 10 4 m 3 NH 3 Vair
VH 2 O
4
3
4
6
−3
3
3
3
2
6- 30
2
4
6.43 a. Basis: 100 mol feed gas 100 mol 0.10 mol NH3 /mol 0.90 mol G/mol
G = NH 3 -free gas Absorber
n 1 (mol H2 O( l))
n 2 (mol) in equilibrium y A (mol NH 3 /mol) at 10°C(50°F) y W (mol H 2O/mol) and 1 atm y G (mol G/mol) n 3 (mol) x A (mol NH 3 /mol) (1 – x A) (mol H 2O/mol)
Composition of liquid effluent . Basis: 100 g solution
Perry, Table 2.32, p. 2-99: T = 10o C (50o F), ρ = 0.9534 g/mL ⇒ 0.120 g NH3 /g solution ⇒
12.0 g NH 3 88.0 g H 2 O = 0.706 mol NH 3 , = 4.89 mol NH 3 (17.0 g / 1 mol) (18.0 g / 1 mol)
⇒ 12.6 mole% NH 3 (aq), 87.4 mole% H 2 O(l)
Composition of gas effluent
g U| = 0.155 psia bTable 2 - 21gV |W = 14 .7 psia b
T = 50 F, x A = 0.126 → o
Perry
p NH 3 = 121 . psia Table 2 - 23 pH 2O p total
y A = 1.21 / 14.7 = 0.0823 mol NH 3 mol
⇒ y W = 0.155 / 14.7 = 0.0105 mol H 2 O mol y G = 1 − y A − yW = 0.907 mol G mol
b gb g b gb g b 0.907 g = 99.2 mol absorbed = b100gb0.10 g − b99.2gb0.0823g = 184 . mol NH
G balance: 100 0.90 = n 2 y G ⇒ n 2 = 100 0.90 NH 3
in
% absorption =
3
out
1.84 mol absorbed × 100% = 18.4% 100 0.10 mol fed
b gb g
b. If the slip stream or densitometer temperature were higher than the temperature in the contactor, dissolved ammonia would come out of solution and the calculated solution composition would be in error.
6.44 a. 15% oleum: Basis - 100kg 15 kg SO 3 +
85 kg H 2SO 4
1 kmol H 2SO 4
1 kmol SO 3
80.07 kg SO 3
98.08 kg H 2SO 4
1 kmol H 2 SO 4
1 kmol SO 3
⇒ 84.4% SO 3
6- 31
= 84 .4 kg
6.44 (cont’d) b.
Basis 1 kg liquid feed n o (mol), 40oC, 1.2 atm
n1 (mol), 40oC, 1.2 atm
0.90 mol SO3 /mol 0.10 mol G/mol
y1 mol SO3/mol (1-y 1) mol G/mol Equilibrium @ 40o C
1 kg 98% H2SO4
m1 (kg) 15% oleum
0.98 kg SO3 0.02 kg H2O
0.15 kg SO3 /kg 0.85 kg H2SO4 /kg
b
p SO3 40° C, 84.4%
g = 115 . = 151 . × 10
−3 mol SO 3 mol P 760 ii) 0.98 kg H 2SO 4 2.02 kg H 0.02 kg H 2 O 2.02 kg H H balance: + 98.08 kg H 2SO 4 18.02 kg H 2 O 0.85 m 1 H 2 SO 4 2.02 kg H = ⇒ m1 = 1.28 kg 98.08 kg H 2SO 4 But since the feed solution has a mass of 1 kg, 0.28 kg SO 3 10 3 g 1 mol SO 3 absorbed = 1.28 − 1.0 kg = = 3.50 mol kg 80.07 g ⇒ 3.5 mol = n 0 − n1 G balance: 0.10n 0 = 1 − 151 . × 10 −3 n1 1444444444444442444444444444443
i)
y1 =
b
g
d
i
E
n 0 = 3.89 mol n1 = 0.39 mol V=
b g
3.89 mol
22.4 L STP
1 kg liquid feed
mol
313K
1 atm
1 m3
273K 1.2 atm 10 3 L
= 8.33 × 10 -2 m 3 kg liquid feed
6.45 a. Raoult’s law can be used for water and Henry’s law for nitrogen. b. Raoult’s law can be used for each component of the mixture, but Henry’s law is not valid here. c. Raoult’s law can be used for water, and Henry’s law can be used for CO2. 6.46 p ∗B (100°C) = 10 ∗∗ ( 6.89272 − 1203.531 (100 + 219.888 ) ) = 1350.1 mm Hg pT∗ (100 °C ) = 10∗∗ ( 6.95805 − 1346.773 (100 + 219.693) ) = 556.3 mm Hg
Raoult's Law: y B P = x B p B∗ ⇒
yB = yT =
0.40 (1350.1) 10 (760 )
0.60 ( 556.3) 10 ( 760)
= 0.0711 mol Benzene mol
= 0.0439 mol Toluene mol
y N2 = 1 − 0.0711 − 0.0439 = 0.885 mol N 2 mol
6- 32
6.47 N 2 - Henry' s law: Perry' s Chemical Engineers' Handbook, Page. 2 - 127, Table 2 -138
b
g
⇒ H N 2 80° C = 12 .6 × 10 4 atm mole fraction
b gd i 3551 . mm Hg 1 atm H O - Raoult's law: p b80° Cg = = 0.467 atm 760 mm Hg ⇒ p = d x id p i = b 0.997gb0.467 g = 0.466 atm ⇒ p N 2 = x N 2 H N2 = 0.003 12 .6 × 10 4 = 378 atm ∗ H 2O
2
H 2O
H 2O
∗ H2 O
Total pressure: P = p N 2 + p H2 O = 378 + 0.466 = 378.5 atm Mole fractions:
y H 2O = p H2 O P = 0 / 466 / 3785 . = 1.23 × 10 −3 mol H 2 O mol gas y N 2 = 1 − y H2 O = 0.999 mol N 2 mol gas
b
g
6.48 H 2 O - Raoult's law: p ∗H2 O 70° C =
2337 . mm Hg
1 atm 760 mm Hg
b
gb
= 0.3075 atm
g
⇒ p H 2O = xH 2 O p ∗H2 O = 1 − xm 0.3075 Methane − Henry' s law: p m = x m ⋅ Hm
Total pressure: P = p m + p H2 O = x m ⋅ 6.66 × 10 4 + (1 − x m )(0.3075) = 10 ⇒ x m = 1.46 × 10 −4 mol CH 4 / mol
6.49 a.
Moles of water : n H 2 O =
1000 cm 3
1g cm
3
mol 18.02 g
= 55.49 mol
Moles of nitrogen: n N2 =
(1 - 0.334) × 14.1 cm 3 (STP )
1 mol 1L = 4.192 × 10 −4 mol 3 22.4 L (STP) 1000 cm
Moles of oxygen: n O2 =
(0.334) ⋅ 14.1 cm3 (STP)
mol L = 2.102 × 10 −4 mol 22.4 L (STP) 1000 cm3
Mole fractions of dissolved gases: n N2 4 .192 × 10 −4 x N2 = = n H 2O + n N 2 + n O2 55.49 + 4.192 × 10 −4 + 2.102 × 10 −4
x O2
= 7.554 × 10 −6 mol N 2 / mol nO2 2.102 × 10 −4 = = n H2 O + n N2 + nO2 55.49 + 4.192 × 10 −4 + 2 .102 × 10 −4 = 3.788 × 10 −6 mol O 2 / mol
6- 33
6.49 (cont’d) Henry' s law p N2
Nitrogen: HN 2 =
0.79 ⋅ 1 = 1046 . × 10 5 atm / mole fraction 7.554 × 10 −6
0.21 ⋅ 1 = 5544 . × 10 4 atm / mole fraction x O2 3.788 × 10 −6 Mass of oxygen dissolved in 1 liter of blood: p O2
Oxygen: HO2 =
b.
x N2
= =
2.102 × 10 -4 mol 32.0 g = 6.726 × 10 −3 g mol 0.4 g O2 1 L blood Mass flow rate of blood: m& blood = = 59 L blood / min min 6.72 × 10 -3 g O 2 c. Assumptions: (1) The solubility of oxygen in blood is the same as it is in pure water (in fact, it is much greater); (2) The temperature of blood is 36.9°C. m O2 =
6.50 a.
bg
Basis: 1 cm 3 H 2 O l H2O =1.0 (SG) →
1 g H 2 O 1 mol = 0.0555 mol H 2 O 18.0 g
= 0.0901
CO2 →
( SC)
b g
0.0901 cm3 STP CO 2
1 mol = 4.022 × 10 − 6 mol CO 2 3 22,400 cm STP
b g
d4.022 × 10 i mol CO = 7.246 × 10 mol CO d0.0555 + 4.022 × 10 i mol 1 atm ⇒ H b20° Cg = = 13800 atm mole fraction 7.246 × 10 −6
p CO2 = x CO2 HCO2
b.
−5
2
p CO2 = 1 atm ⇒ x CO 2 =
−6
2
mol
−5
CO2
b g
For simplicity, assume n total ≈ n H 2 O mol
b
xCO2 = p CO2 H = 3.5 atm
g b13800 atm mole fractiong = 2.536 × 10
−4
mol CO 2 mol
1L 10 3 g H 2 O 1 mol H 2 O 2.536 × 10 − 4 mol CO 2 33.8 oz 1L 18.0 g H 2 O 1 mol H 2 O = 0.220 g CO 2
n CO2 =
c. V =
12 oz
0.220 g CO 2
44.0 g CO 2 1 mol CO 2
b g b273 + 37gK = 0127 . L = 127 cm
1 mol CO2
22.4 L STP
44.0 g CO 2
1 mol
6- 34
273K
3
6.51 a. – SO2 is hazardous and should not be released directly into the atmosphere, especially if the analyzer is inside.
– From Henry’s law, the partial pressure of SO2 increases with the mole fraction of SO2 in the liquid, which increases with time. If the water were never replaced, the gas leaving the bubbler would contain 1000 ppm SO2 (nothing would be absorbed), and the mole fraction of SO2 in the liquid would have the value corresponding to 1000 ppm SO2 in the gas phase.
b
g
b
g
b. Calculate x mol SO 2 mol in terms of r g SO 2 100 g H 2 O
b g r (g SO )b1 mol 64.07 gg = 0.01561r (mol SO ) 0.01561r FG mol SO IJ ⇒x = 5.55 + 0.01561r H mol K
Basis: 100 g H 2O 1 mol 18.02 g = 5.55 mol H2 O 2
2
2
SO2
From this relation and the given data, pS O2 (torr) 0 42 xSO2 (mol SO2 /mol)
85 –3
0
1.4x10
129 –3
176 –3
2.8x10
5.6x10–3
4.2x10
A plot of pSO 2 vs. xS O2 is a straight line. Fitting the line using the method of least squares (Appendix A.1) yields mm Hg pSO2 = H SO2 xS O2 , H SO 2 = 3.136 × 10 4 mole fraction
d
i
c. 100 ppm SO 2 ⇒ ySO = 100 mol SO 2 = 1.00 × 10 −4 mol SO 2 /mol gas 2 106 mols gas
(
)
⇒ pSO2 = ySO2 P = 1.0 ×10−4 ( 760 mm Hg ) = 0.0760 mm Hg
Henry' s law ⇒ xSO 2 =
Since xS O2
H SO2
=
0.0760 mm Hg 3.136 × 104 mm Hg mole fraction
= 2.40 × 10 −6 mol SO 2 mol is so low, we may assume for simplicity that Vfinal ≈ Vinitial = 140 L , and
n final ≈ ninitial = ⇒ nS O2 =
pSO 2
bg
140 L 103 g H 2 O l 1L
1 mol 18 g
= 7.78 × 10 3 moles
7.78 × 10 3 mol solution 2.40 × 10 −6 mol SO 2 1 mol solution
= 0.0187 mol SO 2 dissolved
0.0187 mol SO2 dissolved = 1.34 × 10 −4 mol SO 2 L 140 L Gas-phase composition ySO2 = 1.0 × 10
−4
mol SO 2 mol
yH 2 O =
xH2 O p*H2O (30o C) P
=
(1)(31.824 torr) 760 torr
= 4.19 × 10 −2
mol H 2 O(v) mol
yair = 1 − ySO2 − yH2O = 0.958 mol dry air/mol
d. Agitate/recirculate the scrubbing solution, change it more frequently. Add a base to the solution to react with the absorbed SO 2.
6- 35
6.52 Raoult’s law + Antoine equation (S = styrene, T = toluene): − ( T +214.985) y S P = xS p∗S ⇒ x S = 0.650(150 mm Hg)/10 7.066231507.434
yT P = xT pT∗ ⇒ xT = 0.350(150 mm Hg)/106.95805−1346.773 (T + 219.693 )
0.65(150)
1 = x S + xT =
+
7.066231507.434 − (T + 214.985)
0.35(150) 6.958051346.773 − ( T +219.693)
10 10 ⇒ T = 86.0°C (Determine using E-Z Solve or a spreadsheeet)
xS =
0.65(150) 10
7.066231507.434 − ( 86.0 +214.985)
= 0.853 mol styrene/mol ⇒ xT = 0.147 mol toluene/mol
6.892721203.531 − ( 85+ 219.888)
6.53 PB∗ ( 85°C ) = 10
= 881.6 mm Hg
6.95805−1346.773 (85.0+ 219.693 )
PT∗ ( 85°C ) = 10
= 345.1 mm Hg
yB = 0.35 (881.6 ) /[10(760)] = 0.0406 mol Benzene mol
Raoult's Law: y B P = x B PB∗ ⇒
yT = 0.65 (345.1) /[10(760)] = 0.0295 mol Toluene mol y N2 = 1 − 0.0406 − 0.0295 = 0.930 mol N2 mol
6.54 a. From the Cox chart, at 77° F, p *P = 140 psig , p *nB = 35 psig, p *iB = 51 psig * * Total pressure P=x p ⋅ p*p +x nB ⋅pnB +x iB ⋅ piB
= 0.50(140) + 0.30(3 5)+ 0.20(51)= 91 psia ⇒ 76 psig
P < 200 psig, so the container is technically safe.
b. From the Cox chart, at 140° F, pP* = 300 psig , p *nB = 90 psig, piB* = 120 psig Total pressure P = 0.50( 300) + 0.30 (90) + 0.20(120 ) ≅ 200 psig The temperature in a room will never reach 140o F unless a fire breaks out, so the container is adequate. 6.844711060.793 − (120 +231.541)
6.55 a. Antoine: Pnp∗ (120 °C ) = 10 Pip∗ (120°C) = 10
6.73457 −992.019 (120 + 231.541)
= 6717 mm Hg
= 7883 mm Hg
Note: We are using the Antoine equation at a temperature well above the validity ranges in Table B.4, so that all calculated values must be considered rough estimates. When the first bubble of vapor forms, x np = 0.500 mol n - C5H12 (l) / mol
xip = 0.500 mol i -C5H 12 (l)/mol
Total pressure: P=xnp ⋅ p *np +x ip ⋅ pip* = 0.50(6717) + 0.50(7883) = 7300 mm Hg
6- 36
6.55 (cont’d)
ynp =
* xnp ⋅ pnp
P
=
0.500(6725) = 0.46 mol n-C5 H12 (v)/mol 7342
yip = 1 − ynp = 1 − 0.46 = 0.546 mol i-C5H 12 (v)/mol When the last drop of liquid evaporates, ynp = 0.500 mol n - C5H 12 (v) / mol x np + xip =
xnp =
y np P * pnp (120 o C)
+
yip P * pip (120o C)
yip = 0.500 mol i-C5H12 (v)/mol =
0.500 P 0.500P + = 1 ⇒ P = 7250 mm Hg 6717 7883
0.500*7250 mm Hg = 0.54 mol n-C5H12 (l)/mol 6717 mm Hg
xip = 1 − x np = 1 − 0.54 = 0.46 mol i -C5H12 (l)/mol
b. When the first drop of liquid forms, ynp = 0.500 mol n - C5H 12 (v) / mol
yip = 0.500 mol i - C12 H12 (v) / mol
P = (1200 + 760) = 1960 mm Hg x np + xip =
0.500 P 0.500 P 980 980 + * = 6.844711060.793/( + 6.73457− 992.019/(T + 231.541) = 1 * − Tdp + 231.541) dp pnp (Tdp ) pip (Tdp ) 10 10
⇒ Tdp = 63.1o C
p ∗np = 10 pip∗
6.844711060.793 − ( 63.1+ 231.541 )
6.73457 −992.019 ( 63.1+ 231.541 )
= 10
x np =
= 1758 mm Hg
= 2215 mm Hg
0.5*1960 mm Hg = 0.56 mol n -C5 H12 /mol * pnp (63.1o C)
xip = 1 − xnp = 1 − 0.56 = 0.44 mol i -C5 H12/mol
When the last bubble of vapor condenses, x np = 0.500 mol n - C5H12 (l) / mol
x ip = 0.500 mol i - C5H 12 (l) / mol
Total pressure: P= xnp ⋅ p *np +x ip ⋅ p*ip − ( T + 231.541) ⇒ 1960 = (0.5)106.844711060.793 + (0.5)10
6.73457 − 992.019/(Tb p + 231.541)
⇒ T = 62.6°C
(Obtained with E-Z Solve)
xnp ⋅ p*np (62.6o C)
0.5(1731) = 0.44 mol n -C5H12 (v)/mol P 1960 yip = 1 − y np = 1 − 0.44 = 0.56 mol i -C5 H12 (v)/mol ynp =
=
6- 37
6.56 B = benzene, T = toluene n&v (mol / min) at 80o C, 3 atm
n& N2 =
10 L(STP)/min
xB [mol B(l)/mol]
yN2 (mol N2 /mol) yB [mol B(v)/mol]
n& N2 ( mol / min)
xT [mol T(l)/mol]
yT [mol T(v)/mol]
10.0 L(STP) / min = 0.4464 mol N 2 / min 22.4 L(STP) / mol 6.89272−1203.531 (80+ 219.888 )
= 757.6 mm Hg
6.958051346.773 − (80+ 219.693 )
= 291.2 mm Hg
Antoine: p ∗B (80 °C ) = 10 pT∗ ( 80°C ) = 10
a. Initially, xB = 0.500, xT = 0.500. Raoult's law: yB =
x B p∗B (80o C) 0.500 ( 757.6 ) = = 0.166 mol B(v) mol P 3(760)
yT =
xT p∗T (80o C) 0.500 ( 291.2 ) = = 0.0639 mol T(v) mol P 3(760)
N 2 balance: 0.4464 mol N 2 / min = n&v (1 − 0.166 − 0.0639) ⇒ n&v = 0.5797 mol / min
FG H F = G 0.5797 H
⇒ n&B 0 = 0.5797 n&T0
mol BI mol B(v) IJ FG 0166 . = 0.0962 J KH mol K min mol I F mol BI mol T(v) 0.0639 = 0.0370 J G J min K H mol K min mol min
b. Since benzene is evaporating more rapidly than toluene, xB decreases with time and xT (= 1–xB) increases. c. Since xB decreases, yB (= xBp B*/P) also decreases. Since xT increases, yT (= xTp T*/P) also increases. 6.57 a. P =
∗ xhex phex
dT i + bp
x hep p∗hep
dT i bp
, yi =
d i
x i pi∗ Tbp P
760 mm Hg = 0.500 10
6.885551175.817/( − Tbp + 224.867)
, Antoine equation for p ∗i
+ 0.500 10 6.90253−1267.828 /(Tbp + 216.823)
E-Z Solve or Goal Seek ⇒ Tbp = 80.5° C ⇒ yhex = 0.713, y hep = 0.287
b.
xi =
yi P ⇒ Tdp
pi∗
d i
∑x
i i
∑
=P
i ∗ pi
yi =1 Tdp
d i
6- 38
6.57 (cont’d) 0.30 0.30 760 mmHg 6.88555−1175.817/(T + 224.867) + 6.902531267.828 =1 − /( T + 216.823) dp dp 10 10
E-Z Solve or Goal Seek ⇒ Tdp = 711 . ° C ⇒ x hex = 0.279 , x hep = 0.721
6.58
a.
f (T ) = P −
N
∑
xi pi* (T )
= 0 ⇒ T , where
pi* (T )
FG A − B IJ = 10H T + C K i
i
i
i =1
yi (i = 1,2,L , N ) =
x i pi* (T ) P
b. Calculation of Bubble Points A B Benzene 6.89272 1213.531 Ethylbenzene 6.95650 1423.543 Toluene 6.95805 1346.773
C 219.888 213.091 219.693
P(mmHg)= 760 xB 0.226 0.443 0.226
xEB 0.443 0.226 0.226
T bp (o C) 108.09 96.47 104.48
xT 0.331 0.331 0.548
b
g
d i
When x B = 1 pure benzene , Tbp = Tbp
b
g
= 80.1o C
C6 H 6
d i
When x EB = 1 pure ethylbenzene , Tbp = Tbp
b
g
d i
When x T = 1 pure toluene , Tbp = Tbp
C 7 H8
pB pEB pT f(T) 378.0 148.2 233.9 -0.086 543.1 51.6 165.2 0.11 344.0 67.3 348.6 0.07
C 8H 10
= 136.2 o C ⇒ Tbp , EB > Tbp ,T > Tbp, B
= 1106 . oC
Mixture 1 contains more ethylbenzene (higher boiling point) and less benzene (lower bp) than Mixture 2, and so (Tbp )1 > (Tbp )2 . Mixture 3 contains more toluene (lower bp) and less ethylbenzene (higher bp) than Mixture 1, and so (Tbp )3 < (Tbp )1 . Mixture 3 contains more toluene (higher bp) and less benzene (lower bp) than Mixture 2, and so (Tbp )3 > (Tbp )2
6- 39
6.59
a. Basis: 150.0 L/s vapor mixture
n& 1 (mol/s) @ T(o C), 1100 mm Hg 0.600 mol B(v)/mol 0.400 mol H(v)/mol &n2 (mol/s) x2 [mol B(l)/mol] (1- x2 ) [mol H(l)/mol]
n& 0 (mol/s)@120°C, 1 atm 1atm 0.500 mol B(v)/mol 0.500 mol H(v)/mol
Gibbs phase rule: F=2+c-π =2+2-2=2 Since the composition of the vapor and the pressure are given, the information is enough. Equations needed: Mole balances on butane and hexane, Antoine equation and Raoult’s law for butane and hexane
b. Molar flow rate of feed: n& 0 =
150.0 L 273 K mol = 4.652 mol/s s 393 K 22.4 L (STP)
Raoult's law for butane: 0.600(1100)=x 2 ⋅10 6.82485 −943.453/(T +239.711)
(1)
Raoult's law for hexane: 0.400(1100)=(1-x 2 ) ⋅ 10 Mole balance on butane: 4.652(0.5)=n& 1 ⋅ 0.6+ n& 2 ⋅ x 2 Mole balance on hexane: 4.652(0.5)=n& 1 ⋅ 0.4 + n& 2 ⋅ (1 − x 2 )
(2) (3) (4)
6.88555− 1175.817/(T + 224.867)
c. From (1) and (2), 1= ⇒ T = 57.0 ° C
x2 =
1100(0.6) 1100(0.4) + 943.453 1175.817 10**(6.82485 − ) 10**(6.88555 − ) T + 239.711 T + 224.867
1100(0.6) 6.82485943.453/(57.0 − + 239.711)
10
= 0.149 mol butane /mol
Solving (3) and (4) simultaneously ⇒ n&1 = 3.62 mol C4 H10 /s; n&2 = 1.03 mol C6 H14 /s d.
Assumptions: (1) Antoine equation is accurate for the calculation of vapor pressure; (2) Raoult’s law is accurate; (3) Ideal gas law is valid.
6.60 P = n-pentane, H = n-hexane 170.0 kmol/h, T1a (o C), 1 atm
85.0 kmol/h, T1b (o C), 1
&n0 (kmol/h)
0.98 mol P(l)/mol 0.02 mol H(l)/mol
0.45 kmol P(l)/kmol 0.55 kmol H(l)/kmol
&n2 (kmol/h) (l), o
x2 (kmol P(l)/kmol) (1- x2 ) (kmol H(l)/kmol)
6- 40
6.60 (cont’d) a. Molar flow rate of feed: n& 0 (0.45)(0.95) = 85( 0.98) ⇒ n& 0 = 195 kmol / h Total mole balance : 195 = 85.0 + n& 2 ⇒ n& 2 = 110 kmol / h Pentane balance: 195( 0.45) = 85.0 (0.98) + 110 ⋅ x 2 ⇒ x 2 = 0.0405 mol P / mol
b. Dew point of column overhead vapor effluent: Eq. 6.4-7, Antoine equation ⇒
0.98(760) 10
6.844711060.793/( − T1a + 231.541)
+
0.02(760) 10
6.88555−1175.817 /( T1a + 224.687)
= 1 ⇒ T1a = 37.3o C
Flow rate of column overhead vapor effluent. Assuming ideal gas behavior, 170 kmol 0.08206 m3 ⋅ atm (273.2 + 37.3) K & Vvapor = = 4330 m 3 / h h kmol ⋅ K 1 atm
Flow rate of liquid distillate product. Table B.1 ⇒ ρ P = 0.621 g / mL, ρ H = 0.659 g / mL 0.98(85) kmol P 72 .15 kg P L V&distillate = h kmol P 0.621 kg P 0.02 (85) kmol H 86.17 kg H L = 9 .9 × 10 3 L / h h kmol H 0.659 kg H c. Reboiler temperature. +
6.84471−1060.793/(T2 + 231.541)
+ 0.96 ⋅106.88555−1175.817/(T2 + 224.867) = 760 ⇒ T2 =66.6°C
0.04 ⋅10
Boilup composition. y2 =
− + 231.541) x2 pP* (66.6 o C) 0.04 ⋅ 106.844711060.793/(66.6 = = 0.102 mol P(v)/mol P 760
⇒ (1 - y 2 ) = 0.898 mol H(v) / mol d. Minimum pipe diameter
F I GH JK
m3 V& s
= u max
⇒ D min =
FG m IJ × πD HsK 4
4V&vapor π ⋅ u max
2 min
=
( m2 )
4 4330 m 3 / h 1 h = 0.39 m (39 cm) π 10 m / s 3600 s
Assumptions : Ideal gas behavior, validity of Raoult’s law and the Antoine equation, constant temperature and pressure in the pipe connecting the column and the condenser, column operates at steady state.
6- 41
Condenser
6.61 a. F (mol) x 0 (mol butane/mol)
V (mol) 0.96 mol butane/mol R (mol) x 1 (mol butane/mol)
T P
Partial condenser: 40° C is the dew point of a 96% C 4 H 10 − 4% C 5H 12 vapor mixture at P = Pmin Total condenser: 40° C is the bubble point of a 96% C 4 H 10 - 4% C5 H 12 liquid mixture at P = Pmin Dew Point: 1 = (Raoult's Law)
∑x = ∑ p i
∗ i
yi P ⇒ Pmin = 40° C
b
Antoine Eq. for
g
pi∗
i
b
pi∗ 40° C
943.453 6.82485 − 40 +239.711
( C4 H10 ) = 10
Antoine Eq. for pi∗ ( C5 H12 ) = 10 ⇒ Pmin =
∑y
1
1060.793 6.84471− 40 + 231.541
g
= 2830.7 mm Hg = 867.2 mm Hg
1 = 2596 mm Hg ( partial condenser ) 0.96 2830.7 + 0.04867.2
Bubble Point: P =
∑ y P = ∑ x p b40° Cg i
i
∗ i
P = 0.96 ( 2830.7 ) + 0.04 (867.2 ) = 2752 mm Hg ( total condenser )
b. V& = 75 kmol / h , R& V& = 15 . ⇒ R& = 75 × 15 . kmol / h = 112.5 kmol / h Feed and product stream compositions are identical: y = 0.96 kmolbutane kmol Total balance: F& = 75 + 112.5 = 187.5 kmol / h
c.
Total balance as in b.
R& = 1125 . kmol / h
UV P = 2596 mm Hg gb gW x = 0.8803 mol butane mol = 112.5b0.8803g + 0.96b75g ⇒ x = 0.9122 mol butane mol reflux
b b
g
Equilibrium: 0.96 P = x1 2830.70 Raoult' s law 0.04 P = 1 − x1 867.22
b
g
Butane balance: 187 .5 x 0
6.62 a.
F& = 187.5 kmol / h
1
0
y i p ∗i y x p ∗ P p ∗A = ⇒ α AB = A A = ∗A = = α AB xi P y B xB p B P p ∗B
Raoult's law:
b.
(
)
1507.434 7.06623 − 85 + 214.985
pS 85 C = 10 *
o
1423.543 6.95650 − 85 + 213.091
pEB (85 C) = 10 *
= 109.95 mm Hg
o
1213.531 6.89272− 85 + 219.888
p*B (85o C) = 10
= 151.69 mm Hg = 881.59 mm Hg
6- 42
6.62 (cont’d) α S,EB
p*S 109.95 p * 881.59 = * = = 0.725 , α B,EB = *B = = 5.812 pEB 151.69 pEB 151.69
Styrene − ethylbenzene is the more difficult pair to separate by distillation because α S,EB is closer to 1 than is α B,EB .
c.
α ij =
yi xi yj xj
y j =1 − yi x j =1− xi
⇒α
ij
d. α B, EB = 5.810 ⇒ y B =
α ij x i yi xi ⇒ yi = (1 − y i ) 1 − xi 1 + α ij − 1 x i
b
g
x Bα B, EB 1 + (α B , EB − 1) x B
d
=
i
5.81 x B , P = x B p *B + (1 − x B ) p *EB 1 + 4.81x B
bg bg
xB yB P
6.63 a.
=
0.0 0.2 0.4 0.6 0.8 1.0 mol B l mol 0.0 0.592 0.795 0.897 0.959 1.0 mol B v mol 152 298 444 5900 736 882 mmHg
Since benzene is more volatile, the fraction of benzene will increase moving up the column. For ideal stages, the temperature of each stage corresponds to the bubble point temperature of the liquid. Since the fraction of benzene (the more volatile species) increases moving up the column, the temperature will decrease moving up the column.
b. Stage 1: n& l = 150 mol / h, n& v = 200 mol / h ; x 1 = 0.55 mol B mol ⇒ 0.45 mol S mol ; y 0 = 0.65 mol B mol ⇒ 0.35 mol S mol
Bubble point T : P =
∑x p i
∗ i
bT g
− T + 214.985) P1 = (0.400 × 760) mmHg = ( 0.55 )10 6.89272−1203.531/(T + 219.888) + ( 0.45 )107.066231507.434/( E-Z Solve → T1 = 67.6 o C
⇒ y1 =
b g=
x1 p ∗B T
b g
0.55 508
= 0.920 mol B mol ⇒ 0.080 mol S mol P 0.400 × 760 B balance: y 0 n& v + x 2 n& l = y 1n& v + x 1n&l ⇒ x 2 = 0.910 mol B mol ⇒ 0.090 mol S mol E-Z Solve Stage 2: (0.400 × 760) mmHg = 0.910 p *B (T2 ) + 0.090 p *S ( T 2) →T2 = 55.3o C
y2 =
b
0.910 3310 .
g = 0.991 mol B mol ⇒ 0.009 mol S mol
760 × 0.400 B balance: y 1n& v + x 3n& l = y 2n&v + x 2 n&l ⇒ x 3 ≈ 1 mol B mol ⇒ ≈ 0 mol S mol
c. In this process, the styrene content is less than 5% in two stages. In general, the calculation of part b would be repeated until (1–yn ) is less than the specified fraction.
6- 43
6.64 Basis: 100 mol/s gas feed. H=hexane. 200 mol oil/s
n F (mol/s) y F (mol H/mol) 1 – y F (mol N 2/mol)
100 mol/s 0.05 mol H/mol 0.95 mol N /mol 2
a.
n l (mol/s) x +i 1 (mol H/mol)
n 2 (mol/s) x 2 (mol H/mol) 1 – x 2 (mol Oil/mol)
n v (mol/s) y i (mol H/mol) Stage i
n l (mol/s) x i (mol H/mol)
n v (mol/s) y i– 1 (mol H/mol)
99.5% of H in feed.
U|V |W
b g b g b gb g
n F = 95.025 mol s N 2 balance: 0.95 100 = 1 − y F n F ⇒ 99.5% absorption: 0.05 100 0.005 = y F n F y F = 2 .63 × 10 −4 mol H(v) mol Mole Balance: 100 + 200 = 95.025 + n 2 ⇒ n2 = 205 mol s
b g
b
g b g 1 = b100 + 95.025g ⇒ n& 2
Hexane Balance: 0.05 100 = 2.63 × 10 −4 95.025 + x 1 204.99 ⇒ x 1 = 0.0243 mol H(l) mol n& L =
b
g
1 200 + 205 ⇒ n& L = 202.48 mol s , n&G 2
B
G
= 97 .52 mol s
Antoine
b.
b
g
b
g
y1 = x1 p ∗H 50° C / P = 0.0243 403.73 / 760 = 0.0129 mol H(v) mol
H balance on 1 st Stage: y0 n&v + x2 n& l = y1 n&v + x1n& l ⇒ x2 = 0.00643 mol H(l) mol
c. The given formulas follow from Raoult’s law and a hexane balance on Stage i. d. Hexane Absorption P= y0= nGf= A=
760 0.05 95.025 6.88555
T 30
p*(T) 187.1
i 0 1 2 3
x(i) 2.43E-02 3.10E-03 5.86E-04
PR= x1= nL1= B=
y(i) 5.00E-02 5.98E-03 7.63E-04 1.44E-04
1 0.0243 204.98 1175.817
ye= 2.63E-04 nG= 97.52 nL= C= 224.867
T 50
p*(T) 405.3059
i 0 1 2 3 4 5
x(i) 2.43E-02 6.46E-03 1.88E-03 7.01E-04 3.99E-04
6- 44
y(i) 5.00E-02 1.30E-02 3.45E-03 1.00E-03 3.74E-04 2.13E-04
202.48
T 70
p*(T) 790.5546
i 0 1 2 3 4 5 ... 21
x(i) 2.43E-02 1.24E-02 6.43E-03 3.44E-03 1.94E-03 ... 4.38E-04
y(i) 5.00E-02 2.53E-02 1.29E-02 6.69E-03 3.58E-03 2.02E-03 ... 4.56E-04
6.64 (cont’d) e. If the column is long enough, the liquid flowing down eventually approaches equilibrium with the entering gas. At 70o C, the mole fraction of hexane in the exiting liquid in equilibrium with the mole fraction in the entering gas is 4.56x10–4 mol H/mol, which is insufficient to bring the total hexane absorption to the desired level. To reach that level at 70o C, either the liquid feed rate must be increased or the pressure must be raised to a value for which the final mole fraction of hexane in the vapor is 2.63x10–4 or less. The solution is Pmin = 1037 mm Hg. 6.65 a. Intersection of vapor curve with y B = 0.30 at T = 104° C ⇒ 13% B(l), 87%T(l)
b
g
b
b. T = 100° C ⇒ x B = 0.24 mol B mol liquid , y B = 0.46 mol B mol liquid
g
n V (mol vapor) 0.46 mol B(v)/mol n L (mol liquid) 0.24 mol B(l)/mol
Basis: 1 mol 0.30 mol B(v)/mol
Balances
UV W
Total moles: 1 = nV + n L n L = 0.727 mol nV mol vapor ⇒ ⇒ = 0.375 B: 0.30 = 0.46 nV + 0.24 n L nV = 0.273 mol n L mol liquid
c. Intersection of liquid curve with x B = 0.3 at T = 98° C ⇒ 50% B(v), 50%T(v) 6.66 a.
P = 798 mm Hg, y B = 0.50 mol B(v) mol
b.
P = 690 mm Hg, xB = 0.15 mol B(l) mol
c.
P = 750 mm Hg, y B = 0.43 mol B(v) mol, xB = 0.24 mol B(l) mol nV (mol) 0.43 mol B/mol nL (mol) 0.24 mol B/mol
3 mol B 7 mol T
UV W
Mole bal.: 10 = nV + n L nV = 3.16 mol n mol vapor ⇒ ⇒ v = 0.46 B bal.: 3 = 0.43nV + 0.24 n L n L = 6.84 mol n l mol liquid
Answers may vary due to difficulty of reading chart. d. i)
P = 1000 mm Hg ⇒ all liquid . Assume volume additivity of mixture components. V=
3 mol B 78.11 g B mol B
10 −3 L 0.879 g B
+
7 mol T 92.13 g T
ii) 750 mmHg. Assume liquid volume negligible
6- 45
mol T
10 −3 L 0.866 g T
= 1.0 L
6.66 (cont’d) V=
3.16 mol vapor 0.08206 L ⋅ atm
373 K
760 mm Hg
mol ⋅ K 750 mm Hg
1 atm
− 0.6 L = 97.4 L
(Liquid volume is about 0.6 L) iii) 600 mm Hg v=
10 mol vapor
0.08206 L ⋅ atm
373K
760 mm Hg
mol ⋅ K 600 mm Hg
1 atm
= 388 L
6.67 a. M = methanol n V (mol) y (mol M(v)mol) n L (mol) x (mol M(l)/mol)
n f (mol) x F (mol M(l)/mol)
UV W
Mole balance: n f = nV + n L n x −x ⇒ x F nV + x F n L = ynV + xn L ⇒ f = V = F MeOH balance: x F n f = ynV + xn L nL y−x
x F = 0.4 , x = 0.23, y = 0.62 ⇒ f =
0.4 − 0.23 = 0.436 0.62 − 0.23
b. Tmin = 75o C, f = 0 , Tmax = 87 o C, f = 1 6.68 a.
Txy diagram (P=1 atm) 80 75
o
T( C)
70
Vapor
65 liquid
60 55 50 0
0.2
0.4
0.6
Mole fraction of Acetone
b.
x A = 0.47; y A = 0.66
6- 46
0.8
1
6.68 (cont’d) c. (i) x A = 0.34; y A = 0.55 (ii) Mole bal.: 1 = nV + nL A bal.:
⇒ nV = 0.762 mol vapor, nL = 0.238 mol liquid 0.50 = 0.55nV + 0.34 nL ⇒ 76.2 mole% vapor
(iii) ρ A (l ) = 0.791 g/cm 3 , ρ E(l) = 0.789 g/cm 3 ⇒ ρ l ≈ 0.790 g/cm3 (To be more precise, we could convert the given mole fractions to mass fractions and calculate the weighted average density of the mixture, but since the pure component densities are almost identical there is little point in doing all that.) M A = 58.08 g/mol, M E = 46.07 g/mol
⇒ Ml = ( 0.34 )( 58.08) + (1 − 0.34 )( 46.07 ) = 50.15 g/mol Basis: 1 mol liquid ⇒ (0.762 mol vapor / 0.238 mol liquid) = 3.2 mol vapor (1 mol)(50.15 g / mol) Liquid volume: Vl = = 63.48 cm 3 3 (0.790 g / cm ) Vapor volume: 3.2 mol 22400 cm 3 (STP) (65 + 273)K = 88 ,747 cm 3 mol 273K 88,747 Volume percent of vapor = × 100% = 99.9 volume % vapor 88747 + 63.48 Vv =
d. For a basis of 1 mol fed, guess T, calculate n V as above; if n V ≠ 0.20, pick new T. T 65 °C 64.5 °C e.
xA 0.34 0.36
yA 0.55 0.56
fV 0.333 0.200
Raoult's law: y i P = xi p i* ⇒ P = x A p *A + x E p E* 760 = 0.5 × 10
7.11714 −1210.595/(Tbp +229.664)
8.11220 −1592.864/( Tbp +226.184)
+ 0.5 × 10
⇒ Tbp = 66.16 o C
xp*A 0.5 ×107.11714 −1210.595/(66.25+ 229.664) = = 0.696 mol acetone/mol P 760 ∆Tbp 66.25 − 61.8 The actual Tbp = 61.8o C ⇒ = ×100% = 7.20% error in Tbp Tbp (real) 61.8 y=
y A = 0.674 ⇒
∆y A 0.696 − 0.674 = ×100% = 3.26% error in yA yA (real) 0.674
Acetone and ethanol are not structurally similar compounds (as are, for example, pentane and hexane or benzene and toluene). There is consequently no reason to expect Raoult’s law to be valid for acetone mole fractions that are not very close to 1.
6- 47
6.69 a. B = benzene, C = chloroform. At 1 atm, (Tbp )B = 80.1o C, (Tbp)C = 61.0o C The Txy diagram should look like Fig. 6.4-1, with the curves converging at 80.1o C when xC = 0 and at 61.0o C when xC = 1. (See solution to part c.) b. Txy Diagram for an Ideal Binary Solution A B C Chloroform 6.90328 1163.03 227.4 Benzene 6.89272 1203.531 219.888 P(mmHg)= 760 x 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
T 80.10 78.92 77.77 76.66 75.58 74.53 73.51 72.52 71.56 70.62 69.71 68.82 67.95 67.11 66.28 65.48 64.69 63.93 63.18 62.45 61.73
y 0 0.084 0.163 0.236 0.305 0.370 0.431 0.488 0.542 0.593 0.641 0.686 0.729 0.770 0.808 0.844 0.879 0.911 0.942 0.972 1
p1 0 63.90 123.65 179.63 232.10 281.34 327.61 371.15 412.18 450.78 487.27 521.68 554.15 585.00 614.02 641.70 667.76 692.72 716.27 738.72 760
p2 760 696.13 636.28 580.34 527.86 478.59 432.30 388.79 347.85 309.20 272.79 238.38 205.83 175.10 145.94 118.36 92.17 67.35 43.75 21.33 0
p1+p2 760 760.03 759.93 759.97 759.96 759.93 759.91 759.94 760.03 759.99 760.07 760.06 759.98 760.10 759.96 760.06 759.93 760.07 760.03 760.05 760
Txy diagram (P=1 atm) 85
75
Vapor
o
T( C)
80
70 Liquid 65 60 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Mole fraction of chloroform
6- 48
0.8
0.9
1
6.69 (cont’d) d. Txy diagram (P=1 atm) 85
T(oC)
80
yc
xc
75 70 x
y
0.2 0.3 0.4
0.5
65 60 0
0.1
0.6
0.7
0.8 0.9
1
Mole fraction of choloroform
∆T 71 − 75.3 = × 100% = −5.7% error in Tbp Tactual 75.3
Raoult’s law: Tbp = 71o C, y = 0.58 ⇒
∆y 0.58 − 0.60 = × 100% = −3.33% error in y y actual 0.60
Benzene and chloroform are not structurally similar compounds (as are, for example, pentane and hexane or benzene and toluene). There is consequently no reason to expect Raoult’s law to be valid for chloroform mole fractions that are not very close to 1.
d i b
g d i
6.70 P ≈ 1 atm = 760 mm Hg = x m p *m Tbp + 1 − x m p *P Tbp 7.878631473.11/( − Tbp + 230)
760 = 0.40 ×10
7.74416 −1437.686/(Tbp + 198.463)
+ 0.60 × 10
E-Z Solve →T = 79.9 o C
We assume (1) the validity of Antoine’s equation and Raoult’s law, (ii) that pressure head and surface tension effects on the boiling point are negligible. The liquid temperature will rise until it reaches 79.9 °C, where boiling will commence. The escaping vapor will be richer in methanol and thus the liquid composition will become richer in propanol. The increasing fraction of the less volatile component in the residual liquid will cause the boiling temperature to rise.
6- 49
6.71 Basis: 1000 kg/h product nH4 (mol H 2 /h) E = C2 H5 OH ( M = 46.05) A = CH 3 CHO ( M = 44.05)
scrubber n3 (mol/h) y A3 (mol A/mol), sat'd y E3 (mol E/mol), sat'd y H3 (mol H 2 /h) vapor, –40°C
P = 760 mm Hg Fresh feed n0 (mol E/h) nA1 (mol A/h) nE1 (mol E/h) 280°C
reactor
nA2 (mol A/h) nE2 (mol E/h) nH2 (mol H 2/h)
condenser
nC (mol/h) 0.550 A 0.450 E liquid, –40°C
Scrubbed Hydrocarbons nA4 (mol A/h) nE4 (mol E/h)
still
Product 1000 kg/h np (mol/h) 0.97 A 0.03 E
nr (mol/h) 0.05 A 0.95 E
Strategy
a.
d i
•
Calculate molar flow rate of product n& p from mass flow rate and composition
•
Calculate y A3 and y E3 from Raoult’s law: y H3 = 1 − y A3 − y E3 . Balances about the still involve fewest unknowns ( n&c and n& r )
•
Total mole balance about still
•
A, E and H 2 balances about scrubber ⇒ n&A4 , n& E4 , and n& H4 in terms of n& 3 . Overall atomic balances on C, H, and O now involve only 2 unknowns ( n& 0 , n&3 )
•
Overall C balance
• • • •
A balance about fresh feed-recycle mixing point ⇒ n& A1 E balance about fresh feed-recycle mixing point ⇒ n& E1 A, E, H 2 balances about condenser n& A2 , n& E2 , n& H2 All desired quantities may now be calculated from known molar flow rates.
UV ⇒ n& , n& W c
A balance about still
r
UV ⇒ n& , n& Overall H balanceW 0
3
Molar flow rate of product
b gb
g b gb
g
M = 0.97 M A + 0.03 M E = 0.97 44.05 + 0.03 46.05 = 44.11 g mol n& p =
1000 kg 1 kmol = 22.67 kmol h h 44.11 kg
Table B.4 (Antoine) ⇒ p *A ( −40 °C ) = 44.8 mm Hg
p *E ( −40 °C ) = 0.360 mm Hg
Note: We are using the Antoine equation at a temperature below the ranges of validity in Table B.4, so that all calculated values must be considered rough estimates. Raoult’s law ⇒ yA3 =
0.550 p*A ( −40 °C ) P
=
0.550(44.8) = 0.03242 kmol A/kmol 760
6- 50
6.71 (cont’d) yE3 =
0.450 p*E ( −40 °C) yH3
P = 1 − y A3 − y E3
0.450(0.360) = 2.13 × 10 −4 kmol E kmol 760 = 0.9674 kmol H 2 kmol
=
Mole balance about still: n& c = n& p + n& r ⇒ n&c = 22.67 + n&r
A balance about still: 0.550n&c = 0.97 (22 .67) + 0.05n& r
UV ⇒ n& W n&
r
= 29.5 kmol / h recycle
c
= 52.1 kmol / h
A balance about scrubber: n& A4 = n& 3 y A3 = 0.02815n& 3
(1)
E balance about scrubber: n& E4 = n&3 y E3 = 2.03 × 10 −4 n& 3
(2)
H 2 balance about scrubber: n& H4 = n&3 y H3 = 0.9716n&3
(3)
Overall C balance: n& 0 (mol E) 2 mol C h
1 mol E
b gb g b gb g d
ib g d
ib g
= n& A4 2 + n& E4 2 + 0.97n& p 2 + 0.03n& p 2
⇒ n& 0 = n& A 4 + n& E 4 + 22.67
(4)
Overall H balance:
b gb g b gb g
6n& 0 = 2 n& H4 + 4 n& A4 + 6n& E4 + n& p 0.97 4 + 0.03 6
(5)
Solve (1)–(5) simultaneously (E-Z Solve): n&0 = 23.4 kmol E/h (fresh feed), n&H 4 = 22.8 kmol H2 /h (in off-gas) n&3 = 23.5 kmol/h, n&A 4 = 0.76 kmol A/h, n&E 4 = 0.0050 kmol E/h
A balance about feed mixing point: n& A1 = 0.05n& r = 1.47 kmol A h E balance about feed mixing point: n& E1 = n&0 + 0.95 n&r = 51.5 kmol E h E balance about condenser: n& E2 = n&3 y E3 + 0.450 n&c = 23.5 kmol E h Ideal gas equation of state: Vreactor feed =
(1.47 + 51.5 ) kmol
b. Overall conversion =
h n&0 − 0.03 n& p
22.4 m 3 (STP ) ( 273+280 ) K = 2.40 × 10 3 m 3 h 1 kmol 273K 23.4 − ( 0.03 )( 22.67 )
× 100% = 97% n&0 23.4 n& − n& 51.5 − 23.5 Single-pass conversion = E1 E2 × 100% = × 100% = 54% n&E1 51.5 Feed rate of A to scrubber: n&A4 =0.76 kmol A/h
Feed rate of E to scrubber:
× 100% =
n& E4 = 0.0050 kmolE h
6- 51
6.72 a. G = dry natural gas, W = water n& 3 (lb - mole G / d) n& 4 (lb - mole W / d) 10 lb m W /106 SCF gas 90 o F, 500 psia
Absorber
n& 7 ( lb - mole W / d)
FG lb - mole TEG IJ H d K F lb - mole W IJ n& G H d K
4.0 × 106 SCF / d 4 × 80 = 320 lb m W / d n& 1 ( lb - mole G / d) n& 2 [lb - mole W(v) / d]
n& 5
Distillation Column
6
FG lb - mole TEGIJ H d K F lb- mole WIJ n& G H d K n&5 8
Overall system D.F. analysis:
Water feed rate : n& 2 =
5 unknowns (n&1 , n& 2 , n& 3 , n& 4 , n&7 ) − 2 feed specifications (total flow rate, flow rate of water) − 1 water content of dried gas −2 balances (W, G) 0 D.F.
320 lb m W 1 lb - mole d
18.0 lb m
= 17 .78 lb - moles W / d
Dry gas feed rate: 4.0 × 106 SCF 1 lb- mole lb - moles W &n1 = − 17.78 = 1112 . × 104 lb - moles G/ d d 359 SCF d Overall G balance: n&1 = n& 3 ⇒ n&3 = 1112 . × 10 4 lb - moles G / d Flow rate of water in dried gas: n& 4 =
(n& 3 + n&4 ) lb - moles 359 SCF gas 10 lb m W 1 lb - mole W d lb - mole 10 6 SCF 18.0 lb m n& =1.112 ×104
3 → n& 4 = 2.218 lb - mole W(l) / d
Overall W balance: n& 7 =
(17.78 − 2.218) lb - moles W
18.0 lb m
d
1 lb - mole
6-52
= 280
lb m W × d
F 1 ft I = 4.5 ft W GH 62.4 lb JK d 3
3
m
6.72 (cont’d) b. Mole fraction of water in dried gas = yw =
n&4 2.218 lb - moles W / d lb - moles W(v) = = 1.99 × 10 −4 4 n&3 + n& 4 (2.218 + 1.112 × 10 ) lb - moles / d lb - mole
Henry’s law: yw P = Hwxw ⇒ (1.99 × 10 −4 )(500 psia)(1 atm / 14.7 psia) lb - mole dissolved W = 0.0170 0.398 atm / mole fraction lb - mole solution c. Solvent/solute mole ratio ( x w ) max =
37 lb m TEG 1 lb - mole TEG 18.0 lbm W n&5 lb - mole TEG = = 4.434 lb m W 150.2 lb m TEG 1 lb m W n&2 − n&4 lb - mole W absorbed ⇒ n&5 = 4.434(17.78 − 2.22) = 69.0 lb - moles TEG / d
xw = 0.80(0.0170) = 0.0136
lb - mole W n&6 &5 = 69 .0 = n → n&6 = 0951 . lb - mole W/ d lb - mole n&5 + n&6
Solvent stream entering absorber m= &
0.951 lb- moles W 18.0 lb m 69.0 lb - moles TEG + d lb - mole d
150.2 lb m lb - mole
= 1.04 × 104 lb m / d
W balance on absorber n&8 = (17.78 + 095 . − 2.22) lb -moles W/ d = 16.51 lb - moles W/ d 16.51 lb - moles W / d ⇒ xw = = 019 . lb - mole W / lb - mole (16.51 + 69.9) lb - moles / d
c. The distillation column recovers the solvent for subsequent re-use in the absorber. 6.73 Basis: Given feed rates G1 G2 G3 100 mol/h 200 mol air/h n1 (mol/h) 0.96 H2 0.999 H 2 0.04 H2 S, sat'd 0.001 H 2S 1.8 atm absorber stripper L2 0°C L1 40°C n3 (mol/h) n4 (mol/h) 0.002 H 2S x 3 (mol H 2S/mol) (1 – x 3) (mol solvent/mol) 0.998 solvent 0°C heater
6-53
G4 200 mol air/h n2 mol H 2S/mol 0.40°C, 1 at m
n3 (mol/h) x 3 (mol H 2S/mol) (1 – x 3) (mol solvent/mol) 40°C
6.73 (cont’d)
b gb
g
Equilibrium condition: At G1, p H 2S = 0.04 18 . atm = 0.072 atm ⇒ x3 =
p H 2S H H2 S
=
0.072 atm = 2.67 × 10 −3 mole H 2 S mole 27 atm mol fraction
Strategy: Overall H 2 and H 2 S balances ⇒ n&1 , n& 2 n& 2 + air flow rate ⇒ volumetric flow rate at G4 H 2 S and solvent balances around absorber ⇒ n&3 , n& 4 0.998n& 4 = solvent flow rate
b gb g Overall H S balance: b100gb0.04 g = 0.001n& + n&
Overall H 2 balance: 100 0.96 = 0.999 n1 ⇒ n&1 = 96.1 mol h 2
1
n&1 = 96.1
⇒ n&2 = 3.90 mol H 2S h
2
Volumetric flow rate at stripper outlet
b
g
b g b273 + 40gK = 5240 L hr
200 + 3.90 mol 22.4 liters STP V&G4 = h 1 mol
273 K
H 2 S and solvent balances around absorber:
b100gb0.04g + 0.002n& = 0.001n& 0.998n& = n& d1 − 2.67 × 10 i 4
1
−3
4
U| ⇒ n& V| W
+ n& 3 x 3 ⇒ n& 4 = 1.335n& 3 − 1952
3
3
Solvent flow rate = 0.998n&4 = 5820 mol solvent h 6.74 Basis: 100 g H 2 O Sat'd solution @ 60°C 100 g H 2 O 16.4 g NaHCO 3
Sat'd solution @ 30°C 100 g H 2 O 11.1 g NaHCO3 ms (g NaHCO3 ( s))
bg
NaHCO 3 balance ⇒ 16.4 = 111 . + ms ⇒ m s = 5.3 g NaHCO 3 s
% crystallization =
5.3 g cryst allized × 100% = 32.3% 16.4 g fed
6.75 Basis: 875 kg/h feed solution m1 (kg H2 O(v )/h)
875 kg/h x 0 (kg KOH/kg) (1 – x0) (kg H 2O/kg)
Sat'd solution 10°C m2 (kg H2 O(1)/h) 1.03 m2 (kg KOH/h)
m3 (kg KOH-2H 2O( s)/h) 60% of KOH in feed
6-54
≈ n& 4 = 5830 mol h
6.75 (cont’d) Analysis of feed: 2KOH + H 2 SO 4 → K 2 SO 4 + 2H 2 O
bg
22.4 mL H 2 SO 4 l 1L 0.85 mol H 2 SO 4 2 mol KOH 3 5 g feed soln 10 mL L 1 mol H 2 SO 4 = 0.427 g KOH g feed
x0 =
56.11 g KOH 1 mol KOH
60% recovery: 875 (0.427 )( 0.60 ) = 224.2 kg KOH h m3 =
224.2 kg KOH 92.15 kg KOH ⋅ 2H2 O = 368.2 kg KOH ⋅ 2H2 O h (143.8 kg H 2 O h ) h 56.11 kg KOH
KOH balance: 0.427 (875 ) = 224.2 + 1.03m2 ⇒ m2 = 145.1 kg h Total mass balance: 875 = 368.2 + 2.03 (145.1) + m1 ⇒ m1 = 212kg H 2 O h evaporated 6.76 a. R 0 30 45 g A dissolved mL solution CA 0 0.200 0.300 Plot CA vs. R ⇒ CA = R / 150
CA =
500 mol 1.10 g
= 550 g (160 g A, 390 g S) ml The initial solution is saturated at 10.2 °C. 160 g A Solubility @ 10.2 °C = = 0.410 g A g S = 41.0 g A 100 g S @ 10.2° C 390 g S 17.5 150 g A 1 mL soln At 0°C, R = 17.5 ⇒ CA = = 0106 . g A g soln mL soln 110 . g soln Thus 1 g of solution saturated at 0°C contains 0.106 g A & 0.894 g S. 0106 . gA Solubility @ 0°C = 0.118 g A g S = 118 . g A 100 g S @ 0° C 0.894 g S 390 g S 11.8 g A Mass of solid A: 160 g A − = 114 g A s 100 g S
b. Mass of solution:
bg
c.
A remaining in sol n 6 4 g44 47444 4 8 g A initia l 6 44 7448 0.5 × 390 g S 11.8 g A 160 − 114 g A − = 23.0 g A s 100 g S
b
g
bg
6.77 a. Table 6.5-1 shows that at 50o F (10.0o F), the salt that crystallizes is MgSO 4 ⋅ 7 H 2 O , which contains 48.8 wt% MgSO 4. b. Basis: 1000 kg crystals/h. m& 0 (g/h) sat’d solution @ 130o F
m& 1 (g/h) sat’d solution @ 50o F
0.35 g MgSO4 /g 0.65 g H2 O/g
0.23 g MgSO4 /g 0.77 g H2 O/g
6-55
1000 kg MgSO4 ·7H2 O(s)/h
6.77 (cont’d) & 0 = 2150 kg feed / h m Mass balance: m& 0 = m& 1 + 1000 kg / h MgSO 4 balance: 0.35m 0 = 0.23m & 1 + 0.488(1000) kg MgSO 4 / h ⇒ m& = 1150 kg soln / h 1 The crystals would yield 0.488 × 1000 kg / h = 488
kg anhydrous MgSO 4 h
6.78 Basis: 1 lb m feed solution. Figure 6.5-1 ⇒ a saturated KNO3 solution at 25o C contains 40 g KNO3 /100 g H2 O ⇒ x KNO3 =
40 g KNO3 = 0.286 g KNO3 / g = 0.286 lb m KNO3 / lb m x (40 + 100) g solution 1 lb m solution @ 80o C 0.50 lb m KNO3 /lb m 0.50 lb m H2 O/lb m
m1 (lb m) sat’d solution @ 25o C 0.286 lb m KNO3 /lb m soln 0.714 lb m H2 O/lb m soln m2 [lb m KNO3 (s)]
m1 = 0.700 lb m solution / lb m feed Mass balance: 1 lb m = m1 + m2 ⇒ m2 = 0.300 lb m crystals/ lb m feed KNO3 balance: 0.50 lb m KNO 3 = 0286 . m1 + m2 Solid / liquid mass ratio =
0.300 lb m crystals/ lb m feed = 0.429 lb m crystals/ lb m solution 0.700 lb m solution / lb m feed
6.79 a. Basis: 1000 kg NaCl(s)/h. Figure 6.5-1 ⇒ a saturated NaCl solution at 80o C contains 39 g NaCl/100 g H2 O ⇒ x NaCl =
39 g NaCl = 0.281 g NaCl / g = 0.281 kg NaCl / kg (39 + 100) g solution m& 2 [kg H 2 O(v) / h]
m& 0 (kg/h) solution
m& 1 (kg/h) sat’d solution @ 80o C
0.100 kg NaCl/kg 0.900 kg H2 O/kg
0.281 kg NaCl/kg soln 0.719 kg H2 O/kg soln 1000 kg NaCl(s)/h
& 0 = m&1 + m &2 Mass balance: m NaCl balance: 0.100 kg NaCl = 0281 . m& 1 + m &2 Solid / liquid mass ratio =
⇒
& 1 =0.700 lbm solution / lb m feed m m& 2 = 0.300 lbm crystals/ lb m feed
0.300 lbm crystals/ lb m feed =0.429 lbm crystals / lb m solution 0.700 lbm solution / lb m feed
The minimum feed rate would be that for which all of the water in the feed evaporates to produce solid NaCl at the specified rate. In this case
6-56
6.79 (cont’d)
0100 . (m& 0 ) min = 1000 kg NaCl / h ⇒ ( m & 0 ) min = 10,000 kg / min & 2 = 9000 kg H 2O / h Evaporation rate: m &1 = 0 Exit solution flow rate: m m& 2 [kg H 2 O(v) / h]
b. m& 0 (kg/h) solution
m& 1 (kg/h) sat’d solution @ 80o C 0.281 kg NaCl/kg soln 0.719 kg H2 O/kg soln 1000 kg NaCl(s)/h
0.100 kg NaCl/kg 0.900 kg H2 O/kg
40% solids content in slurry ⇒ 1000
kg NaCl kg & 1 ) max ⇒ (m& 1 ) max = 2500 = 0.400(m h h
NaCl balance: 0.100m & 0 = 0.281(2500) ⇒ m & 0 = 7025 kg / h Mass balance: m & 0 = 2500 + m &2 ⇒ m & 2 = 4525 kg H2O evaporate / h
6.80 Basis: 1000 kg K 2 Cr2 O 7 (s) h . Let K = K 2 Cr2 O 7 , A = dry air, S = solution, W = water. Composition of saturated solution: 0.20 kg K 0.20 kg K ⇒ = 01667 . kg K kg soln kg W 1 + 0.20 kg soln
b
g
n& 2 (mol/ h) y2 (mol W(v) / mol)
m& e [kg W(v) / h)
(1 − y2 )(mol A / mol) 90 C, 1 atm, Tdp = 392 . C o
& f (kg/ h) m
m& f + m& r (kg / h)
CRYSTALLIZERCENTRIFUGE
0.210 kg K / kg 0.790 kg W(l)/ kg
m& 1 ( kg / h) 0.90 kg K(s) / kg
DRYER
o
1000 kg K(s) / h
0.10 kg soln / kg 0.1667 kg K/ kg 0.8333 kg W/ kg n& a (mol A / h)
& r (kg recycle / h) m 0.1667 kg K / kg 0.8333 kg W / kg
b
g
Dryer outlet gas: y2 P = p *W 39.2° C ⇒ y2 =
53.01 mm Hg = 0.0698 mol W mol 760 mm Hg
6-57
& f = 1000 kg K h ⇒ m & f = 4760 kg h feed solution Overall K balance: 0.210m 6.80 (cont’d)
b
gb
g
. 010 . m & 1 = 1000 kg h ⇒ m& 1 = 1090 kg h K balance on dryer: 0.90m& 1 + 01667 Mass balance around crystallizer-centrifuge
m& f + m& r = m & e + m& 1 + m & r ⇒ me = 4760 − 1090 = 3670 kg h water evaporated 95% solution recycled ⇒ m& r =
b0.10 × 1090g kg
h not recycled
95 kg recycled 5 kg not recycled
= 2070 kg h recycled
Water balance on dryer
. gb1090g kg W h b0.8333gb010 = 0.0698n& 1801 . × 10−3 kg mol
2
⇒ n&2 = 7.225 × 104 mol h
Dry air balance on dryer na =
b1 − 0.0698g7.225 × 10
4
b g
b g
mol 22.4 L STP = 151 . × 10 6 L STP h h 1 mol
6-58
6.81. Basis : 100 kg liquid feed. Assume P atm=1 atm 100 kg Feed 0.07 kg Na 2CO 3 / kg 0.93 kg H 2 O / kg
n2w (kmol H2 O )(sat' d)
Reactor Reactor
n2c (kmol CO2 )
e
n2a (kmol Air)
n1(kmol) 0.70 kmol CO 2 / kmol 0.30 kmol Air / kmol
70o C, 3 atm(absolute)
m3 ( kg NaHCO 3( s))
R|m (kg solution) U| S|0.024 kg NaHCO / kgV| T0.976 kg H O / kg W 4
3
2
Filtrate m5 (kg) Filter 0.024 kg NaHCO 3 / kg 0.976 kg H 2 O / kg Filter cake m6 (kg) 0.86 kg NaHCO 3 (s) / kg
R|0.14 kg solution U| S|0.024 kg NaHCO / kgV| T0.976 kg H O / kg W 3
2
Degree of freedom analysis : Reactor 6 unknowns (n1 , n2 , y2w , y2c, m3 , m4 ) –4 atomic species balances (Na, C, O, H) –1 air balance –1 (Raoult's law for water) 0 DF
Filter 2 unknowns –2 balances 0 DF
Na balance on reactor 100 kg 0.07 kg Na 2 CO3 46 kg Na (m + 0.024m4 ) kg NaHCO3 23 kg Na = 3 kg 106 kg Na2 CO3 84 kg NaHCO3 ⇒ 3.038 = 0.2738(m3 + 0.024m4 ) (1)
Air balance: 0.300 n1 = n2a
(2 )
C balance on reactor : n1 (kmol) 0.700 kmol CO2 12 kg C 100 kg 0.07 kg Na 2CO3 12 kg C + kmol 1 kmol CO2 kg 106 kg Na 2CO3 12 = (n2c )(12) + (m3 + 0.024m2 )( ) ⇒ 8.40n1 + 0.7924 = 12n2c + 01429 . (m3 + 0.024m4 ) (3) 84
H balance : 2 1 2 ) = ( n2 w )(2 ) + (m3 + 0.024 m4 )( ) + 0.976m4 ( ) 18 84 18 ⇒ 10.33 = 2 n2 w + 0.01190( m3 + 0.024 m4 ) + 0.1084 m4 (4 ) (100)(0.93)(
6-59
6.81(cont'd) O balance (not counting O in the air): 48 16 n1 (0.700)(932 ) + 100 (0.07 )( ) + 100 ( 0.93)( ) 106 18 48 16 = (n 2w )(16) + n2c (32) + (m3 + 0.024 m4 )( ) + 0.976m4 ( ) 84 18 ⇒ 22.4 n1 + 85.84 = 16n 2w + 32n 2c + 0.5714(m 3 + 0.024 m4 ) + 0.8676 m4
(5)
Raoult's Law : y w P = p *w (70 o C) ⇒ ⇒ n2 w = 0.1025( n2 w
n2 w 233.7 mm Hg = n 2w + n 2c + n 2a (3 * 760) mm Hg + n 2c + n 2a )
(6)
Solve (1)-(6) simultaneously with E-Z solve (need a good set of starting values to converge). n1 = 0.8086 kmol,
n2a = 0.2426 kmol air,
n 2w = 0.0848 kmol H 2 O(v),
n2c = 0.500 kmol CO 2 ,
m3 = 8.874 kg NaHCO3 (s),
m4 = 92.50 kg solution
NaHCO3 balance on filter: m3 + 0.024 m4 = 0.024 m5 + m6 [0.86 + (0.14 )( 0.024 )] m3 =8.874
11.09 = 0.024m 5 + 0.8634m6
(7 )
m4 =92 .50
Mass Balance on filter: 8.874 + 92.50 = 1014 . = m5 + m6 Solve (7) & (8) ⇒ Scale factor =
m5 = 91.09 kg filtrate m6 = 10.31 kg filter cake
(8)
⇒ (0.86)(10.31) = 8.867 kg NaHCO3 (s)
500 kg / h = 56.39 h −1 8.867 kg
(a) Gas stream leaving reactor
U| V| W
R| |S | |T
46.7kmol / h n& 2w = (0.0848)(56.39) = 4.78 kmol H 2 O(v) / h 0.102 kmol H 2 O(v) / kmol n& 2c = (0.500)(56.39) = 28.2 kmol O 2 / h ⇒ 0.604 kmol CO 2 / kmol n& 2a = (0.2426)(56.39) = 13.7 kmol air / h 0.293 kmol Air / kmol
n& RT V&2 = 2 = P
(46.7 kmol / h)(0.08206 3 atm
m 3atm )(343 K) kmol ⋅ K = 438 m 3 / h
56.39 × 0.8086 kmol 22.4 m 3 (STP) 1h (b) Gas feed rate : V&1 = = 17.0 SCMM h kmol 60 min
6-60
6.81(cont'd) (c) Liquid feed: (100)(56.39) = 5640 kg / h To calculate V& , we would need to know the density of a 7 wt% aqueous Na 2 CO3 solution. (d) If T dropped in the filter, more solid NaHCO 3 would be recovered and the residual solution would contain less than 2.4% NaHCO 3. (e) Henry's law
Benefit: Higher pressure ⇒ greater pCO2
higher concentration of CO2 in solution
⇒ higher rate of reaction ⇒ smaller reactor needed to get the same conversion ⇒ lower cost Penalty: Higher pressure ⇒ greater cost of compressing the gas (purchase cost of compressor, power consumption)
6.82 600 lb m / h Dissolution Dissolution 0.90 MgSO4 ⋅ 7H 2 O Dissolution Tank Tank Tank 010 . I m & 1 ( lb m H 2O / h)
Filter I
R|m& (lb soln / h) U| S|0.32 kg MgSO / kgV| T0.68 kg H O / kg W 2
4
4
6000 lb m I / h 110o F
m & 6 (lb m / h) 0.23 lb m MgSO 4 / lb m 0.77 lbm H 2O / lb m
& 4 ( lb m MgSO 4 ⋅ 7 H 2 O / h) m
Filter II
R|m& (lb |S0.23 lb ||0.77 lb T 5
m
soln )
m
MgSO 4 / lb m
m
H 2O / lb m
U| |V || W
R|300 lb soln / hU| S|0.32 MgSO V| T0.68 H O W m
m
2
6000 lbm I / h
2
m & 3 ( lb m so ln/ h) 0.32 MgSO4 0.68 H 2 O
Crystallizer
& 4 ( lb m MgSO 4 ⋅ 7H 2 O) m
R|0.05m& S|0.23 lb T0.77 lb
( lb m soln ) m MgSO 4 / lb m m H 2 O / lb m
4
U| V| W
a. Heating the solution dissolves all MgSO 4 ; filtering removes I, and cooling recrystallizes MgSO 4 enabling subsequent recovery. (b) Strategy: Do D.F analysis.
6-61
6.82(cont'd)
UV ⇒ m& , m& balanceW
Overall mass balance Overall MgSO 4
1
Diss. tank overall mass balance 4
UV ⇒ m& , m& W 2
Diss. tank MgSO 4 balance
6
( MW) MgSO4 = (2431 . + 32.06 + 6400 . ) = 12037 . , ( MW) MgSO4 ⋅7H2 O = (12037 . + 7 *1801 . ) = 24644 . Overall MgSO 4 balance: 60,000 lb m
0.90lb m MgSO4 ⋅ 7H 2 O
h
lb m
= (300 lb m / h)(0.32 lb m
120.37 lb m MgSO 4
246.44 lb m MgSO 4 ⋅ 7H 2 O MgSO 4 / lb m ) + m& 4 (12037 . / 246.44 ) + 0.05m& 4 (0.23)
& 4 = 5.257 x10 4 lb m crystals / h ⇒ m
&4 Overall mass balance: 60,000 + m& 1 = 6300 + 1.05m
c. Diss. tank overall mass balance: Diss. tank MgSO 4 balance: ⇒
& 4 = 5.257 x10 4 lb m / h m
m& 1 = 1494 lb m H 2 O / h
60,000 + m &1 + m & 6 = m& 2 + 6000 54 ,000(120.37 / 24644 . ) + 0.23m & 6 = 0.32 m& 2
m & 2 = 1512 . x10 5 lb m / h
UV W
& 6 = 9.575x10 4 lb m / h recycle m
Recycle/fresh feed ratio =
9.575x10 4 lb m / h = 64 lb m recycle / lb m fresh feed 1494 lb m / h
6.83 a. n& 1 (kmol CO 2 / h) Cryst Filter
1000 kg H 2SO 4 / h (10 wt%) 1000 kg HNO 3 / h m & w (kg H 2 O / h)
m & 2 (kg CaSO4 / h) m & 3 (kg Ca(NO3) 2 / h) m & 4 (kg H2O / h)
Filter cake & 5 (kg / h) m 0.96 kg CaSO 4 (s) / kg 0.04 kg soln / kg
& 0 (kg CaCO3 / h) m & 0 (kg solution / h) 2m
m & 0 (kg CaCO 3 / h)
m & 8 (kg soln / h)
2m & 0 (kg solution / h)
R| X (kg CaSO / kg) U| Solution composition: S 500 X (kg H O / kg) |T(1 − 501X )(kg Ca(NO ) / kg)V|W a
4
a
a
6-62
2
3 2
6.83 (cont’d) b. Acid is corrosive to pipes and other equipment in waste water treatment plant. c. Acid feed:
1000 kg H 2SO 4 / h = 0.10 ⇒ m & w = 8000 kg H 2 O / h (2000 + m& w ) kg / h
Overall S balance: 1000 kg H 2SO 4
32 kg S
h
98 kg H 2SO 4
+
=
& 5 (kg / h) (0.96 + 0.04 X a ) (kg CaSO 4 ) m kg
32 kg S 136 kg CaSO 4
& 8 (kg / h) X a (kg CaSO 4 ) m 32 kg S kg 136 kg CaSO 4
⇒ 3265 . = 0.2353m & 5 (0.96 + 0.04 X a ) + 0.2353m& 8 X a
(1)
Overall N balance: 1000 kg HNO 3
14 kg N
h
63 kg HNO 3
+
=
0.04m& 5 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 ) kg
m& 8 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 ) kg
28 kg N 164 kg Ca(NO 3 ) 2
28 kg N 164 kg Ca(NO 3 ) 2
⇒ 222.2 = 0.00683m & 5 (1 − 501X a ) + 0.171m& 8 (1 − 501X a )
(2)
Overall Ca balance: & 0 (kg / h) m
& 5 (kg / h) (0.96 + 0.04X a ) (kg CaSO 4 ) 40 kg Ca m 40 kg Ca = 100 kg CaCO 3 kg 136 kg CaSO 4 & 5 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 ) 0.04m 40 kg Ca + kg 164 kg Ca(NO 3 ) 2 & 8 (kg / h) X a (kg CaSO 4 ) m 40 kg Ca + kg 136 kg CaSO 4 & 8 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 ) m 40 kg Ca + kg 164 kg Ca(NO 3 ) 2
& 0 = 0.294m& 5 (0.96 + 0.04 X a ) + 0.00976m & 5 (1 − 501 X a ) ⇒ 0.40m & 8 X a + 0.244m& 8 (1 − 501X a ) + 0.294 m
( 3)
Overall C balance : m & 0 (kg / h)
12 kg C 100 kg CaCO 3
⇒ 0.01m & 0 = n&1
=
n& 1 (kmol CO 2 / h)
(4 )
6-63
1 kmol C
12 kg C
1 kmol CO 2
1 kmol C
6.83 (cont’d) Overall H balance : 1000 (kg H 2SO4 )
2 kg H
+
1000 kg HNO3
h
98 kg H 2SO 4
⇒ 92517 . = 2.22m & 5 X a + 5556 . m &8 Xa
(5)
1 kg H
h
+
m & w (kg / h)
2 kg H
63 kg HNO 3 18 kg H 2O & 5 (kg / h) 500 X a (kg H 2 O) & (kg / h) 500 X a (kg H 2 O) 2 kg H 0.04m 2 kg H m = + 8 kg 18 kg H 2 O kg 18 kg H 2 O
Solve eqns. (1)-(5) simultaneously, using E-Z Solve. m& 0 = 1812.5 kg CaCO 3 (s) / h,
m& 5 = 1428.1 kg / h,
& 8 = 9584.9 kg soln / h, m
n&1 = 18.1 kmol CO 2 / h(v),
X a = 0.00173 kg CaSO 4 / kg
Recycle stream = 2 * m& 0 = 3625 kg soln / h
. CaSO U R| 0.00173(kg CaSO / kg) U| R||0173% || S| 500 * 0.00173(kg H O / kg) V| ⇒ S|86.5% H O V| T(1 − 501 * 0.00173)(kg Ca(NO ) / kg)W |T13.3% Ca(NO ) |W 4
4 2
2
3 2
d.
3 2
From Table B.1, for CO2 : Tc = 304.2 K, Pc = 72.9 atm T (40 + 273.2 ) K ⇒ Tr = = = 103 . , Tc 304.2
Pr =
30 atm = 0.411 72.9 atm
From generalized compressibility chart (Fig. 5.4-2): z = 0.86 ⇒ V$ =
zRT 0.86 0.08206 L ⋅ atm 313.2 K L = = 0.737 mol ⋅ K 30 atm P mol CO2
Volumetric flow rate of CO2 : 18.1kmol CO 2 V& = n&1 * V$ = h
e.
0.737 L
1000 mol
mol CO 2
1 kmol
= 1.33x104 L / h
Solution saturated with Ca(NO3 )2 : ⇒
1 − 501X a (kg Ca(NO 3 ) 2 / kg) = 1.526 ⇒ X a = 0.00079 kg CaSO4 / kg 500Xa (kg H 2 O / kg)
Let m& 1 (kg HNO3 /h) = feed rate of nitric acid corresponding to saturation without crystallization.
6-64
6.83 (cont’d) Overall S balance: 1000kg H 2SO 4
32 kg S
h
98 kg H 2SO 4 +
=
m & 5 (kg / h) (0.96 + (0.04)(0.00079)) (kg CaSO 4 )
32 kg S
kg
136 kg CaSO 4
& 8 (kg / h) 0.00079(kg CaSO4 ) m kg
32 kg S 136kg CaSO 4
& 5 + 0.000186m &8 ⇒ 326.5 = 0.226m
(1' )
Overall N balance: m &1 (kg HNO3 )
14 kg N
h
63 kg HNO3
=
0.04m & 5 (kg / h) (1 − (501)(0.00079)) (kg Ca(NO 3) 2 ) kg
+
164kg Ca(NO 3) 2
& 8 (kg / h) (1 − (501)(0.00079)) (kg Ca(NO3 )2 ) m kg
28 kg N 28 kg N
164 kg Ca(NO3 )2
&1 = 000413 & 5 + 0103 &8 ⇒ 0222 . m . m . m
(2')
Overall H balance: 1000 (kg H 2SO 4 )
2 kg H
h
98 kg H 2SO 4
+
8000 (kg / h)
2 kg H 18 kg H2 O
= +
+
m& 1 kg HNO3
1 kg H
h
63 kg HNO3
& 5 (kg / h) 500(0.00079) (kg H 2 O) 0.04m kg
m& 8 (kg / h) 500(0.00079) (kg H 2O) kg
& 1 = 0.00175m& 5 + 0.0439m& 8 ⇒ 909.30 + 0.0159m
2 kg H 18 kg H2 O
2 kg H 18 kg H2 O (3')
Solve eqns (1')-(3') simultaneously using E-Z solve: m& 1 = 1155 . x10 4 kg / h;
m & 5 = 1.424 x10 3 kg / h;
Maximum ratio of nitric acid to sulfuric acid in the feed =
1155 . x10 4 kg / h = 115 . kg HNO 3 / kg H 2 SO 4 1000 kg / h
6-65
m & 8 = 2.484x10 4 kg / h
6.84 Moles of diphenyl (DP): Moles of benzene (B): ⇒ x DP =
bg
U| |V || W
56.0 g = 0.363 mol 154.2 g mol 550.0 ml 0.879 g 1 mol = 6.19 mol ml 78.11 g
0.363 = 0.0544 mol DP mol 6.19 + 0.363
bg
b
g
p *B T = (1 − x DP ) p *B T = 0.945 120.67 mm Hg = 114.0 mm Hg
b
g b0.0554 g = 3.6 K = 3.6 C ⇒ T 8.314b 273.2 + 801 .g = . K = 1.85 C b0.0554 g = 185
2 8.314 2732 . + 5.5 RTm0 ∆Tm = x DP = $ 9837 ∆H m
∆Tbp =
RTb02 x DP ∆H$
2
o
m
= 5.5 − 3.6 = 1.9 ° C
2
o
30,765
v
⇒ Tb = 80.1 + 185 . = 82.0 ° C
6.85 Tm0 = 0.0 o C, ∆Tm = 4.6 o C=4.6K ∆Tm∆Hˆ m (4.6K)(600.95 J/mol) Eq. 6.5-5 → x = = = 0.0445 mol urea/mol u Table B.1 2 R(Tm 0 ) (8.314 J/mol ⋅ K)(273.2K)2 Eq. (6.5-4) ⇒ ∆Tb =
RTb 0 2 (8.314)(373.2) 2 xu = 0.0445 = 1.3K = 1.3o C ∆Hˆ v 40,656
1000 grams of this solution contains mu (g urea) and (1000 – mu ) (g water) nu 1 (mol urea) =
mu1 (g) 60.06 g/mol
nw 1 (mol water) =
(1000 − mu1 )(g) 18.02 g/mol
mu1 (mol urea) 60.06 xu 1 = 0.0445 = ⇒ mu 1 = 134 g urea, mw1 = 866 g water mu1 + (1000 − mu1 ) (mol solution) 60.06 18.02 ∆Tb = 3.0 o C = 3.0K ⇒ xu 2 =
∆Tb ∆Hˆ v R (Tb 0 )
2
=
(3.0K)(40,656 J/mol) (8.314 J/mol ⋅ K)(373.2K) 2
mu 2
= 0.105 mol urea/mol
(mol urea) 60.06 xu 2 = 0.105 = ⇒ mu 2 = 339 g urea 866 mu 2 60.06 + 18.02 (mol solution) ⇒ Add (339-134) g urea = 205 g urea
6-66
6.86 x aI =
b0.5150 gg b1101. g mol g b0.5150 g g b1101. g molg + b100.0 gg b94.10 g molg = 0.00438 mol solute mol
∆Tm =
⇒
b1 − 0.00523g mol solvent
94.10 g solvent
0.4460 g solute
0.00523 mol solute
1 mol solvent
95.60 g solvent
∆H$ m =
6.87 a.
RTm20 ∆T I xI ∆T II 0.49 ° C mol solute x s ⇒ mII = IIs ⇒ xsII = x Is mI = 0.00438 = 0.00523 $ 0.41° C mol solution ∆Tm xs ∆Tm ∆Hm
b
g b0.00523g = 6380 J mol = 6.38 kJ / mol
8.314 2732 . − 5.00 RTm20 xs = ∆Tm 0.49
b g
ln ps* Tb0 = −
2
∆H vI ∆H vII + B , ln ps* Tbs = − +B RTb0 RTbs
b g
Assume ∆HvI ≅ ∆HvII ; T0 Ts ≅ T02
b g
b g
⇒ ln Ps* Tb 0 − ln P0* Tbs = −
b. Raoult’s Law:
= 8350 . g solute mol
p *s
F1 GH T
∆Hv R
b0
b T g = b1 − xg b T g b0
p *0
−
bs
I JK
1 ∆Hv Tbs − Tb 0 ≅ Tbs R Tb20
∆Hv ∆Tb RTb20 ⇒ ln 1 − x ≈ − x = − ⇒ ∆Tb = x ∆Hv RTb20
b g
6.88 m1 (g styrene) 90 g ethylbenezene 100 g EG 90 g ethylbenzene 30 g styrene
m2 (g styrene) 100 g EG
Styrene balance: m1 + m2 = 30 g styrene Equilibrium relation:
F GH
m2 m1 = 0.19 100 + m2 90 + m1
I JK
solve simultaneously
m1 = 25.6 g styrene in ethylbenzene phase m 2 = 4.4 g styrene in ethylene glycol phase
6-67
6.89 Basis: 100 kg/h.
A=oleic acid; C=condensed oil; P=propane
100 kg / h
95.0 kg C / h m & 2 kg A / h
0.05 kg A / kg 0.95 kg C / kg
& 3 kg A / h m & 1 kg P / h m
m & 1 kg P / h
a. 90% extraction: m& 3 = (0.09 )(0.05)(100 kg / h) = 4.5 kg A / h Balance on oleic acid : ( 0.05)(100) = m& 2 + 4.5 kg A / h ⇒ m& 2 = 0.5 kg A / h Equilibrium condition:
0.15 =
0.5 / ( n&1 + 0.5) ⇒ n&1 = 73.2 kg P / h 4.5 / (4 .5 + 95)
b. Operating pressure must be above the vapor pressure of propane at T=85o C=185o F Figure 6.1-4 ⇒ p *propane = 500 psi = 34 atm c. Other less volatile hydrocarbons cost more and/or impose greater health or environmental hazards. 6.90 a. Benzene is the solvent of choice. It holds a greater amount of acetic acid for a given mass fraction of acetic acid in water. Basis: 100 kg feed.
A=Acetic acid, W=H 2 O, H=Hexane, B=Benzene
100 (kg) 0.30 kg A / kg 0.70 kg W / kg
m1 (kg) 0.10 kg A / kg 0.90 kg W / kg m2 (kg A) m H (kg H) or m B (kg B)
m H (kg H) or mB (kg B)
Balance on W:
100 * 0.70 = m1 * 0.90 ⇒ m1 = 77.8 kg
Balance on A:
100 * 0.30 = m2 + 77.8 * 0.10 ⇒ m 2 = 22.2 kg
Equilibrium for H: KH =
m2 / ( m2 + mH ) 22.2 / (22 .2 + mH ) = = 0.017 ⇒ mH = 1.30 x10 4 kg H xA 010 .
Equilibrium for B: KB =
m2 / (m2 + mB ) 22.2 / (22.2 + mB ) = = 0.098 ⇒ mB = 2.20 x10 3 kg B xA 0.10
(b) Other factors in picking solvent include cost, solvent volatility, and health, safety, and environmental considerations.
6-68
6.91
a. Basis: 100 g feed ⇒ 40 g acetone, 60 g H 2 O. A = acetone, H = n - C6 H 14 , W = water 40 g A 60 g W
e1 (g A) 60 g W
25°C
100 g H
100 g H r 1 (g A)
b
25°C
75 g H
75 g H r 2 (g A)
xA in H pha se / xA in W phase = 0.343 x = mass fraction Balance on A − stage 1: Equilibrium condition − stage 1:
g
U| e = 27.8 g acetone b g = 0.343V ⇒ r = 12.2 g acetone |W e b60 + e g 27.8 = e + r U| r = 7.2 g acetone r b75 + r g ⇒ = 0.343V| e = 20.6 g acetone e b60 + e g W 40 = e1 + r1 r1 100 + r1
1
1
1
Balance on A − stage 2:
e 2 (g A) 60 g W
1
2
2
2
Equilibrium condition − stage 2:
% acetone not extracted =
2
2
2
2
2
20.6 g A remaining × 100% = 515% . 40 g A fed
b. 40 g A 60 g W
e1 g A 60 g W
r1 g A 175 g H
175 g H
Balance on A − stage 1: Equilibrium condition − stage 1:
% acetone not extracted = c.
U| r = 17.8 g acetone b g = 0.343V ⇒ e = 22.2g acetone |W b60 + e g
40.0 = e1 + r1 r1 175 + r1 e1
1
1
22.2 g A remaining × 100% = 55.5% 40 g A fed
40 g A 60 g W
19.4 g A 60 g W
20.6 g A m (g H)
m (g H)
Equilibrium condition:
1
20.6 / ( m + 20.6) = 0.343 ⇒ m = 225 g hexane 19.4 / ( 60 + 19.4)
d. Define a function F=(value of recovered acetone over process lifetime)-(cost of hexane over process lifetime) – (cost of an equilibrium stage x number of stages). The most costeffective process is the one for which F is the highest.
6-69
6.92 a. P--penicillin; Ac--acid solution; BA--butyl acetate; Alk--alkaline solution
Broth Mixing tank
100 kg 0.015 P 0.985 Ac
m1 (kg BA)
Extraction Unit I
Acid
D.F. analysis : Extraction Unit I 3 unknown (m1 , m2p , m3p ) –1 balance (P) –1 distribution coefficient – 1 (90% transfer) 0 DF
m3P (kg P) 98.5 (kg Ac) pH=2.1 m4 (kg Alk)
Extraction II m6P (kg P) m1 (kg BA)
m5P (kg P) m4 (kg Alk) pH=5.8
Extraction Unit II (consider m1 , m3p ) 3 unknowns – 1 balance (P) – 1 distribution coefficient – 1 (90% transfer) 0 DF
b. In Unit I, 90% transfer ⇒ m 3P = 0.90(15 . ) = 135 . kg P P balance: 1.5 = m2 P + 1.35 ⇒ m 2 P = 0.15 kg P 1.35 / (1.35 + m1 ) pH=2.1 ⇒ K = 25.0 = ⇒ m1 = 34.16 kg BA 015 . / ( 015 . + 98.5) In Unit II, 90% transfer: m5 P = 0.90(m3 P ) = 1.215 kg P P balance: m 3P = 1.215 + m6 P ⇒ m 6 P = 0135 . kg P m / ( m6 P + 34.16) pH=5.8 ⇒ K = 0.10 = 6 P ⇒ m4 = 29.65 kg Alk 1.215 / (1215 . + m4 ) m1 34.16 kg BA = = 0.3416 kg butyl acetate / kg acidified broth 100 100 kg broth m 4 29.65 kg Alk = = 0.2965kg alkaline solution / kg acidified broth 100 100 kg broth Mass fraction of P in the product solution: m5P 1.215 P xP = = = 0.394 kg P / kg m4 + m5P (29.65 + 1.215) kg c. (i). The first transfer (low pH) separates most of the P from the other broth constituents, which are not soluble in butyl acetate. The second transfer (high pH) moves the penicillin back into an aqueous phase without the broth impurities. (ii). Low pH favors transfer to the organic phase, and high pH favors transfer back to the aqueous phase. (iii).The penicillin always moves from the raffinate solvent to the extract solvent.
6-70
6.93
W = water, A = acetone, M = methyl isobutyl ketone x W = 0.20 x A = 0.33 x M = 0.47
U| V| ⇒ W
Figure 6.6-1
Phase 1: x W = 0.07 , x A = 0.35, x M = 0.58 Phase 2: x W = 0.71, x A = 0.25, x M = 0.04
Basis: 1.2 kg of original mixture, m1 =total mass in phase 1, m2 =total mass in phase 2. H 2 O Balance:
1.2 * 0.20 = 0.07m1 + 0.71m2
Acetone balance:
1.2 * 0.33 = 0.35m1 + 0.25m 2
⇒
R|m = 0.95 kg in MIBK - rich phase S|m = 0.24 kg in water - rich phase T 1
2
6.94 Basis: Given feeds: A = acetone, W = H2 O, M=MIBK Overall system composition:
U| b g V 3500 g b 20 wt% A, 80 wt% Mg ⇒ 700 g A, 2800 g M |W 2200 g A U | ⇒ 3500 g WV ⇒ 25.9% A, 41.2% W, 32.9% M 2800 g M |W 5000 g 30 wt% A, 70 wt% W ⇒ 1500 g A, 3500 g W
Fig. 6.6-1
Phase 1: 31% A, 63% M, 6% W Phase 2: 21% A, 3% M, 76% W
Let m1 =total mass in phase 1, m2 =total mass in phase 2. H 2 O Balance:
3500 = 0.06 m1 + 0.76m 2
Acetone balance:
2200 = 0.31m1 + 0.21m 2
⇒
R|m = 4200 g in MIBK - rich phase S| m = 4270 g in water - rich phase T 1
2
6.95 A=acetone, W = H 2 O, M=MIBK 41.0 lb m / h
32 lb m / h x AF (lb m A / lb m ) x WF (lb m W / lb m )
xA,1 , x W,1, 0.70
& 2 lbm / h m
m & 1 (lb m M / h)
xA ,2 , xW ,2 , x M ,2
Figure 6.6-1⇒ Phase 1: x M = 0.700 ⇒ xw ,1 = 0.05; x A,1 = 0.25 ; Phase 2: x w, 2 = 0.81; x A, 2 = 0.81; x M , 2 = 0.03 Overall mass balance: MIBK balance:
m& 1 = 28.1 lb m MIBK / h 32.0 lb m / h + m& 1 = 41.0 lb m h + m& 2 ⇒ & 1 = 41.0 * 0.7 + m& 2 * 0.03 m m& 2 = 19.1lb m h
UV W
6-71
6.96 a. Basis: 100 kg; A=acetone, W=water, M=MIBK System 1: x a,org = 0.375 mol A, x m,org = 0.550 mol M, x w,org = 0.075 mol W x a,aq = 0.275 mol A, x m,aq = 0.050 mol M, x w,aq = 0.675 mol W
UV W
maq,1 = 417 . kg Mass balance: maq ,1 + morg,1 = 100 ⇒ Acetone balance: maq ,1 * 0.275 + morg ,1 * 0.375 = 33.33 morg,1 = 58.3 kg System 2: x a,org = 0.100 mol A, x m,org = 0.870 mol M, x w,org = 0.030 mol W x a,aq = 0.055 mol A, x m,aq = 0.020 mol M, x w,aq
= 0.925 mol W
UV W
m aq,2 = 22 .2 kg Mass balance: maq ,2 + morg, 2 = 100 ⇒ Acetone balance: maq ,2 * 0.055 + morg , 2 * 0.100 = 9 morg,2 = 77.8 kg
b. K a,1 =
xa ,org,1 xa ,aq,1
=
0.375 = 136 . ; 0.275
K a, 2 =
xa, org ,2 x a, aq , 2
=
0.100 = 182 . 0.055
High Ka to extract acetone from water into MIBK; low Ka to extract acetone from MIBK into water. c.
β
aw,1
=
xa, org / x w ,org
=
xa ,aq / x w ,aq
0.375 / 0.075 0.100 / 0.040 = 12.3; β = = 418 . aw,2 0.055 / 0.920 0.275 / 0.675
If water and MIBK were immiscible, x w, org = 0 ⇒ β aw → ∞ d.
Organic phase= extract phase; aqueous phase= raffinate phase β a, w =
( xa / x w ) org ( xa / x w ) aq
=
( x a ) org / ( xa ) aq ( xw ) org / ( x w ) aq
=
Ka Kw
When it is critically important for the raffinate to be as pure (acetone-free) as possible. 6.97 Basis: Given feed rates: A = acetone, W = water, M=MIBK
e& 2 (kg / h) x2A (kg A / kg) x2W (kg W / kg) x2M (kg M / kg)
e& 1 (kg / h) x1A (kg A / kg) x1W (kg W / kg) x1M (kg M / kg)
200 kg / h 0.30 kg A / kg 0.70 kg M / kg
Stage I
&r1 (kg / h) y1A (kg A / kg) y1W (kg W / kg) y1M (kg M / kg)
300 kg W / h
6-72
Stage II Stage IIStage
300 kg W / h
&r2 (kg / h) y2A (kg A / kg) y2W (kg W / kg) y2M (kg M / kg)
6.97(cont'd) Overall composition of feed to Stage 1:
b200gb0.30g = 60 kg A h U| 500 kg h 200 − 60 = 140 kg M h V ⇒ 12% A, 28% M, 60% W 300 kg W h |W Figure 6.6-1 ⇒
Extract: x1A = 0.095, x1W = 0.880, x1M = 0.025 Raffinate: y1A = 0.15, y1W = 0.035, y1M = 0.815
R| S| T
e&1 = 273 kg / h ⇒ & r1 = 227 kg / h 60 = 0.095e&1 + 015 . r&1
500 = e&1 + r&1
Mass balance Acetone balance:
Overall composition of feed to Stage 2:
U| 527 kg h b227gb0.15g = 34 kg A h b227gb0.815g = 185 kg M h V ⇒ 6.5% A, 35.1% MIBK, 58.4% W b227gb0.035g + 300 = 308 kg W h|W Figure 6.6-1 ⇒
Extract: x2 A = 0.04, x 2W = 0.94, x2 M = 0.02 Raffinate: y2 A = 0.085, y2 W = 0.025, y2 M = 0.89
Mass balance: Acetone balance:
R| S| T
e&2 = 240 kg / h ⇒ & r2 = 287 kg / h 34 = 0.04 e2 + 0.085r2
527 = e&2 + r&2
Acetone removed: [60 − (0.085)(287 )] kg A removed / h = 0.59 kg acetone removed / kg fed 60 kg A / h in feed
Combined extract: Overall flow rate = e&1 + e&2 = 273 + 240 = 513 kg / h Acetone:
( x1 A e&1 + x 2 A e&2 ) kg A
=
0.095 * 273 + 0.04 * 240 = 0.069 kg A / kg 513
Water :
( x1w e&1 + x 2w e&2 ) kg W 0.88 * 273 + 0.94 * 240 = = 0.908 kg W / kg e&1 + e&2 513
MIBK:
( x1 M e&1 + x 2 M e&2 ) kg M 0.025 * 273 + 0.02 * 240 = = 0.023 kg M / kg (e&1 + e& 2 )kg 513
6-73
6.98. a. 1.50 L / min 25o C, 1atm, rh = 25% n&0 (mol / min) y 0 (mol H2O / mol) (1- y 0 ) (mol dry air / mol) n& 0 =
M (g gel) M a (g H2 O)
(1 atm)(1.50 L / min) PV& = = 0.06134 mol / min RT (0.08206 L ⋅ atm / mol ⋅ K)(298 K)
r.h.=25%⇒
pH 2 O pH* 2O (25o C)
= 025 .
Silica gel saturation condition: X * = 12 .5 Water feed rate : ⇒ m& H 2O =
y0 =
0.25 p *H2 O ( 25 o C) p
p H2 O p *H2 O
=
= 12 .5 * 0.25 = 3125 .
0.25( 23.756 mm Hg) mol H 2 O = 0.00781 760 mm Hg mol
0.06134 mol 0.00781 mol H 2 O 18.01g H 2 O min
mol
g H 2 O ads 100 g silica gel
mol H 2 O
= 0.00863 g H 2 O / min
Adsorption in 2 hours = (0.00863 g H 2 O / min)(120min) = 1.035 g H 2 O Saturation condition:
1.035 g H 2O 3.125 g H 2 O = ⇒ M = 33.1g silica gel M (g silica gel) 100 g silica gel
Assume that all entering water vapor is adsorbed throughout the 2 hours and that P and T are constant. b. Humid air is dehumidified by being passed through a column of silica gel, which absorbs a significant fraction of the water in the entering air and relatively little oxygen and nitrogen. The capacity of the gel to absorb water, while large, is not infinite, and eventually the gel reaches its capacity. If air were still fed to the column past this point, no further dehumidific ation would take place. To keep this situation from occurring, the gel is replaced at or (preferably) before the time when it becomes saturated. 6.99 a.
Let c = CCl4 Relative saturation = 0.30 ⇒
pc * p c (34 o C)
⇒ pc = 0.30 * (169 mm Hg) = 50.7 mm Hg
b. Initial moles of gas in tank: n0 =
1 atm 50.0 L P0V 0 = = 1.985 mol RT0 0.08206 L ⋅ atm / mol ⋅ K 307 K
Initial moles of CCl4 in tank: n c 0 = yc 0 n0 =
p c0 50.7 mm Hg n0 = × 1.985 mol = 0.1324 mol CCl 4 P0 760 mm Hg
6-74
6.99 (cont’d) 50% CCl4 adsorbed ⇒ n c = 0.500n c0 = 0662 mol CCl 4 (= n ads) Total moles in tank: n tot = n 0 − n ads = (1.985 − 0.0662) mol = 1.919 mol Pressure in tank. Assume T = T0 and V = V0 . P=
FG H
IJ FG 760 mm Hg IJ = 735 mm Hg K H atm K
n tot RT0 (1919 . )( 0.08206)(307) = atm V0 50.0
yC =
nc 0.0662 mol CCl 4 mol CCl 4 = = 0.0345 n tot 1.919 mol mol
⇒ p c = 0.0345(760 mm Hg) = 26.2 mm Hg c. Moles of air in tank: n a = n0 − nc 0 = (1.985 − 01324 . ) mol air = 1.853 mol air
yc =
nc mol CCl 4 = 0.001 ⇒ nc = 1.854 × 10 −3 mol CCl 4 n c + 1853 . mol
⇒ n tot = nc + n air = 1.854 mol
LM n RT OP = 1854 . × 10 mol 50.0 L N V Q −3
p c = y c P = 0.001
tot
0
0
0.08206 L ⋅ atm 307 K mol ⋅ K
760 mm 1 atm
= 0.710 mm Hg
X*
F g CCl I = 0.0762 p GH g carbon JK 1 + 0.096 p 4
⇒ X* =
c
c
0.0762( 0.710) g CCl 4 adsorbed = 0.0506 1 + 0.096 (0.710) g carbon
Mass of CCl4 adsorbed m ads = (n c0 − nc )( MW ) c =
(0.1324 − 0.001854 ) mol CCl 4
153.85 g 1 mol CCl 4
= 20.3 mol CCl 4 adsorbed 20.3g CCl 4 ads Mass of carbon required: m c = = 400 g carbon g CCl 4 ads 0.0506 g carbon β X * = K F p βNO2 ⇒ ln X * = ln K F + β ln p NO 2
ln(PNO2)
6.100 a.
y = 1.406x - 1.965
2 1.5 1 0.5 0 -0.5 -1 -1.5 0
1
2 ln(X*)
6-75
3
6.100 (cont’d) .406 .406 ln X * = 1406 . ln p NO2 − 1.965 ⇒ X * = e −1.965 p 1NO = 0.140p 1NO 2 2
K F = 0.140 (kg NO 2 / 100 kg gel)(mm Hg)−1.406 ; β = 1.406
b. Mass of silica gel : m g =
π * (0.05m) 2 (1 m) 10 3 L 0.75kg gel 1m 3
L
= 5.89 kg gel
Maximum NO2 adsorbed : p NO2 in feed = 0.010(760 mm Hg) = 7.60 mm Hg m ads =
0.140(7.60) 1.406 kg NO 2
5.89 kg gel
100 kg gel
= 0.143 kg NO 2
Average molecular weight of feed : MW = 0.01( MW ) N O2 + 0.99 ( MW ) air = ( 0.01)(46.01) + ( 0.99)(29.0) = 29.17
kg kmol
Mass feed rate of NO 2 : m& =
8.00 kg
1 kmol
0.01 kmolNO 2
46.01 kg NO 2
h
29.17 kg
kmol
kmol NO2
Breakthrough time :
tb =
= 0.126
kg NO 2 h
0.143 kg NO 2 = 1.13 h = 68 min 0.126 kg NO 2 / h
c. The first column would start at time 0 and finish at 1.13 h, and would not be available for another run until (1.13+1.50) = 2.63 h. The second column could start at 1.13 h and finish at 2.26 h. Since the first column would still be in the regeneration stage, a third column would be needed to start at 2.26 h. It would run until 3.39 h, at which time the first column would be available for another run. The first few cycles are shown below on a Gantt chart. Run
Regenerate
Column 1 0
1.13
2.63
3.39
4.52
6.02
Column 2 1.13
2.26
3.76
4.52
5.65
Column 3 2.26
6-76
3.39
4.89
5.65
6.78
Let S=sucrose, I=trace impurities, A=activated carbon
Add m A (kg A)
m S (kg S)
mS (kg S) mI (kg I) R (color units / kg S) V (L)
m I0 (kg I) R0 (color units / kg S)
Come to equilibrium
V (L)
mA (kg A) mI A (kg I adsorbed)
Assume
no sucrose is adsorbed • solution volume (V) is not affected by addition of the carbon m a. R(color units/kg S) = kCi (kg I / L) = k I (1) V •
k ⇒ ∆R = k ( Ci 0 − Ci ) = ( mI 0 − mI ) V
mIA = mI 0 −mI
∆R =
kmIA V
kmIA / V m ∆R x100% = x100 = 100 IA R0 kmI 0 / V m I0 m Equilibrium adsorption ratio : X i* = I A mA Normalized percentage color removal:
% removal of color =
υ=
m m % removal ( 3) 100 m IA / mI 0 = = 100 IA S m A / mS mA / mS mA mI 0 m mI 0 ⇒ υ = 100X *i S ⇒ X i* = υ mI 0 100 mS ( 1),(5)
Freundlich isotherm X i* = K F Ci β ⇒ υ=
100 mS K F mI 0 k
β
9.500 9.000
(2) (3) (4)
(5)
mI 0 R υ = KF ( )β 100mS k
R β = K F' R β
A plot of ln υ vs. ln R should be linear: slope = β ;
ln v
6.101
y = 0.4504x + 8.0718
8.500 8.000 0.000 1.000 2.000 3.000 ln R
6-77
intercept = lnK'F
6.101 (cont’d) ln υ = 0.4504 ln p NO2 + 8.0718 ⇒ υ = e8.0718 R 0.4504 = 3203R 0.4504
⇒ K F' = 3203, β = 0.4504 b. 100 kg 48% sucrose solution ⇒ m S = 480 kg 95% reduction in color
⇒ R = 0.025(20.0) = 0.50 color units / kg sucrose
υ = K F' R β = 3203(0.50) 0.4504 = 2344 % color reduction 97.5 ⇒ 2344 = = ⇒ m A = 20.0 kg carbon m A / mS m A / 480
6-78
CHAPTER SEVEN
7.1
0.80 L 3.5 × 10 4 kJ 0.30 kJ work h
L
1 kJ heat
7.2
1 kW
3600 s 1 k J s
2.33 kW 10 3 W 1.341 × 10 −3 hp 1 kW
1h
1W
= 2.33 kW ⇒ 2.3 kW
= 312 . hp ⇒ 3.1 hp
All kinetic energy dissipated by friction mu 2 2 5500 lbm 552 miles2 = 2 h2 = 715 Btu
(a) E k =
52802 ft 2 1 2 mile 2
12 h 2 36002 s2
1 lbf 9 .486 × 10 − 4 Btu 32.174 lbm ⋅ ft / s2 0.7376 ft ⋅ lb f
(b) 3 ×108 brakings 715 Btu 1 day day
braking
24 h
1h
1W −4
3600 s 9.486 × 10
1 MW Btu/s 1 06 W
= 2617 MW
⇒ 3000 MW
7.3
(a) Emissions: 1000 sacks Paper ⇒ Plastic ⇒
2000 sacks
1000 sacks
(0.0045 + 0.0146) oz
(724 + 905) Btu sack
Plastic ⇒
2000 sacks
1 lbm
sack 16 oz sack
Energy: Paper ⇒
(0.0510 + 0.0516) oz
= 6.41 lb m
1 lbm 16 oz
= 2 .39 lb m
= 1.63 × 10 6 Btu
(185 + 464 ) Btu sack
= 1.30 × 10 6 Btu
(b) For paper (double for plastic)
Materials for 400 sacks
Raw Materials Acquisition and Production
Sack Production and Use
7- 1
1000 sacks
400 sacks
Disposal
7.3 (cont’d) Emissions: Paper ⇒
400 sacks
Plastic ⇒
0.0510 oz 1 lb m 1000 sacks + sack 16 oz
800 sacks
0.0045 oz sack
0.0516 oz 1 lb m = 4.5 lb m sack 16 oz ⇒ 30% reduction
1 lb m 2000 sacks + 16 oz
0.0146 oz 1 lb m = 2.05 lb m sack 16 oz ⇒ 14% reduction
Energy: Paper ⇒
400 sacks
Plastic ⇒
(c) .
724 Btu sack
800 sacks
3 × 10 8 persons
+
185 Btu sack
1000 sacks
+
905 Btu
= 119 . × 10 6 Btu; 27% reduction
sack 2000 sacks
464 Btu
= 1.08 × 10 6 Btu; 17% reduction
sack
1 sack 1 day 1h person - day 24 h 3600 s
649 Btu 1J 1 MW -4 1 sack 9.486 × 10 Btu 10 6 J / s
= 2 ,375 MW
Savings for recycling: 0.17 (2 ,375 MW) = 404 MW (d) Cost, toxicity, biodegradability, depletion of nonrenewable resources.
7.4
1 ft 3
(a) Mass flow rate: m &=
3.00 gal min
7.4805 gal
Stream velocity: u =
3.00 gal
1728 in 3
Kinetic energy: E k =
min
1 ft
3
2
2
b g
. g ft b1225 2
2
2
s
in
1
1 min 60 s
1
7.4805 gal Π 0.5
mu 2 0.330 lb m = 2 s
d
(0.792)(62.43) lb m
1
1 ft
1 min
12 in
60 s lb f
2 32.174 lb m ⋅ ft / s2
. × 10 hp I = 140 . × 10 iFGH 01341 .7376 ft ⋅ lb / sJK
= 7.70 × 10 −3 ft ⋅ lb f / s
−3
−5
f
(b) Heat losses in electrical circuits, friction in pump bearings.
7- 2
= 0.330 lb m s
= 1.225 ft s
= 7.70 × 10 −3 hp
ft ⋅ lb f s
7.5
(a) Mass flow rate: m& =
42.0 m π ( 0.07 m ) s
2
103 L 673 K 1 m3
4
mu & 2 127.9 g 1 kg E& k = = 2 2 s 1000 g
(b)
130 kPa
1 mol
29 g
273 K 101.3 kPa 22.4 L (STP )
42.0 2 m 2 1N 1J 2 s 1 kg ⋅ m / s2 N ⋅ m
mol
= 127.9 g s
= 113 J s
127.9 g 1 mol 22.4 L (STP ) 673 K 101.3 kPa 1 m3 4 = 49.32 m s s 29 g 1 mol 273 K 130 kPa 10 3 L π (0.07)2 m2 & 2 127.9 g 1 kg mu E& k = = 2 2 s 1000 g
49.32 2 m 2 s
1N
1J
= 1558 . J/s
1 kg ⋅ m / s2 N ⋅ m
2
∆E& k = E& k (400 o C) - E& k (300 o C) = (155.8 - 113) J / s = 42.8 J / s ⇒ 43 J / s
(c) Some of the heat added goes to raise T (and hence U) of the air 7.6
(a)
∆E p = mg∆z =
1 gal
1 ft 3 62.43 lbm 32.174 ft −10 ft 1 lbf = −83.4 ft ⋅ lb f 3 2 7.4805 gal 1 ft s 32.174 lbm ⋅ ft / s2
b g = LMN2FGH 32.174 sft IJK b10 ft gOPQ
b g
mu 2 (b) E k = − ∆E p ⇒ = mg − ∆z ⇒ u = 2 g − ∆z 2
12
12
= 25.4
2
ft s
(c) False 7.7 (a)
∆E& k ⇒ positive When the pressure decreases, the volumetric flow rate increases, and
hence the velocity increases. ∆E& ⇒ negative The gas exits at a level below the entrance level. p
(b)
&= m
b g
5 m π 1.5 cm 2 s 2
1 m3 10 4 cm 2
273 K 10 bars 1 kmol 303 K 1.01325 bars 22.4 m 3 STP
b g
16.0 kg CH 4 1 kmol
= 0.0225 kg s 2 & PoutV&out nRT V& P u (m/s) ⋅ A(m) P = ⇒ out = in ⇒ out = in & & 2 & PinVin nRT Vin Pout uin (m/s) ⋅ A ( m ) Pout P 10 bar ⇒ u out = uin in = 5 ( m s ) = 5.555 m s Pout 9 bar
1 2 ∆E&k = m& (uout − u in2 ) =
0.5(0.0225) kg (5.5552 − 5.000 2 )m 2 s
2
s
1N
1W
1 kg ⋅ m/s
2
2
1 N ⋅ m/s
= 0.0659 W & ( zout − zin ) = ∆E& p = mg
0.0225 kg 9.8066 m -200 m s
s
= − 44.1 W
7- 3
1N kg ⋅ m/s
1W 2
1 N ⋅ m/s
7.8
∆E& p = mg & ∆z =
105 m3 103 L 1 kg H2O 981 . m −75 m 1 m3
h
2.778 × 10−7 kW ⋅ h
1J
1 kg ⋅ m / s2 1 N⋅ m
s2
1L
1N
1J
= −204 . × 10 kW ⋅ h h 4
The maximum energy to be gained equals the potential energy lost by the water, or 2.04 × 10 4 kW ⋅ h h
7.9
24 h
7 days
1 day 1 week
= 3.43 × 10 6 kW ⋅ h week (more than sufficient)
(b) Q − W = ∆U + ∆E k + ∆E p
b g = 0 b no height change g
∆E k = 0 system is stationary ∆E p
Q − W = ∆U , Q < 0,W > 0
(c) Q − W = ∆U + ∆E k + ∆E p
b
g
b
g
Q = 0 adiabatic , W = 0 no moving parts or generated currents ∆E k = 0 system is stationary ∆E p = 0 no height change
b b
g
g
∆U = 0
(d). Q − W = ∆U + ∆E k + ∆E p
b
g
W = 0 no moving parts or generated currents ∆E k = 0 system is stationary ∆E p = 0 no height change Q = ∆U , Q < 0 Even though the system is isothermal, the occurrence of a chemical reaction assures that ∆U ≠ 0 in a non-adiabatic reactor. If the temperature went up in the adiabatic reactor, heat must be transferred from the system to keep T constant, hence Q < 0 .
b b
g
g
7.10 4.00 L, 30 °C, 5.00 bar ⇒ V (L), T (°C), 8.00 bar (a). Closed system:
∆U + ∆E k + ∆E p = Q − W
RS∆E |T∆E
k p
b = 0 b by assumption g
= 0 initial / final states stationary
g
∆U = Q − W
(b)
Constant T ⇒ ∆U = 0 ⇒ Q = W =
−7.65 L ⋅ bar
8.314 J 0.08314 L ⋅ bar
= −765 J
(c) Adiabatic ⇒ Q = 0 ⇒ ∆U = −W = 7.65 L ⋅ bar > 0, Tfinal > 30° C
7- 4
transferred from gas to surroundings
bg
π 3
2
1 m2 = 2.83 × 10 −3 m 2 10 4 cm 2 (a) Downward force on piston:
7.11 A =
cm 2
Fd = Patm A + mpiston+weight g =
1 atm 1.01325 × 10 5 N / m2 2.83 × 10 −3 m2 atm
+
24.50 kg 9.81 m
d
s
i d
Upward force on piston: Fu = APgas = 2.83 × 10 −3 m 2 Pg N m 2
1N 1 kg ⋅ m / s2
2
= 527 N
i
Equilibrium condition: Fu = Fd ⇒ 2.83 × 10 −3 m2 ⋅ P0 = 527 ⇒ P0 = 1.86 × 10 5 N m 2 = 186 . × 10 5 Pa
V0 =
303 K 1.01325 × 105 Pa 0.08206 L ⋅ atm nRT 1.40 g N 2 1 mol N2 = = 0.677 L P0 28.02 g 1.86 × 105 Pa 1 atm mol ⋅ K
(b) For any step, ∆U + ∆E k + ∆E p = Q − W ⇒ ∆U = Q − W ∆Ek = 0 ∆E p = 0
Step 1: Q ≈ 0 ⇒ ∆U = −W Step 2: ∆U = Q − W As the gas temperature changes, the pressure remains constant, so that V = nRT Pg must vary. This implies that the piston moves, so that W is not zero. Overall: Tinitial = Tfinal ⇒ ∆U = 0 ⇒ Q − W = 0 In step 1, the gas expands ⇒ W > 0 ⇒ ∆U < 0 ⇒ T decreases
b gd
i b gb gb g
id
(c) Downward force Fd = 100 . 101325 . × 10 5 2.83 × 10 −3 + 4 .50 9.81 1 = 331 N (units as in Part (a)) F 331 N = = 116 . × 10 5 N m 2 A 2.83 × 10 − 3 m 2 P 1.86 × 10 5 Pa Since T0 = T f = 30° C , Pf V f = P0V0 ⇒ V f = V0 0 = 0.677 L = 108 . L Pf 116 . × 105 Pa
Final gas pressure Pf =
b
g
∆V 1.08 − 0.677 L Distance traversed by piston = = A
b
gb
g
b
g
1 m3 103 L
2.83 × 10 −3 m2
= 0142 . m
⇒ W = Fd = 331 N 0.142 m = 47 N ⋅ m = 47 J Since work is done by the gas on its surroundings, W = +47 J ⇒ Q = +47 J Q −W = 0
(heat transferred to gas) 32.00 g 4.684 cm3 103 L 7.12 V$ = = 01499 . L mol mol g 106 cm3 41.64 atm 0.1499 L 8.314 J / (mol ⋅ K) H$ = U$ + PV$ = 1706 J mol + = 2338 J mol mol 0.08206 L ⋅ atm / (mol ⋅ K)
7- 5
7.13
d
i
Ref state U$ = 0 ⇒ liquid Bromine @ 300 K, 0.310 bar
(a)
(b) ∆U$ = U$ final − U$ initial = 0.000 − 28.24 = −28.24 kJ mol
d i
∆ H$ = ∆U$ + ∆ PV$ = ∆U$ + P∆V$ (Pressure Constant)
∆ Hˆ = −28.24 kJ mol +
b
0.310 bar
(0.0516 − 79.94) L
gb
8.314 J
1 kJ
mol 0.08314 L ⋅ bar 103 J
g
= −30.7 kJ mol
∆ H = n∆ H$ = 5.00 mol −30.7 kJ / mol = −15358 . kJ ⇒ − 154 kJ
b
g b
g
(c) U$ independent of P ⇒ U$ 300 K, 0.205 bar = U$ 300 K , 0.310 bar = 28.24 kJ mol U$ 340 K, Pf = U$ 340 K, 1.33 bar = 29.62 kJ mol
d
∆U$ = U$ final − U$ initial
i b
g
E
∆U$ = 29.62 − 28.24 = 1.380 kJ mol
$ = P' V' $ ⇒ V'= $ PV $ / P' V$ changes with pressure. At constant temperature ⇒ PV $ (T = 300K, P = 0.205 bar) = (0.310 bar)(79.94 L / mol) = 120.88 L / mol V' 0.205 bar 5.00 L 1 mol n= = 0.0414 mol 120.88 L ∆U = n∆U$ = 0.0414 mol 1.38 kJ / mol = 0.0571 kJ
b
gb
g
∆U + ∆E k + ∆E p = Q − W ⇒ Q = 0.0571 kJ 0
0
0
(d) Some heat is lost to the surroundings; the energy needed to heat the wall of the container is being neglected; internal energy is not completely independent of pressure. 7.14 (a) By definition H$ = U$ + PV$ ; ideal gas PV$ = RT ⇒ H$ = U$ + RT
b g bg
b g bg
bg
U$ T , P = U$ T ⇒ H$ T , P = U$ T + RT = H$ T independent of P
. cal 50 K cal 1987 (b) ∆ H$ = ∆U$ + R∆T = 3500 + = 3599 cal mol mol mol ⋅ K ∆ H = n∆ H$ = 2.5 mol 3599 cal / mol = 8998 cal ⇒ 9.0 × 10 3 cal
b
gb
7.15 ∆U + ∆E k + ∆E p = Q − Ws
b b
g
g
∆ E k = 0 no change in m and u ∆ E p = 0 no elevation change Ws = P∆V since energy is transferred from the system to the surroundings
g
b
∆U = Q − W ⇒ ∆U = Q − P∆V ⇒ Q = ∆U + P∆V = ∆(U + PV ) = ∆H
7- 6
g
b b
g
7.16. (a) ∆ E k = 0 u1 = u 2 = 0 ∆ E p = 0 no elevation change
g
∆P = 0 (the pressure is constant since restraining force is constant, and area is constrant) Ws = P∆V the only work done is expansion work H$ = 34980 + 35.5T (J / mol), V1 = 785 cm3, T1 = 400 K, P = 125 kPa, Q = 83.8 J
b
.
g
125 × 103 Pa 785 cm3 1 m3 PV = = 0.0295 mol RT 8.314 m3 ⋅ Pa / mol ⋅ K 400 K 10 6 cm3 $ -H $ ) = 0.0295 mol 34980 + 35.5T - 34980 - 35.5(400K) (J / mol) Q = ∆H = n(H 2 1 2 n=
83.8 J = 0.0295 35.5T2 - 35.5(400) ⇒ T2 = 480 K nRT 0.0295 mol 8.314 m 3 ⋅ Pa 10 6 cm3 480 K = = 941 cm 3 125 × 105 Pa mol ⋅ K 1 m3 P 125 × 105 N (941- 785)cm3 1 m 3 ii ) W = P∆V = = 19.5 J m2 10 6 cm 3 iii ) Q = ∆U + P∆V ⇒ ∆U = Q − ∆PV = 83.8 J − 19.5 J = 64.3 J i) V =
(b) ∆Ep = 0 7.17 (a) "The gas temperature remains constant while the circuit is open." (If heat losses could occur, the temperature would drop during these periods.) (b) ∆U + ∆ E p + ∆ E R = Q& ∆t − W& ∆t ∆ E p = 0, ∆ E k = 0, W& = 0 , U$ ( t = 0) = 0 Q& =
0.90 × 1.4 W
1 J s 1W
= 1.26 J s
U (J ) = 1.26 t Moles in tank: n = PV RT =
1 atm
b
2.10 L 25 + 273 K
g
1 mol ⋅ K = 0.0859 mol 0.08206 L ⋅ atm
U 1.26 t (J) U$ = = = 14.67 t n 0.0859 mol Thermocouple calibration: T = aE + b
T = 0 , E =−0.249 T =100 , E =5 .27
b g
b g
T ° C = 181 . E mV + 4.51
U$ = 14.67 t 0 440 880 1320 T = 181 . E + 4.51 25 45 65 85
(c) To keep the temperature uniform throughout the chamber. (d) Power losses in electrical lines, heat absorbed by chamber walls. (e) In a closed container, the pressure will increase with increasing temperature. However, at the low pressures of the experiment, the gas is probably close to ideal ⇒ U$ = f T only. Ideality could be tested by repeating experiment at several initial pressures ⇒ same results.
bg
7- 7
7.18 (b) ∆H& + ∆E& k + ∆E& p = Q& − W&s (The system is the liquid stream.)
c c
h
∆ E& k =0 no change in m and u ∆ E& p =0 no elevation change W& s = 0 no moving parts or generated currents
c
h
∆ H& = Q& , Q& > 0
h
(c) ∆H& + ∆E& k + ∆E& p = Q& − W&s (The system is the water) ∆ H& =0 T a nd P ~ constant ∆ E& k =0 no change in m and u Q& =0 no ∆ T between system and surroundings
c
c
h
c
h
b
∆E& p = −W& s , W& s > 0 for water system
g
h
(d) ∆H& + ∆E& k + ∆E& p = Q& − W&s (The system is the oil) ∆ E& k =0 no velocity change
c
h
∆H& + ∆ E& p = Q& − W& s Q& < 0 (friction loss); W& s < 0 (pump work).
(e) ∆H& + ∆E& k + ∆E& p = Q& − W&s (The system is the reaction mixture) ∆ E& k = ∆ E& p =0 given ∆W& s = 0 no moving parts or generated current
c
c
h
∆ H& = Q& , Q& pos. or neg. depends on reaction
7.19 (a) molar flow:
1.25 m3 273 K 122 kPa min 423 K 101.3 kPa
h
1 mol 22.4 L STP
b g
103 L = 43.4 mol min 1 m3
∆ H& + ∆ E& k + ∆ E& p = Q& − W& s ∆ E& k = ∆ E& p =0 given W& s = 0 no moving parts
c
c
h
h
43.37 mol 1 min Q& = ∆ H& = n&∆ H$ = min 60s
3640 J
kW
mol 10 3 J / s
= 2.63 kW
(b) More information would be needed. The change in kinetic energy would depend on the cross-sectional area of the inlet and outlet pipes, hence the internal diameter of the inlet and outlet pipes would be needed to answer this question.
7- 8
b g
7.20 (a) H$ = 1.04 T ° C − 25 H$ in kJ kg H$ out = 1.04 34.0 − 25 = 9.36 kJ kg H$ = 104 . 30.0 − 25 = 5.20 kJ kg
n& (m ol/ s) N2 30 o C
in
∆ H$ = 9.36 − 5.20 = 4.16 kJ kg ∆ H& + ∆ E& k + ∆ E& p = Q& − W& s ∆ E& k = ∆ E& p =0 assumed W& s = 0 no moving parts
c
c
h h
Q& = ∆ H& = n& ∆ H$ ⇒ n& =
P= 11 0 kP a
34 o C
Q& =1 .25 k W
Q& 1.25 kW kg 1 kJ / s 10 3 g 1 mol = = 10.7 mol s ∆ H& 4.16 kJ kW 1 kg 28.02 g
b g
10.7 mol 22.4 L STP ⇒ V& = s mol
303 K 101.3 kPa 273 K
110 kPa
= 245.5 L / s ⇒ 246 L s
(b) Some heat is lost to the surroundings, some heat is needed to heat the coil, enthalpy is assumed to depend linearly on temperature and to be independent of pressure, errors in measured temperature and in wattmeter reading.
U| V| W
$ $ 7.21 (a) H$ = aT + b a = H 2 − H 1 = 129.8 − 25.8 = 5.2 ⇒ H$ kJ kg = 5.2 T ° C − 1302 . T2 − T1 50 − 30 $ b = H 1 − aT1 = 258 . − 5.2 30 = −1302 .
b gb g
b
g
b g
1302 . H$ = 0 ⇒ Tref = = 25° C 5.2
b g
Table B.1 ⇒ S . G.
b
g
bg
C 6 H 14 l
b
1 m3 = 0.659 ⇒ V$ = = 1.52 × 10 −3 m 3 kg 659 kg
g
U$ kJ kg = H$ − PV$ = 5.2 T − 1302 . kJ / kg −
b
1 atm 1.0132 × 105 N / m2 1.52 × 10 −3 m3 1J 1 kJ 1 atm 1 kg 1 N ⋅ m 103 J
g
⇒ U$ kJ kg = 5.2 T − 1304 .
(b) Energy balance: Q = ∆U = ∆Ek , ∆E p , W =0
20 kg [(5.2 × 20 - 130.4) - (5.2 × 80 -130.4)] kJ = −6240 kJ 1 kg
Average rate of heat removal =
6240 kJ 1 min 5 min
60 s
7- 9
= 20.8 kW
7.22
m& (kg/s) 260°C, 7 bars H$ = 2974 kJ/kg u0 = 0
& (kg/s) m 200°C, 4 bars H$ = 2860 kJ/kg u (m/s)
∆H& + ∆E& k + ∆E& p = Q& − W&s ∆ E& p = Q& = W&s = 0
& 2 mu ∆E& k =− ∆H& ⇒ =− m & H$ out − H$ in 2
d
b
i
g
(2) 2974 − 2860 kJ 10 3 N ⋅ m 1 kg ⋅ m / s2 m2 u 2 = 2 H$ in − H$ out = = 2.28 × 10 5 2 ⇒ u = 477 m / s kg 1 kJ 1N s
d
i
7.23 (a) 5 L/min
5 L/min 100 mm Hg (gauge)
0 mm Hg (gauge)
Qin
Qout
Since there is only one inlet stream and one outlet stream, and m& in = m& out ≡ m& , Eq. (7.4-12) may be written m & & ∆ PV$ + ∆ u 2 + mg & ∆z = Q& − W& s m& ∆U$ + m 2 ∆U$ = 0 (given )
d i
d i
a
f
& $ Pout − Pin = V&∆P m& ∆PV$ = mV
∆ u = 0 (assume for incompressible fluid ) 2
∆z = 0 W& s = 0 (all energy other than flow work included in heat terms ) Q& = Q& in − Q& out
V&∆P = Q& in − Q& out
b
g
5 L 100 − 0 mm Hg 1 atm 8.314 J (b) Flow work: V&∆P = = 66.7 J min min 760 mm Hg 0.08206 liter ⋅ atm 5 ml O2 20.2 J Heat input: Q& in = = 101 J min min 1 ml O 2 Efficiency:
V& ∆ P 66.7 J min = ×100% = 66% Q& in 101 J min
7.24 (a) ∆H& + ∆E& k + ∆E& p = Q& − W&s ; ∆E& k , ∆E& p , W& s = 0 ⇒ ∆H& = Q&
b b
g
H$ 400° C, 1 atm = 3278 kJ kg (Table B.7) H$ 100° C, sat' d ⇒ 1 atm = 2676 kJ kg (Table B.5)
g
7- 10
7.24 (cont’d) 100 kg H 2 O(v) / s
100 kg H 2 O(v) / s
o
o
100 C, saturated
400 C, 1 atm Q& (kW)
100 kg Q& = s
b3278 − 2676.0gkJ
10 3 J
kg
1 kJ
= 6.02 × 10 7 J s
(b) ∆U + ∆E k + ∆E p = Q − W ; ∆E k , ∆E p , W = 0 ⇒ ∆U = Q kJ m3 Table B.5 ⇒ Uˆ (100 °C, 1 atm ) = 2507 , Vˆ(100 °C, 1 atm ) = 1.673 = Vˆ ( 400°C, Pfinal ) kg kg
Interpolate in Table B.7 to find P at which Vˆ =1.673 at 400o C, and then interpolate again to find Uˆ at 400o C and that pressure:
3.11 − 1.673 3 o ˆ Vˆ = 1.673 m /g ⇒ Pfinal = 1.0 + 4.0 = 3.3 bar , U (400 C, 3.3 bar) = 2966 kJ/kg 3.11 − 0.617 ⇒ Q = ∆U = m ∆Uˆ = 100 kg [( 2966 − 2507 ) kJ kg ] 10 3 J kJ = 4.59 × 10 7 J
(
)
The difference is the net energy needed to move the fluid through the system (flow work). (The energy change associated with the pressure change in Part (b) is insignif icant.)
c bg h H$ bsteam, 20 bars, sat'd g = 2797.2 kJ kg (Table B.6)
7.25 H$ H 2 O l , 20° C = 83.9 kJ kg (Table B.5)
m& [kg H 2 O(l) / h]
m& [kg H 2 O(v) / h]
o
20 C
20 bar (sat'd) Q& = 0.65(813 kW) = 528 kW
(a) ∆H& + ∆E& k + ∆E& p = Q& − W&s ; ∆E& k , ∆E& p , W& s = 0 ⇒ ∆H& = Q& ∆H& = m& ∆H$
m& =
528 kW Q& = ∆ H$
b
gd
b
kg 1 kJ / s 3600 s = 701 kg h 2797.2 − 83.9 kJ 1 kW 1 h
g
i
(b) V& = 701 kg h 0.0995 m 3 kg = 69.7 m 3 h sat'd steam @ 20 bar
A
Table B.6
701 kg / h 10 3 g / kg 485.4 K 0.08314 L ⋅ bar 1 m3 & nRT (c) V& = = = 78.5 m 3 / h P 18.02 g / mol 20 bar mol ⋅ K 103 L The calculation in (b) is more accurate because the steam tables account for the effect of pressure on specific enthalpy (nonideal gas behavior). (d) Most energy released goes to raise the temperature of the combustion products, some is transferred to the boiler tubes and walls, and some is lost to the surroundings.
7- 11
c bg
h
7.26 H$ H 2 O l , 24° C, 10 bar = 100.6 kJ kg (Table B.5 for saturated liquid at 24o C; assume H$ independent of P).
b
g
H$ 10 bar, sat'd steam = 27762 . kJ kg (Table B.6) ⇒ ∆ H$ = 2776.2 − 100.6 = 2675.6 kJ kg & [kg H2O(l)/h] m
& [kg HO(v)/h] m 2 3
o
24 C, 10 bar
15,000 m /h @10 bar (sat'd) &(kW) Q
m& =
15000 m 3
kg 4 0.1943 m 3 = 7.72 × 10 kg h
h
A
b Table 8.6g
d
i
Energy balance ∆E& p , W& s = 0 : ∆H& + ∆E& k = Q&
∆E& k = E&k final − E& k initial ∆E& k =
mu & f
E& kinitial ≈0
∆ E& k = E& kfinal
d15,000 m hi
7.72 × 10 4 kg
2
=
2
3
2
2
0.15 2 π 4 m 2
h
1
1
h3
1J 1 kg ⋅ m 2 / s2
2 3600 3 s3
A
A=π D2 4 = 5.96 × 10 5 J / s 7.72 × 10 4 kg 2675.6 kJ 1h 5.96 × 10 5 J 1 kJ & ∆H$ + ∆E& k = Q& = m + h kg 3600 s s 10 3 J = 57973 kJ s = 5.80 × 10 4 kW
7.27 (a)
228 g/min 25o C
228 g/min T(o C) Q& ( kW)
b g
Energy balance: Q& = ∆H& ⇒ Q& W = ∆ E& x , ∆ E& p , W&s =0
b g b g T b° C g 25 H$ bJ g g = 0.263Q& bW g 0
228 g 1 min ( H$ out − H$ in) J min
60 s
=0
g
⇒ H$ out J g = 0.263Q& W
b
g
(b) H$ = b T − 25
b g
26.4 27.8 29.0 32.4 4.47 9.28 13.4 24 .8
Fit to data by least squares (App. A.1)
b=
b g
⇒ H$ J g = 3.34 T ° C − 25 7.27 (cont’d)
7- 12
∑ H$ i
i
bT − 25g ∑ bT − 25g i
i
i
2
= 3.34
b
g
350 kg 10 3 g 1 min 3.34 40 − 20 J (c) Q& = ∆ H& = min kg 60 s g
kW ⋅ s 10 3 J
= 390 kW heat input to liquid
(d) Heat is absorbed by the pipe, lost through the insulation, lost in the electrical leads. 7.28
m& w [ kg H 2 O(v) / min] 3 bar, sat' d
m& w [ kg H 2 O(l) / min] 27 o C
Q& ( kW )
m& e [ kg C 2 H 6 / min] o 16 C, 2.5 bar
m& e [ kg C 2 H 6 / min] o 93 C, 2.5 bar
3 3 (a) C H mass flow: m& = 795 m 10 L 2.50 bar 2 6 e min m 3 289 K = 2.487 × 103 kg min
1 K - mol 30.01 g 1 kg 0.08314 L - bar mol 1000 g
H$ ei = 941 kJ kg , H$ ef = 1073 kJ kg Energy Balance on C 2 H 6 : ∆E& p , W& s = 0, ∆E& k ≅ 0 ⇒ Q& = ∆H&
LMb N
g OPQ
2.487 × 103 kJ 1 min kJ kg Q& = 2.487 × 103 min 1073 − 941 = = 5.47 × 103 kW min 60 s kg
b b
g
(b) H$ s1 3.00 bar, sat' d vapor = 2724.7 kJ kg (Table B.6) H$ s2 liquid, 27° C = 1131 . kJ kg (Table B.5)
g
Assume that heat losses to the surroundings are negligible, so that the heat given up by the
d
condensing steam equals the heat transferred to the ethane 5.47 × 10 3 kW
d
Energy balance on H 2O: Q& = ∆H& = m & H$ s2 − H$ s1
−5.47 × 10 3 kJ Q& &= ⇒m = s H$ s2 − H$ s1
b
gd
i
i
i
kg = 2.09 kg s steam 1131 . − 2724.7 kJ
b
g
⇒ V&s = 2.09 kg / s 0.606 m 3 kg = 1.27 m 3 s
A
Table B.6
Too low. Extra flow would make up for the heat losses to surroundings.
(c) Countercurrent flow Cocurrent (as depicted on the flowchart) would not work, since it would require heat flow from the ethane to the steam over some portion of the exchanger. (Observe the two outlet temperatures)
7- 13
7.29
250 kg H2 O(v )/min 40 bar, 500°C H$ 1 (kJ/kg)
250 kg/min 5 bar, T 2 (°C), H$ 2(kJ/kg)
Turbine
Heat exchanger
250 kg/min 5 bar, 500°C H$3 (kJ/kg) Q(kW)
W s =1500 kW
b b
g
H 2 O v , 40 bar, 500° C : H$ 1 = 3445 kJ kg (Table B.7) H O v , 5 bar, 500° C : H$ = 3484 kJ kg (Table B.7) 2
g
3
(a) Energy balance on turbine: ∆E& p = 0, Q& = 0, ∆E& k ≅ 0
d
i
& ∆H& = −W&s ⇒ m& H$ 2 − H$ 1 = −W& s ⇒ H$ 2 = H$ 1 − W&s m =
3445 kJ 1500 kJ − kg s
min
60 s
250 kg
1 min
= 3085 kJ kg
H$ = 3085 kJ kg and P = 5 bars ⇒ T = 310° C (Table B.7)
(b) Energy balance on heat exchanger: ∆E& p = 0, W&s = 0, ∆E& k ≅ 0
d
i
250 kg Q& = ∆H& = m& H$ 3 − H$ 2 = min
b3484 − 3085gkJ
1 min 1 kW = 1663 kW kg 60 s 1 kJ / s
(c) Overall energy balance: ∆E& p = 0 , ∆E& k ≅ 0
d
i
∆H& = Q& − W& s ⇒ m& s H$ 3 − H$ 1 = Q& − W&s
b3484 − 3445g kJ
250 kg Q& = ∆H& + ∆W& s = min
kg
1 min 1 kW 1500 kJ 1 kW + 60 s 1 kJ / s s 1 kJ / s
= 1663 kW √
b g H Obv , 5 bar, 310° Cg: V$
(d) H 2 O v , 40 bar, 500° C : V$1 = 0.0864 m 3 kg (Table B.7) 2
u1 = u2 =
2
= 0.5318 m 3 kg (Table B.7)
250 kg 1 min 0.0864 m3 min 250 kg min
60 s
kg
1 0.5 π 4 m 2 2
min 0.5318 m3 60 s
kg
1 0.5 π 4 m 2 2
m& 2 250 kg 1 1 min ∆E& k = u 2 − u12 = 2 min 2 60 s = 0.26 kW 0. yv would be less (less water evaporates) because some of the energy that would have vaporized water instead is converted to kinetic energy.
(c) Pf = 39.8 bar (pressure at which the water is still liquid, but has the same enthalpy as the feed)
7-17
7.36 (cont’d) (d) Since enthalpy does not change, then when Pf ≥ 39 .8 bar the temperature cannot increase, because a higher temperature would increase the enthalpy. Also, when Pf ≥ 39 .8 bar , the product is only liquid ⇒ no evaporation occurs. 0.4 Tf (C)
y
0.3 0.2 0.1 0 0
20
40
60
300 250 200 150 100 50 0
80
1
5
10
Pf (bar)
15
20
25
30
36 39.8 60
Pf (bar)
7.37 10 m3 , n moles of steam(v), 275°C, 15 bar ⇒ 10 m3 , n moles of water (v+l), 1.2 bar 10.0 m3 H2 O (v)
10.0 m3
m in (kg) 275 oC, 1.5 , 15bar bar
mv [kg H2 O (v)] ml [kg H 2O (l)] Q
(a) P=1.2 bar, saturated,
1.2 bar, saturated
Table B.6
(b) Total mass of water: min =
10 m3
T2 = 104.8 o C 1 kg = 55 kg 0.1818 m 3
Mass Balance: m v + m l = 55.0 Volume additivity:
Vv + Vl = 10.0 m 3 = m v (1.428 m 3 / kg) + m l (0.001048 m 3 / kg)
⇒ mv = 7.0 kg, ml = 48.0 kg condensed
(c) Table B.7 ⇒ U$ in = 2739.2 kJ / kg; V$in = 0.1818 m3 / kg
R| S| T
3 $ $ Table B.6 ⇒ U l = 439.2 kJ / kg; Vl = 0.001048 m / kg U$ v = 2512.1 kJ / kg; V$v = 1.428 m3 / kg
Energy balance: Q = ∆U = mvU$ v + ml U$ l − minU$ in ∆E p ,∆Ek , W = 0
= [( 7.0)(2512.1 kJ / kg) + (48.0)(439.2 ) - 55 kg (2739.2)] kJ = −1.12 × 10 5 kJ
7.38 (a) Assume both liquid and vapor are present in the valve effluent. 1 kg H 2 O( v ) / s 15 bar, Tsat + 1 5 0 o C
m& l [ kg H 2 O (l ) / s] m& v [ kg H 2 O( v ) / s] 1.0 bar, saturated
7-18
7.38 (cont’d) (b) Table B.6 ⇒ Tsat'n (15 bar) = 198.3o C ⇒ Tin = 3483 . oC Table B.7 ⇒ H$ in = H$ ( 348.3o C, 15 bar) ≈ 3149 kJ / kg Table B.6 ⇒ H$ l (1.0 bar, sat'd) = 417.5 kJ / kg; H$ v (1.0 bar, sat'd) = 2675.4 kJ / kg Energy balance: ∆H& = 0 ⇒ m & H$ + m& H$ − m& H$ = 0 l
&s =0 ∆E& p ,∆E& k ,Q& , W
⇒m & in H$ in = m & l H$ l + m & v H$ v
l
v
m& v + m& l
v
in
in
3149 kJ / kg = m& l (417.5) + (1 − m& l )(2675.4 )
There is no value of m& l between 0 and 1 that would satisfy this equation. (For any value in this range, the right-hand side would be between 417.5 and 2675.4). The two-phase assumption is therefore incorrect; the effluent must be pure vapor. & in = m& out = 1 m (c) Energy balance ⇒ m & out H$ out = m& in H$ in 3149 kJ / kg = H$ (1 bar, Tout ) Tout ≈ 337o C
Table B.7
(This answer is only approximate, since ∆E& k is not zero in this process). 7.39 Basis: 40 lb m min circulation (a) Expansion valve R = Refrigerant 12 40 lb m R(l)/min
40 lb m / min x v lb m R (v ) / lb m
93.3 psig, 86°F H$ = 27.8 Btu/lb m
(1 − x v ) lb m R(l ) / lb m H$ v = 77 .8 Btu / lb m , H$ l = 9.6 Btu / lb m
Energy balance: ∆E& p , W& s , Q& = 0 , neglect ∆E& k ⇒ ∆H& =
bg
40 X v lb m R v min
b
g
bg
77.8 Btu 40 1 − X v lb m R l + lb m min
E
∑ n& H$ − ∑ n& H$ i
i
out
9.6 Btu 40 lb m − min lb m
b
X v = 0.267 26.7% evaporates
i
i
=0
in
27.8 Btu =0 lb m
g
(b) Evaporator coil 40 lbm /min 0.267 R( v ) 0.733 R( l ) 11.8 psig, 5°F H$ v = 77.8 Btu/lbm , H$ l = 9.6 Btu/lbm
40 lb m R(v )/min 11.8 psig, 5°F H$ = 77.8 Btu/lbm
Energy balance: ∆E& p , W&s = 0, neglect ∆E& k ⇒ Q& = ∆H&
40 lb m Q& = min
b gb
g
bg
77.8 Btu 40 0.267 lb m R v − lb m min
= 2000 Btu min
7-19
b gb
g
bg
77.8 Btu 40 0.733 lb m R l − lb m min
9.6 Btu lb m
7.39 (cont’d) (c) We may analyze the overall process in several ways, each of which leads to the same result. Let us first note that the net rate of heat input to the system is Q& = Q& evaporator − Q& condenser = 2000 − 2500 = −500 Btu min
and the compressor work Wc represents the total work done on the system. The system is
b g
closed (no mass flow in or out). Consider a time interval ∆t min . Since the system is at steady state, the changes ∆U , ∆E k and ∆E p over this time interval all equal zero. The total heat input is Q& ∆t , the work input is W& ∆t , and (Eq. 8.3-4) yields c
−500 Btu 1 min 1.341 × 10 −3 hp Q& ∆t − W&c ∆t = 0 ⇒ W&c = Q& = = 11.8 hp min 60 s 9.486 × 10 −4 Btu s
7.40 Basis: Given feed rates n&1 (mol / h)
n&C 3 H8 (mol C 3 H 8 / h) n&C 4 H10 (mol C 4 H 10 / h) 227o C
0.2 C3 H 8 0.8 C 4 H 10 0 o C, 1.1 atm
n& 2 (mol / h) 0.40 C3 H 8 0.60 C 4 H 10 25 o C, 1.1 atm
Q& (kJ / h)
Molar flow rates of feed streams: n&1 =
300 L 1.1 atm 1 mol = 14.7 mol h hr 1 atm 22.4 L STP
n& 2 =
200 L 273 K 1.1 atm 1 mol hr 298 K 1 atm 22.4 L STP = 9.00 mol h
b g
b g
14.7 mol 0.20 mol C 3 H 8 9.00 mol 0.40 mol C 3 H 8 + h mol h mol = 6.54 mol C3 H 8 h Total mole balance: n&C 4 H10 = (14.7 + 9 .00 − 6.54 ) mol C 4 H 20 h = 17.16 mol C 4 H 20 h Propane balance ⇒ n&C 3 H8 =
Energy balance: ∆E& p , W&s = 0, neglect ∆E& k ⇒ Q& = ∆H&
Q& = ∆H& =
∑ N& H$ − ∑ N& H$ i
i
i
out
in
b0.40 × 9.00g mol C H − 3
h
8
i
=
6.54 mol C 3H 8 h
b
20.685 kJ 17.16 mol C4 H 10 + mol h
g
1.772 kJ 0.60 × 9.00 mol C 4 H10 − mol h
( H$ i = 0 for components of 1st feed stream)
7-20
27.442 kJ mol
2.394 kJ = 587 kJ h mol
510 m3 273 K 10 3 L 1 mol min 291 K m 3 22.4 L STP
b g
7.41 Basis: (a)
1 kmol = 21.4 kmol min 10 3 mol
Q& (kJ/min)
n. 0 ( kmol/min) 38°C, h r = 97% y 0 ( mol H 2 O(v)/mol ) (1 – x 0) ( mol dry air/ mol)
21.4 kmol/min 18°C, sat'd y 1 ( mol H 2 O(v)/ mol ) (1 – y 1) (mol dry air)
.
n 2 ( kmol H 2O(l )/ mol) 18°C
Inlet condition: yo =
hr PH∗2O ( 38°C ) P PH∗2 O
(18 °C )
=
0.97 ( 49.692 mm Hg ) = 0.0634 mol H2 O mol 760 mm Hg
15.477 mm Hg = 0.0204 mol H2 O mol P 760 mm Hg Dry air balance: 1 − 0.0634 n& o = 1 − 0.0204 214 . ⇒ n& o = 22.4 kmol min
Outlet condition: y1 =
b
=
g b
b
g
g
b
g
Water balance: 0.0634 22.4 = n& 2 + 0.0204 21.4 ⇒ n&2 = 0.98 kmol min 0.98 kmol 18.02 kg = 18 kg / min H 2 O condenses min kmol
b
g
b
g
(b). Enthaphies: H$ air 38° C = 0.0291 38 − 25 = 0.3783 kJ mol H$ 18° C = 0.0291 18 − 25 = −0.204 kJ mol
b
air
b b b
g
b
g
g g g
U| || V| || W
2570.8 kJ 1 kg 18.02 g H$ H2 O v , 38° C = = 46.33 kJ mol kg 10 3 g mol 25345 . kJ 1 kg 18.02 g H$ H2 O v , 18° C = = 45.67 kJ mol Table B.5 kg 10 3 g mol 75.5 kJ 1 kg 18.02 g H$ H2 O l , 18° C = = 136 . kJ mol kg 10 3 g mol Energy balance: ∆ E& , W& = 0, ∆E& ≅0 p
s
gd ib g +b0.0204gd21.4 × 10 ib45.67g + d0.98 × 10 ib136 . g − b1 − 0.0634gd22.4 × 10 ib0.3783g −b0.0634gd22.4 × 10 ib46.33g = −5.67 × 10 kJ min
Q& = ∆H& = k
∑ n& H$ − ∑ n& H$ i
i
i
out
i
b
⇒ Q& = 1 − 0.0204 21.4 × 10 3 −0.204
in
3
3
3
3
4
4 ⇒ 5.67 × 10 kJ 60 min 0.9486 Btu 1 ton cooling = 270 tons of cooling min h kJ 12000 Btu
7-21
7.42 Basis: 100 mol feed n2 (mol), 63.0°C 0.98 A(v ) 0.02 B(v )
A - Acetone B - Acetic Acid
Qc (cal)
0.5n 2 (mol) 0.98 A(l ) 0.02 B(l )
100 mol, 67.5°C 0.65 A(l ) 0.35 B(l )
56.8°C
n 5 (mol), 98.7°C 0.544 A(v ) 0.456 B(v )
0.5n 2 (mol) 0.98 A(l ) 0.02 B(l )
n5 (mol), 98.7°C 0.155 A(l ) 0.845 B(l )
Qr (cal)
(a) Overall balances: Total moles: 100 = 0.5n 2 + n 5 n 2 = 120 mol A: 0.65 100 = 0.98 0.5n 2 + 0.155n 5 n 5 = 40 mol
b g
b
g
b g b g 0.155b40g = 6.2 mol A 0.845b40g = 33.8 mol B
Product flow rates: Overhead Bottoms
UV W
0.5 120 0.98 = 58.8 mol A 0.5 120 0.02 = 1.2 mol B
Overall energy balance: Q = ∆H = ∆ E , W =0 , ∆E ≅ 0 p
2
∑ n H$ − ∑ n H$ i
out
x
i
i
i
in
interpolate in table
interpolate in table
↓
↓
⇒ Q = 58.8 (0 ) + 1.2 ( 0 ) + 6.2 (1385 ) + 33.8 (1312 ) − 65 ( 354 ) − 35 ( 335 ) = 1.82 × 10 4 cal
b g 2b12 . g = 2 .4 mols B
(b) Flow through condenser: 2 58.8 = 117.6 mols A Energy balance on condenser: Q c = ∆H ∆E , W = 0 , ∆E ≅ 0 p
b
3
k
g
b
g
Qc = 117 .6 0 − 7322 + 2.4 0 − 6807 = −8.77 × 10 5 cal heat removed from condenser
Assume negligible heat transfer between system & surroundings other than Q c & Q r
(
)
Qr = Q −Qc = 1.82× 10 4 − −8.77 ×10 5 = 8.95 × 105 cal heat added to reboiler
7.43
1.96 kg, P1= 10.0 bar, T1
2.96 kg, P 3= 7.0 bar, T3=250o C
1.00 kg, P2= 7.0 bar, T2 Q= 0
7-22
7.43 (cont’d) (a) T2 = T ( P = 7.0 bar, sat'd steam) = 165.0 o C H$ 3 ( H 2 O(v), P = 7.0 bar, T = 250 o C) = 2954 kJ kg (Table B.7) H$ (H O( v), P = 7.0 bar, sat'd) = 2760 kJ kg ( Table B.6) 2
2
Energy balance ∆E , Q, W , ∆E ≅0 p
s
k
∆H = 0 = 2.96 H$ 3 − 196 . H$ 1 − 1.0 H$ 2 ⇒ 1.96 H$ 1 = 2 .96 kg(2954 kJ / kg) -1.0 kg(2760 kJ / kg) ⇒ H$ (10.0 bar, T ) = 3053 kJ / kg ⇒ T ≅ 300 o C 1
1
1
(b) The estimate is too low. If heat is being lost the entering steam temperature would have to be higher for the exiting steam to be at the given temperature. 7.44
T1 = T ( P = 3.0 bar, sat' d. ) = 133.5o C
(a)
Vapor
V$l ( P = 3.0 bar, sat' d. ) = 0.001074 m3 / kg V$ ( P = 3.0 bar, sat'd.) = 0.606 m 3 / kg
P=3 bar
v
Liquid
0.001074 m 3 1000 L 165 kg Vl = = 177.2 L kg m3 Vspace = 200.0 L -177.2 L = 22.8 L mv =
22.8 L
m=165.0 kg V=200.0 L Pmax=20 bar
1 m3 1 kg = 0.0376 kg 1000 L 0.606 m 3
(b) P = Pmax = 20.0 bar;
m total = 165.0 + 0.0376 = 165.04 kg
T1 = T ( P = 20.0 bar, sat'd.) = 212.4o C
V$l ( P = 20.0 bar, sat' d.) = 0.001177 m3 / kg; V$v ( P = 20.0 bar, sat'd.) = 0.0995 m3 / kg V = m V$ + m V$ ⇒ m V$ + (m − m )V$ total
l
l
v v
l
l
total
l
v
1 m 3 = m kg (0.001177 m 3 / kg) + (165.04 - m ) kg (0.0995 m 3 / kg) l l 1000 L ⇒ ml = 164.98 kg; mv = 0.06 kg ⇒ 200.0 L
Vl =
0.001177 m 3 kg
m evaporated =
1000 L 164.98 kg = 194.2 L; m3
V space = 200.0 L - 194.2 L = 5.8 L
( 0.06 - 0.04) kg 1000 g = 20 g kg
(c) Energy balance Q = ∆U = U ( P = 20.0 bar, sat'd) − U ( P = 3.0 bar, sat'd) ∆E , W , ∆E ≅ 0 p
s
k
U$ l ( P = 20.0 bar, sat'd.) = 906.2 kJ / kg; U$ v ( P = 20.0 bar, sat'd. ) = 2598.2 kJ / kg U$ ( P = 3.0 bar, sat'd. ) = 561.1 kJ / kg; U$ ( P = 3.0 bar, sat'd.) = 2543 kJ / kg l
v
Q = 0.06 kg(2598.2 kJ / kg) + 164.98 kg(906.2 kJ / kg) - 0.04 kg(2543 kJ / kg) − 165.0 kg (561.1 kJ / kg) = 5.70 × 10 4 kJ Heat lost to the surroundings, energy needed to heat the walls of the tank
7-23
7.44 (cont’d) (d) (i) The specific volume of liquid increases with the temperature, hence the same mass of liquid water will occupy more space; (ii) some liquid water vaporizes, and the lower density of vapor leads to a pressure increase; (iii) the head space is smaller as a result of the changes mentioned above. (e) – Using an automatic control system that interrupts the heating at a set value of pressure – A safety valve for pressure overload. – Never leaving a tank under pressure unattended during operations that involve temperature and pressure changes. 7.45 Basis: 1 kg wet steam (a) 1 kg H2 O 20 bars 0.97 kg H2 O(v) 0.03 kg H2 O(l) H$ 1 (kJ/kg)
1 kg H 2 O,(v) 1 atm H$ 2 (kJ/kg)
Q=0
b b
1 kg H2 O Tamb , 1 atm
U|b V| W
g g
Q
Enthalpies: H$ v , 20 bars, sat'd = 2797 .2 kJ kg Table B.7 H$ l , 20 bars, sat' d = 908.6 kJ kg Energy balance on condenser: ∆H = 0 ⇒ H$ 2 = H$ 1 = 0.97 2797.2 + 0.03 9086 .
g
b
∆E , ∆E , Q, W =0 p
K
g
b
g
3
⇒ H$ 2 = 2740 kJ / kg
Table B.7
T ≈ 132o C
(b) As the steam (which is transparent) moves away from the trap, it cools. When it reaches its saturation temperature at 1 atm, it begins to condense, so that T = 100° C . The white plume is a mist formed by liquid droplets. 7.46 Basis:
bg
8 oz H 2O l
1 m3
1 quart
1000 kg
32 oz 1057 quarts m3 (For simplicity, we assume the beverage is water) 0.2365 kg H 2O (l) 18°C m (kg H 2O (s)) 32°F (0°C)
bg
= 0.2365 kg H 2O l
(m + 0.2365) (kg H 2O (l)) 4°C
Assume P = 1 atm
Internal energies (from Table B.5):
b
g
b
g
b
g
U$ H 2 O(l ), 18° C = 755 . kJ / kg; U$ H 2 O(l ), 4° C = 16.8 kJ / kg; U$ H 2 O(s), 0° C = -348 kJ / kg
b
g
Energy balance closed system : ⇒ ∆U = ∆E p , ∆E k , Q , W = 0
∑ n U$ − ∑ n U$ i
out
i
i
i
=0
in
⇒ (m + 0.2365) kg (16.8 kJ / kg) = 0.2365 kg(75.5 kJ / kg) + m kg (-348 kJ / kg) ⇒ m = 0.038 kg = 38 g ice
7-24
7.47 (a) When T = 0 o C, H$ = 0, ⇒ Tref = 0 o C (b) Energy Balance-Closed System: ∆U = 0 ∆E , ∆E , Q, W = 0 k
p
25 g Fe, 175°C 25 g Fe 1000 g H 2O T f (°C)
1000 g H 2O(l) 20°C
d i
b
d i
g
b
g
U Fe T f + U H2 O T f − U Fe 175° C − U H 2 O 20° C, 1 atm = 0 or ∆U Fe + ∆U H2 O = 0 ∆U Fe =
d
i
25.0 g 4.13 T f − 175 cal 4.184 J = 432 Tf − 175 J g cal
e
d i
j
1.0 L 10 3 g U$ H2 O T f − 83.9 J = 1000 U$ H 2 O T f − 83.9 J 1 L g 5 ⇒ 432T f + 1000U$ H 2 O T f − 1.60 × 10 = f T f = 0
Table B.5 ⇒ ∆U H 2O =
d i
⇒
Tf ° C
30
f Tf
−2.1 × 10
d i
d i
j
d i
40 4
e
+2 .5 × 10
7-25
35 4
34
1670 −2612
Interpolate
T f = 34.6° C
7.48
I
II
H 2 O( v) 760 mm Hg 100°C
H 2 O( v) (760 + 50.1) mm Hg Tf
⇒
H 2 O( l), 100 °C
⇒ 1.08 bar sat'd ⇒ Tf = 101.8°C (Table 8.5)
H 2 O( l), Tf
T0
Tf
Energy balance - closed system: ∆E p , ∆ E K , W , Q = 0
∆U = 0 =
mvIIU$ vII
+
mlIIU$ lII
+
mbII U$ bII
b
I 1.01 bar, 100° C 1.044
b g b g b g b g
V$l L kg V$v L kg U$ l L kg U$ v L kg
−
mvI U$ vI
−
mlIU$ lI
−
mbI U$ bI
v-vapor l -liquid b-block
g II b1.08 bar, 101.8° Cg 1046 .
1673 419.0 2506.5
1576 426.6 2508.6
Initial vapor volume: VvI = 20.0 L − 5.0 L −
b
50 kg
g
1L 8.92 kg
bg
= 14.4 L H 2 O v
bg
Initial vapor mass: mvI = 14.4 L 1673 L kg = 8.61 × 10 −3 kg H 2 O v
b g = 0.36b101.8g = 36.6 kJ kg
bg
Initial liquid mass: mlI = 5.0 L 1.044 L kg = 4.79 kg H 2 O l Final energy of bar: U$ bII
Assume negligible change in volume & liquid ⇒ VvII = 14.4 L
b
g
bg
g b g
b
Final vapor mass: mvII = 14.4 L 1576 L kg = 9.14 × 10 − 3 kg H 2 O v Initial energy of the bar: U$ bI =
d
b
g
b
g
b
gi
1 9.14 × 10 −3 2508.6 + 4.79 426.6 + 50 . 36.6 − 8.61 × 10 −3 25065 . − 4.79 419.0 5.0 kg = 44.1 kJ kg 44.1 kJ / kg (a) Oven Temperature: To = = 122 .5° C 0.36 kJ / kg ⋅o C H 2 O evaporated = mvII − mvI = 9.14 × 10 −3 kg - 8.61 × 10 −3 kg = 5.30 × 10 −4 kg = 0.53 g
(b) U$ bI = 44 .1 + 8.3 5.0 = 458 . kJ kg To = 45.8 0.36 = 127.2° C (c) Meshuggeneh forgot to turn the oven on ( To < 100° C )
7-26
7.49 (a) Pressure in cylinder = P=
30.0 kg
weight of piston + atmospheri c pressure area of piston
b100 cmg 1 bmg
9.807 N
400.0 cm 2
2
2
kg
2
10 . bar 10 5 N m 2
+
1 atm 1.013 bar atm
= 108 . bar
⇒ Tsat = 1018 . °C
Heat required to bring the water and block to the boiling point
d b
g
b
gi
d b g
b
Q = ∆U = mw U$ wl 1.08 bar, sat'd − U$ wl l , 20° C + mAl U$ Al Tsat − U$ Al 20° C
b426.6 − 83.9gkJ + 3.0 kg
gi
[0.94 (1018 . − 20) ]kJ = 2630 kJ kg kg 2630 kJ < 3310 kJ ⇒ Sufficient heat for vaporization =
7.0 kg
V$ = 1046 . L kg , U$ l = 426.6 kJ kg (b) T f = Tsat = 1018 . ° C . Table B.5 ⇒ $ l Vv = 1576 L kg , U$ v = 2508.6 kJ kg 7.0 kg H 2O(l ) H$ = 426.6 kJ / kg V $ = 1.046 L / kg
mv (kg H2 O( v )) 1576 L/kg, 2508.6 kJ/kg
T ≡ 101.8°C P ≡ 1.08 bars
Q (kJ)
1.046 L/kg, 426.6 kJ/kg m l (kg H 2 O( l)) W (kJ)
(Since the Al block stays at the same temperature in this stage of the process, we can ignore it -i.e., U$ in = U$ out ) Water balance: 7.0 = m l + mv (1) Work done by the piston: W = F ∆ z = w piston + Patm A ∆ z =
LM w + P OPb A∆ z g = P∆V ⇒ W = b108 . bar g 1576m + 1046 . m − b1.046gb 7.0g L NA Q 8.314 J / mol ⋅ K 1 kJ × = b170.2m + 0.113m − 0.7908gkJ 0.08314 liter - bar / mol ⋅ K 10 J atm
v
l
v
3
l
Energy balance: ∆U = Q − W Q ∆U444448 W 6444447 6447 44 8 6444447 444448 ⇒ 25086 . mv + 426.6mL − 426.6 7 = ( 3310 − 2630 ) − (1702 . mv + 0113 . mL − 0.7908) ⇒ 2679mv + 4267 . mL − 3667 = 0 (2) Solving (1) and (2) simultaneously yields m v = 0.302 kg , m l = 6.698 kg
bg
b gb g Vapor volume = b0.302 kg gb1576 L kgg = 476 L vapor . gL ∆V 7.01 + 476 − b7.0gb1046 Piston displacement: ∆z = = Liquid volume = 6.698 kg 1.046 L kg = 7.01 L liquid
10 3 cm 3
1
= 1190 cm 1 L 400 cm 2 ⇒ All 3310 kJ go into the block before a measurable amount is transferred to the A
(c) Tupper
b
g b
g
water. Then ∆U AL = Q ⇒ 3.0 kg 0.94 Tu − 20 kJ kg = 3310 ⇒ Tu = 1194° C if melting is neglected. In fact, the bar would melt at 660o C.
7-27
7.50 1.00 L H 2 O( v ), 25 o C
not all the liquid UV Assume is vaporized. Eq. at W T , P . m = kg H O vaporized.
m v2 [kg H2 O( v)] = m v1 + me
m v1 (kg)
f
o
f
e
2
m L2 [kg H 2O( l) ] = m L1 + m e
4.00 L H 2 O(l ), 25 C m L1 (kg)
Q=2915 kJ
Initial conditions: Table B.5 ⇒ U$ L1 = 104.8 kJ kg , V$L1 = 1.003 L kg P = 0.0317 bar T = 25° C, sat' d ⇒ U$ = 2409.9 kJ kg , V$ = 43,400 L kg
b
m v1 = 1.00 l
v1 −5
g b 43400 l kgg = 2.304 × 10
L1
b
kg , m LI = 4.00 l
g b1.003 l kgg = 3.988 kg
Energy balance:
d
i d i b
g d i d
ib
∆U = Q ⇒ 2.304 × 10 −5 + me U$ v T f + 3.988 − me U$ L T f − 2.304 × 10 −5 2409.9
b
g
− 3.988 (1048 . ) = 2915 kJ
d
g d i i dEi b 3333 − d2.304 × 10 iU$ − 3.988U$ =
⇒ 2.304 × 10 −5 + me U$ v T f + 3.988 − me U$ v T f = 3333 −5
⇒ me
F GG H
U$ v − U$ L
v
L
I J d i b A JK A 5.00 − d2.304 × 10 iV$ − 3.988V$ =
(1)
g d i
V L + Vv = Vtan k ⇒ 2.304 × 10 −5 + m e V$L T f + 3.988 − me V$L T f = 5.00 L kg
liters kg
−5
⇒ me
b1g − b2g ⇒ f d T i =
v
b 2g
L
V$v − V$L 3333 − 2.304 × 10 −5 U$ v T f − 3.988U$ L T f
d
i d i
f
d
U$ v − U$ L
d i
i
5.00 − 2.304 × 10 −5 V$v − 3.988V$L − =0 V$ − V$ v
L
d i Find T
⇒ U$ v , U$ L , V$v , V$L ⇒ f T f
Table 8.5
Procedure: Assume T f Tf U$ v U$ L 201.4 25938 . 856.7 198.3 2592.4 842.9 195.0 2590.8 828.5 196.4 25915 . 834 .6
bg
Eq 1
bg
V$v 123.7 1317 . 140.7 136.9
f
d i
such that f T f = 0
V$L f 1159 . − 512 . × 10 −2 1154 . −193 . × 10 −2 1149 . 134 . × 10 −2 1151 . −4.03 × 10 − 4 ⇒ T f ≅ 196.4° C, Pf = 14.4 bars
me = 2.6 × 10 −3 kg ⇒ 2.6 g evaporated
or Eq 2
7-28
g
7.51.
Basis: 1 mol feed B = benzene T = toluene
nV (mol vapor) y B(mol B(v)/mol) (1 – y B ) (mol T(v)/mol)
1 mol @ 130°C z B (mol B(l)/mol) (1 – z B )(mol T(l)/mol)
(a)
in equilibrium at T(°C), P(mm Hg)
nL (mol liquid) x B(mol B(l)/mol) (1 – x B ) (mol T(l)/mol)
7 variables: ( nV , y B , n L , x B , Q , T , P) –2 equilibrium equations –2 material balances –1 energy balance 2 degrees of freedom. If T and P are fixed, we can calculate nV , y B , n L , x B , and Q.
(b) Mass balance: nV + n L = 1 ⇒ nV = 1 − n 2 Benzene balance: z B = nV y B + n L x B
(1) (2)
bg d i d i C H bv g: dT = 80, H$ = 41.61i , dT = 120, H$ = 45.79 i ⇒ H$ = 01045 . T + 33.25 C H bl g: dT = 0, H$ = 0i , dT = 111, H$ = 18.58i ⇒ H$ = 0.1674 T C H bv g: dT = 89 , H$ = 49.18 i , dT = 111, H$ = 52 .05i ⇒ H$ = 0.1304T + 37 .57 C 6 H 6 l : T = 0, H$ = 0 , T = 80, H$ = 10.85 ⇒ H$ BL = 0.1356 T 6
6
7
8
7
8
BV
TL
TV
Energy balance: ∆E p , Ws = 0, neglect ∆E k Q = ∆H = nV y B H$ BV + nV 1 − y B H$ TV + n L x B H$ BL + n L 1 − x B H$ TL − 1 z B H$ BL TF − 1 1 − zB H$ TL TF
b gb
Raoult's Law:
b
g g b g
b
g
bg
b g
y B P = x B p *B
p *B (90o C) = 10[6.89272−1203.531/(90+ 219.888)] = 1021 mmHg pT* (90 o C) = 10[6.95805−1346.773/(90+219.693)] = 406.7 mmHg
Adding equations (8) and (9) ⇒
yB =
P − p*T p *B
−
x B p *B P
(4) (5) (6)
(7) (8)
(1 - y B ) P = (1 − x B ) pT* Antoine Equation. For T= 90°C and P=652 mmHg:
P = x B p *B + (1 − x B ) pT* ⇒ x B =
(3)
pT* =
=
P − p*T p *B
−
pT*
=
652 − 4067 . = 0.399 mol B(l) / mol 1021 - 406.7
0.399(1021 mmHg) = 0.625 mol B(v ) / mol 652 mmHg
z B − xB 0.5 − 0.399 = = 0.446 mol vapor y B − x B 0.625 − 0.399 n L = 1 − nV = 1 − 0.446 = 0.554 mol liquid
Solving (1) and (2) ⇒ nV =
7-29
(9)
7.51 (cont’d) Substituting (3), (4), (5), and (6) in (7) ⇒ Q = 0.446(0.625)[ 01045 . (90) + 33.25] + 0.446(1 − 0.625)[0.1304( 90) + 37.57 ] + 0.554 (0.399 )[0.1356(90)] + 0.554 (1 − 0.399 )[0.1674( 90)] − 0.5[0.1356(130)] − 0.5[0.1674 (130)] ⇒ Q = 8.14 kJ / mol
(c). If PP max, all the output is liquid. (d) At P=652 mmHg it is necessary to add heat to achieve the equilibrium and at P=714 mmHg, it is necessary to release heat to achieve the equilibrium. The higher the pressure, there is more liquid than vapor, and the liquid has a lower enthalpy than the equilibrium vapor: enthalpy out < enthalpy in. zB 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
T 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90
P 652 714 582 590 600 610 620 630 640 650 660 670 680 690 700 710
pB 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021
pT 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7
xB 0.399 0.500 0.285 0.298 0.315 0.331 0.347 0.364 0.380 0.396 0.412 0.429 0.445 0.461 0.477 0.494
(e). Pmax = 714 mmHg, Pmin = 582 mmHg nV vs. P 1
nV
0.8 0.6 0.4 0.2 0 582
632
682
732
P (mm Hg)
nV = 0.5 @ P ≅ 640 mmHg
7-30
yB 0.625 0.715 0.500 0.516 0.535 0.554 0.572 0.589 0.606 0.622 0.638 0.653 0.668 0.682 0.696 0.710
nV 0.446 -0.001 0.998 0.925 0.840 0.758 0.680 0.605 0.532 0.460 0.389 0.318 0.247 0.176 0.103 0.029
nL 0.554 1.001 0.002 0.075 0.160 0.242 0.320 0.395 0.468 0.540 0.611 0.682 0.753 0.824 0.897 0.971
Q 8.14 -6.09 26.20 23.8 21.0 18.3 15.8 13.3 10.9 8.60 6.31 4.04 1.78 -0.50 -2.80 -5.14
∆P ∆u 2 + + g∆z = 0 ρ 2
7.52 (a). Bernoulli equation:
d
i
∆P 0.977 × 10 −5 − 1.5 × 10 5 Pa 1 N / m 2 = ρ Pa
bg
m3 1.12 × 10 3 kg
= −46.7
m2 s2
g∆z = (9.8066 m / s2 ) 6 m = 58.8 m 2 / s 2
Bernoulli ⇒
∆u = 46.7 − 58.8 m 2 / s2 ⇒ u 22 = u12 + 2 −12.1 m 2 / s2 2 2
b g
b
g
d
i
2
= 5.00 m 2 / s2 − (2)(12.1) m 2 / s2 = 0.800 m 2 / s 2 ⇒ u 2 = 0.894 m / s
d
i
(b). Since the fluid is incompressible, V& m 3 s = π d 12 u1 4 = π d 22 u 2 4
b g
⇒ d1 = d 2
u2 = 6 cm u1
0.894 m s = 2 .54 cm 5.00 m s
d
d i b g
d i b g
i
A 7.53 (a). V& m 3 s = A1 m 2 u 1 m s = A2 m 2 u 2 m s ⇒ u 2 = u1 1 A2 (b). Bernoulli equation ( ∆z = 0)
d
ρ u 22 − u 12 ∆ P ∆u 2 + = 0 ⇒ ∆ P = P2 − P1 = − ρ 2 2
A1 =4 A2
u2 = 4u 1
i
Multiply both sides by − 1 Substitute u 2 = 16u1 2
2
2
Multiply top and bottom of right - hand side by A1
note V& 2 = A12 u12
P1 − P2 =
d
i
(c) P1 − P2 = ρ Hg − ρ H2 O gh =
b g
15ρ H 2 OV& 2 2 A12
⇒ V& 2 =
F GH
9.8066 m 38 cm
1m
s2
102 cm
7-31
I JK
2 A12 gh ρ Hg −1 15 ρ H 2O
2
2 π 7.5 cm4 1 m4 V& 2 = 15 108 cm 4 ⇒ V& = 0.044 m 3 s = 44 L s 2
15ρV& 2 2 A12
b13.6 − 1g = 1955 . × 10
−3
m6 s2
b g
7.54 (a). Point 1- surface of fluid . P1 = 31 . bar , z1 = +7 m , u1 = 0 m s
b g
Point 2 - discharge pipe outlet . P2 = 1 atm , z2 = 0 m , u 2 = ?
b
b =1.013 bar g
g
∆ρ 1.013 − 3.1 bar = ρ
g∆z =
9.8066 m s2
10 5 N 1 m3 = −263.5 m 2 s 2 m 2 ⋅ bar 0.792 × 10 3 kg
−7 m
Bernoulli equation ⇒
= −68.6 m 2 s2
∆u 2 ∆P =− − g ∆z = 263.5 + 68.6 m 2 s2 = 332.1 m 2 s 2 2 ρ
b
g
∆u 2 = u 22 − 0 2
u 22 = 2 (332.1 m 2 s2 ) = 664.2 m 2 s 2 ⇒ u2 = 25.8 m / s π (1.00 2 ) cm 2 V& = 4
2580 cm 1 L 60 s = 122 L / min 3 3 1 s 10 cm 1 min
(b) The friction loss term of Eq. (7.7-2), which was dropped to derive the Bernoulli equation, becomes increasingly significant as the valve is closed. 7.55 Point 1- surface of lake . P1 = 1 atm , z1 = 0 , u1 = 0
bg
Point 2 - pipe outlet . P2 = 1 atm , z2 = z ft u2 =
V& 95 gal = A min
1 ft 3 1 144 in 2 1 min = 35.3 ft s 2 7.4805 gal π 0.5 × 1.049 in 2 1 ft 2 60 s
b
bP = P g F = 0.041b2 zg ft ⋅ lb lb
Pressure drop: ∆ P ρ = 0
Friction loss:
FG L = Z = 2zIJ H sin 30° K
Shaft work:
1
g
2
f
m
= 0.0822 z (ft ⋅ lb f lb m )
W& s -8 hp 0.7376 ft ⋅ lb f / s = m& 1.341 × 10 − 3 hp
1 min 7.4805 gal 1 ft 3
95 gal
1 ft 3
60 s
62.4 lb m
1 min
= −333 ft ⋅ lb f lb m
Kinetic energy: ∆ u
2
b35.3g 2=
2
2
Potential energy: g ∆z =
b
g
Eq. 7.7 - 2 ⇒
32.174 ft s2
− 0 2 ft 2 s
2
bg
z ft
1
lb f
32.174 lb m ⋅ ft / s2
= 19.4 ft ⋅ lb f lb m
b
1 lb f = z ft ⋅ lb f lb m 32.174 lb m ⋅ ft / s 2
g
∆P ∆u 2 −W& s + + g∆z + F = ⇒ 19.4 + z + 0.082 z = 333 ⇒ z = 290 ft & ρ 2 m
7-32
7.56 Point 1 - surface of reservoir . P1 = 1 atm (assume), u1 = 0 , z1 = 60 m Point 2 - discharge pipe outlet . P2 = 1 atm (assume), u 2 = ? , z2 = 0 ∆P ρ = 0
d h
V& A ∆u 2 u22 = = 2 2 2
2
=
V& 2 (m6 / s2 )
= 3.376V& 2
g∆z =
b g b N ⋅ m kgg π 35
(2)
1 2 2
cm 4
10 8 cm 4
1 N
1 m4
1 kg ⋅ m / s2
9.8066 m −65 m 1N = −637 N ⋅ m kg 2 s 1 kg ⋅ m / s2
s W&s 0.80 × 10 6 W 1 N ⋅ m / s = & W V m3 m &
b
g
1 m3 = 800 V& N ⋅ m kg 1000 kg
d i Mechanical energy balance: neglect F bEq. 7.7 - 2g
∆P ∆u 2 −W&s 800 T + E 127 . m 3 60 s + + g ∆z = ⇒ 3.376V& 2 − 637 = − & ⇒ V& = = 76.2 m 3 min & ρ 2 m V s 1 min
Include friction (add F > 0 to left side of equation) ⇒ V& increases.
b g
7.57 (a). Point 1: Surface at fluid in storage tank, P1 = 1 atm , u1 = 0 , z1 = H m Point 2 (just within pipe): Entrance to washing machine. P2 = 1 atm , z2 = 0 u2 =
600 L min π 4.0 cm
b
∆u 2 u22 ∆P =0; = 2 2 ρ g∆z =
9.807 m s
10 3 cm 3 1 min 1m = 7.96 m s 4 1 L 60 s 100 cm
g 7.96 m sg =b 2
2
2
c0 − Hb mgh
2
Bernoulli Equation:
1 J = 31.7 J / kg 1 kg ⋅ m 2 / s2 1
J
1 kg ⋅ m / s2 2
= −9 .807H (J / kg)
∆P ∆u 2 + + g ∆z = 0 ⇒ H = 3.23 m ρ 2
(b). Point 1: Fluid in washing machine. P1 = 1 atm , u1 ≈ 0 , z1 = 0 Point 2: Entrance to storage tank (within pipe). P2 = 1 atm , u 2 = 7.96 m s , z2 = 3.23 m ∆u 2 J J J ∆P =0; = 31.7 ; g∆z = 9.807 3.23 − 0 = 31.7 ; F = 72 2 kg kg kg ρ
b
LM N
g
∆P ∆u 2 Mechanical energy balance: W& s = −m& + + g∆z + F ρ 2
600 L 0.96 kg 1 min ⇒ W&s = − min L 60 s
b31.7 + 31.7 + 72g J
OP Q
1 kW = −1.30 kW kg 10 3 J s (work applied to the system)
Rated Power = 1.30 kW 0.75 = 1.7 kW
7-33
7.58 Basis: 1000 liters of 95% solution . Assume volume additivity. x i 0.95 0.05 1 l Density of 95% solution: = = + = 0.804 ⇒ ρ = 1.24 kg liter ρ ρ i 1.26 1.00 kg b Eq. 6.1-1g
∑
1 0.35 0.65 l = + = 0.9278 ⇒ ρ = 1.08 kg liter ρ 126 . 100 . kg 1000 liters 1.24 kg Mass of 95% solution: = 1240 kg liter Density of 35% solution:
G = glycerol W = water 1240 kg (1000 L) 0.95 G 0.05 W
m 2 (kg) 0.60 G 0.40 W 23 m
m1 (kg) 0.35 G 0.65 W
5 cm I.D.
UV ⇒ m = 1740 kg 35% solution b gb g b gb g b gb gW m = 2980 kg 60% solution
Mass balance: 1240 + m1 = m2 Glycerol balance: 0.95 1240 + 0.35 m1 = 0.60 m2 Volume of 35% solution added =
1740 kg
b
1 2
1L = 1610 L 1.08 kg
g
⇒ Final solution volume = 1000 + 1610 L = 2610 L
Point 1. Surface of fluid in 35% solution storage tank. P1 = 1 atm , u1 = 0 , z1 = 0 Point 2. Exit from discharge pipe. P2 = 1 atm , z2 = 23 m u2 =
1610 L 13 min
1 m3 1 min 1 2 3 10 L 60 s π 2.5 cm 2
b g
10 4 cm 2 = 1.051 m s 1 m2
b g
∆u 2 ∆u 22 1.051 2 m 2 / s 2 1 N ∆P ρ =0, = = = 0.552 N ⋅ m kg 2 2 (2) 1 kg ⋅ m / s 2
g∆z =
9.8066 m s2
Mass flow rate: m& =
23 m
1N = 225.6 N ⋅ m kg , F = 50 J kg = 50 N ⋅ m kg 1 kg ⋅ m / s 2
1740 kg 1 min = 2.23 kg s 13 min 60 s
b
Mechanical energy balance Eq. 7.7 - 2
LM N
g
OP Q
2 .23 kg ∆P ∆u 2 W& s = −m& + + g∆z + F = − ρ 2 s
b0.552 + 225.6 + 50gN ⋅ m
= −0.62 kW ⇒ 0.62 kW delivered to fluid by pump.
7-34
kg
1J
1 kW
1 N ⋅ m 10 3 J s
CHAPTER EIGHT 8.1
a.
U$ ( T ) = 25.96T + 0.02134 T 2 J / mol U$ ( 0o C) = 0 J / mol U$ (100 o C) = 2809 J / mol
b.
We can never know the true internal energy. U$ (100 o C) is just the change from U$ ( 0 o C) to U$ (100 o C) .
c.
Q − W = ∆U + ∆E k + ∆E p
$ o C) = 0) Tref = 0o C (since U(0
∆E k = 0, ∆E p = 0, W = 0
Q = ∆ U = ( 30 . mol )[( 2809 − 0) J / mol] = 8428 J ⇒ 8400 J d.
Cv =
F ∂U$ I GH ∂T JK
∆U$ =
z
V$
=
dU$ = [ 25.96 + 004268 . T ] J / (mol⋅ o C) dT
T2
z
F GG H
100
Cv (T ) dT =
T1
(2596 . + 0.04268T ) dT = 2596 . T + 004268 .
0
∆U = ( 30 . mol) ⋅ ∆U$ ( J / mol)
T2 2
OP PQ
100
0
I JJ J / mol K
= ( 30 . mol) ⋅ [2596 . (100 − 0) + 0.02134(1002 − 0)] (J / mol) = 8428 J ⇒ 8400 J 8.2
a.
b
g
b
gb
Cv = C p − R ⇒ Cv = 353 . + 0.0291T [ J / (mol⋅° C)] − 8.314 [J / (mol ⋅ K)] 1 K 1° C
g
⇒ C v = 27 .0 + 0.0291T [ J / (mol⋅° C)] 100
b.
∆Hˆ =
∫
C p dT = 35.3T ]25 + 0.0291
z
Cv dT =
100
25
c.
∆U$ =
100
25
d. 8.3
a.
z
100
z
100
C p dT −
25
100
T2 = 2784 J mol 2 25
b
gb
g
RdT = ∆H$ − R∆T = 2784 − 8.314 100 − 25 = 2160 J mol
25
H$ is a state property
C v [ kJ / (mol ⋅ o C) ] = 0.0252 + 1547 . × 10 −5 T − 3012 . × 10 −9 T 2 PV ( 2.00 atm)( 3.00 L ) n= = = 0245 . mol RT ( 0.08206[ atm⋅ L / (mol ⋅ K) ]( 298 K) Q1 = n∆U$ 1 = ( 0.245 mol) ⋅ Q2 = n∆U$ 2 = ( 0.245) ⋅ Q3 = n∆U$ 3 = ( 0245 . )⋅
z z
z
1000
00252 . dT ( kJ / mol) = 6.02 kJ
25
1000
[ 0.0252 + 1547 . × 10 −5 T ] dT = 7.91 kJ
25 1000
[ 0.0252 + 1547 . × 10 −5 T − 3012 . × 10 −9 T 2 ] dT = 7.67 kJ
25
6.02- 7.67 × 100% = −215% . 7.67 7.91- 7.67 % error in Q2 = × 100% = 313% . 7.67 % error in Q1 =
8- 1
8.3 (cont’d) b.
C p = Cv + R C p [ kJ / (mol⋅ o C)] = ( 0.0252 + 1547 . × 10 −5 T − 3012 . × 10 −9 T 2 ) + 0.008314 = 0.0335 + 1547 . × 10 −5 T − 3.012 × 10 −9 T 2
z
T2
Q = ∆ H = n C P dT T1
z
1000
= ( 0.245 mol) ⋅
[ 0.0335 + 1547 . × 10 −5 T − 3012 . × 10 −9 T 2 ] dT [kJ / (mol⋅ o C)] = 9.65 × 10 3 J
25
Piston moves upward (gas expands). c. 8.4
a. b.
The difference is the work done on the piston by the gas in the constant pressure process.
b g b313 K g = 0.06255 + 23.4 × 10
. [kJ / (mol ⋅ K)] b313g = 01360 b g b40° Cg = 0.07406 + 32.95 × 10 b40g − 25.20 × 10 b40g + 77.57 × 10 b40g
dC i dC i
−5
p C H l 6 6
−5
p C H v 6 6
−8
− 12
2
3
= 0.08684 [kJ / (mol⋅ o C)] c.
. + 1095 . × 10 b313g − 4.891 × 10 b 313 g dC i b g b313 Kg = 001118
d.
∆H$ C H
−5
6
e.
8.5
2
p C s
6
b vg
= 0.07406T +
∆H$ C b sg = 0.01118 T +
−2
= 0.009615 [ kJ / (mol ⋅ K)]
32.95 × 10 −5 2 25.20 × 10−8 3 77.57 × 10−12 4 T − T + T 2 3 4
1.095 × 10− 5 2 T + 4.891 × 10 2 T −1 2
OP PQ
OP PQ
300
= 31.71 kJ mol 40
573
= 3.459 kJ / mol 313
H 2 O (v, 100 o C, 1 atm) → H 2 O (v, 350 o C, 100 bar) a. H$ = 2926 kJ kg − 2676 kJ kg = 250 kJ kg b.
H$ =
z
350
003346 . + 06886 . × 10 −5 T + 0.7604 × 10− 8 T 2 − 3593 . × 10 −12 T 3 dT
100
= 8845 . kJ mol ⇒ 491.4 kJ kg Difference results from assumption in (b) that H$ is independent of P. The numerical difference is ∆H$ for H 2 O v, 350 ° C, 1 atm → H 2 O v, 350° C, 100 bar
b
8.6
b.
g
b
g
z
80
dC i
= 02163 . kJ / (mol⋅ C) ⇒ ∆H$ = [ 0.2163 ] dT = 11.90 kJ / mol o
p n− C H (l) 6 14
25
The specific enthalpy of liquid n-hexane at 80o C relative to liquid n-hexane at 25o C is 11.90 kJ/mol c.
dC i
p n− C H (v) [ kJ / 6 14
∆H$ =
z
−5
−8
(mol⋅ C)] = 013744 . + 40.85 × 10 T − 2392 . × 10 T + 57.66 × 10 o
2
−12
T
3
0
[ 0.13744 + 40.85 × 10 −5 T − 23.92 × 10 −8 T 2 + 57.66 × 10− 12 T 3 ] dT = −110.7 kJ / mol
500
The specific enthalpy of hexane vapor at 500o C relative to hexane vapor at 0o C is 110.7 kJ/mol. The specific enthalpy of hexane vapor at 0o C relative to hexane vapor at 500o C is –110.7 kJ/mol.
8- 2
8.7
b g 181. T ′b° Fg − 32 = 05556 . T ′ b° Fg − 17.78 C b cal mol⋅° Cg = 6.890 + 0.001436 0.5556 T ′ b° Fg − 17 .78 = 6.864 + 0.0007978 T ′b° Fg cal 453.6 mol 1 Btu 1° C C ′ b Btu lb - mole⋅° F g = C = b100 . gC mol ⋅° C 1 lb mole 252 cal 1.8 °F E T °C = p
p
p
b
g
p
drop primes
b g
C p Btu lb - mole⋅° F = 6.864 + 0.0007978T ° F
8.8
. − 01031 . b01588 g T = 01031 bT g = 01031 . + . + 0000557 . T [kJ / (mol⋅ 100 550 . L 789 g 1 mol F 0.000557 O Q = ∆H = 01031 . T+ T P G s 1 L 46.07 g H 2 Q3 144444244444
dC i
p C H CH OH(l) 3 2
o
C)]
78 .5
2
20
kJ mol
= 941.9 × 7.636 kJ / s = 7193 kW 8.9
a.
b
g
Q& = ∆ H& = 5,000 mol s ⋅
6444444444444474444444444444 8
z
k J mol
200
0.03360 + 1367 . × 10 −5 T − 1607 . × 10−8 T 2 + 6.473 × 10−12 T 3 dT
100
= 17,650 kW b.
b
gb
gb
Q = ∆ U = ∆H − ∆ PV = ∆H − nR∆T = 17,650 kJ − 5.0 kmol ⋅ 8.314 [kJ / (kmol ⋅ K)] ⋅ 100 K = 13,490 kJ
c.
8.10 a. b.
The difference is the flow work done on the gas in the continuous system. Qadditional = heat needed to raise temperature of vessel wall + heat that escapes from wall to surroundings.
C p is a constant, i.e. C p is independent of T. Q m∆T Q (16.73- 6.14) kJ 1 L 86.17 g 10 3 J Cp = = = 0.223 kJ / (mol ⋅ K) m∆T (2.00 L)(3.10 K) 659 g 1 mol 1 kJ Q = mC p ∆ T ⇒ C p =
Table B.2 ⇒ C p = 0.216 kJ / (mol⋅ o C) = 0.216 kJ / (mol ⋅ K)
FG ∂H$ IJ = FG ∂U$ IJ H ∂T K H ∂T K F ∂U$ I = dU$ = F ∂U$ I But since U$ depends only on T, G H ∂T JK dT GH ∂T JK PV$ = RT
8.11
a∂ ∂T f
H$ = U$ + PV$ =====> H$ = U$ + RT =====> P
p
p
8- 3
+ R ⇒ Cp = p
V$
FG ∂U$ IJ H ∂T K
+R p
≡ Cv ⇒ C p = Cv + R
g
8.12 a.
dC i n=
= 754 . kJ / (kmol ⋅ C) =75.4 kJ/(kmol.o C) V = 1230 L , o
p H O(l) 2
Vρ 1230 L 1 kg 1 kmol = = 68.3 kmol M 1 L 18 kg
zd
T2
n⋅ Q Q& = = t b.
i
Cp
H2 O(l)
dT
T`
=
t
683 . kmol 75.4 kJ ( 40 − 29) o C 1 h = 1967 . kW 8h 3600 s kmol⋅ o C
Q& total = Q& to the surroundings + Q& to water , Q& to the surroundings = 1967 . kW
z
40
Q& to water
n ⋅ C P( H2 O) dT
Q = to water = t
=
29
t
68.3 kmol 754 . kJ / (kmol⋅ o C) 11 o C = 5245 . kW 3h 3600 s / h
Q& total = 7.212 kW ⇒ E total = 7.212 kW × 3 h = 21.64 kW ⋅ h c.
Cost heating up from 29 o C to 40 o C = 21.64 kW ⋅ h × $0.10 / (kW ⋅ h) = $2.16
Costkeeping temperature constant for 13 h = 1.967 kW × 13 h × $0.10/(kW ⋅ h)=$2.56 Costtotal = $2.16 + $2.56 = $4.72 d.
If the lid is removed, more heat will be transferred into the surroundings and lost, resulting in higher cost.
8.13 a.
∆H$ N
b.
∆H$ H
o o 2 (25 C) → N 2 (700 C)
2 (800
c.
∆H$ CO
d.
∆H$ O
8.14 a.
o
F)→ H 2 (77 o F)
= H$ N
= H$ H
o o 2 (300 C)→ CO2 (1250 C)
2 (970
o
= H$ O
F) →O 2 (0o F)
& = 300 kg / min n& = m Q& = n& ⋅ ∆H$ = n& ⋅
z
T2
2 (700
2 (77
o
o
F)
= H$ CO 2 (0
o
F)
C)
− H$ N
− H$ H
2 (1250
− H$ O
o
b g = b0 − 5021g = − 5021 Btu / lb - mol − H$ = b 63.06 − 1158 . g = 5148 . kJ mol = b −539 − 6774 g = −7313 Btu / lb- mol
2 (25
2 (800
C)
o
o
C)
= 2059 . − 0 = 2059 . kJ mol
F)
o
C O2 (300 C)
o 2 (970 F)
300 kg 1 min 1000 g 1 mol = 1785 . mol / s min 60 s 1 kg 28.01 g
C p dT
T1
z
50
= (178.5 mol / s) ⋅
b
[0.02895 + 0.411 × 10 −5 T + 0.3548 × 10 −8 T 2 − 2.22 × 10− 12 T 3 ] dT [kJ / mol]
450
g
= (178.5 mol / s) − 12.076 [kJ / mol] = −2,156 kW b. 8.15 a.
Q& = n& ⋅ ∆H$ = n& ⋅ H$ (50o C) − H$ (450o C)
= (178.5 mol / s)(0.73-12.815[kJ / mol]) = − 2 ,157 kW
n& = 250 mol / h
250 mol ( 2676 − 3697 ) kJ 1 kg 1 h 18.02 g Q& = n& ∆H$ = = −1.278 kW h 1 kg 1000 g 3600 s 1 mol
i)
Q& = n& ∆H$ = n& ⋅ ii)
=
z
T2
T1
Cp dT
250 mol 1 h h 3600 s
z
100
600
[0.03346 + 0.6880 × 10−5T + 0.7604 × 10−8 T 2 − 3.593× 10−12 T 3 ] = −1.274 kW
8- 4
8.15 (cont’d)
b
g
250 mol Q& = ⋅ 2.54 − 20.91 [kJ / mol] = − 1276 . kW 3600 s Method (i) is most accurate since it is not based on ideal gas assumption. The work done by the water vapor. iii)
b. c.
8.16 Assume ideal gas behavior, so that pressure changes do not affect ∆H$ .
200 ft 3 492 o R 12 . atm 1 lb - mol = 0.6125 lb- mole / h o h 537 R 1 atm 359 ft 3 (STP) lb - mole Q& = n& ∆H$ = ( 0.6125 ) ⋅ ( 2993 − 0) [Btu / lb - mole] = 1833 Btu / h h n& =
b
8.17 a.
50 kg 1.14 kJ kg⋅° C
g
b50 −10g° C = 2280 kJ
b.
(C )
p Na C O 2 3
≈ 2 (C p ) Na + ( C p)C + 3 ( C p )O = 2 ( 0.026 ) + 0.0075 + 3 ( 0.017 ) = 0.1105 kJ mol ⋅°C
b
g
50,000 g 0.1105 kJ 1 mol 50 − 10 ° C = 2085 kJ mol ⋅° C 105.99 g 2085 − 2280 % error = × 100% = −8.6% error 2280
8.18
dC i dC i
p C H O(l) 6 14
b
g b
g b
g
= 6 0.012 + 14 0.018 + 1 0025 . = 0.349 kJ / (mol⋅ C) (Kopp’s Rule) o
−5
p C H COCH (l) 3 3
= 01230 . + 18.6 × 10 T kJ (mol⋅° C)
Assume ∆Hmix ≅ 0 ↓ CH 3 COCH 3
C pm =
↓ C 6 H 14 O
0.30 ( 0.1230+18.6 × 10 −5T ) kJ 1 mol 0.70 ( 0.349 ) kJ 1 mol + mol ⋅°C 58.08 g mol ⋅°C 102.17 g
= [0.003026 + 9.607 ×10−7 T] kJ (g ⋅ °C) 20 ∆Hˆ = ∫ [0.003026 + 9.607 ×10−7 T] dT = −0.07643 kJ g 45
8.19 Assume ideal gas behavior, ∆Hmix ≅ 0
g b g ∆H$ = dC i dT = 10.08 kJ / mol, ∆H$ = dC i dT = 14.49 kJ / mol 2 L1 O F 1000 g IJ FG 1 mol IJ = 433 kJ kg H$ = M b14.49 kJ / mol g + b10.08 kJ / molgP G 3 N3 Q H 1 kg K H 26.68 g K Mw = O2
b
1 2 g 16.04 + 32.00 = 26.68 3 3 mol
z
350
25
p O 2
CH4
8- 5
z
350
25
p CH 4
8.20
1000 m 3 1 min 273 K 1 kmol = 0.6704 kmol s = 670.4 mol / s min 60 s 303 K 22.4 m3 STP Energy balance on air: Table B.8 for ∆ H$ 670.4 mol 0.73 kJ 1 kW Q = ∆ H = n∆H Q= = 489.4 kW s mol 1 kJ s n=
b g
Solar energy required = Area required =
8.21
489.4 kW heating 1 kW solar energy = 1631 kW 0.3 kW heating
1627 kW 1000 W 1 m2 = 1813 m 2 1 kW 900 W
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O n& fuel = n& air =
1.35 × 105 SCFH 1 lb - mol lb - mol = 376 h h 359 ft 3
376 lb − mol 5 lb - mol O 2 1 lb - mol air 115 . lb − mol = 103 . × 104 h 1b - mol C3 H 8 0.211b - mol O2 h
z
T2
Q = ∆H = n& ⋅ C p dT T1
FG H
= 1.03 × 104
IJ K
lb − mol ⋅ h
z
302
[ 002894 . + 0.4147 × 10 −5 T + 0.3191 × 10 −8 T 2 − 1965 . × 10 −12 T 3 ] dT
0
103 . × 104 lb - mol 8.954 kJ 453.593 mol 9.486 × 10 -1 Btu = = 3.97 × 10 7 Btu / h h mol lb - mol kJ 8.22 a.
Basis : 100 mol feed (95 mol CH4 and 5 mol C2 H6 ) 7 CH 4 + 2O 2 → CO 2 + 2H 2 O C 2 H 6 + O 2 → 2CO 2 + 3H 2 O 2
nO 2 = 125 . ⋅
LM 95 mol CH MN
4
OP PQ
2 mol O2 5 mol C2 H 6 35 . mol O 2 + = 259.4 mol O 2 1 mol CH 4 1 mol C 2 H 6
Product Gas: CO 2 : 95(1) + 5(2) = 105 mol CO 2
H 2 O: 95(2) + 5(3) = 205 mol H 2 O O 2 : 259.4 - 95(2) - 5(3.5) = 51.9 mol O 2 N 2 : 3.76(259.4) = 975 mol N 2 Energy balance (enthalpies from Table B.8) $ $ ∆H =H − H$ = 18.845 − 42.94 = −24.09 kJ / mol o o CO2
(CO2 , 450 C)
$ $ ∆H H 2 O = H (H $ $ ∆H O2 = H (O $ $ ∆H N 2 = H (N
2 O,
o
450 C)
(CO2 , 900 C)
− H$ (H
2,
450 o C)
− H$ (O
2,
450o C)
$ −H (N
2 O,
o
900 C)
= 1512 . − 3332 . = −18.20 kJ / mol
2,
900 o C)
= 13375 . − 2889 . = −15.51 kJ / mol
2,
900 o C)
= 12.695 − 2719 . = −14.49 kJ / mol
Q = ∆H = 105(-24.09) + 205(-18.20) + 51.9(-15.51) + 975(-14.49) Q = 21,200 kJ / 100 mol feed
8- 6
8.22 (cont’d) $ (40 o C) = 167.5 kJ / kg; H$ b. From Table B.5: H liq vap (50 bars) = 2794.2 kJ / kg;
$ = n(2794.2 -167.5) = 21200 ⇒ n = 8.07 kg /100 mol feed Q = n ⋅ ∆H c.
From part (b), 8.07 kg steam is produced per 100 mol feed
n& feed =
d.
1250 kg steam 01 . kmol feed 1 h = 4.30 × 10−3 kmol / s h 8.07 kg steam 3600 s
4.30 mol feed 1336.9 mol product gas 8.314 Pa ⋅ m 3 723 K V&product gas = = 341 . m3 / s 5 s 100 mol feed mol ⋅ K 1.01325 × 10 Pa Steam produced from the waste heat boiler is used for heating, power generation, or process application. Without the waste heat boiler, the steam required will have to be produced with additional cost to the plant. Assume ∆Hmix ≅ 0 ⇒ ∆H = ∆H C10 H12 O2 + ∆H C6 H6
8.23
d i
Kopp’s rule: C p
∆H C10 H12 O2 = ∆H C6 H6 =
C10 H12 O2
e
j
e
= 10(12) + 12(18) + 2( 25) = 386 J mol ⋅ C = 2.35 J g⋅ C o
20.0 L 1021 g 1 kJ 2.35 J ( 71 − 25) o C = 2207 kJ L 10 3 J g⋅ o C
15.0 L 879 g 1 mol ⋅ L 78.11 g
z
LM N
348
298
o
j
OP Q
[ 0.06255 + 23.4 × 10 −5 T] dT = 1166 kJ
∆H = 2207 + 1166 = 3373 kJ 8.24 a.
100 mol C3 H8 @ 40 o C, 250 kPa
100 mol C3 H8 @ 240 o C, 250 kPa
VP 1(m3 )
VP 2(m3 ) mw kg H2 O(v) @ 300 o C, 5.0 bar
mw kg H2 O(l, sat‘d) @ 5.0 bar Vw2 (m3 )
b.
Vw1 (m3 )
References: H2 O (l, 0.01 o C), C3 H8 (gas, 40 o C)
z
240
C3 H8 : H$ in = 0 kJ / mol; H$ out = C pC H dT = 19.36 kJ mol (Cp from Tabl e B.2) 40
3 8
H 2 O : Hˆ in = 3065 kJ/kg (Table B.7); Hˆ out = 640.1 kJ/kg (Table B.6) c.
∆Hˆ C3H 8 = 19.36 kJ/mol, ∆Hˆ w = (640.1 − 3065) kJ/kg = −2425 kJ/kg Q = ∆ H = 100 ∆H$ C 3 H8 + m w ∆H$ w = 0 ⇒ mw = 0.798 kg
b
g
From Table B.7: V$steam 50 . bar, 300 ° C = 0522 . m 3 kg
0.008314 m 3 ⋅ kPa (mol ⋅ K) 313 K V$C3 H8 40° C, 250 kPa = = 0.0104 m 3 mol C 3 H 8 250 kPa
b
g
0.798 kg steam 0.522 m3 steam 100 mol C3 H 8
1 kg steam
1 mol C3 H8 0.0104 m 3 C 3 H 8
= 0.400 m 3 steam m 3 C 3H 8
d.
Q = mw ∆Hˆ w = 0.798 kg × (-2425 kJ/kg)=-1935 kJ
e.
A lower outlet temperature for propane and a higher outlet temperature for steam.
8- 7
8.25
a. 5500 L(STP)/min CH 3 OH (v) 65o C
n 2 mol/min CH 3 OH (v) 260 o C
n 2 (mol/min) mw kg /min H 2 O(l, sat‘d) @ 90o C
mw kg /min H 2 O(v, sat‘d) @ 300o C
Vw 2 ( m3 /min )
n2 =
Vw 1 ( m3 /min )
5500 L(STP) 1 mol = 245.5 mol CHOH(v)/min 3 min 22.4 L(STP)
An energy balance on the unit is then written, using Tables B.5 and B.6 for the specific enthalpies of the outlet and inlet water, respectively, and Table B.2 for the heat capacity of methanol vapor. The only unknown is the flow rate of water, which is calculated to be 1.13 kg H2 O/min. b.
kg kJ 1 min 1 kW Q& = 1.13 2373.9 = 44.7 kW min kg 60 sec 1 kJ/s
8.26 a. 100 mol/s (30o C) 0.100 mol H2 O(v)/mol 0.100 mol CO/mol 0.800 mol CO2 /mol
n 2 mol/s (30o C) 0.020 mol H2 O(v)/mol y 2 mol (molCO/s CO/mol) (molCO CO (0.980-y 2 ) mol 2/mol) 2 /s
m3 kg humid air/s (50o C)
m4 kg humid air/s (30 (48o oC) C)
H 2O(v) only
(0.002 /1.002 ) kg H2 O(v)/kg humid air (1.000 /1.002 ) kg dry air/kg humid air
y 4 kg H2 O(v)/kg humid air (1-y 4 ) kg dry air/kg humid air
Basis : 100 mol gas mixture/s 5 unknowns: n 2 , m3 , m4 , y2 , y4 – 4 independent material balances, H2 O(v), CO, CO2 , dry air – 1 energy balance equation 0 degrees of freedom (all unknowns may be determined) b.
(1) CO balance: (2) CO 2 balance:
U|V ⇒ n& (100)(0.800) = n (1 − y ) | W
(100)(0.100) = n& 2 y 2 2
2
2
= 9184 . mol / s, x 2 = 0.1089 mol CO / mol
1.000 = m 4 (1 − y 4 ) 1002 . (100)(0.100)(18) 0.002 ( 0.020)(18) (4) H 2 O balance: + m& 3 = 91.84 + m& 4 y 4 1000 1.002 1000 References: CO, CO2 , H2 O(v), air at 25o C ( H$ values from Table B.8 ) substance n& in ( mol / s) n& out ( mol / s) H$ (kJ / mol) (3) Dry air balance: m 3
0.169 0.146
91.84(0.020) 10
0.193
80
H$ out (kJ / mol) 0.169 0.146 0.193
0.847 0.727
m4 y 4 (1000/18 ) m4 (1-y 4 ) (1000 /29 )
0.779 0.672
in
H2 O(v) CO
10 10
CO2 H2 O(v) dry air
80 0.002
1000
m3 ( /1.002 )( /18 ) m3 (1.000 /1.002 ) (1000/29 )
8- 8
8.26 (cont’d) (5) Energy balance:
FG 0.002 IJ FG 1000IJ (0.847) + m FG 1.000 IJ FG 1000 IJ (0.727) H 1.002 K H 18 K H 1.002 K H 29 K F 1000IJ + m (1 − y )(0.672)FG 1000IJ = 91.84 ( 0.020 )( 0.169) + m y ( 0.779 ) G H 18 K H 29 K
10(0.169) + m 3
3
4
4
4
4
Solve Eqs. (3)–(5) simultaneously ⇒ m3 = 2.55 kg/s, m4 = 2.70 kg/s, y4 = 0.0564 kg H2 O/kg
2.55 kg humid air / s kg humid air = 0.0255 100 mol gas / s mol gas
Mole fraction of water :
00564 . kg H2 O
c.
8.27 a.
=.0963
(1-.0564) kg dry air kmol DA 18 kg H 2 O ⇒
Relative humidity:
29 kg DA 1 kmol H2 O
0.0963 kmol H2 O (1 + 00963 . ) kmol humid air
pH 2 O p*H2 O
e48 Cj o
=
= 0.0878
kmol H2 O kmol DA
kmol H2 O kmol humid air
( 0.0878)( 760 mm Hg ) × 100% = 79.7% 83.71 mm Hg
The membrane must be permeable to water, impermeable to CO, CO2 , O2 , and N2 , and both durable and leakproof at temperatures up to 50o C.
y H2 O =
b
p* 57° C P
b g
28.5 m 3 STP h
g = 12982 . mm Hg = 0171 . mol H O mol 2
760 mm Hg ↓ 1 mol = 1270 mol h ⇒ 217.2 mol H 2 O h 0.0224 m 3 STP
b g
. kg H b391
R| 89.5 mol CO h mol dry gas |110.5 mol CO h 1270 − 217 .2 = 1053 =======> S h || 5.3 mol O h T847.6 mol N h given
2
percentages
2
2
1270 mol/h, 620°C 425°C m (kg H2 O( l )/h), 20°C
References for enthalpy calculations:
e
j
CO, CO 2 , O 2 , N 2 at 25°C (Table B.8); H 2 O l, 0.01 o C (steam tables) substance CO CO 2
O2 N2
bg H Ob lg H2O v
nin 89.5 110.6 5.3 847.6 3.91 m
H$ in 18.22 27.60 19.10 18.03 3749 83.9
nout 89.5 110.6 5.3 847.6
H$ out 12.03 17.60 12.54 11.92
3.91 + m 3330 ---
2
8- 9
U| n in mol h V| H$ in kJ mol W
UV n in kg h W H$ in kJ kg
2O
h
g
8.27 (cont’d) ∆H =
∑ n H$ − ∑ n H$ i
i
i
out
b.
i
= 0 ⇒ − 8504 + 3246m = 0 ⇒ m = 2.62 kg h
in
When cold water contacts hot gas, heat is transferred from the hot gas to the cold water lowering the temperature of the gas (the object of the process) and raising the temperature of the water.
b gb
g
8.28 2°C, 15% rel. humidity ⇒ pH 2 O = 015 . 5294 . mm Hg = 0.7941 mm Hg
dy i H2 O
inhaled
n& inhaled =
b0.7941g b760g = 1.045 ×10
−3
mol H 2 O mol inhaled air
5500 ml 273 K 1 liter 1 mol = 0.2438 mol air inhaled min 3 min 275 K 10 ml 22.4 liters STP
Saturation at 37 °C ⇒ y H2 O =
b
p 37° C *
g
760 mm Hg
b g
=
47.067 = 0.0619 mol H 2 O mol exhaled dry gas 760
0.2438 mol/min 2o C
n2 kmol/min 37oC
1.045 x 10-3 H2O 0.999 dry gas
0.0619 H2 O 0.9381 dry gas n 1 mol H2O(l)/min 22o C
Mass of dry gas inhaled (and exhaled) =
b0.2438 gb0.999gmol dry gas
29.0 g
min
mol
b0.999gb0.2438g = 0.9381 n&
Dry gas balance:
2
= 7.063 g min
⇒ n&2 = 0.2596 mols exhaled min
. . × 10 j + n& = b0.2596 gb0.0619 g ⇒ n& = 0.0158 mol H O min b02438 ge1045 References for enthalpy calculations: H Ob lg at triple point, dry gas at 2 °C −3
H 2 O balance:
1
1
2
2
substance
m& in
Dry gas H2O v
7.063 0.00459 0.285
bg H Ob lg 2
Q = ∆H =
H$ in 0 2505 92.2
m& out
7.063 0.290 —
H$ out 36.75 2569 —
m& in g min H$ in J g
m& H2 O = 18.02n&H 2O Hˆ H2 O from Table 8.4 Hˆ dry gas = 1.05 ( T − 2 )
∑ m& H$ − ∑ m& H$ i
out
i
i
in
i
=
966.8 J 60 min 24 hr = 1.39 × 10 6 J day min 1 hr 1 day
8- 10
8.29 a.
bg
75 liters C 2 H 5OH l
789 g
1 mol
liter
46.07 g
bg
= 1284 mol C 2 H 3OH l
e
j 1 mol = 3054 mol H O b lg 18.01 g
( C p ) C H3 OH = 01031 . + 0557 . × 10 −3 T kJ / (mol⋅ o C) (fitting the two values in Table B.2)
bg
55 L H 2 O l
1000 g liter
2
b
( C p ) H2 O = 0.0754 kJ mol⋅° C
g
1284 mol C2 H5 OH(l) (70.0oC) 1284 mol C2H5 OH (l) (To C) 3054 mol H2 O(l) (20.0oC)
3054 mol H2O(l) (To C) T
T
0 = 1284 ∫ ( 0.1031 + 0.557 × 10 −3 T ) dT + 3054 ∫ ( 0.0754 ) dT
70 Q = ∆ U ≅ ∆ H ( liquids ) ⇒ ⇓ Integrate, solve quadratic equation Q = 0 (adiabatic ) T=44.3 o C
b.
1. 2. 3. 4. 5. 6. 7.
25
Heat of mixing could affect the final temperature. Heat loss to the outside (not adiabatic) Heat absorbed by the flask wall & thermometer Evaporation of the liquids will affect the final temperature. Heat capacity of ethanol may not be linear; heat capacity of water may not be constant Mistakes in measured volumes & initial temperatures of feed liquids Thermometer is wrong
8.30 a. 1515 L/s air 500oC, 835 tor, Td p=30 o C
1515 L/s air , 1 atm 110 g/s H2O(v)
110 g/s H2O, T=25oC
Let n& 1 (mol / s) be the molar flow rate of dry air in the air stream, and n& 2 (mol / s) be the molar flow rate of H2 O in the air stream. 1515 L 835 mm Hg mol ⋅ K n& 1 + n& 2 = = 262 . mol / s s 773 K 62.36 L ⋅ mm Hg
n& 2 p * (30 o C) 31824 . mmHg = y= = = 00381 . mol H 2 O / mol air &n1 + n& 2 Ptotal 835 mmHg ⇒ n& 1 = 252 . mol dry air / s; n& 2 = 10 . mol H 2 O / s
8- 11
8.30 (cont’d) References: H2 O (l, 25o C), Air (v, 25o C) substances n& in (mol / s) H$ in (kJ / mol) dry air
25.2
H2 O(v)
14.37
z z
1.0
dC i dC i
100
p
25 500
100
H2 O(l)
p
H2O (l )
H2 O( v )
6.1
25.2
dT + H$ vap
7.1
z
zd zd
--
z
Cp
25
100
100
0
zd T
25 T
dT
∆H = 0 = n& out ⋅ H$ out − n& in ⋅ H$ in
z
H$ out (kJ / mol)
n& out (mol / s)
i i
Cp
Cp
i
air
H2O (l )
H2 O( v )
dT
dT + H$ vap dT
--
b25.2gFGH dC i dT IJK + b71. gFGH dC i dT + H$ + dC i dTIJK F dC i dT + H$ + dC i dTIJ = 0 −b 25.2gb14.37g − b100 . gG H K T
25
p air
z
100
25
p H O( l ) 2
vap
z
100
25
p
vap
H2 O( l )
T
100
p H O( v) 2
500
100
p
H2 O( v )
Integrate, solve : T = 139 o C b.
139 139 Q& = − ( 25.2 ) ∫ ( C p )air dT − (1.00 ) ∫ ( C p )H O( v) dT = −290 kW 500
500
2
This heat goes to vaporize the entering liquid water and bring it to the final temperature of 139o C. c.
When cold water contacts hot air, heat is transferred from the air to the cold water mist, lowering the temperature of the gas and raising the temperature of the cooling water.
8- 12
8.31
520 kg NH 3 103 g 1 mol 1h = 8.48 mol NH 3 s h 1 kg 17.03 g 3600 s
Basis:
8.48 mol NH 3/s 25°C n 1 (mol air/s) T °C Q = –7 kW
n2 (mol/s) 0.100 NH 3 0.900 air 600°C
NH 3 balance: 848 . = 0100 . n2 ⇒ n2 = 848 . mol s
b
gb g
Air balance: n1 = 0.900 84.8 = 76.3 mol air s
bg
References for enthalphy calculations: NH 3 g , air at 25°C NH 3
H$ in = 0.0
. kJ mol dC i dT ⇒ H$ = 2562 Air: C b J mol ⋅° Cg = 28.94 + 0.4147 × 10 T b ° Cg L F T − 25 I OP J × 1 kJ H$ = C dT = M28.94b T − 25g + 0.004147 G H 2 2 JK PQ mol 10 J MN = e2.0735 × 10 T + 0.02894T − 0.7248jb kJ molg H$ out =
Cp from
600
25
p NH 3
out
Table B.2
−2
p
2
T
in
25
3
−6
H$ out =
100 25
2
p
2
Cp dT = 17.39 kJ mol
Energy balance: Q = ∆ H = ∑ n i H$ i − ∑ ni H$ i
E out
in
b gb g b gb − b8.48gb 00 . g − b76.3ge 2.0735 × 10 T + 0.02894 T − 07248 . j
−7 kJ s = 8.48 mols NH 3 s 25.62 kJ mol + 763 . mols air s 17.39 kJ mol −6
g
2
E
1582 . × 10−4 T 2 + 2.208 T − 1606 = 0 ⇒ T = 693° C (–14,650°C) 8.32 a.
Basis: 100 mol/s of natural gas. Let M represent methane, and E for ethane
100 mol/s 0.95 mol M/mol 0.05 mol E/mol
Furnace
Stack gas (900oC)
Stack gas (ToC)
n3 n4 n5 n6
n3 n4 n5 n6
mol CO2/s mol H2O/s mol O2/s mol N2/s
Heat Exchanger
mol CO2/s mol H2O/s mol O2/s mol N2/s
air (245oC)
20 % excess air (20oC)
n1 mol O2/s n2 mol N2/s
n1 mol O2/s n2 mol N2/s
CH 4 + 2O 2 → CO2 + 2H 2O
b g
C 2 H 6 + 7 / 2 O 2 → 2CO2 + 3H 2 O
8- 13
8.32 (cont’d)
95 mol M 2 mol O2 4.76 mol air 5 mol E 3.5 mol O2 4.76 mol air n&air = 1.2 + s 1 mol M mol O 2 s 1 mol E mol O2 n&air = 1185 mol air/s n&1 = 0.21× 1185 = 249 mol O 2 /s, n&2 = 0.79 ×1185 = 936 mol N 2 /s n&3 =
95 mol M 1 mol CO 2 5 mol E 2 mol CO 2 + = 105 mol CO2 /s s 1 mol M s 1 mol E
n&4 =
95 mol M 2 mol H 2 O 5 mol E 3 mol H 2 O + = 205 mol H 2 O/s s 1 mol M s 1 mol E
95 mol M 2 mol O2 5 mol E 3.5 mol O2 + = 41.5 mol O 2 /s s 1 mol M s 1 mol E n&6 = n&2 = 936 mol N 2 /s n&5 = 249 −
Energy balance on air:
Q& = n&air
∫
o
245 C
20 o C
(C ) p
air
mol air kJ kJ dT = 1185 6.649 = 7879 ⇒ 7879 kW s mol s
Energy balance on stack gas: 6
∑ n& ∫ ( C ) dT
Q& = − ∆H = −
T
i
i= 3
−7879 = n&3
900
∫ (C ) T
p CO 2
900
p i
dT + n&4
∫ (C ) T
p H O (v ) 2
900
dT + n&5
∫ (C ) T
900
p O 2
dT + n&6
∫ (C ) T
900
p N 2
dT
Substitute for the heat capacities (Table B.2), integrate, solve for T using E-Z Solve⇒ T = 732 C o
b.
350 m 3 (STP) mol 1000 L 1 h = 4.34 mol / s h 22.4 L(STP) m 3 3600 s 4.34 mol / s = 0.0434 100 mol / s Q& ′ = 00434 . 7851 = 341 kW Scale factor =
b g
8.33 a.
b.
∆H$ =
b
g b
g
100 335 . + 4 351 . + 38.4 + 42.0 + 2 367 . + 40.2 439 . = 23100 J mol 3 150 mol 23100 J 1 kW Q = ∆H = n∆H$ = = 3465 kW s mol 1000 J / s 600
0
C p dT =
The method of least squares (Equations A1-4 and A1-5) yields (for X = T , y = C p )
b g
Cp = 0.0334 + 1732 . × 10 −5 T ° C kJ (mol ⋅° C) ⇒ Q = 150
600 0
0.0334 + 1732 . × 10 −5 T dT = 3474 kW
The estimates are exactly identical; in general, (a) would be more reliable, since a linear fit is forced in (b). 8.34 a.
e
j
ln C p = bT 1 2 + ln a ⇒ C p = a exp bT 1 2 , b=
ln C p2 C p1
T1 = 71 . , C p1 = 0.329 ,
= 00473 .
T2 − T1 ln a = ln C p1 − b T1 = −14475 . ⇒ a = e −1.4475
8- 14
U| |V ⇒ C = 0.235 | |W
p
T2 = 17.3 , C p2 = 0533 .
e
= 0235 . exp 0.0473T 1 2
j
8.34 (cont’d) b.
150
e
0235 . exp 0.0473T
1800
1
20
30
40
2 200
12
.235gb 2g R . T STexpe0473 jdT = b000473 .
12
jLMNT
12
1 − .0473
OP UV QW
150
= −1730 cal g
1800
DIMENSIONS CP(101), NPTS(2) WRITE (6, 1) FORMAT (1H1, 20X'SOLUTION TO PROBLEM 8.37'/) NPTS(1) = 51 NPTS(2) = 101 DO 200K = 1, 2 N = NPTS (K) NM1 = N – 1 NM2 = N – 2 DT = (150.0 – 1800.0)/FLOAT (NM1) T = 1800.0 DO 20 J = 1, N CP (J) = 0.235*EXP(0.0473*SQRT(T)) T = T + DT SUMI = 0.0 DO 30 J = 2, NM1, 2 SUMI = SUMI + CP(J) SUM2 = 0.0 DO 40 J = 3, NM2, 2 SUM2 = SUM2 + CP (J) DH = DT*(CP(1) + 4.0 = SUM1 + 2.0 = SUM2 + CP(N))/3.0 WRITE (6, 2) N, DH FORMAT (1H0, 5XI3, 'bPOINT INTEGRATION bbbDELTA(H)b= ', E11.4,'bCAL/G') CONTINUE STOP END
Solution: N = 11 ⇒ ∆H$ = −1731 cal g N = 101 ⇒ ∆H$ = −1731 cal g
Simpson's rule with N = 11 thus provides an excellent approximation 8.35 a.
U| V| W
m& = 175 kg / min 175 kg 1000 g 1 mol 56.9 kJ 1 min M . W. = 62.07 g / mol ⇒ Q& = ∆H = = 2670 kW min kg 62.07 g mol 60 s $ ∆H = 56.9 kJ / mol v
b.
The product stream will be a mixture of vapor and liquid.
c.
The product stream will be a supercooled liquid. The stream goes fro m state A to state B as shown in the following phase diagram.
P B
A
T
8- 15
Table B.1 ⇒ Tb = 68.74o C, ∆H$ v (Tb ) = 28.85 kJ / mol Assume: n - hexane vapor is an ideal gas, i.e. ∆H$ is not a function of pressure
8.36 a.
bC H g B ∆H$ bC H g
1
∆H$ 1 =
68 .74 20 200
∆H$ 2 =
6
14 v, 200o C 2
$ T ∆H v b
14 l, 68.74o C
6
bC H g A ∆H$ b g → b C H g $ ∆H
Total →
14 l, 20 o C
6
68.74
14 v, 68.74 o C
6
0.2163 dT = 10.54 kJ / mol 013744 . + 4085 . × 10 −5 T − 2392 . × 10 −8 T 2 + 57.66 × 10−9 T 3 dT
∆H$ 2 = 24.66 kJ / mol ∆H$ Total = ∆H$ 1 + ∆ H$ 2 + ∆ H$ v Tb = 1054 . + 24.66 + 28.85 = 64.05 kJ / mol
b g
b.
∆H$ = −64.05 kJ / mol
c.
U$ 200o C, 2 atm = H$ − PV$
e
j
Assume ideal gas behavior ⇒ PV$ = RT = 393 . kJ / mol U$ = 64.05 − 3.93 = 6012 . kJ / mol 8.37
bg
∆H$ v tb = 40.656 kJ mol
Tb = 100.00° C
e j B ∆H$ H O e l, 100 Cj H2 O l, 50o C
$ 50 o C ∆H v
e
j
→
1
o
2
∆H$ 1 =
100
C pH
2O
e j A ∆H$ H O e v, 100 Cj
H 2 O v, 50o C 2
$ 100o C ∆H v
e j →
o
2
. kJ mol bl g dT = 377
25
∆H$ 2 =
25
C pH
2O
. kJ mol bv g dT = − 169
100
b
g
B
Table B.1
∆H$ v 50° C = 377 . + 40.656 − 169 . = 42.7 kJ mol Steam table:
( 2547.3 − 104.8 ) kJ kg
18.01 g 1 kg = 44.0 kJ mol 1 mol 1000 g
The first value uses physical properties of water at 1 atm (Tables B.1, B.2, and B.8), while the heat of vaporization at 50o C in Table B.5 is for a pressure of 0.1234 bar (0.12 atm). The difference is ? H for liquid water going from 50o C and 0.1234 bar to 50o C and 1 atm plus ? H for water vapor going from 50o C and 1 atm to 50o C and 0.1234 bar. 8.38
n& =
1.75 m3
879 kg
2.0 min
3
m
1 kmol 78.11 kg
103 mol 1 min 1 kmol
b g
Tb = 801 . ° C , ∆H$ v Tb = 30765 . kJ mol
8- 16
60 s
= 164.1
mol s
8.38 (cont’d)
e j → B ∆H$ −∆ µ H C H e v, 80.1 Cj → C 6 H 6 v, 580 o C 1
6
2
v
o
C6 H 6
6
∆H$ 1 =
80 .1
C pC
6H 6
e j $ A ∆H el, 80.1 Cj
C6 H 6 l, 25 o C
o
b vg dT = −77.23
kJ mol
bl gdT = −7.699
kJ mol
580
∆H$ 2 =
298
Cp C
6 H6
3531 .
d i Q = ∆H = n ∆H$ = b164 .1 mol / sgb−115.7 kJ / mol g = −1.90 x 10 ∆H$ = ∆H$ 1 − ∆H$ v 80.1o C + ∆H$ 2 = −1157 . kJ / mol
B
−4
kW
Antoine
8.39
UV ⇒ y 15% relative saturation W 35° C
( ∆ H$ v ) CCl 4
Table B.1
=
300 .
CCl 4
= 0.15
b
PV∗ 25° C 1 atm
g = 0.15 176.0 mm Hg = 0.0347 mol CCl 760 mm Hg
10 mol 0.0347 mol CCl 4 kJ ⇒ Q = ∆H = mol min mol
4
mol
30.0 kJ = 104 . kJ min mol CCl 4
Time to Saturation
6 kg carbon 0.40 g CCl 4 g carbon 8.40 a.
b
1 mol CCl 4 153.84 g CCl 4
g
b
1 mol gas 0.0347 mol CCl 4
g
CO2 g, 20° C → CO2 s, − 78.4 ° C : ∆H$ =
−78.4 20
1 min = 45.0 min 10 mol gas
b g dT − ∆Hsub b− 78.4° Cg
dC i
$
p CO g 2
In the absence of better heat capacity data; we use the formula given in Table B.2 (which is strictly applicable only above 0°C ). − 78 .4 kJ ∆H$ ≈ .03611 + 4.233 × 10−5 T − 2.887 × 10 −8 T 2 + 7.464 × 10 −12 T 3 dT 20 mol
z
FG IJ H K
− 6030
cal 4184 . × 10 −3 kJ = − 28.66 kJ mol mol 1 cal
300 kg CO 2 Q = ∆ H = n∆H$ = h
10 3 g 1 mol 28.66 kJ removed = 195 . × 105 k J h 1 kg 44.01 g mol CO 2
(or 6.23 × 10 7 cal hr or 72.4 kW ) b.
According to Figure 6.1-1b, T fusion =-56o C
Q& = ∆ H = n& ∆H$ where, ∆H$ =
Q& = n&
−56 20
LM dC i N
dC i p
−56
20
p
e j dC i dT +∆ H$ e− 56 Cj + dC i dTOPQ CO2 (v)
dT +∆H$ v − 56o C + o
CO2 (v)
v
−78. 4
−56
p
−78 .4
−56
8- 17
p
CO2 (l)
CO2 (l)
dT
8.41 a.
C p = a + bT
U| . + 001765 . T b Kg V| ⇒ C bJ mol ⋅ Kg = 4512 a = 5394 . − b 0.01765gb 500g = 4512 . |W NaCl b s, 300 Kg → NaClb s, 1073 Kg → NaClbl , 1073 Kg L b4512 O J + 30.21 kJ ∆H$ = C dT + ∆H$ b1073 K g = M . + 0.01765T gdT P N Q mol mol b=
5394 . − 50.41 = 0.01765 500 − 300
p
z
ps
300
= 7.44 × 10 Q = ∆U = n
b.
z
1073
z
m
4
300
103 J 1 kJ
J mol
b
C v dT + ∆U$ m 1073 K
1073
300
1073
g
Cv ≈ Cp ∆ U m ≅ ∆H m
200 kg 10 3 g
2.55 × 10 8 J
s
t=
c. 8.42
Q ≈ ∆H = n∆H$ =
1 kg
1 mol
74450 J
58.44 g
mol
1 kJ
0.85 × 3000 kJ 103 J
= 2.55 × 108 J
= 100 s
∆H$ v = 35.98 kJ mol , Tb = 136.2° C = 409.4 K , Pc = 37.0 atm , Tc = 619.7 K (from Table B.1)
b
gb
g
b
Trouton's rule: ∆H$ v ≈ 0.088Tb = 0.088 409.4 K = 360 . kJ mol 01% . error Chen's rule:
LM MN
g
FG T IJ − 0.0327 + 0.0297 log P OP HT K PQ = 35.7 kJ mol (–0.7% error) ∆H$ ≈ FT I 107 . −G J HT K F 619.7 − 373.2 IJ = 38.2 kJ mol Watson’s correlation : ∆H$ b100° Cg ≈ 3598 . G H 619.7 − 409.4 K Tb 0.0331
v
b
10
c
c
b c
0.38
v
8.43
b g b g Trouton's Rule ⇒ ∆H$ b 200 ° Cg = 0.088 b200+ 273.2 g = 41.6 kJ mol C H Nb l, 25° Cg → C H Nb l, 200° Cg → C H Nbv , 200° Cg
C 7 H 2 N : Kopp's Rule ⇒ C p ≈ 7 0.012 + 12 0.018 + 0.033 = 0.333 k J (mol ⋅° C) v
7
12
7
200
∆Hˆ =
12
7
kJ
∫ C dT + ∆Hˆ ( 200 °C ) ≈ 0.333(200 − 25) mol p
v
25
8- 18
12
+ 41.6
kJ = 100 kJ mol mol
8.44 a.
b g
Antoine equation: Tb ° C = Watson Correction:
b.
b g F 562.6 − 299.3IJ ∆H$ b261 . ° Cg = 30.765G H 562.6 − 353.1 K
0. 38
= 336 . kJ mol
v
b
g
b
g
Antoine equation: Tb 50 mm Hg = 118 . ° C ; Tb 150 mm Hg = 352 . °C $ ln p2 p1 ∆H v Clausius-Clapeyron: ln p = − + C ⇒ ∆H$ v = − R RT 1 T2 − 1 T1
∆H$ v = −0.008314 c.
1211.033 − 220.790 = 261 . °C 6.90565 − log 100
R|S T|
b
U|V W|
g
C6 H 6 (v, 26.1°C)
∆ H$1
∆H$ v (80.1°C)
C6 H 6 ( l , 80.1°C)
∆ H$2
C6 H 6 (v, 80.1°C)
80 .1
dC i dT = 7.50 kJ mol
26 .1
∆H$ 2 =
g
ln 150 50 kJ = 34.3 kJ mol mol ⋅ K 1 3084 . K − 1 285.0 K
C6 H 6 ( l , 26.1°C)
∆H$ 1 =
b
p l
26 .1
dC i dT = −4.90 kJ mol
80.1
b
p v
g
∆H$ v 261 . ° C = 7.50 + 30.765 − 4.90 = 334 . kJ mol 8.45 a. Tout = 49.3o C. The only temperature at which a pure species can exist as both vapor and liquid at 1 atm is the normal boiling point, which from Table B.1 is 49.3o C for cyclopentane.
b. Let n& f , n& v , and n& l denote the molar flow rates of the feed, vapor product, and liquid product streams, respectively. Ideal gas equation of state
n& f =
1550 L 273 K s
1 mol
423 K 22.4 L(STP)
= 44.66 mol C 5 H 10 (v) / s
55% condensation: n& l = 0550 . ( 44.66 mol / s) = 24.56 mol C 5 H10 ( l) / s Cyclopentane balance ⇒ n& v = ( 44.66 − 24.56) mol C5H 10 / s = 20.10 mol C 5H 10 (v) / s Reference: C5 H10 (l) at 49.3o C
n& in (mol/s)
H$ in (kJ/mol)
n& out (mol/s)
H$ out (kJ/mol)
C5 H10 (l)
—
—
24.56
0
C5 H10 (v)
44.66
H$ f
20.10
H$ v
Substance
Hi = ∆H$ v +
z
Ti o
49. 3 C
8- 19
C p dT
8.45 (cont’d)
Substituting for ∆H$ v from Table B.1 and for C p from Table B.2 ⇒ H$ f = 38.36 kJ / mol, H$ v = 27.30 kJ / mol
Energy balance: Q& =
8.46 a.
∑n
$
out H out
−
∑n
$
in H in
= −116 . × 10 3 kJ / s = −116 . × 10 3 kW
Basis: 100 mol humid air fed n 2 (mol), 20o C, 1 atm
Q(kJ)
y 2 (mol H2 O/mol), sat’d 1-y 2 (mol dry air/mol)
100 mol y 1 (mol H2 O/mol) 1-y 1 (mol dry air/mol) 50o C, 1 atm, 2o superheat
n 3 (mol H2 O(l))
There are five unknowns (n 2 , n 3 , y 1 , y 2 , Q) and five equations (two independent material balances, 2o C superheat, saturation at outlet, energy balance). The problem can be solved. b.
2° C superheat ⇒ y1 =
b
p∗ 48° C p
saturation at outlet ⇒ y2 =
g
b
p∗ 20° C p
g
b100gb1 − y g = n b1 − y g H O balance: b100gb y g = bn gb y g + n
dry air balance:
1
2
1
2
2
2
b
c.
2
3
g
b
g
References: Air 25° C , H 2O l, 20° C
nin
Air
100 ⋅ 1 − y1
H2O v
100 ⋅ y1
H$ 2
−
−
bg H Obl g 2
H$ 1 = H$ 2 = =
H$ in H$
Substance
d i d i
b
g
H$ out H$
n in mol
n2 ⋅ y2
H$ 4
H$ in kJ mol
n3
0
nout
b
n 2 ⋅ 1 − y2
1
g
3
50 50 C p dT = 0.02894 + 0.4147 × 10−5 T + 0.3191× 10−8 T 2 25 air 25 100 50 Cp dT + ∆ H$ v 100o C + Cp dT 20 100 H 2O(l) H 2 O(v) 100 20 50 100
H$ 3 =
25
H$ 2 =
20
e
− 1965 . × 10 −12 T 3 dT
d i
j
0.0754 dT + 40.656 + 0.03346 + 0.688 × 10−5 T + 0.7604 × 10− 8 T 2 − 3593 . × 10 −12 T 3 dT
dC i dT dC i
20
p air
100
p
H 2 O(l)
e
j
dT + ∆H$ v 100 o C +
20 100
dC i
8- 20
p
H 2O(v)
dT
8.46 (cont’d) c.
Q = ∆ H = ∑ ni H$ i − ∑ ni H$ i out
Vair =
in
100 mol 8314 . Pa ⋅ m 3 323 K 5 mol ⋅ K 101325 . × 10 Pa
∑ n H$ − ∑ n H$ i
⇒
d.
i
i
i
Q out in = V air 100 mol 8.314 Pa ⋅ m 3 323 K mol ⋅ K 101325 . × 105 Pa
2° C superheat ⇒ y1 =
b
g
p∗ 48° C 8371 . mm Hg = = 0110 . mol H 2 O mol p 760 mm Hg
b
g
p∗ 20° C 17.535 mm Hg = = 0023 . mol H 2O mol p 760 mm Hg
saturation at outlet ⇒ y2 =
. g = n b1 − 0.023g ⇒ n = 9110 . mol b100gb1 − 0110 890 . mol H O H O balance: b100 gb0110 . g = b 9110 . gb0.023 g + n ⇒ n =
dry air balance:
2
2
2
2
3
3
0.018 kg 1 mol
= 0160 . kg H2 O condensed Q = ∆ H = ∑ n i H$ i − ∑ ni H$ i = − 4805 . kJ out
Vair =
100 mol 8.314 Pa ⋅ m 3 323 K = 2.65 m 3 mol ⋅ K 1.01325 × 105 Pa
⇒ ⇒ e. f.
in
0.160 kg H 2 O condensed 2.65 m 3 air fed − 4805 . kJ 3
2.65 m air fed
= 0.0604 kg H 2 O condensed / m 3 air fed
= −181 kJ / m 3 air fed
Solve equations with Maple.
Q=
8.47 Basis:
−181 kJ 250 m 3 air fed 1 h 1 kW = −12.6 kW h 3600 s 1 kJ / s m 3 air fed
226 m3
273 K
10 3 mol
min
309 K
22.415 m 3 STP
b g
= 8908 mol humid air min . DA = Dry air
Q& ( kJ / min)
8908 mol / min y 0 [ mol H 2 O(v) / mol] (1- y 0 ) (mol DA / mol)
n& 1 ( mol / min) y1[ mol H 2 O(v) / mol] (1- y1 ) (mol DA / mol)
36 o C, 1 atm, 98% rel. hum.
10 o C, 1 atm, saturated
8- 21
n& 2 [ mol H 2 O(l) / min], 10 o C
8.47 (cont’d) a. Degree of freedom analysis 5 unknowns – (1 relative humidity + 2 material balances + 1 saturation condition at outlet + 1 energy balance) = 0 degrees of freedom.
44.563 mm Hg) = 0.0575 mol H O(v) mol b36B° Cg ⇒ y = 0.98(760 mm Hg Outlet air: y = p (10 C) / P = b9.209 mm Hg g b 760 mm Hgg = 0.0121 mol H O(v) mol Air balance: b1 − 0.0575g(8908 mol / min) = b1 − 00121 . gn& ⇒ n& = 8499 mol / min F mol IJ = 0.0121(8499 mol ) + n& ⇒ n& = 409 mol H O(l) min H O balance: 0.0575G 8908 H min K min References: H Ob l, triple point g, air b 77° Fg Table B.3
b. Inlet air: y0 P =
098 . p *w ∗
1
0
2
o
2
1
2
1
2
2
2
2
n& in
Substance Air
bg H Obl g
H$ in
8396 0.3198 8396
H2O v
512
462
103
−
−
409
2
H$ out
n& out
−0.4352 n& in mol min 453 H$ in kJ / mol 0.741
Air: H$ from Table B.8 H 2 O: H$ ( kJ / kg) from Table B.5 × (0.018 kg / mol) Energy balance:
Q = ∆H =
∑
ni H$ i −
out
∑
ni H$ i =
−196 . × 105 kJ
60 min 9486 . × 10−4 Btu
min
in
1h
0.001 kJ
1 ton −12000 Btu h
= 930 tons
8.48 Basis:
746.7 m3 outlet gas / h 3 atm
1 kmol
b g
1 atm 22.4 m3 STP
= 100.0 kmol / h
100 kmol/h at 0°C, 3 atm yout (kmol C 6 H 14( v)/kmol), saturated (1 – yout) (kmol N 2/kmol) n 2 kmols/h nC 6 H 14( v), 0°C
nn&11 (kmol/h) at 75°C, 3 atm yin (kmol C 6 H 14( v)/kmol), 90% sat'd (1 – yin) (kmol N 2/kmol)
o n&2 [kmol n-CH ( v)/h],0 C 6 14
Antoine:
log p∗v = 6.88555 −
1175.817 224.867 + T
p∗v (0 °C ) = 45.24 mm Hg, p∗v (75°C) = 920.44 mm Hg
8- 22
b g = 45.24 = 0.0198 kmol C H kmol , P 3b 760 g 0.90 p b75° Cg b0.90gb 920.44g kmol C H = = = 0.363 P 3b 760g kmol
yout = yin
p∗v 0° C
6
14
∗ v
6
14
8.48 (cont’d)
b
g
b
g
N 2 balance: n& 1 1 − 0.363 = 100 1 − 00198 . ⇒ n& 1 = 153.9 kmol h
b
gb g b gb g bg Percent Condensation: b5389 . kmol h condenseg b 0363 . × 1539 . gb kmol h in feedg × 100% = 96.5% C 6 H 14 balance: 1539 . 0363 . = 100 0.0198 + n& 2 ⇒ n& 2 = 5389 . kmol C 6 H 14 l h
References: N2 (25o C), n-C6 H14 (l, 0o C) Substance n H$ n in
N2
bg b lg
n - C 6 H 14 r n - C 6 H 14
b
98000
in
146 .
out
98000 −0.726
55800 44.75 −
g
−
H$ out
2000
33.33
53800
0.0
N 2 : H$ = C p T − 25 , n − C 6 H 14 (v): H$ =
z
68 .7
n& in mol h H$ in kJ mol
b g
C pl dT + ∆ H$ v 68.7 +
0
z T
C pv dT
68.7
Energy balance: Q = ∆ H = ( −2.64 × 10 6 kJ h)(1 h / 3600 s) ⇒ −733 kW
∑ n H$ − ∑ n H$ i
i
out
i
i
in
8.49 Let A denote acetone.
Q& ( kW)
W& s = − 25.2 kW
n& 1 (mol / s) @ − 18o C, 5 atm y1[mol A(v) / mol], sat'd (1 − y 1 )( mol air / mol)
142 L/ s @ 150 o C, 1.3 atm n& 0 ( mol / s) y 0 [mol A(v) / mol], sat'd (1 − y 0 )( mol air / mol)
n& 2 [ mol A(l) / s]@−18 o C, 5 atm a.
Degree of freedom analysis :
6 unknowns ( n& 0 , n& 1 , n& 2 , y0 , y1 , Q& ) –2 material balances –1 equation of state for feed gas –1 sampling result for feed gas –1 saturation condition at outlet –1 energy balance 0 degrees of freedom
8- 23
b.
Ideal gas equation of state
Raoult’s law
P0V&0 RT0
(1) n& 0 =
(2) y1 =
p*A ( −18o C) 5 atm
(Antoine equation for p *A )
Feed stream analysis (3)
y0
FG mol A IJ = H mol K
[(4.973 − 4.017) g A][1 mol A /58.05 g] [( 300 . L) P0 / RT0 ] mol feed gas
8.49 (cont’d) Air balance (4) n& 1 =
Acetone balance
n& 0 (1 − y0 ) (1− y1 )
(5) n& 2 = n& 0 y 0 − n&1 y1 o
o
Reference states : A(l, –18 C), air(25 C)
n&in (mol/s)
Substance A(l)
−
A(v)
n&0 y0
air (6) H$ A(v) ( T ) =
z
(kJ/mol)
−18 C o
(7)
Hˆ a1
z
T
56 o C
Tab le B.1
( C p ) A(v) dT Ta ble B.2
H$ air ( T ) from Table B.8
(8) Q& = W& s + c.
n&1 (1 − y1 )
( C p ) A(l) dT + ( ∆H$ v ) A + Table B.2
(kJ/mol) 0 Hˆ A1
n&1 y1
Hˆ a 0
Hˆ out
n&2
− Hˆ A0
n&0 (1 − y 0 ) 56 o C
n&out (mol/s)
Hˆ in
∑ n&
out
H$ out −
∑ n&
(1) ⇒ n&0 = 5.32 mol feed gas/s
$
in H in
(W& s = −252 . kJ / s)
(2) ⇒ y1 = 6.58 × 10 −3 mol A(v)/mol outlet gas
(3) ⇒ y 0 = 0.147 mol A(v)/mol feed gas (4) ⇒ n&1 = 4.57 mol outlet gas/s
(5) ⇒ n&2 = 0.75 mol A(l)/s
(6) ⇒ Hˆ A 0 = 48.1 kJ/mol, Hˆ A1 = 34.0 kJ/mol (7) ⇒ Hˆ a 0 = 3.666 kJ/mol, Hˆ a1 = −1.245 kJ/mol (8) ⇒ Q& = −84.1 kW
8- 24
8.50 a.
b g
3 m π × 35
2
cm 2
1 m2
273 K
b273 + 40gK
104 cm 2
s
850 mmHg
1 kg ⋅ mol
b g
760 mmHg 22.4 m 3 STP
10 3 mol 1 kg ⋅ mol
= 50.3 mol s H = n-hexane 50.3 mol/s, 850 mmHg
assume P=850 mmHg n2 mol H(v)/mol, sat’d @ To C n3 mol air/mol n 1 (mols H( l )/s) (90% feed) (60% of of H in feed)
x0 mol H/mol (1-x0) mol air/mol 40o C, Tdp =20o C
8.50 (cont’d) Degree-of-freedom analysis 5 unknowns (n 1 , n 2 , n 3 , x0 and T) – 2 independent material balances – 1 saturation condition – 1 60% recovery equation – 1 energy balance 0 degrees of freedom All unknowns can be calculated. b.
Antoine equation, Table B.4
dT i dp
feed
= 25 ° C ⇒ x 0 =
b
p*H 25° C
⇒ n1 =
60% recovery
P 0.600
b
g = 151 mm Hg = 0178 . mol H mol
850 mm Hg 50.3 0.178 mols H feed = 5.37 mols H l s s
b gb
g
gb gb
g
bg
bg
n2 = 0.400 50.3 0178 . = 358 . mols H v s
b gb
g
Air balance: n3 = 50.3 1 − 0178 . = 413 . mols air s Mole fraction of hexane in outlet gas:
bg
p*H T n2 3.58 = = ⇒ p*H T = 67.8 mm Hg n2 + n3 3.58 + 41.3 850 mm Hg
b
g
bg
Antoine equation: p *H = 67 .8 mm Hg ⇒ T = 7 .8 ° C
b
g
Reference states: C 6 H 14 l, 7.8° C , air (25°C)
8.95
H$ in 37.5
3.58
H$ out 32.7
—
—
5.37
0
41.3
0.435
41.3
–0.499
Substance
n& in
C6 H14 ( v ) C6 H14 (l ) Air
n& out
8- 25
n& in mol/s $ H in kJ/mol
C6 H14 ( v ) : H$ =
z
68 . 74
b
g
C pl dT + ∆ H$ v 6 8 .74 ° C +
7 .8
z T
C pv d T ,
68 .7 4
C p from Table B.2
∆ H$ v from Table B.1
Air: H$ from Table B.8 Energy balance: Q = ∆ H =
∑ n& H$ − ∑ n& H$ i
i
i
out
c.
u ⋅ A = u'⋅ A '; A =
i
=
−257 kJ s 1 kW cooling
in
U|V |W
π ⋅ D2 1 ; D' = D ⇒ u' = 4 ⋅ u = 12.0 m / s 4 2
8- 26
−1 kJ s
= 257 kW
8.51
n& v ( mol / min) @ 65o C, P0 ( atm) y[ mol P(v) / mol], sat'd (1- y) (mol H(v) / mol)
100 mol / s @80o C, 5.0 atm 0.500 mol P(l) / mol 0.500 mol H(l) / mol
Q& ( kJ / s)
n& l ( mol / min) @ 65o C, P0 ( atm) 0.41 mol P(l) / mol 0.59 mol H(l) / mol a. Degree of freedom analysis 5 unknowns – 2 material balances – 2 equilibrium relations (Raoult’s law) at outlet – 1 energy balance = 0 degrees of freedom Antoine equation (Table B.4) ⇒ p *P ( 65o C) = 1851 mm Hg, p *H ( 65o C) = 675 mm Hg
Raoult' s law for pentane and hexane 0.410 p *P (65o C) = yP0
y = 0.656 mol P(v) / mol
⇒
0.590 p *H ( 65o C) = (1 − y) P0
P0 = 1157 mm Hg (1.52 atm)
Total mole balance: 100 mol = n& v + n& l Pentane balance: 50 mole P = 0.656n& v + 0.410n& l Ideal gas equation of state : Vv = Fractional vaporization: f =
nv RT 36.6 mol = P0 s
⇒
n& v = 36.6 mol vapor / s n& l = 634 . mol liquid / s
0.08206 L ⋅ atm mol ⋅ K
36.6 mol vapor / s mol vaporized = 0.366 100 mol / s mol fed
References: P(l), H(l) at 65 o C
H$ in
Substance n& in P(v)
−
P(l)
50
H(v)
−
H(l)
50
H$ out
n& out
−
24.0 24.33
2.806 26.0 −
0
12.6 29.05
3.245 37.4
Vapor: H$ ( T ) =
Liquid: H$ ( T) =
n& in mol s H$ in kJ / mol
z z
Tb o
65 C
0
C pl dT + ∆H$ v ( Tb ) +
z
T
Tb
C pv dT
T o
65 C
C pl dT
Tb and ∆H$ v from Table B.1, C p from Table B.2 Energy balance:
Q& =
∑ n&
$
out H out
−
∑ n&
$
in H in
8- 27
= 1040 kW
(65 + 273)K 152 . atm
= 667 L / s
8.52 a.
B=benzene; T=toluene n 2 mol/s 95o C 1320 mol/s 25o C
0.735 mol B/mol 0.265 mol T/mol
0.500 mol B/mol 0.500 mol T/mol
n 3 mol/s 95o C 0.425 mol B/mol 0.575 mol T/mol
Q
UV RS W T
Total mole balance: 1320 = n2 + n3 n = 319 mol / s ⇒ 2 Benzene balance: 1320(0.500) = n2 ( 0.735) + n3 ( 0.425) n 3 = 1001 mol / s References: B(l, 25o C), T(l, 25o C)
n& in (mol / s) H$ in ( kJ / mol) n& out (mol / s) H$ out ( kJ / mol)
Substance B(l) B(v) T(l) T(v)
Q=
660 -660 --
∑ n Hˆ − ∑ n Hˆ i
i
i
out
b.
0 -0 -i
425 234 576 85
9.838 39.91 11.78 46.06
= 2.42 × 10 4 kW
in
e
j
e
j
Antoine equation (Table B.4) ⇒ p *B 95o C = 1176 torr , pT* 95o C = 476.9 torr Raoult's law
b gb g b g |UV ⇒ P ≠ P' b0.575gb476.9g = b0.265gP ' ⇒ P ' = 1035 torr|W
Benzene: 0.425 1176 = 0.735 P ⇒ P = 680 torr Toluene:
⇒ Analyses are inconsistent.
Possible reasons: The analyses are wrong; the evaporator had not reached steady state when the samples were taken; the vapor and liquid product streams are not in equilibrium; Raoult’s law is invalid at the system conditions (not likely).
bg b gb g b gb g C H Ob lg — C = b 5gb12g + b12gb18g + 25 = 301 J mol Trouton’s rule — Eq. (8.4-3): ∆H = b0109 . gb113 + 273g = 421 . kJ mol Eq. (8.4-5) ⇒ ∆H$ = b0.050 gb52 + 273g = 16.25 k J mol
8.53 Kopp’s rule (Table B.10):
C 5 H12 O s — C p = 5 7.5 + 12 9.6 + 17 = 170 J mol 5
12
p
v
m
Basis :
235 m
3
273 K
1 kmol 3
h
b g
389 K 22.4 m STP
103 mol
1h
1 kmol
3600 s
= 2.05 mol s
Neglect enthalpy change for the vapor transition from 116°C to 113°C.
b
g
b
g Ob s, 25° Cg
b
C 5 H 12 O v , 113° C → C 5H 12 O l, 113° C → C 5 H 12 O v, 52° C
b
g
→ C 5 H 12 O s, 52° C → C 5 H12
8- 28
g
8.53 (cont’d)
b
g
b
g b gb g b gb g
∆H$ = − ∆H$ v + C pl 52 − 113 − ∆H$ m + C ps 25 − 52 kJ kJ J 1 kJ = −421 . − 16.2 − 301 61 + 170 27 × 3 = −813 . kJ mol mol mol mol 10 J Required heat transfer: Q = ∆H = n∆H$ =
2.05 mol −813 . kJ s
1 kW
mol
1 kJ s
= −167 kW
8.54 Basis: 100 kg wet film ⇒ a.
95 kg dry film
0.5 kg acetone remain in film
90% A evaporation
5 kg acetone
95 kg DF 5 kg C3 H 6O( l) Tf 1 = 35°C n 1 mol air Ta1 , 1.01 atm
4.5 kg acetone exit in gas phase
95 kg DF 0.5 kg C 3 H 6O( l) Tf 2 n 1 mol air 4.5 kg C3 H 6O( v) (40% sat'd) Ta2 = 49°C, 1.0 atm
Antoine equation (Table B.4) ⇒ p *C3 H 6O = 59118 . mm Hg
1 kmol
10 3 mol
58.08 kg
kmol
4.5 kg C3 H 6 O
⇒y=
b
bg
= 77.5 mol C 3 H 6O v in exit gas
b g
g
171.6 mol 22.4 L STP 040 . 59118 . mm Hg 775 . = ⇒ n1 = 775 . + n1 760 mm Hg mol
b
g
b
g b
95 kg DF
b g
L STP
kg DF
g
References: Air 25° C , C 3 H 6O l , 35° C , DF 35° C
b.
Substance
nin
H$ in
DF
95
0
95
86.1
0
8.6
—
—
77.5
32.3
dC i
171.6
0.70
bg Ob vg
C 6 H 14 O l C 6 H 14
171.6
Air
z
Ta1
25
H$ A(v) =
zd
86
Cp
i
l
dT + ∆H$ v +
35
nout
p air dT
zd
49
Cp
i dT , v
H$ out
d 0.129 d T
i − 35i
1.33 T f 2 − 35
b
f2
∆H =
i
i
out
⇒
z
i
g
d
25
dC i
Ta1 = 120° C ⇒
p
z
air
Ta1
25
dT =
dC i
z
i
= 126.4 T f 2 − 35 + 111 . (T f 2 − 35) + 26234 . − 171.6
in
Ta1
n in mol H$ in kJ/mol
86
∑ n H$ − ∑ n H$ i
n in kg H$ in kJ/kg
H$ DF = C p T − 35
Energy balance
c.
= 405 .
d
i
1275 . T f 2 − 35 + 2623.4
p air dT
1716 .
d
i
= 2.78 kJ mol ⇒ T f 2 − 35 ° C = −168 . °C
8- 29
Ta 1
25
dC i
p air dT
=0
8.54 (cont’d) T&E
T&E
d. T f 2 = 34° C ⇒ Ta1 = 506° C , T f 2 = 36° C ⇒ Ta1 = 552° C e.
8.55
In an adiabatic system, when a liquid evaporates, the temperature of the remaining condensed phase drops. In this problem, the heat transferred from the air goes to (1) vaporize 90% of the acetone in the feed; (2) raise the temperature of the remaining wet film above what it would be if the process were adiabatic. If the feed air temperature is above about 530 °C, enough heat is transferred to keep the film above its inlet temperature of 35 °C; otherwise, the film temperature drops.
b
g
Tset p = 200 psia ≈ 100° F (Cox chart – Fig. 6.1-4) a. Basis:
3.00 × 10 3 SCF 1 lb - mole = 8.357 lb ⋅ mole h C 3 H 8 h 359 SCF 8.357 lb-mole C3 H8 (v)/h 200 psia, 100o F
8.357 lb-mole C3 H8 (l)/h 200 psia, 100o F
Q&
m(lb & - mole H 2O(l) / h 70o F
m(lb & - mole H 2O(l) / h 85o F
The outlet water temperature is 85o F. It must be less than the outlet propane temperature; otherwise, heat would be transferred from the water to the propane near the outlet, causing vaporization rather than condensation of the propane. b.
Energy balance on propane:
B & = − n& ∆H$ = 8.357 lb − moles −18.77 kJ 0.9486 Btu 453.593 mol = −6.75 × 104 Btu Q& = ∆ H v h h mol kJ 1 lb ⋅ mole Table B.1
Energy balance on cooling water: Assume no heat loss to surroundings.
& p ∆T ⇒ m& = Q& = ∆H& = mC
6.75 ×104 Btu
lb m ⋅°F 1.0 Btu 15 °F
h
8.56
= 4500
lbm cooling water h
m& 2 [kg H2 O(v)/h]@100o C, 1 atm
o
1000 kg/h, 30 C 0.200 kg solids/kg 0.800 kg H2 O(l)/kg
m& 3 (kg/h) @ 100 o C 0.350 kg solids/kg 0.650 kg H2 O(l)/kg
m& 1 [ kg H 2 O(v) / h], 1.6 bar, sat'd a.
Solids balance: 200 = 0.35m3
b
H 2 O balance: 800 = m2 + 0.65 5714 .
m& 1 [ kg H 2 O(l) / h], 1.6 bar, sat'd
g
⇒ m3 = 571.4 kg h slurry
bg
⇒ m2 = 428.6 kg h H 2 O v
8- 30
8.56 (cont’d) References: Solids (0.01°C), H 2 O (l, 0.01o C) Substance
m & in
Solids H2O l
200 800 —
bg H Ob v g 2
H 2 O , 1.6 bar E.B. Q = ∆H =
H$ in 62.85 125.7 —
m &1
200 571.4 428.6
m &1
2696.2
∑ m& Hˆ − ∑ m& Hˆ i
i
i
out
H$ out 209.6 419.1 2676
m & out
i
& ( kg h ) H$ H 2O from steam tables m
b
H$ kJ kg
g
475.4
& 1 = 592 kg steam h = 0 ⇒ 1.315 × 106 − 2221m& 1 = 0 ⇒ m
in
b.
( 592.0 − 428.6 ) = 163
c.
The cost of compressing and reheating the steam vs. the cost of obtaining it externally.
8.57 Basis: 15,000 kg feed/h.
kg h additional steam
A = acetone, B = acetic acid, C = acetic anhydride Q c (kJ/h) 2 n 1 (kg A(v )/h) n (kg A(l )/h) condenser 1 329 K 303 K
15000 kg/h 0.46 A 0.27 B 0.27 C 348 K, 1 atm
n 1 (kg A(l )/h) 303 K still
1% of A in feed n 2 (kg A(l )/h) n 3 (kg B(l )/h) Q r (kJ/h) n 4 (kg C(l )/h) 398 K reboiler
a.
b gb gb
g
n& 2 = 001 . 0.46 15,000 kg h = 69 kg A h
b gb g Acetic anhydride balance: n& = b 027 . gb15, 000g = 4050 kg h Acetone balance: b 046 . gb15,000g = n + 69 ⇒ n = 6831 kg h ` Acetic acid balance: n& 3 = 0.27 15,000 = 4050 kg B h 4
1
1
⇓ Distillate product: 6831 kg acetone h
8169 kg h 0.8% acetone Bottoms product: 69 + 4050 + 4050 kg h = 49.6% acetic acid 49.6% acetic anhydride
b
b.
g
Energy balance on condenser
8- 31
8.57 (cont’d)
b
g
b
g
b
C 3 H 6 O v , 329 K → C 3 H 6 O l, 329 K → C 3 H 6 O l, 303 K
b
z
g
303
g C dT = −520.6 + b2.3gb−26g = −580.4 kJ kg b2 × 6831gkg −580.4 kJ = −7.93 × 10 kJ h = ∆H& = n& ∆H$ =
∆H$ = − ∆H$ v 329 K +
pl
329
Q& c c.
6
h
kg
Overall process energy balance Reference states : A(l), B(l), C(l) at 348 K (All H$ m = 0 )
H$ in
n& in
Substance
n& out
H$ out
b g — 0 6831 –103.5 — 0 69 115.0 A b l, 398 Kg — 0 4050 109.0 B b l, 398 Kg — 0 4050 113 C b l, 398 Kg J C ≈ b 4 × 12g + b 6 × 18g + b 3 × 25g Acetic anhydride (l): mol⋅° C A l, 303 K
p
n& in kg/h H$ in kJ/kg
1 mol 103 g 1 kJ 102.1 g 1 kg 103 J
= 2.3 kJ kg⋅° C $ H T = C p T − 348 (all substances)
bg
b
g
Q& = ∆H& ⇒ Q& c + Q& r =
∑ n& H$ − ∑ n& H$ i
out
i
i
in
i
⇒ Q& r = −Q& c +
∑ n& H$ = e7.93 × 10 i
i
6
out
A=0
= 813 . × 106 kJ h
(We have neglected heat losses from the still.) d.
H 2 O (saturated at ≈ 11 bars): ∆H$ v = 1999 kJ kg (Table 8.6)
813 . × 106 kJ h Q& r = n& H2 O ∆H$ v ⇒ n& H 2O = = 4070 kg steam h 1999 kJ kg 8.58
Basis : 5000 kg seawater/h a.
S = Salt n 3 (kg H2 O(l )/h @ 4 bars) 2738 kJ/kg
5000 kg/h @ 300 K 0.035 S 0.965 H 2O( l) 113.1 kJ/kg
n 5 (kg H2 O(l )/h @ 4 bars) 605 kJ/kg
b.
S balance on 1st effect:
n 4 kg H 2 O(v )/h @ 0.2 bars 2610 kJ/kg
n 2 (kg H 2 O(v )/h @ 0.6 bars) 2654 kJ/kg n 1 (kg/h @ 0.6 bars) 0.055 S 0.945 H 2O( l) 360 kJ/kg
. gb5000g = 0055 . n& b0035
1
n 3 (kg/h @ 0.2 bars) x (kg S/kg) (1 – x) (kg H2 O(l )/hr) 252 kJ/kg n 2 (kg H 2O(l )/h @ 0.6 bars) 360 kJ/kg
⇒ n& 1 = 3182 kg h
Mass balance on 1st effect: 5000 = 3182 + n& 2 ⇒ n& 2 = 1818 kg h
8- 32
j
+ 2.00 × 105 kJ h
8.58 (cont’d) Energy balance on 1st effect:
b gb
g b gb g b gb = 2534 kg H Ob v g h
g b
gb g
∆H& = 0 ⇒ n& 2 2654 + n& 1 360 + n& 5 605 − 2738 − 5000 1131 . =0 n& 5
n&1 = 3182 n&2 =1818
c.
2
Mass balance on 2nd effect: 3182 = n& 3 + n& 4 (1)
b∆H = 0g bn gb2610g + bn gb252g + bn gb360 − 2654g − bn gb360g = 0 E n = 3182, n = 1818
Energy balance on 2nd effect: 4
3
2
1
1
2
5.316 × 106 = 252 n3 + 2610n4
(2)
Solve (1) and (2) simultaneously:
n& 3 = 1267 kg h brine solution
bg
n& 4 = 1915 kg h H 2 O v
b
g
Production rate of fresh water = n& 2 + n& 4 = 1818 + 1915 = 3733 kg h fresh water
b
gb g
Overall S balance: 0035 . 5000 = 1267 x ⇒ x = 0138 . kg salt kg
d.
e.
The entering steam must be at a higher temperature (and hence a higher saturation pressure) than that of the liquid to be vaporized for the required heat transfer to take place.
n 5 (kg H 2 O(v )/h) 2738 kJ/kg
5000 kg/h 0.035 S 0.965 H 2O( l) 113.1 kJ/kg
3733 kg/h H 2 O(v ) @ 0.2 bar 2610 kJ/kg n 1 (kg brine/h @ 0.2 bar 252 kJ/kg
Q3
n 5 (kg H 2 O(l )/h) 605 kJ/kg
Mass balance: 5000 = 3733 + n& 1 ⇒ n&1 = 1267 kg h
d∆H& = 0i b3733gb2610 g + b1267gb252g + n& b605 − 2738g − b5000gb113.1g = 0 ⇒ n& = 4452 kg H O bv g h
Energy balance:
5
5
2
Which costs more: the additional 1918 kg/hr fresh steam required for the single-stage process, or the construction and maintenance of the second effect?
8- 33
8.59 a.
b0.035gb5000g = 583 kg h
Salt balance: x L7 n& L 7 = x L1n& L1 ⇒ n& L1 =
030 . Fresh water produced: n L 7 − n L1 = 5000 − 583 = 4417 kg fresh water h
b.
Final result given in Part (d).
c.
Salt balance on i th effect:
b g bx g
n& Li x Li = n& L
i+ 1
bn& g bx g L i +1
⇒ x Li =
L i +1
n& θ Li
L i +1
(1)
Energy balance on i th effect:
b g e H$ j
∆H& = 0 ⇒ n& vi H$ vi + n& v ⇒
bn& g
v L −1
=
b g e H$ j + n& H$ − b n& g e H$ j e H$ j − e H$ j
L −1
n& vi H$ vi
v
Li v
n& Li
Li
v i −1
L i −1
+ n& Li H$ Li − n& L L i +1
L
i −1
b g effect: = b n& g + bn& g ⇒ bn& g
Mass balance on i − 1
L −1
L
L +1
L
L +1
b g e H$ j
− n& v
L −1
v
L −1
=0 (2)
i +1
L −1
th
L i −1
b g
= n& Li − n& v
(3)
i −1
d.
Fresh steam Effect 1 Effect 2 Effect 3 Effect 4 Effect 5 Effect 6 Effect (7)
P (bar) 2.0 0.9 0.7 0.5 0.3 0.2 0.1 1.0
T (K) 393.4 369.9 363.2 354.5 342.3 333.3 319.0 300.0
nL (kg/h) --584 1518 2407 3216 3950 4562 5000
8- 34
xL --0.2997 0.1153 0.0727 0.0544 0.0443 0.0384 0.0350
nV (kg/h) 981 934 889 809 734 612 438 ---
HL (kJ/kg) 504.7 405.2 376.8 340.6 289.3 251.5 191.8 113.0
HV (kJ/kg) 2706.3 2670.9 2660.1 2646.0 2625.4 2609.9 2584.8 ---
8.60 a.
dC i = dC i p
p
v
l
b g ≈ dC i
= 20 cal (mol⋅° C) ; Cv
p v
v
b
− R ≈ 10 − 2
g molcal⋅° C = 8 cal (mol⋅° C)
b. n 0 (mol N2 )
n 0 (mol N2 )
o
3.00 L@ 93 C, 1 atm
n 2 [mol A(v)] 85o C, P(atm)
n 1 (mol A(l)
n 3 [mol A(l)]
o
85o C, P(atm)
0.70 mL, 93 C
n0 =
3.00 L
n1 =
70.0 mL
b
273 K 1 mol = 0100 . mol N 2 273 + 93 K 22.4 L STP
g
b g
bg
0.90 g 1 mol = 15 . mol A l mL 42 g
Energy balance ⇒ ∆U = 0 ⇒
∑ n U$ − ∑ n U$ i
i
i
out
c.
i
=0
in
b g b gb
References: N 2 g , A l 85° C, 1 atm
g
nin U$ in nout U$ out 010 . 398 . 010 . 0 n in mol 15 . 160 n3 0 U$ in cal mol − − n2 20050
Substance N2 Al Av
bg bg
b g b g b g Ab v, 85° Cg: U$ = 20b 90 − 85g + 20,000 + 10b85 − 90g = 20050 cal mol ∆U = 0 ⇒ n b 20050 g − b010 . gb 398 . g − b15 . gb160 g = 0 ⇒ n = 0.012 mol A evaporate A l, 93° C and N 2 g, 93° C : U$ = Cv 93 − 85 A( v )
v1
0.012 mol A 42 g A
⇒ d.
v1
mol A
= 051 . g evaporate
Ideal gas equation of state
P=
bn
0
g
+ n2 RT V
=
0.112 mol 3.00 liters
b273 + 85gK
0.08206 L ⋅ atm = 1.097 atm mol ⋅ K
Raoult’s law
b
g
p ∗A 85° C = y A P =
n2 0.012 mol 1097 . atm P= = 0.117 atm 0.112 mol n0 + n2
8- 35
b= 89.3 mmHgg
8.61 (a) i)
Expt 1 ⇒
FG mIJ HVK
=
b 4.4553 − 3.2551gkg = 0.600 kg ⇒ b SG g 2.000 L
liquid
L
b
liquid
= 0.600
g
ii) Expt 2 ⇒ Mass of gas = 3.2571 − 3.2551 kg = 0.0020 kg = 2.0 g Moles of gas =
b
Molecular weight = 2.0 g
iii) Expt. 1 ⇒ n =
b liquidg
b763 − 500gmm Hg
2.000 L 273 K 363 K
1 mol = 0.0232 mol 22.4 liters STP
b g
760 mm Hg
g b0.0232 mol g = 86 g mol
2.000 liters 10 3 cm 3 1 liter
0.600 g 1 mol = 14 mol cm 3 86 g
Energy balance: The data show that Cv is independent of temperature Q = ∆U = nC v ∆T
b g
liquid
=
b g
liquid
≡ 24 J mol ⋅ K
⇒ Cv
⇒ Cv
Q 800 J = = 24 J mol ⋅ K@ 284.2 K n∆T 14 mols 2.4 K
b gb g 800 J = b14 molsgb 2.4 Kg = 24 J mol ⋅
[email protected] K
bg
Expt. 2 ⇒ n = 0.0232 mol from ii
b vapor g
L MN
T
Cv = a + bT ⇒ Q = 0.0232 T 2 (a + bT ) dT = 0.0232 a ( T2 − T1) + 1
b 2 (T2 − T12 ) 2
b LM OU (3669 . − 363.0 ) P | 2 N Q |V ⇒ a = −4.069 b L O b = 0.05052 1.30 J = 0.0232 Ma(492.7 - 490.0) + (492.7 − 490.0 )P | 2 N Q |W ⇒ bC g (J / mol ⋅ K) = −4.069 + 0.05052 T b Kg 1.30 J = 0.0232 a(366.9 - 363.0) +
2
2
2
2
v vapor
iv) Liquid: C p ≈ Cv ≡ 24 J mol ⋅ K Vapor: Assuming ideal gas behavior, C p = Cv + R = Cv + 8.314 J mol ⋅ K
b
g
bg
⇒ C p J mol ⋅ K = 4.245 + 0.05052 T K v)
Expt. 3 ⇒ T T T T
b
g
= 315K , p ∗ = 763 − 564 mm Hg = 199 mm Hg = 334K , p ∗ = 401 mm Hg = 354K , p ∗ = 761 mm Hg = 379K , p ∗ = 1521 mm Hg
8-36
O PQ
8.61 (cont’d) Plot p ∗ (log scale) vs. 1 T (linear scale); straight line fit yields −3770 ln p ∗ = + 17.28 or p ∗ = 3196 . × 10 7 exp − 3770 T T K
b
bg
g
b g
1 17 .28 − ln 760 = = 2 .824 × 10 − 3 K −1 ⇒ Tb = 354 K A T 3770 Part v b
vi) p ∗ = 760 mm Hg ⇒
vii)
∆H$ v = 3770 K ⇒ ∆H$ v = 3770 K 8.314 J mol ⋅ K ⇒ ∆H$ v = 31,300 J mol R PartA v
b g
b
gb
g
3.5 L feed 273 K 1 mol = 0.0836 mol s feed gas s 510 K 22.4 l STP Let A denote the drug
b g
(b) Basis:
.
0.0836 mol/s @ 510 K 0.20 A 0.80 N 2
n 1 [mol A(v)/s] n. 2 [mol N 2/s] T(K), saturated with A
.
Q(kW)
n 3 (mols A(l )/s), 90% of A in feed T(K)
b
gb g 90% condensation: n& = b0.900gb0.200 × 0.0836 g = 0.01505 mol Abl g s n& = b0.100gb0.200 × 0.0836g = 1.67 × 10 mol Abv g s N 2 balance: n&2 = 0.800 0.0836 mol s = 0.0669 mol N 2 s 3
−3
1
Partial pressure of A in outlet gas: pA =
b
n&1 1.67 × 10 −3 mol P= ( 760 mm Hg) = 18.5 mm Hg = p∗A T n&1 + n& 2 0.0686 mol
bg
g
E Part (a) - (v) 1 17.28 − lnb18.5g = = 3.81 × 10
T
−3
K −1
3770
⇓ T = 262 K
bg
(c) Reference states: N2 , A l at 262 K substance n&in H$ in n& out H$ out N2 0.0669 7286 0.0669 0 n& in mol s −3 Av 0.0167 37575 1.67 × 10 31686 H$ in J mol Al − − 0.01505 0
bg bg
8-37
8.61 (cont’d)
b
g
N 2 510 K : H$ N2 (510 K) - H$ N 2 ( 262 K) = H$ N 2 (237o C) - H$ N 2 (−11o C) Table B.8
B
= [6.24 - (-1.05)] kJ / mol = 7.286 kJ / mol = 7286 J / mol
b
g
b
g
A(v, 262K): H$ = C pl Tb − 262 + ∆H$ v 359K + Part (a) results for Tb , C pl , C pv , ∆H$ v
262 Tb
Cpv dT
L T O H$ = 24b354 − 262g + 31300 + M4.245 + 0.05052 P = 31686 J mol 2 Q N A(v, 510K): H$ = C bT − 262g + ∆H$ b354K g + C dT = 37575 J mol 2
262
354
510
pl
b
v
Tb
pv
−1060 J s 1 kW cooling Energy balance: Q& = ∆H& = ∑ ni H$ i − ∑ ni H$ i = = 1.06 kW −103 kJ s out in
8.62 a. Basis: 50 kg wet steaks/min D.M. = dry meat m1 (kg H 2 O(v )/min) (96% of H 2 O in feed) 60°C
50 kg/min @ –26°C 0.72 H 2O( s) 0.28 D.M. Q(kW)
m2 (kg D.M./min) m3 (kg H 2 O(l )/min) 50°C
96% vaporization: m& 1 = 0.96 0.72 × 50 kg min = 34 .56 kg H 2 O v min
b g bg m& = 0.04b0.72 × 50 kg ming = 1.44 kg H O bl g min Dry meat balance: m& = b0.28gb50g = 14.0 kg D.M. min Reference states: Dry meat at −26° C , H Ob l, 0° Cg 3
2
2
2
substance
m& in
H$ in
m & out
H$ out
dry meat
14.0
0
14.0
105
−
−
b
H 2 O s, − 26° C
g
36.0 −390
m & in kg min H$ in kJ kg
b g − − 1.44 209 H Obv , 60° Cg − − 34.56 2599 1.38 kJ 76° C Dry meat: H$ b50° Cg = C 50 − b− 26g = = 105 kJ kg kg ⋅ C° H Obs, − 26° Cg: H Obl , 0° Cg → H Obs, 0° Cg → H Obs, − 26° Cg H 2 O l , 50° C 2
p
2
2
2
8-38
2
8.62 (cont’d)
z
b g
−26
∆H$ = − ∆H$ m 0° C + C p dT =
−6.01 kJ
1 mol 10 3 g 2.17 kJ −26° C = −390 kJ kg 18.02 g 1 kg + kg⋅° C
mol
A
0
Table B.1
b
g
b
g b b50 − 0g° C
H 2 O l, 50° C : H 2 O l , 0° C → H 2 O l , 50° C
z
0.0754 kJ 50 $ ∆H = C pdT = mol ° C
1 mol
b
g
b
1000 g 1 kg = 209 kJ kg
18.02 g
A Table B.2
0
g
g
b
g
b
g
b
H 2 O v , 60° C : H 2 O l , 0° C → H 2 O l , 100° C → H 2 O v , 100° C → H 2O v, 60° C 0.0754 kJ ∆H$ = mol⋅° C
b100 − 0g° C
A
+ 40.656
Ad
Table B.2
=
kJ + mol
Table B.1 ∆H$ v
46.830 kJ
1 mol
1000 g
mol
18.02 g
1 kg
zd 60
100
i
Cp
i
H 2 O(v)
A
g
dT
Table B.2
= 2599 kJ kg
Energy balance: Q = ∆H =
∑ m H$ − ∑ m H$ i
out
i
i
i
=
1.06 × 105 kJ 1 min min
in
60 s
1 kW 1 kJ s
= 1760 kW
8.63 Basis: 20,000 kg/h ic e crystallized. S = solids in juice. W = water .
Qf
preconcentrate . . m 1 (kg/h) juice m 2 (kg/h) freezer 0.12 solids(S) x 2 (kg S/kg) 0.88 H 2O( l )(W) (1 – x 2) (kg W/kg) 20°C
.
Slurry(10% ice), –7°C m 5 (kg/h) product filter 20,000 kg W(s )/h 0.45 kg S/kg . m4 kg residue/h 0.55 kg W/kg 0.45 kg S/kg 20,000 kg W(s )/h . 0.55 kg W( l )/kg m 4 (kg/h), 0.45 S, 0.55 W . m 3 (kg/h), 0°C separator 0.45 kg S/kg 20,000 kg W( s )/h 0.45 kg W( l )/kg
20000 10 = ⇒m & 4 = 180000 kg h concentrate leaving freezer &4 m 90 & 1 = 27273 kg h feed m Overall S balance: 012 . m& 1 = 0.45m &5 ⇒ & m5 = 7273 kg h concentrate product &1 = m & 5 + 20000 Overall mass balance: m
(a) 10% ice in slurry ⇒
UV W
Mass balance on filter: 20000 + m& 4 + m& 5 + 20000 + m& 6
⇒
& 4 =180000 m m & 5 = 7273
m& 6 = 172730 kg h recycle
Mass balance on mixing point: & 2 = 2.000 × 105 kg h preconcentrate 27273 + 172730 = m& 2 ⇒ m
8-39
8.63 (Cont’d) S balance on mixing point: 0.12 27273 + 0.45 172730 = 2.000 × 10 5 X 2 ⇒ X 2 ⋅ 100% = 40.5% S
b gb
g b gb
g
(b) Draw system boundary for every balance to enclose freezer and mixing point (Inputs: fresh feed and recycle streams; output; slurry leaving freezer)
bg
Refs: S, H2 O l at −7° C & out substance m& in H$ in m H$ out & kg h 12% soln 27273 108 − − m $ 45% soln 172730 28 180000 0 H kJ kg H 2O s − − 20000 −337
b b
bg
bg b
g g
b g g b g
Solutions : H$ T = 4.00 T − −7 kJ kg Ice: H$ = − ∆H$ m −T ° C ≈ − ∆H$ m 0° C = −6.0095 kJ mol ⇒ −337 kJ kg
D Table B.1
−1.452 × 10 7 kJ 1h 1 kW E.B. Q& c = ∆H& = ∑ m & i H$ i − ∑ m & i H$ i = = −4030 kW h 3600 s 1 kJ s out in
d
8.64 a. B=n-butane, I=iso-butane, hf=heating fluid. ( C ) = 2.62 kJ / kg⋅o C p hf 24.5 kmol/h @ 10o C, P (bar) 0.35 kmol B(l)/h
i
24.5 kmol/h @ 180o C 0.35 kmol B(l)/h
Q& ( kW) o m& (kg HF / h), T( C)
m& (kg HF / h), 215 o C
From the Cox chart (Figure 6.1-4)
d
i
d
i
p *B 10 o C = 22 psi, pI* 10 o C = 32 psi
FG 1.01325 bar IJ = 196 H 14.696 psi K . bar
p min = p B + p I = x B p *B + x I p I* = 28.5 psi
b.
d
i
$
d
i
$
d
i
Hv H1 B l, 10 o C ∆ → B v, 10 o C ∆ → B v, 180 o C
d
i
$
d
i
$
d
Hv H2 I l, 10 o C ∆ → I v, 10 o C ∆ → I v, 180o C
i
Assume temperature remains constant during vaporization. Assume mixture vaporizes at 10o C i.e. won’t vaporize at respective boiling points as a pure component.
8-40
8.64 (cont’d) References: B(l, 10o C), I(l, 10o C) substance n& in mol / h H$ in kJ / mol B (l) 8575 0 B (v) --I (l) 15925 0 I (v) ---
b
z z
g
b
b
g
g
n& out mol / h -8575 -15925
b
H$ out kJ / mol -42.21 -41.01
g
d H$ i = d∆H$ i + d C i = 42.21 kJ / mol d H$ i = d∆H$ i + dC i = 41.01 kJ / mol ∆H = ∑ n H$ − ∑ n H$ = 8575b42.21g − 15825b41.01g out
v
B
out
v
I
i
B
p
10 180
I
p
10
i
out
180
i
B
I
i
in
∆H& = 1.015 × 10 6 kJ / h
c.
d
& hf 2.62 kJ / kg⋅o C Q = 1015 . × 10 6 kJ / h = m
i b215 − 45g C o
m& hf = 2280 kg / h
d.
. × 10 b2540 kg / hg 2.62 kJ / d kg⋅ Ci b215 − 45g C = 1131 o
o
6
kJ / h
Heat transfer rate = 1131 . × 10 6 − 1015 . × 10 6 = 116 . × 10 5 kJ / h e. The heat loss leads to a pumping cost for the additional heating fluid and a greater heating cost to raise the additional fluid back to 215o C. f. Adding the insulation reduces the costs given in part (e). The insulation is probably preferable since it is a one-time cost and the other costs continue as long as the process runs. The final decision would depend on how long it would take for the savings to make up for the cost of buying and installing the insulation.
8.65 (a) Basis: 100 g of mixture, SGBenzene=0.879: SG Toluene=0.866 50 g 50 g + = (0.640 + 0.542 ) mol = 1183 . mol 78.11 g / mol 92.13 g / mol 50 g 50 g = + = 114.6 cm 3 3 3 0.879 g / cm 0.866 g / cm
n total = Vtotal
dx i f
C6 H 6
=
0.640 mol C 6H 6 = 0.541 mol C6 H6 mol 1.183 mol
Actual feed:
32.5 m 3 106 cm 3 h
1 m
3
1183 . mol mixture 3
1h
114.6 cm mixture 3600 s
= 93.19 mol / s
T = 90° C ⇒ p ∗C6 H 6 = 1021 mm Hg , pC∗ 7 H8 = 407 mm Hg (from Table 6.1-1)
b
gb g b
gb g
Raoult's law: p tot = x C 6 H6 pC∗ 6 H6 + x C7 H8 p ∗C7 H8 = 0.541 1021 + 0.459 407 =
739.2 mmHg
1 atm = 0.973 atm ⇒ P0 > 0.973 atm 760 mmHg
8-41
8.65 (cont’d) (b) T = 75° C ⇒ pC∗ 6 H6 = 648 mm Hg , pC∗ 7 H8 = 244 mm Hg (from Table 6.1-1)
b
gb g b
gb g
Raoult's law ⇒ p tank = xC 6 H6 pC∗ 6 H6 + xC7 H 8 p ∗C7 H 8 = 0.439 648 + 0.561 244
b
g
= 284 + 137 mm Hg = 421 mmHg ⇒ Ptank = 0.554 atm
yC 6 H6 =
bg
284 mm Hg = 0.675 mol C6 H 6 v mol 421 mm Hg n v (mol/s), 75°C 0.675 C 6H 6 (v ) 0.554 atm 0.325 C 7H 8 (v ) n L (mol/s), 75°C 0.439 C6 H6 (l ) 0.541 C7H8 (l )
93.19 mol/s 0.541 C 6H 6 (l ) 0.459 C 7 H 8 (l ) 90°C, P0 atm
UV W
nv = 40.27 mol vapor s Mole balance: 93.19 = nv + n L ⇒ n L = 52.92 mol liquid s C 6H 6 balance: 0.541 93.19 = 0.675nv + 0.439 n L
b
gb
g
bg
bg
(c) Reference states: C 6H 6 l , C6 H6 l at 75° C Substance
n&in
H$ in
H$ out
n& out
b g − − 27.18 31.0 n& in mol s C H bl g 50.41 2.16 23.23 0 H$ in kJ mol C H b vg − − 13.09 35.3 C H bl g 42.78 2.64 29.69 0 C H bl , 90° Cg: H$ = b0144 . gb90 − 75g = 2.16 kJ mol C H bl , 90° Cg: H$ = b 0176 . gb90 − 75g = 2.64 kJ mol C H bv , 75° Cg: H$ = b0.144gb801 . − 75g + 30.77 + 0.074 + 0.330 × 10 A C6H6 v 6
6
7
8
7
8
6
6
7
8
6
6
75
b
∆H$ v 80 .1 °C
80.1
g
−3
T dT
= 31.0 kJ mol
b
g
b
gb
g
C 7 H 8 v , 75° C : H$ = 0.176 110.6 − 75 + 33.47 +
75 110.6
0.0942 + 0.380 × 10 −3 T dT
= 35.3 kJ mol
Energy balance: Q& = ∆H& =
∑ n& H$ − ∑ n& H$ i
out
i
i
in
i
=
1082 kJ 1 kW = 1082 kW s 1 kJ s
(d) The feed composition changed; the chromatographic analysis is wrong; the heating rate changed; the system is not at steady state; Raoult’s law and/or the Antoine equation are only approximations; the vapor and liquid streams are not in equilibr ium. (e) Heat is required to vaporize a liquid and heat is lost from any vessel for which T>Tambient. If insufficient heat is provided to the vessel, the temperature drops. To run the experiment isothermally, a greater heating rate is required.
8-42
8.66 a. Basis: 1 mol feed/s nV mol vapor/s @ T, P 1 mol/s @ TFo C
y mol A/mol (1-y) mol B/mol
xF mol A/mol (1-xF) mol B/mol
nL mol vapor/s @ T, P
vapor and liquid streams in equilibrium
x mol A/mol (1-x) mol B/mol
bg b g bg x ⋅ p bT g = y ⋅ P = x ⋅ p bTg ⇒ y =
Raoult's law ⇒ x ⋅ p ∗A T + 1 − x ⋅ p ∗B T = P ⇒ x =
pA
bg bT g − p bT g
P − p ∗B T p ∗A
(1)
∗ B
∗ A
∗ A
(2)
P
Mole balance: 1 = n& L + n&V ⇒ n&V = 1 − n& L
(4)
b gb g
for n v from (4) A balance: x F 1 = y ⋅ n&V + x ⋅ n& L Substitute → n& L =
Energy balance: ∆H& =
∑ n& H$ − ∑ n& H$ i
i
i
out
in
A 6.84471 6.88555
B 1060.79 1175.82
C 231.541 224.867
xF Tf(deg.C) P(mm Hg) HAF(kJ/mol) HBF(kJ/mol)
0.5 110 760 16.6 18.4
0.5 110 1000 16.6 18.4
0.5 150 1000 24.4 27.0
T(deg.C) pA*(mm Hg) pB*(mm Hg) x y nL(mol/s) nV(mol/s) HAL(kJ/mol) HBL(kJ/mol) HAV(kJ/mol) HBV(kJ/mol) DH(kJ/s)
51.8 1262 432 0.395 0.656 0.598 0.402 5.2 5.8 31.4 42.4 -0.01
60.0 1609 573 0.412 0.663 0.649 0.351 6.8 7.6 32.5 43.7 -0.01
62.3 1714 617 0.349 0.598 0.394 0.606 7.3 8.0 32.8 44.1 -0.01
i
y − xF y− x
(3)
=0
(5)
b. Tref(deg.C) = 25 Compound n-pentane n-hexane
al 0.195 0.216
8-43
av 0.115 0.137
bv 3.41E-04 4.09E-04
Tbp 36.07 68.74
DHv 25.77 28.85
8.66 (cont’d) c. C* C* 1 C* C* C*
C* 2
20 25 3 30
PROGRAM FOR PROBLEM 8.66 IMPLICIT REAL (N) READ (5, 1) A1, B1, C1, A2, B2, C2 ANTOINE EQUATION COEFFICIENTS FOR A AND B FORMAT (8F10.4) READ (5, 1) TRA, TRB ABRITARY REFERENCE TEMPERATURES (DEG.C.) FOR A AND B READ (5, 1) CAL, TBPA, DHVA, CAV1, CAV2 READ (5, 1) CBL, TBPB, DHVB, CBV1, CBV2 CP(LIQ, KS/MBL-DEG.C.), NORMAL BOILING POINT (DEG.C), HEAT OF VAPORIZATION (KJ/MOL), COEFFICIENTS OF CP(VAP., KJ/MOL-DEG.C) = CV1 + CV2*T(DEG.C) READ (5, 1) XF, TF, P MOLE FRACTION OF A IN FEED, FEED TEMP.(DEG.C), EVAPORATOR PRESSURE (MMHG) WRITE (6, 2) TF, XF, P FORMAT (1H0, 'FEEDbATb', F6.1, 'bDEG.CbCONTAINSb', F6.3,' bMOLESbA/MOLEbT *OTAL'//1X'EVAPORATOR bPRESSUREb=', E11.4, 'bMMbHG'/) ITER = 0 DT = 0.5 HAF = CAL*(TF – TRA) HBF = CBL*(TF – TRB) F1 = XF*HAF + (1.0 – XF)*HBF F2 = CAL*(TBPA – TRA) + DHVA – CAV1*TBPA – 0.5*CAV2*TBPA**2 F3 = CBL*(TBPB – TRB) + DHVB – CBV1*TBPB – 0.5*CBV2*TBPB**2 T = TF INTER = ITER + 1 IF(ITER – 200) 30, 30, 25 WRITE (6, 3) FORMAT (1H0, 'NO CONVERGENCE') STOP PAV = 10.0** (A1 – B1/(T + C1)) PAV = 10.0** (A2 – B2/(T + C2)) XL = (P – PBV)/(PAV – PBV) XV = XL*PAV/P NL = (XV – XF)/(XV – XL) NV = 1.0 – NL IF (XL.LE.00.OR.XL.GE.1.0.OR.NL.LE.0.0.OR.NL.GE.1.0) GO TO 45 HAL = CAL*(T – TRA) HBL = CBL*(T – TRB) HAV = F2 + CAV1*T + 0.5*CAV2*T**2 HBV = F3 + CBV1*T + 0.5*CBV2*T**2
8-44
8.66(cont’d)
DELH = NL *(XL*HAL + (1.0 – XL)*HBL) + NV*(XV*HAV + (1.0 – XV)*HBV) – F1 WRITE (6, 4) T, NL, NV, DELH 4 FORMAT (1H b, 5X' Tb=', F6.1, 3X' NLb=', F7.4, 3X' NVb=', F7.4, 3X'DELHb =',* E11.4) WRITE (6, 5) PAV, PBV, XL, HAL, HBL, XV, HAV, HBV 5 FORMAT (1H b, 5X' PAV, PBVb=', 2F8.1, 3X' XL, HAL, HBLb=', F7.4, 2E13.4,3X' XV, HAV, HBVb=', F7.4, 2E13.4/) IF (DELH) 50, 50, 40 40 DHOLD = DELH TOLD = T 45 T = T – DT GO TO 20 50 T = (T*DHOLD – TOLD*DELH)/(DHOLD – DELH) PAV = 10.0**(A1 – B1/(T + C1)) PBV = 10.0**(A2 – B2/(T + C2)) XL = (P – PBV)/(PAV – PBV) XV = XL * PAV/P NL = (XV – XF)/(XV – XL) NV = 1.0 – NL WRITE (6, 6) T, NL, XL, NV, XV 6 FORMAT (1H0, 'PROCEDUREbCONVERGED'//3X'EVAPORATORb TEMPERATUREb=', F6. *1//3X' LIQUIDbPRODUCTb--', F6.3, 'bMOLEbCONTAININGb', F6.3, 'bMOLEbA/ *MOLEbTOTAL'//3X' VAPORbPRODUCTb--', F6.3, MOLEbCONTAININGb,' F6.3, *'bMOLEbA/MOLEb TOTAL') STOP END $DATA (Fields of 10 Columns) 6.84471 1060.793 231.541 6.88555 1175.817 224.867 25.0 25.0 0.195 36.07 25.77 0.115 0.000341 0.216 68.74 28.85 0.137 0.000409 0.500 110.0 760.0 Solution: Tevaportor = 52.2° C
d
i
d
i
n L = 0.552 mol, xC 5H 12 n v = 0.448 mol, x C5 H12
liquid
= 0.383 mol C5 H 12 mol liquid
vapor
= 0.644 mol C5 H 12 mol liquid
8-45
8.67 Basis:
2500 kmol product 1 kmol condensate = 10,000 kmol h fed to condenser h .25 kmol product
. m m& 11(kg/h) , (kg/h)atatT1 T1 1090 kmol/h C3 H8 (v ) 7520 kmol/h i -C 4H10 ( v) 1390 kmol/h n -C 4H10 (v ) saturated vapor at T f, P
1090 kmol/h C 3 H 8 ( l ) 7520 kmol/h i -C H ( l ) P (mm Hg) 1390 kmol/h -C 4H10 ( ) n 4 10 l T out
.
m1 (kg/h) at 2 T2
(a) Refrigerant: Tout = 0 o C , T1 = T2 = −6o C . Antoine constants C 3H 8 i − C 4H 10 n − C 4 H 10
A 7.58163 6.78866 6.82485
B 1133.65 899.617 943.453
C 283.26 241.942 239.711
Calculate P for Tout = Tbubble pt. P=
∑ x p ( 0°C ) = 0.109( 3797 mm Hg ) + 0.752(1176 mm Hg ) + 0.139( 775 mm Hg) i
* i
i
⇒ P = 1406 mm Hg Dew pt. T f = Tdp ⇒ f (T f ) = 1 − P
∑ p (T ) = 0 yi
i
* i
trial & error to find T f
f
Substitute Antoine expressions, use E-Z Solve ⇒ T f = 5.00°C
bg
bg
Refs: C 3H8 l , C 4H10 l at 0 °C, Refrigerant @ –6°C substance C 3H 8 i − C 4H 10 n − C 4 H 10
Refrigerant
n&in
H$ in
n&out
1090 19110 1090 7520 21740 7520 1390 22760 1390 m& 1
0
m& 1
H$ out
b g
Assume: ∆H$ v Tb , Table B.1 ↓
0 0 0
n& (kmol/h) H$ (kJ/kmol)
151
m& (kg/h) H$ (kJ/kmol)
U| H$ b vapor g = ∆H$ b0° Cg + V C dTbTable B.2g |W 2
z
v
4 .95
p
0
UVH$ = ∆H$ W
v
E.B.: ∆H = ∑ n&i H$ i − ∑ n&i H$ i = 0 ⇒ 151m& 1 − 2.16 × 106 = 0 ⇒ m& 1 = 1.43 × 106 kg h refrigerant out
in
8-46
8.67 (cont’d) (b) Cooling water: Tout = 40° C , T2 = 34 °C , T1 = 25°C P=
∑ x p ( 40°C) = 0.109 (11877 ) + 0.752( 3961) + 0.139( 2831) = 4667 mm Hg i
* i
i
( )
f Tf = 1 − P
bg
∑ ( ) i
E-Z Solve yi = 0 ⇒ T f = 45.7° C p*i T f
bg
bg
Refs: C 3H8 l , C 4H10 l @ 40°C, H2 O l @ 25°C. & 1 = 5.74 × 10 6 kg H 2 O / h ∆H& = 0 ⇒ 37 .7m& 1 − 2.17 × 108 = 0 ⇒ m (c) Cost of refrigerant pumping and recompression, cost of cooling water pumping, cost of maintaining system at the higher pressure of Part (b). 8.68 Basis: 100 mol leaving conversion reactor H 2 O(v ) 3.1 bars, sat'd
n3 (mol O 2 ) 3.76n 3 (mol N 2 )
H 2 O( l ) 45°C
conversion 100 mol, 600°C, 1 atm 145°C 100°C reactor 0.199 mol HCHO/mol n4 (mol H 2 O(v )) 0.0834 mol CH 3OH/mol 0.303 mol N 2/mol n 1 (mol CH 3 OH(l )) n 2 (mol CH3 OH(l )) 0.0083 mol O 2/mol 0.050 mol H 2/mol m w1 (kg H 2 O(l )) m w2 (kg H 2 O(l )) n 8 (mol CH3 OH(l )) 0.356 mol H 2 O( v)/mol 3.1 bars, sat'd 30°C Q(kJ) CH 3 OH(l ), 1 atm, sat'd
2.5n8 (mol CH3 OH) n 6a (mol HCHO) distillation absorption (l ) n 6b (mol CH3 OH(l )) sat'd, 1 atm n 6c (mol H2 O(l )) Product solution 88°C, 1 atm n 7 (mol) 0.37 g HCHO/g ( x 1 mol/min) Absorber off-gas m w3 (kg H 2 O( l )) 0.01 g CH 3OH/g ( x 2mol/min) n5 a (mol N 2 ) 30°C 20oC 0.82 g H 3 O/g (x 3 mol/min) n5 b (mol O 2 ) n5 c (mol H 2 ) n5 d (mol H 2 O(v )), sat'd n5 e (mol HCHO(v )), 200 ppm 27°C, 1 atm
a. Strategy C balance on conversion reactor ⇒ n 2 , N 2 balance on conversion reactor ⇒ n 3 H balance on conversion reactor ⇒ n 4 , (O balance on conversion reactor to check consistency) N 2 balance on absorber ⇒ n5a , O 2 balance on absorber ⇒ n5b H 2 balance on absorber ⇒ n5e H 2 O saturation of absorber off - gas ⇒ n5d , n 5b 200 ppm HCHO in absorber off - gas
UV W
8-47
8.68 (cont’d) HCHO balance on absorber ⇒ n6a , CH 3OH balance on absorber ⇒ n6b Wt. fractions of product solution ⇒ x 1 , x 2 , x 3 HCHO balance on distillation column ⇒ n 7 CH 3OH balance on distillation column ⇒ n8 CH 3OH balance on recycle mixing point ⇒ n1 Energy balance on waste heat boiler ⇒ m w1 , E.B. on cooler ⇒ mw2 Energy balance on reboiler ⇒ Q C balance on conversion reactor: n 2 = 19.9 mol HCHO + 8.34 mol CH 3 OH = 28.24 mol CH 3OH N 2 balance on conversion reactor: 3.76n 3 = 30.3 ⇒ n 3 = 8.06 mol O 2 , 3.76 × 8.06 = 30.3 mol N 2 feed
H balance on conversion reactor:
bg
bg
bg
bg bg
bg
n 4 2 + 28.24 4 − 19.9 2 + 8.34 4 + 5 2 + 35.6 2 ⇒ n 4 = 20.7 mol H 2O fed
O balance: 65.1 mol O in, 65.5 mol O out. Accept (precision error) N 2 balance on absorber: 30.3 = n 5a ⇒ n5a = 30.3 mol N 2 O 2 balance on absorber: 0.83 = n5b ⇒ n5b = 0.83 mol O 2 H 2 balance on absorber: 5.00 = n5c ⇒ n5c = 5.00 mol H 2 H 2 O saturation of off - gas:
b
g LM N
p w* 27° C n5d 26.739 mm Hg yw = = = P 760 mm Hg 30.3 + 0.83 + 5.00 + n5d + n 5e
g U| | 200 ppm HCHO in off gas: V| n 200 ⇒ = 2| 36.13 + n + n |W 10 b
OP Q
⇒ n5d = 0.03518 3613 . + n5d + n5e 1
solve
⇒
n5d = 1.318 mol H 2 O n5e = 7.49 × 10
−3
mol HCHO
5e
6
5d
5e
Moles of absorber off-gas = n5a + n5b + n 5c + n 5e = 37.46 mol off - gas HCHO balance on absorber: 19.9 = n 6a + 7 .49 × 10 −3 ⇒ n 6a − 19.89 mol HCHO CH 3 OH balance on absorber: 8.34 = n 6b ⇒ n 6b = 8.34 mol CH 3OH Product solution
U| |V || W
%MW
Basis-100 g ⇒ 37.0 g HCHO ⇒ 1.232 mol HCHO x 1 = 0.262 mol HCHO mol 1.0 g CH 3 OH ⇒ 0.031 mol CH 3OH ⇒ x 2 = 0.006 mol CH 3OH mol 62.0 g H 2 O ⇒ 3.441 mol H 2 O x 3 = 0.732 mol H 2 O mol
8-48
8.68 (cont’d) HCHO balance on distillation column (include the condenser + reflux stream within the system for this and the next balance): 19.89 = 0.262 n 7 ⇒ n7 = 75.9 mol product
CH 3 OH balance on distillation column:
b g
8.34 = 0.006 75.9 + n8 ⇒ n 8 = 7.88 mol CH 3OH CH 3 OH balance on recycle mixing point: n1 + n8 = n2 ⇒ n1 = 28.24 − 7.83 = 20.36 mol CH 3 OH fresh feed
Summary of requested material balance results:
bg
n1 = 20.4 mol CH 3 OH l fresh feed n 2 = 75.9 mol p roduct solution
bg
n 3 = 7.88 mol CH 3 OH l recycle n 4 = 37.5 mol a bsorber off - gas
Waste heat boiler:
b
g
b
g
bg
Refs: HCHO v, 145° C , CH 3 OH v ,145° C ; N 2 , O 2 , H 2 , H 2 O v at 25°C for product
b
g
gas, H 2 O l, triple point for boiler water substance
n in
H$ in
n out
H$ out
HCHO CH 3OH
19.9 8.34 30.3 0.83 5.0 35.6
22.55 32.02 17.39 18.41 16.81 20.91
19.9 8.34 30.3 0.83 5.0 35.6
0 0 n (mol) 3.51 3.60 H$ (kJ/mol) 3.47 4.09
m w1
566.2
m w1 2726.32 m (kg) H$ (kJ/kg)
N2 O2 H2 H2O H2O (boiler)
E.B. ∆H =
∑ n H$ − ∑ n H$ i
out
i
i
i
UVH$ = C dT W U| |VH$ = C bT g T − 25 | |W UVH$ from steam tables W
z T
p
145
p
= 0 ⇒ −1814 + 2160mw1 = 0 ⇒ mw1 = 0.84 kg 3.1 bar steam
in
8-49
8.68 (cont’d)
b
g
Gas cooler: Same refs. as above for product gas, H 2 O l, 30° C for cooling water substance
n in
H$ in
n out
H$ out
HCHO CH 3OH
19.9 8.34 30.3 0.83 5.0 35.6
0 0 3.51 3.60 3.47 4.09
19.9 8.34 30.3 0.83 5.0 35.6
–1.78 –2.38 2.19 2.24 2.16 2.54
m w2
0
m w2
62.76
N2 O2 H2 H2O H2O (coolant)
E.B. ∆H =
∑ n H$ − ∑ n H$ i
i
i
out
i
n (mol) H$ (kJ/mol)
m (kg) H$ (kJ/kg)
H$ = 4.184
b
g
kJ T − 30 ° C kg⋅° C
= 0 ⇒ −1581 . + 62 .6mw 2 = 0 ⇒ mw 2 = 2.52 kg cooling water
in
b gb g Q = −n ∆H$ b1 atm g = − b27.58 molgb35.27 kJ molg
Condenser: CH 3OH condensed = n8 + 2.5n8 = 35 . 7.88 = 27 .58 mol CH 3 OH condensed E.B.:
b.
v
= −973 kJ (transferred from condenser)
3.6 × 10 4 tonne / y
10 6 g
1 yr
1d
1 metric ton 350 d 24 h
. × 10 b0.37gd 4.286 × 10 i = 1586 ⇒ b0.01gd4.286 × 10 i = 4.286 × 10 b0.62gd 4.286 × 10 i = 2.657 × 10
= 4.286 × 10 6 g h product soln
6
6
g HCHO h ⇒ 5.281 × 10 4 mol HCHO h
6
6
g CH 3 OH h ⇒ 1338 mol CH 3OH h
6
6
g H 2 O h ⇒ 1.475 × 10 5 mol H 2 O h
⇒ 2.016 × 10 5 mol h ⇒ Scale factor =
U| V| |W
2.016 × 10 5 mol h = 2657 h −1 75.9 mol
8.69 (a) For 24°C and 50% relative humidity, from Figure 8.4-1, Absolute humidity = 0.0093 kg water / kg DA , Humid volume ≈ 0.856 m 3 / kg DA Specific enthalpy = (48 - 0.2) kJ / kg DA = 47.8 kJ / kg DA , Dew point = 13o C, Twb = 17 o C
(b) 24 o C (Tdb ) (c) 13o C (Dew point) (d) Water evaporates, causing your skin temperature to drop. Tskin ≈ 13 o C (Twb ). At 98% R.H. the rate of evaporation would be lower, Tskin would be closer to Tambient , and you would not feel as cold.
8-50
Vroom = 141 ft 3 . DA = dry air.
8.70
mDA = ha =
140 ft 3
lb - mol⋅ o R 29 lb m DA 1 atm = 10.1 lb m DA lb - mol 550 o R 0.7302 ft 3 ⋅ atm
0.205 lb m H 2 O = 0.0203 lb m H 2 O / lb m DA 10.1 lb m DA
From the psychrometric chart, Tdb = 90 o F, ha = 0.0903 h r = 67%
Twb = 80.5 o F
Tdew point = 77 .3o F
8.71
Tdb = 35° C Tab = 27 ° C
8.72 a.
$ = 44.0 − 0.11 ≅ 43.9 Btu / lb DA H m
⇒ h r = 55% He wins
Mass of water:
hr = 33%, h a = 0.0148 kg H 2 O kg dry air
Fig. 8.4-1
Tdb = 40° C, Tdew point = 20° C
b. Mass of dry air: m da =
c.
$ = 14.3 ft 3 / lb DA V m
⇒
Twb = 25.5° C
1 m3
2.00 L
1 kg dry air
= 2.2 × 10 −3 kg dry air 0.92 m3 ↑ from Fig. 8.4-1
3
10 L
2.2 × 10 −3 kg dry air 0.0148 kg H 2O 10 3 g 1 kg dry air
1 kg
= 0.033 g H 2 O
b g b g H$ b20° C, saturated g ≈ 57.5 kJ kg dry air (both values from Fig. 8.4-1) 2.2 × 10 kg dry air b57 .5 − 77.4g kJ 10 J ∆H = = −44 J
H$ 40° C, 33% relative humidity ≈ 78.0 − 0.65 kJ kg dry air = 77.4 kJ kg dry air
−3
3
40→ 20
kg dry air
1 kJ
d. Energy balance: closed system n=
2.2 × 10 −3 kg dry air 10 3 g 1 mol 1 kg
d
i
29 g
+
0.033 g H 2 O 1 mol 18 g
= 0.078 mol
Q = ∆U = n∆U$ = n ∆H$ − R∆T = ∆H − nR∆T = −44 J −
0.078 mol 8.314 J
b20 − 40g° C
mol ⋅ K
1K 1° C
8-51
= −31 J(23 J transferred from the air)
400 kg 2.44 kg water
= 10.0 kg water evaporates / min 97 .56 kg air 10 kg H 2O min (b) ha = = 0.025 kg H2 O kg dry air , Tdb = 50° C 400 kg dry air min
8.73 (a)
min
Fig. 8.4-1
b
g
H$ = 116 − 11 . = 115 kJ kg dry air , Twb = 33° C, hr = 32%, Tdew point = 28.5° C
(c) Tdb = 10° C , saturated ⇒ ha = 0.0077 kg H 2O kg dry air , H$ = 29.5 kJ kg dry air
(d)
b0.0250 − 0.0077g kg H O = 6.92 kg H O min condense
400 kg dry air
2
min
2
kg dry air
bg
References: Dry air at 0° C, H 2O l at 0° C substance m& in m& out H$ in H$ out Air
bg
H2 O l
b
400
115
400
29.5
—
—
6.92
42
g
b
g
m& air in kg dry air/min, m& H 2 O in kg/min
H$ air in kJ/kg dry air, H$ H 2 O in kJ/kg
H2 O l, 0° C → H 2O l , 20° C : H$ =
75.4
Q = ∆H =
J
1 mol
mol⋅° C
18 g
∑ m& Hˆ − ∑ i
i
out
(e)
in
b10 − 0g° C
1 kJ
103 g
= 42 kJ kg 10 3 J 1 kg −34027.8 kJ 1 min 1 kW m& i Hˆ i = = −565 kW min 60 s 1 kJ/s
T>50°C, because the heat required to evaporate the water would be transferred from the air, causing its temperature to drop. To calculate (Tair)in , you would need to know the flow rate, heat capacity and temperature change of the solids.
8.74 a. Outside air: Tdb = 87 ° F , hr = 80% ⇒ ha = 0.0226 lb m H 2 O lb m D.A. , H$ = 455 . − 0.01 = 455 . Btu lb D.A. m
Room air: Tdb = 75° F , hr = 40% ⇒ ha = 0.0075 lb m H 2 O lb m D.A. , H$ = 26.2 − 0.02 = 26.2 Btu lb D.A. m
Delivered air: Tdb = 55° F , ha = 0.0075 lb m H 2 O lb m D. A. ⇒ H$ = 214 . − 0.02 = 21.4 Btu lb m D.A. , V$ = 13.07 ft 3 lb m D.A. Dry air delivered:
1,000 ft 3 1 lb m D.A.
H 2 O condensed: 76.5 lb m D.A. min
min
13.07 ft 2
= 76.5 lb m D.A. min
b0.0226 − 0.0075g lb
m
H2O
lb m D.A.
8-52
= 1.2 lb m H 2 O min condensed
8.74 (cont’d) The outside air is first cooled to a temperature at which the required amount of water is condensed, and the cold air is then reheated to 55°F. Since ha remains constant in the second step, the condition of the air following the cooling step must lie at the intersection of the ha = 0.0075 line and the saturation curve ⇒ T = 49° F
b
g
References: Same as Fig. 8.4-2 [including H 2 O l, 32° F ] substance m& in H$ in m& out H$ out Air
b
g
H 2 O l, 49° F
76.5 45.5 76.5 21.4 m& air in lb m D.A./min —
—
1.2 17.0 H$ in Btu/ lb D.A. air m m& in lb /min, H$ H2 O
Q = ∆H =
b76.5g 214. − 455. + 1.2(17.0) (Btu) min
m
H2 O
in Btu/ lb m
60 min 1 ton cooling 1h −12 ,000 Btu h
= 9.1 tons cooling
b. 6 7
(76.5 lb m DA/min) hr = 40%, ha = 0.0075 lb m H 2O/lbm DA o 75F, 26.2 Btu/lb m DA
1 7
(76.5 lb m DA/min)
hr = 80%, ha = 0.0226 lb m H 2O/lb m DA
Coolerreheater
o 87F, 45.5 Btu/lb m DA
76.5 lb m DA/min ha = 0.0075 lbm H 2O/lb m DA o 55F, 21.4 Btu/lbm DA
m& H2 O (kg H2 O(l)/min)
Q& lab
Q& (tons)
Water balance on cooler-reheater (system shown as dashed box in flow chart) 1 76.5 7
Lab
lb m H 2O 6 lb m DA 0.0226 + ( 76.5 min lbm DA 7
)( 0.0075 ) = (76.5)(0.0075) + m& H O
⇒ m& H2 O = 0.165 kg H2 O condensed/min
Energy balance on cooler-reheater References: Same as Fig. 8.4-2 [including H2 O(l, 32o F)]
8-53
2
m& in Hˆ in 10.93 45.5 65.57 26.2
Substance Fresh air feed Recirculated air feed Delivered air o
Condensed water (49 F)
Hˆ out — —
21.4
—
—
76.5
—
—
0.165 17.0
& i Hˆ i − ∑ m& i Hˆ i = Q& = ∆H& = ∑ m out
m& out — —
in
m& DA in lb m dry air/min Hˆ air in Btu/lb m dry air m& H2O ( l ) in lb m /min
Hˆ HO(l) in Btu/lb m 2
−575.3 Btu 60 min 1 ton cooling = 2.9 tons min 1h -12,000 Btu/h
Percent saved by recirculating =
(9.1 tons − 2.9 tons) ×100% = 68% 9.1 tons
Once the system reaches steady state, most of the air passing through the conditioner is cooler than the outside air, and (more importantly) much less water must be condensed (only the water in the fresh feed). c. Total recirculation could eventually lead to an unhealthy depletion of oxygen and buildup of carbon dioxide in the laboratory. 8.75 Basis: 1 kg wet chips. DA = dry air, DC = dry chips Outlet air: Tdb =38o C, Twb =29o C m2a (kg DA) m2w [kg H2 O(v)]
Inlet air: 11.6 m3 (STP), Tdb =100o C m1a (kg DA)
1 kg wet chips, 19o C 0.40 kg H2 O(l)/kg 0.60 kg DC/kg
m3c (kg dry chips) m3w [kg H2 O(l)] T (o C)
(a) Dry air: m1a =
b g
11.6 m3 STP DA
1 kmol
29.0 kg
3
1 kmol
b g
22.4 m STP
= 15.02 kg DA = m 2a
Outlet air: Fig. 8.4-1 → Hˆ 2 = (95.3 − 0.48) = 94.8 kJ kg D.A. (Tdb = 38°C, Twb = 29°C) ha2 = 0.0223 kg H 2O kg D.A.
b
g
Water in outlet air: m2 w = ha2 m2a = 0.0223 15.02 = 0.335 kg H 2 O (b) H2 O balance: 0.400 kg = 0.335 kg + m3w ⇒ m3w = 0.065 kg H 2O Moisture content of exiting chips: 0.065 kg water × 100% = 9.8%< 15% ∴ meets design specification 0.600 kg dry chips + 0.065 kg water
8-54
8.75 (cont’d)
bg
(c) References: Dry air, H 2 O l , dry chips @ 0°C. H$ in
substance
min
Air H2 O l dry chips
15.02 0.400 0.600
bg
H$ out
mout
100.2 15.02 94.8 mair in kg DA, H$ air in kJ/kg DA 79.5 0.065 4.184T m in kg DC, H$ in in kJ/kg DC 39.9 0.6 2.10T
Energy Balance: ∆H =
∑m
ˆ
out Hout
Tdb = 45° C hr = 10% b. Twb = 21.0° C hr = 60%
8.76 a.
H 2 O added:
8.77
in
= 0 ⇒ −136.8 + 1.532T = 0 ⇒ T = 89.3°C
ha = 0.0059 kg H 2 O kg DA
Tdb = 26.8° C h a = 0.0142 kg H 2 O kg DA Fig. 8.4-1
15 kg air
1 kg D. A.
min
1.0059 kg air
0.92 m 3
2
1 kg D.A.
= 12.3 kg D.A. min
12.3 kg D.A. min
UV W
2
h a = 0.0050 kg H 2 O kg D.A.
saturated
Tdb = 45° C Tdew point = 4° C
b0.0142 − 0.0059g kg H O = 0.12 kg H O min
V$ = 0.92 m 3 kg D.A. , Twb = 22 ° C
Fig. 8.4-1
Outlet air: Twb = Tas = 22° C
8.78 a.
in
Tas = Twb = 210 . °C
11.3 m3 1 kg D.A.
Evaporation:
∑ m Hˆ
Fig. 8.4 -1
Inlet air: Tdb = 50° C Tdew pt. = 4° C
min
−
⇒ T = 22° C ha = 0.0165 kg H 2O kg D. A.
b0.0165 − 0.0050g kg H O = 0.14 kg H O min 2
kg D.A.
bh g
a in
Fig. 8.4-1
2
= 0.0050 kg H 2 O kg D. A.
Twb = 20.4 ° C, V$ = 0.908 m 3 kg D.A.
b g
Twb = Tas = 20.4° C, saturated ⇒ ha
out
= 0.0151 kg H 2 O kg D.A.
8-55
8.78 (cont’d) b. Basis: 1 kg entering sugar (S) solution m1 (kg D.A.) 0.0050 kg H2 O/kg DA
m1 (kg D.A.) 0.0151 kg H2 O(v)/kg
1 kg 0.05 kg S/kg 0.95 kg H2 O/kg
m2 (kg) 0.20 kg S/kg 0.80 kg H2 O/kg
b gb g b g Water balance: bm gb0.0050 g + b1gb 0.95g = bm gb0.0151g + b0.25gb 0.80g Sugar balance: 0.05 1 = 0.20 m2 ⇒ m2 = 0.25 kg 1
1
m1 = 74 kg dry air
⇒
8.79
A 1 lb mD.A. h a1 (lb mH 2O) T d = 20°F h r = 70%
Inlet air (A):
Coil bank
Outlet air (D):
1 kg D. A.
B C 1 lb mD.A. Spray 1 lb mD.A. h a2 (lb mH 2O) chamber h a3 (lb mH 2O) Td = 75°F H2 O
UV W
Tdb = 20° F hr = 70%
V=
0.908 m3
74 kg dry air
Fig. 8.4-2
UV W
Tdb = 70° F hr = 35%
Fig. 8.4-2
a. Inlet of spray chamber (B):
= 67 m 3
D 1 lb mD.A. h a3 (lb m H 2O) T d = 70°F h r = 35%
Coil bank
ha1 ≈ 0.0017 lb m H 2 O lb m D.A. V$ ≈ 12.2 ft 3 lb m D.A. h a3 = 0.0054 lb m H 2 O lb m D.A.
UV ⇒ T W
ha = 0.0017 lb m H 2 O lb m D.A. Tdb = 75° F
wb
= 49.5° F
The state of the air at (C) must lie on the same adiabatic saturation curve as does the state at (B), or Twb = 49.5° F . Thus, h = 0.0054 lb m H 2 O lb m D.A. Outlet of spray chamber (C): a ⇒ hr = 52% Twb = 49.5° F
UV W
At point C, Tdb = 58.5° F
b.
bh
a3
g
− ha1 lb m H 2O evaporate lb m DA
lb m DA
= V$A ft 3 inlet air
d
8-56
i
b0.0054 − 0.0017 g = 3.0 × 10 12.2
−4
lb m H 2 O ft 3 air
8.79 (cont’d) (20 - 6.4) Btu / lb m dry air c. QBA = ∆H = H$ B − H$ A ≅ = 1.1 Btu / ft 3 3 12.2 ft / lb m dry air (23 - 20) Btu / lb m dry air QDC = ∆H = H$ D − H$ C ≅ = 0.25 Btu / ft 3 12.2 ft 3 / lb m dry air d. 70% 52% 35%
C
D
A
B
58.5
20
70
75
8.80 Basis: 1 kg D.A. a.
1 kg D.A. h a1 (kg H2 O/kg D.A.) Tdb= 40°C, Tab = 18°C
1 kg D.A. ha2 (kg H2 O/kg D.A.) 20°C,
m w kg H2 O
Tdb = 40° C
⇒ ha1 = 0.0039 kg H 2 O kg D.A. Twb = 18° C Tdb = 20° C Outlet air: ⇒ h a2 = 0.0122 kg H 2O kg D.A. Twb = 18° C adiabatic humidification
Inlet air:
b
b gb g b gb g
g
b
g
Overall H 2 O balance: mw + 1 ha1 = 1 h a2 ⇒ m n = 0.0122 − 0.0039 kg H 2 O kg D. A. = 0.0083 kg H 2 O kg D. A.
b.
1250 kg/h T=37 o C, h r=50%
ma (lb m H2 O/h) T=15 o C, sat’d
mc (lb m H2O/h) liquid, 12°C Qc (Btu/h)
8-57
8.80 (cont’d) Inlet air:
Tdb = 37° C hr = 50%
Moles dry air: m& a =
RSh = 0.0198 kg H O kg DA TH$ = b88.5- 0.5g kJ kg DA = 88.0 kJ kg DA
Fig. 8.4-1
⇒
a1
2
1
1250 kg
1 kg DA
h
1.0198 kg
Outlet air: Tdb = 15° C, sat'd
= 1226 kg DA h
RSh = 0.0106 kg H O kg DA TH$ = 42.1 kJ kg DA 1226 kg DA b0.0198 − 0.0106 g kg H O = Fig. 8.4-1
a
⇒
2
2
Overall water balance ⇒ m &c
2
h
kg DA
= 11.3 kg H 2 O h withdrawn
bg
Reference states for enthalpy calculations: H 2 O l , dry air at 0o C. kJ ⇒ H2O ( l , 12°C ): Hˆ = kg ⋅o C Overall system energy balance: (C p )H O2 ( l ) = 4.184
Q& c = ∆H& =
∑ m& H$ − ∑ m& H$ i
i
out
=
i
∫
12
0
C p dT = 50.3 kJ/kg
i
in
LM 11.3 kg H O 50.3 kJ + 1226 kg DA b42.1 − 88g kJ OPFG 1 h IJ FG 1 kW IJ h kg DA QH 3600 s K H 1 kJ / sK N h kg H O 2
2
= −155 . kW
∆H =
8.81
8.82 a.
b.
400 mol NH 3
− 78.2 kJ mol NH3
b
g
b
= −31,280 kJ
g
b
g
HCl g , 25° C , H 2 O l , 25° C → HCl 25° C, r = 5 . Table B.11 ∆H$ = ∆H$ s 25° C, r = 5 → ∆H$ = −64.05 kJ mol HCl
b
g HClbaq, r = ∞ g → HClbr = 5g, H Obl g ∆H$ = ∆H$ b25° C, n = 5g − ∆H$ b25° C, n = ∞ g = b −64.05 + 75.14g kJ mol HCl = 11.09 kJ / mol HCl 2
s
s
8.83 Basis: 100 mol solution ⇒ 20 mol NaOH, 80 mol H2 O ⇒r=
80 mol H2O mol H2O = 4.00 20 mol NaOH mol NaOH
8-58
8.83 (cont’d)
bg
Refs: NaOH(s), H2 O l @25° C H$ in
substance
nin
NaOH s
20.0 0.0
bg H Obl g NaOH br = 4.00g 2
H$ out
−
−
n in mol − − H$ in kJ mol 20.0 − 34.43 ← n in mol NaOH
80.0 0.0 − −
b
n out
g
H$ NaOH, r = 4.00 = −34.43 kJ mol NaOH (Table B.11)
−688.6 kJ 9.486 × 10 −4 Btu ∆H = ∑ ni H$ i − ∑ ni H$ i = (20)(−34.43) = = −6532 . Btu 10 −3 kJ out in
−6532 . Btu 103 g Q= = −132.3 Btu lb m product solution 20.0 40.00 + 80.0 18.01 g 2.20462 lb m
b
g
b
g
8.84 Basis: 1 liter solution n H 2SO 4 = mtotal = nH 2O =
1 L 8 g - eq L
1 mol 2 g - eq
1 L 1.230 kg L
= 4 mol H 2 SO 4 ×
2
1000 mol H 2 O
n H 2SO4
=
46.49 mol H 2 O mol H 2 O = 11.6 4 mol H 2 SO 4 mol H 2 SO 4
d
i
d
i
d
H 2SO 4 aq, r = ∞ ,25 o C → H 2 SO 4 aq, r = 11.6, 25 o C + H 2 O l , 25 o C ∆H$ 1 = ∆H$ s (r = 11.6) − ∆H$ s (r = ∞ )
LMn N H$ bH SO , r = 11.6, 60° Cg = 2
=
4
= 46.5 mol H 2 O
18.02 kg H 2 O
n H2 O
2
= 1230 . kg solution
b1.230 − 0.392gkg H O
⇒r=
FG 0.09808 kg IJ = 0.392 kg H SO H 1 mol K
4
RS T
Table B.11
H 2S O4 ∆H 1
=
( −67.6 + 96.19 ) = 28.6
+m
z
OP Q
60
25
kJ mol H 2 SO 4
C pdT kJ
n H2 SO ( mol H 2SO 4 )
4 mol H 2 SO 4 1 4 mol H 2 SO 4
4
28.6 kJ mol H 2 SO 4
= 60.9 kJ mol H 2 SO 4
8-59
i
+
1.230 kg
3.00 kJ kg⋅° C
b60 − 25g° CUV W
d
i
8.85 2 mol H2SO 4 = 0.30 2.00 + nH 2 O ⇒ n H2 O = 4.67 mol H2 O ⇒ r = a.
For this closed constant pressure system, 2 mol H 2SO 4 Q = ∆H = n H2 SO4 ∆H$ s 25° C, r = 2.33 =
b
b. msolution =
2 mol H 2SO 4
g
98.08 g H 2SO 4 mol
+
b
g b280.6 + 150gg −88.6 kJ +
4.67 mol H 2O = 2.33 2 mol H 2SO 4
−44.28 kJ mol H 2SO 4
= −88.6 kJ
4.67 mol H2 O 18.0 g H 2 O mol
= 2802 . g
∆H = 0 ⇒ n H2 SO4 ∆H$ s 25° C, r = 2.33 + m 25 C pdT = 0
8.86 a.
Basis:
T
3.3 J
bT − 25g° C
g⋅° C
1 kJ 1000 J
= 0 ⇒ T = 87° C
e j
1 L product solution 1.12 103 g = 1120 g solution L
1 L 8 mol HCl 36.47 g HCl = 292 g HCl L mol HCl 46.0 mol H2O(l, 25°C)
8.0 mol HCl(g , 20°C, 790 mm Hg)
1 L HCl (aq)
1120 g − 292 g = 828 g H 2 O 828 g H 2 O
n=
mol = 46.0 mol H 2O 18.0 g
46.0 mol H 2O = 575 . mol H 2O mol HCl 8.0 mol HCl
Assume all HCl is absorbed Volume of gas: 8 mol 293 K
b g = 185 liter bSTP g gas feed L HCl solution
760 mm Hg 22.4 L STP
273 K
790 mm Hg
mol
b. Ref: 25°C substance H 2O l HCl g HCl n = 5.75
b
bg bg
g
nin H$ in nout 46.0 00 . − 8.0 −015 . − − − 8.0
H$ out − − −59.07
n in mol $ H in kJ mol
8-60
8.86 (cont’d)
b
g
b
g
1 H$ HCl, n = 5.75 = ∆H$ s 25° C, n = 575 . + nHCl = −64.87 kJ mol +
e
j
H$ HCl, 20o C =
20 25
40
mC pdT 25
1120 g 0.66 cal 8 mols g⋅° C
b40 − 25g° C
4184 . J kJ cal 103 J
0.02913 − 01341 . × 10 −5 T + 0.9715 × 10−8 T 2 − 4.335 × 10−12 T 3 dT
= -0.15 kJ / mol Q = ∆ H = − 471 kJ L product
c.
e j b
Q = 0 = ∆H = 8 H$ − 8 −015 .
g b
g
o 1120 g 0.66 cal T − 25 C 4.184 J 1 kJ −015 . = H$ = − 64.87 + 8 mol g⋅ o C cal 1000 J
T = 192 o C
8.87 Basis: Given solution feed rate
.
n. a (mol air/min) . n1 (mol H 2O( v)/min) saturated @ 50°C, 1 atm
na (mol air/min) 200°C, 1.1 bars 150 mol/min solution 0.001 NaOH 0.999 H 2O 25°C
.
n2 (mol/min) @ 50°C 0.05 NaOH 0.95 H 2O
b
gb g H O balance: b0.999gb150g = n& + 0.95b3.0g ⇒ n& = 147 mol H O min n& Raoult’s law: y P = P = p b50° Cg = 92.51 mm Hg ⇒ n& n& + n& NaOH balance: 0.001 150 = 0.05n&2 ⇒ n&2 = 3.0 mol min 2
1
H 2O
2
Table B.4
∗ H 2O
1
1
1
n&1 =147 P = 760
a
b g
1061 mol 22.4 L STP V&inlet air = min 1 mol
bg
473 K 1.013 bars 273 K
1.1 bars
a
= 1061
mol air min
= 37 ,900 L min
bg ∆H$ b25° Cg = −42.47 kJ mol NaOH
References for enthalpy calculations: H 2O l , NaOH s , air @ 25° C
999 mol H2 O Table B.11 ⇒ s 1 mol NaOH 95 mol H 2O 19 mol H 2O kJ 5% solution @ 50°C: r = = ⇒ ∆H$ s 25° C = −42 .81 5 mol NaOH mol NaOH mol NaOH
0.1% solution @ 25°C: r =
b
Solution mass: m=
b
g
b
1 mol NaOH 40.0 g 19 mol H 2 O 18.0 g g solution + = 382 1 mol 1 mol mol NaOH
g
50 H$ 50° C = ∆H$ s 25° C + m 25 C pdT
= −42.81
g
382 g 4.184 J kJ + mol NaOH mol NaOH 1 g⋅° C
8-61
b50 − 25g° C
1 kJ = −2.85 kJ 10 3 J
8.87 (cont’d) Air @ 200°C: Table B.8 ⇒ H$ = 5.15 kJ mol Air (dry) @ 50°C: Table B.8 ⇒ H$ = 0.73 kJ mol
b
g
2592 − 104.8 kJ 1 kg 18.0 g H2 O v , 50° C : Table B.5 ⇒ H$ = = 44.81 kJ mol kg 10 3 g 1 mol
b
g
n&in H$ in n&out H$ out 015 . −42.47 0.15 −2.85 n& in mol min − − 147 44 .81 H$ in kJ mol 1061 5.15 1061 0.73
substance NaOH aq H2 O v Dry air
b g bg
Energy balance: Q& = ∆H& = ∑ ni H$ i − ∑ ni H$ i = 1900 kJ min transferred to unit
b neglect ∆E g
out
n
in
8.88 a. Basis: 1 L 4.00 molar H2 SO4 solution (S.G. = 1.231) 4.00 mol H 2 SO 4 1231 − 392.3 = 8387 . g H2 O 1 L 1231 g = 1231 g ⇒ ⇒ = 392.3 g H 2SO 4 L = 46.57 mol H 2 O B.11 ⇒ r = 11.64 mol H 2 O / mol H 2SO 4 Table → ∆H$ s = −67.6 kJ / mol H 2 SO 4
b
g
b
Ref: H 2 O l , 25° C , H 2SO 4 25° C substance H2O l H 2 SO 4 l H 2 SO 4 25° C, n = 11.64
bg bg
b
b
g
n in H$ in 46.57 0.0754 T − 25 4.00 0 − −
b
g
g
b
n out H$ out − − − − 4.00 −67.6
g
gb
n in mol $ H in kJ mol
g
Q = ∆H = 0 = 4.00 −67.6 − 46.57 0.0754 T − 25 ⇒ T = −52° C
(The water would not be liquid at this temperature ⇒ impossible alternative!)
b
g
b
b. Ref: H 2 O l , 25° C , H 2SO 4 25° C substance
nin
H$ in
H2O l
nl
0.0754 0 − 25
ns 4.00
bg H O b sg 2
H 2SO 4 ( l )
b
H 2 SO 4 25° C, n = 11.64
b
g
g
g
n out
H$ out
−
−
−6.01 + 0.0754 0 − 25
−
−
0
−
−
b
b
g
g
n in mols $ H in kJ mol
4.00 −67.61
∆H$ m H 2 O, 0° C = 6.01 kJ mol
A
UV ⇒ n = 1618 . mol liquid H O ∆H = 0 = 4 .00b−67.61g − n b −1885 . g − b46.57 − n gb−7.895gW n = 30.39 mol ice ⇒ 291.4 g H Obl g + 547 .3 g H Obsg@0° C Table B.1
n l + n s = 46.57
l
l
2
l
2
8-62
s
2
8.89 P2O 5 + 3H 2O → 2H 3PO 4
B } 2n
mol H 3 PO 4
a.
wt% P2 O 5 =
b
g × 100% ,
n 14196 . mt
wt% H 3 PO 4 =
B
g H3 PO 4 mol
b98.00g × 100%
mc
A
g total
where n = mol P2 O 5 and mt = total mass . wt% H 3 PO 4 =
b
g
2 98.00 wt% P 2 O 5 = 1381 . wt% P2O 5 14196 .
b. Basis: 1 lb m feed solution 28 wt% P2O 5 ⇒ 38.67 wt% H 3PO 4 m1 (lb m H2 O(v )), T , 3.7 psia 1 lb msolution, 125°F 0.3867 lb mH 3PO 4 0.6133 lb mH 2O
m2 (lbm solution),T 0.5800 lb mH 3PO 4/lb 0.4200 lb mH 2O/lb m
m
H3 PO 4 balance: 0.3867 = 0.5800m2 ⇒ m2 0.667 lb m solution
bg
Total balance: 1 = m1 + m2 ⇒ m1 = 0.3333 lb m H 2O r
bg
Evaporation ratio: 0.3333 lb m H 2 O v lb m feed solution c. Condensate:
b
P = 3.7 psia 0.255 bar
g
Table B.6
⇒ Tsat = 654 . o C= 149 o F, Vliq =
m& =
100 tons feed day
46.3 lb m V& = min
000102 . m3 353145 . ft 3 / m 3 ft 3 = 0.0163 kg 2.205 lb m / kg lb m H 2O(l)
2000 lb m 1 lb m H 2O 1 ton
1 day (24 × 60) min
3 lb m
0.0163 ft 3
7.4805 gal
lb m
ft 3
= 46.3 lb m / min
= 5.65 gal condensate / min
Heat of condensation process:
46.3lbm H2O(v)/min
46.3 lbm H2O(l)/min
(149+37)°F, 3.7 psia
149°F, 3.7psia
.
Q (Btu/min)
8-63
8.89 (cont’d)
R| ||H$ Table B.6 ⇒ S |H$ || T
Btu F I lb G JJ = 1141 Btu / lb (186 F = 85.6 C) = (2652 kJ / kg) 0.4303 kJ GH kg K (149 F = 65.4 C) = (274 kJ / kg)b 0.4303 g = 118 Btu / lb o
H2 O (v )
o
o
H2 O (l )
m
m
o
m
LM N
lb Btu Q& = m & ∆H$ = ( 46.3 m ) (118 − 1141) min lb m
OP = −47,360 Btu / min Q
⇒ 4.74 × 10 4 Btu min available at 149 o F
bg
bg
d. Refs: H3 PO 4 l , H 2O l @77° F substance H 3PO 4 28% H 3PO 4 42% H 2O v
b g b g bg
b
min H$ in mout H$ out 1.00 13.95 − − m in lb m $ − − 0.667 34.13 H in Btu lb m − − 0.3333 1099
g
H$ H 3 PO 4 , 28% = +
g
H$ H 3 PO 4 , 42% = +
0.705 Btu lb m ⋅° F
1 lb - mole H 3 PO 3 98.00 lb m H 3 PO 4
b125 − 77g° F = 13.95 Btu lb
0.705 Btu lb m ⋅° F
b
−5040 Btu lb - mole H 3PO 4
−5040 Btu lb - mole H 3 PO 4
m
soln
1 lb - mole H 3PO 4 98.00 lb m H 3 PO 4
b186.7 − 77g° F = 34.13 Btu lb
b g b
0.3867 lb m H 3 PO 4 1.00 lb m soln
g b
m
0.5800 lb m H 3 PO 4 1.00 lb m sol.
soln
g b
g
H$ H 2O = H$ 3.7 psia, 186° F − H$ l , 77° F = 2652 − 104.7 kJ kg ⇒ 1096 Btu lb m
At 27.6 psia (=1.90 bar), Table B.6 ⇒ ∆H$ v = 2206 kJ / kg = 949 Btu / lb m ∆H = ∑ ni H$ i − ∑ ni H$ i = 375 Btu = m steam ∆H$ v ⇒ m steam = out
⇒ ⇒
in
375 Btu = 0.395 lb m steam 949 Btu / lb m
0.395 lb m steam 100 × 2000 lb m H 3 PO 4
1 day
lb m 28% H 3PO 4
24 h
day
3292 lb m steam (46.3 × 60) lb m H 2 O evaporated / h
= 118 .
8-64
= 3292 lb m steam / h
lb m steam lb m H2 O evaporated
8.90 Basis: 200 kg/h feed solution. A = NaC2 H 3O 2
.
n1 (kmol H 2O( v)/h) 50°C, 16.9% of H O 2 in feed 200 kg/h @ 60°C . n0 (kmol/h) 0.20 A 0.80 H O 2
Product slurry @ 50°C . n2 (kmol A-3H 2O( v)/h) . n3 (kmol solution/h) 0.154 A 0.896 H 2O
Q (kJ/hr)
a. Average molecular weight of feed solution: M = 0.200 M A + 0.800 M H 2 O
b
gb g b
gb g
= 0.200 82.0 + 0.800 18.0 = 30.8 kg k
Molar flow rate of feed: n 0 =
200 kg
1 kmol
h
30.8 kg
= 6.49 kmol h
b
gb gb g n b kmol A ⋅ 3 H O g A balance: b0.20gb6.49 kmol hg =
bg
b. 16.9% evaporation ⇒ n1 = 0.169 0.80 6.49 kmol h = 0.877 kmol H 2 O v h 2
E
1 mole A ⋅ 3 H 2 O
h
⇒ n2 + 0.154 n3 = 1.30
b gb
1 mole A
2
g
H 2 O balance: 0.80 6.49 kmol h = 0.877 +
(1)
b
n 2 kmol A ⋅ 3 H 2 O
g
3 moles H 2 O 1 mole A ⋅ 3 H 2 O
h
+ 0.846n 3 ⇒ 3n 2 + 0.846n 3 = 4.315
bg
+ 0154 . n3
bg
bg
b 2g
Solve 1 and 2 simultaneously ⇒ n 2 = 113 . kmol A ⋅ 3H 2 O s h n 3 = 1.095 kmol solution h
Mass flow rate of crystals
bg
1.13 kmol A ⋅ 3H 2 O 136 kg A ⋅ 3H 2 O 154 kg NaC2 H 3O 2 ⋅ 3H 2 O s = h 1 kmol h
Mass flow rate of product solution
b
gb g
bg
bg
bg
200 kg feed 154 kg crystals 0.877 18.0 kg H 2 O v − − = 30 kg solution h h h h
c.
References for enthalpy calculations: NaC2 H 3O 2 s , H 2 O l @25° C
b
g
z
Feed solution: nH$ = n A ∆H$ s 25° C + m
b g
60
25
C p dT (form solution at 25° C , heat to 60° C )
0.20 6.49 kmol A −1.71 × 10 4 kJ 200 kg 3.5 kJ nH$ = + h kmol A hr kg⋅° C
8-65
b60 − 25g° C = 2300 kJ h
8.90 (cont’d)
b
g
z
Product solution: nH$ = n A ∆H$ s 25° C + m =
b0.154g1.095 kmol A
50
25
−1.71 × 10 4 kJ 30 kg 3.5 kJ + h kg⋅° C kmol A
h = −259 kJ h
z
Crystals: nH$ = n A ∆H$ hydration + m
25
C p dT (hydrate at 25° C , heat to 50° C )
bg
1.13 kmol A ⋅ 3H 2 O s h = −36700 kJ h
b
LM N
g
=
z
50
25
b neglect ∆E g
OP Q
C p dT (vaporize at 25° C , heat to 50° C )
b gb
i
g
4.39 × 10 4 + 32.4 50 − 25 kJ
∑ n H$ − ∑ n H$ i
out
R
b50 − 25g° C
−3.66 × 10 4 kJ 154 kg 1.2 kJ + h kg⋅° C kmol
0.877 kmol H 2 O h
Energy balance: Q = ∆H =
b50 − 25g° C
50
=
H 2 O v , 50° C : n∆H = n ∆H$ v +
C p dT
i
i
b
= 39200 kJ h
g b g
= −259 − 36700 + 39200 − 2300
kJ h
in
= −60 kJ h (Transfer heat from unit)
U| mL |V ⇒ r = 5.00 mol H O mol H SO 84.2 mL H Oblg 100 . g = 84.2 g H Oblg ⇒ 4.678 mol H Oblg| |W mL
8.91 50 mL H2SO4
1834 . g
= 917 . g H2SO4 ⇒ 0.935 mol H2SO4
2
2
2
2
2
Ref: H 2 O , H 2SO 4 @ 25 °C H$ ( H 2 O(l ), 15 o C) = [0.0754 kJ / (mol ⋅o C)](15 − 25) o C = − 0.754 kJ / mol
b
g
(91.7 + 84.2) g 2.43 J kJ H$ H 2 SO 4 , r = 5.00 = −58.03 + mol 0.935 mol H 2 SO 4 g⋅° C = ( −69.46 + 0.457 T )( kJ / mol H 2SO 4 )
substance
nin
bg
H$ in
b T − 25g° C
H$ out
nout
4.678 –0.754 — — n in mol 0.935 0.0 — — H$ in kJ/mol — 0.935 −69.46 + 0.457 T n mol H 3SO 4 H2SO 4 r = 4 .00 — H2 O l H 2SO 4
b
g b
b
g
1 kJ 10 3 J
b
g
b
g
g
Energy Balance: ∆H = 0 = 0.935 −69.46 + 0.457T − 4.678 −0.754 ⇒ T = 144 ° C Conditions: Adiabatic, negligible heat absorbed by the solution container. 8-66
4
8.92 a.
mA (g A) @ TA0 (o C) n A (mol A)
n S (mol solution) @ Tmax (o C)
mB (g B) @ TB0 (o C) n B (mol B) Refs: A(l), B(l) @ 25 °C substance n in H$ in n out
H$ out
B
n A H$ A n B H$ B
—
—
S
— —
nA
H$ S (J mol A )
A
—
—
n in mol H$ in J / mol
mA (g A) m , nB = B M A (g A / mol A) MB
Moles of feed materials: n A (mol A) = Enthalpies of feeds and product
H$ A = m A C pA ( TA 0 − 25o C), H$ B = m B C pB ( TB 0 − 25o C) r (mol B mol A) = n B n A =
H$ S
FG J IJ = H mol A K n
A
mB / M B mA / M A
Ln MM 1 (mol A) M M+ MN
A ( mol
FG J IJ H mol A K F J I × (T )(g soln) × C G H g soln ⋅ C JK
A) × ∆ H$ m (r )
( mA + mB
ps
max
o
− 25)( o
O PP P C) P PQ
1 ⇒ H$ S = n A ∆ H$ m (r ) + ( m A + m B )C ps (Tmax − 25) nA
Energy balance ∆H = n A H$ S − n A H$ A − n B H$ B = 0
bg
b
g
b
g
mA $ ∆Hm r + (mA + mB )C ps (Tmax − 25) − mA CpA TA0 − 25 − mB CpB TB0 − 25 = 0 MA m m AC pA TA0 − 25 + mB C pB TB0 − 25 − A ∆H$ m r MA ⇒ Tmax = 25 + (mA + mB )Cps ⇒
b
g
b
g
bg
Conditions for validity: Adiabatic mixing; negligible heat absorbed by the solution container, negligible dependence of heat capacities on temperature between 25o C and TA0 for A, 25o C and TB0 for B, and 25o C and Tmax for the solution. b.
m A = 100.0 g
M A = 40.00 TA0 = 25° C
m B = 2250 . g
M B = 18.01 TB 0 = 40° C C pB
C ps = 3.35 J (g⋅° C)
b
g
g UV ⇒ r = 5.00 mol H O = 4.18 J (g⋅° C) |W mol NaOH b
C pA = ? irrelevant
∆H$ m n = 5.00 = −37,740 J mol A ⇒ Tmax = 125° C
8-67
2
8.93 Refs: Sulfuric acid and water @ 25 °C b.
substance
nin
H$ in
H2 SO4 H2 O H 2SO 4 aq
1 r —
M A CpA T0 − 25 M w C pw T0 − 25 —
b g
b b
g g
H$ out
nout
— — 1
— n in mol — H$ in J/mol ∆H$ m r + M A + rM w C ps Ts − 25 (J/mol H2 SO4 )
bg b
g b
bg b g b g b g = ∆H$ br g + b98 + 18rgC bT − 25g − (98C + 18 rC )bT 1 ⇒ T = 25 + (98C + 18rC )bT − 25g − ∆H$ br g (98 + 18r) C
b
g
g
∆H = 0 = ∆H$ m r + M A + rM w C ps Ts − 25 − M A C pa T0 − 25 − rM w C pw T0 − 25 m
ps
s
s
pa
pa
pw
pw
0
0
g
− 25
m
ps
c. H2 O(l) H2 SO4 r 0.5 1 1.5 2 3 4 5 10 25 50 100
Cp (J/mol-K) 75.4 185.6
Cp (J/g-K) 4.2 1.9
Cps 1.58 1.85 1.89 1.94 2.1 2.27 2.43 3.03 3.56 3.84 4
∆H$ m (r ) -15,730 -28,070 -36,900 -41,920 -48,990 -54,060 -58,030 -67,030 -72,300 -73,340 -73,970
Ts 137.9 174.0 200.2 205.7 197.8 184.0 170.5 121.3 78.0 59.6 50.0
250
Ts
200 150 100 50 0 0.1
1
10
100
r
d. Some heat would be lost to the surroundings, leading to a lower final temperature. 8-68
8.94 a.
Ideal gas equation of state n A0 = P0V g / RT0 Total moles of B: n B0 (mol B) =
b
(1)
gd
i
Vl ( L) × SG B × 1 kg / L 10 3 g / kg
(2)
M B (g / mol B)
Total moles of A: n Ao = n Av + n Al Henry’s Law: r
(3)
FG mol A(l)IJ = k p H mol B K s
A
⇒
b
n Al = c 0 + c1T nB 0
g n V RT Av
(4)
g
Solve (3) and (4) for nAl and n Av.
b
n B0 RT c0 + c1T Vg
n Al =
g
LM1 + n RT bc + c T gOP MN V PQ n = LM1 + n RT bc + c T gOP MN V PQ
(5)
B0
0
1
g
n Av
Ao
(6)
B0
0
1
g
Ideal gas equation of state P=
n Av RT ( 6) n A 0 RT = Vg Vg + nB 0 RT c0 + c1T
b
g
(7)
b g bg
Refs: A g , B l @ 298 K substance
bg Bb l g
A g
Solution
n eq
U$ eq
M A CvA T0 − 298
n Av
M A CvA T − 298
M B Cv B
—
—
n Al
U$ 1 (kJ/mol A)
n in
U$ in
n Ao n B0
—
b bT
0
g − 298g
—
b
g b
b
g
n in mol $ U in kJ/mol
g
1 U$ 1 = ∆U$ s + n Al M A + n B0 M B Cvs T − 298 n Al
E.B.: ∆U = 0 =
∑ n U$ − ∑ n U$ i
out
c
i
i
i
in
b
g hb g b d− ∆U$ i + bn C + n C gb T − 298g n C + b n M + n M gC
gb
0 = n Av CvA + n Al M A + n B M B Cvs T − 298 + n Al ∆U$ s − n AoCvA + nB CvB T0 − 298 ⇒ T = 298 +
n Al
s
Av
Ao
vA
Al
vA
B
A
8-69
vB
B
0
B
vs
g
8.94 (cont’d) b. Vt 20.0
MA 47.0
CvA 0.831
MB 26.0
CvB 3.85
SGB 1.76
Vl 3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0
T0 300 300 300 300 330 330 330 330
P0 1.0 5.0 10.0 20.0 1.0 5.0 10.0 20.0
Vg 17.0 17.0 17.0 17.0 17.0 17.0 17.0 17.0
nB0 203.1 203.1 203.1 203.1 203.1 203.1 203.1 203.1
nA0 0.691 3.453 6.906 13.811 0.628 3.139 6.278 12.555
c0 c1 0.00154 -1.60E-06 T 301.4 307.0 313.9 327.6 331.3 336.4 342.8 355.3
nA(v) 0.526 2.624 5.234 10.414 0.473 2.359 4.709 9.381
Dus -174000
Cvs 3.80
nA(l) 0.164 0.828 1.671 3.397 0.155 0.779 1.569 3.174
P 0.8 3.9 7.9 16.5 0.8 3.8 7.8 16.1
c. C*
REAL R, NB, T0, P0, VG, C, D, DUS, MA, MB, CVA, CVB, CVS REAL NA0, T, DEN, P, NAL, NAV, NUM, TN INTEGER K R = 0.08206 1 READ (5, *) NB IF (NB.LT.0) STOP READ (1, *) T0, P0, VG, C, D, DUS, MA, MB, CVA, CVB, CVS WRITE (6, 900) NA0 = P0 * VG/R/T0 T = 1.1 * T0 K=1 10 DEN = VG/R/T/NB + C + D * T P = NA0/NB/DEN NAL = (C + D * T) * NA0/DEN NAV = VG/R/T/NB * NA0/DEN NUM = NAL * (–DUS) + (NA0 * CVA + NB * CVB) * (TO – 298) DEN = NAV * CVA + (NAL * MA + NB * MB) * CVS TN = 298 + NUM/DEN WRITE (6, 901) T, P, NAV, NAL, TN IF (ABS(T – TN).LT.0.01) GOTO 20 K=K+1 T = TN IF (K.LT.15) GOTO 10 WRITE (6, 902) STOP 20 WRITE (6, 903) GOTO 1 900 FORMAT ('T(assumed) P Nav Nal T(calc.)'/ * ' (K) (atm) (mols) (mols) (K)') 901 FORMAT (F9.2, 2X, F6.3, 2X, F7.3, 2X, F7.3, 2X, F7.3, 2) 902 FORMAT (' *** DID NOT CONVERGE ***') 903 FORMAT ('CONVERGENCE'/) END $ DATA 300 291 10.0 15.0 1.54E–3 –2.6E–6 –74 35.0 18.0 0.0291 0.0754 4.2E–03
8-70
Tcalc 301.4 307.0 313.9 327.6 331.3 336.4 342.8 355.3
8.94 (cont’d) 300 291 35.0 –1
50.0 18.0
Program Output T (assumed) (K) 321.10 296.54 296.57
15.0 0.0291
1.54E–3 0.0754
–2.6E–6 4.2E–03
–74
P (atm) 8.019 7.415 7.416
Nav (mols) 4.579 4.571 4.571
Nal (mols) 1.703 1.711 1.711
T(calc.) (K) 296.542 296.568 296.568
P (atm) 40.093 39.676 39.680
Nav (mols) 22.895 22.885 22.885
Nal (mols) 8.573 8.523 8.523
T(calc.) (K) 316.912 316.942 316.942
Convergence T (assumed) (K) 320.10 316.91 316.94
8.95 350 mL 85% H2 SO 4 ma(g), 60 oF, ρ=1.78
Q=0 30% H2 SO 4 ms(g), T( oF)
H2O, Vw(mL), mw(g), 60 oF
a. Vw =
350 mL feed
178 .
g
0.85(70 / 30) − 0.15 g H 2O added 1 mL water
1 mL feed
g feed
1 g water
= 1140 mL H 2O
b. Fig. 8.5-1 ⇒ Hˆ a ≈ −103 Btu/lb m;
Water: Hˆ water ≈ 27 Btu/lb m
Mass Balance: mp =mf+mw =(350 mL)(1.78 g/mL)+(1142 mL)(1 g/mL)=623+1142=1765 g Energy Balance:
m Hˆ + mw Hˆ w ∆H = 0 = mp Hˆ product − ma Hˆ a − mw Hˆ w ⇒ Hˆ s = f f mp (623)(−103) + (1140)(27) ⇒ Hˆ product = = −18.9 Btu/lb m 1765
c. T (Hˆ = −18.9 Btu/lb m ,30%) ≈ 130o F d. When acid is added slowly to water, the rate of temperature change is slow: few isotherms are crossed on Fig. 8.5-1 when xacid increases by, say, 0.10. On the other hand, a change from xacid=1 to xacid=0.9 can lead to a temperature increase of 200°F or more.
8-71
8.96 a.
2.30 lb m 15.0 wt% H 2SO 4 $ = −10 Btu / lb @ 77 o F ⇒ H 1
m
U| || V| → m ( lb | |W adiabatic mixing
3
m2 (lb m ) 80.0 wt% H 2SO 4 $ = −120 Btu / lb @ 60 o F ⇒ H 2
$ 60.0 wt% H 2SO 4 @ T o F, H 3
U| ⇒ R|m mass balance: 2.30b0.150g + m b0.800g = m (0.600 )VW| S|m T
Total mass balance: H 2SO 4
m
m)
2.30 + m2 = m3
2
2
3
3
= 5.17 lb m (80%) = 7.47 lb m (60%)
b. Adiabatic mixing ⇒ Q = ∆H = 0
. gb −120g = 0 ⇒ H$ b7.47g H$ − b2.30gb −10g − b517 3
3
= −861 . Btu / lb m
E Figure 8.5 - 1
T = 140o F
c.
d
i
H$ 60 wt%, 77 o F = −130 Btu / lb m
d
i
b
gb
g
Q = m3 H$ 60 wt%, 77 o F − H$ 3 = 7 .475 −130 + 86.1 = −328 Btu
d. Add the concentrated solution to the dilute solution . The rate of temperature rise is much lower (isotherms are crossed at a lower rate) when moving from left to right on Figure 8.5-1. 8.97 a.
b.
x NH 3 = 0.30
Fig. 8.5-2
y NH 3 = 0.96 lb m NH 3 lb m vapor , T = 80° F
Basis: 1 lb m system mass
⇒ 0.90 lb m liquid
⇒ 010 . lb m vapor
Mass fractions: zNH 3 =
b0.27 + 0.096glb
m
NH 3
1 lb m
x NH 3 =0.30
0.27 lb m NH 3 0.63 lb m H 2 O
xNH 3 = 0.96
0.096 lb m NH 3 0.004 lb m H 2 O
= 0.37 lb m NH 3 lb m
1 − 0.37 = 0.63 lb m H2 O lb m
0.90 lb m liquid −25 Btu 0.10 lb m vapor 670 Btu Enthalpy: H$ = + = 44 Btu lb m 1 lb m 1 lb m liquid 1 lb m 1 lb m vapor
8-72
8.98
T = 140° F
Fig. 8.5-2
Vapor: 80% NH 3 , 20% H 2 O
C
Liquid: 14% NH 3 , 86% H 2 O
A B
Basis: 250 g system mass ⇒ mv (g vapor), m L (g liquid)
.14
.60
.80 xNH3
Mass Balance: mv + m L = 250 NH3 Balance: 0.80m g + 014 . mL = (0.60)( 250) ⇒ mv = 175 g, mL = 75 g
b gb g = b0.14 gb75 g g = 10.5 g NH , 64.5 g H O Liquid
Vapor: m NH3 = 0.80 175 g = 140 g NH 3 , 35 g H 2 O Liquid: mNH 3
3
2
8.99 Basis: 200 lb m feed h m& v (lb m h) xv(lb m NH3 (g)/lb m) H$ v ( Btu lb m )
200 lb m/h 0.70 lb m NH3 (aq)/lb m 0.30 lb m H2 O(l)/lb m
m& l (lb m h)
H$ f = −50 Btu lb m
in equilibrium at 80o F
xl [lb m NH3 (aq)/lb m] H$ l ( Btu l b m )
Q& ( Btu h)
Figure 8.5-2 ⇒ Mass fraction of NH 3 in vapor: xv = 0.96 lb m NH 3 lb m Mass fraction of NH3 in liquid: x l = 0.30 lb m NH 3 lb m
Specific enthalpies: H$ v = 650 Btu lb m , H$ l = −30 Btu lb m
UV W
& v = 120 lb m h vapor m &l Mass balance: 200 = m& v + m ⇒ & ml = 80 lb m h liquid & v + 0.30m& l Ammonia balance: 0.70 200 = 0.96 m
b gb g
Energy balance: Neglect ∆E& k . Q& = ∆H& =
∑ m& H$ i
out
i
120 lb m & f H$ f = −m h = 86,000
650 Btu 80 lb m + lb m h
Btu h
8-73
−30 Btu 200 lb m − lb m h
−50 Btu lb m
CHAPTER NINE 4 NH 3 ( g) + 5O2 ( g) → 4NO(g) + 6H 2O(g) ∆H$ o = −904 .7 kJ / mol
9.1
r
a.
When 4 g-moles of NH3 (g) and 5 g-moles of O2 (g) at 25°C and 1 atm react to form 4 g-moles of NO(g) and 6 g-moles of water vapor at 25°C and 1 atm, the change in enthalpy is -904.7 kJ.
b.
Exothermic at 25°C. The reactor must be cooled to keep the temperature constant. The temperature would increase under adiabatic conditions. The energy required to break the molecular bonds of the reactants is less than the energy released when the product bonds are formed.
c.
5 2 NH 3 ( g) + O 2 ( g) → 2NO(g) + 3H 2O(g) 2 Reducing the stoichiometric coefficients of a reaction by half reduces the heat of reaction by half. 904.7 ∆H$ ro = − = −452.4 kJ / mol 2 3 5 NO(g) + H 2O(g) → NH 3 ( g) + O 2 ( g) 2 4 Reversing the reaction reverses the sign of the heat of reaction. Also reducing the stoichiometric coefficients to one-fourth reduces the heat of reaction to one-fourth. ( − 904.7) ∆H$ ro = − = +2262 . kJ / mol 4
d.
e.
& NH3 = 340 g / s m 340 g 1 mol = 20.0 mol / s s 17.03 g $o n& ∆H r 20.0 mol NH 3 −904.7 kJ & = ∆H & = NH3 Q = s 4 mol NH 3 ν NH n& NH 3 =
= −4.52 × 10 4 kJ / s
3
The reactor pressure is low enough to have a negligible effect on enthalpy. f.
9.2
Yes. Pure water can only exist as vapor at 1 atm above 100°C, but in a mixture of gases, it can exist as vapor at lower temperatures.
C 9 H20 ( l) + 14O2 ( g) → 9CO 2 (g) + 10H2 O(l) ∆H$ o = −6124 kJ / mol r
a.
b.
c.
When 1 g-mole of C9 H20 (l) and 14 g-moles of O2 (g) at 25°C and 1 atm react to form 9 g-moles of CO2 (g) and 10 g-moles of water vapor at 25°C and 1 atm, the change in enthalpy is -6124 kJ. Exothermic at 25°C. The reactor must be cooled to keep the temperature constant. The temperature would increase under adiabatic conditions. The energy required to break the molecular bonds of the reactants is less than the energy released when the product bonds are formed.
& = Q& = ∆ H
$0 n& C9 H 20 ∆ H r ν C 9 H 20
=
25.0 mol C9 H 20 s
−6124 kJ 1 mol C 9 H20
9- 1
1 kW = −153 . × 105 kW 1 kJ / s
9.2 (cont'd) Heat Output = 1.53×105 kW. The reactor pressure is low enough to have a negligible effect on enthalpy. d.
C 9 H 20 ( g) + 14O2 ( g) → 9CO 2 (g) +10H 2O(l) ∆H$ o = −6171 kJ / mol
(1)
C 9 H 20 ( l) + 14O2 (g) → 9CO 2 (g) +10H 2O(l) ∆H$ o = −6124 kJ / mol
(2)
r
r
(2) − (1) ⇒ C9 H 20 ( l) → C 9 H 20 (g) ∆H$ o (C H , 25o C) = − 6124 kJ / mol − ( −6171 kJ / mol) = 47 kJ / mol v
e.
9.3
a.
9
20
Yes. Pure n-nonane can only exist as vapor at 1 atm above 150.6°C, but in a mixture of gases, it can exist as a vapor at lower temperatures.
Exothermic. The reactor will have to be cooled to keep the temperature constant. The temperature would increase under adiabatic conditions. The energy required to break the reactant bonds is less than the energy released when the product bonds are formed.
b.
b g 192 O bgg → 6CO bgg + 7 H Obgg b1g ∆H$ = ? 19 C H blg + O bgg → 6CO bgg + 7 H Oblg b2g ∆H$ = ∆H$ = −1791 . × 10 Btu lb- mole 2 C H bgg → C H blg b 3g ∆H$ = − e ∆H$ j = −13,550 Btu lb - mole H Ob lg → H Obgg b4g ∆H$ = e ∆H$ j = 18,934 Btu lb - mole . × 10 Btu lb - mole b1g = b2g + b3g + 7 × b4g ⇒ ∆H$ = ∆H$ + ∆H$ + 7 ∆H$ = −1672 C 6H 14 g + 6
14
6
14
2
2
2
6
2
o r
2
2
2
14
2
3
2
v
4
v
o r
C 2 H14
H2 O
Hess's law
6
1
c.
m& = 120 lb m / s
MO2 =32.0
⇒
6
2
3
4
n& = 375 . lb - mole / s.
n&O ∆Hˆ ro 3.75 lb-mole/s −1.672 ×106 Btu Q& = ∆H& = 2 = = −6.60 × 105 Btu/s (from reactor ) 9.5 1 lb-mole O 2 v O2
bg
bg
bg
bg
bg
CaC 2 s + 5H 2O l → CaO s + 2CO 2 g + 5H 2 g , ∆H$ ro = 69.36 kJ kmol
9.4 a.
Endothermic. The reactor will have to be heated to keep the temperature constant. The temperature would decrease under adiabatic conditions. The energy required to break the reactant bonds is more than the energy released when the product bonds are formed.
b.
∆U$ ro
=
∆H$ ro
LM − RT M ∑ ν MM N
gaseous products
= 52.0 kJ mol
i
−
OP kJ 8.314 J ∑ ν PP = 69.36 mol − mol ⋅ K PQ i
gaseous reactants
9- 2
1 kJ 103 J
298 K
b7 − 0g
9.4 (cont’d) ∆U$ ro is the change in internal energy when 1 g- mole of CaC2 (s) and 5 g - moles of H 2O(l) at 25o C and 1 atm react to form 1 g - mole of CaO(s), 2 g- moles of CO 2 (g) and 5 g - moles of H 2 (g) at 25o C and 1 atm. c.
Q = ∆U =
nCaC 2 ∆ U$ ro vCaC 2
=
150 g CaC2
1 mol 52.0 kJ = 121.7 kJ 64.10 g 1 mol CaC2
Heat must be transferred to the reactor. 9.5 a.
Given reaction = (1) – (2)
Hess's law
⇒
b
g
∆ H$ ro = ∆H$ ro1 − ∆H$ ro2 = 1226 − 18, 935 Btu lb- mole = −17 ,709 Btu lb - mole
b.
Given reaction = (1) – (2)
Hess's law
⇒
b
g
∆H$ ro = ∆H$ ro1 − ∆ H$ ro2 = −121740 , + 104,040 Btu lb - mole = −17, 700 Btu lb - mole
Hess's law
9.6
a. b.
9.7
Reaction (3) = 0.5× (1) − ( 2 ) ⇒ ∆Hˆ r = 0.5 −326.2 o
Reactions (1) and (2) are easy to carry out experimentally, but it would be very hard to decompose methanol with only reaction (3) occurring.
bg
bg
bg
e
a.
b.
n − C5 H12 g +
bg
F B I = 2G 90.37 kJ mol J = 180.74 kJ mol GH JK Table B.1
N2 g + O2 g → 2NO g , ∆H$ ro = 2 ∆H$ of
j
NO(g)
bg bg bg ∆H$ = 5e ∆H$ j e j bg e j b g = b5gb −11052 . g + b6gb −285.84g − b −1464 . g kJ mol = −21212 . kJ mol 19 C H blg + O b gg → 6CO b gg + 7H Obgg 2 ˆ ∆H = 6 ( ∆Hˆ ) + 7 ( ∆ Hˆ ) − ( ∆ Hˆ ) ( ) () o r
c.
kJ kJ kJ − −285.8 = 122.7 mol mol mol
6
o f
11 O 2 g → 5CO g + 6H 2O l 2 + 6 ∆ H$ fo − ∆H$ fo
CO(g)
14
2
o r
o f
n − C5 H12 g
H 2O l
2
2
o f
CO 2
o f
H2 O g
C6 H 14 l
= ( 6 )( −393.5 ) + 7 ( −241.83 ) − ( −198.8) kJ mol = −3855 kJ mol d.
Na 2SO 4 (l) + 4CO( g) → Na 2S( l) + 4CO 2 ( g) ∆H$ o = ∆H$ o + 4 ∆H$ o − ∆H$ o r
e j
− 4e ∆H$ j e j e j = ( − 3732 . + 6.7) + b 4gb −393.5g − b − 13845 . + 24.3g − 4( − 11052 . f
Na 2S( l )
f
CO2 (g )
f
Na 2S O4 (l )
9- 3
o f
CO(g)
kJ mol = − 138.2 kJ mol
9.8
e j = e ∆H$ j
⇒ e ∆H$ j = −385.76 + 52.28 = −333.48 kJ mol e j $ $ ∆H$ = − 276.2 − 92.31 + 33348 . = −3503 . kJ mol b g + e∆H j b g − e∆H j Given reaction = b1g + b2g ⇒ −385.76 − 35.03 = −420.79 kJ mol ∆H$ ro1 = ∆H$ fo
a.
o r2
C 2 H 2 Cl 4 (l )
o f
− ∆H$ fo o f
C 2 HCl3 l
o f
C 2 H4 ( g)
o f
HCl g
C2 H 2 Cl 4 (l )
C2 H 2 Cl 4 (l )
b.
c.
9.9 a.
b.
300 mol C2 HCl 3 Q& = ∆ H& = h Heat is evolved.
e j
∆H$ co = 2 ∆H$ of
B
=
g
CO2 ( g )
b
e j
+ ∆H$ fo
g b
b g − e∆H f jC H b gg $o
H2 O l
2
g b
2
kJ kJ = −1299.6 g mol mol
2 − 3935 . + −28584 . − 226.75
d i
( i) ∆H$ ro = ∆H$ fo
B
Table B.1
=
C2 H6 ( g )
d i
− ∆H$ fo
bg
C2 H2 g
kJ = −3114 . b −84.67g − b 226.75g mol
d i
( ii) ∆H$ ro = ∆H$ co
B
Table B.1
= d.
b
5 C 2 H2 ( g) + O 2 ( g) → 2CO 2 ( g) + H 2O(l) ∆H$ co = −1299.6 kJ mol 2 The enthalpy change when 1 g-mole of C2 H2 (g) and 2.5 g-moles of O2 (g) at 25°C and 1 atm react to form 2 g-moles of CO2 (g) and 1 g-mole of H2 O(l) at 25°C and 1 atm is -1299.6 kJ.
Table B.1
c.
− 420.79 kJ = −126 . × 105 kJ h = − 35 kW mol
C 2 H2 ( g )
d i
+ 2 ∆H$ co
H2 (g )
d
− ∆H$ co
kJ mol
i
bg
C2H 6 g
kJ kJ = −3114 . b −1299.6g + 2( −285.84) − b −1559.9g mol mol (1) ∆H$ co1 = − 1299.6 kJ mol
5 C 2 H2 ( g) + O 2 ( g) → 2CO2 ( g) + H 2 O( l) 2 1 H2 ( g) + O 2 ( g) → H 2 O( l) 2 7 C 2 H6 (g) + O 2 ( g) → 2CO 2 ( g) + 3H 2 O( l) 2
(2) ∆H$ c 2 = − 28584 . kJ mol o
(3) ∆H$ co3 = − 1559.9 kJ mol
The acetylene dehydrogenation reaction is (1) + 2 × (2) − (3) Hess's law
⇒
∆ H$ ro = ∆H$ co1 + 2 × ∆H$ co2 − ∆H$ co3
b
g
= −1299.6 + 2( − 28584 . ) − ( − 1559.9) kJ mol = −3114 . kJ / mol
9- 4
9.10
bg
bg
bg
∆H$ ro = −4850 kJ / mol
25 O2 (g) → 8CO 2 g + 9H 2 O g 2
a.
C 8H18 l +
b.
When 1 g-mole of C8 H18 (l) and 12.5 g-moles of O2 (g) at 25°C and 1 atm react to form 8 g-moles of CO2 (g) and 9 g-moles of H2 O(g), the change in enthalpy equals -4850 kJ. Energy balance on reaction system (not including heated water):
b
g
b
g
∆E k , ∆E p , W = 0 ⇒ Q = ∆U = n mol C8 H 18 consumed ∆U$ co kJ mol
( Cp ) H 2 O(l) from Table B.2 = 75.4 × 10 −3 kJ / mol.o C
−Q = mH2 O ( Cp ) H 2 O(l) ∆T = Q = ∆ U ⇒ − 89.4 kJ =
1 mol 75.4 × 10−3 kJ 21.34° C = 89.4 kJ 18.0 × 10−3 kg mol.o C
1.00 kg
2.01 g C 8H18 consumed 1 mol C8H 18 114.2 g
⇒ ∆U$ co = −5079 kJ mol ∆H$ co = ∆U$ co + RT
LM OP MM ∑ ν − ∑ ν PP MN PQ i
gaseous products
=− 5079 kJ mol +
∆U$ co (kJ) 1 mol C8 H18
i
gaseous reactants
8.314 J
1 kJ
298 K
b8 + 9 − 12.5g
mol ⋅ K 10 J 3
⇒ ∆Hˆ co = −5068 kJ mol c.
% difference =
(−5068) −( −4850) ×100 = − 4.3 % −5068
e j b g + 9e∆H$ j b g − e∆H$ j b g ⇒ ( ∆Hˆ ) = 8 ( −393.5) + 9 ( −241.83 ) + 5068 kJ/mol = −256.5 kJ/mol ()
∆H$ co = 8 ∆H$ of
CO2 g
o f
o ff
H 2O g
o f
C8 H18 l
C 8H18 l
There is no practical way to react carbon and hydrogen such that 2,3,3-trimethylpentane is the only product.
9- 5
9.11
a.
bg
bg
n − C 4H10 g → i − C 4H10 g Basis : 1 mol feed gas
0.930 mol n -C4 H10
(n n- C4H10 )out
0.050 mol i-C4 H10
( n i-C4H10)out
0.020 mol HCl
0.020 mol HCl
149°C
Q(kJ/mol)
149°C
(n n-CH4 H10 ) out = 0.930(1 − 0.400) = 0.560 mol (n i-C H4 H10 ) out = 0.050 + 0. 930 × 0.400 = 0.420 mol ξ =
(n n-C4 H10 ) out − ( n n-C4 H10 ) in ν n-C
e j
4 H 10
b.
∆H$ ro = ∆H$ fo
c.
References: n − C 4H10
i − C 4 H10
substance
e j ⇒ ∆ H$ = −134.5 − b−124.7g bgg, i − C H bgg at 25° C
− ∆H$ fo
Table B.1
H$ in
n in
bkJ molg
n − C 4 H 10
1
H$ 1
i − C 4 H 10
−
−
LM MN
z
B
Table B.2
149
25
Cp
o r
n − C4 H10 4
(mol)
H$ 1 =
0.560 − 0.930 = 0.370 mol 1
=
kJ mol = −9.8 kJ mol
10
H$ out
n out (mol)
bkJ molg
0.600
H$ 1
0.400
H$ 2
O kJ dT P PQ mol = 14.29 kJ mol
H$ 2 =
LM MN
z
B
Table B.2
149
25
C p dT
OP kJ . kJ mol PQ mol = 1414
ˆ ro + ∑ ni H ˆ i − ∑ ni Hˆ i ] = 0.370 −9.8 + (1)(14.142 ) − (1)(14.287 ) kJ Q = ∆H = ξ [ ∆ H out
in
= − 3.68 kJ
&= For 325 mol/h fed, Q d.
−9.8 kJ 325 mol feed 1h 1 kW = −0.90 kW 1 mol feed h 3600 s 1 kJ/s
−3.68 kJ ∆Hˆ r (149 °C ) = = −9.95 kJ/mol 0.370 mol
9- 6
9.12
a. 1 m3 at 298K, 3.00 torr
Products at 1375K, 3.00 torr
n 0 (mol) 0.111 mol SiH4 /mol 0.8889 mol O2 /mol
n 1 (mol O2 ) n 2 (mo l SiO2 ) n 3 (mol H2 )
SiH 4 ( g) + O 2 ( g) → SiO 2 (s) + 2H 2 ( g)
Ideal Gas Equation of state : no =
1 m3
273 K 3.00 torr 298 K
760 torr
1 mol 22.4 × 10-3 m 3
= 0.1614 mol
ni = nio + ν iξ SiH4 : 0=0.1111(0.1614 mol) − ξ ⇒ ξ = 0.0179 mol O2 : n1 =0.8889(0.1614 mol) − ξ = 0.1256 mol O 2 SiO2 : n2 = ξ = 0. 0179 mol SiO2 H 2 : n3 =2ξ =0.0358 mol H 2 b.
∆H$ ro = ( ∆H$ fo ) SiO 2 ( s) − ( ∆H$ fo ) SiH 4 (g) = [− 851 − ( −619 . )] kJ mol = −789.1 kJ / mol References : SiH 4 ( g),O2 (g), SiO2 (g), H 2 ( g) at 298 K
Substance SiH 4 O2
nin nout Hˆ in Hˆ out (mol h) (kJ mol) (mol h) (kJ mol) 0.0179 0 − − 0.1435 0 0.1256 Hˆ 1
SiO2
−
−
0.0179
H2
−
−
0.0358
Hˆ 2 Hˆ 3
B
Table B.8
O 2 (g,1375K): H$ 1 = H$ O2 (1102 o C) = 3614 . kJ / mol SiO 2 (s,1375K): H$ 2 =
z
1375
( C p ) SiO2 ( s) dT = 79.18 kJ / mol
298
B
Table B.8
H 2 (g,1375K): H$ 3 = H$ H 2 (1102o C) = 32.35 kJ / mol c.
Q = ∆H = ξ ∆Hˆ ro + ∑ ni Hˆ i − ∑ ni Hˆ i = −7.01 kJ/m 3 feed out
−7.01 kJ 27.5 m 3 & Q= m3 h
in
1h 1 kW = −0.0536 kW (transferred from reactor) 3600 s 1 kJ/s
9- 7
9.13
a.
bg bg
bg
bg
Fe 2O3 s + 3C s → 2 Fe s + 3CO g , ∆H$ r ( 77 o F) = 2111 . × 105 Btu lb - mole 2000 lb m Fe 1 lb - mole = 3581 . lb - moles Fe produced 55.85 lb m 53.72 lb - moles CO produced 17.9 lb- moles Fe2 O3 fed 53.72 lb - moles C fed
Basis :
17.9 lb -moles Fe2 O3 (s) 77° F
35.81 lb-moles Fe (l) 2800° F
53.72 lb -moles C 77° F
53.72 lb-moles CO(g) 570° F Q (Btu/ton Fe)
b.
bg bg bg
Substance
b
Fe2 O3 s,77° F C s,77° F Fe l,2800° F CO g,570° F
b b b
g
g g
g
nin nout H$ in H$ out (lb - moles) (Btu lb - mole) (lb - moles) (Btu lb - mole) 17.91 0 − − 5372 . 0 − − − − 3581 . H$ 1 − − 5372 . H$ 2
Fe(l,2800o F): H$ 1 =
z
2794
77
dC i
b gdT + ∆Hm b 2794° Fg + $
p Fe s
CO(g,570 o F): H$ 2 = H$ C O (570o F)
c.
bg
References: Fe 2O3 s , C s , Fe s , CO g at 77° F
Q = ∆H =
nFe ∆H$ ro + ν Fe
z
dC i
2800
2794
b gdT = 28400
p Fe l
Btu lb - mole
= A I 3486 Btu lb - mole F interpolating H from Table B.9K
∑ n H$ − ∑ n H$ i
out
i
i
i
in
. ge 2111 . × 10 j b3581 = + b 3581 . gb28400g + b53.72gb 3486g − 0 = 4.98 × 10 5
d.
2 Effect of any pressure changes on enthalpy are neglected.
6
Btu / ton Fe produced
Specific heat of Fe(s) is assumed to vary linearly with temperature from 77°F to 570°F. Specific heat of Fe(l) is assumed to remain constant with temperature. Reaction is complete. No vaporization occurs.
9- 8
9.14
a.
bg
C 7 H16 g → C 6H 5CH3 ( g) + 4 H 2 ( g) Basis : 1 mol C7 H16 1 mol C7 H16
1 mol C6 H5 CH3
400°C
4 mol H2 400° C
Q (kJ/mol)
bg bg
References: C s , H 2 g at 25° C
b.
H$ in
substance nin
H$ out
nout
bmolg bkJ molg bmolg bkJ molg
C7 H 16
1
H$ 1
−
C7 H 8
−
−
1
H2
−
−
4
− H$ 2 H$ 3
400 ↓ C7 H16 ( g,400°C ): Hˆ 1 = (∆Hˆ fo ) C7 H16 (g) + C p dT 25 = ( − 187.8 +91.0) kJ/mol= −96.8 kJ/mol 0.2427
∫
b
g
C 6 H 5CH 3 g,400° C : H$ 2 = ( ∆H$ fo ) C6 H 5CH 3 ( g) +
LM MN
z
B
Table B.2
400
25
C p dT
OP PQ
= (+50 + 60.2) kJ / mol = 110.2 kJ / mol
b
B
Table B.8
g
H2 g,400 ° C : H$ 3 = H$ H 2 ( 400 o C) = 1089 . kJ mol c.
Q = ∆H =
∑ n Hˆ − ∑ n Hˆ i
out
i
i
i
in
= [(1)(110.2) + (4)(10.89) - (1)(-96.8)] kJ = 251 kJ (transferred to reactor) d.
∆Hˆ r (400o C)=
251 kJ = 251 kJ/mol 1 mol C7 H16 react
9- 9
9.15
a.
bCH g Obgg → CH bgg + H bgg + CObgg 3 2
4
2
Moles charged: (Assume ideal gas)
b g = 0.01286 mol bCH g O
2.00 liters 273 K 350 mm Hg
1 mol
3 2
873 K 760 mm Hg 22.4 liters STP
b g O decomposed (Clearly x CS1 ). Then CS2 decreases with increasing t as well as CS1 . Finally dCS2 /dt approaches zero as t→∞. Therefore, CS2 increases until it reaches a maximum value, then it decreases. The plot of CS3 vs. t begins at (t=0, CS3 =0). When t=0, the slope (=dCS3 /dt) is 0.04 (0 − 0) = 0 . As t increases, CS2 increases (CS3 < CS2 )⇒ d CS3 /dt =0.04(CS2 -CS3 ) becomes positive ⇒ CS2 increases with increasing t until dCS3 /dt changes to negative (CS3 > CS1 ). Finally dCS3 /dt approaches zero as t→∞. Therefore, CS3 increases until it reaches a maximum value then it decreases. c. 3
CS1 , CS2 , CS3 (g/L)
2.5 2 CS1 1.5 CS2
1
CS3
0.5 0 0
20
40
60
80
100
120
140
160
t (s)
11.29 a. (i) Rate of generation of B in the 1st reaction: rB1 = 2 r1 = 0.2 CA (ii) Rate of consumption of B in the 2nd reaction: −rB2 = r2 = 0.2CB2 b. Mole Balance on A: Accumulation=-Consumption
E
d( C AV ) dC = −01 . C AV ⇒ A = −01 . CA dt dt t = 0, C A0 = 100 . mol / L Mole Balance on B: Accumulation= Generation-Consumption
E
d ( CB V ) dC = 0.2CAV − 0.2CB2V ⇒ B = 0.2CA − 0.2CB2 dt dt t = 0, CB0 = 0 mol / L
11-23
11.29 (cont’d) c. 2
C A, CB, CC
CC
1 CB CA 0 t
The plot of CA vs. t begins at (t=0, CA =1). When t=0, the slope (=dCA /dt) is −01 . × 1 = −01 . . As t increases, CA decreases ⇒ dCA /dt=-0.1CA becomes less negative, approaches zero as t→∞. CA →0 as t→∞. The curve is therefore concave up. The plot of CB vs. t begins at (t=0, CB =0). When t=0, the slope (=dCB/dt) is 0.2 (1 − 0) = 0.2 . As t increases, CB increases, CA decreases ( C 2B < CA )⇒ d CB/dt =0.2(CA - C 2B ) becomes less positive until dCB/dt changes to negative ( C 2B > CA ). Then CB decreases with increasing t as well as CA . Finally dCB/dt approaches zero as t→∞. Therefore, CB increases first until it reaches a maximum value, then it decreases. CB→0 as t→∞. The plot of CC vs. t begins at (t=0, CC =0). When t=0, the slope (=dCC/dt) is 0.2 ( 0) = 0 . As t increases, CB increases ⇒ dCc/dt =0.2 C 2B becomes positive also increases with increasing t ⇒ CC increases faster until CB decreases with increasing t ⇒ dCc/dt =0.2 C 2B becomes less positive, approaches zero as t→∞ so CC increases more slowly. Finally CC→2 as t→∞. The curve is therefore S-shaped.
CA, CB , CC (mol/L)
d. 2.2 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0
CC
CB CA
0
10
20
30 t (s)
11-24
40
50
11.30 a. When x =1, y =1 .
y=
ax x =1, y=1 a ⇒ 1= ⇒ a =1+ b x +b 1+ b
pC5 H12 = yP = xp *C5 H12 ( 46o C) ⇒ y =
b. Raoult’s Law:
p *C5 H12 ( 46 o C ) = 10
Antoine Equation:
⇒ y=
xp *C5 H12 ( 46o C ) P
=
(6 .85221−
1064 .63 ) 46 + 232. 00
xp *C5 H12 ( 46o C ) P
= 1053 mm Hg
0.7 × 1053 = 0.970 760
0.70a R| y = ax x=0.70, y=0.970 0970 . . = LL (1) |Ra = 1078 ⇒S 0.70 + b S| x + b TFrom part (a), a = 1+ bLLLLLLLLLLL(2) |Tb = 0.078
c. Mole Balance on Residual Liquid: Accumulation=-Output
E
dN L = − n&V dt t = 0, N L = 100 mol Balance on Pentane: Accumu lation=-Output
E
d( N L x ) dN L dx ax = − n&V y ⇒ x + NL = − n&V dt dt dt x+b dN L / dt = − n&V
E
FG H
dx n& ax =− V −x dt NL x + b
IJ K
t = 0, x = 0.70
d. Energy Balance: Consumption=Input
E
) n&V ∆H vap = Q&
) ∆H vap =27.0 kJ/mol
t = 0, N L = 100 mol
From part (c),
dN L = − n& V dt n&V Q& 27.0 = & NL Qt 100 27.0
n&V =
b
Q& 270 . kJ / mol
N L = 100 − n&V t = 100 −
Substitute this expression into the equation for dx/dt from part (c):
11-25
g
& Qt 27.0
11.30 (cont’d)
FG H
IJ K
FG H
IJ K
& 270 dx n& ax Q . ax =− V −x = − −x & dt NL x + b Qt x+b 100 27.0 x(0) = 0.70 e. 1 0.9 0.8
y (Q=1.5 kJ/s)
0.7 x, y
0.6 x (Q=1.5 kJ/s)
0.5 0.4
x (Q=3 kJ/s)
y (Q=3 kJ/s)
0.3 0.2 0.1 0 0
200
400
600
800
1000 1200 1400 1600 1800
t(s)
f. The mole fractions of pentane in the vapor product and residual liquid continuously decrease over a run. The initial and final mole fraction of pentane in the vapor are 0.970 and 0, respectively. The higher the heating rate, the faster x and y decrease.
11-26
CHAPTER THIRTEEN Problem
13.1
Methanol
Production
Rate
430,000 metric tons/year 51,114 kg/h 1,597 kmol/h Process stoichiometry: CH4 + H20 ---> CH30H + H2 So that the required feed rates (with given assumptions) are CH4 Feed Rate = Steam Feed Rate =
Problem
1,597 kmol/h 1,597 kmol/h
cubic
meters/min
13.2
1 P = r.h. = (pH20/p*H20 p*H20 = 0.0424 pH20 = 0.02968 yH20 0.029292 Basis: 1 mole of component moles N2 0.79 0.21 02 0.030176 Water Total 1.03 component N2 02 Total
Basis:
atm @ 30C) x 100% = bar bar
70%
dry air (79 mole% N2, 21 mole% 02) Mw mass mole frac 28 22.12 32 6.72 0.0293 18 0.54 29.38 1.0000
moles Mw 0.79 0.21 1.00
mass 22.12 6.72 . 28.84 Difference in avg molecular weight is due to presence of water; the difference is slight. 1 km01 of CH4 burned
flow rate of air/km01 Problem
596 standard 28,75lkg/h
28 32
nat gas burned
13.3
Composition of effluent gas from burners mole frac mass frac Component km01 kg 3.2 0.0103 02 0.100 0.0088 N2 7.900 0.6990 221.2 0.7139 44.0 0.1420 co2 1.000 0.0885 0.1337 H20 2.302 0.2037 41.4 1.0000 309.8 1.0000 total 11.302 p*H20 @ 150C = 4.74 bar < pH20 Therefore, there is no condensation in cooling the exhaust gases to 15OC, which means the effluent gas and stack gas have the same composition. Volumetric
flow rate effluent gas stack gas density of air = density of stack gas = specific gravity =
1,143 m3/kmol CH4 burned 392 m3/kmol CH4 burned 1.1471 kg/m3 0.7899 kg/m3 1 0.6886 relative to ambient air
13-1
I
Problem
13.4
Reformer Temperature Reformer Temperature Reformer Pressure Equilibrium Temperature
855 1128 15.8 1128
C K atm K
1.6 Mpa
CH4 + Hz0 ---> CO + 3Hz Production
rates CH4 H20 CO CO2 H2
feed rate CH4 x (1 - fractional conversion) feed rate H20 - fractional conversion x feed rate CH4 fractional conversion x feed rate CH4 feed rate CO2 feed rate + 3 x feed rate CH4 x fractional conversion
(a) methane:steam of 3:l Stoichiometric
Feed I (kmol/h)( 1600 4800 0 0 0 6400
Table: CH4 H20 co co2 H2 Total
(kmol/h) 170 3,370 1,430 0 4,291 9,261
574.3668 KP~ Ratio1 574.366846 fractional conversion of CH4* =
=
Product MolFrac (kg/h) 0.0183 2,716 0.3639 60,655 0.1544 40,047 0.0000 0 0.4633 8,582 1.0000 112,000
MassFrac 0.0242 0.5416 0.3576 0.0000 0.0766 1.0000
lO"(-(11,769/T(K))+l3.1927)
( (Yco x YH2A3) / (YCH4 x YH20) 6.8939 -3.57E-07 Converge* = ((Kpl/Ratiol)-1)lOO * the Goal Seek tool is used to adjust the fractional conversion until Converge is close to zero Methane Conversion =
) p2
1
Product Flow Rate =-I (b) methane:steam of 1:l Stoichiometric Table: CH4 H20 co co2 H2 Total
fractional
Feed I (kmol/h)l 1600 1600 0 0 0 3200
574.3668 Kpl Ratio1 1653.61852 conversion of CH4 =
(kmol/h) 471 471 1,129 0 3,387 5,458 =
lO"(-(11,769/T(K))+l3.1927)
=
( kc0 x YH2A3)
/ (YCH4
x YH20)
Jp
2
0.7055 -65.266061Converge = ((Kpl/Ratiol)-1)lOO
Methane Conversion --I-[ Product Flow Rate ='
Product MolFrac (kg/h) MassFrac 0.0863 7,538 0.1386 0.0863 8,480 0.1559 0.5810 0.2068 31,608 0.0000 0 0.0000 0.6205 6,773 0.1245 54.400 1.0000 1.0000
5,458 kmol/h 54,400 kg/h 13-2
Problem 13.4 (cont'd) methane:steam
of 2:l
Stoichiometric
Table: CH4 H20 co co2 H2 Total
Feed I (kmol/h)l, 1600 3200 0 0 0 4800
Product (kmol/h) MolFrac MassFrac (kg/h) 0.0476 248 0.0330 3,960 1,848 0.2462 33,256 0.3997 0.1802 37,869 0.4552 1,352 0 0.0000 0 .o.oooo 4,057 0.5406 8,115 0.0975 7,505 1.0000 83,200 1.0000
574.3668 = lO"(-(11,769/T(K))+l3.1927) Kpl Ratio1 874.51201 = ( (ycO x YH2^3) / (YCH4 x YH20) )p2 fractional conversion of CH4 = 0.84529344 -34.321446 Converge = ((Kpl/Ratiol)-I)100 Methane
Conversion
=1
Product Flow Rate =
methane:steam
of
Stoichiometric
7,505 kmol/h 83,200 kg/h
4:l
Table: CH4 H20 co co2 H2 Total
Feed I (kmol/h)l 1600 6400 0 0 0 8000
Product MassFrac (kmol/h) MolFrac (kg/h) 0.0147 130 0.0118 2,074 4,930 0.4506 88,733 0.6302 1,470 0.1344 41,171 0.2924 0.0000 0 0.0000 0 0.4032 8,822 0.0627 4,411 10,941 1.0000 140,800 1.0000
= 574.3668 lO"(-(11,769/T(K))+l3.1927) Kpl = Ratio1 411.494126 ( (Yco x YH2^3) / (YCH4 x YH~o) jp2 fractional conversion of CH4 = 0.91899524 39.5808125 Converge = ((Kpl/Ratiol)-I)100 Methane
Conversion = 1
Product Flow Rate =I[
Summary
Moles Steam:Mole CH4 1 2 3 4
Methane H2 Conversion Produced 6,773 70.6% 84.5% 8,115 89.4% 8,582 91.9% 8,822
13-3
H2:CO 3.0000 3.0000 3.0000 3.0000
Problem
13.5
Reformer Temperature Reformer Temperature Reformer Pressure Equilibrium Temperature
Production
1
855 1128 15.79 1128
C K atm K
1.6 Mpa
CH, + Hz0 ---> CO + 3H, CO f Hz0 ---> COz f H, rates CH4 feed rate CH4 x (1 - fractional conversion) H20 feed rate H20 - fractional conversion x feed rate CH4 - production rate of CO2 co fractional conversion x feed rate CH4 - production rate of CO2 co2 feed rate CO2 + production rate of CO2 H2 feed rate + 3 x feed rate CH4 x fractional conversion + production rate of CO2
methane:steam of 3:l Stoichiometric Feed I (kmol/h)[ CH4 1600 H20 4800 co 0 co2 0 H2 0 Total 6400
(kmol/h) 176 3,156 1,205 219 4,492 9,249
Product MolFrac (kg/h) 0.0190 2,811 0.3413 56,814 0.1303 33,740 0.0237 9,650 0.4857 8,985 1.0000 112,000
MassFrac 0.0251 0.5073 0.3013 0.0862 0.0802 1.0000
574.4 = lO"(-(11,769/T(K))+l3.1927) Kpl Ratio1 574.4 = ( (Yco x YHZh3) / (YCHQ x YH20) )P2 0.2590 = 10"(1,197.8/T(K)-1.6485) KP~ 0.2590 = Ratio2 (Y co2 x YtI2) / (Yco x YH20) fractional conversion of CH4* 0.89021022 Converge 1" -4.3538E-05 = ((Kpl/Ratiol)-1)*100 moles of CO2 formed** 219.322711 = ((Kp2/Ratio2)-l)*lOO Converge 2"" -0.00028801 * Use Goal Seek tool to adjust CH4 conversion so that Converge 1 goes to zero. ** Use Goal Seek tool to adjust CO2 formation so that Converge 2 goes to zero. CH4 Conv = CH4 to CO =
-1 Product Flow Rate =mI
CO with water-gas shift reaction:CO
without water-gas shift reaction = 184.381
13-4
Problem 13.5 (cont'd)
methane:steam of 1:l Stoichiometric Feed I (Iunol/h)l CH4 1600 H20 1600 co 0 co2 0 H2 0 Total 3200
*(kmol/h) 482 445 1,082 37 3,392 5,437
Product MolFrac (kg/h) 0.0886 7,706 0.0818 8,007 0.1990 30,286 0.0068 1,617 0.6239 6,784 1.0000 54,400
MassFrac 0.1416 0.1472 0.5567 0.0297 0.1247 1.0000
= KP~ 574.4 lO"(-(11,769/T(K))+l3.1927) = Ratio1 1662.1 ( (Yco x Y,2^3 11 (YCH4 x YH20) ) p2 = KP~ 0.2590 10A(1,197.8/T(K)-1.6485) = Ratio2 0.2590 (Y co2 x Y,,) / (Yco x YIi20) fractional conversion of CH4* 0.69900166 Converge 1" -65.4436238 = ((Kpl/Ratiol)-1)*100 moles of CO2 formed** 36.7477842 Converge 2** -5.79043-05 = ((Kp2/Ratio2)-I)*100 CH4 Conv = CH4 to CO = Product Flow Rate =I1
methane:steam of 2:l Stoichiometric Feed I (kmol/h)l CH4 1600 3200 H20 co 0 0 co2 H2 0 Total 4800
(kmol/h) 257 1,726 1,213 130 4,160 7,487
= 574.4 Kpl = 876.6 Ratio1 = 0.2590 KP~ = Ratio2 0.2590 fractional conversion of CH4* Converge l* -34.4757385 moles of CO2 formed** Converge 2** -1.52263-05
5:;4: g;Fh
Product MolFrac (kg/h) 0.0343 4,108 0.2306 31,074 0.1620 33,961 0.0174 5,737 0.5557 8,320 1.0000 83,200
lO^(-(11,769/T(K))+13.1927) ( (yco x ~~2~3) 1 (YcH4 x YHZO) ) P2 10"(1,197.8/T(K)-1.6485) (Yco2 x YH2) / (Yco x YH20) 0.83954105 = ((Kpl/Ratiol)-I)*100 130.381595 = ((Kp2/Ratio2)-l)*lOO
CH4 Conv = CH4 to CO = Product Flow Rate -Im[
13-5
MassFrac 0.0494 0.3735 0.4082 0.0690 0.1000 1.0000
Problem 13.5 (cont'd)
methane:steam of 4:l Stoichiometric Feed (kmol/h)
CH4 H20 co co2 H2 Total
1600 6400 0 0 0 8000
I
1
,(kmol/h) 133 4,634 1,169 299 4,700 10,934
Product MolFrac (kg/h) 0.0122 2,126 0.4238 83,418 0.1069 32,722 0.0273 13,134 0.4298 9,400 140,800 1.0000
MassFrac 0.0151 0.5925 0.2324 0.0933 0.0668 1.0000
574.4 = lO"(-(11,769/T(K))+l3.1927) Kpl Ratio1 410.9 = ( (Yco x ~~2~x1 ! (YCH4 x YH20) ) p2 0.2590 = 10"(1,197.8/T(K)-1.6485) Kp2 Ratio2 0.2590 = (Y co2 x YH2) / (Yco x YH20) fractional conversion of CH4* 0.91695885 Converge 1* 39.77214561 = ((Kpl/Ratiol)-l)*lOO moles of CO2 formed** 298.507525 Converge 2** -l.O4E-05 = ((Kp2/Ratio2)-l)*lOO CH4 Conv = CH4 to CO = Product Flow Rate =mI
Moles Methane H2 H2:CO Steam:Mole CH4 Conversion Production Production CO:H2 1 69.9% 3,392 3.1359 0.3188883 84.0% 4,160 3.4300 0.2915462 2 91.7% 4,492 3.7280 0.2682379 3 4,700 4.0217 0.2486487 4 89.0%
13-6
Problem
13.6
Reformer Reformer
Temperature Temperature Pressure
855 C 1128 K ,15.79 atm
1.6 MPa
Production
rates CH4 feed rate CH4 x (1 - fractional conversion) H20 feed rate H20 - fractional conversion x feed rate CH4 - production rate of CO2 CO fractional conversion x feed rate CH4 - production rate of CO2 CO2 feed rate CO2 + production rate of CO2 H2 feed rate + 3 x feed rate CH4 x fractional conversion + production rate of CO2 Stoichiometric Table: COMPONENT FEED PRODUCT MolFrac CH4 1600 176 0.0190 4800 H20 3,156 0.3413 co 0 1,205 0.1303 co2 0 219 0.0237 H2 0 4,492 0.4857 Total 6400 9,249 1.0000 574.4 = Kpl 574.4 = Ratio1 0.2590 = Kp2 Ratio2 0.2590 = fractional conversion of CH4* Converge 1* -4.35353-05 moles of CO2 formed** Converge 2** -0.00028798 Reformer Reformer Pressure Stoichiometric CH4 H20 co co2 H2 Total
T T
lO"(-(11,769/T(K))+l3.1927) ( (Yco x Y,,^3) / (YCHI x YHZO) )P2 10"(1,197.8/T(K)-1.6485) (Y CO2 x YH2) / (YCO x YH20) 0.8902102 ((Kpl/Ratiol)-l)*lOO 219.=32271 = ((Kp2/Ratio2)-l)*lOO
750 c 1023 K 15.79 atm
1.6 MPa
Table: FEED 1600 4800 0 0 0 6400
PRODUCT 576 3,510 759 266 3,338 8,448
MOLFRAC 0.0681 0.4155 0.0898 0.0314 0.3951
Kpl 48.79 Ratio1 48.79 Kp2 0.3329 Ratio2 0.3329 fractional conversion of CH4* 0.64015 Converge 1" - 6 . 4 5 1 9 3 - 0 5 = moles of CO2 formed** 265.59039 Converge 2** 0.000608333 =
13-7
nCO/nCH4 nH2/nCH4 nCO/nC02
0.474156 2.086444 2.856465
((Kpl/Ratiol)-l)*lOO ((Kp2/Ratio2)-1)*100
Problem 13.6 (cont'd)
Reformer Reformer
Stoichiometric
Temperature Temperature Pressure
800 C 1073 K .15.79 atm
1.6 MPa
Table: FEED 1600 4800 0 0 0 6400
CH4 H20 co co2 H2 Total
PRODUCT 357 3,312 999 244 3,975 8,887
MOLFRAC 0.0401 0.3727 0.1124 0.0275 0.4472
Kpl 167.6 Ratio1 167.6 Kp2 0.2936 Ratio2 0.2936 fractional conversion of CH4* 0.7771058 Converge l* 0.000238464 ((Kpl/Ratiol)-l)*lOO moles of CO2 formed** 244: 4398 Converge 2** -2.48073-05 = ((Kp2/Ratio2)-l)*lOO
Reformer Reformer
Stoichiometric CH4 H20 co co2 H2 Total
Temperature Temperature Pressure
900 c 1173 K 15.79 atm
1.6 MPa
Table: FEED 1600 4800 0 0 0 6400
PRODUCT 88 3,086 1,311 201 4,738 9,424
MOLFRAC 0.0093 0.3275 0.1391 0.0214 0.5027 1.0000
Kpl 1443.6 Ratio1 1443.6 KP~ 0.2359 Ratio2 0.2359 fractional conversion of CH4* 0.9451022 Converge l* -4.1853-05 = moles of CO2 formed** 201.39252 Converge 2"" -0.00020159 =
13-8
((Kpl/Ratiol)-I)*100 ((Kp2/Ratio2)-l)*lOO
Problem 13.6 (cont'd)
Reformer Reformer Stoichiometric
Temperature Temperature Pressure
950 c ' 1223 K 15.79 atm
1.6 MPa
Table: FEED 1600 4800 0 0 0 6400
CH4 H20 co co2 H2 Total
PRODUCT 39 3,054 1,377 185 4,869 9,523
MOLFRAC 0.0040 0.3207 0.1445 0.0194 0.5113 1.0000
Kpl 3712.3 Ratio1 3712.3 KP~ 0.2142 Ratio2 0.2142 fractional conversion of CH4* 0.9759079 Converge l* 0.000662212 = moles of CO2 formed** 184.93694 Converge 2** -2.0226E-05 = 1.60 MPa T (Cl nCO/nCH4
nH2/nCH4 nCO/nCO2
750 0.474 2.086 2.856
800 0.624 2.484 4.087
13-9
855 0.753 2.808 5.494
((Kpl/Ratiol)-l)*lOO ((Kp2/Ratio2)-l)*lOO 900 0.819 2.961 6.509
950 0.860 3.043 7.443
Problem 13.6 (cont'd)
Reformer Reformer
Temperature Temperature Pressure
855 C 1128 K *11.84 atm
1.2 MPa
Production
rates CH4 feed rate CH4 x (1 - fractional conversion) H20 feed rate H20 - fractional conversion x feed rate CH4 - production rate of Co2 CO fractional conversion x feed rate CH4 - production rate of CO2 CO2 feed rate CO2 + production rate of CO2 H2 feed rate + 3 x feed rate CH4 x fractional conversion + production rate of CO2 Stoichiometric Table: COMPONENT FEED PRODUCT MolFrac CH4 1600 116 0.0124 H20 4800 3,098 0.3307 co 0 1,266 0.1352 co2 0 218 0.0232 H2 0 4,670 0.4985 6400 9,368 Total 1.0000 574.4 = Kpl Ratio1 574.4 = 0.2590 = Kp2 Ratio2 0.2590 = fractional conversion of CH4* Converge 1* 5.08455E-05 moles of CO2 formed** Converge 2** -4.7046E-05
Reformer T Pressure Stoichiometric CH4 H20 co co2 H2 Total
lO"(-(11,769/T(K))+13.1927) ( (Yco x YH2&3) / (YCH4 x YH20) )P2 10"(1,197.8/T(K)-1.6485) (Y CO2 x yi-12) / (YCO x YH20) 0.9275941 ((Kpl/Ratiol)-1)*100 217.=65358 = ((Kp2/Ratio2)-1)*100
1023 K 11.84 atm
1.2 MPa
Table: FEED 1600 4800 0 0 0 6400
PRODUCT 472 3,405 861 267 3,651 8,656
MOLFRAC 0.0545 0.3934 0.0994 0.0309 0.4218
KP~ 48.79 Ratio1 48.79 0.3329 KP~ 0.3329 Ratio2 fractional conversion of CH4* 0.7049568 Converge 1* -0.00022618 moles of CO2 formed** 267 .=23789 Converge 2** 0.000779089 =
13-10
nCO/nCH4 0.537933 nH2/nCH4 2.281894 3.2207 nCO/nCO2
((Kpl/Ratiol)-l)*lOO ((Kp2/Ratio2)-l)*lOO
Problem 13.6 (cont'd)
Reformer Temperature Reformer Temperature Pressure Stoichiometric
800 c 1073 K *11.84 atm
Table: FEED 1600 4800 0 0 0 6400
CH4 H20 co co2 H2 Total
PRODUCT 265 3,221 1,092 243 4,249 9,071
MOLFRW 0.0292 0.3551 0.1204 0.0268 0.4685
KP~ 167.6 Ratio1 167.6 KP~ 0.2936 Ratio2 0.2936 fractional conversion of CH4* 0.8346471 Converge 1* -0.00043805 moles of CO2 formed** 243.14362 Converge 2** -2.3364E-05 =
Reformer Temperature Reformer Temperature Pressure Stoichiometric CH4 H20 co co2 H2 Total
1.2 MPa
900 c 1173 K 11.84 atm
((Kpl/Ratiol)-l)*lOO ((Kp2/Ratio2)-l)*lOO
1.2 MPa
Table: FEED 1600 4800 0 0 0 6400
PRODUCT 54 3,054 1,346 200 4,839 9,492
MOLFRAC 0.0057 0.3217 0.1418 0.0211 0.5098 1.0000
KP~ 1443.6 Ratio1 1443.6 Kp2 0.2359 Ratio2 0.2359 fractional conversion of CH4* 0.966348 Converge l* -0.00083839 = moles of CO2 formed** 200.31077 Converge 2** -2.46013-05 =
13-11
((Kpl/Ratiol)-l)*lOO ((Kp2/Ratio2)-l)*lOO
