Flow Between Parallel Plates

Solution: First we start with the continuity equation in Cartesian coordinates: ∂ρ ∂ (ρv x ) ∂ (ρv y ) ∂ (ρv z ) + + + =0 ∂t ∂x ∂y ∂z

For incompressible fluid, the density is constant: ∂v x ∂v y ∂v z + =0 + ∂y ∂z ∂x

Since the flow is in the x-direction only, then we have only one component of the velocity v x ≠ 0 and v y = v z = 0 , continuity equation simplifies to: ∂v x =0 ∂x

Conclusion the simplified continuity equation implies that v x is not a function of x v x ≠ f(x)

CHE 204, Prepared by: Dr. Usamah Al-Mubaiyedh

Page 1

Also, for wide plates in z-direction, v x ≠ f (z ) Second we use the Navier-Stokes equations in Cartesian coordinates: ρ 

 ∂ 2v ∂ 2vx ∂ 2vx ∂v ∂v ∂v   ∂v x ∂p + v x x + v y x + v z x  = − + µ  2x + + ∂x ∂x ∂y ∂z  ∂y 2 ∂z 2  ∂t  ∂x

  + ρg x , 

 ∂ 2v y ∂ 2v y ∂ 2v y ∂v y ∂v y ∂v y   ∂v y ∂p  = − + µ  2 + + vx + vy + vz + ρ  2  ∂x ∂ t ∂ ∂y x ∂ y ∂ z ∂ ∂z 2 y   

  + ρg y ,  

 ∂v z ∂v ∂v ∂v + vx z + v y z + vz z ∂x ∂y ∂z  ∂t

ρ 

 ∂ 2v  ∂ 2vz ∂ 2vz ∂p  = − + µ  2z + + 2 ∂z ∂y 2 ∂z   ∂x

  + ρg z . 

To simplify Navier-Stokes equations we can utilize the following results: 1. Steady state:

∂ (any thing) =0 ∂t

∂ (any thing) =0 ∂z 3. We have one component of the velocity v x ≠ 0 and v y = v z = 0

2. Plates are wide in z-direction:

4. v x ≠ f(x, z) it is only a function of y: v x = f(y). 5. g x = g z = 0. Only we have gravity in y direction g y = -g. Therefore the N-S equations simplify to: 0=−

d 2vx ∂p +µ ∂x dy 2

(1)

0=−

∂p − ρg ∂y

(2)

0=0

(3)

Pressure Profile: Integrate equation (2): ∂p = − ρg ∂y

⇒

p = − ρ g y + f (x )

Note for we integrated the pressure with respect to y, hence the constant of integration is not a function of y but could be a function of x. In pressure driven flows like this problem the pressure changes linearly along the direction of the flow: CHE 204, Prepared by: Dr. Usamah Al-Mubaiyedh

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f (x ) = a x + b ⇒ p = − ρ g y + (a x + b )

y d P = P1

x

Pressure Driven Flow P1 > P2

P = P2

Now we apply boundary conditions for the pressure: @ x = 0 and y = 0 →

p = p1

@ x = L and y = 0 →

p = p2

p1 = − ρ g (0 ) + (a 0 + b )  p 2 − p1 ∆P = and b = p1 ⇒a = p 2 = − ρ g (0 ) + (a L + b ) L L

Therefore, the pressure profile is given by:   ∆P p = −ρ g y +  x + p1    L

(4)

Velocity Profile: Recall equation (1): 0=−

From equation (4):

d 2vx ∂p +µ ∂x dy 2

∂p ∆p , substitute in equation (1): = ∂x L

d 2 v x 1 ∆p = µ L dy 2

Integrate the above equation once: CHE 204, Prepared by: Dr. Usamah Al-Mubaiyedh

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dv x 1 ∆p y + C1 = dy µ L

Integrate second time: vx =

1 ∆p 2 y + C1 y + C 2 2µ L

To find the constants of integration apply the following Boundary Conditions: y=0

dv x =0 dy

(Velocity is maximum at center between the two plates)

y=d

vx = 0

(No - Slip Boundary Condition )

0=

1 ∆p 0 + C1 µ L

⇒

C1 = 0

0=

1 ∆p 2 d + C1 y + C 2 2µ L

⇒

C2 = −

vx =

1 ∆p 2 d 2µ L

1 ∆p 2 1 ∆p 2 y − d 2µ L 2µ L

Rearrange: vx =

1 ∆p 2 y −d2 2µ L

(

)

The above equation is similar to an equation of a parabola and hence the velocity profile is called a parabolic velocity profile, see figure below: Parabolic Velocity Profile y d P = P1

x

Pressure Driven Flow P1 > P2

CHE 204, Prepared by: Dr. Usamah Al-Mubaiyedh

P = P2

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Volumetric Flow Rate: Q = ∫ v x dA A

In this problem: dA = dy dz W +d

Q=

∫ ∫ v dy dz x

0 −d

If we consider the volumetric flow rate per unit width W = 1: +d

Q = ∫ v x dy −d

d

 1 ∆p 2 1 ∆p  y 3 2  Q= ∫ y − d dy = − d 2 y  2µ L 2µ L  3  −d −d 3 2d − ∆p Q= 3µ L d

(

)

Maximum Velocity: v Max = v x

y =0

=

1 − ∆p 2 d 2µ L

Mean Velocity: 2d 3 − ∆p Q(perunit width ) 3µ L d 3 − ∆p 2 vm = = = = v Max A(perunit width ) 2d 3µ L 3

Shear Stress:

τ xx τ xy  τ yx τ yy τ zx τ zy 

  ∂v x    2 ∂x    τ xz   ∂v y ∂v x    τ yz  = µ  + ∂y  ∂x   τ zz   ∂v z ∂v x  +   ∂z   ∂x

CHE 204, Prepared by: Dr. Usamah Al-Mubaiyedh

 ∂v x ∂v y  + ∂x  ∂y  ∂v y   2   ∂y   ∂v z ∂v y  + ∂z  ∂y

  ∂v x ∂v z   + ∂x   ∂z  ∂v y ∂v z  + ∂y  ∂z   ∂v z    2  ∂z  

         Page 5

Recall simplifications: ∂ (any thing) =0 ∂z 2. v x ≠ 0 and v y = v z = 0

1.

3. v x ≠ f(x, z) it is only a function of y: v x = f(y). This leads to the following simplifications:

τ xx τ xy  τ yx τ yy τ zx τ zy 

  0  τ xz   ∂v  τ yz  = µ  x  ∂y τ zz   0  

  

   0    0 0  0 0  

 ∂v x   ∂y

 ∂v 

Therefore, the only nonzero stresses are: τ xy = τ yx = µ  x  .  ∂y  Recall, v x =

1 ∆p 2 y −d2 2µ L

(

⇒ τ yx =

)

∆p y L

∆p = p 2 − p1 = −ve if y is + ve ⇒ τ yx is − ve if y is - ve ⇒ τ yx is + ve

Shear Stress Profile y d P = P1

x

Pressure Driven Flow P1 > P2

CHE 204, Prepared by: Dr. Usamah Al-Mubaiyedh

P = P2

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