Flows With More Than One Dependent
Variable - 2D Example
Juan M. Lopez
BIEN 501
Wednesday, March 21, 2007
Louisiana Tech University
Ruston, LA 71272
Recall - Generalized Newtonian
T  pI  2  D where   2tr D  D
Recall that:
D

1
T
v  v 
2
tr stands for “trace,” which is
the sum of the diagonal
elements. Tr(T)=Tii
While the expression looks complicated, it will look
much simpler once a given form for  is found.
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
Generalized Newtonian
T  pI  2  D where   2tr D  D
Recall that:
D

1
T
v  v 
2

tr stands for “trace,” which is
the sum of the diagonal
elements. Tr(T)=Tii

u
 2 i
x1

1 0 0
 u
u
T   P    v  0 1 0    1  2


 x 2 x1
0 0 1
 u1  u3
 x
 3 x1
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u2 u1

x1 x2
u
2 2
u2
u2 u3

x3 x2
u3 u1 


x1 x3 
u3 u2 

x2 x3 
u
2 3 
x3 
Parallel Plate Poiseuille Flow
•
Given: A steady, fully developed, laminar flow of a Newtonian fluid in a
rectangular channel of two parallel plates where the width of the channel is
much larger than the height, h, between the plates.
•
Find: The velocity profile and shear stress due to the flow.
•
•
•
•
Assumptions:
Entrance Effects Neglected
No-Slip Condition
No vorticity/turbulence
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Additional and Highlighted
Important Assumptions
• The width is very large compared to the
height of the plate.
• No entrance or exit effects.
• Fully developed flow.
• THEREFORE…
– Velocity can only be dependent on vertical
location in the flow (vx)
– (vy) = (vz) = 0
– The pressure drop is constant and in the xdirection only. p  Constant  p , where L is a length in x.
x
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L
Boundary Conditions
• No Slip Condition Applies
– Therefore, at y = -h/2 and y = +h/2, v = 0
• The bounding walls in the z direction are
often ignored. If we don’t ignore them we
also need:
– z = -w/2 and z = +w/2, v = 0, where w is the
width of the channel.
• For this problem we include this, and
make the width finite to make this
dependent on two variables.
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Incompressible Newtonian
Stress Tensor
Adapted from Table 3.3 in the text.

 u y u x 
u x
 u z u x  



2







x

x

y

x

z





  u
u y
 u z u y  
u x 
y

 
τ    

2
 

y 
y
z  
  x
 y
  u u 

 u z u y 
u z
x
z
 



2
  x  z    y  z 

z





 0

Now, we cancel terms out
  u
based on our assumptions.
τ     x
This results in our new
  y
tensor:
  u x
  
  z
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





 u x 
 u x  
  
 


y

z





0
0 


0
0 


Navier-Stokes Equations
In Vector Form:
 v 
    v  v  p   2 v  g
 t 
(Eq. 3.3.25)
Which we expand to component form from table 3.4:
x - component :
  2vx  2vx  2vx 
 vx
vx
v x
v x 
p
      2  2  2   g x
 
 vx
 vy
 vz
x
y
z 
x
y
z 
 t
 x
y - component :
  2v y  2v y  2v y 
v y
v y
v y 
 v y
p
      2  2  2   g y
 
 vx
 vy
 vz
x
y
z 
y
y
z 
 x
 t
z - component :
  2vz  2vz  2vz 
 vz
v z
v z
v z 
p
      2  2  2   g z
 
 vx
 vy
 vz
x
y
z 
z
y
z 
 t
 x
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Reducing Navier-Stokes
x - component :
  2vx  2vx  2vx 
 vx
vx
v x
v x 
p
      2  2  2   g x
 
 vx
 vy
 vz
x
y
z 
x
y
z 
 t
 x
y - component :
  2v y  2v y  2v y 
v y
v y
v y 
 v y
p
      2  2  2   g y
 
 vx
 vy
 vz
x
y
z 
y
y
z 
 x
 t
z - component :
  2vz  2vz  2vz 
 vz
v z
v z
v z 
p
      2  2  2   g z
 
 vx
 vy
 vz
x
y
z 
z
y
z 
 t
 x
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Reducing Navier-Stokes
• ThisModified
reduces to:

 p
0
x
Pressure
  2vx  2vx 
  2  2 
z 
 y
• Including our constant pressure drop:
  2vx  2vx 
p
0
  2  2 
L
z 
 y
• Oops! Now we have a nonhomogenous
higher-order differential equation that is
inseparable. How do we deal with it?
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DiffEq Assumptions
• We will assume this solution is a combination of simple parallel plate
Poiseuille flow plus some perturbation that is dependent on the walls
and finite width.
• Extracting the 1D Poiseuille flow, we can rewrite the equation as:
0
  2vx  2vx 
p

  2  2 
L
z 
 y
where vx
 vx  y, z 
 Vx  y     y , z 
Therefore :
0
 d 2Vx 
  2  2 
p

   2     2  2 
L
z 
 y
 dy 
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DiffEq Solution - Setup
• We can separate this into two equations, each of which equals zero.
• Why?
– 0=0+0
0
 d 2Vx 
  2  2 
p

   2     2  2 
L
z 
 y
 dy 
Separated :
0
 d 2Vx 
p

   2 
L
 dy 
0
  2  2 
 2  2 
z 
 y
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DiffEq Solution - Poiseuille
 d 2Vx 
p
   2 
L
 dy 
because this is the simple Poiseuille solution, we can see from
Eq. 2.7.18 that the above equation is equivalent to :
p
L
 d 2Vx 
   2 
 dy 
 d yx 

 
 dy 
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DiffEq Solution - Poiseuille
From our stress tensor definition , yx
 du x 

  
 dy 
Therefore, we can follow the solution from
Section 2.7.2 to end up with :
ph 2  4 y 2 
1  2 
ux 
8L 
h 
Now we can focus our remaining efforts on the perturbation
function.
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Perturbation Function - Reduction
0
  2  2 
 2  2 
z 
 y
Where
  y, z   Y  y Z z 
• We can approach this
perturbation function by
a separation of
variables method, as it
is homogeneous.
Therefore
0
  2Y  y Z  z   2Y  y Z  z  
 


2
2

y

z




 2Y  y 
 2 Z z 
1
 2Y  y 
 2 Z z 
  Z z 
 Y y

 Z z 
 Y y
2
2 
2
2 





y

z
Y
y
Z
z

y

z




 1  2Y  y 
1  2 Z z 
0 

2
2 





Y
y

y
Z
z

z


0
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Perturbation Function - Separation
• Because each term is independent of the other term, the
ONLY way this can be true is if each of the expressions is
equal to a constant. Thus we define a constant as follows:
2
2


1

Y
y
1

Z z 
2  

Y  y  y 2
Z  z  z 2
We can now use our standard homogeneou s
general solution :
Y  y   A1 sin y   A2 cosy 
Z  z   B1 sinh y   B2 cosh y 
• Now we can use our boundary conditions to solve for
these constants.
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Perturbation Function – B.C.’s
At y  0, we have a symetric region in our flow
(the point of maximum velocity on our parabola)
dY

 0|y  0
dy
dY
 A1 cos 0  A2  sin  0   0
dy
Because cos0  1, A1 must be 0.
At the walls (y   / - h/2) :
dY
 0 cos h 2  A2  sin  h 2   0
dy
To be a nontrivial solution, A2 cannot  0
Therefore, sin  h 2  0 (error in text ?)
Thus  can only have values of n  2n  1
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h
Perturbation Function – B.C.’s
For the Z portion of our separated function :
dZ
dZ

 0|z  0 
 B1n cosh  0   B2 n sinh  0  0
dz
dz
Because cosh 0  1, B1 must be 0.
dZ
  B2  sinh h   0
dz
To be a nontrivial solution, B2 cannot  0
We can now combine our equation for Y and our equation for Z to give us  .

  Y  y Z  z     A1 cosn y B2 cosh n z 
n 1
Because constants are just constants, they combine

   An cosh n z  cosn y 
n 1
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Perturbation Function – B.C.’s
We use our Boundary Conditions one more time to obtain :

  An cosh n z  cosn y 
  y, z 




 y , w 2
 y , w 2
n 1









  An cosh n w cosn y   u x
2
n 1


n 1
2
2



p
h
4
y
1  2 
An cosh n w cosn y  
2
8L 
h 
We now have an equation purely in terms of one variable (y). We can integrate to solve for
the coefficien t An . At this point the textbook multiplies both sides of the equation by
 2m  1y 
cos
. This makes both sides of the equation appropriat ely periodic.
h


This solution is nontrivial only when n  m, so this can be rewritten as :


n 1

2
2

  2n  1y 


2
n

1

y

p
h
4
y


1  2  cos
An cosh n w cosn y  cos



2
h
h  
h

 8L 



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Perturbation Function – Integration
We can now integrate :
h/2
 2n  1y 
An    cosh n w cosn y  cos
dy 
2



h
/
2
 
h

n 0 



ph 2  4 y 2   2n  1y 
h / 2 8L 1  h 2  cos h dy
DID YOU CATCH THAT?
h/2
Rearrangin g, we can re - write with the coefficien t isolated :
 h / 2 cosh  w cos y  cos 2n  1y dy

n
n
2
  h / 2
 
h

n 0 
2
h / 2 ph 
4 y 2   2n  1y 
h / 2 8L 1  h 2  cos h dy

An 


Which results in :

 ph 2 
32


 1 
3 3 
 8L  2n  1  
An 
 2n  1w 
cosh 

2
h


n
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for
n  0,1,2,..., 
This is a form of the
Fourier Transform.
Express a function as a
series of sin and cosine
terms, and then you can
integrate and
Perturbation Function – Integration
We can plug this into our original equations :
 2n  1z   2n  1y 
n


32

1
cosh

 cos

ph 2  4 y 2  ph 2 
h
h

 

1  2  
v x  y, z  

8L 
h  8L n 0
2n  13  3 cosh  2n  1w 
2h


The textbook covers a way of calculating the shear stress. However, we have the
stress tensor, so we can go to this tensor directly to calculate this from our equation
above.

 u x 
u x  

 0
  
 


 z  
 y 
  u 

x

τ    
0
0 
  y 

  u x 

0
0 

  


z




Louisiana Tech University
Ruston, LA 71272
You should be able to start
spotting the similarities between
our velocity equation, above, and
the stress tensor on the left.
Discussion
• Why would it be useful to run an analysis
like this?
– Helps select critical design dimensions for a
flow channel.
– If there is a controlling dimension, we can
design a workaround.
• Where else do you think they run this type
of analysis in engineering?
Louisiana Tech University
Ruston, LA 71272
Announcements
• Office hours today, let me know if you
need them
• Tutorial lab tonight…will go over more
problems and answer questions about the
current assignment.
• New assignment to be posted soon.
Louisiana Tech University
Ruston, LA 71272
• QUESTIONS?
Louisiana Tech University
Ruston, LA 71272