Transcript Flows With More Than One Dependent Variable - 2D Example Juan M.
Flows With More Than One Dependent Variable - 2D Example
Juan M. Lopez Steven A. Jones BIEN 501 Wednesday, April 4, 2008
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Recall - Generalized Newtonian
T
p
I
2
D
where 2 tr
D
D
Recall that:
D
1 2
v
T
tr stands for “trace,” which is the sum of the diagonal elements. Tr(
T
)=
T ii
While the expression looks complicated, it will look much simpler once a given form for is found.
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Generalized Newtonian
T
p
I
2
D
where 2 tr
D
D
Recall that:
D
1 2
v
T
tr stands for “trace,” which is the sum of the diagonal elements. Tr(
T
)=
T
ii
T
P
v
1 0 0 0 1 0 0 0 1
u
1
x u
1 2
x
3 2
u i
x
1
u
x u
1 3
x
2 1
u
2
x
1
u
2
x
3 2
u
2
u
x
1 2
u
2
u
3
x
2
u
3
x u
1 3
x
2 2
u
3
u x u
x
1 3 2 3
x
3
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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 (
v x
) – (
v y
) = (
v z
) = 0 – The pressure drop is constant and in the x direction only.
p
x
Constant p
L
, where
L
is a length in
x
.
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Boundary Conditions
• No Slip Condition Applies – Therefore, at y = -h/2 and y = +h/2,
v
– z = -w/2 and z = +w/2,
v
width of the channel.
= 0 • The bounding walls in the z direction are often ignored. If we don’t ignore them we also need: = 0, where w is the • 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.
τ
2
u y
x
u
x z
u
x x
u
y x
u
z x
u y
x
2
u
y z
u
y x
u y
y
u y
z
u
x z
u
y z
2
u
z z
u
z x
u y
z
Now, we cancel terms out based on our assumptions.
This results in our new tensor:
τ
0
u
y x
u
z x
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u
y x
0 0
u
z x
0 0
Navier-Stokes Equations
In Vector Form:
v
t
v
v
p 2
v
g
Which we expand to component form from table 3.4: x component :
v x
t
v x
v
x x
v y
v
y x
v z
v x
z
p
x
2
x v x
2 2
y v x
2 2
z v x
2
g x
( Eq.
3.3.25) y component :
v y
t
v x
v y
x
v y
v y
y
v z
v y
z
p
y
2
v y
x
2 2
v y
y
2 2
v y
z
2
g y
z component :
v z
t
v x
v z
x
v y
v
y z
v z
v
z z
p
z
2
x v z
2 2
y v z
2 2
z v z
2
g z
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Reducing Navier-Stokes
x component :
v
t x
v x
v
x x
v y
v x
y
v z
v
z x
p
x
2
x v x
2 2
v x
y
2 2
z v x
2
g x
y component :
v y
t
v x
v y
x
v y
v y
y
v z
v y
z
p
y
2
v
x
2
y
2
v y
y
2 2
v y
z
2
g y
z component :
v
t z
v x
v
x z
v y
v
y z
v z
v
z z
p
z
2
x v z
2 2
y v z
2 2
z v z
2
g z
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Reducing Navier-Stokes
• This reduces to: Modified 0 p
x
2
y v
2
x
2
z v
2
x
• Including our constant pressure drop: 0 p
L
2
v x
y
2 2
v x
z
2 • 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 where
v x
p
L
v
V x x
2
v
y
2
x
2
v
z
2
x
Therefore : 0 p
L
d
2
V x dy
2
y
2 2 2
z
2
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DiffEq Solution - Setup
• We can separate this into two equations, each of which equals zero.
• Why?
– 0=0+0 0 p
L
d
2
V x dy
2
y
2 2 2
z
2 Separated : 0 p
L
d
2
V x dy
2 0
y
2 2 2
z
2
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DiffEq Solution - Poiseuille
p
L
d
2
V x dy
2 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
2
V x dy
2
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 :
u x
p
h
2 8
L
1 4
y
2
h
2 Now we can focus our remaining efforts on the perturbation function.
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Perturbation Function - Reduction
0 2
y
2 2
z
2 Where
y
,
z Y y Z z
Therefore 0 0 0 2
Y
y
2 2
Y
z
2
Z
Y
1 2
Y
2
Y
y
1
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• We can approach this perturbation function by a separation of variables method, as it is homogeneous.
Y y
1
Z
2
Y
y
2
y Y y
2
Z
z
2
z
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
Y
1 2
Y
y
2
Z
1 2
Z
z
2 We can now use our standard homogeneou s general
Y Z
solution
A
1
B
1 sin sinh
y
:
A
2
B
2 cos 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 dy
0 |
y
0
dY
A
1 cos 0
A
2 sin 0
dy
Because cos 1 ,
A
1 must be 0.
At the walls (y / h/2) : 0
dY
0 cos
h
2
A
2 sin
h
2 0
dy
To be a nontrivial Therefore, sin
h
solution, 2 0
A
2 cannot (
error in text
0 ?) Thus can only have values of
n
2
n
1
h
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Perturbation Function – B.C.’s
For the Z portion of our separated function :
dZ dz
0 |
z
0 Because cosh
dZ
1 ,
dz B
1
B
1
n
must cosh be 0.
B
2
n
sinh 0
dZ
B
2 sinh 0
dz
To be a nontrivial solution,
B
2 cannot 0 We can now combine our equation for Y and our equation for Z to give us .
Y
Z
n
1
A
1 cos
n
B
2 cosh
n
Because constants are just constants, they combine
n
1
A n
cosh
n
n
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Perturbation Function – B.C.’s
We use our Boundary Conditions one more time to obtain :
y
,
y
,
w
2
y
,
w
2
n
1
A n n
1
A n n
1
A n
cosh cosh
n
cosh
n
n
n
w
2 cos
n
u x w
2 cos
n
p
h
2 8
L
1 4
y
2
h
2 We now have an equation purely in terms of one variable (y).
We can integrate to solve for the coefficien cos 2
m
1
y
t
h A n
.
At this .
This point the textbook multiplies makes both sides of the equation both sides of the equation appropriat ely periodic.
by This
n
1
A n
solution cosh
n
is nontrivial
w
2 cos
n
only when cos 2
n
n m, so this can be rewritten 1
y
h
p
h
2 8
L
1 4
y
2
h
2 cos as : 2
n
1
y h
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Perturbation Function – Integration
We can now
A n
n 0
h h
/ / 2 2 integrate cosh
n w
2 : cos
h h
/ / 2 2 p
h
2 8
L
1 4
y
2
h
2
n
cos cos 2
n
2
n
1
y h
1
y dy h dy
Rearrangin
A n
g, we can re n 0
h h
/
h h
/ / 2 2 / 2 2 cosh p
h
8
L
2
n
write 1 2 4 with the cos
h y
2 2
n
cos coefficien 2
n
cos 2
n
h
1
y
h
t isolated 1
y dy dy
: Which results in :
A n
1
n
8 p
h
L
2 2 cosh
n
2
h
32 2
n
1
w
1 3
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3
Ruston, LA 71272
for n
0 , 1 , 2 ,...,
DID YOU CATCH THAT?
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
v x
p
h
2 8
L
1 4
y
2
h
2 p
h
2 8
L
equations n 0 32 1 :
n
2 cosh
n
1 3 2 3
n
h
1
z
2 cos
n
1 2
n
w
cosh 2
h
1
y h
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
τ
above.
0
u
y x
u
z x
u
y x
0 0
u
z x
0 0 You should be able to start spotting the similarities between our velocity equation, above, and the stress tensor on the left.
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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?
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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.
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• QUESTIONS?
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