Transcript The Rectangular Channel Steven A. Jones BIEN 501 Friday, April 4th, 2008 Louisiana Tech University Ruston, LA 71272 Slide 1

The Rectangular Channel
Steven A. Jones
BIEN 501
Friday, April 4th, 2008
Louisiana Tech University
Ruston, LA 71272
Slide 1
The Rectangular Channel
Major Learning Objectives:
1. Deduce boundary conditions for a 2dimensional laminar internal flow problem.
2. Reduce continuity and momentum for the
problem.
3. Divide the momentum equation into its
homogeneous and non-homogeneous
components.
4. Transform the boundary conditions for the new
momentum equation.
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Ruston, LA 71272
Slide 2
The Rectangular Channel
Major Learning Objectives (continued):
• Use separation of variables to deduce the form
of the solutions.
• Apply boundary conditions along three
boundaries.
• Use superposition to deduce the series form of
the complete solution.
• Use orthogonality (from Sturm-Liouville) to
deduce the coefficients in the infinite series.
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Ruston, LA 71272
Slide 3
Rectangular Channel
y
y
h
2
z
w
2
x
z
Look at the half-channel. z=0 is the midline of the channel.
z ranges from –w/2 to +w/2
y ranges from –h/2 to +h/2
Fully Developed Flow
No-Slip Boundary Conditions
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Slide 4
Boundary Conditions
y
y
h
2
z
w
2
x
z
We can write down 6 boundary conditions, (but we only
need 4).
vx   h / 2, z   0
vx  y ,  w / 2   0
vx / y  0, z   0
vx / dz  y, 0   0
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Slide 5
Navier-Stokes Equations
Look at v v Term for Fully Developed Flow
vx
vx
vx
vx
 vy
 vz
x  momentum
x
y
z
v y
v y
v y
vx
 vy
 vz
y  momentum
x
y
z
vz
vz
vz
vx
 vy
 vz
z  momentum
x
y
z
With vx = vy = 0 and no z gradients, which terms go to
zero?
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Slide 6
Navier-Stokes Equations
vx
vx
vx
vx
 vy
 vz
x  momentum
x
y
z
v y
v y
v y
vx
 vy
 vz
x  momentum
x
y
z
vz
vz
vz
vx
 vy
 vz
x  momentum
x
y
z
All of them!
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Slide 7
Continuity
vx v y vz


0
x y z
All terms in the continuity equation are zero.
This result tells us that our assumptions are
consistent with continuity.
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Slide 8
y-momentum
Before we look at x-momentum, look at y-momentum.
v y
t
 vx
v y
x
 vy
v y
y
 vz
v y
z

2
2
2


v

v

vy
1 p
y
y

  2  2  2
 x
 y
y
z


  g y

With no y and z velocities, this equation tells us that the y
pressure gradient is cancelled by gravity.
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Slide 9
Constant Pressure Gradient
 2vx  2vx 1 p
 2 
2
y
z
 x
We learned from the y and z momentum that pressure
did not depend on y and z. So the right hand side of the
above equation can only depend on x, but the left hand
of the equation cannot depend on x because of the fully
developed flow assumption. Therefore, p / x cannot
depend on x, y or z and must be constant.
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Slide 10
Particular Solution
The equation:
 vx  vx 1 p
 2 
2
y
z
 x
2
2
Is non-homogeneous, meaning that the right hand side
is not zero. A standard method for solving this type of
equation is to first find a “particular solution” and
subtract that solution from the equation to find a new
homogeneous equation.
1 p  4 y 2 
It is easy to show that the solution: vx 
1  2 
 x 
h 
Satisfies the equation (hint: substitute this function back
into the equation).
Louisiana Tech University
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Slide 11
Particular Solution
The function:
1 p  4 y 2 
v x  Vx 
1  2 
 x 
h 
Also satisfies the boundary conditions at y  h / 2 .
It does not satisfy the boundary conditions for z   w / 2
so our work is not quite done yet. However, we have
made progress.
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Slide 12
Comments on the Particular Solution
1 p  4 y 2 
Instead of: vx 
1  2 
 x 
h 
1 p  4 z 2 
We could have used: vx 
1  2 
 x  w 
This solution would have reversed the roles of y and z, but
the procedure would otherwise be the same.
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Slide 13
Comments on Particular Solution
When you have an equation like:
3vx 3vx  2vx  2vx vx
 3  2  2 
 C0
3
y
z
y
z
y
Or something more complicated, it is generally easy to find a
particular solution. Choose one of the variables, say y, and ask
if there is a function f(y) that will yield a constant when
differentiated an amount of times equal to the lowest order
differential. Since it is not a function of z, the derivatives in z do
not contribute, nor do the higher order derivatives in y (because
the derivative of a constant is zero).
For example, a particular solution to the above equation is
vx  y   C0 x
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Slide 14
Exercise
Find particular solutions to the following:
 4 vx  4 vx
 4  C0
4
y
z
3vx
 3 vx
 2 vx
 2 vx
  3   2   2  C0
3
y
z
y
z
 5 vx  3vx

 vx  C0
5
2
y
zy
 3 vx
 3 vx
 2 vx

  2  C0
3
2
y
zy
y
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Slide 15
Exercise Answers
 4vx  4vx
C0 4
C0 4
 4  C0  vx 
y or
z
4
y
z
24
24
 3 vx
 3vx
 2 vx
 2 vx
C0 2
C0 2
  3   2   2  C0  vx 
y or
z
3
y
z
y
z
2
2
 vx  vx

 vx  C0  vx  C0
5
2
y
zy
5
3
3vx
 3 vx
 2 vx
C0 2

  2  C0  vx 
y
3
2
y
zy
y
2
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Slide 16
Complete Solution
The complete solution will be of the form:
1 p  4 y 2 
vx  Vx  y   f  y, z  
1  2   f  y, z 
 x 
h 
Where the particular solution part will handle the
nonhomogeneity in the partial differential equation and
the second part, f (y, z), will satisfy the homogeneous
equation and satisfy the boundary conditions.
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Slide 17
Reduce the Equation
Plug:
1 p  4 y 2 
vx 
1  2   f  y, z 
 x 
h 
 vx  vx 1 p
 2 
2
y
z
 x
2
Into:
To get:
2
These two
terms cancel
2
 2  1 p  4 y 2    f  y, z 
1  2   
2 

y   x 
h 
y 2
2
 2  1 p  4 y 2    f  y, z  1 p
 2 

1  2   
2
z   x 
h 
z
 x
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Slide 18
Reduce the Equation (continued)
We are left with Laplace’s equation:
 2f  y, z 
y
But remember
2

 2f  y, z 
z
2
0
vx  y, z   Vx  y   f  y, z 
so
f  y, z   vx  y, z  Vx  y 
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Slide 19
Reduce the Equation (continued)
If
f  x, y   vx  x, y  Vx  y 
And if vx must satisfy the boundary conditions:
vx   h / 2, z   0
vx  y,  w / 2   0
vx / y  0, z   0
vx / dz  y, 0   0
Then f must satisfy the boundary conditions:
f   h / 2, z   Vx   h / 2   0
f  y,  w / 2   Vx  y   0
f / y  0, z   Vx  0  / y  0
f / dz  y, 0   Vx  y  z  0
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Slide 20
Exercise
Why is each of the indicated terms below zero?
f   h / 2, z   Vx   h / 2   0
f  y,  w / 2   Vx  y   0
f / y  0, z   Vx  0  / y  0
f / dz  y, 0   Vx  y  z  0
Vx   h / 2   0
Vx  0  / y  0
Vx  y  z  0
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Slide 21
Exercise Answers
Why is each of the indicated terms below zero?
f   h / 2, z   Vx   h / 2   0
f  y,  w / 2   Vx  y   0
f / y  0, z   Vx  0  / y  0
f / dz  y, 0   Vx  y  z  0
Vx   h / 2   0 From Couette flow (plug h/2 into Vx(y))
Vx  0  / y  0 From symmetry of Couette flow
Vx  y  z  0
Because Vx(y) does not depend on z.
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Slide 22
Summary of Equations
We must therefore solve Laplace’s equation:
 2f  y, z 
y
2

 2f  y, z 
z
2
0
Subject to the following boundary conditions:
f   h / 2, z   0
f / y  0, z   0
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1 p  4 y 2 
f  y,  w / 2   Vx  y  
1  2   0
 x 
h 
f / dz  y, 0   0
Slide 23
Visual
Vz  y 
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Slide 24
Separable Solution to
Homogeneous Equation
f  f  y, z   Y  y  Z  z 
  2YZ  2YZ 
 2  2 0
z 
 y
2
2
Y
 Z
Z 2 Y 2  0
y
z
1Y
1Z
1d Z
2
2 1 d Y
2

 
 ,
 
2
2
2
2
Y y
Z z
Z dz
Y dy
2
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2
2
2
Slide 25
Solution to ODEs
2
d 2Y
d
Z
2
2


Y

0,


Z 0
2
2
dy
dz
Z  z   C cosh  z  D sinh  z
Y  y   A cos  y  B sin  y
It may help to remember that the sin and sinh (cos and
cosh) functions can be written as:
eia  e i
ei  e i
cos  
, sin  
2
2i
e a  e 
e  e 
cosh  
, sinh  
2
2
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Slide 26
Solution to ODEs
Z  z   C cosh  z  D sinh  z
If
Y  y   A cos  y  B sin  y
And
f  y, z   Y  y  Z  z 
Then anything that has this form:
f  y, z   C cosh  z  D sinh  z  A cos  y  B sin  y 
Satisfies the homogeneous equation.
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Slide 27
Superposition
Since there may be multiple values of l that work, a
complete solution must consider all possible such
solutions. We also note that the equations are linear, so
that we can add solutions and still have a solution.
Thus, we can write:
f  y, z    A  C cosh  z  D sinh  z  A cos  y  B sin  y 
all 
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Slide 28
Boundary Conditions in y
f  y, z     C cosh  z  D sinh  z  A cos  y  B sin  y 
all 
First consider the boundary condition along the centerline
where y = 0.
f  0, z 
y
 0   A  sin   0   B  cos   0   0  B  0
Next, the boundary condition at y = h/2 requires:
f  h / 2, z     C cosh  z  D sinh  z  A cos h / 2   0
all 
This equation is true for all values of z only if: A cos  h / 2  0

i.e.  h / 2    n  12      2n  1 (The book uses n’=2n+1)
h
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Slide 29
Boundary Conditions in z
We now have the following:

2n  1 
2n  1 


f  y, z     Cn cosh
z  Dn sinh
h
h
n=0 


2n  1  

z  An cos
y
h


But it is a lot to write, so we will continue to write it in terms
of  for now.
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Slide 30
Boundary Conditions in z

f  y, z     C cosh  z  D sinh  z  A cos  y 
n=0
Consider the boundary condition along the centerline where
z = 0.
f  y, 0 
z
 0  C  sinh   0   D  cosh   0   0  D  0
Next, the boundary condition at z = w/2 requires:
1 p  4 y 2 
f  y, w / 2    A cosh   w / 2  cos   y  
1  2 
 x 
h 
all 
This boundary condition is the one that requires the most
work.
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Slide 31
Boundary Conditions in z
Notice that this equation is a function of y only.
1 p  4 y 2 
f  y, w / 2    A cosh   w / 2  cos   y  
1  2 
 x 
h 
all 
It can be interpreted to mean that the right hand side is
being expanded as a Fourier cosine series within the interval
of interest (i.e. –h/2 < y < h/2).
   2n  1 y 
p  4 y 2  

1  2     n cos 
x 
h  n 0
h


We use the standard approach that was used to derive Fourier
series.
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Slide 32
Orthogonal Expansion
Multiply both sides of the equation by cos (m y ).
cos  m y  p  4 y 2 
cos  m y   A cosh  n w / 2  cos  n y  
1  2 

x 
h 
all n
cos  m y  p  4 y 2 
A cosh  n w / 2  cos  n y  cos  m y  

1  2 

x 
h 
all n
Integrate from –h/2 to h/2
cos  m y  p  4 y 2 
h / 2 alln A cosh  n w / 2 cos  n y  cos  m y dy  h / 2  x 1  h2  dy
h/2
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h/2
Slide 33
Orthogonality
cos  m y  p  4 y 2 
h / 2 alln A cosh  n w / 2 cos  n y  cos  m y dy  h / 2  x 1  h2  dy
h/2
h/2
The left hand side will be zero for all m except n = m so:
cos  m y  p  4 y 2 
h / 2 Am cosh  m w / 2 cos  m y  dy  h / 2  x 1  h2  dy
h/2
2
h/2
Note that the sum disappeared because only the
value of n that is equal to m is needed.
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Slide 34
Orthogonality
2
cos

y





p
4
y
m
2
h / 2 Am cosh  m w / 2 cos  m y  dy  h / 2  x 1  h2  dy
h/2
h/2
But

h/2
h / 2
cos2  n y dy  h / 2
So
cos  n y  p  4 y 2 
cosh  n w / 2   
1  2  dy

h
/
2
2

x 
h 
An h
Or
h/2
cos  m y  p  4 y 2 
Am 
1  2  dy


h
/
2
h cosh  m w / 2 

x 
h 
Louisiana Tech University
Ruston, LA 71272
2
h/2
Slide 35
Integrating Cosine Squared
cos (with 1/2)
cos squared (with 1/2)
cos (with 3/2)
cos squared (with 3/2)
zero line
1.5
cosines
1
0.5
0
-0.5
-1
-1.5
-25
-20
-15
-10
-5
0
5
10
15
20
25
y
The area of the rectangle is h. The area under cos2 is h/2.
Louisiana Tech University
Ruston, LA 71272
Slide 36
Orthogonality
cos  m y  p  4 y 2 
Am 
1  2  dy


h
/
2
h cosh  m w / 2 

x 
h 
2
h/2
So
h
p 
4
2
Am 
sin

y

 m  y h2 2

h cosh  m w / 2  x 
h
2

2
y
cos

y
dy
 m  
h / 2

h/2
The final integral can be obtained with integration by parts
(twice).
Louisiana Tech University
Ruston, LA 71272
Slide 37
Derived Information
The final form of the solution is:
1 dp  4 y 2 
vx  y , z   
1  2    A cosh  z / 2  cos   y 
 dx 
h  all 
We can obtain the shear stress from the stress tensor.
Louisiana Tech University
Ruston, LA 71272
Slide 38
The Stress Tensor
For fluids:

u1

x1

 
1 u u 
τ  2   2  1 
 2  x1 x2 

 1  u3 u1 
 2  x  x 
3 
  1
1  u1 u2 



2  x2 x1 
u2
x2
1  u3 u2 



2  x2 x3 
1  u1 u3  



2  x3 x1  

1  u2 u3  



2  x3 x2  

u3


x3

The shear stress has 4 non-zero components.
Louisiana Tech University
Ruston, LA 71272
Slide 39
Shear Stress, Bottom Surface

 0

 1  v
τ  2   x
 2  y

 1  vx
 2  z





0
0 





0
0 



Along the bottom surface, we are concerned only with txy.
1  vx 


2  y 
1  vx
2  z
vx
t xy  
y
Louisiana Tech University
Ruston, LA 71272
Slide 40
Shear Stress, Bottom Surface
f  y, z    A cosh  z / 2  cos   y 
all 
vx
t xy  
y
t xy    A cosh  z / 2  sin   y 
all 
Louisiana Tech University
Ruston, LA 71272
Slide 41
Relationship of Flow Rate to
Pressure Gradient
To obtain flow rate in terms of pressure gradient, we
must integrate the velocity over the cross-section.
Q
w/ 2

h/2
w/ 2 h / 2
vx  y, z  dy dz
1 dp  4 y 2 
  
1  2    A cosh  z / 2  cos   y dydz

w/ 2 h / 2
 dx 
h  all 
w/ 2
h/2
This relationship could be used, for example, to determine
how much pressure is required to drive blood through a
microchannel device at a given flow rate.
Louisiana Tech University
Ruston, LA 71272
Slide 42
Example
You are interested in designing a microdevice that samples
blood from a vein and causes it to flow with a shear rate of
15 dynes/cm2 over a microchannel that is coated with
fibrinogen. The pressure difference driving the flow is the
venous pressure. If you use a vacuum container at the
downstream end of the device, can you obtain the required
shear stress, and if so, what should be the dimensions of
the channel?
Louisiana Tech University
Ruston, LA 71272
Slide 43