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Transcript of Louisiana Tech University Ruston, LA 71272 Slide 1 The Rectangular Channel Steven A. Jones BIEN 501...
Louisiana Tech UniversityRuston, LA 71272
Slide 1
The Rectangular Channel
Steven A. Jones
BIEN 501
Friday, April 4th, 2008
Louisiana Tech UniversityRuston, LA 71272
Slide 2
The Rectangular Channel
Major Learning Objectives:1. Deduce boundary conditions for a 2-
dimensional 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.
Louisiana Tech UniversityRuston, LA 71272
Slide 3
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 4
Rectangular Channel
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
x
y
z
2
hy
2
wz
Louisiana Tech UniversityRuston, LA 71272
Slide 5
Boundary Conditions
We can write down 6 boundary conditions, (but we only need 4).
x
y
z
2
hy
2
wz
/ 2, 0
, / 2 0
/ 0, 0
/ ,0 0
x
x
x
x
v h z
v y w
v y z
v dz y
Louisiana Tech UniversityRuston, LA 71272
Slide 6
Navier-Stokes Equations
Look at Term for Fully Developed Flow
With vx = vy = 0 and no z gradients, which terms go to zero?
momentum
momentum
momentum
x x xx y z
y y yx y z
z z zx y z
v v vv v v x
x y z
v v vv v v y
x y z
v v vv v v z
x y z
v v
Louisiana Tech UniversityRuston, LA 71272
Slide 7
Navier-Stokes Equations
All of them!
momentum
momentum
momentum
x x xx y z
y y yx y z
z z zx y z
v v vv v v x
x y z
v v vv v v x
x y z
v v vv v v x
x y z
Louisiana Tech UniversityRuston, LA 71272
Slide 8
Continuity
0yx zvv v
x y z
All terms in the continuity equation are zero. This result tells us that our assumptions are consistent with continuity.
Louisiana Tech UniversityRuston, LA 71272
Slide 9
y-momentum
2 2 2
2 2 2
1
y y y yx y z
y y yy
v v v vv v v
t x y z
v v vpg
y x y z
With no y and z velocities, this equation tells us that the y pressure gradient is cancelled by gravity.
Before we look at x-momentum, look at y-momentum.
Louisiana Tech UniversityRuston, LA 71272
Slide 10
Constant Pressure Gradient2 2
2 2
1x xv v p
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, cannot depend on x, y or z and must be constant.
/p x
Louisiana Tech UniversityRuston, LA 71272
Slide 11
Particular Solution
2 2
2 2
1x xv v p
y z x
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.
It is easy to show that the solution:
Satisfies the equation (hint: substitute this function back into the equation).
2
2
1 41x
p yv
x h
The equation:
Louisiana Tech UniversityRuston, LA 71272
Slide 12
Particular Solution
Also satisfies the boundary conditions at .
It does not satisfy the boundary conditions for
so our work is not quite done yet. However, we have made progress.
2
2
1 41x x
p yv V
x h
The function:
/ 2y h
/ 2z w
Louisiana Tech UniversityRuston, LA 71272
Slide 13
Comments on the Particular Solution
We could have used:
2
2
1 41x
p yv
x h
Instead of:
2
2
1 41x
p zv
x w
This solution would have reversed the roles of y and z, but the procedure would otherwise be the same.
Louisiana Tech UniversityRuston, LA 71272
Slide 14
Comments on Particular Solution
3 3 2 2
03 3 2 2x x x x xv v v v v
Cy 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
When you have an equation like:
0xv y C x
Louisiana Tech UniversityRuston, LA 71272
Slide 15
Exercise
4 4
04 4x xv v
Cy z
Find particular solutions to the following:
3 3 2 2
03 3 2 2x x x xv v v v
Cy z y z
5 3
05 2x x
x
v vv C
y z y
3 3 2
03 2 2x x xv v v
Cy z y y
Louisiana Tech UniversityRuston, LA 71272
Slide 16
Exercise Answers
4 44 40 0
04 4or
24 24x x
x
v v C CC v y z
y z
3 3 2 2
2 20 003 3 2 2
or2 2
x x x xx
v v v v C CC v y z
y z y z
5 3
0 05 2x x
x x
v vv C v C
y z y
3 3 220
03 2 2 2x x x
x
v v v CC v y
y z y y
Louisiana Tech UniversityRuston, LA 71272
Slide 17
Complete Solution
Where the particular solution part will handle the nonhomogeneity in the partial differential equation and the second part, (y, z), will satisfy the homogeneous equation and satisfy the boundary conditions.
2
2
1 4, 1 ,x x
p yv V y y z y z
x h
The complete solution will be of the form:
Louisiana Tech UniversityRuston, LA 71272
Slide 18
Reduce the Equation
2 2
2 2
1x xv v p
y z x
Into:
Plug: 2
2
1 41 ,x
p yv y z
x h
To get:
22 2
2 2 2
22 2
2 2 2
,1 41
,1 4 11
y zp y
y x h y
y zp y p
z x h z x
These two terms cancel
Louisiana Tech UniversityRuston, LA 71272
Slide 19
Reduce the Equation (continued)
We are left with Laplace’s equation:
2 2
2 2
, ,0
y z y z
y z
, ,x xv y z V y y z But remember so
, ,x xy z v y z V y
Louisiana Tech UniversityRuston, LA 71272
Slide 20
Reduce the Equation (continued)
If
And if vx must satisfy the boundary conditions:
, ,x xx y v x y V y
/ 2, 0 , / 2 0
/ 0, 0 / ,0 0
x x
x x
v h z v y w
v y z v dz y
Then must satisfy the boundary conditions:
/ 2, / 2 0 , / 2 0
/ 0, 0 / 0 / ,0 0
x x
x x
h z V h y w V y
y z V y dz y V y z
Louisiana Tech UniversityRuston, LA 71272
Slide 21
Exercise
/ 2, / 2 0 , / 2 0
/ 0, 0 / 0 / ,0 0
x x
x x
h z V h y w V y
y z V y dz y V y z
Why is each of the indicated terms below zero?
/ 2 0
0 / 0
0
x
x
x
V h
V y
V y z
Louisiana Tech UniversityRuston, LA 71272
Slide 22
Exercise Answers
/ 2, / 2 0 , / 2 0
/ 0, 0 / 0 / ,0 0
x x
x x
h z V h y w V y
y z V y dz y V y z
Why is each of the indicated terms below zero?
/ 2 0
0 / 0
0
x
x
x
V h
V y
V y z
From Couette flow (plug h/2 into Vx(y))
From symmetry of Couette flow
Because Vx(y) does not depend on z.
Louisiana Tech UniversityRuston, LA 71272
Slide 23
Summary of Equations
We must therefore solve Laplace’s equation:
2 2
2 2
, ,0
y z y z
y z
Subject to the following boundary conditions:
2
2
1 4/ 2, 0 , / 2 1 0
/ 0, 0 / ,0 0
x
p yh z y w V y
x h
y z dz y
Louisiana Tech UniversityRuston, LA 71272
Slide 24
Visual
zV y
Louisiana Tech UniversityRuston, LA 71272
Slide 25
Separable Solution to Homogeneous Equation
2 2
2 2
2 2
2 2
,
0
0
y z Y y Z z
YZ YZ
y z
Y ZZ Y
y z
2 2 2 22 2 2
2 2 2 2
1 1 1 1,
Y Z d Z d Y
Y y Z z Z dz Y dy
Louisiana Tech UniversityRuston, LA 71272
Slide 26
Solution to ODEs
2 22 2
2 20, 0
cosh sinh
cos sin
d Y d ZY Z
dy dz
Z z C z D z
Y y A y B y
It may help to remember that the sin and sinh (cos and cosh) functions can be written as:
cos , sin2 2
cosh , sinh2 2
ia i i i
a
e e e e
i
e e e e
Louisiana Tech UniversityRuston, LA 71272
Slide 27
Solution to ODEs
cosh sinh
cos sin
Z z C z D z
Y y A y B y
And
If
,y z Y y Z z
Then anything that has this form:
, cosh sinh cos siny z C z D z A y B y
Satisfies the homogeneous equation.
Louisiana Tech UniversityRuston, LA 71272
Slide 28
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:
all
, cosh sinh cos siny z A C z D z A y B y
Louisiana Tech UniversityRuston, LA 71272
Slide 29
Boundary Conditions in y
all
, cosh sinh cos siny z C z D z A y B y
First consider the boundary condition along the centerline where y = 0.
0,0 sin 0 cos 0 0 0
zA B B
y
Next, the boundary condition at y = h/2 requires:
all
/ 2, cosh sinh cos / 2 0h z C z D z A h
This equation is true for all values of z only if: cos / 2 0A h
12. . / 2 2 1i e h n n
h
(The book uses n’=2n+1)
Louisiana Tech UniversityRuston, LA 71272
Slide 30
Boundary Conditions in z
n=0
2 1 2 1 2 1, cosh sinh cosn n n
n n ny z C z D z A y
h h h
We now have the following:
But it is a lot to write, so we will continue to write it in terms of for now.
Louisiana Tech UniversityRuston, LA 71272
Slide 31
Boundary Conditions in z
n=0
, cosh sinh cosy z C z D z A y
Consider the boundary condition along the centerline where z = 0.
,00 sinh 0 cosh 0 0 0
yC D D
z
Next, the boundary condition at z = w/2 requires:
2
2all
1 4, / 2 cosh / 2 cos 1
p yy w A w y
x h
This boundary condition is the one that requires the most work.
Louisiana Tech UniversityRuston, LA 71272
Slide 32
Boundary Conditions in z
Notice that this equation is a function of y only.
2
2all
1 4, / 2 cosh / 2 cos 1
p yy w A w y
x h
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).
We use the standard approach that was used to derive Fourier series.
2
20
2 141 cosn
n
n yp y
x h h
Louisiana Tech UniversityRuston, LA 71272
Slide 33
Orthogonal Expansion
Multiply both sides of the equation by cos (m y ).
2
2all
cos 4cos cosh / 2 cos 1
n
mm n n
y p yy A w y
x h
2
2all
cos 4cosh / 2 cos cos 1
n
mn n m
y p yA w y y
x h
Integrate from –h/2 to h/2
2/ 2 / 2
2/ 2 / 2all
cos 4cosh / 2 cos cos 1
n
h h mn n mh h
y p yA w y y dy dy
x h
Louisiana Tech UniversityRuston, LA 71272
Slide 34
Orthogonality
The left hand side will be zero for all m except n = m so:
Note that the sum disappeared because only the value of n that is equal to m is needed.
2/ 2 / 2
2/ 2 / 2all
cos 4cosh / 2 cos cos 1
n
h h mn n mh h
y p yA w y y dy dy
x h
2/ 2 / 222/ 2 / 2
cos 4cosh / 2 cos 1
m
h h mm mh h
y p yA w y dy dy
x h
Louisiana Tech UniversityRuston, LA 71272
Slide 35
Orthogonality
But
/ 2 2
/ 2cos / 2
h
nhy dy h
So
2/ 2
2/ 2
cos 4cosh / 2 1
2n
h nn h
A h y p yw dy
x h
2/ 2 / 222/ 2 / 2
cos 4cosh / 2 cos 1
m
h h mm mh h
y p yA w y dy dy
x h
Or
2
/ 2
2/ 2
cos2 41
cosh / 2m
h m
hm
y p yA dy
h w x h
Louisiana Tech UniversityRuston, LA 71272
Slide 36
Integrating Cosine Squared
-1.5
-1
-0.5
0
0.5
1
1.5
-25 -20 -15 -10 -5 0 5 10 15 20 25
y
cosi
nes
cos (with 1/2)cos squared (with 1/2)cos (with 3/2)cos squared (with 3/2)zero line
The area of the rectangle is h. The area under cos2 is h/2.
Louisiana Tech UniversityRuston, LA 71272
Slide 37
Orthogonality
So
The final integral can be obtained with integration by parts (twice).
2
/ 2
2/ 2
cos2 41
cosh / 2m
h m
hm
y p yA dy
h w x h
2
2
/ 2 22 / 2
2 4sin cos
cosh / 2
h
hm
h
m my hm
pA y y y dy
h w x h
Louisiana Tech UniversityRuston, LA 71272
Slide 38
Derived Information
The final form of the solution is:
We can obtain the shear stress from the stress tensor.
2
2all
1 4, 1 cosh / 2 cosx
dp yv y z A z y
dx h
Louisiana Tech UniversityRuston, LA 71272
Slide 39
The Stress Tensor
For fluids:
31 1 2 1
1 2 1 3 1
32 1 2 2
1 2 2 3 2
3 3 31 2
1 3 2 3 3
1 1
2 2
1 12
2 2
1 1
2 2
uu u u u
x x x x x
uu u u u
x x x x x
u u uu u
x x x x x
τ
The shear stress has 4 non-zero components.
Louisiana Tech UniversityRuston, LA 71272
Slide 40
Shear Stress, Bottom Surface
1 10
2 2
12 0 0
2
10 0
2
x x
x
x
v v
y z
v
y
v
z
τ
Along the bottom surface, we are concerned only with xy.
xxy
v
y
Louisiana Tech UniversityRuston, LA 71272
Slide 41
Shear Stress, Bottom Surface
xxy
v
y
all
, cosh / 2 cosy z A z y
all
cosh / 2 sinxy A z y
Louisiana Tech UniversityRuston, LA 71272
Slide 42
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.
/ 2 / 2
/ 2 / 2
2/ 2 / 2
2/ 2 / 2all
,
1 41 cosh / 2 cos
w h
xw h
w h
w h
Q v y z dy dz
dp yA z y dydz
dx h
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 UniversityRuston, LA 71272
Slide 43
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?