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?