Louisiana Tech University Ruston, LA 71272 Flows With More Than One Dependent Variable - 2D Example...

Louisiana Tech UniversityRuston, LA 71272

Flows With More Than One Dependent

Variable - 2D Example

Juan M. Lopez

BIEN 501

Wednesday, March 21, 2007


Recall - Generalized Newtonian

TvvD 2

1

DIT 2p where

Recall that:

tr stands for “trace,” which is the sum of the diagonal elements. Tr(T)=Tii

DD tr2

While the expression looks complicated, it will look much simpler once a given form for is found.


Generalized Newtonian

TvvD 2

1

DIT 2p

3

3

2

3

3

2

1

3

3

1

3

2

2

3

2

2

1

2

2

1

3

1

1

3

2

1

1

2

1

2

2

2

100

010

001

x

u

x

u

x

u

x

u

x

ux

u

x

u

u

u

x

u

x

ux

u

x

u

x

u

x

u

x

u

P

i

vT

where

Recall that:

tr stands for “trace,” which is the sum of the diagonal elements. Tr(T)=Tii

DD tr2


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


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 x-

direction only. .in length a is where,p

Constantp

xLLx


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.


Incompressible Newtonian Stress Tensor

z

u

z

u

y

u

z

u

x

u

z

u

y

u

y

u

y

u

x

uz

u

x

u

y

u

x

u

x

u

zyzxz

yzyxy

xzxyx

2

2

2

τ

Adapted from Table 3.3 in the text.

Now, we cancel terms out based on our assumptions.

This results in our new tensor:

00

00

0

z

uy

u

z

u

y

u

x

x

xx

τ


3.3.25) Eq.(p 2 gvvvv

t

Navier-Stokes EquationsIn Vector Form:

zzzzz

zz

yz

xz

yyyyy

zy

yy

xy

xxxxx

zx

yx

xx

gz

v

y

v

x

v

zz

vv

y

vv

x

vv

t

v

gz

v

y

v

x

v

yz

vv

y

vv

x

vv

t

v

gz

v

y

v

x

v

xz

vv

y

vv

x

vv

t

v

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

p

:component-z

p

:component-y

p

:component-x

Which we expand to component form from table 3.4:


Reducing Navier-Stokes

zzzzz

zz

yz

xz

yyyyy

zy

yy

xy

xxxxx

zx

yx

xx

gz

v

y

v

x

v

zz

vv

y

vv

x

vv

t

v

gz

v

y

v

x

v

yz

vv

y

vv

x

vv

t

v

gz

v

y

v

x

v

xz

vv

y

vv

x

vv

t

v

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

p

:component-z

p

:component-y

p

:component-x


Reducing Navier-Stokes• This reduces to:

• Including our constant pressure drop:

• Oops! Now we have a nonhomogenous higher-order differential equation that is inseparable. How do we deal with it?

2

2

2

2

Pressure Modified

p0

z

v

y

v

xxx

2

2

2

2p0

z

v

y

v

Lxx


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:

2

2

2

2

2

2

2

2

2

2

p0

:Therefore

,

,where

p0

zydy

Vd

L

zyyV

zyvv

z

v

y

v

L

x

x

xx

xx


DiffEq Solution - Setup

• We can separate this into two equations, each of which equals zero.• Why?

– 0=0+0

2

2

2

2

2

2

2

2

2

2

2

2

0

p0

:Separated

p0

zy

dy

Vd

L

zydy

Vd

L

x

x


DiffEq Solution - Poiseuille

dy

d

dy

Vd

L

dy

Vd

L

yx

x

x

2

2

2

2

p

: toequivalent isequation above that the2.7.18 Eq.

from seecan wesolution, Poiseuille simple theis thisbecause

p


DiffEq Solution - Poiseuille

2

22 41

8

p

: withup end to2.7.2Section

fromsolution thefollowcan weTherefore,

,definition tensor stressour From

h

y

L

hu

dy

du

x

xyx

Now we can focus our remaining efforts on the perturbation function.


Perturbation Function - Reduction

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

110

0

1

0

Therefore

,

Where

0

z

zZ

zZy

yY

yY

z

zZyY

y

yYzZ

zZyYz

zZyY

y

yYzZ

z

zZyY

y

zZyY

zZyYzy

zy

• We can approach this

perturbation function by a separation of variables method, as it is homogeneous.


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:

yByBzZ

yAyAyY

z

zZ

zZy

yY

yY

coshsinh

cossin

:solution general

shomogeneou standardour use nowcan We

11

21

21

2

2

2

22

• Now we can use our boundary conditions to solve for these constants.


Perturbation Function – B.C.’s

hn

textinerrorh

A

hAhdy

dY

A

AAdy

dY

ydy

dY

n

12 of valueshaveonly can Thus

?)(02sin Therefore,

0 cannot solution, nontrivial a be To

02sin2cos0

:h/2) -/ (y wallsAt the

0. bemust ,10cos Because

00sin0cos

0|0

parabola)our on velocity maximum ofpoint (the

flowour in region symetric a have we0,yAt

2

2

1

21



1

121

2

2

1

21

coscosh

combine they constants,just are constants Because

coshcos

. us give tofor Zequation our and Yfor equation our combine nowcan We

0 cannot solution, nontrivial a be To

0sinh

0. bemust ,10cosh Because

00sinh0cosh0|0

:function separatedour ofportion ZFor the

nnnn

nnn

nn

yzA

zByAzZyY

B

hBdz

dZ

B

BBdz

dZz

dz

dZ



12

22

12

22

1

1

12cos

41

8

p12coscos2cosh

:asrewritten becan thisso m, n only when nontrivial issolution This

periodic.ely appropriatequation theof sidesboth makes This .12

cos

by equation theof sidesboth multiplies textbook point the At this .t coefficien the

for solve tointegratecan We(y). variableone of in termspurely equation an have now We

41

8

pcos2cosh2,

cos2cosh2,

coscosh,

:obtain to timemore one ConditionsBoundary our use We

nnnn

n

nnnn

nxnnn

nnnn

h

yn

h

y

L

h

h

ynywA

h

ym

A

h

y

L

hywAwy

uywAwy

yzAzy


Perturbation Function – Integration

,...,2,1,0

212

cosh

12

328p

1

:in resultsWhich

12cos

41

8p

12coscos2cosh

:isolatedt coefficien with thewrite-recan weg,Rearrangin

12cos

41

8

p

12coscos2cosh

:integrate nowcan We

33

2

2/

2/ 2

220n

2/

2/

2/

2/ 2

22

0n

2/

2/

nfor

hwn

nLh

A

dyh

ynhy

Lh

dyh

ynyw

A

dyh

yn

h

y

L

h

dyh

ynywA

n

n

h

h

h

h nn

n

h

h

h

h nnn

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

0n 33

2

2

22

212

cosh12

12cos

12cosh132

8

p41

8

p,

:equations originalour into thisplugcan We

hwn

n

hyn

hzn

L

h

h

y

L

hzyv

n

x

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.

00

00

0

z

uy

u

z

u

y

u

x

x

xx

τ

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?


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.


• QUESTIONS?

Louisiana Tech University Ruston, LA 71272 Flows With More Than One Dependent Variable - 2D Example...

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