creeping flows

Louisiana Tech UniversityRuston, LA 71272

Slide 1

Creeping Flows

Steven A. Jones

BIEN 501

Wednesday, March 21, 2007

Start on Slide 53


Slide 2

Creeping Flows

Major Learning Objectives:

1. Compare viscous flows to nonviscous flows.

2. Derive the complete solution for creeping flow around a sphere (Stokes’ flow).

3. Relate the solution to the force on the sphere.


Slide 3

Creeping Flows

Minor Learning Objectives:1. Examine qualitative inertial and viscous effects.2. Show how symmetry simplifies the equations.3. Show how creeping and nonviscous flows

simplify the momentum equations.4. Give the origin of the Reynolds number.5. Use the Reynolds number to distinguish

creeping and nonviscous flows.6. Apply non-slip boundary conditions at a wall

and incident flow boundary conditions at .


Slide 4

Creeping Flows

Minor Learning Objectives (continued):7. Use the equations for conservation of mass and

conservation of momentum in spherical coordinates.8. Use the stream function to satisfy continuity.9. Eliminate the pressure term the momentum

equations by (a) taking the curl and (b) using a sort of Gaussian elimination.

10. Rewrite the momentum equations in terms of the stream function.

11. Rewrite the boundary conditions in terms of the stream function.

12. Deduce information about the form of the solution from the boundary conditions.


Slide 5

Important Concepts

• Flow Rate

• Cross-sectional average velocity

• Shear Stress (wall shear stress)

• Force caused by shear stress (drag)

• Pressure loss


Slide 6

Creeping Flows

Minor Learning Objectives (continued):13. Discuss the relationship between boundary conditions

and the assumption of separability.14. Reduce the partial differential equation to an ordinary

differential equation, based on the assumed shape of the solution.

15. Recognize and solve the equidimensional equation.16. Translate the solution for the stream function into the

solution for the velocity components..17. Obtain the pressure from the velocity components.18. Obtain the drag on the sphere from the stress

components (viscous and pressure).


Slide 7

Creeping vs. Nonviscous Flows

Nonviscous Flows

Viscosity goes to zero (High Reynolds Number)

Left hand side of the momentum equation is important.

Right hand side of the momentum equation includes pressure only.

Inertia is more important than friction.

Creeping Flows

Viscosity goes to (Low Reynolds Number)

Left hand side of the momentum equation is not important.

Left hand side of the momentum equation is zero.

Friction is more important than inertia.


Slide 8

Creeping vs. Nonviscous FlowsNonviscous Flow Solutions

Use flow potential, complex numbers.

Use “no normal velocity.”

Use velocity potential for conservation of mass.

Creeping Flow Solutions

Use the partial differential equations. Apply transform, similarity, or separation of variables solution.

Use no-slip condition.

Use stream functions for conservation of mass.

In both cases, we will assume incompressible flow.


Slide 9

Flow around a Sphere

Nonviscous FlowCreeping Flow

Larger velocity near the sphere is an inertial effect.

Velocity


Slide 10

Flow around a Sphere

A more general case

Increased velocity as a result of inertia terms.

Incident velocity is approached far from the sphere.

Shear region near the sphere caused by viscosity and no-slip.


Slide 11

Stokes Flow: The Geometry

3ev v

r

Use Standard Spherical Coordinates, (r, , and )

Far from the sphere (large r) the velocity is uniform in the rightward direction. e3 is the

Cartesian (rectangular) unit vector. It does not correspond to the spherical unit vectors.


Slide 12

The Objective

1. Obtain the velocity field around the sphere.

2. Use this velocity field to determine pressure and drag at the sphere surface.

3. From the pressure and drag, determine the force on the sphere as a function of the sphere’s velocity, or equivalently the sphere’s velocity as a function of the applied force (e.g. gravity, centrifuge, electric field).


Slide 13

Some Applications

1. What electric field is required to move a charged particle in electrophoresis?

2. What g force is required to centrifuge cells in a given amount of time.

3. What is the effect of gravity on the movement of a monocyte in blood?

4. How does sedimentation vary with the size of the sediment particles?

5. How rapidly do enzyme-coated beads move in a bioreactor?


Slide 14

Symmetry of the Geometry

The flow will be symmetric with respect to .

r


Slide 15

Components of the Incident Flow

3ev v

r

cosv

Incident Velocity

Component of incident velocity in the radial direction,

sinvComponent of incident velocity in the - direction,


Slide 17

Creeping Momentum Equation

To see how creeping flow simplifies the momentum equation, begin with the equation in the following form (Assume a Newtonian fluid):

For small v, 2nd term on the left is small. It is on the order of v2. (v appears in the right hand term, but only as a first power).

Dv

2P

t

Dvvv

2P

t


Slide 18

Convective Term in Spherical Coordinates


Slide 19

Reynolds Number

The Reynolds number describes the relative importance of the inertial terms to the viscous terms and can be deduced from a simple dimensional argument.

dominant.aretermswhichdetermines

thatalone,velocityorthanratherratio,thisisItor

,isratioThe)termtypicalaofthinktohelp

may(Itlikegoeslength. sticcharacteri a is and

velocity sticcharacteri a iswhere,likegoes

.

.

.V

2

2

2

2

2

2

VL

L

V

L

V

x

v

LL

VL

V

D

vv


Slide 20

Reynolds Number

VLVL

ReRe or

Different notations are used to express the Reynolds number. The most typical of these are Re or Nr.

Also, viscosity may be expressed as kinematic ( ) or dynamic () viscosity, so the Reynolds number may be

In the case of creeping flow around a sphere, we use v for the characteristic velocity, and we use the sphere diameter as the characteristic length scale. Thus,

DuRe


Slide 22

Boundary Conditions (B.C.s) for Creeping Flow around a Sphere

rv

Rr

for

for

3

0

ev

v

There is symmetry about the -axis. Thus (a) nothing depends on , and (b) there is no - velocity.

0

,

,

v

rvv

rvv rr


Slide 23

Summary of Equations to be Solved

0 v

0sin

1sin

sin

11 22

vr

vr

vrrr r

We must solve conservation of mass and conservation of momentum, subject to the specified boundary conditions.

Conservation of mass in spherical coordinates is:

Which takes the following form in spherical coordinates (Table 3.1):

0&00sinsin

11 22

vv

rvr

rr r WhenOr


Slide 24

Summary of Equations (Momentum)

0sin

1

0cot222

22

222

r

vvv

p

r

vr

v

rv

rv

r

p

r

rr

H

H

sinsin

11 22 rr

rrr

HWhere

Because there is symmetry in , we only worry about the radial and circumferential components of momentum.

Which takes the following form in spherical coordinates (Table 3.4):

0P τ (Incompressible, Newtonian Fluid)

Radial

Azimuthal


Slide 25

Simplified Differential Equations

22 2 2

1 1 1sin 0

sin sinrv v vvp

rr r r r r r

Yikes! You mean we need to solve these three partial differential equations!!?

Conservation of Radial Momentum

Conservation of Azimuthal Momentum

22 2 2 2

1 1 2 2 2sin cot 0

sinr r

r

vv vpr v v

r r r r r r r r

22

1 1sin 0

sinrr v vr r r

Conservation of Mass


Slide 26

Comments

22 2 2

1 1 1sin 0

sin sinrv v vvp

rr r r r r r

Three equations, one first order, two second order.

Three unknowns ( ).

Two independent variables ( ).

Equations are linear (there is a solution).

22 2 2 2

1 1 2 2 2sin cot 0

sinr r

r

vv vpr v v

r r r r r r r r

22

1 1sin 0

sinrr v vr r r

, andrv v P

andr


Slide 27

Stream Function Approach

We will use a stream function approach to solve these equations.

The stream function is a differential form that automatically solves the conservation of mass equation and reduces the problem from one with 3 variables to one with two variables.


Slide 28

Stream Function (Cartesian)

Cartesian coordinates, the two-dimensional continuity equation is:

0u v

x y

If we define a stream function, , such that:

, ,, 0

x y x yu v

y x

Then the two-dimensional continuity equation becomes:

2 2

0u v

x y x y y y x y y x


Slide 29

Summary of the Procedure

1. Use a stream function to satisfy conservation of mass.

a. Form of is known for spherical coordinates.

b. Gives 2 equations (r and momentum) and 2 unknowns ( and pressure).

c. Need to write B.C.s in terms of the stream function.

2. Obtain the momentum equation in terms of velocity.

3. Rewrite the momentum equation in terms of .

4. Eliminate pressure from the two equations (gives 1 equation (momentum) and 1 unknown, namely ).

5. Use B.C.s to deduce a form for (equivalently, assume a separable solution).


Slide 30

Procedure (Continued)

6. Substitute the assumed form for back into the momentum equation to obtain an ordinary differential equation.

7. Solve the equation for the radial dependence of .

8. Insert the radial dependence back into the form for to obtain the complete expression for .

9. Use the definition of the stream function to obtain the radial and tangential velocity components from .

10. Use the radial and tangential velocity components in the momentum equation (written in terms of velocities, not in terms of ) to obtain pressure.


Slide 31

Procedure (Continued)

11. Integrate the e3 component of both types of forces (pressure and viscous stresses) over the surface of the sphere to obtain the drag force on the sphere.


Slide 32

Stream Function

Recall the following form for conservation of mass:

rrv

rvr

sin

1,

sin

12

If we define a function (r,) as:

then the equation of continuity is automatically satisfied. We have combined 2 unknowns into 1 and eliminated 1 equation.

Note that other forms work for rectangular and cylindrical coordinates.

0sinsin

11 22

vr

vrrr r Slide 22


Slide 33

Exercise

With:

rrv

rvr

sin

1,

sin

12

0sinsin

11 22

vr

vrrr r

Rewrite the first term in terms of .


Slide 34

Exercise

With:

rrv

rvr

sin

1,

sin

12

0sinsin

11 22

vr

vrrr r

Rewrite the second term in terms of .


Slide 35

Momentum Eq. in Terms of

Userr

vr

vr

sin

1,

sin

12

Substitute these expressions into the steady flow momentum equation (Slide 23) to obtain a partial differential equation for from the momentum equation (procedure step 2):

0sin

1sin2

22

2

rr

and conservation of mass is satisfied (procedure step 1).


Slide 36

Elimination of Pressure

0 vp

The final equation on the last slide required several steps. The first was the elimination of pressure in the momentum equations. The second was substitution of the form for the stream function into the result. The details will not be shown here, but we will show how pressure can be eliminated from the momentum equations. We have:

We take the curl of this equation to obtain:

v p


Slide 37


But it is known that the curl of the gradient of any scalar field is zero (Exercise A.9.1-1). In rectangular coordinates:

0eee

eee

312

2

21

2

213

2

31

2

123

2

32

2

321

321

321

xx

p

xx

p

xx

p

xx

p

xx

p

xx

p

x

p

x

p

x

pxxx

p


Slide 38


ikj

ijk

kkk

k

ij

kijk

x

p

xp

x

pp

x

pp

x

v

e

e

ev

,

Alternatively:

So, for example, the e1 component is:

012332

123

13232

12311

e

ee

x

p

xx

p

x

x

p

xx

p

xx

p

x kjjk


Slide 39

Exercise: Elimination of Pressure

One can think of the elimination of pressure as being equivalent to doing a Gaussian elimination type of operation on the pressure term.

This view can be easily illustrated in rectangular coordinates:

2 2

2 2

2 2

2 2

0 momentum

0 momentum

Take of the first equation and of the second and subtract.

x x

y y

v vpx

x x y

v vpy

y x y

y x


Slide 40


momentum

momentum

yyx

v

x

v

yx

p

xy

v

xy

v

xy

p

yy

xx

2

3

3

32

3

3

2

32

0

0

2

3

3

3

3

3

2

3

0yx

v

x

v

y

v

xy

v yyxx

This view can be easily illustrated in rectangular coordinates:

23

3

2

3

2

3

3

3

0.,. Ex

v

yx

v

xy

v

y

vei yyxx


Slide 41

Exercise: 4th order equation

With:

3 33 3

3 2 2 30y yx x

v vv v

y y x x y x

What is the momentum equation:

, ,, 0x y

x y x yv v

y x

in terms of ?


Slide 42

Exercise: 4th order equation

Answer:

4 4 4 4

4 2 2 2 2 40

y y x x y x

or

22 2

2 20

y x


Slide 43


Fortunately, the book has already done all of this work for us, and has provided the momentum equation in terms of the stream function in spherical coordinates (Table 2.4.2-1). For v=0:

4

2

22

22 sin

1cos

sin

2

,

,

sin

1E

rrr

E

r

E

rE

t

Admittedly this still looks nasty. However, when we remember that we have already eliminated all of the left-hand terms, the result for the stream function is relatively simple.


Slide 44

Momentum in terms of

4

2

22

22 sin

1cos

sin

2

,

,

sin

1E

rrr

E

r

E

rE

t

How does this simplify for our problem?

Recall:

Steady state

Low Reynolds number

If:


Slide 45

Stream Function, Creeping Flow

When the unsteady (left-hand side) terms are eliminated:

.0sin

1sin

.sin

1sin,0

2

22

2224

rr

rrEE

Thus

where

This equation was given on slide 35.


Slide 46

Boundary Conditions in Terms of

From

0 atrv r R 0 at ,v r R

2

1

sinrvr

1

sinv

r r

and

Exercise: Write these boundary conditions in terms of .


Slide 47


From

Rr at0

Rrr

vr

at0sin

12

andr

must be zero for all at r=R. Thus,

must be constant along the curve r=R. But since it’s constant of integration is arbitrary, we can take it to be zero at that boundary. I.e.

,0sin

1Rr

rrv

at


Slide 48

QuestionConsider the following curves. Along which of these curves must velocity change with position?


Slide 49

Comment

does not change as changes.

As r changes, however, we move off of the curve r=R, so can change.

A key to understanding the previous result is that we are talking about the surface of the sphere, where r is fixed.

curve.thatalongconstantbemust

becausesoAndBecause

,

0.0sin

1,0

2

allforr

vr


Slide 50


From

3cos, e vvr rAt

sin,

sin

1 22

rvr

v rr

Thus, in contrast to the surface of the sphere, will change with far from the sphere.

32

32 sincossincos, ee

rvrvr

asThus,

(See Slide 14)


Slide 51


From

rgvr

dvrdrvd rr

22

0 0

2

0

2

sin2

1

sincossin

sin,

sin

1 22

rvr

v rr

which suggests the -dependence of the solution.

2sin vrf


Slide 52

Comment on Separability

For a separable solution we assume that the functional form of is the product of one factor that depends only on r and another that depends only on .

rr R,

Whenever the boundary conditions can be written in this form, it will be possible to find a solution that can be written in this form. Since the equations are linear, the solution will be unique. Therefore, the final solution must be written in this form.


Slide 53

Comment on Separability

0, RR R

In our case, the boundary condition at r=R is:

Both of these forms can be written as a function of r multiplied by a function of . (For r=R we take R(r)=0). The conclusion that the dependence like sin2 is reached because these two boundary conditions must hold for all . A similar statement about the r-dependence cannot be reached. I.e. we only know about two distinct r locations.

2 21, sin

2v r

and the boundary condition at r is:


Slide 54

Separability

0, RR RAgain, at r=R:

2 21, sin

2v r and at r :

For a separable solution, we look for a form:

,R R R

Because the -dependence holds for all , but the r-dependence does not, we must write:

2 21, sin

2r f r v r


Slide 55

Momentum Equation

The momentum equation:

is 2 equations with 3 unknowns (P, vr and v). We have used the stream function to get 2 equations and 2 unknowns (P and ). We then used these two equations to eliminate P.

D 20 P


Slide 56

Substitute Back into Momentum

With

0884

432

2

24

4

r

f

dr

df

rdr

fd

rdr

fd

(slide 45) becomes:

2sin vrf

0sin

1sin2

22

2

rr

Note the use of total derivatives.


Slide 57

Exercise: Substitute

2 2

2 2 2 2

sin 1 sin 10

sin sinr r r r

2sin vrf

into


Slide 58

Exercise: Substitute

22

2 2

2 22

2 2

22

2 2

22

2 2

22

2

sin 1sin

sin

sin 1 sinsin

sin

sin 2sin cossin

sin

sin cossin 2

sinsin 2

f r vr r

f rv f r

r r

f rv f r

r r

f rv f r

r r

f rv

r

2

2f r

r


Slide 59

Exercise: So we now need

22 22

2 2 2 2

4 2 222 2

4 2 2 2 2 4

4 22

4 2 2

sin 1 sinsin 2

sin

2sin sin 1 2sinsin sin

sin

sinsin 2sin

f rv f r

r r r r

f r f r f rv f r

r r r r r r

f r f rv

r r r

2

2

4 2 2 22

4 2 2 2 2

2sin cos 2cos

sin sin

sin cossin 2sin 2sin 2 1

sin

f rf r

r

f r f r f rv f r

r r r r


Slide 60

Substitute Back into Momentum

The student should recognize the differential equation as an equidimensional equation for which:

08842

22

4

44 f

dr

dfr

dr

fdr

dr

fdr

narrf

Substitution of this form back into the equation yields:

0,2

1,

4

3,

4

1 3

42

DvCRvBRvA

DrCrBrr

Arf with


Slide 61

Equidimensional Equation

The details are like this:

08814321

8814321

0884

12244

2

22

4

44

nnnnnnn

ar

ararnrarnnrarnnnnr

ardr

darr

dr

ardr

dr

ardr

n

nnnn

nnnn

bydivide

This is a 4th order polynomial, i.e. there are 4 possible values for n which happen to turn out to be -1, 1, 2 and 4.


Slide 62

Solution for Velocity Components

Once the boundary conditions are evaluated, the solution is:

sin4

1

4

31

cos2

1

2

31

3

3

r

R

r

R

v

v

r

R

r

R

v

vr


Slide 63

Pressure

DP 2

To obtain pressure, we return to the momentum equation:

This form was 2 equations with 3 unknowns, but now vr and v have been determined. Once the forms for these two velocity components are substituted into this equation, one obtains:

23

sin

2

3cos3

rRv

P

rRv

r

P

Integrate to get P.


Slide 64

Pressure

The result of this exercise is:

20

cos

2

3cos

rRvgrPP


Slide 65

Force

To obtain force on the sphere, we must remember that force is caused by both the pressure and the viscous stress.

2

0 0

23 sinsincos ddTTRF

Rrrrr

z3 is the direction the sphere is moving relative to the fluid.

Used to get the z3 component.


Slide 66

Potential Flow

0 v

Potential flow derives from the viscous part of the momentum equation.

vIf we write:

Then the viscous part of the momentum equation will automatically be zero.


Slide 67

Potential Flow

0 v

The continuity equation:

02

Becomes:

Therefore potential flow reduces to finding solutions to Laplace’s equation.

creeping flows

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