lecture 7 on stokes flow-3

Louisiana Tech UniversityRuston, LA 71272

Slide 1

Creeping Flows

Steven A. JonesBIEN 501

Wednesday, March 21, 2007Start 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 FlowsNonviscous 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 SphereNonviscous FlowCreeping Flow

Larger velocity near the sphere is an inertial effect.

Velocity


Slide 10

Flow around a SphereA 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 vr

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 EquationTo 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 NumberThe 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

VLLV

LV

xv

LL

VLV

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,,

vrvvrvv rr


Slide 23

Summary of Equations to be Solved

0 v

0sin1sin

sin11 2

2

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&00sinsin11 2

2

vv

rvr

rr r WhenOr


Slide 24

Summary of Equations (Momentum)

0sin

1

0cot222

22

222

rvvvp

r

vr

vr

vr

vrp

r

rr

H

H

sinsin11 2

2 rrr

rrHWhere

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 1 sin 0sin sin

rv 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 0sin

r rr

vv vp r v vr r r r r r r r

22

1 1 sin 0sinrr v v

r r r

Conservation of Mass


Slide 26

Comments

22 2 2

1 1 1 sin 0sin sin

rv 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 0sin

r rr

vv vp r v vr r r r r r r r

22

1 1 sin 0sinrr v v

r 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 vx 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 vx y x y y y x y y x


Slide 29

Summary of the Procedure1. 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

sin1,

sin1

2

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.

0sinsin11 2

2

vr

vrrr r Slide 22


Slide 33

Exercise

With:

rrv

rvr

sin1,

sin1

2

0sinsin11 2

2

vr

vrrr r

Rewrite the first term in terms of .


Slide 34

Exercise

With:

rrv

rvr

sin1,

sin1

2

0sinsin11 2

2

vr

vrrr r

Rewrite the second term in terms of .


Slide 35

Momentum Eq. in Terms of

Userr

vr

vr

sin1,

sin1

2

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

Elimination of PressureBut 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

xxp

xxp

xxp

xxp

xxp

xxp

xp

xp

xp

xxxp


Slide 38


ikj

ijk

kkk

k

ij

kijk

xp

xp

xpp

xpp

xv

e

e

ev

,

Alternatively:

So, for example, the e1 component is:

012332

123

13232

12311

e

ee

xp

xxp

x

xp

xxp

xxp

x kjjk


Slide 39

Exercise: Elimination of PressureOne 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 vp xx x y

v vp yy x y

y x


Slide 40


momentum

momentum

yyx

vxv

yxp

xyv

xyv

xyp

yy

xx

2

3

3

32

3

3

2

32

0

0

2

3

3

3

3

3

2

3

0yx

vxv

yv

xyv yyxx

This view can be easily illustrated in rectangular coordinates:

23

3

2

3

2

3

3

3

0.,. Exv

yxv

xyv

yvei yyxx


Slide 41

Exercise: 4th order equationWith:

3 33 3

3 2 2 3 0y yx x v vv vy 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 equationAnswer:

4 4 4 4

4 2 2 2 2 4 0y y x x y x

or

22 2

2 2 0y x


Slide 43

Elimination of PressureFortunately, 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 sin1cos

sin2

,,

sin1 E

rrrE

rE

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 sin1cos

sin2

,,

sin1 E

rrrE

rE

rE

t

How does this simplify for our problem?

Recall:

Steady state

Low Reynolds number

If:


Slide 45

Stream Function, Creeping FlowWhen 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

1sinrv

r

1sin

vr r

and

Exercise: Write these boundary conditions in terms of .


Slide 47


From

Rr at0

Rrr

vr

at0sin1

2

and

r 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

1 Rrrr

v

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

,

0.0sin1,0 2

allforr

vr


Slide 50


From

3cos, e vvr rAt

sin,

sin1 2

2 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

sin21

sincossin

sin,

sin1 2

2 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, sin2

v r

and the boundary condition at r is:


Slide 54

Separability 0, RR RAgain, at r=R:

2 21, sin2

v 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, sin2

r f r v r


Slide 55

Momentum EquationThe 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

0884432

2

24

4

r

fdrdf

rdrfd

rdrfd

(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 1 0sin 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 1 sinsin

sin 1 sinsinsin

sin 2sin cossinsin

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

2 f rr


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 2sin

2sin sin 1 2sinsin sinsin

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 2cossin sin

sin cossin 2sin 2sin 2 1sin

f rf r

r

f r f r f rv f r

r r r r


Slide 60

Substitute Back into MomentumThe student should recognize the differential equation as an

equidimensional equation for which:

0884 2

22

4

44 f

drdfr

drfdr

drfdr

narrf

Substitution of this form back into the equation yields:

0,21,

43,

41 3

42

DvCRvBRvA

DrCrBrrArf with


Slide 61

Equidimensional Equation

The details are like this:

08814321

8814321

0884

12244

2

22

4

44

nnnnnnnar

ararnrarnnrarnnnnr

ardr

darrdrardr

drardr

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:

sin41

431

cos21

231

3

3

rR

rR

vv

rR

rR

vvr


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

sin23cos3

rRvP

rRv

rP

Integrate to get P.


Slide 64

Pressure

The result of this exercise is:

20cos

23cos

rRvgrPP


Slide 65

ForceTo 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.

lecture 7 on stokes flow-3

Documents

creeping flow solutionsuse

momentum equations

creeping flowsviscosity

creeping flowssteven

conservation of momentum

use velocity potential

incompressible flow

sphere stokes flow