Creeping Flows
Steven A. Jones
BIEN 501
Wednesday, March 21, 2007
Start on Slide 53
Louisiana Tech University
Ruston, LA 71272
Slide 1
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.
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Ruston, LA 71272
Slide 2
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 .
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Slide 3
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 from 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.
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Slide 4
Important Concepts
•
•
•
•
•
Flow Rate
Cross-sectional average velocity
Shear Stress (wall shear stress)
Force caused by shear stress (drag)
Pressure loss
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Slide 5
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).
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Slide 6
Creeping vs. Nonviscous Flows
Creeping Flows
Nonviscous Flows
Viscosity goes to  (Low
Reynolds Number)
Viscosity goes to zero
(High Reynolds Number)
Left hand side of the
momentum equation is not
important.
Left hand side of the
momentum equation is
important.
Left hand side of the
momentum equation is
zero.
Right hand side of the
momentum equation
includes pressure only.
Friction is more important
than inertia.
Inertia is more important
than friction.
Louisiana Tech University
Ruston, LA 71272
Slide 7
Creeping vs. Nonviscous Flows
Creeping Flow Solutions
Nonviscous Flow Solutions
Use the partial differential
equations. Apply
transform, similarity, or
separation of variables
solution.
Use flow potential, complex
numbers.
Use no-slip condition.
Use “no normal velocity.”
Use stream functions for
conservation of mass.
Use velocity potential for
conservation of mass.
In both cases, we will assume incompressible flow.
Louisiana Tech University
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Slide 8
Flow around a Sphere
Creeping Flow
Nonviscous Flow
Velocity
Larger velocity near the sphere is an inertial
effect.
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Slide 9
Flow around a Sphere
A more general case
Incident velocity is
approached far
from the sphere.
Increased velocity
as a result of inertia
terms.
Shear region near the
sphere caused by
viscosity and no-slip.
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Slide 10
Stokes Flow: The Geometry
Use Standard Spherical Coordinates, (r, , and f)
r
v  v e 3
f

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.
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Slide 11
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).
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Slide 12
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?
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Slide 13
Symmetry of the Geometry
r
f

The flow will be symmetric with respect to f.
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Slide 14
Components of the Incident Flow
Component of incident velocity
in the radial direction, v cos 
r
f


Incident Velocity v  v e 3
Component of incident velocity
in the  - direction, v sin 
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Slide 15
Creeping Momentum Equation
To see how creeping flow simplifies the momentum
equation, begin with the equation in the following form
(Assume a Newtonian fluid):
v
r
 rv  v  P  2 m  D
t
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).
v
r
 P  2 m  D
t
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Slide 17
Convective Term in Spherical
Coordinates
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Slide 18
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.
rv  v goes like
rV 2
L
, where V is a characteri stic velocity
and L is a characteri stic length. m  D goes like
mV
2
L
. (It may
 rV 2   mV 
 2v
  2 ,
help to think of a typical term m 2 ). The ratio is 
x
 L   L 
rVL
or
. It is this ratio, rather than m or velocity alone, that
m
determines which terms are dominant.
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Slide 19
Reynolds Number
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 (m) viscosity, so the Reynolds number may be
rVL
Re 
or Re 

m
VL
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,
ru D
Re 
m
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Slide 20
Boundary Conditions (B.C.s) for
Creeping Flow around a Sphere
v  0 for r  R
v  v e 3 for r  
There is symmetry about the f axis. Thus (a) nothing depends
on f, and (b) there is no f velocity.
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v r  v r r ,  
v  v r ,  
vf  0
Slide 22
Summary of Equations to be Solved
We must solve conservation of mass and conservation of
momentum, subject to the specified boundary conditions.
Conservation of mass in spherical coordinates is:
 v  0
Which takes the following form in spherical coordinates
(Table 3.1):
1 
1

1

2




rvf   0
r
r
v

r
v
sin


r

r 2 r
r sin  
r sin  f
Or
1 
1


2




r
r
v

r
v
sin


0
When
v

0
&
0
r

f
r 2 r
r sin  
f
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Slide 23
Summary of Equations (Momentum)
Because there is symmetry in f, we only worry about the
radial and circumferential components of momentum.
P   τ  0
(Incompressible, Newtonian Fluid)
Which takes the following form in spherical coordinates
(Table 3.4):
Radial
Azimuthal
Where H 
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p
2
2 v
2


 m  Hvr  2 vr  2   2 v cot    0
r
r
r  r


v
1 p
v



 m  Hv  r  2  2   0
r 
 r sin  


1   2 
1
 
 
r

sin





r 2 r  r  r sin   
 
Slide 24
Simplified Differential Equations
Yikes! You mean we need to solve these three partial
differential equations!!?
Conservation of Mass
1 
1

2
r
r
v

 r v sin    0
r
2
r r
r sin  


Conservation of Radial Momentum

 1   v 

vr  2
p
1
 
2 v 2
 m  2  r2 r  
sin


v


v
cot

0

 2 r
2
2 
r
r

r

r
r
sin





r
r


r






Conservation of Azimuthal Momentum

 1   v 
v  vr
v
1 p
1
 
 m  2  r2   
sin




r 
   r 2 sin 2 
 r r  r  r sin   
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
0

Slide 25
Comments
1 
1

2
r
r
v

 r v sin    0
r
2
r r
r sin  


 1   v
p
 m  2  r2 r
r
 r r  r


vr
1
 


sin




 r sin   

2 v 2
 2

v


v
cot

0
 2 r
2
2 
r  r
 r

 1   v 
v  vr
v
1 p
1
 
 m  2  r2   
sin




r 
   r 2 sin 2 
 r r  r  r sin   

0

Three equations, one first order, two second order.
Three unknowns ( vr , v and P ).
Two independent variables ( r and  ).
Equations are linear (there is a solution).
Louisiana Tech University
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Slide 26
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.
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Slide 27
Stream Function (Cartesian)
Cartesian coordinates, the two-dimensional continuity
equation is:
u v

0
x y
If we define a stream function, y, such that:
u
y  x, y 
y
, v
y  x, y 
x
0
Then the two-dimensional continuity equation becomes:
u v   y

 
x y x  y
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   y
 
 y  y
  2y  2y

0

 xy yx
Slide 28
Summary of the Procedure
1. Use a stream function to satisfy conservation of mass.
a. Form of y is known for spherical coordinates.
b. Gives 2 equations (r and  momentum) and 2
unknowns (y 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 y.
4. Eliminate pressure from the two equations (gives 1
equation (momentum) and 1 unknown, namely y).
5. Use B.C.s to deduce a form for y (equivalently, assume
a separable solution).
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Slide 29
Procedure (Continued)
6. Substitute the assumed form for y back into the
momentum equation to obtain an ordinary differential
equation.
7. Solve the equation for the radial dependence of y.
8. Insert the radial dependence back into the form for y to
obtain the complete expression for y.
9. Use the definition of the stream function to obtain the
radial and tangential velocity components from y.
10. Use the radial and tangential velocity components in
the momentum equation (written in terms of velocities,
not in terms of y) to obtain pressure.
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Slide 30
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.
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Slide 31
Stream Function
Recall the following form for conservation of mass:
1 
1

2


rv sin    0
r
r
v

r
2
r r
r sin  
Slide 22
If we define a function y(r,) as:
1
y
1 y
vr  2
, v 
r sin  
r sin  r
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.
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Slide 32
Exercise
With:
1 
1

2


rv sin    0
r
r
v

r
2
r r
r sin  
1
y
1 y
vr  2
, v 
r sin  
r sin  r
Rewrite the first term in terms of y.
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Slide 33
Exercise
With:
1 
1

2


rv sin    0
r
r
v

r
2
r r
r sin  
1
y
1 y
vr  2
, v 
r sin  
r sin  r
Rewrite the second term in terms of y.
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Slide 34
Momentum Eq. in Terms of y
1
y
1 y
, v 
Use v r  2
r sin  
r sin  r
and conservation of mass is satisfied (procedure step 1).
Substitute these expressions into the steady flow
momentum equation (Slide 23) to obtain a partial
differential equation for y from the momentum equation
(procedure step 2):
2

sin    1  
 r 2  r 2   sin    y  0



2
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Slide 35
Elimination of Pressure
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:
 p  m  v   0
We take the curl of this equation to obtain:
  p  m    v 
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Slide 36
Elimination of Pressure
But it is known that the curl of the gradient of any scalar
field is zero (Exercise A.9.1-1). In rectangular coordinates:
e1

  p 
x1
p
x1
e2

x2
p
x2
e3

x3
p
x3
 2 p
 2 p
 2 p
2 p 
2 p 
2 p 
e1  
e 2  
e3  0
 



 x1x2 x2 x1 
 x2 x3 x3x2 
 x1x3 x3x1 
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Slide 37
Elimination of Pressure
Alternatively:   v   ijk
vk
ei
x j
p
p
p 
e k ,  p k 
xk
xk

 p

  p    ijk
 xk
So, for example, the e1 component is:
 1 jk

x j

x j
 p

 xk




e1   123

x2


 
 
 x2
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
e i

 p 


   132
x3
 x3 
 p  

 
 x3  x3
 p  

 e1
 x2  
 p  

 e1  0
 x2  
Slide 38
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 vx  2 vx 
p
0    m  2  2  x  momentum
x
y 
 x
  2vy  2vy 
p
0    m  2  2  y  momentum
 x

y

y


Take


of the first equation and
of the second and subtract.
y
x
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Slide 39
Elimination of Pressure
This view can be easily illustrated in rectangular coordinates:
  3v x
 3v x 
2 p
0
 m 
 3  x  momentum
2
yx
y 
 yx
  3v y
 3v y 
2 p
 y  momentum
0
 m 3 
2
 x

xy

x

y


  3v y  3v y 
  3v x  3v x 

0  m 
 3   m  3 
2
2 

y 
xy 
 yx
 x
  3v x  3v x
 3v y  3v y
i.e.,  3 

 3
2
2
 y
yx
xy
x

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
  0  E2


Slide 40
Exercise: 4th order equation
With:
vx 
y  x, y 
y
, vy  
y  x, y 
x
0
What is the momentum equation:
  3vx  3vx
 3v y  3v y

 3
 3 
2
2
yx xy
x
 y

  0

in terms of y?
Louisiana Tech University
Ruston, LA 71272
Slide 41
Exercise: 4th order equation
Answer:
  4y
 4y
 4y
 4y
 4  2 2 2 2 4
y x x y
x
 y

0

or
2
 
 
 2  2  y 0
x 
 y
2
Louisiana Tech University
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2
Slide 42
Elimination of Pressure
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
vf=0:
 2
1
 y , E 2y  2 E 2y

E y  2
 2
t
r sin  r,  
r sin 
1 y
 y

cos


sin    E 4y

r 
 r

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.
Louisiana Tech University
Ruston, LA 71272
Slide 43
Momentum in terms of y
If:
 2
1  y , E 2y  2 E 2y  y
1 y

4

E y  2
 2
cos


sin


  E y
t
r sin  r,  
r sin   r
r 

How does this simplify for our problem?
Recall:
Steady state
Low Reynolds number
Louisiana Tech University
Ruston, LA 71272
Slide 44
Stream Function, Creeping Flow
When the unsteady (left-hand side) terms are eliminated:
  sin    1  
 2

.
2
r   sin   
 r
E 4y  0, where E 2  
2
  sin    1  
Thus  2  2

  y  0.
r   sin   
 r
This equation was given on slide 35.
Louisiana Tech University
Ruston, LA 71272
Slide 45
Boundary Conditions in Terms of y
From
v  0 at r  R,
vr  0 at r  R
and
1 y
v 
r sin  r
1 y
vr  2
r sin  
Exercise: Write these boundary conditions in
terms of y.
Louisiana Tech University
Ruston, LA 71272
Slide 46
Boundary Conditions in Terms of y
From
1
y
 1 y
v

 0 at r  R
v 
 0 at r  R, r
2
r sin  
r sin  r
y
y
and
must be zero for all  at r=R. Thus, y
r

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.
y  0 at r  R
Louisiana Tech University
Ruston, LA 71272
Slide 47
Question
Consider the following curves. Along which of these curves
must velocity change with position?
Louisiana Tech University
Ruston, LA 71272
Slide 48
Comment
A key to understanding the previous result is that we are
talking about the surface of the sphere, where r is fixed.
1
y
y
Because vr  0, 2
 0. And so because
0
r sin  

for all  , y must be constant along that curve.
y does not
change as 
changes.
Louisiana Tech University
Ruston, LA 71272
As r changes,
however, we
move off of the
curve r=R, so y
can change.
Slide 49
Boundary Conditions in Terms of y
1
y y
2
,
 v r r sin 
From v r  2
r sin   
At r  , vr  v cos  e 3
Thus, as r  ,
(See Slide 14)
y
 v cos   r 2 sin  e 3  v r 2 cos  sin  e 3

Thus, in contrast to the surface of the sphere, y will
change with  far from the sphere.
Louisiana Tech University
Ruston, LA 71272
Slide 50
Boundary Conditions in Terms of y
1
y y
2
,
 v r r sin 
From v r  2
r sin   
y

r 

0


y
2
2
d   v r r sin  d  r  v cos   sin  d
0
0

1 2
 r v  sin 2   g r 
2
which suggests the  -dependence of the solution.
y  f r v sin 
2
Louisiana Tech University
Ruston, LA 71272
Slide 51
Comment on Separability
For a separable solution we assume that the functional
form of y is the product of one factor that depends only
on r and another that depends only on .
y r,   R 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.
Louisiana Tech University
Ruston, LA 71272
Slide 52
Comment on Separability
In our case, the boundary condition at r=R is:
y R,   R R   0
and the boundary condition at r is:
1
2
y  ,   v r 2 sin 2 
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.
Louisiana Tech University
Ruston, LA 71272
Slide 53
Separability
Again, at r=R:
y R,   R R   0
and at r :
y  ,   v r 2 sin 2 
1
2
For a separable solution, we look for a form:
y  R,   R  R    
Because the -dependence holds for all , but the rdependence does not, we must write:
y  r ,  
Louisiana Tech University
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1
f  r  v r 2 sin 2 
2
Slide 54
Momentum Equation
The momentum equation:
0  P  2 m  D
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 y). We then used these two equations
to eliminate P.
Louisiana Tech University
Ruston, LA 71272
Slide 55
Substitute Back into Momentum
With
y  f r v sin 
2
2

sin    1  
 r 2  r 2   sin    y  0



2
4
(slide 45) becomes:
2
d f
4 d f
8 df 8 f
 2
 3
 4 0
4
2
dr
r dr
r dr r
Note the use of total derivatives.
Louisiana Tech University
Ruston, LA 71272
Slide 56
Exercise: Substitute
y  f r v sin 
2
into
  2 sin    1      2 sin    1   
 2 2

  2  2

y  0
r   sin      r
r   sin    
 r
Louisiana Tech University
Ruston, LA 71272
Slide 57
Exercise: Substitute
  2 sin    1   
2

f
r
v
sin




 2


2
r   sin    
 r
 2 2 f  r 
sin    1  sin 2 
 v sin 
 f r  2

2

r
r


 sin  

 2 2 f  r 
sin   2sin  cos  
 v sin 
 f r  2

2
r
r 
sin 


 2 2 f  r 
sin   cos  
 v sin 
 2 f r  2

2

r
r




 2 2 f  r 

sin 2 
 v sin 
 2 2 f  r 
2
r
r



Louisiana Tech University
Ruston, LA 71272




Slide 58
Exercise: So we now need

  2 sin    1     2  2 f  r 
sin 2 
v  2  2
 2 2 f  r  

  sin 
2

r
r


sin




r
r

 


4
2
2
2



 

f
r

f
r

f  r  2sin 




2sin

sin


1

2
2
 v sin 

 2

f  r   

 sin 
4
2
2
2
4
 


r
r

r
r


sin




r
r


 

4
2
2



 

f
r

f
r

f  r  2 cos 




sin


2sin

cos

2
 v sin 
 2sin 
 2

f  r    
 
4
2
2



r

r
r


sin


r
sin


  

 2 4 f  r 
 2 f  r  sin  
2 f r 
 cos 2    
 v sin 
 2sin 
 2  2sin 
 2 f  r  1 
  
4
2
2
2

r

r
r

r
sin


  


Louisiana Tech University
Ruston, LA 71272
Slide 59
Substitute Back into Momentum
The student should recognize the differential equation as an
equidimensional equation for which:
4
2
d f
df
2 d f
r
 4r
 8r
8f  0
4
2
dr
dr
dr
4
f r   ar n
Substitution of this form back into the equation yields:
A
f r    Br  Cr 2  Dr 4 with
r
1
3
1
A  v R 3 , B 
v  R, C  v  , D  0
4
4
2
Louisiana Tech University
Ruston, LA 71272
Slide 60
Equidimensional Equation
The details are like this:
2
n
d 4 ar n
dar n
2 d ar
n
r

4
r

8
r

8
ar
0
4
2
dr
dr
dr
r 4 n n  1n  2 n  3ar n 4  4r 2 n n  1ar n 2  8r n ar n 1  8ar n
4
divide by ar n
n n  1n  2n  3  4n n  1  8n  8  0
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.
Louisiana Tech University
Ruston, LA 71272
Slide 61
Solution for Velocity Components
Once the boundary conditions are evaluated, the solution is:
3

vr
3 R 1R 
 1 
    cos 
v  2 r 2  r  
3

v
3 R 1R 
 1 
    sin 
v  4 r 4  r  
Louisiana Tech University
Ruston, LA 71272
Slide 62
Pressure
To obtain pressure, we return to the momentum equation:
P  2 m  D
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:
P
cos 
 3mv  R 3
r
r
P 3
sin 
 mv  R 2
 2
r
Integrate to get P.
Louisiana Tech University
Ruston, LA 71272
Slide 63
Pressure
The result of this exercise is:
3
cos 
P  P0  rgr cos   mv R 2
2
r
Louisiana Tech University
Ruston, LA 71272
Slide 64
Force
To obtain force on the sphere, we must remember
that force is caused by both the pressure and
the viscous stress.
F3  R
2
2

  T
0
0
rr
cos   Tr sin   r R sin  d df
Used to get the z3 component.
z3 is the direction the sphere
is moving relative to the fluid.
Louisiana Tech University
Ruston, LA 71272
Slide 65
Potential Flow
Potential flow derives from the viscous part of the
momentum equation.
m  v  0
If we write:
v  f
Then the viscous part of the momentum equation will
automatically be zero.
Louisiana Tech University
Ruston, LA 71272
Slide 66
Potential Flow
The continuity equation:
 v  0
Becomes:
  f   2f  0
Therefore potential flow reduces to finding solutions to
Laplace’s equation.
Louisiana Tech University
Ruston, LA 71272
Slide 67