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Professor Fred Stern
Fall 2009
Chapter 8
1
Chapter 8: Inviscid Incompressible Flow: a Useful Fantasy
8.1 Introduction
For high Re external flow about streamlined bodies viscous effects are confined to
boundary layer and wake region. For regions where the B.L is thin i.e. favorable pressure
gradient regions, Viscous/Inviscid interaction is weak and traditional B.L theory can be
used. For regions where B.L is thick and/or the flow is separated i.e. adverse pressure
gradient regions more advanced boundary layer theory must be used including
viscous/Inviscid interactions.
For internal flows at high Re viscous effects are always important except near the
entrance. Recall that vorticity is generated in regions with large shear. Therefore, outside
the B.L and wake and if there is no upstream vorticity then ω=0 is a good approximation.
Note that for compressible flow this is not the case in regions of large entropy gradient.
Also, we are neglecting noninertial effects and other mechanisms of vorticity generation.
Potential flow theory
1) Determine  from solution to Laplace equation
 2  0
B.C:
at S B : V .n  0 
at S  : V  

0
n
Note:
F: Surface Function
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Professor Fred Stern
Chapter 8
2
Fall 2009
DF
F
1 F
0
 V .F  0  V .n  
Dt
t
F t
for steady flow V .n  0
2) Determine V from V   and p(x) from Bernoulli equation
Therefore, primarily for external flow application we now consider inviscid flow theory
(   0 ) and incompressible flow (   const )
Euler equation:
.V  0
DV
 p   g
Dt
V

 V .V  ( p   z )
t
V2
V .V  
V 
2
Where :    V  vorticity  2  fluid angular velocity
V
1

 ( p  V 2   z )  V  
t
2
If   0 ie  V  0 then V  :



1
 p      z  B(t )
t
2
Bernoulli’s Equation for unsteady incompressible flow, not f(x)
Continuity equation shows that GDE for  is the Laplace equation which is 2nd order
linear PDE ie superposition principle is valid. (Linear combination of solution is also a
solution)
  V      2  0
  1  2
  2  0
 2  0   2 (1  2 )  0   21   22  0   2 1
 2  0
Techniques for solving Laplace equation:
1) superposition of elementary solution (simple geometries)
2) surface singularity method (integral equation)
3) FD or FE
4) electrical or mechanical analogs
5) Conformal mapping ( for 2D flow)
6) Analytical for simple geometries (separation of variable etc)
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Fall 2009
Chapter 8
3
8.2 Elementary plane-flow solutions:
Recall that for 2D we can define a stream function such that:
u  y
v   x
z  vx  u y 


( x )  ( y )   2  0
x
y
i.e.  2  0
Also recall that  and  are orthogonal.
u   y  x
v   x   y
d   x dx   y dy  udx  vdy
d   x dx   y dy  vdx  udy
i.e.
dy
u
1
 
dx  const
v dy
dx  const
8.2 Elementary plane flow solutions
Uniform stream
u  U    y   x  const
v  0   x   y
  Ux
  U y
Note:  2   2  0 is satisfied.
V    Uiˆ
i.e.
Say a uniform stream is at an angle  to
the x-axis:
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u  U  cos  
Chapter 8
4
Fall 2009
 

y x
v  U  sin   
 

x y
After integration, we obtain the following expressions for the stream function and
velocity potential:
  U  y cos   x sin  
  U  x cos   y sin  
2D Source or Sink:
Imagine that fluid comes out radially at origin with uniform rate in all directions.
(singularity in origin where velocity is infinite)
Consider a circle of radius r enclosing this source. Let vr be the radial component of
velocity associated with this source (or sink). Then, form conservation of mass, for a
cylinder of radius r, and unit width perpendicular to the paper,
 L3 
Q   V d A  
A
S
Q   2 r    b   vr
Or ,
vr 
Q
2 br
m
, v  0
r
Q
Where: m 
is the convenient constant with unit velocity × length
2 b
(m>0 for source and m<0 for sink). Note that V is singular at (0,0) since vr  
 vr 
In a polar coordinate system, for 2-D flows we will use:
 1 
V   

r r 
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Chapter 8
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Fall 2009
And:
.V  0
1 
1 
(rvr ) 
(v )  0
r r
r 
i.e.:
 1 

r r 
1 

v  Tangential velocity 

r 
r
Such that V  0 by definition.
v r  Radial velocity 
Therefore,
m  1 


r
r r 
1 

v  0 

r 
r
vr 
  m ln r  m ln x 2  y 2
i.e.
  m  m tan 1
y
x
Doublets:
The doublet is defined as:
ψ   m(  1   2 )   1   2  
sink
source
sink
source

m
tan θ  tan θ

1
2
tan (θ  θ )  tan (  ) 
1 2
m
1  tan θ tan θ
1
2
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Professor Fred Stern
Chapter 8
6
Fall 2009
r sin 
r sin 
; tan  2 
r cos   a
r cos   a
r sin 
r sin 


2ar sin 
tan( )  r cos   a r cos   a  2
r sin 
m 1  r sin 
r  a2
r cos   a r cos   a
tan 1 
For small value of a
2ar sin  
2amy
y
)   2   2
2
2
r a 
r
x  y2
 cos 
  2am ( Doublet Strength)


a 0

  lim  m tan 1 (
r
By rearranging:
y
 2

ψ 2
 x2  (y 
)  ( )2
2
2
2
x y
It means that streamlines are circles with radius R 

and center at (0, -R) i.e. circles
2
are tangent to the origin with center on y axis. The flow direction is from the source to
the sink.
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Chapter 8
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Fall 2009
2D vortex:
Suppose that value of the  and  for the source are reversal.
vr  0
1 
 K


r 
r
r
Purely recirculating steady motion, i.e. v  f  r  . integration results in:
  Kθ
ψ   K ln r
K=constant
2D vortex is irrotational everywhere except at the origin where V and V  are infinity.
v 
Circulation
Circulation is defined by:
For irrotational flow
Γ   V d s
c
C  closed contour
Or by using Stokes theorem: ( if no singularity of
the flow in A)
Γ   V  d s    V  d A   .ndA  0
c
A
A
Therefore, for potential flow   0 in general.
However, this is not true for the point vortex due to the singular point at vortex core
where V and V  are infinity.
If singularity exists: Free vortex
2
   v eˆ  rd eˆ  
0
2
0
V
ds
 
K
r
K

(rd )  2 K and K 
r
2
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Professor Fred Stern
Chapter 8
8
Fall 2009
Note: for point vortex, flow still irrotational everywhere except at origin itself where
V, i.e., for a path not including (0,0)   0
Also, we can use Stokes theorem to show the existence of  :
 V ds  
ABC
V  d s  C Since
AB'C

V .d s  0
ABCB' A
Therefore in general for irrotational motion:
 V .d x  
V  d x  d
d x d
dx

  .
ds ds
ds
dx
eˆs 
ds
(V   ).eˆs  0
V
Where: e s =unit tangent vector along curve x
Since e s is not zero we have shown:
V  
i.e. velocity vector is gradient of a scalar function  if the motion is irrotational.
(  V ds  0 )
The point vortex singularity is important in aerodynamics, since, extra source and sink
can be used to represent airfoils and wings and we shall discuss shortly. To see this,
consider as an example:
an infinite row of vortices:

1
2y
2x 
1
   K  ln ri   K ln  (cosh
 cos
)
2
a
a 
2
i 1
Where ri is radius from origin of ith vortex.
Equally speed and equal strength (Fig 8.11 of Text book)
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Professor Fred Stern
For y
Fall 2009
Chapter 8
9
a the flow approach uniform flow with

K

y
a
+: below x axis
-: above x axis
Note: this flow is just due to infinite row of vortices and there isn’t any pure uniform
flow
u
Vortex sheet:
From afar (i.e. y
a ) looks that thin sheet occurs which is a velocity discontinuity.
2K
 strength of vortex sheet
a
2K
d  u l dx  u u dx  (u l  u u )dx 
dx
a
d
i.e.  
=Circulation per unit span
dx
Note: There is no flow normal to the sheet so that vortex sheet can be used to simulate a
body surface. This is beginning of airfoil theory where we let    (x ) to represent body
geometry.
Define  
Vortex theorem of Helmholtz: (important role in the study of the flow about wings)
1) The circulation around a given vortex line is constant along its length
2) A vortex line cannot end in the fluid. It must form a closed path, end at a
boundary or go to infinity.
3) No fluid particle can have rotation, if it did not originally rotate
Circular cylinder (without rotation):
In the previous we derived the following
equation for the doublet:
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Professor Fred Stern
Chapter 8
10
Fall 2009
y
 Doublet   

 sin 
x y
r
When this doublet is superposed over a uniform flow parallel to the x- axis, we get:
2
2

 1
 U  1 
r sin 
2 
r
 U r 
Where:   doublet strength which is determined from the kinematic body boundary
condition that the body surface must be a stream surface. Recall that for inviscid flow it is
no longer possible to satisfy the no slip condition as aresult of the neglect of viscous
terms in PDEs.
  U  r sin  
 sin 
The inviscid flow boundary condition is:
F: Surface Function
DF
F
1 F
0
 V .F  0  V  n  
 0 (for steady flow)
Dt
t
F t
Therefore at r=R, V.n=0 i.e. vr  0 r  R .
F
F
eˆr 
eˆ
F

 r
 eˆr
V  vr eˆr  v eˆ , n 
F
Fr2  F2
V  n  vr 

1 
 
 U  1 
cos  =0
2 
r 
U
r



   U R2
If we replace the constant

by a new constant R2, the above equation becomes:
U

R2 
r sin 
2 
 r 
This radial velocity is zero on all points on the circle r=R. That is, there can be no
velocity normal to the circle r=R. Thus this circle itself is a streamline.
  U  1 
We can also compute the tangential component of velocity for flow over the circular
cylinder. From equation,
 R2 

v  
 U  1  2  sin 
r
r 

On the surface of the cylinder r=R, we get the following expression for the tangential and
radial components of velocity:
v  2U  sin 
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Professor Fred Stern
Chapter 8
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Fall 2009
vr  0
How about pressure p? look at the Bernoulli's equation:
p
p 1 2
1
  vr  v 2     U 2
 2
 2
After some rearrangement we get the following non-dimensional form:
vr 2  v 2
p  p
C p  r ,  
 1
1
U 2
U 2
2
For the circular cylinder being studied here, at the surface, the only velocity component
that is non-zero is the tangential component of velocity. Using v  2U  sin  , we get at
the cylinder surface the following expression for the pressure coefficient:
C p  1  4 sin 2 
Where  is the angle measured from the rear stagnation point (at the intersection of the
back end of the cylinder with the x- axis).
Cp
Cp over a Circular Cylinder
1.5
1
0.5
0
-0.5 0
-1
-1.5
-2
-2.5
-3
-3.5
30
60
90
120 150 180 210 240 270 300 330 360
Theta, Degrees
From pressure coefficient we can calculate the fluid force on the cylinder:
1
F    ( p  p )nds   U 2  C p ( R,  )nds
2
A
A
ds  ( Rd )b b=span length
2
1
F   U 2 bR  (1  4sin 2  )(cos  iˆ  sin  ˆj )d
2
0
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Professor Fred Stern
Lift
Fall 2009
Chapter 8
12
2
F  ˆj
=   (1  4 sin 2  ) sin  d  0 (due to symmetry of flow

1
1
0
U 2 bR
U 2 bR
2
2
around x axis)
2
Drag
F  iˆ
=   (1  4 sin 2  ) cos  d  0 (dÁlembert paradox)
CF 

1
1
0
U 2 bR
U 2 bR
2
2
We see that C L  C D  0 is due to symmetry of Cp. But how realistic is this potential
flow solution. We started explain that viscous effects were confined to thin boundary
layer and it has negligible effect on the inviscid flow for high Rn flow about streamlined
bodies. A circular cylinder is not a streamlined body (like as airfoil) but rather a bluff
body. Thus, as we shall determine now the potential flow solution is not realistic for the
back portion of the circular cylinder flows (  90) .
For general:
CD  CD
 C DSKIN FRICTION
CL 
FORM DRAG
CD
: drag due to viscous modification of pressure destruction
FORM DRAG
CDSKIN FRICTION : Drag due to viscous shear stress
 Streamlined Body : CDSK  CD FO

 Bluff Body :CD FO  CDSK
Circular cylinder with circulation:
The stream function associated with the flow over a circular cylinder, with a point vortex
of strength  placed at the cylinder center is:
 sin  
  U  r sin  

ln r
r
2
From V.n=0 at r=R:   U  R 2
Therefore,   U  r sin  
U  R 2 sin  

ln r
r
2
The radial and tangential velocity is given by:
 R2 
1 
vr 
 U  1  2  cos 
r 
 r 
 R2 


 U  1  2  sin  
r
r 
2 r

On the surface of the cylinder (r=R):
 R2 
1 
vr 
 U  1  2  cos   0
r 
 R 
v  
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Professor Fred Stern
Chapter 8
13
Fall 2009
v  


 2U  sin  
r
2 R
Next, consider the flow pattern as a function of  . To start lets calculate the stagnation
points on the cylinder i.e.:

v  2U  sin  
0
2 R

K
sin  

 /2
4 U  R 2U  R

K

Note: K 
2
U R
So, the location of stagnation point is function of  .
K


U  R 2 U  R
0 ( sin   0 )
1( sin   0.5 )
2 ( sin   1 )
>2( sin   1 )

 s (stagnation point)
0,180
30,150
90
Is not on the circle but where vr  v  0
For flow patterns like above (except a), we should expect to have lift force in –y
direction.
Summary of stream and potential function of elementary 2-D flows:
In Cartesian coordinates:
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Professor Fred Stern
Chapter 8
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Fall 2009
u   y  x
v   x   y
In polar coordinates:
 1 
vr 

r r 
1 

v 

r 
r
Flow
Uniform Flow
Source (m>0)
Sink (m<0)
Doublet

U x
m ln r
 cos 
r
Vortex
90 Corner flow
Solid-Body rotation
K
1 / 2 A( x 2  y 2 )
Doesn’t exist

U y
m

 sin 
r
- K ln r
Axy
1 2
r
2
These elementary solutions can be combined in such a way that the resulting solution can
be interpreted to have physical signification;that is. Represent the potential flow solution
for various geometers. Also, methods for arbitrary geometries combine uniform stream
with distribution of the elementary solution on the body surface.
Some combination of elementary solutions to produce body geometries of practical
importance
Body name
Elemental combination
Flow Patterns
Rankine Half Body
Uniform stream+source
Rankine Oral
Uniform stream+source+sink
Kelvin Oral
Uniform stream+vortex point
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Chapter 8
15
Fall 2009
Circular Cylinder
without circulation
Uniform stream+doublet
Circular Cylinder with
circulation
Uniform
stream+doublet+vortex
Keep in mind that this is the potential flow solution and may not well represent the real
flow especially in region of adverse px.
The Kutta – Joukowski lift theorem:
Since we know the tangential component of velocity at any point on the cylinder (and the
radial component of velocity is zero), we can find the pressure field over the surface of
the cylinder from Bernoulli’s equation:
p vr2 v2 p U 2
  

 2 2 
2
Therefore:
2
U  
1  2 
   
1
 2 
2
p  p   U   2U  sin  

p


U

 2 U 2 sin 2  
sin 


 

2 2 
2 
2 R   
2
8 R 
R

 A  B sin   C sin 2 
where

1
2 
2
A   p  U   2 2 
2
8 R 

U  
B
R
C  2 U 2
Calculation of Lift: Let us first consider lift. Lift per unit span, L (i.e. per unit distance
normal to the plane of the paper) is given by:
L   pdx   pdx
Lower
upper
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On the surface of the cylinder, x = Rcos. Thus, dx = -Rsind, and the above integrals
may be thought of as integrals with respect to . For the lower surface,  varies between
 and 2. For the upper surface,  varies between  and 0. Thus,
2


0


L   R  A  B sin   C sin  sin d  R  A  B sin   C sin 2  sin d
2


Reversing the upper and lower limits of the second integral, we get:
2

2



L   R  A  B sin   C sin 2  sin d   R  A sin   B sin 2   C sin 3  d   BR
0
0
Substituting for B ,we get:
L   u 
This is an important result. It says that clockwise vortices (negative numerical values of
) will produce positive lift that is proportional to  and the free stream speed with
direction 90 degrees from the stream direction rotating opposite to the circulation. Kutta
and Joukowski generalized this result to lifting flow over airfoils. Equation L   u 
is known as the Kutta-Joukowski theorem.
Drag: We can likewise integrate drag forces. The drag per unit span, D, is given by:
D   pdy   pdy
Front
rear
Since y=Rsin on the cylinder, dy=Rcosd. Thus, as in the case of lift, we can convert
these two integrals over y into integrals over . On the front side,  varies from 3/2 to
/2. On the rear side,  varies between 3/2 and /2. Performing the integration, we can
show that
D0
This result is in contrast to reality, where drag is high due to viscous separation. This
contrast between potential flow theory and drag is the dÁlembert Paradox.
The explanation of this paradox are provided by Prandtl (1904) with his boundary layer
theory i.e. viscous effects are always important very close to the body where the no slip
boundary condition must be satisfied and large shear stress exists which contributes the
drag.
Lift for rotating Cylinder:
We know that L   U   therefore:
L

CL 

1
U 2 (2 R) U  R
2
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Define: v Average
 CL 
1

2
Chapter 8
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Fall 2009
2
1 
 v d  2  R 
Note: Γ   V  d s   V   d A   V Rd
0
c
c
c
2
v
U  averagfe
Velocity ratio:
a
U
Theoretical and experimental lift and drag of a rotating cylinder
Experiments have been performed that simulate the previous flow by rotating a circular
cylinder in a uniform stream. In this case v  R which is due to no slip boundary
condition.
- Lift is quite high but not as large as theory (due to viscous effect ie flow
separation)
- Note drag force is also fairy high
Fletttern (1924) used rotating cylinder to produce forward motion.
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Chapter 8
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Chapter 8
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8.3 Method of Images
The method of image is used to model “slip” wall effects by constructing appropriate
image singularity distributions.
Plane Boundaries:
2-D:  
m 
2
2
ln  x  1  y 2   x  1  y 2 




2
1
1 

2
2
2
2
3-D:    M   x  1  y 2  z 2    x  1  y 2  z 2  


 

Similar results can be obtained for dipoles and vortices:
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Spherical and Curvilinear Boundaries:
The results for plane boundaries are obtained from consideration of symmetry. For
spherical and circular boundaries, image systems can be determined from the Sphere &
Circle Theorems, respectively. For example:
Flow field
Source of strength M at c outside sphere of
radius a, c>a
Dipole of strength  at l outside sphere of
radius a, l>a
Source of strength m at b outside circle of
radius a, b>a
Image System
2
Sources of strength ma at a and line
c
c
sink of strength m extending from center
a
2
of sphere to a c
3
2
dipole of strength  a  at  a
l
l
2
equal source at a
and sink of same
b
strength at the center of the circle
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Chapter 8
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Multiple Boundaries:
The method can be extended for multiple boundaries by using successive images.
(1) For example, the solution for a source equally spaced between two parallel planes
w( z )  m
 ln z  4n  a   ln z  4n  2  a 
m  0 , 1, 2 ,
 mln z  a   ln z  2  a   ln z  4  a   ln z  6  a   ln z  4  a   ln z  2  a   
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(2) As a second example of the method of successive images for multiple boundaries
consider two spheres A and B moving along a line through their centers at
velocities U1 and U2, respectively:
Consider the kinematic BC for A:
2
F  x, t    x  yt   y 2  z 2  a 2
DF
 0  V  eˆR  U1kˆ  eˆR or R  U1 cos
Dt

2
Ua 3
where    2 cos , R   3 cos   
R
R
2
Similarly for B R  U 2 cos '
This suggests the potential in the form
  U11  U 22
where 1 and 2 both satisfy the Laplace equation and the boundary condition:
 1 
 1 
0

  cos , 

 R  R a
 R '  R 'b
 2 
 2 
 cos '

  0, 

 R  R a
 R '  R 'b
(*)
(**)
1 = potential when sphere A moves with unit velocity towards B, with B at rest
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2 = potential when sphere B moves with unit velocity towards A, with A at rest
If B were absent.
1  
0
a3
a3
,
cos



cos



0
2R2
R2
2
but this does not satisfy the second condition in (*). To satisfy this, we introduce the
image of  0 in B, which is a doublet 1 directed along BA at A1, the inverse point of A
with respect to B. This image requires an image  2 at A2, the inverse of A1 with respect
to A, and so on. Thus we have an infinite series of images A1, A2, … of strengths
1 ,  2 ,  3 etc. where the odd suffixes refer to points within B and the even to points
within A.
Let f n  AAn & AB  c
b2
a2
b2
, f2 
, f3  c 
,…
f1  c 
c  f2
c
f1
 a3 


 b3 
b3
 ,…
1   0   3  ,  2   1   3  ,  3   2  
3 
 c 
 f1 
 c  f 2  
radius 3
where 1 = image dipole strength,  0 = dipole strength 
sistance 3
 cos  cos  cos
1   0 2  1 2 1  2 2 2  with a similar development procedure for 2.
R
R1
R2
Although exact, this solution is of unwieldy form. Let’s investigate the possibility of an
approximate solution which is valid for large c (i.e. large separation distance)
 2c
c2 
R '2  R 2  c 2  2cr cos  R 2 1  cos  2 
R
R 

1 1
c
c 
 1  2 cos  2 
R' R 
R
R 
2

1
2
1
R
R 
 1  2 cos  2 
c
c
c 
2

1
2
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Considering the former representation first defining  

1 1
 1  2u   2
R' R
By the binomial theorem valid for x  1
1  x  2
1


c
and u  cos
R
1
2
  0   1 x   2 x 2   3 x 3   ,  0  1 and  n 
1  3  2n  1
2  4  2n
Hence if 2u   2  1
  2
1


  0   1 2u   2   2 
2    P0 u   P1 u   P2 u  2
After collecting terms in powers of  , where the Pn are Legendre functions of the first
kind (i.e. Legendre polynomials which are Legendre functions of the first kind of order
zero). Thus,
1 1 R
R2
R  C :   2 P1  cos   3 P2  cos  
R' c c
c
1 1 c
c2
R  C :   2 P1  cos   3 P2  cos  
R' R R
R
Next, consider a doublet of strength  at A

 
1
  
3
c
 R 2  c 2  2cR cos
R 2  c 2  2cR cos 2

Thus,
 1 2 RP1  cos  3R 2 P2  cos 

 cos 
R  c : 







2
R '2
c3
c
c

 cos 
 

R '2
  R cos  c 



 1
2c
3c 2
R  c :     2 P1  cos   3 P2  cos   4 P3  cos  
R
R
R



1 
2 




Going back to the two sphere problem. If B were absent
1  
a3
cos
2R2
 R' 
using the above expression for the origin at B and near B   1 , RR’,   '
 c

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 1 a3 a 3 R ' P1  cos 

a3
   2 cos    2 
 
3
2R
c
2 c

3
a cos
R 

c3
which can be cancelled by adding a term to the first approximation, i.e.
1 a3 cos 1 a3b3 cos '
1  

2 R2
2 c3 R '2
to confirm this
1 a3 a3R 'cos ' 1 a3b3 cos
1   2 

2c
c3
2 c3
R
1 '
a 3 cos ' a 3b3 cos '
( R  b)  
 3
 0  hot
3
R '
c3
c
R'
Similarly, the solution for f2 is
2  
1 b3 cos ' 1 a 3b3 cos

2
2 R'
2 c3 R 2
These approximate solutions are converted to   c 3  .
To find the kinetic energy of the fluid, we have

1 
K      n dS   n dS 
2  S A
SB

1

K   A11U12  A12U1U 2  A22U 22   
n dS
2
2 S A  SB
A11     1
A
1



dS A , A22     2 2 dS B , A12     2 1 dS A     1 1 dS B
n
n
n
n
A
B
B
where dS  2 R 2 sin  d
2
2
2 a 3b3
A11  a 3  , A12 
 , A22  b 3  ,
3
3
3
c
3 3
1
2 a b 
1
K  M '1 U12 
U1U 2  M '2 U 22 : using the approximate form of the potentials
3
4
c
4
1
1
where M '1 U12 , M '2 U 22 : masses of liquid displaced by sphere.
4
4
8.4 Complex variable and conformal mapping
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This method provides a very powerful method for solving 2-D flow problems. Although
the method can be extended for arbitrary geometries, other techniques are equally useful.
Thus, the greatest application is for getting simple flow geometries for which it provides
closed form analytic solution which provides basic solutions and can be used to validate
numerical methods.
Function of a complex variable
Conformal mapping relies entirely on complex mathematics. Therefore, a brief review is
undertaken at this point.
A complex number z is a sum of a real and imaginary part; z = real + i imaginary
i  1
The term i, refers to the complex number
i   1, i  1, i  i, i  1
2
so that;
3
4
Complex numbers can be presented in a graphical format. If the real portion of a complex
number is taken as the abscissa, and the imaginary portion as the ordinate, a twodimensional plane is formed.
z = real +i imaginary = x + iy
y, imaginary
x, real
-A complex number can be written in polar
form using Euler's equation;
z = x + iy = rei = r(cos + isin)
Where:
r2 = x2 + y2
- Complex multiplication: z1z2 = (x1+iy1)(x2+iy2) = (x1x2 - y1y2) + i(x1y2 + y1x2)
 r1e i1  r2 e i 2
 r1r2  e i (1  2 )
- Conjugate: z = x + iy z  x  iy z.z  x 2  y 2
-Complex function:
w(z) = f(z)=  (x,y) + i (x,y)
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If function w(z) is differentiable for all values of z in a region of z plane is said to be
regular and analytic in that region. Since a complex function relates two planes, a point
can be approached along an infinite number of paths, and thus, in order to define a unique
derivative f(z) must be independent of path.
PP1( y  0) :
 w w1  w 1  i 1  (  i )


 z z1  z
x1  x

dw
 x  i x
dz 1
Note : w1  1 ( x   x, y )  i 1 ( x   x, y )
 w w2  w 2  i 2  (  i )
PP 2( x  0) :


 z z2  z
i ( y2  y )

dw
 i y   y
dz 2
dw
to be unique and independent of path:
dz
x   y and   y   x Cauchy Riemann Eq.
For
Recall that the velocity potential and stream function were shown to satisfy this
relationship as a result of their othogonality. Thus, complex function w    i
represents 2-D flows.
xx   yx  yy   xy i.e.  xx   yy  0 and similarly for Therefore if analytic and
regular also harmonic, i.e., satisfy Laplace equation.
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Application to potential flow
w( z )    i Complex potential where  : velocity potential,  : stream function
dw
 x  i x  u  iv   ur  iu  e  i Complex velocity
dz
r ' e i '  re i (  ) where r '  r (magnification) and  '     (rotation)
Triangle about z0 is transformed into a similar triangle in the ζ-plane which is
magnified and rotated.
Implication:
-Angles are preserved between the intersections of any two lines in the physical domain
and in the mapped domain.
-The mapping is one-to-one, so that to each point in the physical domain, there is one and
only one corresponding point in the mapped domain.
For these reasons, such transformations are called conformal.
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Usually the flow-field solution in the ζ-plane is known:
W ( )   ( , )  i ( , )
Then
w( z)  W  f z    ( x, y)  i ( x, y) or    &   
Conformal mapping
The real power of the use of complex variables for flow analysis is through the
application of conformal mapping: techniques whereby a complicated geometry in the
physical z-domain is mapped onto a simple geometry in the ζ-plane (circular cylinder) for
which the flow-field solution is known. The flow-field solution in the z-plane is obtained
by relating the ζ-plane solution to the z-plane through the conformal transformation
ζ=f(z) (or inverse mapping z=g(ζ)).
Before considering the application of the technique, we shall review some of the more
important properties and theorems associated with it.
Consider the transformation,
ζ=f(z) where f(z) is analytic at a regular point Z0 where f’(z0)≠0
δζ= f’(z0) δz
  r ' e i ' ,  z  rei , f ' z 0   e i
The streamlines and equipotential lines of the ζ-plane (Φ, Ψ) become the streamlines of
equipotential lines of the z-plane (, ψ).
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
 2z  2  0
 2z  2  0
Chapter 8
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Fall 2009
i.e. Laplace equation in the z-plane transforms into Laplace equation
is the ζ-plane.
The complex velocities in each plane are also simply related
dw dw d dw '


f ( z)
dz d dz d
dw
dW
( z )  u  iv  (U  iV ) f '( z ) 
(  f ( z ))
dz
d
i.e. velocities in two planes are proportional.
Two independent theorems concerning conformal transformations are:
(1) Closed curves map to closed curves
(2) Rieman mapping theorem: an arbitrary closed profile can be mapped onto the unit
circle.
More theorems are given and discussed in AMF Section 43. Note that these are for the
interior problems, but we equally valid for the exterior problems through the inversion
mapping.
Many transformations have been investigated and are compiled in handbooks. The AMF
contains many examples:
1) Elementary transformations:
az  b
, ad  bc  0
a) linear: w 
cz  d
b) corner flow: w  Az n
2
c) Jowkowsky: w    c

d) exponential: w  e
e) w  z s , s irational
n
2) Flow field for specific geometries
a) circle theorem
b) flat plate
c) circular arc
d) ellipse
e) Jowkowski foils
f) ogive (two circular areas)
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g) Thin foil theory [solutions by mapping flat plate with thin foil BC onto unit
circle]
h) multiple bodies
3) Schwarz-Cristoffel mapping
4) Free-streamline theory
The techniques of conformal mapping are best learned through their applications. Here
we shall consider corner flow.
A simple example: Corner flow
1. In ζ-plane, let W ( )   i.e. uniform stream
2. Say   f ( z )  z



3. w( z )  W ( f ( z ))  z  i.e. corner flow
Note that 1-3 are unit uniform stream.
w( z )  Uz n  UR n cos   iUR n sin n , where z  Re i
i.e.   UR n cos n ,   UR n sin n
  UR n sin n =const.=streamlines
  UR n cos n =const.=equipotentials
dw dW d

 nUz n 1  nUR n 1ei ( n 1)  (nUR n 1 cos n  inUR n 1 sin n )e i
dz d dz
 (ur  iu )ei
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33
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u r  nUR n1 cos n
u  nUR n1 sin n
 2n   u  0 , u  0
 2n     n  u  0 , u  0
0   

r

r
i.e. w( z )  Uz
n
represents corner flow: n=1uniform stream, n=290° corner
8.5 Introduction to Surface Singularity methods
(also known as Boundary Integral and Panes Methods)
Next, we consider the solution of the potential flow problem for an arbitrary geometry.
Consider the BVP for a body of arbitrary geometry fixed in a uniform stream of an
inviscid, incompressible, and irrotational fluid.
The surface singularity method is founded on the symmetric form of Greens theorem and
what is known as Greens function.
  G    G  dV 
2
V
2
G 
 

G
 dS
n
n 
 S  S  S B  S S 

(1)
where Φ and G are any two scalar field in V (control volume bounded by s infinity S
body and S inserted to reder the domain simply connected) and for our application.
Φ= velocity potential
G= Green’s function
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Say,
 2 G   ( x  x0 ) in V+V’ (i.e. entire domain) where δ is the Dirac delta function.
G0 on S∞
Solution for G (obtained Fourier Transforms) is: G  ln r , r  x  x0 , i.e. elementary
2-D source at x  x0 of unit strength, and (1) becomes

G 
 

G
dS

n

n


 S SB

First term in integrand represents source distribution and second term dipole distribution,
which can be transformed to vortex distribution using integration by parts. By extending
the definition of Φ into V’ it can be shown that Φ can be represented by distributions of
sources, dipoles or vortices, i.e.


 GdS : source distribution,  : source strength
S SB
or


 S SB

G
dS : dipole distribution,  : dipole strength
n
Also, it can be shown that a source distribution representation can only be used to
represent the flow for a non-lifting body; that is, for lifting flow dipole or vortex
distributions must be used.
As this stage, let’s consider the solution of the flow about a non-lifting body of arbitrary
geometry fixed in a uniform stream. Note that since G0 on S∞ Φ already satisfy the
condition S∞. The remaining condition, i.e. the condition is a stream surface is used to
determine the source distribution strength.
Consider a source distribution method for representing non-lifting flow around a body of
arbitrary geometry.
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35
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V  U    : total velocity
U  : uniform stream,  : perturbation potential due to presence of body
U   U  (cos iˆ  sin ˆj ) : note that for non-lifting flow  must be zero (i.e. for a
symmetric foil   0 or for cambered filed    oLift )

K
 2 ln rds : source distribution on body surface
SB
Now, K is determined from the body boundary condition.

 U   n
V  n  0 i.e. U  n    n  0 or
n
i.e.
normal
velocity
induced
by
sources
 K
stream
ln rds  U   n
n SB 2
must
cancel
uniform
This singular integral equation for K is solved by descretizing the surface into a number
of panels over which K is assumed constant, i.e. we write

ni
where rij 
x
M  no. of panels

j 1

Si
ln rij dSi  U   ni , i=1,M, j=1,M
 x j   z i  z j  =distance from ith panel control point to rj  position
2
i
Kj
2
2
vector along jth panel.
Note that the integral equation is singular since

1 rij
ln rij 
ni
rij ni
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at for rij  0 this integral blows up; that is, when i=j and we trying to determine the
contribution of the panel to its own source strength. Special care must be taken. It can be
shown that the limit does exist at the integral equation can be written
ki 
2 ni
M K
j
Ki

2 i 1 2
j i
where

S j Si
ln rij dS i 
ki
2
1 rij
dS j  U   ni
ij nij
r
Si
rij 
1 rij
1
1  rij
1
  i rij  ni  
n xi 
n zi   2 xi  x j n xi  z i  z j n zi
rij ni rij
rij  xi
z i
 rij


where rij2  xi  x j   z i  z j 
2
2
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Consider the ith panel
S i  cos  i iˆ  sin  i ˆj , ni  S i  ˆj   sin  i iˆ  cos  i ˆj
n xi   sin  i , n zi  cos  i
 K
ln rds  U   n
n SB 2
r j  r j 0  S S i , r j 0 : origin of jth panel coordinate system, S Si : distance along jth panel
V  U   
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K
 2 ln rds

SB
 K
ln rds  U   n
n SB 2
V n  0
Ki M K j

2 i 1 2
j i
1 rij
dS j  U   ni  U  sin    i  , ni   sin  i iˆ  cos  i ˆj
ij nij
r
Si
1 rij
dS j I j and RHS i  U  sin    i  , then
ij nij
r
Let
Si
xi  x j n xi  zi  z j n zi
Ki M K j
dS j

I j  RHS i , I j  
2
2
2 i 1 2




x

x

z

z
Sj
i
j
i
j
j i
x  x  S n n  z  z  S n n
dS
x  x  S n   z  z  S n 
x  x n  z  z n  S n n  n n 

x  x   z  z   2S  n x  x   n z  z   S
Ii  
li
i
j0
j
zj
xi
i
j0
li
i
i

0
j
j0
j
xj
j0
zi
j
2
zj
i
xi
j0
i
2
0
j
j0
zi
xj
j
xj
zi
zj
xi
2
j0
i
CS j  D
li
i
2
0
S 2j  2 AS j  B
j0
j
zi
i
j0
zj
i
j0
2
j
dS j
dS j
where
A   cos  j xi  x j 0   sin  j z i  z j 0 
B  xi  x j 0   z i  z j 0 
2
2
C  sin  i   j 
D  xi  x j 0 sin  i  z i  z j 0 cos  i
li S
1
dS j  C  i dS j  DI i1  CI i 2 where   S 2j  2 AS j  B
0 
0 
I j1 depends on if q<0 or >0 where q  4B  4 A2
I j  D
I j2 
li
1
ln   AI i1
2
Ki M K j

2 i 1 2
j i
x  x n  z
 x  x   z
i
j
xi
i
2
Sj
i
j
 z j n zi
 zj
2
i
dS j  U   ni  RHS i
RHS i  U   cos  sin  i  sin  cos  i   U  sin    i 
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Ki M K j

I i  RHS i : Matrix equation for Ki and can be solved using Standard methods
2 i 1 2
j i
such as Gauss-Siedel Iteration.
In order to evaluate Ij, we make the substitution
x j  x j 0  S j S xj
 rj  rj0  S j S j
z j  z j 0  S j S xj
where Sj= distance along the jth panel 0  S j  l i
S xj  n zi  cos  j
S zj  n xi  sin  j
After substitution, Ij becomes
B  A 2  xi  x j 0   z i  z j 0 
2

2

 cos 2  j xi  x j 0   sin 2  j z i  z j 0   2 cos  j sin  j xi  x j 0 z i  z j 0 
2

2



 xi  x j 0  1  cos 2  j  z i  z j 0  1  sin 2  j  2
2

 
2

q  4 B  A 2  4 xi  x j 0 sin  i  z i  z j 0 cos  i
i.e. q>0 and as a result,
2S  2 A 1
S A
2
I j1 
tan 1 i
 tan 1 i
where q  2 B  A 2  2 E
E
E
q
q
2
C
1

I j  DI j1  C  ln   AI j1   D  CAI j1  ln  l0i
2
2

D  CA tan 1 li  A  tan 1 A   C ln  l j z  2 Al j  B 





E
E
E 2 
B


where   Si2  2 AS j  B
Therefore, we can write the integral equation in the form
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Ki M K j

I j  U  sin    i 
2 i 1 2
j i
 1

 2

kj
 2 I j

Ij
2 


1 
2 
kj
  

 K  =  RHS 
i
 i 
  

which can be solved by standard techniques for linear systems of equations with GaussSiedel Iteration.
Once Ki is known,
V  U   
And p is obtained from Bernoulli equation, i.e.
V V
p  1 2
U
Mentioned potential flow solution only depend on is independent of flow condition, i.e.
U∞, i.e. only is scaled
On the surface of the body Vn=0 so that
V2
C p  1  S2 where VS  V  S =tangential surface velocity
U
VS    S  U   S    S

 U  cos   i   QS
S
where U   S i  U  cos iˆ  sin ˆj  cos iiˆ  sin  i ˆj
VSi  U   S i 

S 



S S
M
S  
i
j 1
j i

K
 2 ln rds
SB

 ln r  dS
2  S
Kj
ji
lj

j
: j=i term is zero since source panel induces no tangential
i
flow on itself. ( 
lj

ln r ji dS j  J j )
S i

ln rij   1 rij  1  i rij  S i
S i
rij S rij
rij
 rij

1
S

S zi   2 xi  x j S xi  z i  z j S zi

xi
z i
 xi
 rij
where S xi  cos  i , S zi  sin  i

1
rij


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 VS
C pi  1   i
U
li
x
li
0
Fall 2009
2

K
 , V  U  cos   i    i J j

j 1 2

i j
xi  x j n xi  z i  z j n zi
1 rij
dS j  
dS j
2
2
rij S
S j x i  x j   z i  z j 
J ij  

Chapter 8
41
 x j 0  S j S xj S xi  z i  z j 0  S j S zxj S zi
i
x
 x j 0  S j S xj   z i  z j 0  S j S zj 
2
i
2
dS j
where S xi  cos  i , S zi  sin  i , xi  x j 0  S j S xj   z i  z j 0  S j S zj   S 2j  2 AS j  B
2
x

2
 x j cos  i  z i  z i 0 sin  i  S j  S xj S xi  S zj S zi   cos  j cos  i  sin  j sin  i
i
i
S
0
CS j  D
2
j
 AS j  C
dS i
D  xi  x j 0 cos  i  z i  z j 0 sin  i
1

C   cos i   j   DI j1  CI j 2  DI j1  C  ln   AI j1 
2

2
C
D  AC  1 l j  A
C l j  2 Al j  B
1 A 
J j  D  AC I j1  ln  
 tan
tan
  ln
2
E 
E
E 2
B
D  AC  xi  x j 0 cos  i  z i  z j 0 sin  i


  xi  x j 0 cos  i  z i  z j 0 sin  i  cos j   i 
x  x cos   z  z sin   sin  sin   cos  cos  
 x  x cos   sin  sin  cos   cos  cos  
z  z sin   sin  sin   cos  cos  sin  
 x  x cos  1  cos    sin  sin  cos  
z  z sin  1  sin    cos  cos  sin  
 x  x sin  cos  sin   sin  cos    z  z cos   sin 
i
j0
i
i
j0
i
i
j
i
j
2
i
j0
i
i
j
j
i
i
2
i
j0
i
i
j
i
j
j
2
i
j0
i
j
i
j
j
2
i
j0
i
where
i
j0
j
j
i
i
j
j
i
j
j
i
j0
j
i
cos  j  cos  i sin  j
D  AC
  sin  i   j 
E
41
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