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Professor Fred Stern
Fall 2009
Chapter 8
1
Chapter 8: Inviscid Incompressible flow: a useful fantasy
Introduction
For high Rn 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. forward 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 Rn 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 w=0 is a good approximation.
Note that for compressible flow this is not the case in regions of large entropy gradient.
Now, we are neglecting noninertial effects and the mechanism of vorticity generation.
Potential flow theory
1) Determine  from solution to Laplace equation
 2  0
B.C:

0
n
at S  : V    0
at S B : V .n  0 
Note:
F: Surface Function
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Fall 2009
Chapter 8
2
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, primary 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 : w    V  vorticity
V
1

 ( p  V 2  z )  V  
t
2
If   0 ie   V  0 then V   :


1
 p    z  B(t ) : Bernoulli’s Equation for unsteady incompressible flow
t
2
Continuity equation shows that PDE 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1  0 
2
2
2
2
   1    2  0   (1   2 )  0     0
2
  2  0
Techniques for solving Laplace equation:
1) superposition of elementary solution
2) surface singularity method (integral equation)
3) Fr or FE
4) electrical or mechanical analogs
5) Conformal mapping ( for 2D flow)
6) Analytic for simple geometry ( separation of variable etc)
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Fall 2009
Chapter 8
3
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
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.
For a uniform stream moving at an angle,
 , relative to the x-axis, we can write
u  Ucos  
 

y x
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Professor Fred Stern
v  Usin   
Chapter 8
4
Fall 2009
  

x y
After integration, we obtain the following expressions for the stream function and
velocity potential:
  U y cos  xsin  
  U x cos  ysin 
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 height perpendicular to the paper,
Q  2r   1  v r
Or ,
vr 
Q
2r
m
, v  0
r
Q
Where: m 
m= convenient constant (m>0 for source and m<0 for sink)
2
 vr 
In a polar coordinate system, for 2-D flows we will use:
 1 
V   

r r 
And:
.V  0
1 
1 
(rv r ) 
(v )  0
r r
r 
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Professor Fred Stern
Chapter 8
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Fall 2009
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
r sin 
r sin 
tan 1 
; tan  2 
r cos   a
r cos   a
r sin 
r sin 


2ar sin 
tan(  )  r cos   a r cos   a  2
r sin 
r sin 
m
r  a2
1
r cos   a r cos   a
2
ar sin 
  m tan 1 ( 2
)
r  a2
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Professor Fred Stern
Chapter 8
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Fall 2009
For small value of a :
ψ
2amr sin θ
2amy
 2
2
2
r a
x  y2  a2
ma  constant (m   ; a  0 )  ψ  
  2am ( Doublet Strength)
 sin θ
r

y
x  y2
2

 cos 
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.
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Professor Fred Stern
Chapter 8
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Fall 2009
2D vortex:
Suppose that value of the  and  for the source are reversal.
  Kθ
ψ  Kθ
r  0
K  constant
1 
 K


r 
r
r
2d vortex is irrotational anywhere except at the origin where v and V   are infinity.
 
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  dA   .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
 
K
r
K

(rd )  2K and K 
r
2
Note: for point vortex, flow still irrotational everywhere except at origin itself where
v=infinity.

2
0
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Professor Fred Stern
Fall 2009
Chapter 8
8
Now, we can use Stokes theorem to show the existence of  :
 V .ds   V .ds  
ABC
'
AB C
C
Since
 V .ds  0
ABCB' A
Therefore in general for irrotational motion:
 V .dx  
V .dx  d
dx d
dx

  .
ds ds
ds
dx
es 
ds
(V   ).e s  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 8.7 of Text book)
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Professor Fred Stern
Fall 2009
Chapter 8
9
For y  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 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 the 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:
 Doublet  
 sin 
r
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Professor Fred Stern
Chapter 8
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Fall 2009
When this doublet is superposed over a uniform flow parallel to the x- axis, we get:

 1
r sin 
 u  1 
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.
 sin 
  u  r 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 .
vr 

1 
 
 cos  =0
 u  1 
2 
r 
 u R 
   u R 2
If we replace the constant

u
by a new constant R2, the above equation becomes:

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 
vr  0
How about pressure p? look at the Bernoulli's equation:
p
p 1 2
1
2
 v r  v    u 2
 2
 2


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Professor Fred Stern
Chapter 8
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Fall 2009
After some rearrangement we get the following non-dimensional form:
2
2
v v
p  p
Cp 
 1 r 2 
1 2
u
u 
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  ( 2Rd )b b=span length
2
1
F   u 2 b(2R)  (1  4 sin 2  )(cos  iˆ  sin  ˆj )d
2
0
CL 
Lift
1 2
u  b(2R)
2
2
=   (1  4 sin 2  ) sin  d  0 (due to symmetry of flow around x
0
axis)
CF 
Drag
2
=   (1  4 sin 2  ) cos  d  0 (dÁlembert paradox)
1 2
0
u  b(2R)
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
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Chapter 8
12
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
FORM DRAG
CD
: drag due to viscous modification of pressure destruction
FORM DRAG
CDSKIN FRICTION : Drag due to viscous shear stress
Streamlined Body : CDSK  CDFO

Bluff Body :CDFO  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
2r
r 

On the surface of the cylinder (r=R):
 R2 
1 
vr 
 u  1  2  cos   0
r 
 R 


v  
 2u  sin  
r
2R
v  
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
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Chapter 8
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Fall 2009
So, the location of stagnation point is function of  .
 s (stagnation point)
K
u R
0 ( sin   0 )
1( sin   0.5 )
2 ( sin   1 )
>2( sin   1 )

0,180
30,150
90
Is not on the circle
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:
u   y  x
v   x   y
In polar coordinates:
 1 
vr 

r r 
1 

v 

r 
r
Flow
Uniform Flow
Source,Sink

U x
K
ln r
2

U y
K
2
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Professor Fred Stern
Dipole
Vortex
90 Corner flow
Solid-Body rotation
Chapter 8
14
Fall 2009
 cos 
r



2
1 / 2 A( x 2  y 2 )
Doesn’t exist

 sin 
r

ln r
2
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
Circular Cylinder
Uniform stream+dipole
Circular Cylinder with
circulation
Uniform
stream+dipole+vortex
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Chapter 8
15
Fall 2009
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
1  2 
   
1 2
 2

p  p    u   2u  sin  

p


u

 


2 
2R   
2
2 2 R 2


u 
  2 u 2 sin 2    sin 
R

 A  B sin   C sin 2 
where,

1 2
 2 
A   p   u   2 2 
2
2 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
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 freestream speed. Kutta and
Joukowski generalized this result to lifting flow over airfoils. Equation L   u  is
known as the Kutta-Joukowski theorem.
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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 Pranttl (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 2
u  (2R) u  R
2
2
1
1 
Define: v Average 
v d 
 

2 0
2  R 
 CL 
 
Note: Γ   V  d s   V  dA   V Rd 
c
A
A
2
v
u  a vera g fe
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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 producee forward motion.
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Chapter 8
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Fall 2009
Complex variable and conformal mapping
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.
1. 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
so that;
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.
2
3
z = real + iimaginary = x + iy
4
y, imaginary
x, real
- A complex number can be written in polar form using Euler's equation;
z = x + iy = 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 + iy z  x  i. y z.z  x 2  y 2
-Complex function:
w(z) = f(z)=  (x,y) + i (x,y)
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If function which 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 always an infinite number of paths, and thus, in order to define a
unique derivative f(z) must 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 ( x 2  x)

dw
 i y   y
dz 2
dw
to be unique and independent of path:
dz
 x   y   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 definitions.
 xx   yx   yy   xy i.e.  xx   yy  0
Thus, complex function w    i represents 2-D flows.
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3. Application to potential flow
W ( z )    i Complex potential where  : velocity potential,  : stream function
dW
  z  i z  u  iv  u r  iu  e i Complex velocity
dz
r ' e i '  re i (  ) where r '  r (magnification) and  '     (rotation)
xxx about z0 is transformed into a similar xxx in the ζ-plane which is magnified and
rotational.
Implication:
-Angles are presented between the intersections of any two lines in the physical domain
xxx 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 xxxx
As mentioned above, usually the flow-field solution in the ζ-plane is known:
W ( )   ( , )  i ( , )
Then
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w( z)  W  f z    ( x, y)  i ( x, y) or    &   
4. Conformal mapping
The real power of the use of complex variables for flow analysis is throughout 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 xxx. The ζ-plane solution to the z-plane through the conformal transformation ζ=f(z)
(or inverse mapping z=g(ζ)).
Before xxx 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 ' , t  re i , 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 (φ, ψ).

 2z    2   0
 2z    2   0
i.e. Laplace equation ins the z-plane transforms into Laplace equation
is the ζ-plane.
The complex velocities in each plane are also simply xxx
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dw dw d

dz d dz
dw
( z )  u  iv  (U  iV ) f ' ( z )
dz
i.e. velocities in two planes are xxx.
Two independent theorems xxx conformal transformations are:
(1) Closed curves map to closed curves
(2) Xxxx mapping theorem: an arbitrary xxx xxx can be mapped onto the unit circle.
More theorems are given and discussed in AMF Section 43. Note that there are for the
interior problems, but we equally xxx for the exterior problems through the inversion
mapping.
Many transformations have been investigated and are xxx in xxx. The xxx (particularly
AMF) contains many examples:
1) Elementary transformations:
az  b
, ad  bc  0
a) linear: w 
cz  d
b) xxx flow: w  Az n
2
c) Jowkowsky: w    c

d) exponential: w  e
e) w  z s , s xxx
n
2) Flow field for specific geometries
a) circle theorem
b) flat plate
c) circular area
d) ellipsoid
e) Jowkowski foils
f) xxx (two circular areas)
g) Thin foil theory [ solutions by mapping flat plate with thin foil b.c. onto xxx
circle (xxx))
h) multiple xxx
3) Schwarx-Cristoffel mapping
4) xxx-streamline theory
The techniques of conformal mapping are best xxx through their applications. Hence we
shall consider two transformations (xxx flow or Joukowski) and briefly the SchwarzCristoffel mapping of free-streamline theory. For further details, consult the course book.
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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 1e i ( n 1)  (nUR n 1 cos n  inUR n 1 sin n )e i
dz d dz
 (u r  iu  )e i
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xxx corner
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Chapter 8
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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 xxx geometry.
Consider the B.V.P for a body of arbitrary geometry fixed in a uniform stream of an xxx,
incompressible, and irrotational fluid.
The surface singularity method is formulated on the xxx form of Greens theorem and
what is known as Greens function.
 G    G dV 
2
2
V
G 
 

G
dS (1)
n
n 

S

S

S

S

b
S


where Φ and G are any two xxx xxx in V (control volume) of xxx xxx application.
Φ= velocity potential
G= Green’s function
Say,
 2 G   ( x  x0 ) in V+V’ (i.e. entire domain) where δ is the xxx xxx function.
G0 on S∞ (xxx xxx Former Transforms)
Solution for G 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

By externally the definition of Φ into V’ xxx can be shown that Φ can be xxx represented
by distributions of sources, defined as
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

 GdS : source distribution,  : source strength
S SB
or

Fall 2009
Chapter 8
27

 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 xxx body; it is , for xxx xxx dipole or vortex distributions must
be used.
As this stage, let’s consider the solution of the flow about a xxx body of xxx gemometry
fixed in a uniform stream. Note that since G0 on S∞ Φ xxx solution the condition S∞.
The xxx condition is the condition that the xxx xxx is a stream xxx need to determine the
k
source distribution. Strength  
is the xxx notation.
2
First, consider a source distribution meeting for representing xxx flow around a body of
arbitrary geometry.
V  U   Q : total velocity
U  : uniform stream, Q : perturbation potential due to xxx of body
U   U  (cos iˆ  sin ˆj ) : note that for xxx flow  must be given (i.e. for a symmetric
field   0 or for xxx filed    o )
K
Q 
ln rds : source distribution on body surface
2
SB
Now, K is determined from the body boundary condition.
Q
 U   n
V  n  0 i.e. U   n  Q  n  0 or
n
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i.e. normal velocity xxx by xxx must cancel xxx
 K
ln rds  U   n
n SB 2
This singular integral equation for K is xxx by xxx the into a number of panelsover which
K is assumed constant, i.e. we write

ni
M  no. of panels

j 1
Ki
2
x
where rij 

Si
ln rij dS i  U   ni , i=1,M, j=1,M
 x j   z i  z j  =distance from ith panel control xxx to ri  position
2
i
2
vector along jth panel.
Note that the integral equation is singular since

1 rij
ln rij 
ni
rij ni
at for rij  0 then integration blows up j that is when j=i at we trying to determine the
contribution of the panel to its own source strength. Speed xxx must be taken that can be
shown that the xxx does exist at the integral xxx can be written
ki 
2 ni

S j Si
Ki M K j

2 i 1 2
j i
where
ln rij dS i 
ki
2
1 rij
 r nij dS j  U   ni
Si ij
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 
Consider the ith panel
2
2
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
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r j  r j 0  S S i , r j 0 : origin of jth panel xxx xxx, S Si : distance along jth panel
V  U   Q
K
Q 
ln rds
2
SB
V n  0
Ki M K j

2 i 1 2
j i
Let
 K
ln rds  U   n
n SB 2
1 rij
S rij nij dS j  U   ni  U  sin    i  , ni   sin  i iˆ  cos  i ˆj
i
1 rij
 r nij dS j I j and RHS i  U  sin    i  , then
Si ij
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
S j xi  x j   z i  z j 
j i
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Fall 2009
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
0
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
S  2 AS j  B
2
j
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 
Ki M K j

I i  RHS i : Matrix equation for Ki xxx can be solved using Standard
2 i 1 2
j i
methods such as Gauss-Sidel 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
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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
Ki M K j

I j  U  sin    i 
2 i 1 2
j i
 1
 2
 
k
 i Ij
 2
ki 
Ij
2 

 
1 


2 

  

 K  =  RHS 
i
 i 
  

Which can be solved by standard xxx for linear xxx of wquations with xxx xxx
coefficient methods (Gauss-Sidel Iteration).
Once Ki is known,
V  U   Q
And p is obtained from Bernoulli equation, i.e.
V V
p  1 2
U
Xxx potential flow solution only xxx xxx in independent of flow condition, i.e. U∞, i.e.
only xxx
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On the xxx of the xxx 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 

QS 
Q


S S
M
QSi  
j 1
j i


K
 2 ln rds
SB

ln r dS

2 S
Kj
ji
j
: j=i term is xxx since source panel induces no tangential
i
lj
flow on its xxx. ( 
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 xi 
S zi   2 xi  x j S xi  z i  z j S zi

z i
 xi
 rij
where S xi  cos  i , S zi  sin  i


1
rij
 VS
C pi  1   i
U
li
li
x
0
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  


i
 x j 0  S j S xj S xi  z i  z j 0  S j S zxj S zi
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

i
i
 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
S
0
2
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

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2
C
D  AC  1 l j  A
C l j  2 Al j  B
1 A 
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
J j  D  AC I j1 


  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
33

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Chapter 8
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Fall 2009
Method of Image
The method of image is useful for plane, spherical, and circular flow domain. Since we
have already discussed this method through a variety of applications, here we shall
provide a summary.
Plane Boundaries:




m
2
2
ln  x  1  y 2  x  1  y 2
2
1
2
2
2
2  2
3-D: Q  M 
 x  1  y 2  z 2
 x  1  y  z

Similar results can be obtained for dipoles and vortices.
2-D: Q 



1
2



Spherical and Curvilinear Boundaries:
The results for plane boundaries are obtained from consultations of symmetry. For
spherical and circular boundaries, image systems can be determined from the previously
discussed Sphere&Circle Theorems, respectively, e.g.
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Chapter 8
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Fall 2009
Flow field
Sources 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 externally from center
a
2
of sphere to a
c
3
2
dipole pf strength  a  at  a
l
l
2
equal source at a
and sink of some
b
strength at the center of the circle
The method can be extended for multiple boundaries by using successive images. Recall
the xxx for a source equally spaced between two parallel palnes
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w( z )  m
Fall 2009
Chapter 8
45
 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   
As a xxx example of the method of successive images consider two sphere moving along
a line through their centers:
Consider the kinematic b.c. for A:
2
F  x, t    x  yt   y 2  z 2  a 2
DF
 0  V  eˆ R  U 1 kˆ  eˆk or  R  U 1QR '
Dt

2
Ua 3
where Q   2 cos , QR   3 cos   
R
2
R
Similarly for B QR  U 2QR '
This suggests the potential in the form
Q  U1Q1  U 2 Q2
where Q1 and Q2 both satisfy the xxx equation and the boundary xxx:
 Q1 
 Q 
 cos ,  1 
 0 ---(*)


 R  R  a
 R'  R 'b
 Q2 
 Q 
 0,  2 
 cos ' ---(**)


 R  R  a
 R'  R 'b
Q1 = potential where sphere A moves with unit velocity towards B, with B at rest
Q2 = potential when sphere B moves with unit velocity towards A, with A at rest
B were xxx
0
a3
a3
Q1   2 cos   2 cos ,  0 
2
2R
R
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 image requires at
image  2 at A2, the image of A1 with respect to A, and so on. Thus we have an infinite
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Professor Fred Stern
Chapter 8
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Fall 2009
series of images A1, A2, … of strength 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
Q1   0 2  1 2 1  2 2 2   with a similar development procedure for Q2.
R
R1
R2
Although exact, this solution is of xxx form. Let’s investigate the possibility of an
approximate solution which is valid for large C (i.e. large separation distance),
 2C
C2 
R'  R  C  2Cr cos  R 1 
cos  2 
R
R 

2
2
2
2
1 1
C
C2 
 1  2 cos  2 
R' R 
R
R 

1
2
1
R
R2 
 1  2 cos  2 
C
C
C 

1
2
Considering the former representation first,
1

1 1
 1  2u   2 2
R' R
By the binomial theorem valid for x  1

1  x  2
1

  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 xxx terms in powers of  , where the Pn are as before Legendre functions of the
first kind (i.e. Legendre polynomials which are Legendre functions of the first kind of
xxx xxx). Thus,
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Professor Fred Stern
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47
Fall 2009
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
 cos 
Q

R' 2
 R cos  C 
R
2
 C 2  2CR cos

3
2

 
1
 
C  2
2
 R  C  2CR cos



1

2


Thus,
 1 2 RP1 cos  3R 2 P2 cos 

  cos 
R  C :Q 
  2 

 
2
3
C
R'
C
C

 1

2C
3C 2
R  C : Q     2 P1 cos   3 P2 cos   4 P3 cos   
R
R
R

Going back to the two sphere problem. If B were absent
a3
Q1   2 cos
2R
 R' 
using the above expression for the xxx at B and near B   1 , RR’,   '
C

3
3
3
1 a

a R' P1 cos 
a
Q   2 cos   




2
2R
C3
2 C

 a 3 cos

C3
which can be cancelled by adding a term to the first approximation, i.e.
1 a 3 cos 1 a 3b 3 cos '
Q1  

2 R2
2 C 3 R' 2
to confirm this
1 a 3 a3R' cos ' 1 a 3b 3 cos
Q1  


2 C2
2 C3 R
C3
These approximate solutions are convert to  C 3 .
QR 
 
To find the kinetic energy of the fluid, we have

1 
K      QQ n dS   QQ n dS 
2  SA
SB

1

K  A11U 12  A12U 1U 2  A22U 22  
QQ n dS
2
2 S A  S


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Chapter 8
48
Q1
Q2
dS A , A22     Q2
dS B ,

n

n
A
B
Q1
Q1
A12     Q2
dS A     Q1
dS B where dS  2R 2 sin rdr
n
n
A
B
A11     Q1
2 3
2
2a 3b 3
A11  a  , A12 
 , A22  b 3  ,
3
3
3
C
3 3
1
2a b 
1
K  m'1 U 12 
U 1U 2  M ' 2 U 22 : using the approximate form of the potentials
3
4
4
C
1
1
where m'1 U 12 , M ' 2 U 22 : masses of liquid xxx by sphere
4
4
48