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CHAPTER 10
ELEMENTS OF POTENTIAL FLOW
___________________________________________________________________________________________________________
10.1 INCOMPRESSIBLE FLOW
Most of the problems we are interested in involve low speed flow about wings and
bodies. The equations governing incompressible flow are
Continuity
!i U = 0 ( 10. 1 )
Momentum
! U
! t
P
I
2
U ( 10. 2 )
The convective term can be rearranged using !i U = 0 and the identity
U i! U =! " U ( )
" U +!
U i U
The viscous term in ( 10. 2 ) can be rearranged using the identity
! "! " U
( )
=! !i U ( )
2
U ( 10. 4 )
Using these results and !i U = 0 the momentum equation can be written in terms of the
vorticity.
! = " # U ( 10. 5 )
in the form
! U
! t
+ " # U + $
P
U i U
If the flow is irrotational,! = 0 , and the velocity can be expressed in terms of a velocity
potential.
U = !" ( 10. 7 )
The continuity equation becomes Laplace’s equation
!i U = !i!" =!
2
" = 0 ( 10. 8 )
and the momentum equation is fully integrable.
" t
P
U i U
= 0 (^10.^9 )
The quantity in parentheses is at most a function of time
! t
P
U i U
= f (^) ( t ) ( 10. 10 )
The expression ( 10. 10 ) is called the Bernoulli integral and can be used to determine the
pressure throughout the flow once the velocity potential is known from a solution of
Laplace’s equation ( 10. 7 ).
Generally the flow is specified within a volume V surrounded by surface A (Figure
10.1). A solid body defined by the function G ( x , t ) = 0 may be imbedded inside V.
Figure 10.1 Flow volume V with surface A and imbedded solid body G (^) ( x , t ) = 0_._
The solid body has the outward normal
n^ ˆ Body
! G
! G
Laplace’s equation is second order in the spatial derivatives and two conditions must be
known to construct a solution. Generally the value of! is known on the boundary
(Derichlet condition) as well as the derivative of! normal to the boundary !" /! n
U = !" +! # A ( 10. 16 )
The potentials satisfy a system of Poisson equations, a single equation for the scalar
potential
!i!" =!
2
" = Q ( x , t ) ( 10. 17 )
and three equations for the Cartesian components of the vector potential.
2
A = "# (^) ( x , t ) ( 10. 18 )
where the identity!^ "^!^ "^ A^ =^!^ ( !i A ) #^!
2
A (^) has been used with the choice of a
Coulomb gauge on the vector potential (^) !i A = 0. This choice has no effect on the
velocity field generated by the vector potential since to any vector potential we can add
the gradient of a scalar. Let A = A! + " s. Clearly! " A =! " A #. Choose s so that
!i A = !i A " +!
2
s = 0.
This approach allows one to construct fairly complex flow fields that can be rotational
while retaining the simplicity of working in terms of potentials governed by linear
equations and the associated law of superposition. The flow is determined once the
distribution of mass sources and vorticity sources are specified. Notice that this theory of
potential flow is exactly analogous to the theory of potentials in electricity and
magnetism. The mass sources coincide with the distribution of electric charges and the
vorticity coincides with the electric currents.
10.3 USEFUL SPECIAL FUNCTIONS
A function that is highly useful in the development of potential theory is the smooth
version of the Heavyside-theta function
h ( x ; !) =
1 + Erf
x
where! is a small parameter that determines the steepness of the smooth transition from
0 to 1. The Heavyside function is dimensionless. The small parameter has the same units
as the argument x , (^) !ˆ = 1 / x ˆ and the Heaviside-theta function is dimensionless.
The first derivative of h ( x ; !) is a smooth version of the Dirac delta function
! (^) ( x ; ") =
e
x
2
4 "
2
The integral of! ( x ; ") is
e
!
x
2
4 "
2
!$
$
%
dx = 1 ( 10. 21 )
Integrating the product of the Dirac delta function and some function f (^) ( x ) is
lim
! " 0
e
( x # a )
2
4!
2
#%
%
&
f ( x ) dx = f ( a ) ( 10. 22 )
In three dimensions
! ( x ; ")! ( y ; ")! ( z ; ") =
e
r
2
4 "
2
3
$
3 / 2
where r
2
= x
2
2
2
. The units of the Dirac delta function are
! = 1 / x ˆ and its
derivative is
! ' (^) ( x ; ") =
xe
x
2
4 "
2
3
$
x
2
! (^) ( x ; ") ( 10. 24 )
The derivative satisfies
! x
e
"
x
2
4 #
f (^) ( x )
"+
,
dx = f (^) ( x )
! x
e
"
x
2
4 #
! f
! x
e
"
x
2
4 #
"+
,
dx =
e
"
x
2
4 #
f (^) ( x )
"+
In the limit! " 0
f (^) ( x )
! " (^) ( x )
! x #$
$
%
dx = #
! f
! x
" (^) ( x )
#$
$
%
dx ( 10. 26 )
The result ( 10. 26 ) can be used to convert integrals involving derivatives of the Dirac
delta function to the basic form ( 10. 22 ).
10.4 POINT SOURCE SOLUTION OF THE POISSON EQUATION
The figure below shows a smooth spherically symmetric source of mass centered at the
origin of a set of spherical polar coordinates.
Use Gauss’s theorem on the left hand side.
! r
0
2 #
$ 0
$
r
2
Sin ( % ) d & d % =
Q ( t )
3
3 / 2
(
e
)
r
2
4 '
2
0
2 #
$ 0
$ o
r
$
r
2
Sin ( % ) d & d % dr ( 10. 32 )
When ( 10. 30 ) is substituted into ( 10. 31 ) the result is C 2
= 0. This analysis leads to a
smooth version of the source solution of the Poisson equation.
! ( r , t ) = "
Q ( t )
4 #$ r
Erf
r
In the limit! " 0 , the source distribution reduces to a point and! ( 1 / 4 " r ) Erf (^) ( r / ( 2 #))
becomes the classical Green’s function for the Laplace equation,
G x , x s
( ) =! 1 / 4 " x! x s
( ) where^ x s
is the vector position of the source. The
fundamental point source solution of the Laplace equation for a source located as x s
is
! x , x s
( , t ) =^ "^
Q ( t )
4 #$ x " x s
10.5 GENERAL SOLUTION OF THE POISSON EQUATION
Figure 10.3 shows a general smooth distribution of mass sources and vorticity sources in
a finite region near the origin. The source strength outside the finite region is zero and is
said to be compact.
Figure 10.3 Smooth, finite distribution of mass and vorticity sources near the origin.
We can solve the Poisson equation by using the fundamental point solution ( 10. 33 ) and
superposing the scalar potentials produced by an infinite sum of differential sources. The
incremental scalar potential, !" , produced by the mass source contained in the
differential volume shown in the figure above is determined from the Poisson equation
for this source. Namely
2
( "#) =
Q x s
( , t ) " x " y " z
3
%
3 / 2
&
e
'
x ' x s
2
4 $
2
where now Q is the source strength per unit volume with units Mass / Sec! Length
3
( ).
The solution of ( 10. 35 ) with lim
! " 0
Erf x # x s
( )
= 1 is
Q x s
( , t )! x s
! y s
! z s
4 $% x # x s
In the limit where the differential volume becomes infinitesimally small
d! = "
Q x s
( , t )
4 #$ x " x s
dx s
dy s
dz s
and the general solution to the scalar Poisson equation ( 10. 17 ) is
! ( x , t ) = "
Q x s
( , t )
x " x s
dx s
dy s
dz s "%
%
& "%
%
& "%
%
&
The Poisson equation for the vector potential
2
A (^) ( x , y , z , t ) = "# (^) ( x , y , z , t ) ( 10. 39 )
is equivalent to three scalar equations relating the Cartesian components of the vector
potential to the Cartesian components of the vorticity.
2
A x
x
2
A y
y
2
A z
z
The same procedure used to derive ( 10. 38 ) can be applied to each of the Cartesian
components of the vector potential. The fundamental solution of ( 10. 39 ) at vector
position x due to a point source of vorticity of strength! x s
( , t ) dx s
dy s
dz s
at source point
x s
is
dA =
! x s
( , t ) dx s
dy s
dz s
4 " x # x s
! ( x , y , z , t ; a , b ) =
S ( t )
Ln
z # b + x
2
2
2
z # a + x
2
2
2
The potential ( 10. 45 ) can be used to generate the potential due to a semi-infinite line of
sources by expanding ( 10. 45 ) about the point a! " #.
lim
a !"#
$ ( x , y , z , t ; a , b ) =
S (^) ( t )
" Ln (^) ( 2 ) + Ln "
a
2
2
2
( )
S ( t ) y
4 % a
S ( t )
16 % a
2
x
2
2
" 2 z
2
( ) +^
S ( t )
24 % a
3
3 x
2
z + 3 y
2
z " 2 z
3
( ) +^ O^
a
4
Taking the limit, a! "# in ( 10. 46 ) the potential of a semi-infinite line of sources is
lim
a !"#
$ (^) ( x , y , z , t ; a , b ) =
S (^) ( t )
Ln z " b + x
2
2
2
( )
S (^) ( t )
Ln "
2 a
The additive constant which goes to infinity as a! "# is not surprising in view of the
fact that the source distribution extends to infinity and the potential from any segment of
this source distribution only dies off like 1 / r where r is the distance from the segment.
The singularity in the potential has no effect on the velocity field generated from
U = !".
Let the line of sources extend to plus infinity by setting a =! b in ( 10. 45 ). Expand about
the point b! ". The result is
lim
b !"
( x , y , z , t ; b ) =
S (^) ( t )
% 2 Ln ( 2 ) + 2 Ln
b
2
2
( )
S (^) ( t )
8 $ b
2
x
2
2
% 2 z
2
( ) +^ O^
b
4
When the limit b! " is applied the result is
lim
b !"
( x , y , t ; b ) =
S ( t )
Ln x
2
2
( ) +^
S ( t )
Ln
2 b
Again there is a logarithmic infinity in the potential when we add up an infinite line of
sources in a three-dimensional world. Dropping the constant we recover the potential for
a two-dimensional line source of area Q ( t ).
! (^) ( x , y , t ) =
Q (^) ( t )
Ln x
2
2
( )
1 / 2
The radial velocity generated by differentiating ( 10. 50 ) with respect to r is
U
r
Q (^) ( t )
r
If the origin is enclosed by a circle and the area flux from the source is integrated the
result is
U
r 0
2!
"
rd # =
Q (^) ( t )
Using the same differential procedure we used in three-dimensions, the circularly
symmetric source solution ( 10. 50 ) can be used to generate the general solution of the
two-dimensional Poisson equation for a compact distribution of sources. The result is
! ( x , y , t ) =
Q x s
( , t ) Ln x $ x s
1 / 2
dx s
dy s $%
%
& $%
%
&
Example 2 – Vector potential of a vortex monopole
The figure below shows a vortex monopole located at the origin with its counter-
clockwise rotation axis aligned with the z axis.
Figure 10.5 A vortex monopole
The vorticity source term is
! ( x , t ) = 0 , 0 ,
{ W^ ( t^ )"^ (^ x )"^ (^ y )"^ (^ z )} (^10.^54 )
Insert ( 10. 54 ) into the general solution A = 0 , 0 , A z
( ) where
A
z
( x , t ) =
W ( t ) " x s
( )"^ y s
( )^ "^ z s
( )
x # x s
dx s
dy s
dz s #$
$
% #$
$
% #$
$
%
The solution is a vector point source
The vorticity source distribution for this case is
! ( x , t ) = (^) { 0 , 0 , " ( t ) # ( x ) # ( y ) u ( b $ z ) u ( z $ a )} ( 10. 60 )
and the vector potential is A = 0 , 0 , A z
{ } where
A
z
( x ,^ y ,^ z , t ; a , b ) =
" ( t ) # x s
( )^ #^ y s
( ) u^ b^ $^ z s
( ) u^ z s
( $^ a )
x $ x s
( )
2
( )
2
( )
2
1 / 2
dx s
dy s
dz s $%
%
& $%
%
& $%
%
&
" (^) ( t )
x
2
2
( )
2
1 / 2 a
b
&
dz s
$" (^) ( t )
Ln
z $ b + x
2
2
2
z $ a + x
2
2
2
The velocity field is U =! " A.
U =
x
2
2
2
z! b + x
2
2
2
( )
x
2
2
2
z! a + x
2
2
2
( )
- { y ,! x , 0 } ( 10. 62 )
Let the vortex line extend equal distances from the origin, a =! b. The circulation about
a contour that encircles the z-axis on the plane z = 0 is
circle
= U i c ˆ dC =
C
!"
x
2
2
2
$ b + x
2
2
2
( )
x
2
2
2
b + x
2
2
2
( )
x
2
2
( )
0
2 #
"
d + ( 10. 63 )
Consider two limits of ( 10. 63 ). In the first let the radius of the circle of integration
become large.
circle
= lim
R "#
R
2
2
% b + R
2
2
( )
R
2
2
b + R
2
2
( )
R
2
0
2 $
,
d - =
! b
R
The circulation of a finite line of circulation decays with radius from the source and there
is no net circulation at infinity. Now assume the radius of the circle of integration is finite
and let the length of the line of circulation become infinite.
circle
= lim
b "#
R
2
2
% b + R
2
2
( )
R
2
2
b + R
2
2
( )
R
2
0
2 $
,
d - =! ( 10. 65 )
The circulation is constant independent of the radius of the circle of integration consistent
with Helmholtz’ laws for inviscid flow with vorticity.
Now return to the vector potential ( 10. 61 ) let a =! b and take the limit b! ". The
resulting vector potential for an infinite vortex line is A = 0 , 0 , A z
( ) where
lim
b !"
A
z
( x , y , z , t ; b ) =
#$ ( t )
Ln x
2
2
( ) #^
$ ( t )
Ln
2 b
Drop the singular constant. The fundamental source solution for the two-dimensional
Poisson equation for the z component of the vector potential (aka the stream function) is
! ( x , y , t ) =
"# ( t )
Ln x
2
2
( )
1 / 2
( )
Again we apply the same differential procedure we used in three-dimensions to two
dimensions. The circularly symmetric source solution ( 10. 67 ) is used to construct the
general solution of the two-dimensional Poisson equation for a compact distribution of
vorticity sources. The result is
! ( x , y , t ) =
% x s
, y s
( , t ) Ln^ x^ "^ x s
1 / 2
dx s
dy s "&
&
' "&
&
'
Example 3 – Uniform flow past a sphere
The velocity potential of a dipole source of volume in a three-dimensional flow is
Dipole
" x
x
2
2
2
( )
3 / 2
where! is the strength of the dipole. The dipole added to a uniform flow generates the
potential flow about a sphere in uniform flow.
and the velocities are
U
x
( x , y , z ) = U !
3 R
Sphere
( )
3
x
2
2 r
5
R
Sphere
( )
3
2 r
3
U
y
( x , y , z ) = " U !
3 R
Sphere
( )
3
xy
2 r
5
U
z
( x , y , z ) = " U !
3 R
Sphere
( )
3
xz
2 r
5
10.6 ELEMENTARY 2D POTENTIAL FLOWS
Any irrotational, incompressible, 2-D flow can be represented by, either the velocity
potential and/or a stream function.
U =
! x
! y
V =
! y
! x
The relations in ( 10. 76 ) may be familiar as the Cauchy-Riemann equations from the
theory of complex variables. Let
z = x + iy ( 10. 77 )
The complex stream function is
W (^) ( z ) =! (^) ( x , y ) + i " (^) ( x , y ) ( 10. 78 )
An important representation of a complex variable due to Leonhard Euler is
z = x + iy = r (^) ( Cos (! ) + iSin (! )) = re
i!
( 10. 79 )
where
r = x
2
2
( )
1 / 2
and
Tan (! ) =
y
x
Two-dimensional potential flows can be constructed from any analytic function of a
complex variable, W ( z ). From the Cauchy-Riemann conditions ( 10. 76 )
2
"
! x
2
2
"
! y
2
2
! x! y
2
! y! x
2
"
! x
2
2
"
! y
2
2
$
! y! x
2
$
! x! y
Both the velocity potential and stream function satisfy Laplace’s equation.
The derivative of the complex potential (the complex velocity) is independent of the path
along which the derivative is taken.
dW
dz
! x
dx
dz
! x
dx
dz
= U $ iV
dW
dz
! y
dy
dz
! y
dy
dz
V
i
U
i
= U $ iV
Some Elementary Flows with their streamline patterns
1) Uniform flow in the x-direction
W = U
!
z! = U "
x! = U "
y ( 10. 85 )
2) A mass source at the origin
Here we solve the Poisson equation in two dimensions for the velocity potential with a
point source of area at the origin.
2
" = Q # ( x ) ( 10. 86 )
where Q is the strength of the area source.Use the Green’s function solution
! = U
"
x +
Q
Ln ( x + a )
2
2
( )
1 / 2
Q
Ln ( x $ a )
2
2
( )
1 / 2
! = U
"
y +
Q
ArcTan
y
x + a
Q
ArcTan
y
x * a
5) Point vortex
Here we solve the Poisson equation for the stream function with a point source of
circulation at the origin.
2
" = #$ % ( x ) ( 10. 93 )
where! is the strength of the source. The Greens function solution is
! (^) ( x ) =
$ % x s
( ) Ln^ x^ "^ x ( (^) s )
A
&
dA =
$ % r s
( ) Ln^ r^ "^ r ( (^) s )
0
r
& 0
2 #
&
drd ' =
Ln (^) ( r ) ( 10. 94 )
This is the same solution we derived earlier through a limiting process of allowing a
finite vortex line become infinite. The potentials for a point vortex are
W =!
i "
Ln (^) ( z )! =
Ln (^) ( r ) ( 10. 95 )
For any contour C surrounding the origin
! dA
A
"
= # $ U dA
A
"
= U
C
!"
c ˆ dC =
2 & r
0
2 &
"
rd ' = % ( 10. 96 )
6) Vortex doublet
This is constructed from two point vortices of opposite circulation separated by the
distance a. As they are brought together the strength! = a " is held constant.
W =
i
z
i
r
e
) i *
! =
Sin ( $ )
r
Cos ( % )
r
7) Stagnation point flow
W = Az
2
! = A x
2
" y
2
( )!^ =^2 Axy^ (^10.^98 )
8) Flow in a corner
The potentials are ( n = 2 is the stagnation point flow above).
W = Az
n
= A re
i!
( )
n
! = Ar
n
Cos ( n ")! = Ar
n
Sin ( n ") ( 10. 99 )
