Panel Methods
Source and Vortex
• Point Source
• Point Vortex
+
dz in Polar Form
ix
ds
z+dz

z
x
Panels
Singularity distributed along a line
Example: The Source Panel (or Sheet)
Consider a point source +
W ( z) 
q
2 ( z  z1 )
Imagine spreading the source along a line. We would then end up with a
certain strength per unit length q(s) that could vary with distance s along
the line.
z1
z
s
ds
z1
Constant Strength Source Panel
b
ds
vn
vs
z1
z

a
The panel is not a solid boundary to the flow. To make it behave like one you would
set the strength q so that the total vn (due to the panel and the flow) is zero
Vortex?
A Simple Source Panel Method
For flow past an arbitrary body
•
•
•
Break up the body surface into N
straight panels.
Write an expression for the normal
component of velocity at the middle of
the panel from the sum of all the
velocities produced by the panels and
the free stream. Gives N expressions.
Given that each expression must be
equal to zero, solve the N equations
for the N strengths
control
point
panel
Defining the N Panels
• Number the panels anticlockwise, 1 to N
• Define N coordinates za that identify the
start of every panel (going counter
clockwise), and zb that identify the end of
every panel.
za(3)
mth
control
point
nth
panel
dz 

 Im W ( z )  is the component normal
ds  to the panel

1
2
2
N-1
So, if W(z) is the velocity of the whole flow,
zc 
za(2)
1
N
dz i zb  za
• Each panel has a slope ds  e  z  z
b
a
• We pick a control point very close to the
center of the panel at
3
zb(2)
za(N-1)
zb
zb(N-1)
 zc
za  zb   i ( zb  za )
Center point
Displacement 
<<1 (say 0.001)
za
Completing the Method
 z  za ( n )  ds
(n) 1

W ( z )  W   q
log e 
(n) 
2
n 1
 z  zb  dz1
N
Velocity produced by whole
flow is
(n)
1
(n)
(m)
(m)
(n)
Velocity at the control point of
(m)
N


dz
1
z

z
ds
dz
1
 W 1   q ( n )
log e  c( m ) a ( n ) 
the mth panel zc(m) in panel
ds
2
n 1
 zc  zb  dz1 ds
aligned components is
So, velocity normal to
the mth panel is
Velocity parallel to
the mth panel is
 dz1 ( m )  N ( n )
 Im 
W    q Im{C ( m ,n ) }
 ds
 n 1
 dz1 ( m )  N ( n )
 Re 
W    q Re{C ( m ,n ) }
 ds
 n 1
C ( m ,n )
2
We want the normal velocity to be zero, so this is what we use to get the q’s
We write
Or
 dz1 ( m )  N ( n )
 Im 
W    q Im{C ( m ,n ) }
 ds
 n 1
1xN result matrix
(known)
=
1xN matrix
of strengths
(unknown)
x
NxN matrix of
ceoffs (known)
Once we have solved
this for the q’s we can
use eqn. 2 to get the
velocities along the body
surface, or eqn. 1 to get
them anywhere else
Computational Steps
• Define coordinates of start and end of panels za and zb
•
za( n ) , zb( n )
dz i zb  za
e 
Compute the panel slopes
ds
zb  za
• Put the control points next to the panel centers zc  12 za  zb   i ( zb  za )
• Determine the component of W∞ normal to each panel
• Determine the influence coefficients
C ( m ,n )
 dz1 ( m) 
 Im 
W 
 ds

 zc( m )  za ( n )  ds
1


log e  ( m )
(n) 
2
 zc  zb  dz1
• Solve the matrix problem, i.e. matrix divide Im{C ( m,n ) } by
• Compute the flow velocities and pressures
(n)
dz1
ds
(m)
 dz1 ( m) 
 Im 
W 
ds


ConstantSourcePanel.m
Matlab Code
za( n ) , zb( n ) , dz1 ds
(n)
zc(n )
C ( m ,n )
Matrix div.
 dz1 ( m) 
 Im 
W 
ds


Result
matrix
 dz1 ( m )  N ( n )
Re 
W    q Re{C ( m ,n ) }
 ds
 n 1
Velocities
along body
surface
 dz1 ( m )  N ( n )
 Im 
W    q Im{C ( m ,n ) }
 ds
 n 1
 res(1)   Im{C (1,1) } Im{C (1, 2 ) }  Im{C (1,n ) }  Im{C (1, N ) }  q(1) 
 ( 2 ) 

 
( 2 ,1)

 q 
 res( 2)   Im{C }
  
   






( m ,1)
( m ,n )
(n)
 q 
Im{C
}
 res( m)   Im{C }


   






 
 res( N )   Im{C ( N ,1) }
( N , N )  ( N ) 
Im{C
}  q 

 
mth
control
point
3
2
1
N
q(n)
nth
panel
N-1
ConstantSourcePanel.m
Matlab Code Ideas:
1. Non-Uniform Free Stream
E.g. Suppose
free stream
includes, say a
doublet outside
the body at a
location x=5, so
W  1 
10
( z  5) 2
ConstantSourcePanel.m
Matlab Code Ideas:
2. More than one body
E.g. Suppose
we have two
circles
ConstantSourcePanel.m
Matlab Code Ideas:
3. Use more sophisticated panels
E.g. Panels with
linearly varying
strength
LinearSourcePanel.m
3. Linear Source Panel Method
E.g. Panels with
linearly varying
strength
3. Linear Source Panels
Constant
strength panels
Linear strength
panels
qc
c
qb
qa
b
b
qa
a
a
Influence of qa
depends only
panel ab
Influence of qb
depends on panel
ab and panel bc
LinearSourcePanel.m
3. Linear Source Panel Method
E.g. Panels with
linearly varying
strength
LinearSourcePanel.m
Matlab Code Ideas:
4.Change to vortex panel method
W ( z) 
q
2 ( z  z1 )
W ( z) 
 i
2 ( z  z1 )
 dz1 ( m )  N ( n )
 Im 
W    q Im{C ( m ,n ) }
 ds
 n 1
 res(1)   Im{C (1,1) } Im{C (1, 2 ) }  Im{C (1,n ) }  Im{C (1, N ) }  q(1) 
 ( 2 ) 

 
( 2 ,1)

 q 
 res( 2)   Im{C }
  
   






( m ,1)
( m ,n )
(n)
 q 
Im{C
}
 res( m)   Im{C }


   






 
 res( N )   Im{C ( N ,1) }
( N , N )  ( N ) 
Im{C
}  q 

 
mth
control
point
3
2
1
N
q(n)
nth
panel
N-1
LinearVortexPanel.m
(see also
ConstantVortexPanel.m)
Matlab Code Ideas:
4. Vortex panel method
LinearVortexPanel.m
(see also
ConstantVortexPanel.m)
Kutta condition
requires that
surface vorticity
at trailing edge is
zero.
Matlab Code Ideas:
5. Set a Kutta Condition
LinearVortexPanelKutta.m
Matlab Code Ideas:
5. Kutta Condition Code
Kutta condition
requires that
surface vorticity
at trailing edge is
zero.
 dz1 ( m )  N ( n )
 Im 
W    q Im{C ( m ,n ) }
 ds
 n 1
 res(1)   Im{C (1,1) } Im{C (1, 2 ) }  Im{C (1,n ) }  Im{C (1, N ) }  q(1) 
 ( 2 ) 

 
( 2 ,1)

 q 
 res( 2)   Im{C }
  
   






( m ,1)
( m ,n )
(n)
 q 
Im{C
}
 res( m)   Im{C }


   






 
 res( N )   Im{C ( N ,1) }
( N , N )  ( N ) 
Im{C
}  q 

 
mth
control
point
3
2
1
N
q(n)
nth
panel
N-1
Something to watch out for…
• The control-point equation
zc  12 za  zb   i ( zb  za )
Center point
Displacement 
assumes that as you move from a to b you are
progressing counter-clockwise around the body surface.
(For clockwise you need to reverse the sign before i).