Panel Methods

Source and Vortex

• Point Source

• Point Vortex

+ )(2)(

1zzqzW−

=π

)(2)(

1zzizW−Γ−

=π

dz in Polar Form

x

ix

βz

z+dzds

=dz

so

or

Useful! … E.g. velocity components parallel and normal to line are given by ds

dzzWezWivv ins )()( ==− β

Panels

Consider a point source

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 salong the line. Each elemental length of the line ds would behave like a miniature source and so would produce a velocity field:

s ds

Where gives the coordinate of the line at s.The total flowfield produced by the line is then:

)(1 sz

Example: The Source Panel (or Sheet)

z1

z

+z1

Vortex?

Singularity distributed along a line

Constant Strength Source Panel

∫ −=

panel szzdssqzW

)()(

21)(

1π

If strength q is constant with s then

z

a

b

β

dsz1

So

βie−

vn vs

In terms of components aligned with the panel: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

==−b

aens zz

zzqdsdzzWivv log

2)( 1

π

The panel is not a solid boundary. To make it behave like a solid boundary in a flow you would have to set the strength q so that the total vn (due to the panel and the flow) is zero

Note that vn from positive to negative across the panel

Vortex?

βie

• 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 Nexpressions.

• Given that each expression must be equal to zero, solve the N equations for the Nstrengths panel

control point

A Simple Source Panel MethodFor flow past an arbitrary body

Defining the N Panels

• We pick a control point very close to the center of the panel at

mth

control point

1

23

N

N-1

za(2)zb

(2)za(3)

za(N-1)

zb(N-1)

( ) )(21

abbac zzizzz −−+= εCenter point Displacement Δ

ε<<1 (say 0.001)

• 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.

• Each panel has a slopeSo, if W(z) is the velocity of the whole flow,

is the component normal to the panel

zc

za

zb Δ

nth

panel

Completing the MethodVelocity produced by whole flow is ∑

=∞ ⎟⎟

⎠

⎞⎜⎜⎝

⎛

−−

+=N

n

n

nb

na

en

dzds

zzzzqWzW

1

)(

1)(

)()( log21)(π

Velocity at the control point of the mth panel zc

(m) in panel aligned components is

∑=

∞ +⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=N

n

nmnm

CqWdsdz

1

),()()(

1 }Im{Im

We want the normal velocity to be zero, so this is what we use to get the q’s

We write

1xN result matrix (known)

So, velocity normal to the mth panel is

∑=

∞ =⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−N

n

nmnm

CqWdsdz

1

),()()(

1 }Im{Im

Or 1xN matrix of strengths (unknown)

NxN matrix of ceoffs (known)= x

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

1

2

)(1

)(

11)()(

)()()(

)(1 log

21 mnN

nn

bm

c

na

mc

en

m

dsdz

dzds

zzzzq

dsdzW ∑

=∞ ⎟⎟

⎠

⎞⎜⎜⎝

⎛

−−

+=π

),( nmC

Velocity parallel to the mth panel is ∑

=∞ +⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=N

n

nmnm

CqWdsdz

1

),()()(

1 }Re{Re

Summary

• Define coordinates of start and end of panels za and zb

• Compute the panel slopes

• Put the control points next to the panel centers

• Determine the component of W∞ normal to each panel

• Determine the influence coefficients

• Solve the matrix problem, i.e. matrix divide by

• Compute the flow velocities and pressures

ab

abi

zzzze

dsdz

−−

== β

)()( , nb

na zz

( ) )(21

abbac zzizzz −−+= ε

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

− ∞Wdsdz m)(

1Im}Im{ ),( nmC

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

− ∞Wdsdz m)(

1Im

)(1

)(

1)()(

)()(),( log

21 mn

nb

mc

na

mc

enm

dsdz

dzds

zzzzC ⎟⎟

⎠

⎞⎜⎜⎝

⎛

−−

=π

Matlab Code

)(1

)()( ,, nnb

na dsdzzz

)(ncz

),( nmC

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

− ∞Wdsdz m)(

1Im

Matrix div.

ConstantSourcePanel.m

∑=

∞ +⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧ N

n

nmnm

CqWdsdz

1

),()()(

1 }Re{Re

Result matrix

Velocities along body surface

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

2)5(101−

+=∞ zW

Matlab Code Ideas:2. More than one body

E.g. Suppose we have two circles

Change z vector to contain points on both circles. Change ‘a’ and ‘b’vectors to make sure they really point to the start and end of each panel

Update plotting

Matlab Code Ideas:3. Use more sophisticated panels

E.g. Panels with linearly varying strength

Only the influence coefficient equation changes

3. Linear Source Panel MethodE.g. Panels with linearly varying strength

Only the influence coefficient equation changes

LinearSourcePanel.m

q’s are now source panel strengths at points z(b)

3. Linear Source Panels

a

bqa

Constant strength panels

Linear strength panels

Influence of qbdepends on panel ab and panel bc

Influence of qadepends only

panel ab

a

bqa

qb

qc

c

3. Linear Source Panel MethodE.g. Panels with linearly varying strength

Only the influence coefficient equation changes

LinearSourcePanel.m

q’s are now source panel strengths at points z(b)

Matlab Code Ideas:4.Change to vortex panel method

)(2)(

1zzqzW−

=π

)(2)(

1zzizW−Γ−

=π

insert ‘-i*’

Specify a circulation

Change last column of cm so that it sums total panel strength.Set last value of res equal to the circulation

Matlab Code Ideas:4. Vortex panel method

insert ‘-i*’

Specify a circulation

Change last column of cm so that it sums total panel strength.Set last value of res equal to the circulation

LinearVortexPanel.m(see also

ConstantVortexPanel.m)

q’s are now vortex panel strength (circulation/unit length)

Matlab Code Ideas:5. Set a Kutta Condition

Circulation given implicitly by Kuttacondition

Kutta condition requires that surface vorticity at trailing edge is zero.

Change last column of cm so that to be 1 at the index corresponding to the trailing edge.Set last value of res equal to zero

Matlab Code Ideas:5. Kutta Condition Code

k is index of Kutta condition point (e.g‘trailing edge’)

Kutta condition requires that surface vorticity at trailing edge is zero.

Change last column of cm so that to be 1 at the index corresponding to the trailing edge.Set last value of res equal to zero

Kutta condition is applied to the end point of the kth

panel, i.e. at z(b(k))

LinearVortexPanelKutta.m

Things to watch out for…

• Always use an odd number of panels when you have a symmetric body

• The control-point equation

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).

( ) )(21

abbac zzizzz −−+= εCenter point Displacement Δ