topic 18 - panelmethods - studentsdtmook/aoe5104_online/class notes/18_class... · kutta condition...
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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