Содержание

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What are panel methods? Panel methods are techniques for solving incompressible

What are panel methods?

Panel methods are techniques for solving incompressible potential

flow over thick 2-D and 3-D geometries.
In 2-D, the airfoil surface is divided into piecewise straight line segments or panels or “boundary elements” and vortex sheets of strength γ are placed on each panel.
We use vortex sheets (miniature vortices of strength γds, where ds is the length of a panel) since vortices give rise to circulation, and hence lift.
Vortex sheets mimic the boundary layer around airfoils.
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Analogy between boundary layer and vortices Upper surface boundary layer contains,

Analogy between boundary layer and vortices

Upper surface boundary layer contains, in

general, clockwise rotating vorticity
Lower surface boundary layer contains, in general, counter clockwise vorticity.
Because there is more clockwise vorticity than counter clockwise
Vorticity, there is net clockwise circulation around the airfoil.
In panel methods, we replace this boundary layer, which has a small but finite thickness with a thin sheet of vorticity placed just outside the airfoil.
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Panel method treats the airfoil as a series of line segments

Panel method treats the airfoil as a series of line segments

On each

panel, there is vortex sheet of strength ΔΓ = γ0 ds0
Where ds0 is the panel length.
Each panel is defined by its two end points (panel joints)
and by the control point, located at the panel center, where we will
Apply the boundary condition ψ= Constant=C.
The more the number of panels, the more accurate the solution,
since we are representing a continuous curve by a series
of broken straight lines
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Boundary Condition We treat the airfoil surface as a streamline. This

Boundary Condition

We treat the airfoil surface as a streamline.
This ensures that

the velocity is tangential to the airfoil surface, and no fluid can penetrate the surface.
We require that at all control points (middle points of each panel) ψ= C
The stream function is due to superposition of the effects of the free stream and the effects of the vortices γ0 ds0 on each of the panel.
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Stream Function due to freestream The free stream is given by

Stream Function due to freestream

The free stream is given by

Recall


This solution satisfies conservation of mass

And irrotationality

It also satisfies the Laplace’s equation. Check!

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Stream function due to a Counterclockwise Vortex of Strengh Γ

Stream function due to a Counterclockwise Vortex of Strengh Γ

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Stream function Vortex, continued.. Pay attention to the signs. A counter-clockwise

Stream function Vortex, continued..

Pay attention to the signs.
A counter-clockwise vortex is

considered “positive”
In our case, the vortex of strength γ0ds0 had been placed on a panel with location (x0 and y0).
Then the stream function at a point (x,y) will be

Panel whose center
point is (x0,y0)

Control Point whose center
point is (x,y)

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Superposition of All Vortices on all Panels In the panel method

Superposition of All Vortices on all Panels

In the panel method we

use here, ds0 is the length of a small segment of the airfoil, and γ0 is the vortex strength per unit length.
Then, the stream function due to all such infinitesimal vortices at the control point (located in the middle of each panel) may be written as the interval below, where the integral is done over all the vortex elements on the airfoil surface.
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Adding the freestream and vortex effects.. The unknowns are the vortex

Adding the freestream and vortex effects..

The unknowns are the vortex strength

γ0 on each panel, and the value of the
Stream function C.
Before we go to the trouble of solving for γ0, we ask what is the purpose..
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Physical meaning of γ0 Panel of length ds0 on the airfoil

Physical meaning of γ0

Panel of length ds0 on the airfoil
Its circulation

= ΔΓ = γ0 ds0

V = Velocity of the flow just outside the boundary layer

If we know γ0 on each panel, then we know the velocity of the flow
outside the boundary layer for that panel, and hence pressure over that panel.

Sides of our contour
have zero height
Bottom side has zero
Tangential velocity
Because of viscosity

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Pressure distribution and Loads Since V = -γ0

Pressure distribution and Loads

Since V = -γ0

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Kutta Condition Kutta condition states that the pressure above and below

Kutta Condition

Kutta condition states that the pressure above and below the

airfoil trailing edge must be equal, and that the flow must smoothly leave the trailing edge in the same direction at the upper and lower edge.

γ2upper = V2upper
γ2lower = V2lower
F

From this sketch above, we see that pressure will be equal, and the flow
will leave the trailing edge smoothly, only if the voritcity on each panel
is equal in magnitude above and below, but spinning in opposite
Directions relative to each other.

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Summing up.. We need to solve the integral equation derived earlier And, satisfy Kutta condition.

Summing up..

We need to solve the integral equation derived earlier
And, satisfy

Kutta condition.
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Numerical Procedure We divide the airfoil into N panels. A typical

Numerical Procedure

We divide the airfoil into N panels. A typical panel

is given the number j, where J varies from 1 to N.
On each panel, we assume that γ0 is a piecewise constant. Thus, on a panel numbered j, the unknown strength is γ0,j
We number the control points at the centers of each panel as well. Each control point is given the symbol “i”, where i varies from 1 to N.
The integral equation becomes
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Numerical procedure, continued Notice that we use two indices ‘i’ and

Numerical procedure, continued

Notice that we use two indices ‘i’ and ‘j’.

The index ‘I’ refers to the control point where equation is applied.
The index ‘j’ refers to the panel over which the line integral is evaluated.
The integrals over the individual panels depends only on the panel shape (straight line segment), its end points and the control point í’.
Therefore this integral may be computed analytically.
We refer to the resulting quantity as
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Numerical procedure, continued.. We thus have N+1 equations for the unknowns

Numerical procedure, continued..

We thus have N+1 equations for the unknowns γ0,j

(j=1…N) and C.
We assume that the first panel (j=1) and last panel (j=N) are on the lower and upper surface trailing edges.

This linear set of equations may be solved easily, and γ0 found.
Once go is known, we can find pressure, and pressure coefficient Cp.

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Panel code Our web site contains a Matlab code I have

Panel code

Our web site contains a Matlab code I have written,

if you wish to see how to program this approach in Matlab.
See http://www.ae.gatech.edu/people/lsankar/AE3903/Panel.m
And, sample input file http://www.ae.gatech.edu/people/lsankar/AE3903/panel.data.txt
An annotated file telling you what the avrious numbers in the input means is found at
http://www.ae.gatech.edu/people/lsankar/AE3903/Panel.Code.Input.txt