Panel Methods - Aerodynamics - Lecture Slides, Slides of Engineering Dynamics

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Panel Methods

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.

Panel method treats the airfoil as

a series of line segments

On each panel, there is vortex sheet of strength ∆Γ = γ 0 ds 0 Where ds 0 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

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 γ 0 ds 0 on each of the panel.

Stream function due to a Counterclockwise

Vortex of Strengh Γ

r r

v

r

v

r

r

∂ π

∂ψ

∂θ

∂ψ

π

ψ

θ (^2)

0

1

ln 2

Γ =− =

= =

Γ =−

v v (^) r Circulation v rd

r =

= (^) ∫ =

0

θ (^2) θ

π θ

Γ

Γ

Stream function Vortex, continued..

• Pay attention to the signs.
• A counter- clockwise vortex is considered “positive”
• In our case, the vortex of strength γ 0 ds (^0) had been placed on a panel with location (x (^0) and y 0 ).
• Then the stream function at a point (x,y) will be

( )

( ) ( ) ( )

2 0

2 0

0 0 ln 2

r r x x y y

where

r r

ds

o

o

− = − + −

= − − π

γ ψ

Panel whose center point is (x 0 ,y 0 )

Control Point whose center point is (x,y)

Adding the freestream and vortex

effects..

u y − v x − ( r − r ) ds (^) o = C ∞ ∞ ∫ 0 0 ln 2

1 γ π

The unknowns are the vortex strength γ 0 on each panel, and the value of the

Stream function C.

Before we go to the trouble of solving for γ 0 , we ask what is the purpose..

Physical meaning of γ

0

Panel of length ds 0 on the airfoil

Its circulation = ∆Γ = γ 0 ds 0

0

0 0 0

, γ

γ

= −

= (^) ∫ • = = −

Or V

Circulation V ds ds Vds

Contour

^ 

V = Velocity of the flow just outside the boundary layer

If we know γ 0 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

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.

γ γ

Upper lower

= −

γ (^2) upper = V^2 upper γ (^2) lower = V^2 lower 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.

Summing up..

• We need to solve the integral equation derived

earlier

• And, satisfy Kutta condition.

u y − v x −^ (^ r − r ) ds^ o = C ∞ ∞ ∫ 0 0 ln 2

1 γ π

γ Upper = −γ lower

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

, =^ ∫ ln^ (^ − 0 ) 0

Influenceof Panel jonindexi =

A r r ds

where

i j i π

u y i v x i Ai j j C

j

N

∞ ∞

− − (^) ∑ (^) , γ 0 − = 1

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.

N

N

j

u yi v xi Ai j j C

0 , 1 0 ,

1

, 0 0

γ γ

γ

= −

− −∑ − =

=

∞ ∞

This linear set of equations may be solved easily, and γ 0 found. Once go is known, we can find pressure, and pressure coefficient Cp.