Source Panel Method and the Vortex Panel Method

Home , Flow velocity

Applied Aerodynamics

V Def: Mach Number (M), M = ∞ a

Ratio of flow velocity to the speed of sound Compressibility Effects

ρ ∞V∞c Def: Reynolds Number (Re), Re = µ∞

Ratio of inertial forces to viscous forces Viscous Effects

If R is a reaction force acting on a body and CR is its dimensionless coefficient:

R C = R 1 ρ V 2 S 2 ∞ ∞ then dimensional analysis (Buckingham’s PI theorem) tells us

CR = f1 (Re , M ∞ ) which implies

CL = f2 (Re , M ∞ )

C D = f3 (Re , M ∞ )

C M = f4 (Re , M ∞ ) 2

Your Fluid Mechanics Courses:

Basic Fluid Mechanics • Application of F=ma to fluids • Imcompressible Flows • Mostly Inviscid (some pipe flow if you were lucky)

Fundamental Aerodynamics • Incompressible Flow Theory (no compressibility, no viscousity) • Sources, sinks and vortices • Panel Methods (2D only)

Gas Dynamics • Mach Number effects (mostly shock/expansion theory, supersonic flows) • Possible corrections for high subsonic flow

Applied Aerodynamics • 3D Bodies • Viscous Flows • Compressibility Effects • Aerodynamic Design Issues

Inviscid 3D Wing Theory

Viscous Airfoil Codes – Xfoil from MIT

Class Plan: • Review 2D Airfoil Theory (Chapters 3&4 of Anderson) • Introduce 3D Wing Theory (Chapters 5&6) • Navier-Stokes and Boundary Layer Equations • XFOIL • Applied Aerodynamics Design Concepts (along the way)

Review of Airfoil Theory and 2D Aerodynamics

Elemental Solutions

Flows about 2D airfoils are built from four elemental solutions: uniform flow, source/sink, doublet and vortex. These elemental solutions are solutions to the governing equations of incompressible flow, Laplace’s equation.

∇ 2φ = 0 ∇ 2ϕ = 0 (3.1) which relies on the flow being irrotational

r ∇ ×V = 0 (3.2)

Equations (3.1) are solved for

N - the velocity potential R - the stream function

The velocity potential is such that

r ∂φ ∂φ V = ∇φ , i.e., u = ,v = (3.3) ∂x ∂y whereas the stream function is always orthogonal (perpendicular) to the velocity potential,

∂ϕ ∂ϕ u = ,v = − (3.4) ∂y ∂x

Convenient Coordinate Systems

Cartesian: φ = φ(x, y, z)

∂ 2φ ∂ 2φ ∂ 2φ ∇ 2φ = + + = 0 (3.5) ∂x 2 ∂y 2 ∂z 2

Cylindrical: φ = φ(r,θ , z)

2 2 2 1 ∂  ∂φ  1 ∂ φ ∂ φ ∇ φ = r  + + = 0 (3.6) r ∂r  ∂r  r 2 ∂θ 2 ∂z 2

Spherical: φ = φ(r,θ ,Φ)

2 1  ∂  2 ∂φ  ∂  ∂φ  ∂  1 ∂φ  ∇ φ = r sinθ  + sinθ  +   = 0 r 2 sinθ ∂r  ∂r  ∂θ  ∂θ  ∂Φ  sinθ ∂Φ  (3.7)

Big Advantage: Laplace’s Equation is LINEAR

∴ Superposition is possible 6

Boundary Conditions

Solution about a specific body requires that the body be a streamline, i.e., flow tangency condition.

r V ⋅ nˆ = 0 (3.8)

or ∂φ = 0 ∂n where n- surface normal, s – surface tangent ∂ϕ = 0 ∂s

Another way to say it is that the surface has to be a streamline:

dy v b = streamline equation dx u

Uniform Flow

φ = C1 x + C2 ϕ = C1 y + C3 (3.9)

∂φ ∂ϕ u = = C = V = ∂x 1 ∞ ∂y

∂φ ∂ϕ v = = − = 0 ∂y ∂x

In cylindrical coordinates

φ = V∞ r cosθ ϕ = V∞ rsinθ (3.10)

Source/Sink

c V = , V = 0 (3.11) r r θ

v Source Strength Λ ≡ & = 2πrV (3.12) l r m& Where v& is the volume flow rate (sometimes called Q) v& = ρ Λ So V = (3.13) r 2πr

Λ Λ φ = ln r ϕ = θ 2π 2π ∂φ Λ 1 ∂ϕ Λ V = = V = = (3.14) r ∂r 2πr r r ∂θ 2πr 1 ∂φ ∂ϕ V = = 0 V = − = 0 θ r ∂θ θ ∂r

Def: Circulation ': r r Γ = −∫∫ (∇ ×V )⋅ ds = −∫V ⋅ds (3.15) C

if S is an area that includes the source, Γsource / sin k = 0 9

Superposition of Source and Uniform Flow

Λ ϕ = V r sinθ + θ ∞ 2π 1 ∂ϕ Λ V = = V cosθ + (3.16) r r ∂θ ∞ 2πr ∂ϕ V = − = −V sinθ θ ∂r ∞

Stagnation Point – Set Vr = Vθ = 0

 Λ   ,π  (3.17)  2πV∞ 

The stagnation point can now be used to define the body streamline, i.e., insert the stagnation point coordinates into the stream function equation to get its constant value. That new equation defines the body streamline.

Superposition of Source, Sink and Uniform Flow Λ Λ ϕ = V r sinθ + θ − θ ∞ 2π 1 2π 2 (3.18)

Doublet Flow

The source and sink of the last section were separated by a distance l and they had identical strengths. If one imagines that their stream function is variable and their distance can be altered a new condition would result if they are moved together while holding constant the product l7. In the limit, as l 0 a doublet is formed.

K cosθ K sinθ φ = ϕ = − 2π r 2π r (3.19)

where K= l7. Streamlines are found by choosing R=constant so that we get K r = − sinθ = d sinθ 2πc (3.20)

Doublet + Uniform Flow

When a doublet is superposed together with a uniform flow we get the flow over a circular cylinder.

K sinθ ϕ = V r sinθ − ∞ 2π r (3.21)

or if we let R2 = K 2πV∞

 R2  ϕ = V∞ r sinθ 1− 2   r  (3.22)

Velocities become:

1 ∂ϕ 1  R2  Vr = = V∞ r cosθ 1− 2  r ∂θ r  r   R2  Vr = 1− 2 V∞ cosθ  r  ∂ϕ  2R2  R2   Vθ = − = −V∞ r sinθ 3 + 1− 2 V∞ sinθ  ∂r  r  r    R2  Vθ = −1+ 2 V∞ sinθ  r  (3.23-3.24)

Stagnation Points are again found by setting the velocities to zero.

 R2   R2  1− 2 V∞ cosθ = 0 1+ 2 V∞ sinθ = 0  r   r  (3.25)

Which gives (R,0) and (R,B). The stagnation streamline then becomes

 R2  ϕ = V∞ rsinθ 1− 2  = 0  r  , i.e., r=R, a circle (3.26)

Pressure Coefficient

From the Bernoulli Equation we get:

2  V  C = 1−   p   (3.27) V∞ 

Which becomes on the cylinder

C = 1− 4sin2 θ p (3.28)

From this we see that the axial and normal force coefficients become: 1 c Cn = ∫ ()C p,l − C p,u dx c 0 1 TE Ca = ∫ ()C p, f − C p,b dy (3.29) c LE

Both turn out to be zero since the pressure coefficient is symmetric. , i.e., D’Alembert’s paradox!

Vortex Flow Γ Γ φ = − θ ϕ = ln r 2π 2π (3.30)

1 ∂φ Γ V = = − θ r ∂θ 2πr ∂φ V = = 0 θ ∂r

r Circulation Γ = −∫V ⋅ds = −Vθ 2πr (3.31) C

r But we also know that Γ = −∫∫ (∇ ×V )⋅ ds (3.32)

How does this square if the flow is everywhere irrotational? Answer, it doesn’t, the flow is irrotational everywhere except at the center of the vortex. The angular velocity becomes infinite at that singularity.

Doublet + Uniform Flow + Vortex

 R2  Γ r ϕ = ()V∞ r sinθ 1− 2  + ln  r  2π R (3.33)

1 ∂ϕ  R2  Vr = = 1− 2 V∞ cosθ r ∂θ  r  (3.34) ∂ϕ  R2  Γ Vθ = − = −1+ 2 V∞ sin θ − ∂r  r  2πr

This time the stagnation points depend upon '.

Γ < 4πV R  −1  − Γ  ∞  R,sin     4πV∞ R  Γ = 4πV R R,3π ∞ ()2 2  Γ  Γ  3π  Γ > 4πV R  ±   − R2 ,  ∞  2πV  4πV  2   ∞  ∞  

2  2Γsinθ  Γ   C = 1− 4sin2 θ + +    p πRV  2πRV   ∞  ∞   Γ (3.35) Cl = RV∞

Cd = 0 Lift is directly proportional to circulation L′ = q∞ SCl = ρ∞V∞ Γ

Panel Methods

The primary technique for low speed analysis (incompressible flow) is the panel method. We will discuss two, the source panel method and the vortex panel method.

Source Panel Method

To begin consider what would happen if you stretched a point source/sink out over some curve in space

Defined by the equation

λds dφ = ln r (3.36) 2π where λ = λ(s), s being the direction along the “source sheet.” To determine the potential at some point P(x,y) induced by the source sheet, one integrates Eq. 3.36 along the sheet.

b λds φ(x, y) = ∫ ln r (3.37) a 2π 18

When combined with a uniform flow this yields the flow over an arbitrary body if λ = λ(s)is prescribed in such a way that the desired body shape becomes a streamline.

In general, it is difficult to find an analytical expression for λ = λ(s), so it is approximated either by a series of constant source “panels” or by linearly varying source “panels.”

The equation for the potential induced by the jth panel is then

λ ∆φ = j ln r ds (3.38) j 2π ∫j ij j

Which gives the contribution to N at point P from panel j. The potential for the entire flow field then comes by summing the contributions from all the panels. At the center of each panel is a control point at which the basic equation for the surface normal velocity can be found from

∂ λ n λ ∂ V = []φ ()x , y = i + j ln r ds (3.39) ni i i ∑ ∫ ()ij j ∂ni 2 j=1 2π j ∂ni j≠i

This is the normal velocity contribution at the control point from only the source sheet. However, what we want is the body to be a streamline, so we must have:

V +V = 0 (3.40) ∞n ni

This leads to the equation that can be solved for the λ j unknowns

n λi λ j + ∑ I ij +V∞ cos β i = 0 (3.41) 2 j=1 2π j≠i where I is the integral in Eq. 3.39.

The surface velocity on the body at the control point can then be found from

∂ n λ ∂ V = []φ ()x , y = j ln r ds (3.42) si i i ∑ ∫ (ij )j ∂s j=1 2π j ∂s

once the λ j are found. Lastly, the integral can be evaluated as on

page 256 of the Anderson text and C p found from the Bernoulli equation.

Unfortunately, since we are dealing only with sources and sinks the lift and drag are both zero!!! 20

Lifting Airfoils

The problem of lift in general is bound up in the idea of circulation and hence requires vortices in the panel method, which we showed earlier to be the only elemental solution to exhibit circulation. To accommodate this we use an idea similar to the source sheet, the vortex sheet

Whose potential is given by the equation

1 b φ()x, z = − ∫θγds (4.1) 2π a

where γ = γ (s) .

Circulation comes about from the vortices and is related directly to the distribution of γ , via:

b Γ = ∫γds (4.2) a

From this point the entire thin airfoil theory can be developed (which should be reviewed) or one can follow the same approach 21

used to develop the source panel method using the vortex sheet equation to produce a vortex panel method for arbitrary lifting bodies. However, one thing missing is the specification of stagnation points.

Which requires the Kutta Condition.

Kutta Condition

The basic idea is related to the fact that an infinite number of stagnation point locations can exist on a lifting airfoil if only the earlier theory is applied, and because of that, virtually any circulation (lift) is possible. However, Wilhelm Kutta observed that the flow leaves a pointed trailing edge airfoil smoothly from the trailing edge, placing the stagnation point at the trailing edge. This is the so-called Kutta Condition and it provides the additional information we need to determine the circulation and hence lift of an arbitrary airfoil.

Start-Up Vortex

Another interesting, yet apparently contradictory, idea associated with airfoils comes about because of Kelvin’s Theorem, which says circulation does not change along a fixed fluid element. If it starts out zero around an airfoil at rest, it must remain zero along that fluid element.

For circulation to exist about the airfoil (and hence lift) it must coexist with an equal strength negative circulation shed at airfoil start-up and convecting downstream bounded by the original fluid element.

Vortex Panel Method

So just as in the source panel method, the vortex sheet can be approximated by a series of discrete vortex panels of constant or varying strength. Again the body must be a streamline and hence the flow tangency condition can be applied at the control points. The resulting equation for vortex strength is

n γ j ∂θ ij V∞ cos β i − ∑ ∫ ds j = 0 (4.3) j=1 2π j ∂ni This produces a system of n unknowns and n equations. We think “good” but not so, because it still doesn’t incorporate the Kutta Condition. A way to do this is to make the trailing edge panels very small and enforce

γ i = −γ i−1 (4.4) as illustrated in the Figure below.

Unfortunately, now the system is over determined. To obtain a determinate system one must eliminate one of the panels from the linear equation set. Then, once the system is solved, the lift can be determined directly from the known coefficients, i.e.,

n L′ = ρ ∞V∞ ∑γ j s j (4.5) j=1

Next up…Wing Theory!!!!