Aerodynamics Lecture 4: Panel methods G. Dimitriadis Introduction • Until now we’ve seen two methods for modelling wing sections in ideal flow: –Conformal mapping: Can model only a few classes of wing sections –Thin airfoil theory: Can model any wing section but ignores the thickness. • Both methods are not general. • A more general approach will be presented here. 2D Panel methods • 2D Panel methods refers to numerical methods for calculating the flow around any wing section. • They are based on the replacement of the wing section’s geometry by singularity panels, such as source panels, doublet panels and vortex panels. • The usual boundary conditions are imposed: – Impermeability – Kutta condition Panel placement • Eight panels of length Sj • For each panel, sj=0-Sj. • Eight control (collocation) points (xcj,ycj) located in the middle of each panel. • Nine boundary points (xbj,ybj) • Normal and tangential unit vector on each panel, nj, tj • Vorticity (or source strength) on each panel j(s) (or j(s)) Problem statement • Use linear panels • Use constant singularity strength on each panel, j(s)=j (j(s)=j). • Add free stream U at angle . • Apply boundary conditions: – Far field: automatically satisfied if using source or vortex panels – Impermeability: Choose Neumann or Dirichlet. • Apply Kutta condition. • Find vortex and/or source strength distribution that will satisfy Boundary Conditions and Kutta condition. Panel choice • It is best to chose small panels near the leading and trailing edge and large panels in the middle: x 1 = ()cos +1 , = 0 2 c 2 Notice that now x/c 17 29 begins at 1, passes 16 30 through 0 and then goes 2 1 back to 1. 15 3 The usual numbering 9 scheme is: lower trailing edge to lower leading edge, upper leading edge to upper trailing edge. Panel normal and tangent • Consider a source panel on an airfoil’s upper surface, near the trailing edge. y+yci i ni (xbi,ybi) (xci,yci) x+xc ti i i (xbi+1,ybi+1) • In the frame x+xci,y+yci, n is a linear function: x y n = x, n = y n n x y • So that = cos = sin , = cos n 2 i i n i NACA four digit series airfoils • The NACA 4-digit series is defined by four digits, e.g. NACA 2412, m=2%, p=40%, t=12% • The equations are: t y = 0.2969 x - 0.126x - 0.35160x 2 +0.2843x 3 - 0.1015x 4 t 0.2 () m 2 2 ()2px - x for x < p p yc = m 2 2 ()()1-2p +2px - x for x > p 1 p • Where t () is the maximum thickness as a percentage of the chord, m is the maximum camber as a percentage of the chord, p is the chordwise position of the maximum camber as a tenth of the chord. • The complete geometry is given by y=yc+yt . NACA 4-digit trailing edge • It should be stressed that the NACA 4-digit thickness equation specifies a trailing edge with a finite thickness: The equation can be Gap because of Zero thickness- finite thickness modified so no gap that yt(1)=0 Actual trailing edge Modified trailing edge NACA 4-digit with thin airfoil theory • Thin airfoil theory solutions for NACA four- digit series airfoils can be readily obtained. • The camber slope is obtained by differentiating the camber line and substituting for =cos-1(1-2x/c): dz m = ()2p 1+ cos for p dx p2 dz m = 2 ()2p 1+ cos for p dx ()1 p 1 dz A = d Substitute 0 0 dx these in: 2 dz A = cosnd n 0 dx NACA 4-digit with thin airfoil theory • To obtain: m m A 2p 1 sin 2p 1 sin 0 = 2 ()() p + p + 2 ()() () p p p ()1 p 2m 1 p A1 = 2 ()2p 1 sin p + sin2 p + p 4 2 2m 1 1 2 ()2p 1 sin p + sin2 p () p ()1 p 4 2 • Which are easily substituted into: cl = 2A0 + A1 Source panel airfoils • Consider an airfoil idealized as m linear source panels with constant strength. • The potential induced at any point (x,y) in the flowfield by the jth panel is: 2 2 j S j j ()x,y = ln x x j s j + y y j s j ds j 2 0 ()() ()() • Including the free stream and summing the contributions of all the panels, the total potential at point (x,y) is: m 2 2 j S j x,y = Uxcos + y sin + ln x x j s j + y y j s j ds j () () 0 ()() ()() j =1 2 Source panel airfoils • As the panels are linear, then xb j +1 xb j x j ()s j = s j + xb j = cos j s j + xb j S j yb j +1 yb j y j ()s j = s j + yb j = sin j s j + yb j S j • So that the total potential becomes: m j S j 2 2 x,y = Uxcos + y sin + ln x xb j cos j s j + y yb j sin j s j ds j () () 0 ()() j =1 2 • There is no obvious expression for this integral… Boundary condition • Try the Neumann boundary condition: = 0 n • This condition is applied on the control point of each panel so that: ()xci,yci x y = U cos + sin ni ni ni m j S j 2 2 + ln xci xb j cos j s j + yci yb j sin j s j ds j 0 ()() j =1 2 ni x y • Notice that: = sin i, = cosi and then ni ni m j S j 2 2 ln xci xb j cos j s j + yci yb j sin j s j ds j = U sin i 0 ()()() j =1 2 ni Differentiation • Carrying out the differentiation in the integral: 2 2 ln ()xci xb j cos j s j + ()yci yb j sin j s j = ni 2 2 xc xb cos s + yc yb sin s 1 n ()()i j j j ()i j j j i = 2 2 2 ()xci xb j cos j s j + ()yci yb j sin j s j x y 2()xci xb j cos j s j + 2()yci yb j sin j s j ni ni 2 2 = ()xci xb j cos j s j + ()yci yb j sin j s j ()xci xb j cos j s j cosi ()yci yb j sin j s j sini 2 2 ()xci xb j cos j s j + ()yci yb j sin j s j Integration • After this differentiation, it is now possible to evaluate the integral. • The boundary condition becomes: m j Cij Fij + DijGij = U sin()i j =1 2 2 Where: Aij = ()xci xb j cos j ()yci yb j sin j 2 2 Bij = ()xci xb j + ()yci yb j 2 S j + 2Aij S j Cij = sin()i j , Dij = cos()i j , F = ln 1 + Bij 1 E ij S j E ij = ()xci xb j sin j ()yci yb j cos j , Gij = tan Aij S j + Bij C F ii ii + D G = 2 ii ii System of equations • Therefore, the problem of choosing the correct source strengths to enforce impermeability has been reduced to: m j Cij Fij + DijGij = U sin()i j =1 2 2 • Or, in matrix notation, Dn=Usin(-) • Where Dn=(-CF/2+DG)/2 • Which can be solved directly for the unknown . Tangential Velocities • The velocities tangential to the panels are given by: ()xci,yci x y = U cos + sin ti ti ti m j S j 2 2 + ln xci xb j cos j s j + yci yb j sin j s j ds j 0 ()() j =1 2 ti • So that: m D F v = U cos j ij ij + C G ti ()i ij ij j =1 2 2 v 2 ti c p = 1 • As usual: i U Cartesian velocities • For plotting the velocity field around the airfoil, the cartesian velocities, u, v are needed. • These can be obtained from the normal and tangential expressions: m j Dij Fij u = U cos()i + CijGij j =1 2 2 m j Cij Fij v = U sin()i + + DijGij j =1 2 2 • for i=0. Example: • NACA 2412 airfoil at 5o angle of attack • 50 panels Full flowfield Near trailing edge Discussion • The usual problem: the Kutta condition was not enforced. The flow separates on the airfoil’s upper surface. • Additionally, the lift must be equal to zero, since there is no ciurculation in the flow. But is it? • Calculate c c dx x = p c c dy y = p Lift definition • Lift is the force perpendicular to the free stream L Fy D Fx Therefore: L=Fycos-Fxsin D=Fysin+Fxcos Or: cl=cycos-cxsin cd=cysin+cxcos Pressure distribution Stagnation point Stagnation point cl=0.1072 cd=-0.0082 The aerodynamic forces are not zero. Very high velocity at trailing edge. Hence very low pressure. Increasing number of panels • Increasing the number of panels also increases the accuracy. • The forces move very slowly towards 0. • The problem is the infinite velocity at the trailing edge. Enforcing the Kutta condition • The number of equations was equal to the number of unknowns • Therefore, the Kutta condition could not be enforced anyway, it would have been an additional equation. • More equations than unknowns means a least squares solution. • Conclusion: we need an additional equation (Kutta condition) and an additional unknown. Vortex panels • Vortex panels with exactly the same geometry as the source panels are added.