1 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 3 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) 4 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 5 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 7 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) 8 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. 10 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) 11 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) 12 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) 13 Pressure Coefficient From the Bernoulli Equation we get: 2 V C p = 1− (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! 14 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. 15 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 ∞ ∞ 16 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 17 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 19 ∂ 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.