Prandtl Lifting Line Theory PDF

PRANDTL LIFTING LINE THEORY A theory that models the effects of the trailing vortices by a su- perposition of a continuous distribution of circulation from “bound vortices” attached to the trailing edge (called the “lifting line”). Using ideas for superposition of a continuous distrib...

PRANDTL LIFTING LINE THEORY A theory that models the effects of the trailing vortices by a su- perposition of a continuous distribution of circulation from “bound vortices” attached to the trailing edge (called the “lifting line”). Using ideas for superposition of a continuous distribution of infinites- imal contribution, downwards velocity at position y0 is ˆ b/2 1 dΓ/dy w(y0) = − dy (5.15) 4π −b/2 y0 − y 1 Note 4π here as half of an infinite line vortex. Then   −w(y0) −w(y0) αi(y0) = tan−1 ≈ (5.16) V∞ V∞ In the text (Sec 5.3), these equations are combined with K-J theorem for cross sections of the wing into a longer equation (Eq. 5.23) which governs how the function Γ(y) may behave. We focus on getting some results for “reasonable” forms of Γ(y). 1 General formulas If the circulation distribution Γ(y) is known across the wing-span (−b/2 ≤ y ≤ b/2), then we can determine lift and induced drag: ˆ b/2 ˆ b/2 L= L′(y)dy = ρV∞Γ(y)dy −b/2 −b/2 ˆ b/2 ˆ b/2 ′ Di = L (y) sin αi(y)dy = ρV∞Γ(y)αi(y)dy −b/2 −b/2 Where the induced angle of attack αi(y) is also determined by Γ(y) via Eq 5.15 for the downwash velocity. q 1. Case of an elliptical lift distribution: Γ(y) = Γ0 1 − (y/( 2b )2) Find some simple functional form which is (i) smooth, (ii) zero at y = ±b/2 (wing tips) and (iii) has a maximum at y = 0 (root). Lift ˆ ˆ 1/2 b/2 b/2 4y 2  L = ρV∞ Γ(y)dy → = ρV∞Γ0 1− 2 dy −b/2 −b/2 b Change of variables: let y = 2b cos θ ˆ 0 −b 2 bπ L = ρV∞Γ0 sin θdθ =...... = ρV∞Γ0 π 2 4 Usually we aim at a targeted amount of lift. (Think of an airplane already in the air). Better to express Γ0 in terms of L or CL. 2 Note that L = 12 ρV∞2 CLS where S is the area of the wing planform. Setting this expression for L equal to the one above gives: 2V∞SCL Γ0 = (5.40) bπ Downwash velocity q y 2 Substitution of Γ(y) = Γ0 1 − ( b/2 ) into 5.15 (see text) gives −Γ0 w(y) = (5.35) 2b which is negative, as we expect. It is interesting that, for this “ellip- tic” case, w, and hence αi is independent of spanwise location. −w Γ0 αi = = (5.36) V∞ 2bV∞ This result in terms of lift (substitute 5.40 into 5.36) gives: CL αi = (5.42) πAR Where AR = b2/S is the wing aspect ratio. This is equivalent to AR = b/c if the wing is rectangular. For airfoils: as b → ∞, AR → ∞, αi → 0 (Di → 0) Induced Drag Recall the general formulas ˆ b/2 Di = L′(y)αi(y)dy −b/2 Induced drag coefficient: ˆ b/2 Di 2 CDi = 1 2 = ρV∞Γ(y)αi(y)dy 2 ρV∞ S ρV∞2S −b/2 3 In the elliptic case since αi turns out to be independent of y, the math becomes simple. We obtain: Γ02 bπ S CL 2 CDi = = 2 bSV∞ 4 b π Which when Γ0 is substituted via 5.40, yields CL 2 CDi = (5.43) πAR Some Remarks 1. To reduce induced drag (“cost” of producing lift) make AR and hence wingspan large. But this causes some structural and air- port operation issues. For a B747, AR ≈7. 2. For untwisted wings, α and αL=0 are independent of y. Now if αi is also independent of y, the wing section lift coeff Cl (y) = 2π(α − αi − αL=0) is also independent of y. But L′(y) = ρV∞Γ(y) = 12 ρV∞2Cl C(y). This means that the chord length distributions C(y) ∝ Γ(y). Hence the wing planform is also elliptic in shape. (Or, this is an “elliptic wing”!) 3. It will be shown later that the elliptic wing has minimum induced drag for given lift. This is nice. But of course, induced drag is not the only consideration in designing wing surfaces. 4 GENERAL (NON-ELLIPTIC) LIFT DISTRIBUTION How do we “generalize” upon the elliptical case, in some systematic yet manageable manner. When we used a change of variables y = (b/2) cos θ, the elliptic case for Γ(y) took on the simple form Γ(θ) = Γ0 sin θ. Try a sum of sine waves of different wavelengths: let ∞ X Γ(θ) = 2bV∞ An sin nθ (5.48) n=1 where A1, A2,... are to be specified. These are different from those coefficients in cambered airfoils. Lift ˆ b/2 L = ρV∞Γ(y) dy ˆ−b/2 π b = ρV∞Γ(θ)(− sin θ) dθ 0 2 ˆ π ∞ ! 1 X = ρV∞b 2bV∞ An sin nθ sin θ dθ 2 0 n=1 ∞ X  ˆ π  = ρb2V∞2 An sin nθ sin θ dθ n=1 0 = ρb2V∞2A1π/2 (use orthogonality relation) Hence L CL = 1 2 = A1 πAR (5.53) 2 ρV ∞ S Thus only A1 contributes to lift. In fact we re-define A1 to be equal to CL/(πAR). 5 Induced angle of attack ∞ X sin nθ αi(θ) = nAn (5.37) n=1 sin θ Unlike elliptic case, here αi is not constant. (More Math..) Induced drag ˆ π b Di = ρV∞Γ(θ)αi(θ) sin θdθ 0 2 ˆ π X ∞ ! ∞ ! b X = ρV∞2bV∞ Am sin mθ nAn sin nθ dθ 2 0 m=1 n=1 where the subscripts m, n must be kept distinctly. Since the integral of a sum is equal to the sum of the integrals, X∞ X ∞  ˆ π  Di = b2V∞2 nAmAn sin mθ sin nθ dθ m=1 n=1 0 The Fourier orthogonality relation tells us only those terms with m = n contribute. Hence ∞ ∞  2 X π X An π Di = ρb2V∞2 nAn2 , or ρb2V∞2A12 n n=1 2 n=1 A1 2 where we isolate A1 because of its direct connection to CL. Next, ∞  2 Di b2 π 2 X An CDi = 1 2 = 2 A 1 n 2 ρV ∞ S S 2 n=1 A1  2 CL = πAR (1 + δ) πAR where δ = ∞ An 2 P n=2 n( A1 ) ≥ 0; = 0 if elliptic. This proves the elliptic case gives minimum CDi for given CL and AR. 6  ACTUAL LIFT CURVES: Airfoils vs. Wings) Notation: let a0 and a denote lift slopes for airfoils and wings. Thin airfoil theory says a0 = 2π. In reality, since viscous effects are never completely absent, it will be slightly less (such as 6.0, 6.1). For airfoils, let’s write Cl = a0(α − αL−0); where dCl /dα = a0 For 3D wings, since α is “effectively” reduced by an amount equaling αi, we can extend the above to CL = a0(α − αi − αL−0) But — since αi depends on CL and hence on α, the slope is not a0. Lift slope for the wing Note the following, assuming untwisted wing(s): Untwisted: a single value each for α and αL=0 At condition of zero lift there is no downwash, and αi = 0. Hence both wing airfoil (as the wing cross-section) share the same αL=0. If wing is elliptic, use αi = CL/(πAR) (Eq. 5.42)   CL C L = a0 α − − αL=0 πAR h a0 i CL 1 + = a0(α − αL=0) πAR CL = a(α − αL=0) Lift slope of wing: dCL a0 a= = < a0 dα [1 + a0/(πAR)] 7  If wing is not elliptic, αi will be slightly larger than that in Eq. 5.42 used above. The denominator in expression for a above will thus be slightly larger. We can write it as dCL a0 a= = a0 dα [1 + πAR (1 + τ )] where τ ≥ 0 is a “correction factor” similar in magnitude to δ used earlier for the induced drag coefficient. In many cases both δ and τ are in the range of 5 to 10% (approximately). To summarize our studies of lift in this course, we can draw a graph of Cl and CL versus α in the space below: (And we should always be aware of the assumptions and limita- tions in the theory...) We will work out in class Example 5.2 from the textbook. (For convenience, the problem statement is reproduced on the next page.) 8 Consider a rectangular wing with an aspect ratio of 6, an induced drag factor δ = 0.055, and a zero-lift angle of attack of −2o. At an angle of attack of 3.4o, the induced drag coefficient for this wing is 0.01. Calculate the induced drag coefficient for a similar wing (a rectangular wing with the same airfoil section) at the same angle of attack, but with an aspect ratio of 10. Assume that the induced factors for drag and lift slope, δ and τ , respectively, are equal to each other (i.e. δ = τ ). Also, for AR=10, δ = 0.105. 9 VISCOUS EFFECTS IN INCOMPRESSIBLE FLOWS Some basic ideas of viscosity that have been touched upon in this course so far and in AE 2010 include: 1. Importance of viscous effects Starting vortex (which provided circulation) Flow separation (which causes stall at high angle of attack) Frictional resistance (drag, pipelines, human heart) Transfer of momentum, and of heat, between adjacent fluid element, or at a solid boundary. 2. The nature of viscous flow solutions Fundamentally different from inviscid flow The equations of motion contains viscous terms, which ac- count for momentum diffusion “No-slip” boundary conditions: no relative velocity between a boundary and the fluid touching it, tangential to the surface The no-slip BC applies as long as the viscosity is nonzero (no matter how small). It also explains the formation of a boundary layer, which is a thin layer next to a boundary or surface, where most of the velocity changes occur. To describe how the flow properties (V⃗ , P , etc) vary with position and time, we need to: (i) develop differential forms of the equations of motion, (ii) see if they can be simplified, and (iii) obtain some solutions of those simplified equations, somehow. 10 CONSERVATION LAWS IN DIFFERENTIAL FORM In the text (Secs. 2.4 and 2.5), and probably in AE 2010, the for- mulation of equations for conservation of mass and momentum is approached in the following way: (a) Select an arbitrary control volume (CV), of finite size (b) Formulate conservation law for fluid in CV, using volume and area integrals (the latter over the control surface, CS). ´´´ (c) Convert to (...) dVol = 0, then let the CV shrink to a point. If the final integrand is zero everywhere, we get a PDE. Here we use differential approach directly, which will facilitate treat- ment of viscous terms. Consider infinitesimal “box” of dimensions dx, dy, dz The CV has volume dVol = dx dy dz The CS consists of three pairs of opposing plane surfaces Rate of increase of mass of fluid inside CV (with fixed dx, dy, dz) ∂ ∂ρ [ρ dVol] = dx dy dz ∂t ∂t 11 Taylor series (from Math, very important) In the typical notation in math textbooks: 1 1 f (x + ∆x) = f (x) + ∆xf ′(x) + (∆x)2f ′′(x) + (∆x)3f ′′′(x) +.... 2 6 ′ ≈ f (x) + ∆xf (x) assuming ∆x is small and all higher-order derivatives are well-behaved. We apply this relation to partial derivatives in relating mass flow rates at the 3 pairs of opposing faces in our CS The left face has area dy dz, and velocity component ⊥ to it is u. Rate of mass flow entering at the left = ρudydz. Rate of mass flow leaving at the right = (ρu + ∂ρu ∂x dx)dydz. By subtraction, net rate of inflow counting the left and right faces is thus ∂ρu ∂ρu − dxdydz = − dVol. ∂x ∂x The other two faces will give ∂ρv ∂ρw − dydxdz and − dzdxdy ∂y ∂z Counting all 6 faces thus gives the net rate of mass inflow over the entire CS as (noting the common factor, dVol = dxdydz)   ∂ρu ∂ρv ∂ρw − + + dxdydz ∂x ∂y ∂z Hence mass conservation requires   ∂ρ ∂ρu ∂ρv ∂ρw =− + + = −∇ · (ρV⃗ ) ∂t ∂x ∂y ∂z or ∂ρ/∂t + ρ∇ · V⃗ + V⃗ · ∇ρ = 0 which reduces ∇ · V⃗ = 0 if ρ = const. 12