Thin Airfoil Theory Airfoil Characteristics l y α x m V ∞ 2 • Lift coeffcient: C = l / 1 ρV c l 2 ∞ • Moment coeficient: • Center of pressure: • Aerodynamic Center: 2 Thin Airfoil Theory – Setup Symbols: v=V sinα+v' Assumptions: y ∞ 1. Airfoil is thin η<< c u=V cosα+u' 2. Angles/slopes are small e.g. ∞ sinα≈ α, cosα≈ 1, slope ≈ angle η η t 3. Airfoil only slightly disturbs u η c free stream u', v' << V x ∞ α η (<0) l V Kutta condition? (cusped TE) ∞ Chord c Camber η = 1 (η +η) η =η +η c 2 u l or u c t ½ Thickness η = 1 (η −η) η =η −η t 2 u l l c t Non-penetration condition? Bernoulli? 3 Thin Airfoil Theory - Simplifications Bernoulli: Assumptions: 1. Airfoil is thin η<< c C =1−(u2 + v2)/V 2 p ∞ 2. Angles/slopes are small e.g. =1−[(V cosα+u')2 + (V sinα+ v')2]/V 2 ∞ ∞ ∞ sinα≈ α, cosα≈ 1, slope ≈ angle 2u' 2v' u'2 v'2 = − cosα− sinα− − 3. Airfoil only slightly disturbs free V V V 2 V 2 ∞ ∞ ∞ ∞ stream u', v' << V ∞ “Linearized Pressure C ≈ Coefficient” p v=V sinα+v' ∞ u=V cosα+u' y ∞ Kutta Condition: V(c,0 ) =V(c,0 ) V(c,0 ) + − + x c V(c,0 ) - 4 Thin Airfoil Theory - Simplifications Exact: v=V sinα+v' Exact: y ∞ u=V cosα+u' ∞ η dη dη v'(x,η ) η t c + t ≈α+ u u η Small disturbances dx dx V c x ∞ and angles: v'(x,η ) dη dη α η =η +η u ≈ −α+ c + t η(<0) u c t l V V dx dx η =η −η ∞ ∞ l c t v'(x,0 ) dη dη Linearized: Airfoil thin: + ≈ −α+ c + t C (x,0 ) ≈ y V dx dx p ± ∞ v'(x,0 ) dη dη Likewise for − ≈ −α+ c − t y=0 + lower surface: V dx dx ∞ x α v'(x,0 ) dη dη Or: ± ≈ −α+ c ± t V∞ y=0- u(c,0 )=u(c,0 ) + - V dx dx ∞ 5 A Source Sheet Jump in normal velocity component A Vortex Sheet Jump in tangential velocity component 6 Solving for the Flow Linearized problem: Proposed Ideal Flow Solution: y iy C (x,0 ) ≈ −2u(x,0 )/V p ± ± ∞ z v'(x,0 ) dη dη ± ≈ −α+ c ± t y=0 V dx dx y=0 + ∞ + x x α α x dx 1 1 Source sheet V y=0 V y=0 ∞ - u(c,0+)=u(c,0-) ∞ - + vortex sheet Complex velocity at z 1 q(x )−iγ(x ) Complex velocity at z 1 c q(x )−iγ(x ) 1 1 dx W'(z) = ∫ 1 1 dx due to element dx : 2π (z − x ) 1 due to whole sheet: 2π (x+iy − x ) 1 1 1 0 1 c 1 q(x )−iγ(x ) Complex velocity at W'(x,0 ) = ∫ 1 1 dx K y=0 due to sheet: ± 2π (x−x ) 1 ± 0 1 So what? v'(x,0 ) ⎧W'(x,0 )⎫ 1 c γ(x ) q(x) ± = −Im⎨ ± ⎬ = ∫ 1 dx ± 1 V V 2πV (x− x ) 2V ⎩ ⎭ ∞ ∞ ∞ 0 1 ∞ N.B. Karamcheti defines the vortex u'(x,0 ) ⎧W'(x,0 )⎫ 1 c q(x ) γ(x) ± = Re⎨ ± ⎬ = ∫ 1 dx m 7 sheet strength with the opposite sign 1 V V 2πV (x − x ) 2V ⎩ ⎭ ∞ ∞ ∞ 0 1 ∞ Solving for the Flow Linearized problem: Proposed Ideal Flow Solution: y iy C (x,0 ) ≈ −2u(x,0 )/V v'(x,0 ) 1 c γ(x ) q(x) p ± ± ∞ ± = ∫ 1 dx ± V 2πV (x− x ) 1 2V v'(x,0 ) dη dη ∞ ∞ 0 1 ∞ ± ≈ −α+ c ± t y=0 V dx dx y=0 u'(x,0 ) 1 c q(x ) γ(x) + ∞ + ± = ∫ 1 dx m V 2πV (x− x ) 1 2V ∞ ∞ 0 1 ∞ x x α α x dx 1 1 Source sheet V y=0 V y=0 ∞ - u(c,0+)=u(c,0-) ∞ - + vortex sheet Kutta u'(c,0 ) u'(c,0 ) 1 c q(x ) γ(c) 1 c q(x ) γ(c) + = − so ∫ 1 dx − = ∫ 1 dx + so γ(c) = 0 condition: V V 2πV (c− x ) 1 2V 2πV (c− x ) 1 2V ∞ ∞ ∞ 0 1 ∞ ∞ 0 1 ∞ c Non- dη dη 1 iγ(x ) q(x) −α+ c ± t = ∫ 1 dx ± 1 penetration: dx dx 2πV (x− x ) 2V ∞ 0 1 ∞ c 1 q(x ) γ(x) Pressure: C (x,0 ) = − ∫ 1 dx ± p ± 1 πV (x − x ) V ∞ 0 1 ∞ Solution Pressure Difference: order? 8 Need q(x)? General Algebraic Solution Non-penetration c dη dη 1 iγ(x ) q(x) −α+ c + t = ∫ 1 dx + 1 dx dx 2πV (x− x ) 2V c ∞ 0 1 ∞ dη 1 iγ(x ) add −α+ c (x) = ∫ 1 dx 1 dx 2πV (x− x ) c dη dη 1 iγ(x ) q(x) ∞ 0 1 −α+ c − t = ∫ 1 dx − 1 dx dx 2πV (x− x ) 2V ∞ 0 1 ∞ +Kutta γ(c) = 0 dη How to solve for γ(x) w/o specifying c ( x ) ? dx •Write γ as x/c = 1 (1+cosθ) 2 •Solve integral for each term in x /c = 1 (1+cosθ) 1 2 1 dη •Relate to for c dx 0 c α x V π 0 ∞ θ 9 General Algebraic Solution Fourier Series Solution gives: x/c = 1 (1+cosθ) 1−cosθ ∞ 2 l γ(θ)/V = −(2α− B ) + 2∑B sin(nθ) ∞ o sinθ n m ΔC O n=1 p 0 c 2 π dη α x where: B = ∫ c (θ)cos(nθ)dθ n π dx V π 0 0 ∞ θ and we have that ΔC (θ) = 2γ(θ)/V p ∞ l C = = l 1 ρV 2c 2 ∞ Substitute for γ(θ) and evaluate: C = 2πα−π(B + B ) l 0 1 c 1 0 m −1 x x C = O = ∫ xΔpdx = −∫ ΔC (x)d = − 1 ∫ ΔC (θ)(1+ cosθ)sinθdθ mO 1 ρV 2c2 1 ρV 2c2 c p c 4 p 2 ∞ 2 ∞ 0 0 π Substitute for γ(θ) and evaluate: C = − 1πα+ 1π(B + 2B + B ) mO 2 4 0 1 2 = − 1 C + 1π(B + B ) 10 4 l 4 1 2