Linearized Lifting Surface Theory Thin-Wing Theory
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1 Marine Hdrodnamics Lecture 23 Copright c 2001 MIT - Department of Ocean Engineering, All rights reserved Marine Hdrodnamics Lecture 23 Linearized Lifting Surface Theor Thin-Wing Theor =() u U - /2 /2 =() L Assume thin wing, small angle of attack: U l, L l << 1 d U, d L << 1 Since wing is slender, mean sream is onl slightl perturbed, total potential Φ = U + φ. Here φ is perturbation potential. Linearize equations for small φ (sa φ << U) 1
2 gov. eqt. 2 φ = 0 (1) k.b.c. on foil φ n = Un ( Φ = 0 = U n n + φ n ) (2) φ 0 r (3) φ < at T.E. Kutta Jonkowski (4) (streamlines leave T.E. smoothl) linearize (2) : φ = (u, v) v u - U φ = u also v = (u U) d U v U = u d U d U } U {{ } 0 on = U on = U v(0) v( U ) = v(0) + U + }{{} 0 finall : v(, 0 + ) = U d U on = 0 + Linearized problem: similarl : v(, 0 ) = U d L on = 0 2
3 }{{} φ < T.E φ < T.E φ 0 v 2 φ = 0 2 φ = 0 φ du = U φ du = U φ dl = U φ dl = U φ 0 v t() General solution procedure: decompose problem into lifting and thickness problems then add (linear super pose). Camber line t() t() η() Camber line t() η() Define then half thickness : camber : t() = 1 2 ( U() L ()) η() = 1 2 ( U() + L ()) t() 3 U
4 I. thickness problem(η = 0) U () = η() + t() L () = η() t() t() U v(, 0 ± ) = φ (, 0 ±) = U II. lifting problem (t = 0)(camber including angle of attack) η() U v(, 0 ± ) = φ (, 0 ±) = U dη Solution to thickness problem: v T OT = v I + v II = U U dη = U d (η ± t) = U d ( U L ) on = 0 ± 4 v = U v = U
5 ( ) v = U v = U v = U v = U distribute sources/sinks along -ais: 2 φ = 0 0 φ r v(, = 0 ± ) = U jump in v. (,). (,) m(ξ) m(ξ) ξ ξ dφ(, ) = m(ξ)dξ 1 4π log(( ξ)2 + 2 ) φ(, ) = l/2 l/2 m(ξ) 1 4π log(( ξ)2 + 2 )dξ 5
6 satisfing,... ± 1 2 m() = v(, 0 ±) ± m m m() = 2U () = U d ( U() L ()) thin wing theor { () { +m in front = m in rear m() t + δ t V EL U thin wing theor t m() δ t + δ U U δ U(t + m()δ δ) + 2 m() = 2U () 6 dη v = U = Ut φ 0 v dη v = U
7 Solution to lifting problem: φ(, ) = U 2π l/2 l/2 (ξ) dξ log[( ξ) ]dξ v = U dη 2 φ = 0 0 φ r v(, = 0 ± ) = U dη distribute (2D) vortices along -ais: no jump in v (in general, jump in u : u + u = p + p = lift). (,) γ(ξ) ξ 7
8 Two kinds of problems to solve: dφ(, ) = γ(ξ)dξ 1 2π tan 1 ( ξ ) φ(, ) = [ φ = l/2 =0 l/2 l/2 γ(ξ) 1 l/2 (i) analsis problem given dη, find γ Γ lift... γ(ξ) 1 2π tan 1 ( ξ )dξ ] 2π tan 1 ( ξ )dξ =0 = U dη (ii) design problem given load distribution γ() ( Γ lift), find camber slope dη Lift due to circulation: v(, = 0 ± ) = F (γ()) = U dη (i) analsis problem? (ii) design problem? load distribution mean camber slope U stead flow: p p = 1 2 ρ( v 2 U 2 ), v = (u U, v) earized: for stead flow: p p = 1 2 ρ( v 2 U 2 ), v = (u U, v) linearized: p p = 1 2 ρ((u U)2 + v 2 U 2 ) = 1 2 ρ(u2 + v 2 2uU) p p = ρuu }{{} p p = 1 small 2 ρ((u U)2 + v 2 U 2 ) = 1 2 ρ(u2 }{{ + v } 2 2uU) small p p = ρuu total lift L = (p p )n ds = l/2 total lift l/2 [(p(, 0 ) p ) 0 ) = L = (p p )n ds = l/2 [(p(, 0 l/2 ) p ) (p(, 0 + ) p )] l/2 l/2 (p(, 0 ) p(, 0 + )) = ρu = l/2 (p(, 0 l/2 ) p(, 0 + )) l/2 = ρu l/2 (u(, l/2 0(u(, ) 0 l/2 u(, ) 0 u(, + )) 0 + )) 5 8
9 γ() γ() γ() u(, 0 ± ) = γ() (recall v ± for m()) 2 l/2 L = ρu γ() = ρuγ }}{{{{}} l/2 dl = ρuγ Γ dγ = γ dl = ρuγ dl = ρuγ dγ = γ dγ = γ recall Kutta-Joukowski law(for nonlinear foil) L = ρu Γ. Moment: (with respect to mid-chord) M cp L M = L cp M M cp cp L L M = L cp M = L cp 9
10 M = δl = ρuγ()δ l/2 l/2 ρu γ() C M = M 1 ρu 2 2 l 2 center of pressure cp = M L, M = L cp Lifting design problem: given the loading/circulation distribution γ() required, calculate the mean camber η(). Eample: uniform loading γ() = γ 0 η() = η(0) + γ 0 2πU [( l 2 + ) log(1 2 + l ) + ( l 2 ) log(1 2 l )] mean camber slope dη() is infinite at = ±1/2 10
11 at LE, = l OK (+ thickness) 2 at T E, = l 2 violate Kutta condition! separation at TE need γ 0 at TE to satisf Kutta condition. Remed NACA a series foils (for eample) 11
12 Lifting analsis problem: given the mean camber line η(), calculate the loading/circulation γ() it would produce (solution must satisf γ T E = 0 to meet Kutta condition). with γ(), can obtain Two simple closed-form solutions: Γ = l/2 γ(ξ)dξ ; L = ρuγ ; C l/2 L = L 1 M = ρu l/2 ξγ(ξ)dξ ; C l/2 M = M 1 (1) flat plate at angle of attack α, i.e., dη = α = const. 2 ρu 2 l ; 2 ρu 2 l 2 12
13 solution: γ() = 2αU ( l + )/( l ) 2 2 blows up at LE OK 0 at TE Kutta condition C L = 2πα (eact nonlinear hdrofoil: C L = 2π sin α) C M = 1πα center of pressure 2 cp = l at quarter chord: 4 (2) parabolic camber: η = η 0 [1 ( 2 l (sa α = 0 for now) 2 ] dη = 8η 0 l 2 solution: γ() = 16Uη 0 ( l l 2 2 )2 2 0 at both = ±l/2 due to fore-aft smmetr C L = 4π η 0 η 0 camber ratio l l C M = 0 Linear superposition: E. (1) + (2) : dη = α 8η 0 l 2 13
14 C L = 2πα + 4π η 0 l C L (α) = 2πα + C L (0) }{{} 4π η 0 l Summar - linearized lifting theor use superposition: general f oil L, U t, η t() = 1 2 ( U() + L ()) η() = 1 2 ( U() L ()) = thickness problem (t()) no lif t (can ignore in calculating L) + effect of camber onl (α = 0) lif ting problem (η()) + effect of α onl (no camber) η(le) η(t E) α = η or l η c = η α 14
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