Drag of a thin wing and optimal shape to minimize it

Drag of a thin wing and optimal shape to minimize it Alejandro Pozo December 21 st, 211

Outline 1 Statement of the problem 2 Inviscid compressible flows 3 Drag for supersonic case 4 Example of optimal shape

Statement of the problem A thin wing is placed at a small angle of incidence α in a steady supersonic stream, so that the wing is given by: y = εf ±(x) αx, for < x < l where l is the length of the wing and α = O(ε).

Statement of the problem A thin wing is placed at a small angle of incidence α in a steady supersonic stream, so that the wing is given by: y = εf ±(x) αx, for < x < l where l is the length of the wing and α = O(ε). Our aim is to: 1 Show that the drag can be approximated by D = ρu2 [ (εf +(x) α ) 2 + ( εf (x) α ) 2 ] dx 2 Confirm that the drag on a flat plate of length l at small angle of incidence α is (2ρ U 2 /)α 2 l. 3 Obtain the optimal shape for a kite-form cross-section wing with given length and thickness.

Statement of the problem A thin wing is placed at a small angle of incidence α in a steady supersonic stream, so that the wing is given by: y = εf ±(x) αx, for < x < l where l is the length of the wing and α = O(ε). Our aim is to: 1 Show that the drag can be approximated by D = ρu2 [ (εf +(x) α ) 2 + ( εf (x) α ) 2 ] dx 2 Confirm that the drag on a flat plate of length l at small angle of incidence α is (2ρ U 2 /)α 2 l. 3 Obtain the optimal shape for a kite-form cross-section wing with given length and thickness. H. Ockendon, J. R. Ockendon, Waves and Compressible Flow, Text in Applied Mathematics, vol. 47, Springer, 24

Equations of the model We will consider that the wing is placed in an inviscid compressible flow of a perfect gas. Therefore, we will start from the following equations: Continuity equation: Euler s equation: Energy equation: Perfect gas law: ρ t + (ρu) = u t + (u )u = 1 ρ p ρ de dt = p dρ dq + (k T ) + ρ ρ dt dt p = ρrt (Cont.) (Euler) (Energy) (Gas) esides the necessary initial state, we could also have some boundary condition (e.g., thing wing): f (x, t) = = df dt = f t + u f =

Acoustics Let us take pressure, density and temperature as perturbations: p = p + p ρ = ρ + ρ T = T + T Assuming that the barred quantities are small and neglecting squares: (Cont.) ρ + ρ u + ( ρu) = t (Euler) (ρ + ρ) u + (ρ + ρ)(u )u + p = t Suppose gas is in a state of uniform motion along the x axis with some small disturbance, i.e. u = Ui + ū. Then: ( t + U ) ( ρ + ρ ū = t + U ) ū + 1 p = ρ Note that c p c v = R and denote γ = cp c v. Assuming also that k (no heat conduction): (Energy) + (Gas) p ρ γ = p ρ γ p = γp ρ }{{} c 2 ρ

Acoustics Therefore, using the three simplified equations: 2 ϕ = 1 ( c 2 t + U ) 2 ϕ, for ϕ = p, ρ, ū In particular, for steady flows we have: 2 ϕ = U2 c 2 2 ϕ (Wave) We denote M = U/c, which is called Mach number. Depending on the value of M, we distinguish two main cases: If M > 1, the equation is hyperbolic and the flow is said to be supersonic. If M < 1, the equation is elliptic and the flow is said to be subsonic.

Compressible flow past a thin wing Now, let us consider the perturbation of u of the type ū = ε φ. Then, φ also satisfies (Wave) and from (Euler) we have: p = p ερ U φ Let us apply the flow to a two-dimensional thin wing, which is nearly aligned with the flow. The wing is given by: y = εf ±(x) αx, for < x < l Therefore, the boundary condition is: f ±(x) = φ/ y U + ε φ / + α ε, on y = εf±(x) αx And, if we suppose α = O(ε) (i.e. α Cε), we have the linear approximation: Uf ±(x) = φ + UC, on y = ±, for < x < l (C) y

Supersonic case (M > 1) When M > 1, the solution to is of the form: M 2 2 φ = 2 φ φ = F (x y) + G(x + y) where 2 = M 2 1. Assuming that there will be no upstream influence due to the wing, we impose: φ = φ x= = x= Therefore: { φ = φ+ = F (x y), y > φ = φ = G(x + y), y < and applying boundary condition (C) on the wing: { F (x) + UC = Uf +(x) G (x) + UC = Uf (x), for < x < l

Supersonic case (M > 1) So the solution to that is: φ + = U UC f+(x y) + x, for < x y < l, y > φ = U UC f (x + y) x, for < x y < l, y < So now we can easily compute the drag: ( p + dy + ) dx D = dx dy p dx [( ) (εf = p ερ U φ+ +(x) α ) ( p ερ U φ ( = εp f + (x) f (x) ) ( ) dx + ε 2 ρ U φ+ f +(x) + φ f (x) dx ( ) αερ U φ+ + φ dx ) (εf (x) α )] dx

Drag of the wing Recovering the expressions for φ, we finally obtain: D = ε 2 ρ U αερ U = ε2 ρ U 2 ε2 ρ U 2 αερu2 = ρu2 [( U f +(x) UC [( U f ) ( f +(x) U + f ) ( U + f +(x) UC [ (f +(x) ) 2 ( + f (x) ) ] 2 dx [ Cf + (x) + Cf (x) ] dx (x) UC (x) UC [( f + (x) C ) + ( f (x) C )] dx [ (εf +(x) α ) 2 + ( εf (x) α ) 2 ] dx )] dx ) ] f (x) dx Therefore, if we take ε, the drag on a flat plate of length l at a small angle of incidence α is (2ρ U 2 /)α 2 l.

Example Suppose now that α = and, for a given thickness h, let us consider a wing with a cross section given by: mx, f + = f = h(l x) l h/m, < x < h m h m < x < l Using the formula obtained before, we have that the drag of this wing is: D = ε2 ρ U 2 = 2ε2 ρ U 2 = 2ε2 ρ U 2 [ (f +(x) ) 2 + ( f (x) ) 2 ] dx [ hm m 2 hl ml h m 2 dx + h m ( ) 2 h dx] l h/m

Minimum drag For l and h fixed, the minimum drag is achieved for: ( ) 2ε 2 ρ U 2 m 2 hl = = h m ml h m = l 2 that corresponds to a diamond shaped wing.

Thanks for your attention!