Drag of a thin wing and optimal shape to minimize it

Transcription

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

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

3 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).

4 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.

5 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

6 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 =

7 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

8 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.

9 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

10 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

11 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

12 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.

13 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

14 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.

15 Thanks for your attention!