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Example Problem: Diamond-Wedge Airfoil in Supersonic Flow A diamond –wedge airfoil with a half-angle ε = 10° is placed at an angle of attack α = 15° in a Mach 3 freestream. Calculate the lift and wave-drag coefficients for the airfoil. Solution

For region 2 TableA.5 M1 = 3 → ν 1 = 49.76 TableA.5 M 2 = 3.27 ν 2 = ν 1 + θ = 49.76 + 5 = 54.76 → TableA.1 M1 = 3 →

po1 = 36.73 p1

TableA.1 M 2 = 3.27 →

po 2 = 54.76 p2

For region 3 TableA.5 ν3 =ν2 +θ = 54.76 + 20 = 74.76 → M3 = 4.78 TableA.1 M3 = 4.78 →

po3 = 407.83 p3

For region 4

M1 = 3  θ −β −M → β = 440    θ = 25  TableA.2 M1n = M1 sin β = 3sin44 = 2.08   →

p p4 = 4.881, M4n = 0.5643, o4 = 0.6835 p1 po1

M4 n 0.5643 TableA.5 = = 1.733   →ν 4 = 18.69 sin(β −θ ) sin(44 − 25) p TableA.1  → o4 = 5.165 M4 = 1.733  p4 M4 =

For region 5 TableA.5 TableA.1 ν 5 = ν 4 + θ = 18.69 + 20 = 38.69   → M5 = 2.48   →

po5 = 16.56 p5

p2 p p o1 p o1 1 = 2 = (1)(36.73) = 0.6707 p1 p o 2 p o1 p1 54.76 p3 p p3 p o 3 p o 2 1   = 2 = 0.6707   (1)(54.76) = 0.09 p1 p1 p o 3 p 3 p 2  407.83  p4 = 4.881 p1 p5 p p o 5 p o 4 p o1 1 = 5 = (1)(0.6835)(36.73) = 1.516 p1 p 5 o p o 4 p 01 p1 16.56

L ' = p4l cos 25 + p5l cos5 − p2l cos5 − p3l cos 25 L ' = ( p4 − p3 )l cos 25 + ( p5 − p2 )l cos5  p5 − p2 L' L' 2 l  p4 − p3 = = cos 25 + cos5  = q∞ S γ p M 2c γ M12 c  p1 p1  1 1 2 2 l l − + − = Cl = (4.881 0.09)cos 25 (1.516 0.6707)cos5 0.823 [ ] 1.4(3)2 c c c/2 l 1 = cos10 ⇒ = where = 0.5077 so l c 2cos10 Cl = 0.418 Cl =

D' = p4l sin25 + p5l sin5 − p2l sin5 − p3l sin25 D' = ( p4 − p3 )l sin25 + ( p5 − p2 )l sin5   p5 p2  D' D' 2 l  p4 p3  Cd = = =  −  sin25 +  −  sin5 q∞S γ p M 2c γ M12 c  p1 p1  p1 p1    1 1 2 2 l 1l Cd = 4.881 0.09 sin25 (1.516 0.6707)sin5 − + − =   ( )  3c 1.4(3)2 c  Cd = 0.169