A wing of span 2.6 m is placed in a supersonic flow at an angle of attack of 0°

Question

A wing of span 2.6 m is placed in a supersonic flow at an angle of attack of 0° The cross section is a rhombic diamond with chord length c = 2 1 m and half-angle = 8®. The upper and lower surfaces of the leading half of the cross section are subjected to a constant pressure of 261 kPa. The upper and lower surfaces of the trailing half of the cross section employ suction and blowing such that the upper surface is subjected to a pressure with distribution Pu 217 (x - c/2) kPa and the lower surface is subjected to a pressure with distribution pI 355 (x - C/2) kPa The shear stress over the upper and lower surfaces of the cross section is given by w-0.31 x 02 kN/m2 What weight (W) can the wing carry in this situation and how much air resistance (R) does it face? (Answer to 4 significant figures)

Explanation / Answer

given
wing cord, c = 2.1 m
span, s = 2.6 m
theta = 8 deg

front upper and lower surfaces, pressure
Pfu = Pfl = 261 kPa

rear surfaces pressure distribution
Pru = 217(x - c/2)kPa
Prl = 355( x - c/2) kPa

shear stress
Tw = 0.31/x^(1/5) kN/m^2

hence total lift = L
Drag = D

consider the rear parts of the diamond airfoil as the front parts have same pressure above and below
dL = (Prl - Pru)*sdx
integrating
L = [355(x^2/2 - cx/2) - 217(x^2/2 - cx/2)]s
L = 69(x^2 - cx)s
for x ranging from x = c/2 to x = c
L = 69(c^2 - c^2 - c^2/4 + c^2/2)s = 69(c^2/4)s = 197.7885 N
hence maximum weight the wing can lift is 197.7885 N

air resistamnce = D
D = due to lift + due to shear
D = integrating((Prl - Pru)*s*tan(theta)dx) + integrating 2((Tw*s*dx)) from x = 0 to x = c
D = 197.7885*tan(8) + 2*0.31*2.6(2.10^0.8)/0.8 = 31.44532 N

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