Answer:
Wave drag = 365.9 lb
Step-by-step explanation:
We are given;
Wing Area = 210 ft²
Weight = 1600 lb
Mach number = 2.2
Altitude = 36,000 ft
At this altitude, from the table i attached and by interpolation, the
Density (ρ) = 7.1 x 10^(-4) slug/ft³
And temperature; T = 390.5°R
Speed of sound at this altitude is given by;
c = √kRT
where k and R are constants with values of;
k = 1.4
R = 1716 (ft lb/slug.°R)
Thus, c = √1.4 x 1716 x 390.5
c = √938137.2
c = 969 ft/s
Speed of aircraft = Mc
Where M is mach number.
Thus,
V = 2.2 x 969 = 2131.8 ft/s
Now, dynamic pressure is given as;
Q=(1/2)ρV²
Q = (1/2)•7.1 x 10^(-4)•2131.8²
Q = 1613.32 lb/ft²
Also,
C_d = weight/((1/2)•ρ•V²•S)
Where, weight = 16,000 lb
S = 210 ft²
C_d = (16,000 x 2)/(7.1 x 10^(-4)•2131.8²•210)
C_d = 0.047
From Prandtl-Glauert equation;
C_d =4α/√(M² -1)
Thus,
0.047 = 4α/√(2.2² -1)
0.047² = 16α²/1.96
Solving for α gives; α = 0.023 rad
Now, C_p = =4α²/√(M² -1)
C_p = (4 x 0.023²)/√(2.2² -1)
C_p = 0.00108
Now, D_w = Q•S•C_p = 1613.32 x 210 x 0.00108 = 365.9 lb