The wing area of the Lockheed F-104 straight-wing supersonic fighter is approximately 210 ft2 . If the airplane weighs 16,000 lb and is flying in level flight at Mach 2.2 at a standard altitude of 36,000 ft, estimate the wave drag on the wings. Assume that all lift is derived from the wings and that the wings can be approximated by a thinplate airfoil (moving at supersonic speeds!).

Wave drag = 365.9 lb

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

this is true

explanation:

i looked it up

answer:

Romans

Explanation:

Hope it helps