Answer:
The percent of power extended against sea-level air drag is 50.15%
Explanation:
Length L is 22 ft, Power P is 590 hp,
Speed of air V = 175 mi/h. To convert to ft, we will multiply by 1.46
V = 175 × 1.46 = 255.5 ft
Frontal Area A is given as 23.5 ft
Coefficient of drag, Cd is 0.35
From text, Density of Air at sea level ρ = 0.00237 slug/ft³
The formula for calculating Drag Force F is
F = Cd × × ρAV²
Substitute the values above into the formula
F = 0.35 × × 0.00237 × 23.5 × 255.5²
= 0.35 × × 3635.78
= 636.26 lbf
But Power P = FV
Therefore we substitute for F and V to get P
P = 636.26 × 255.5
= 162564.43 ft.lbf/s
To convert to horsepower, we multiply by 0.00182 hp
= 162564.43 × 0.00182
= 295.87 hp
The percent of power extended against sea-level air drag is
= (295.87/590) × 100
= 50.15%