“The drag force, F_d, imposed by surrounding air on an automobile moving with velocity V is given by F_d = C_d · A · (1/2) · ρ · V², where C_d is a constant called the drag coefficient. A is the projected frontal area of the vehicle, and ρ is the air density. For C_d = 0.45, A = 22½ ft², and ρ = 1.23 ounce_mass/ft³, calculate the power required (in hp) to overcome drag at a constant velocity of 60 mph.”

The drag force, $F_d$, imposed by surrounding air on an automobile moving with velocity $V$ is given by:

$$F_d=C_d\cdot A\cdot \frac{1}{2}\cdot \rho \cdot V^2$$

where:
• $C_d$ = drag coefficient
• $A$ = projected frontal area
• $\rho$ = air density
• $V$ = velocity

Given:
$C_d$ = 0.45, $A$ = 22½ ft², $\rho$ = 1.23 oz_m/ft³, and $V$ = 60 mph.
Find the power required (in horsepower) to overcome drag at this speed.

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Solution:

Step 1. Convert velocity to ft/s.

$$V=60\times \frac{5280}{3600}=\mathrm{88 ft/s}$$

Step 2. Convert air density to lb_m/ft³.

1 oz_m = (1/16) lb_m

$$\rho =\frac{1.23}{16}$$

= 0.0769 lb_m/ft³

Step 3. Compute drag force.

$$F_d=C_d\cdot A\cdot \frac{1}{2}\cdot \rho \cdot V^2$$

F_d = (0.45)(22½ ft²)(½)(0.0769 lb_m/ft³)(88 ft/s)² ÷ (32.174 lb_m·ft)/(lbf·s²) = 84.6 lbf

Step 4. Compute power.

P = F_d × V = (84.6 lbf)(88 fps) = 7445 ft·lb/s

Convert to horsepower: 1 hp = 550 ft·lb/s

$$P=\frac{7445}{550}=\mathrm{13½}$$

= 13½ hp

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Answer:

$$P=13.5$$

= 13½ hp