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3 years ago

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A truck is subject to a drag force, Fd, from the surrounding air: where Cd = 0.28 is the drag coefficient, A = 30 ft2 is the are

a of the front of the vehicle, and r = 0.075 lb/ft3 is the density of the air. How much power (in hp) is required to overcome drag if the truck is traveling at v=65 miles/hr? If the truck maintains a constant velocity, how much work (Btu) does the truck require for a 1-hour trip?

2 answers:

Leviafan [203]3 years ago

7 0

Answer:(a) 272897.4W (273Kw)

(b) 982.43MJ

Explanation:

See explanation in attachment.

denis-greek [22]3 years ago

6 0

Answer:

Power is 15.41 hp

Work is 39205.18 Btu

Explanation:

The power to overcome the drag force is given by the formula:

P = (1/2)ρ v³ A Cd

where,

P = Power

ρ = Density of air = 0.075 lbm/ft³

v = speed of truck = (65 miles/hr)(1 hr/3600 s)(5280 ft/1 mile) = 95.33 ft/s

A = Area = 30 ft²

Cd = drag coefficient = 0.28

therefore,

P = (1/2)(0.075 lbm/ft³)(95.33 ft/s)³(30 ft²)(0.28)

P = (272926 ft².lbm/s³)(1 lbf/32.2 lbm.ft/s²)

P = (8476 ft. lbf/s)(1 hp/550 ft.lb/s)

<u>P = 15.41 hp</u>

Now, for the work of 1 hour:

Work = W = P x time

W = (15.41 hp)(2544 Btu/h / 1 hp)(1 h)

<u>W = 39205.18 Btu</u>

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