A 10 HP Motor turning 1800 rpm is attached to a 60:1 speed reducing gearbox. The

Question

A 10 HP Motor turning 1800 rpm is attached to a 60:1 speed reducing gearbox. The motor has an efficiency of 80%. It operates at 240V AC. A 12" diameter drum is attached to the gearbox output shaft, and a wire rope is wrapped around the drum to act as a winch. The winch is used to raise bucket of raw material in a factory.

A). What is the rotational speed of the drum in rpm?

B). What is the linear speed of the wire/bucket in fps?

C). What is the maximum weight that the winch can lift in lbs?

D). What is the work done in lifting the maximum weight to a height of 50 feet?

E). What is the current draw of the motor during this operation?

Explanation / Answer

A)

Drum RPM = 1800 / 60 = 30 RPM

In rad/s, we'll have w = 2*pi*N/60 = 2*3.14*30/60 = 3.14 rad/s

B)

Bucket speed = r*w = (12/2)*3.14 = 18.84 in/s = 1.57 ft/s

C)

10 HP = 7457 Watts

For motor, Power = 2*pi*N*T/60

7457 = 2*3.14*1800*T / 60

Torque T = 39.58 Nm = 29.19 ft-lb

Gearbox output torque = Input torque*60 = 39.58*60 = 2374.8 Nm = 1751.6 ft-lb

Torque = W*r

1751.6 = W*[(12/2)/12]

W = 3503 lb

D)

Work done = Potential Energy stored = W*h = 3503*50 = 175156 ft-lb

E)

Motor input power = Output power / efficiency

= 7457 / 0.8

= 9321.2 Watts

For motor, Power = V*I

9321.2 = 240*I

Current I = 38.83 amps

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