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Shaft Torque, Horsepower, And Driver Efficiency

A 4-in (10.2-cm) diameter shaft is driven at 3600 r/min by a 400-hp (298.3-kW) motor. The shaft drives a 48-in (121.9-cm) diameter chain sprocket having an output efficiency of 85 percent. Determine the torque in the shaft, the output force on the sprocket, and the power delivered by the sprocket.

Calculation Procedure:

1. Compute the torque developed in the shaft

For any shaft driven by any driver, the torque developed is T lb·in = 63,000 hp/R, where hp = horsepower delivered to, or by, the shaft; R = shaft rotative speed, r/min. Thus, the torque developed by this shaft is T = (63,000)(400)/3600 = 7000 lb·in (790.9 N·m).

2. Compute the sprocket output force

The force developed at the output surface, tooth, or other part of a rotating member is given by F = T/r, where F = force developed, lb; r = radius arm of the force, in. In this drive the radius is 48/2 = 24 in (61 cm). Hence, F = 7000/24 = 291 lb (1294.4 N).

3. Compute the power delivered by the sprocket

The work input to this shaft is 400 hp (298.3 kW). But the work output is less than the input because the efficiency is less than 100 percent. Since efficiency = work output, hp/work input, hp, the work output, hp = (work input, hp)(efficiency), or output hp = (400)(0.85) = 340 hp (253.5 kW).

Related Calculations: Use this procedure for any shaft driven by any driver--electric motor, steam turbine, internal-combustion engine, gas turbine, belt, chain, sprocket, etc. When computing the radius of toothed or geared members, use the pitch-circle or pitch-line radius.

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