A 300 HP motor is running at 1800 RPM at full load. The shaft torque is approximately what value?

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Multiple Choice

A 300 HP motor is running at 1800 RPM at full load. The shaft torque is approximately what value?

Explanation:
Torque and horsepower are linked by the formula HP = (Torque × RPM) / 5252, with torque in foot-pounds and speed in RPM. To find torque, rearrange to Torque = HP × 5252 / RPM. Plugging in 300 HP and 1800 RPM gives Torque ≈ 300 × 5252 / 1800 ≈ 875 ft-lb. So the shaft torque is about 875 ft-lb. Using 5250 instead of 5252 would yield 875 exactly, which is why this value shows up as the quick approximation. The main idea is that for a fixed horsepower, increasing speed lowers torque proportionally.

Torque and horsepower are linked by the formula HP = (Torque × RPM) / 5252, with torque in foot-pounds and speed in RPM. To find torque, rearrange to Torque = HP × 5252 / RPM. Plugging in 300 HP and 1800 RPM gives Torque ≈ 300 × 5252 / 1800 ≈ 875 ft-lb. So the shaft torque is about 875 ft-lb. Using 5250 instead of 5252 would yield 875 exactly, which is why this value shows up as the quick approximation. The main idea is that for a fixed horsepower, increasing speed lowers torque proportionally.

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