Power–torque–speed relation4 min read

Worked Example: Motor Torque from kW and HP — 37 kW at 1,480 rpm and 50 hp at 1,780 rpm

Full-load torque from power and speed in SI and US units, and why a 50 hp motor at 60 Hz produces less torque than a 37 kW motor at 50 Hz.

Scenario

Motor A (IEC, 50 Hz)37 kW, 4-pole, 1,480 rpm rated speed
Motor B (NEMA, 60 Hz)50 hp, 4-pole, 1,780 rpm rated speed

Step-by-step calculation

Step 1: Motor A in SI units

T (N·m) = 9550 × P (kW) / n (rpm)
9550 × 37 / 1480
T = 238.8 N·m (176.1 lb·ft)

Step 2: Motor B in US units

T (lb·ft) = 5252 × P (hp) / n (rpm)
5252 × 50 / 1780
T = 147.5 lb·ft (200.0 N·m)

Result summary

CheckRequirementActualStatus
Motor A full-load torquen/a238.8 N·m✓ PASS
Motor B full-load torquen/a200.0 N·m✓ PASS
Motor A produces 238.8 N·m and motor B 200.0 N·m. Although 50 hp (37.3 kW) is about the same power as 37 kW, motor B runs 20% faster at 60 Hz, so it develops less torque.

Key insight: Torque, not power, decides whether a motor can drive a load at a given speed. When a 60 Hz motor is replaced by a 50 Hz one of the same kW (or the reverse), check the torque and the driven machine's speed, not just the power rating.

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Frequently asked questions

Should I use synchronous speed or rated speed?

Use the rated (nameplate) speed for full-load torque. Synchronous speed (1,500 or 1,800 rpm for 4 poles) gives a slightly lower, non-physical value because the motor never reaches it under load.

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