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8-pole IM: speed at maximum torque

Short answer

Synchronous speed is 750 rpm. Slip at maximum torque s_m = R2/X2 = 0.8/2 = 0.4, so speed = 750×(1−0.4) = 450 rpm. Maximum torque = 72.5 N·m.

An 8-pole, 50 Hz, 3 phase induction motor develops a starting torque of 50 N-m. The rotor winding has an impedance of (0.8+j2) Ω per phase. At what speed the motor will develop maximum torque and calculate the magnitude of maximum torque.

The rotor impedance per phase at standstill is Z₂ = R₂ + jX₂ = 0.8 + j2 Ω (R₂ = 0.8 Ω, X₂ = 2 Ω).

For an induction motor:

  • T = K s R₂ / (R₂² + s²X₂²)
  • Starting torque (s = 1): T_st = K R₂ / (R₂² + X₂²)
  • Max torque: at s_m = R₂/X₂, T_max = K / (2X₂)

Synchronous speed

N_s = 120 f / P = 120 × 50 / 8 = 750 rpm

Slip at maximum torque

s_m = R₂ / X₂ = 0.8 / 2 = 0.4

Speed at maximum torque

N_m = N_s (1 − s_m) = 750 × (1 − 0.4) = 450 rpm

Maximum torque

Ratio:

T_max / T_st = (R₂² + X₂²) / (2 R₂ X₂) = (0.64 + 4) / (2 × 0.8 × 2) = 4.64 / 3.2 = 1.45

T_max = 1.45 × 50 = 72.5 N·m

QuantityValue
N_s750 rpm
s_m0.4
N at max torque450 rpm
T_max / T_st1.45
T_max72.5 N·m

Answer: maximum torque of 72.5 N·m is developed at 450 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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