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4-pole IM: speed at maximum torque

Short answer

Maximum torque occurs at slip s_m = R₂/X₂ = 0.4, speed 900 rpm. Magnitude T_max = 50 × (0.8²+2²)/(2×2×0.8) = 72.5 N·m.

A 4-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 stand still. At what speed the motor will develop maximum torque and calculate magnitude of the maximum torque.

Given: 4-pole, 50 Hz; Z₂ = R₂ + jX₂ = 0.8 + j2 Ω/phase at standstill; starting torque T_st = 50 N·m. Stator impedance is neglected.

N_s = 120 × 50 / 4 = 1500 rpm.

Slip and speed at maximum torque

s_m = R₂ / X₂ = 0.8 / 2 = 0.4

N_m = N_s (1 − s_m) = 1500 × (1 − 0.4) = 900 rpm

Maximum torque

With the supply voltage fixed (the same E₂ in both cases), the starting and maximum torques are

T_st = k E₂² R₂ / (R₂² + X₂²), T_max = k E₂² / (2X₂), where k = 3/ω_s.

 T_max / T_st = (R2² + X2²) / (2 X2 R2)
              = (0.64 + 4) / (2 × 2 × 0.8)
              = 4.64 / 3.2 = 1.45
 T_max = 1.45 × 50 = 72.5 N·m

(Check using T/T_max = 2 s_m/(1 + s_m²) = 0.8/1.16 = 0.690 at s = 1, which gives 50/0.690 = 72.5 N·m.)

Answer: maximum torque occurs at 900 rpm (slip 0.4), and its magnitude is 72.5 N·m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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