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4-pole motor: speed at max torque (50 N-m, j2)

Short answer

Synchronous speed 1500 rpm. Slip at max torque = 0.8/2 = 0.4, speed = 900 rpm. Maximum torque = 50 / 0.6897 = 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(0.8 + j2)\ \Omega per phase at stand still. At what speed will the motor develop maximum torque and calculate magnitude of the maximum torque.

Given: 4 poles, 50 Hz, Tst=50T_{st} = 50 N-m, rotor impedance 0.8+j2 Ω0.8 + j2\ \Omega per phase at standstill, so R2=0.8 ΩR_2 = 0.8\ \Omega, X2=2 ΩX_2 = 2\ \Omega.

Synchronous speed

Ns=120fP=120×504=1500 rpmN_s = \frac{120 f}{P} = \frac{120\times 50}{4} = 1500\ \text{rpm}

Slip and speed at maximum torque

sm=R2X2=0.82=0.4s_m = \frac{R_2}{X_2} = \frac{0.8}{2} = 0.4 Nm=Ns(1−sm)=1500×(1−0.4)=900 rpmN_m = N_s(1-s_m) = 1500\times(1 - 0.4) = 900\ \text{rpm}

Maximum torque

The torque at slip ss compared to maximum torque is TTmax=2ssms2+sm2\frac{T}{T_{max}} = \frac{2 s s_m}{s^2 + s_m^2}. At starting s=1s = 1:

TstTmax=2sm1+sm2=2×0.41+0.42=0.6897\frac{T_{st}}{T_{max}} = \frac{2 s_m}{1 + s_m^2} = \frac{2\times 0.4}{1 + 0.4^2} = 0.6897 Tmax=Tst0.6897=500.6897=72.5 N-mT_{max} = \frac{T_{st}}{0.6897} = \frac{50}{0.6897} = 72.5\ \text{N-m}

Answer: maximum torque occurs at 900 rpm (slip 0.4), and Tmax=72.5T_{max} = 72.5 N-m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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