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8-pole motor: speed at max torque (50 N-m, j4)

Short answer

Synchronous speed 750 rpm. Slip at maximum torque = R2/X2 = 0.2, so speed = 600 rpm. Using T_st/T_max = 2s_m/(1+s_m²) = 0.3846 gives maximum torque = 130 N-m.

A 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+j4) Ω(0.8 + j4)\ \Omega per phase. At what speed the motor will develop maximum torque and calculate the magnitude of maximum torque.

Given: 8 poles, 50 Hz, Tst=50T_{st} = 50 N-m, rotor impedance 0.8+j4 Ω0.8 + j4\ \Omega per phase at standstill, so R2=0.8 ΩR_2 = 0.8\ \Omega, X2=4 ΩX_2 = 4\ \Omega.

Synchronous speed

Ns=120fP=120×508=750 rpmN_s = \frac{120 f}{P} = \frac{120\times 50}{8} = 750\ \text{rpm}

Slip and speed at maximum torque

sm=R2X2=0.84=0.2s_m = \frac{R_2}{X_2} = \frac{0.8}{4} = 0.2 Nm=Ns(1−sm)=750×(1−0.2)=600 rpmN_m = N_s(1-s_m) = 750\times(1 - 0.2) = 600\ \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=2×0.21+0.22=0.3846\frac{T_{st}}{T_{max}} = \frac{2 s_m}{1 + s_m^2} = \frac{2\times 0.2}{1 + 0.2^2} = 0.3846 Tmax=Tst0.3846=500.3846=130.0 N-mT_{max} = \frac{T_{st}}{0.3846} = \frac{50}{0.3846} = 130.0\ \text{N-m}

Answer: maximum torque occurs at 600 rpm (slip 0.2), and Tmax=130.0T_{max} = 130.0 N-m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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