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8-pole motor: speed at max torque (50 Kg-m)

Short answer

Ns=750 rpm; sm=R2/X2=0.8/4=0.2; speed at Tmax=750(1-0.2)=600 rpm. Tst/Tmax=2 sm/(1+sm²)=0.3846; Tmax=50/0.3846=130 kg·m (≈1275 N·m).

An 8 pole, 50 Hz, three phase induction motor develops a starting torque of 50 Kg-m. The rotor 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 kg-m, Z2=0.8+j4 ΩZ_2 = 0.8 + j4\ \Omega per phase, so R2=0.8 ΩR_2 = 0.8\ \Omega, X2=4 ΩX_2 = 4\ \Omega.

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

Slip at maximum torque:

sm=R2X2=0.84=0.2s_m = \frac{R_2}{X_2} = \frac{0.8}{4} = 0.2

Speed at maximum torque:

Nm=Ns(1−sm)=750×0.8=600 rpmN_m = N_s(1 - s_m) = 750\times 0.8 = 600\ \text{rpm}

Maximum torque: using TstTmax=2sm1+sm2\frac{T_{st}}{T_{max}} = \frac{2 s_m}{1+s_m^2} (stator impedance neglected)

TstTmax=2×0.21+0.22=0.41.04=0.3846\frac{T_{st}}{T_{max}} = \frac{2\times 0.2}{1 + 0.2^2} = \frac{0.4}{1.04} = 0.3846 Tmax=500.3846=130 kg-m(=130×9.81=1275.3 N-m)T_{max} = \frac{50}{0.3846} = 130\ \text{kg-m} \quad (= 130\times 9.81 = 1275.3\ \text{N-m})

Answer: the motor develops maximum torque at 600 rpm, and Tmax=130T_{max} = 130 kg-m (≈1275\approx 1275 N-m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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