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Max-torque speed and value of 8-pole motor

Short answer

Maximum torque occurs at slip 0.2, corresponding to 600 rpm. The maximum torque is 130 kg-m, approximately 1275.3 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) ohm 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, rotor impedance at standstill =0.8+j4 Ω= 0.8 + j4\ \Omega/phase. Stator impedance neglected.

Speed at maximum torque

Ns=120×508=750 rpmsm=R2X2=0.84=0.2Nm=(1−0.2)×750=600 rpm\begin{aligned} N_s &= \frac{120 \times 50}{8} = 750\ \text{rpm} \\ s_m &= \frac{R_2}{X_2} = \frac{0.8}{4} = 0.2 \\ N_m &= (1 - 0.2) \times 750 = 600\ \text{rpm} \end{aligned}

Maximum torque

TmaxTst=R22+X222R2X2=0.82+422×0.8×4=16.646.4=2.6Tmax=2.6×50=130 kg-m\begin{aligned} \frac{T_{max}}{T_{st}} &= \frac{R_2^2 + X_2^2}{2R_2X_2} = \frac{0.8^2 + 4^2}{2 \times 0.8 \times 4} = \frac{16.64}{6.4} = 2.6 \\ T_{max} &= 2.6 \times 50 = 130\ \text{kg-m} \end{aligned}

In SI units (g=9.81g = 9.81 m/s²): Tmax=130×9.81=1275.3T_{max} = 130 \times 9.81 = 1275.3 N·m.

Answer: Maximum torque occurs at 600 rpm (slip 0.2); Tmax=130T_{max} = 130 kg-m ≈1275.3\approx 1275.3 N·m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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