Which ratio gives Tmax/Tst?

(R2^2 + X2^2)/(2 R2 X2) = 2.125

2076 Baisakh · 8 marks

Speed and value of maximum torque of 2-pole motor

Short answer

At slip 0.25, corresponding to 2250 rpm, the motor develops maximum torque. Maximum torque is 2.125 times starting torque, giving 21.25 N·m.

A 3 phase, 400 V, 50 Hz, 2 pole induction motor has rotor circuit resistance of 2 Ω and rotor circuit reactance of 8 Ω at stand still. It develops a starting torque of 10 N-m. At which speed, the motor develops maximum torque and calculate the value of maximum torque.

Given: 2 poles, 50 Hz, $R_2 = 2\ \Omega$ , $X_2 = 8\ \Omega$ (standstill), $T_{st} = 10$ N·m. Stator impedance neglected.

Speed at maximum torque

\begin{aligned} N_s &= \frac{120 f}{P} = \frac{120 \times 50}{2} = 3000\ \text{rpm} \\ s_m &= \frac{R_2}{X_2} = \frac{2}{8} = 0.25 \\ N_m &= (1 - s_m)N_s = 0.75 \times 3000 = 2250\ \text{rpm} \end{aligned}

Maximum torque

T_{st} = \frac{K E_2^2 R_2}{R_2^2 + X_2^2}, \qquad T_{max} = \frac{K E_2^2}{2X_2}

\begin{aligned} \frac{T_{max}}{T_{st}} &= \frac{R_2^2 + X_2^2}{2R_2X_2} = \frac{2^2 + 8^2}{2 \times 2 \times 8} = \frac{68}{32} = 2.125 \\ T_{max} &= 2.125 \times 10 = 21.25\ \text{N·m} \end{aligned}

Answer: The motor develops maximum torque at 2250 rpm (slip 0.25), and $T_{max} = 21.25$ N·m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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