Which expression gives T/Tst?

s(R2^2 + (sX2)^2)/(R2^2 + X2^2)

2075 Bhadra · 8 marks

Torque of 4-pole induction motor at 1400 rpm

Short answer

Slip at 1400 rpm is 0.0667. Using the torque ratio with starting torque, the developed torque is 5.29 N·m.

A 3 phase, 400 V, 50 Hz, 4 pole induction motor has rotor circuit resistance 2 Ω and rotor circuit reactance of 8 Ω at stand still. It develops a starting torque of 5 N-m. The stator to rotor turn ratio is unity. Calculate the torque developed by the motor when it runs at 1400 rpm.

Given: 4 poles, 50 Hz, $R_2 = 2\ \Omega$ , $X_2 = 8\ \Omega$ at standstill, $T_{st} = 5$ N·m, turns ratio 1, $N = 1400$ rpm. Stator impedance neglected.

Slip

\begin{aligned} N_s &= \frac{120 \times 50}{4} = 1500\ \text{rpm} \\ s &= \frac{1500 - 1400}{1500} = 0.0667 \end{aligned}

Torque ratio

T = \frac{K s E_2^2 R_2}{R_2^2 + (sX_2)^2}, \qquad T_{st} = \frac{K E_2^2 R_2}{R_2^2 + X_2^2}

\begin{aligned} \frac{T}{T_{st}} &= \frac{s(R_2^2 + X_2^2)}{R_2^2 + (sX_2)^2} \\ &= \frac{0.0667 \times (4 + 64)}{4 + (0.0667 \times 8)^2} \\ &= \frac{4.5333}{4 + 0.2844} = \frac{4.5333}{4.2844} = 1.0581 \end{aligned}

T = 1.0581 \times 5 = 5.29\ \text{N·m}

(The 400 V supply and unity ratio are not needed when the ratio method is used.)

Answer: Torque at 1400 rpm $\approx 5.29$ N·m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Torque of 4-pole induction motor at 1400 rpm

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