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6-pole IM: torque at 4% slip

Short answer

Torque at 4% slip: T = 200 × 2×0.04×0.15/(0.04²+0.15²) = 99.6 N·m. For starting torque 2/3 of maximum, added rotor resistance = 0.619 Ω per phase (using s_m'=0.382, R=1.019 Ω).

A 6-pole, 50 Hz, three-phase induction motor has rotor resistance of 0.4 Ω/phase, maximum torque is 200 Nm at 850 rpm. Find (i) torque at 4% slip, and (ii) additional rotor resistance to get (2/3)rd of maximum torque at starting.

Given: 6-pole, 50 Hz; R₂ = 0.4 Ω/phase; T_max = 200 N·m at 850 rpm.

N_s = 120 × 50 / 6 = 1000 rpm.

Slip at maximum torque: s_m = (1000 − 850)/1000 = 0.15.

Since s_m = R₂/X₂, standstill reactance X₂ = R₂ / s_m = 0.4 / 0.15 = 2.667 Ω/phase.

(i) Torque at 4 % slip

Neglecting stator impedance, T / T_max = 2 s s_m / (s² + s_m²)

 T/Tmax = 2 × 0.04 × 0.15 / (0.04² + 0.15²)
        = 0.0120 / (0.0016 + 0.0225)
        = 0.0120 / 0.0241 = 0.4979
 T = 0.4979 × 200 = 99.6 N·m

Torque at 4 % slip ≈ 99.6 N·m.

(ii) Extra rotor resistance for starting torque = (2/3) T_max

At starting s = 1, with total rotor resistance R and slip at maximum torque s_m' = R / X₂:

T_st / T_max = 2 s_m' / (1 + s_m'²) = 2/3

 3 s_m' = 1 + s_m'²   ->   s_m'² - 3 s_m' + 1 = 0
 s_m' = (3 ± √5)/2 = 0.382  or  2.618

Taking the practical value s_m' = 0.382 (the peak occurs at a speed within the running range, and the smaller resistance is needed):

R = s_m' X₂ = 0.382 × 2.667 = 1.019 Ω/phase

External resistance per phase = 1.019 − 0.4 = 0.619 Ω.

(The other root, 2.618, would need R = 6.98 Ω, i.e. 6.58 Ω extra. It gives the same starting torque but is not used because it wastes more power and the maximum torque would then lie in the reverse region.)

Answer: (i) 99.6 N·m; (ii) about 0.62 Ω per phase added in the rotor circuit.

✓ Answer checked by Claude Opus 5.5 (high effort)

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