440 V IM with 50 kW rotor input

Given: 440 V, 50 Hz, 3-phase, 6-pole; rotor input P₂ = 50 kW; rotor emf frequency = 120 cycles/min = 2 Hz; friction and windage loss = 2 kW.

Synchronous speed: N_s = 120f/P = 120 × 50 / 6 = 1000 rpm (ω_s = 104.72 rad/s).

(i) Slip

Rotor frequency f₂ = s·f, with f₂ = 120/60 = 2 Hz:

s = f₂ / f = 2 / 50 = 0.04 (4 %)

(ii) Rotor speed

N = N_s(1 − s) = 1000 × (1 − 0.04) = 960 rpm (ω = 100.53 rad/s)

(iii) Rotor copper loss

P_cu = s × P₂ = 0.04 × 50 kW = 2 kW

(iv) Mechanical power developed

P_m = P₂ − P_cu = (1 − s)P₂ = 50 − 2 = 48 kW

(v) Output power

P_out = P_m − friction and windage = 48 − 2 = 46 kW

(vi) Output torque

T_out = P_out / ω = 46 000 / (2π × 960/60) = 46 000 / 100.53 = 457.6 N·m

(For reference, the gross torque is P₂/ω_s = 50 000/104.72 = 477.5 N·m.)