Determine the shaft power output:

A 6-pole, 50 Hz, 3-phase induction motor running on full load develops a useful torque of 180 Nm when the rotor emf frequency is 2 Hz. Determine the shaft power output. If the mechanical torque lost in friction and that for core-loss is 12 Nm, determine

(a) the input to the motor, and

(b) efficiency.

The total stator loss is 900 W.

Solution

f₂  = S f  = 2

S =  2 /50 = 0.04

N _s   = (120 × 50 )/6= 1000 RPM

N_r  = (1 - S ) N _s  = (1 - 0.04) 1000 = 960 RPM

ω _r = (2 π× 960 )/60 = 100.53 rad/s

Shaft power output = 180 × 100.53 = 18.095 kW

Mechanical power developed = (180 + 12) × 100.53 = 19.301 kW

P _m   = Rotor copper loss ( (1/S) - 1)

Rotor copper loss = P_m (S/(1-S) = 19.301 (0.04/(1-0.04) = 0.804 kW

(a) Input to the motor = 19.301 + 0.804 + 0.9 = 21.005 kW

(b) Efficiency of motor = 18.095 × 100 = 86.14%