Question

The rotor impedance of a three-phase, 50 Hz, 1460 rpm, star-connected ring ASM is (0.1+J1.5) Ω and the stator impedance (0.22+J0.35)Ω . With the rotor at rest, the voltage measured between the rings is 90 V. Wind and friction losses of the motor are 350 W. Accordingly, find the copper losses in the rotor when the motor is running at full load.

Answer 1

The synchronous speed of the motor = $\frac{(50 ×120)}{3} =2166.67 rpm$

Slip = (Synchronous speed – Actual speed) = 2166.67 – 1460 = 2006.67 rpm

Percentage slip = [(2006.67 / 21667.67) × 100] = 9.26% = 0.0926

Rotor input = 90- 1 = 89 kW

Rotor copper loss = Rotor input × Slip = 89 × 0.0926 = 8.242 kW