6-pole motor: rotor loss and efficiency

Given: 440 V, 50 Hz, 6-pole, input $P_{in} = 50$ kW at 0.85 lagging p.f., $N_r = 950$ rpm, friction loss $= 2$ kW, stator loss $= 1.5$ kW.

$$N_s = \frac{120\times 50}{6} = 1000\ \text{rpm}, \qquad s = \frac{1000-950}{1000} = 0.05$$

Power flow

Air-gap power (rotor input):

$$P_g = 50 - 1.5 = 48.5\ \text{kW}$$

(i) Rotor loss

$$P_{cu2} = sP_g = 0.05\times 48.5 = 2.425\ \text{kW}$$

(ii) Efficiency

Mechanical power developed: $P_m = 48.5 - 2.425 = 46.075$ kW Output: $P_{out} = 46.075 - 2 = 44.075$ kW

$$\eta = \frac{44.075}{50}\times 100 = 88.15\%$$

(The power factor is not needed.)

Answer: (i) rotor copper loss $= 2.425$ kW; (ii) efficiency $= 88.15\%$.