Question

6. A 3-phase 2.2 kV 25 Hz squirrel cage induction motor draws a line current of 224 A and a total power of 777 kW when operating at full load. The full-load speed is 245 rpm. The stator is wye connected and has an effective resistance per phase of 0.125 Ω. The stator iron (core) loss is 8.88 kW; friction and windage losses are 4.54 kW. Calculate at full-load conditions: b. power supplied to the rotor, a. the power factor; d. rotor IR (copper) losses; e. developed (mechanical) torque; f. efficiency c. slip; Solution:

Answer 1

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2.2 kv 25 HZ line current = 22 A total power = 717 k w fill load speed = 245 r p m Total power = 777 k w = squareroot 3 v_L I_L cos phi = cos phi = 777 k/squareroot 2.2 k times 224 = 0.91 (lag) power factor (cos phi) cos phi = 777 k/squareroot 3 times 2.2 k times 22.4 = 0.91 (lag) power supplied to the rotor factor input power = starter input power - starter losses (Air gas loss) = 777 k w - 4.54 k w - 8.88 k w = 763.58 kw Full load speed (N_s) = 245 rpm The above speed nearly speed n_1 poles will be (12) N_s = 120 times 25/12 = 250 rpm % slip N_s - N_r/N_s times 100 = 250 - 245/250 times 100 = 2 % colours = Rotor input times slip = 763.58 times 1000 times 0.02 = 15.27 k w \r\n Devolved mechanical Torque p = 2 pi N T/60