prob unit 2_3 Q&A - Electric Drives and Control(EEE 402...

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Electric Drives and Control(EEE 402) Introduction to Electric Drives Dr Umashankar, Asso Prof, SELECT-VIT 1. A 220 V, 500 A, 600 rpm separately excited motor has armature and field resistance of 0.02 and 10 respectively. The load torque is given the expression T L = 2000 – 2 N , where N is the speed in rpm. Speeds below the rated are obtained by armature voltage control and speeds above the rated are obtained by field control. i) Calculate motor terminal voltage and armature current when the speed is 450 rpm. ii) Calculate field winding voltage and armature current when the speed is 750 rpm. Assume the rated field voltage is the same as the rated armature voltage. V =220V , I a1 = 500 A, N 1 = 600 rpm, R a = 0.02 ohm, R f = 10 ohm, T L = 2000 _ 2N Nm ( separately excited motor) (i) V =? I a2 = ? (450 rpm) T 2 = 2000_ 2 (450) = 1100 Nm E 1 = V _ I a1 R a = 220 _ 500 (0.02) = 210 V T 1 = P / W m = (E a1 I a1 ) / W m = 210 x 500 / ( 600 x 2 / 60) = 1671 Nm T 2 / T 1 = I a2 / I a1 I a2 = (T 2 / T 1 ) x I a1 = (1100 / 1671 ) x 210 = 138.24 A E 2 = V _ I a2 R a = 220 _ (138.24 x 0.02) = 217.2 V 600 rpm --------------E = 217.2 V 450 rpm --------------E = ? E= (450 / 600) x 217.2 = 163 V V = 163 + (138.24 x 0.02) = 165.8 V (ii) V f = ?, I a2 = ?, N = 750 rpm T 2 = 2000_ 2 (750) = 500 Nm T 2 = T 2 K e ø I a2 = 500 K e ø =(500) / I a2 V = E 2 + I a2 R a 220 = K e ø N 2 (2 /60) + 0.02I a2 220 =(500 / I a2 ) x 750 x (2 /60) + 0.02I a2 0 = 0.02I
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