Question

6.12 A three-phase, Y-connected, 460-V (line-line), 37-kW, 60-Hz, four-pole induction motor has the following equivalent-circuit parameters in ohms-per-phase referred to the statr: R10.070 R2 0.152 X 0.743 X2 0.764 Xm 40.1 The total friction and windage losses may be assumed constant at 390 W, and the core loss may be assumed to be equal to 325 W. With the motor connected directly to a 460-V source, compute the speed, output shaft torque and power, input power, and power factor and efficiency for slips of 1, 2, and 3 percent. You may choose either to represent the core loss by a resistance connected directly across the motor terminals or by resistance Re connected in parallel with the magnetizing reactance X

Answer 1

Per phase voltage V_p h = 460 V/squareroot 3 = 265.58 volt. rated shaft power, P_s h = 37 kW, frequency f = 60 Hz No. of poles P = 4 stator plux speed N_s = 120 t/ = 120 * 60/4 = 1800 Therefore N_s = 1800 rpm. Parameter reffered to stator side perphase R_1 = 0.070 Ohm/ph, R_2 prime = 0.152 Ohm/ph. X_1 = 0.743 Ohm/ph, X_2 prime = 0.764 Ohm/phase. X_m = 40.1 Ohm/phase. given friction and windage losses P_k & W = 390 Watt. & core fosses, P_c = 325 Watt. (i) at slip Delta = 1% = 0.01 per phase equivalent circuit reffered to stator side (A) rotor speed, Delta = N_s - N_r/N_ N_r = N_s (1 - s) N_r = 1800 (1 - 0.01) Therefore N_r = 1782 rpm. \r\n let voltage at Node A is V_a (B) using Node method at Node V_A 'or' A. V bar_A - 265.85/(0.070 + j0.743) + V bar_A - 0/j40.1 + V bar_A - 0/(0.152/0.01 + j0.764) = 0 V bar_A {1/(0.070 + j0.743) + 1/j40.1 + 1/(0.152/0.01 + j0.784)}