Question

A three-phase 60 Hz induction motor 480 v (rms, line- to line), lo tp, spee is 12 amperes (rms), total stator Copper loss is Copper loss is 150 W, total combine with operation at the specified paramders, a) The number of Poles ! induction motor has specifications as follows R- to line), lo tp, speed is 1765 RPM, line Current stal Stator Copper loss is 140 w, total rotor is 15o W , total combined rotational loss is losw the specified parameers, determine; b. The Slip c) the output power in watts d the output passer torque e) The input power in watts of the input apparent power g) the motor powe factor w the efficiency. 5666 105

Answer 1

Given 3-0 TOM - 60 Hz , v = 480 (RMS) Nr = 1765 R.PM II = 12 A Total stator Copper loss (Pst) = 140w Total Rotor copper lan (Pout) = 150w Rotational lon Pull = 105w synchronous speed, Ng= 120 t - P - 120 X 60 Ns Dear the given goren Rotor speed, Nr Rotor speed , Nr=1765 rpm, the speed, Ng = 1800 rp.m [ Ng > Nr] - P = 120X60 = 4 1800 6 slip ,s = NS-Nr . - 0.0194 1800-1765 1800 Ns Pout Protatini Stator о lon lon

Rotor copper lon, Pr=sPag from mechanical output power, Parod = (1-5) Paq - Pama Po - (-s) Pag 5 Pag Pg md = (1-3) Pv = 1765 x 150 = 7564.28W 35 Pout = 1564.28 - Protation - 7564.28 - 105 = 7459.28 W lo Pout 17459.18 w] Pout =7459.28 W & output torque, Neut - Pout Pout = wr 27 Nr 6. = 7459.28860 (271) (1765) - 40.357 Vout = 40.351 N-m ③ The input power, Pinput - Pag + Pastator lon Pag = Pr - 150 = 7114.285 w

Pe 1714,28 +140 - 1854.28 [impul 1954.000 input apprant power, (6) - BV, XIL - B X 480 X 12 - 9976.619 w. input power factor, Cost - Pin 1854.28 9976.612 = 0.1872 (lag) I elticiency, y - - Pout Pin 7459.28 1854.28 = 94.91