Question

2. A 60 Hz 4-pole, 3-phase induction machine operating at s = 0.028 with rotational losses of 380 W has following parameters for a single-phase: Source voltage is 120 Vrms Rs = 0.39 2, Xs = 0.46 2, XM = 25 2, R'R 0.25 2, X'R = 0.82 2 The rotor mechanical speed is given to be 1749.6 rpm. a. (5) Calculate the real input power, Pin. b. (3) Calculate the power delivered to the rotor. c. (3) Calculate the mechanical power at the rotor. d. (2) Calculate the output power at the shaft. e. (2) Calculate the power efficiency of the system.

Answer 1

Given es the 36 Induction machine, It is operating at S = 0.028 E Numbeer of pales = 4 foequieney = 6043 Rotallöner losses = 380 kJ Sacroce voltage Vs = 120 Vame Given parameters over I Rs = 0.39 2 RR = 0.252 xs = 0.462 x = 0.8202 XM = 252 from the above given parameters the Caecialunt impedance can be calculated as 289 = (R$+5xs) + ( +5Xp) smo i tj xp' + 3x Subsliteete the caleces, then ze is 789 = (0:39+5046) + COBE +30-da) (jas) 0.028 + 30.82 + 135 +(6:39+j0:46) + (8.966 (5:24) 2590 27:32 (70.92 = 0.39+30.46 +7.476 +j3:378 0.25

= 120 * 8.450 Toci the Équivalent impedance Zo is given by 1:729 = 8752 262 the source Current can be calculated as the salio of voltage near phase to the Eqrewalent impedance. Is = Yeh ZER =_ 120 . Ex 8.752606 15:1529126* Is = 7:115-3.47% Is = 7:91 4-2519 = 71916-26A @) To Calculal the Sreal inpeet poweet, Pani - the inpect power, Por can be written as Pin - BV, Icosó . = 3x 120 x 1.91 x cos (-a6) = 1644.06 x 0.898 in peel power, Pie = 1.477 kW the stator copper loss can be calculard as copper loss = 3 I'RS =3(7.91 x 0.3 = 73.2 W.

(b) to calculali the scorer delivered to sotos e the power delivered to the scotor can be obtained by SubIracting the statou loss from the expect power. Pa = Pon - Stator loss = (1.477X10") - 73.2 the power delivered to the sotor is. .: B = 1331.8 W co to find the mechanê cal power of rotor. the mechanical powee oleveloped can be written as PM = Po ci-s) S = 1331.8 (1-0.098) = 1331.8 (0.972). e the mechanical power is .: PM = 1294.5 W (d) To calculate output power at the shaft: the outpect poweer can be obtained by sub- beäeling the salalional losses from mechanical Power Pout = Pm - Kalalwnal losses = 1294.5 – 380 W A output power put = 914.5 W

(e) To find Efficiency of the systero: B the Efficiency can be calculated as the ratio of outpect poweer to the expert power Efficiency, 2= Paul x100 2 x 100 = 0.6191 X 100 Fi Efficiency, z = 61191% Pio