Question

2. A 208V, two pole, 60 Hz, y connected wound rotor induction motor is rated at 15hp. Its equivalent circuit parameters are R:-0.200, R:-0.120, X.-X-0.410, X-150. The rotational losses are 430W. The motor drives a mechanical load with a slip of 5 percent. By using simplified equivalent circuit calculate; (30 pts) a) The motor speed in revolutions per minute and radians per second. b) The line current c) The stator copper losses. d) The air gap power. e) The power converted from electrical to mechanical form. f) The induced torque. g) The load torque. h) The overall machine efficiency.

Answer 1

h We solve part (as at least Fest we solve the (6) toch) lo The equivalent circuit of this feate y conector wound rotor induction motor is shown below: - IA jX₂ Rs AN 0. 200 J X1 son JOUIR R2 Att 0. 1202 + N your Vo 31523 * R (1) 2.282 (b The line current i The easiest way to find the line ewerent is to get the equivalent impedance ar of the rotore circait in paralled with j xri, and then calculate the

current as the phase voltage divided by the suns of the sexies impedances, as shown belowo →I ht w 0,202 ix, rest jo. use RE int → The equivalent impedance of the rotor circuit in parallel with gym is, 1+1 I ZE 1 संज t- 915 2.40 tjo.us jXM 22 2.2.220 +0.745 2 = 2.34 €18.50 2 3 The phase voltage is Vz - 208 3 3 Vos 120 V So, the line cuocent It is 도 Vo Ryti X + Retix 120Loo 0,20 10.4*2.22 +1 0.745 = 44.8 -25.5° A I S Ausa (c) The stator copper losses

The stator copper losses are, Pse = 350k Psc = 3 (44.8² (0,20) ber - 1205 W lansa The air gap powercia (d) The aut PAG A کہ gap power is, = 3 I 2 R₂ 3 T2RE PAG= 3(44.83 (2,220) PAG = 13.4kW] Ans. here 3 TARF is equal to 32² Rals), since the only resistance in the original cotone circuit was (R2/3 and the resistance in the the venen equivalent circuit is Rs. The power consumed by the Thevenind equivalent circuit must be the same as the power consumed by the original circuit.] The power converted from electrical to mechanical form is, Pony = (1-3) PAG [Given motor decives a mechanical load with a slipot 5% in ss 5 = 0.05 Joo

Iload 12. 3x63 (3420 almin) (I min 160s) (21radl ta) Zload = 34.3 Nm Anso x 100% (6) The overall efficiency is, Pout n Pout 100% * B 3 V. In cos 12.3 X 103 3(120) (448) Lcos 25.50) = 84.5% pex [calculated as an "Aus. Now finally answer of (a) (a) The matou speed in revolutions menute is 3420 cel mainta in 6)] Now, the motor speed in a radians second is um= (3420 eelmin) 11 (Tradka) (I min 60s) Wm= 358 rad /s, fee Anso