Question

5.7) An induction motor is installed in a factory. Under conditions in which the three-phase 415-volt 50-Hz supply is perfectly balanced, the motor runs at 1492 rpm. A single-phase load is then energised elsewhere in the factory, which has the effect of reducing the supply voltage on phase "6" by 5% without changing its phase angle. It may be assumed that the motor's speed remains unchanged, and that the motor supply has no neutral conductor. The motor equivalent circuit parameters are: Stator resistance 0.08 12, Stator leakage reactance 0.30 12. Rotor resistance 0.062, Rotor leakage reactance 0.30 2. The magnetising reactance may be neglected. Calculate the magnitude of the current in each phase under rated conditions with a balanced supply equal to its rated value. (As it is balanced you can use the standard single-phase equivalent circuit) Calculate the positive, negative and zero sequence voltages associated with the unbalanced supply assuming phase a has zero phase angle. Calculate the magnitude of the line currents with unbalanced supply Calculate the degree of unbalance in the supply voltages, by finding the ratio of the negative sequence to the positive sequence voltage components and the VUF using the deviation mean formula for the line voltages. Repeat this for the supply currents. Calculate the total motor copper losses for the balanced condition (part a) and the unbalanced condition (part b)

Answer 1

SE 0:08 1500 1492 1600 = 0.00533 a R2 - 0.06 = 0.00533 11:25.2 i 21:1178 2-3:03 21788-303 i Ta= COo84 1125+ jo16) IG= 21:11782-123.03' A Te = 21.1178 (+116.97° A (5) Under unbelance condition Vaz 415/0 = 239,6 COV Vo= 415 x 0.95 (-120 = 227:62 (-120 v Ve fils x 1 120° .239.6 /120V. 13 2394 Vai = 6 [vat a Vo + d² Ve] = 5|239.660 +2 27:62 + 239.6] Vay = 235.6 Lo° v Vaga dar vat 8p2V6 + a ve] E T 239.660' +227.624-240° + 239-622407

VaR = 3.996–60° voet Vao = $ [ van +Up + Ve] =[239-620' +226.622_120+ 239.6 (12007 Vao = 3.99 £60° (d) Degree of unbalance - Vaze 3:19 = 0,016935 3.99 0.016035 -= 235-6 val copper (e) Total motor loss (under benced condition) = 312 x R2 = 3x(81.11783 x 0.06 = 80,273 hatt (C) Ia = 21:1178 (-3.03 A Tor You C20347 .25+ 30-6] . 20106 (423.037 teo 21.1178 (+116.97° A capper loss under unbalonce condition. = 2621.1178)*0.06 +(201069°x0.06 = 77-65 watt