Question 6.7: A three-phase, 460-V, 60-Hz, four-pole, 500-hp wound-rotor i......

The solution involves recognition of the fact that the effects of changes in the rotor resistance are seen from the stator in terms of changes in the referred resistance R_2/s. Examination of the equivalent circuit shows that, for specified applied voltage and frequency, everything concerning the stator performance is fixed by the value of R_2/s, the other impedance elements being constant. For example, if R_2 is doubled and s is simultaneously doubled, there will be no indication from the stator that anything has changed. The stator current and power factor, the power delivered to the air gap, and the torque will be unchanged as long as the ratio R_2/s remains constant.

Added physical significance can be given to the argument by examining the effects of simultaneously doubling R_2 and s from the viewpoint of the rotor. An observer on the rotor would see the resultant air-gap flux wave traveling past at twice the original slip speed, generating twice the original rotor voltage at twice the original slip frequency. The rotor reactance therefore is doubled, and since the original premise is that the rotor resistance also is doubled, the rotor impedance is doubled while the rotor power factor is unchanged. Since rotor voltage and impedance are both doubled, the effective value of the rotor current remains the same; only its frequency is changed. The air gap still has the same synchronously rotating flux and mmf waves with the same torque angle. An observer on the rotor would then agree with a counterpart on the stator that the torque is unchanged.

An observer on the rotor, however, would be aware of two changes not apparent in the stator: (1) the rotor I² R loss will doubled, and (2) the rotor is turning more slowly and therefore developing less mechanical power with the same torque. In other words, more of the power absorbed from the stator goes into I² R heat in the rotor, and less is available for mechanical power.

The preceding thought processes can be readily applied to the solution of this example.

a. If the rotor resistance is increased five times, the slip must increase five times for the same value of R_2/s and therefore for the same torque. But the original slip at full load is 0.015. The new slip at full-load torque therefore is 5(0.015) = 0.075.

b. The effective value of the rotor current is the same as its full-load value before addition of the series resistance, and therefore the rotor R_2/s loss is five times the full-load value of 5.69 kW, or

Rotor I^2R = 5 × 5.69 = 28.4  kW

c. The increased slip has caused the per-unit speed at full-load torque to drop from 1 – s = 0.985 down to 1 – s = 0.925. Since the ratio R_2/s is unchanged, the torque is the same and hence the power output has dropped proportionally, or

P_{mech} = \frac{0.925}{0.985}(500) = 470  hp

Because the air-gap power is unchanged, the decrease in electromechanical mechanical shaft power must be accompanied by a corresponding increase in rotor I² R loss.

d. If rotor resistance is increased five times, the slip at maximum torque simply increases five times. But the original slip at maximum torque is 0.060. The new slip at maximum torque with the added rotor resistance therefore is

s_{maxT} = 5(0.060) = 0.30

e. The effective value of the rotor current at maximum torque is independent of rotor resistance; only its frequency is changed when rotor resistance is varied. Therefore,

I_{2,maxT} = 2.82I_{2,\mathrm{fl}}

f. With the rotor resistance increased five times, the starting torque will be the same as the original running torque at a slip of 0.20 and therefore equals the running torque without the series resistors, namely,

T_{start} = 1.20T_{\mathrm{fl}}

g. The rotor current at starting with the added rotor resistances will be the same as the rotor current when running at a slip of 0.20 with the slip rings short-circuited, namely,

I_{2,start}= 3.95I_{2,\mathrm{fl}}