Question 4.13: A three-phase, 20-pole slip-ring induction motor runs at 291...
A three-phase, 20-pole slip-ring induction motor runs at 291 rpm when connected to a 50 Hz supply. Calculate slip for full-load torque if the total rotor-circuit resistance is doubled. Assume R_{2} >> SX_{2}.
N_{s} = \frac{120 f}{p}= \frac{120 ×50 }{20}= 300 rpm
slip,
S=\frac{N_{s}-N_{r}}{N_{s}} × 100= \frac{(300-291)}{300} × 100= 3 per cent
Full-load torque equation is,
T=\frac{K S E_{20}^{2} R_{2} }{R_{2}^{2}+S^{2} X^{2}_{20}}If R^{2}>> S^{2} X^{2}_{20}
T=\frac{K S E_{20}^{2} R_{2} }{R_{2}^{2}}=\frac{K S E_{20}^{2} }{R_{2}}For a given torque, T and rotor-induced EMF at standstill, i.e., E_{20} ,
R_{2} ∝ SIf R_{2} is doubled, S will be doubled. The slip at doubled R_{2} will be 6 per cent.
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