Question 9.13: 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} \gg SX_{2}.
slip, \begin{aligned}&N_{s}=\frac{120 f}{P}=\frac{120 \times 50}{20}=300 rpm \\&S=\frac{N_{s}-N_{r}}{N_{s}} \times 100=\frac{(300-291)}{300} \times 100=3 \text { per cent }\end{aligned}
Full-load torque equation is,
\begin{gathered}T=\frac{K S E_{20}^{2} R_{2}}{R_{2}^{2}+S^{2} X_{20}^{2}} \\If R^{2}>>S^{2} X_{20}^{2} \\T=\frac{K S E_{20}^{2} R_{2}}{R_{2}^{2}}=\frac{K S E_{20}^{2}}{R_{2}}\end{gathered}For a given torque, T and rotor-induced EMF at standstill, i.e., E_{20},
R_{2} \propto SIf R_{2} is doubled, S will be doubled. The slip at doubled R_{2} will be 6 per cent.
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