The impedance per phase of a double cage induction motor are:
Inner cage-(0.06 +j0.5) ohm; Outer cage – (0.6 + j0.12) ohm,
Estimate the toque in synchronous watt per phase at standstill and 4% slip considering that the rotor equivalent induced emf per phase is 110 V at standstill.
The equivalent circuit of a double-cage rotor is shown in Fig. 9.60.
Considering voltage as a reference phasor, E_{2S} = 110 \pm j0 = 110\angle 0^{\circ }
\overline{Z}_{2i} = \left(0.06 + j0.5\right) \Omega ; \overline{Z}_{2o} = \left(0.6 + j0.12\right) \Omega
At standstill \overline{Z}_{2i} = \left(0.06 + j0.5\right) = \sqrt{\left(0.06\right)^{2} + \left(0.5\right)^{2} } \angle \tan ^{-1} \frac{0.5}{0.06} = 0.5036\angle 92.4^{\circ } \\[0.5cm] \overline{Z}_{2o} = \left(0.6 + j0.12\right) = \sqrt{\left(0.6\right)^{2} + \left(0.12\right)^{2} } \angle \tan ^{-1} \frac{0.12}{0.6} = 0.612\angle 12.57^{\circ } \\[0.5cm] \overline{Z}_{2i} + \overline{Z}_{2o} = \left(0.06 + 0.6 +j0.5 + j0.12\right) = \left(0.66 + j0.62\right) = 0.906\angle 48^{\circ }
Total impedance of the two cages
\overline{Z}_{2} = \frac{\overline{Z}_{2i} \overline{Z}_{2o}}{\left(\overline{Z}_{2i} + \overline{Z}_{2o}\right) } = \frac{\left(0.5036\angle 92.4^{\circ }\right) \left(0.612\angle 12.57^{\circ }\right) }{\left(0.906\angle 48^{\circ }\right)} \\[0.5cm] \quad \; = 0.34 \angle 56.97^{\circ } = \left(0.34 \cos 56.97^{\circ } + j 0.34 \sin 56.93^{\circ }\right) \\[0.5cm] \quad \; = \left(0.2127 + j 0.2652\right) ohm
Rotor current at standstill
\overline{I}_{2S} = \frac{\overline{E}_{2S} }{\overline{Z}_{2} } = \frac{110 \angle 0^{\circ }}{0.34 \angle 56.97^{\circ }} = 323.5 \angle \boxtimes 56.97^{\circ }
Synchronising power or torque developed per phase in the rotor at start
T_{syn} = I^{2}_{2S} R_{2S} = \left(323.5\right)^{2} \times 0.2127 = \mathbf{22260 syn. watt}
At a slip of 4% \overline{Z}_{2i} = \left(\frac{0.06}{0.04} + j0.5\right) = \left(1.5 + j0.5\right) = 1.58\angle 20.48^{\circ } \\[0.5cm] \overline{Z}_{2o} = \left(\frac{0.6}{0.04} + j0.12\right) = \left(15 + j0.12\right) = 15 \angle 0.5^{\circ } \\[0.5cm] \overline{Z}_{2i} + \overline{Z}_{2o} = \left(1.5 + j0.5 + 15 + j0.12\right) = \left(16.5 + j0.62\right) = 16.51 \angle 2.39^{\circ }
Total impedance of the two cages
\overline{Z}_{2} = \frac{\overline{Z}_{2i} \overline{Z}_{2o}}{\left(\overline{Z}_{2i} + \overline{Z}_{2o}\right) } = \frac{\left(1.58 \angle 20.48^{\circ }\right) \left(15 \angle 0.5^{\circ }\right) }{16.51 \angle 2.39^{\circ }} = 1.435 \angle 18.59^{\circ } \\[0.5cm] \quad \; = 1.435 \left(\cos 18.59^{\circ } + j \sin 18.59^{\circ } \right) = \left(1.374 = j 0.413\right) ohm
Rotor current during running
\overline{I}_{2} = \frac{S\overline{E}_{2S} }{\overline{Z}_{2} } = \frac{0.04 \times 110 \angle 0^{\circ }}{1.435 \angle 18.59} = 3.07 \angle -18.59^{\circ }
Synchronous power or torque developed per phase in the rotor,
T_{syn} = I^{2}_{2} R_{2} = \left(3.07\right)^{2} \times 1.374 = \mathbf{12.95 syn. watt}