The standstill impedances per phase of the inner cage and outer cage of a 3-phase, 400 V, 50 Hz, double squirrel cage induction motor referred to stator side are given as;
\overline{Z}_{2i} = \left(0.6 + j5\right) \Omega ; \overline{Z}_{2o} = \left(3 + j1\right) \Omega
Determine the ratio of the currents and torques of the two cages (i) at stand-still (ii) at a slip of 4%. Neglect magnetising current.
The simplified equivalent circuit of the motor as per data is shown in Fig. 9.59.
( i) At standstill; slip, S = 1
Impedance of inner-cage referred to stator side;
Z^{\prime }_{2i} = \sqrt{\left(0.6 \right)^{2} + \left(5\right)^{2}} = 5.036 \Omega
Let the voltage impressed across the cage be E_{1}, then
I^{\prime }_{2i} =\frac{E_{1}}{5.036} ampere
Power input to inner-cage,
P^{\prime }_{2i} = \left(I^{\prime }_{2i}\right)^{2} R^{\prime }_{2i} = \left\lgroup\frac{E_{1}}{5.036}\right\rgroup ^{2} \times 0.6 = 0.02366 E^{2}
Impedance of outer-cage referred to stator side;
Z^{\prime }_{2o} = \sqrt{\left(3 \right)^{2} + \left(1\right)^{2}} = 3.162 \OmegaI^{\prime }_{2o} =\frac{E_{1}}{3.162}
Power input to outer-cage,
P^{\prime }_{2o} = \left(I^{\prime }_{2o}\right)^{2} R^{\prime }_{2o} = \left\lgroup\frac{E_{1}}{3.162}\right\rgroup ^{2} \times 3 = 0.3 E^{2}
Ratio of outer to inner cage current,
\frac{I^{\prime }_{2o}}{I^{\prime }_{2i}} = \frac{E_{1}}{3.162} \times \frac{5.036}{E_{1}} = \mathbf{1.593}
Ratio of outer to inner-cage torques or power
\frac{T_{2o}}{T_{2i}} =\frac{P^{\prime }_{2o}}{P^{\prime }_{2i}} = \frac{0.3 E^{2}_{1}}{0.02366 E^{2}_{1}} = \mathbf{12.68}
( ii) At a slip of 4%; Slip, S = 0.04
Impedance of inner-cage referred to stator side;
Z^{\prime }_{2i} = \sqrt{\left\lgroup\frac{R^{\prime }_{2i}}{S} \right\rgroup^{2} + \left(X_{2i}\right)^{2} } = \sqrt{\left\lgroup\frac{0.6}{0.04} \right\rgroup^{2} + \left(5\right)^{2} } = 15.81 \Omega
Current, I^{\prime }_{2i} = \frac{E_{1}}{Z_{2i}} = \frac{E_{1}}{15.81} ampere
Power input, P^{\prime }_{2i} = \left(I^{\prime }_{2i}\right)^{2} \times \frac{R^{\prime }_{2i}}{S} = \left\lgroup \frac{E_{1}}{15.8}\right\rgroup ^{2} \times \frac{0.6}{0.04} = 0.06 E^{2}_{1}
Impedance of outer-cage referred to stator side;
Z^{\prime }_{2i} = \sqrt{\left\lgroup\frac{R^{\prime }_{2o}}{S} \right\rgroup^{2} + \left(X_{2o}\right)^{2} } = \sqrt{\left\lgroup\frac{3}{0.04} \right\rgroup^{2} + \left(1\right)^{2} } = 75 \Omega
Current, I^{\prime }_{2o} = \frac{E_{1}}{Z_{2o}} = \frac{E_{1}}{75} ampere
Power input to outer-cage,
P^{\prime }_{2o} = \left(I^{\prime }_{2o}\right)^{2} \times\frac{ R^{\prime }_{2o}}{S} = \left\lgroup\frac{E_{1}}{75}\right\rgroup ^{2} \times \frac{3}{0.04} = 0.01333 E^{2}_{1}
Ratio of outer to inner-cage current,
\frac{I^{\prime }_{2o}}{I^{\prime }_{2i}} = \frac{E_{1}}{75} \times \frac{15.81}{E_{1}} = \mathbf{0.2108}
Ratio of outer to inner-cage torques or power
\frac{T^{\prime }_{2o}}{T^{\prime }_{2i}} =\frac{P^{\prime }_{2o}}{P^{\prime }_{2i}} = \frac{0.01333 E^{2}_{1}}{0.06 E^{2}_{1}} = \mathbf{0.222}
Conclusion: During starting major part of the power is developed by the outer-cage but during running condition major power is developed by the inner-cage.
