Here, R_{1} = 0.4 \Omega ; X_{1} = 0.5 \Omega;R_{2}= 0.25 \Omega ; X_{2S}= 0.5 \Omega \\[0.5cm] X_{m}=15.5 \Omega ; S= 0.04; V_{L} = 400 V; f = 50 Hz.
Phase voltage, V = \frac{V_{L}}{\sqrt{3} } = \frac{400}{\sqrt{3} } = 231 V
Slip to develop maximum torque, S_{m}= \frac{R_{2}}{X_{2S}} = \frac{0.25}{0.5} = \mathbf{0.5}
For maximum torque development
Equivalent impedance of the motor at S_{m}
\overline{Z} _{eq1\left(m\right) } = \left(R_{1} + \frac{R_{2}}{S_{m}} \right) + j\left(X_{1} + X_{2S}\right) = \left(0.4 + \frac{0.25}{0.5} \right) + j\left(0.5 + 0.5\right) \\[0.5cm] \qquad \quad \, = 0.9 + j 1 = 1.345 \angle 48^{\circ } ohm
Rotor current, \overline{I} _{2} = \frac{\overline{E}_{2S} }{\overline{Z} } =\frac{231}{1.345\angle 48^{\circ }}\left(here, E_{2}=V since K=1\right) \\[0.5cm] \hspace{30 pt} \qquad \quad \quad \, = 171.7\angle -48^{\circ }
I _{2} = \mathbf{171.7 A}
Rotor copper loss = 3 \overline{I}^{2}_{2} R_{2} = 3 \times \left(171.7\right)^{2} \times 0.25 = 22122 W
Power input to rotor, P_{2} = P_{2} = \frac{Rotor copper loss}{S_{m}} = \frac{22122}{0.5} = 44244 W
Synchronous speed of revolving field,
N_{S} = \frac{120 f}{P} = \frac{120 \times 50}{4} = 1500 rpm
Maximum torque developed, T_{m} = \frac{P_{2}}{\omega _{S}} = \frac{P_{2} \times 60}{2 \pi N_{S}} = \frac{44244 \times 60}{2 \pi \times 1500} = \mathbf{281.67 Nm}
For starting torque:
At start, slip, S_{S} = 1.0
Equivalent motor impedance, \overline{Z}_{eq1\left(S\right) } = \left\lgroup R_{1} + \frac{R_{2}}{S_{S}} \right\rgroup + j\left(X_{1}+ X_{2S}\right) \\[0.5cm] \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \qquad \quad \quad = \left(0.4 + \frac{0.25}{1} \right) + j\left(0.5+ 0.5\right) = \left(0.46 + j1\right) ohm \\[0.5cm] \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \qquad \quad \quad = \left(1.1 \angle 65.3^{\circ }\right) ohm.
Rotor current at start, \overline{I}_{2S } = \frac{\overline{E}_{2S} }{\overline{Z}_{S} } = \frac{231}{1.1\angle 65.3^{\circ }} = 210 \angle \boxtimes 65.3^{\circ }
I_{2S } = \mathbf{210 A}
Rotor copper loss = 3 I^{2}_{2S } R_{2}= 3\times \left(210\right)^{2} \times 0.25 = 33075 W
Power input to rotor, P_{2S } = \frac{33075}{1} = 33075 W
Starting torque developed, T_{s} = \frac{P_{2S }}{\omega _{S}} = \frac{33.75 \times 60}{2\pi \times 1500} = \mathbf{210.6 Nm}