A 3- phase 4-pole, 50 Hz, 400 V, star-connected squirrel cage induction has the following data:
Stator impedance, \overline{Z}_{1} = \left(0.07 + j 0.25\right) ohm
Equivalent rotor impedance, \overline{Z}^{\prime }_{2} = \left(0.07 + j 0.3\right) ohm
Determine the maximum power developed (gross output), torque and the slip at which it occurs. Neglect exciting reactance.
Here, V_{L} = 400 V; P= 4 ; f =50 Hz ; \overline{Z}_{1} = \left(0.07 + j 0.25\right) ohm \\[0.5cm] \overline{Z}^{\prime }_{2} = \left(0.07 + j 0.3\right) ohm
Phase voltage, V = \frac{V_{L}}{\sqrt{3} } = \frac{400 }{\sqrt{3} } = 231 V
motor equivalent impedance referred to stator side
\overline{Z}_{eq1} = \left(0.07 + j0.25\right) + \left(0.07 + j0.3\right) = \left(0.14 + j0.55\right) ohm \\[0.5cm] \quad \quad \, = \left(0.5675\angle 75.7^{\circ }\right) ohmZ_{eq1} = 0.5675 ohm
Slip corresponding to maximum power developed
S_{m} = \frac{R_{2} / K^{2}}{Z_{eq1} + R_{2} / K^{2}} \hspace{30 pt} \hspace{30 pt} \left(where K= 1 not given\right) \\[0.5cm] \quad \enspace = \frac{0.07}{0.5675 + 0.07} = 0.11 = \mathbf{11\%}
Maximum power developed, P_{m} = \frac{3 V^{2}}{2 \left(R_{eq1} + Z_{eq1}\right) } = \frac{3 \times \left(231\right)^{2} }{2\left(0.14 + 0.5675\right)} \\[0.5cm] \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \qquad \qquad \enspace =113132 W \cong \mathbf{113 kW}
Synchronous speed, N_{S} = \frac{120 f}{P} = \frac{120 \times 50}{4} = 1500 r.p.m
Maximum torque developed, T_{m} = \frac{P_{m} \times 60}{2 \pi N_{S}} = \frac{113132 \times 60}{2 \pi \times 1500} = \mathbf{2263 Nm}