Question 9.31: A 3-phase induction motor with a unity turn ratio has the fo...

The equivalent circuit for the motor as per data is shown in Fig. 9.37.

Equivalent load resistance, R^{\prime }_{L} = \frac{R_{2}}{K^{2}} \left(\frac{1-S}{S} \right) = \frac{1}{\left(1\right)^{2} }\left(\frac{1 – 0.04}{0.04} \right) = 24  \Omega

Effective impedance per phase, \overline{Z} _{eff} = \left\lgroup R_{1} + \frac{R_{2}}{K^{2}} + R^{\prime }_{L}\right\rgroup + j \left\lgroup X_{1}+ \frac{X_{2S}}{K^{2}} \right\rgroup \\[0.5cm] \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \qquad \quad \quad = \left\lgroup 1 + \frac{1}{\left(1\right) ^{2}} + 24\right\rgroup + j \left\lgroup 3 + \frac{2}{\left(1\right) ^{2}} \right\rgroup = \left(26 + j 5\right)  \Omega

Stator load current, \overline{I} _{1}^{\prime } = \frac{\overline{V} }{\overline{Z}_{eq} } = \frac{240}{\left(26 + j5\right) } \\[0.5cm] \hspace{30 pt} \hspace{30 pt} \qquad \quad \enspace = \frac{240}{26.47\angle 10.9^{\circ }} = \left(9.06\angle -10.9^{\circ }\right) A \\[0.5cm] \hspace{30 pt} \hspace{30 pt} \qquad \quad \enspace = 9.06 \left(\cos 10.9^{\circ } – j\sin 10.9^{\circ }\right) = \left(8.896 – j 1.713\right)  A

Equivalent rotor current, I_{2} = \frac{I^{\prime }_{1}}{K} = \frac{9.06}{1} =\mathbf{9.06  A}

No-load current, \overline{I} _{0} = \frac{\overline{V} }{\overline{Z} _{0}} = \frac{240}{\left(10 + j 50\right) } = \frac{240}{51\angle 78.7^{\circ }} \\[0.5cm] \hspace{30 pt} \qquad \qquad \quad \, = 4.7\angle -78.7^{\circ } = \left(0.921 – j 4.61\right)  A

Stator equivalent current, \overline{I} _{1} = \overline{I}^{\prime }_{1} + \overline{I}_{0} = \left(8.896 – j 1.713\right) + \left(0.921 – j4.61\right) \\[0.5cm] \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \qquad \quad = \left(9.817 – j 6.323\right) = \left(11.7\angle -32.78^{\circ }\right)  A

Power factor, \cos \phi _{1} = \cos 32.78^{\circ }= \mathbf{0.84  lagging}

Motor efficiency, \eta = \frac{Output }{input} = \frac{5910 }{3 \times 240 \times 11.7 \times 0.84} \\[0.5cm] \hspace{30 pt} \qquad \qquad \quad = 0.8352 = \mathbf{83.52\% }