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Question 6.7: A single-phase induction motor takes an input power of 490 W...

A single-phase induction motor takes an input power of 490 W at a power factor of 0.57 lagging from a 110-V supply when running at a slip of 5 percent. Assume that the rotor resistance and reactance are 1.78 Ω and 1.28 Ω, respectively, and that the magnetizing reactance is 25 Ω. Find the resistance and reactance of the stator.

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The equivalent circuit of the motor yields

Zf={[1.78/2(0.5)]+j0.64}(j12.5)[1.78/2(0.05)]+j(0.64+12.5)=5.6818+j8.3057Z_{f} =\frac{\left\{\left[1.78/2\left(0.5\right) \right]+j0.64 \right\}\left(j12.5\right) }{\left[1.78/2\left(0.05\right) \right]+j\left(0.64+12.5\right) } =5.6818+j8.3057

 

Zb={[1.78/2(1.95)]+j0.64}(j12.5)[1.78/2(1.95)]+j(0.64+12.5)=0.4125+j0.6232Z_{b} =\frac{\left\{\left[1.78/2\left(1.95\right) \right]+j0.64 \right\}\left(j12.5\right) }{\left[1.78/2\left(1.95\right) \right]+j\left(0.64+12.5\right) } =0.4125+j0.6232

As a result of the problem specifications

Pi=490WP_{i} =490 W                 cos⁡ϕ=0.57\cos \phi =0.57              V=110V=110

Ii=PiVcos⁡ϕ=590110(0.57)=7.815∠−55.2488°I_{i} =\frac{P_{i}}{V \cos \phi}=\frac{590}{110\left(0.57\right) } = 7.815\angle -55.2488°

Thus the input impedance is

Zi=VIi=1107.815∠55.2498°=8.023+j11.5651Z_{i}=\frac{V}{I_{i}} =\frac{110}{7.815} \angle 55.2498°=8.023+j11.5651

The stator impedance is obtained as

Z1=Zi−(Zf+Zb)=1.9287+j2.636ΩZ_{1}=Z_{i}-\left(Z_{f}+Z_{b}\right) =1.9287+j2.636\Omega

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