Question 18.11: Analysis of Single-Phase Induction Motor Find the following ...
Analysis of Single-Phase Induction Motor
Find the following quantities for the single-phase machine of Example 18.10:
1. Output torque
2. Output power
3. Efficiency
Known Quantities: Motor operating characteristics.
Find: Motor torque, T; output power, P_{\text {out }}; efficiency, \eta.
Schematics, Diagrams, Circuits, and Given Data: Motor operating data: \frac{1}{4} \mathrm{hp} ; 110 \mathrm{~V}; 60 \mathrm{~Hz} ; 4 poles; s=0.05.
Assumptions: The motor is operated at rated voltage and frequency. The combined rotational and core losses are P_{\mathrm{rot}}+P_{\text {core }}=30 \mathrm{~W}.
Analysis:
1. Output power calculation. The motor generated power is the difference between the forward and backward components, as explaind in Example 18.10. Thus,
P=P_{f}-P_{b}=261.8 \mathrm{~W}
The motor power is the difference between the generated power and the losses:
P_{\text {out }}=P-P_{\text {loss }}=261.8-30 \mathrm{~W}=231.8 \mathrm{~W}
2. Shaft torque calculation. The shaft speed is:
\omega=(1-s) \times \omega_{s}=(1-s) \times \frac{4 \pi f}{p}=179 \mathrm{rad} / \mathrm{s}
and the torque is:
T=\frac{P_{\text {out }}}{\omega}=\frac{231.8 \mathrm{~W}}{179 \mathrm{rad} / \mathrm{s}}=1.29 \mathrm{~N}-\mathrm{m}
3. Efficiency calculation. To calculate the overall efficiency of the motor we must account for three loss mechanisms: mechanical losses, core losses, and electrical losses. The first two are given as a lumped number; the electrical losses can be computed by calculating the I^{2} R losses in the stator and forward and backward circuits:
\begin{aligned}P_{S \text { loss }} &=I_{S}^{2} R_{s}=(4.85)^{2} \times 1.5=35.3 \mathrm{~W} \\P_{R_{f} \text { loss }} &=s P_{f}=0.05 \times 279.9=14 \mathrm{~W} \\P_{R_{b} \text { loss }} &=(1-s) P_{b}=1.95 \times 18.1=35.3 \mathrm{~W} \\P_{\text {elec }} &=P_{S \text { loss }}+P_{R_{f} \text { loss }}+P_{R_{b} \text { loss }}=114.6 \mathrm{~W}\end{aligned}
The efficiency can be calculated according to the following expression:
\begin{aligned}\eta &=1-\frac{\sum \text { losses }}{P_{\text {out }}+\sum \text { losses }}=1-\frac{P_{\text {rot }}+P_{\text {core }}+P_{\text {elec }}}{P_{\text {out }}+P_{\text {rot }}+P_{\text {core }}+P_{\text {elec }}} \\&=1-\frac{30+114.16 \mathrm{~W}}{2131.8+30+114.16 \mathrm{~W}}=0.617=61.7 \%\end{aligned}
Comments: Note that the overall efficiency of this machine is fairly low. Multiphase AC machines can achieve significantly higher efficiencies.