Question 18.10: Analysis of Single-Phase Induction Motor Find the input curr...
Analysis of Single-Phase Induction Motor
Find the input current and generated torque for a single-phase induction motor.
Known Quantities: Motor operating characteristics and circuit parameters.
Find: Motor input (stator) current, I_{S}; motor torque, T.
Schematics, Diagrams, Circuits, and Given Data: Motor operating data: \frac{1}{4} hp; 110-\mathrm{V}; 60 \mathrm{~Hz} ; 4 poles
Circuit parameters: R_{S}=1.5 \Omega ; X_{S}=2 \Omega ; R_{R}=3 \Omega ; X_{R}=2 \Omega ; X_{m}=50 \Omega; s=0.05.
Assumptions: The motor is operated at rated voltage and frequency.
Analysis: With reference to the equivalent circuit of Figure 18.26, you can easily show that the impedance seen by the backward emf, E_{b}, is much smaller than that seen by the forward emf, E_{f}. This corresponds to stating that the backward component of the magnetizing impedance (which is in parallel with the backward component of the rotor impedance) is much larger than the backward component of the rotor impedance, and can therefore be neglected. This approximation is generally true for values of slip less than 0.15, and corresponds to stating that
0.5 Z_{b}=0.5 \frac{j X_{m}\left(\frac{R_{R}}{2-s}+j X_{R}\right)}{\frac{R_{R}}{2-s}+j\left(X_{m}+X_{R}\right)} \approx 0.5\left(\frac{R_{R}}{2-s}+j X_{R}\right)
The approximate circuit based on this simplification is shown in Figure 18.28.
Using the approximate circuit, we find that the impedance seen by the backward emf is given by the expression
0.5 Z_{b}=0.5\left(\frac{R_{R}}{2-s}+j X_{R}\right)=0.5(1.538+j 2)=0.5\left(R_{b}+j X_{b}\right)
The impedance seen by the forward emf is, on the other hand, given by the exact expression:
\begin{aligned}0.5 Z_{f} &=0.5 \frac{j X_{m}\left(\frac{R_{R}}{s}+j X_{R}\right)}{\frac{R_{R}}{s}+j\left(X_{m}+X_{R}\right)} \\&=0.5 \frac{j 50(60+j 2)}{60+j(50+2)}=11.9+j 14.69=0.5 R_{f}+j 0.5 X_{f}\end{aligned}
If we let Z_{S}=R_{S}+j X_{S}=1.5+j 2 \Omega, we can write an expression for the total impedance of the motor as follows:
Z=Z_{S}+0.5 Z_{f}+0.5 Z_{b}=14.169+j 17.69=22.66 \angle 51.3^{\circ} \Omega
Knowing the total series impedance we can calculate the stator current:
\mathbf{I}_{S}=\frac{\mathbf{V}_{S}}{Z}=\frac{110 \mathrm{~V}}{22.66 \angle 51.3^{\circ} \Omega}=4.85 \angle-51.3^{\circ} \mathrm{A}
We can now calculate the power absorbed by the motor by separately computing the real power absorbed in the forward and backward fields:
\begin{aligned}P_{f} &=I_{S}^{2} \times 0.5 R_{f}=(4.85)^{2} \times 11.9=279.9 \mathrm{~W} \\P_{b} &=I_{S}^{2} \times 0.5 R_{b}=(4.85)^{2} \times 0.769=18.1 \mathrm{~W}\end{aligned}
The net power is the difference between the two components, thus, P=P_{f}-P_{b}=261.8 \mathrm{~W}, and the torque developed by the motor is equal to the ratio of the power to the motor speed. The synchronous speed can be computed to be:
\omega_{s}=\frac{4 \pi f}{p}=188.5 \mathrm{rad} / \mathrm{s}
and, if we assume negligible rotational losses, we have:
T=\frac{P}{\omega}=\frac{P}{(1-s) \times \omega_{s}}=\frac{261.8 \mathrm{~W}}{0.95 \times 188.5 \mathrm{rad} / \mathrm{s}}=1.46 \mathrm{~N}-\mathrm{m}
Comments: Note that the power factor of the motor is pf =\cos \left(-51.3^{\circ}\right)=0.625. Such low power factors are typical of single-phase motors.
