Question 8.2: For the single-phase induction motor of Example 8.1, it is r......
For the single-phase induction motor of Example 8.1, it is required that the power and torque output, and the efficiency when running at a slip of 5%, be found. Neglect core and rotational losses.
In Example 8.1, we obtained
Z_{i}=26.841\angle43.36~\mathrm{deg}As a result, with V_{1}=110\angle0, we obtain
I_{1}={\frac{110\angle0}{26.841\angle43.36\,\mathrm{deg}}}=4.098\angle-43.36\,\,\mathrm{deg\,\,A}The power factor is thus
\cos\phi_{1}=\cos43.36\,\,\mathrm{deg}=0.727The power input is
P_{1}=V_{1}\,I_{1}\cos\phi_{1}=327.76\,\mathrm{W}We have from Example 8.1 for s = 0.05,
R_{f}=17.294\ \Omega\ \,\ \ \ \ R_{b}=0.721\ \OmegaThus we have
P_{g f}=|I_{1}|^{2}R_{f}=(4.098)^{2}(17.294)=290.46\,\mathrm{W} \\ P_{g b}=|I_{1}|^{2}R_{b}=(4.098)^{2}(0.721)=12.109\,\mathrm{W}The output power is thus obtained as
P_{m}=(1-s)(P_{g f}-P_{g b}) \\ =0.95(290.46-12.109)=264.43\,\mathrm{W}As we have a four-pole machine, we get
n_{s}={\frac{120(60)}{4}}=1800\,\mathrm{rpm} \\ \omega_{s}={\frac{2\pi n_{s}}{60}}=188.5\,\mathrm{rad/s}The output torque is therefore obtained as
T_{m}={\frac{1}{\omega_{s}}}(P_{g f}-P_{g b}) \\ ={\frac{290.46-12.109}{188.5}}=1.4767\,\mathrm{N}\cdot\mathrm{m}The efficiency is now calculated as
\eta={\frac{P_{m}}{P_{1}}}={\frac{264.43}{327.76}}=0.8068It is instructive to account for the losses in the motor. Here we have the static ohmic losses obtained as
P_{l_{\mathrm{s}}}=|I_{1}|^{2}R_{1}=(4.098)^{2}(1.5)=25.193\,\mathrm{W}The forward rotor losses are
P_{I_{rf}}=s P_{g\!f}=0.05(290.46)=14.523\,\mathrm{W}The backward rotor losses are
P_{lb}=(2-s)P_{g b}=1.95(12.109)=23.613\,\mathrm{W}The sum of the losses is
P_{l}=25.193+14.523+23.613=63.329\,\mathrm{W}The power output and losses should match the power input:
P_{m}+P_{l}=264.46+63.33=327.76\,\mathrm{W}which is indeed the case.