Question 6.6: For the single-phase induction motor of Example 6.5, it is n...
For the single-phase induction motor of Example 6.5, it is necessary to find the power and torque output and the efficiency when running at a slip of 5 percent. Neglect core and rotational losses.
In Example 6.5 we obtained
Z_{i} =26.841\angle 43.36°As a result, with V_{1}= 110\angle 0 , we obtain
I_{1}=\frac{110\angle 0}{26.841\angle 43.36°}
= 4.098\angle -43.36° A
The power factor is thus
\cos \phi _{1} =\cos 43.36°=0.727The power input is
P_{1}= V_{1}I_{1}\cos \phi _{1}=327.76 WWe have from Example 6.5 for s = 0.05,
R_{f}= 17.294\Omega R_{b}= 0.721\Omega
Thus we have
P_{gf} =\left|I_{1} \right| ^{2} R_{f} = \left(4.098\right) ^{2} \left(17.294\right) =290.46 W P_{gb} =\left|I_{1} \right| ^{2} R_{b} = \left(4.098\right) ^{2} \left(0.721\right) =12.109 WThe output power is thus obtained as
P_{m} =\left(1-s\right) \left(P_{gf}-P_{gb}\right)
=0.95\left(290.46-12.109\right) =264.43 W
As we have a four-pole machine, we get
n_{s}=\frac{120\left(60\right) }{4} =1800 r/min\omega _{s} =\frac{2\pi n_{s}}{60} =188.5 rad/s
The output torque is therefore obtained as
T_{m}=\frac{1}{\omega _{s}} \left(P_{gf}-P_{gb}\right)
=\frac{290.46-12.109}{188.5} = 1.4767 N.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_{s} } =\left|I_{1} \right| ^{2} R_{1} = \left(4.098\right) ^{2} \left(1.5\right) =25.193 WThe forward rotor losses are
P_{l_{rf} } =sP_{gf} = 0.05 \left(290.46\right) =14.523 WThe backward rotor losses are
P_{l_{rb} }=\left(2-s\right) P_{gb}=1.95\left(12.109\right) =23.613 WThe sum of the losses is
P_{l}=25.193+14.523+23.613=63.329 WThe power output and losses should match the power input
P_{m}+P_{l}=264.43+ 63.33 = 327.76 Wwhich is indeed the case.