A 220 V, 3-phase, 4-pole, 50-Hz, Y-connected induction motor is rated 3.73 kW. The equivalent circuit parameters are:
| \(R_{1}=0.45~\Omega\) | \(X_{1}=0.8~\Omega\) | \(R_{2}^{\prime}=0.4~\Omega\) |
| \(X_{2}^{\prime}=0.8~\Omega\) | \(B_{0}=-1/30~\mbox{mho}\) |
The stator core loss is 50 W and rotational loss is 150 W. For a slip of 0.04, Determine:
input current
power factor
air-gap power
mechanical power
electro-magnetic torque
output power and
efficiency
Since \(R_0\) (or \(G_0\)) is negligible, consider \(B_0\) (or \(X_0\)) only \[\begin{aligned} Z_{A B}&=\frac{j X_{0}\left[\left(R_{2}^{\prime} / s\right)+j X_{2}^{\prime}\right]}{\left(R_{2}^{\prime} / s\right)+j\left(X_{2}^{\prime}+X_{m}\right)}=\frac{j 30(10+j 0.8)}{10+j 30.8}\\ &=8.58+j 3.56=9.29 \angle 22.5^{\circ} \\ Z_{01}&=Z_{1}+Z_{A B}=(0.45+j 0.8)+(8.58+j 3.56)\\ &=9.03+j 4.36=10 \angle 25.8^{\circ} \end{aligned}\]
\[\begin{aligned} V_{p h}&=\frac{220}{\sqrt{3}} \angle 0^{\circ}=127 \angle 0^{\circ}\\ I_{1} &=V_{1} / Z_{01}=127 \angle 0^{\circ} / 10 \angle 25.8^{\circ} \\ &=12.7 \angle-25.8^{\circ} \mathrm{A}\\ \text { p.f. }&=\cos 25.8^{\circ}=0.9\\ \text { air-gap power, } P_{2}&=3 I_{2}^{\prime 2}\left(R_{2}^{\prime} / s\right)=3 I_{1}^{2} R_{A B}\\ &=3 \times 12.7^{2} \times 8.58=4152 \mathrm{~W} \\ P_{m} &=(1-s) P_{2}=0.96 \times 4152=3,986 \mathrm{~W}\\ N_{s}&=1500 \text { r.p.m.. } \\ N&=1500(1-0.04)=1440 \mathrm{r.p.m.} \\ T_{g}&=\frac{P_{m}}{2 \pi N / 60}=9.55 \frac{P_{m}}{N} \mathrm{~N}-\mathrm{m} \\ & = 9.55 \times 3986 / 1440=26.4 \mathrm{~N}-\mathrm{m} \end{aligned}\]
\[\begin{aligned} \text { output power }&=3986-150=3836 \mathrm{~W} \\ \text { stator core loss }&=50 \mathrm{~W} \\ \text { stator Cu loss } &=3 I_{1}^{2} R_{1}=3 \times 12.7^{2} \times 0.45=218 \mathrm{~W} \\ \text { Rotor Cu loss } & =3 I_{2}^{\prime 2} R_{2}^{\prime}=s P_{2}=0.04 \times 4152=166 \mathrm{~W}\\ \text { Rotational losses }&=150 \mathrm{~W} \\ \text { Total loss }&=50+218+166+150=584 \mathrm{~W} \\ \eta&=3836 /(3836+584)=0.868 \text { or } 86.8 \% \end{aligned}\]
A 115 V, 60-Hz, 3-phase, Y-connected, 6-pole induction motor has an equivalent T-circuit consisting of stator impedance of \((0.07+j0.3)~\Omega\) and an equivalent rotor impedance at standstill of \((0.08+j0.3)~\Omega\). Magnetizing branch has G0 = 0.022 mho, B0 = 0.158 mho. Find
Secondary current
primary current
primary p.f.
gross power output
gross torque
input
gross efficiency by using approximate equivalent circuit. Assume a slip of 2%
\[\begin{aligned}
R_{L}^{\prime} &=R_{2}^{\prime}[(1 / s)-1] \\
&=0.88\left(\frac{1}{0.02}-1\right)\\
&=3.92 \Omega / \text { phase }
\end{aligned}\]
The impedance to the right of terminals \(c\) and \(d\) is \[\begin{aligned} Z_{c d} &=R_{01}+R_{L}^{\prime}+j X_{01} \\ &=(0.07+0.08)+3.92+j 0.6 \\ &=4.07+j 0.6 \\ &=4.11 \angle 8.4^{\circ} \Omega / \text { phase } \\ \end{aligned}\]
\[\begin{aligned} V &=115 / \sqrt{3}=66.5 \mathrm{~V}\\ \mathrm{I}_{2}^{\prime}&=\mathrm{I}_{2}\\ &=\frac{66.5}{4.11 \angle 8.4^{\circ}}=16.17 \angle-j 8.4^{\circ}\\ &=16-j 2.36 \mathrm{~A}\\ \mathrm{I}_{0}&=V\left(G_{0}-j B_{0}\right)\\ &=66.5(0.022-10.158)=1.46-j 10.5 \mathrm{~A} \\ P_{g}&=3 I_{2}^{2} R_{L}^{\prime}=3 \times 16.17^{2} \times 3.92=3,075 \mathrm{~W} \\ \end{aligned}\]
\[\begin{aligned} N_{s}&=120 \times 60 / 6=1,200~\mathrm{rpm} \\ N &=(1-s) N_{s}=(1-0.02) \times 1200=1,176 \mathrm{r.p.m} \\ T_{g} &=9.55 \times \frac{P_{m}}{N}=9.55 \times \frac{3075}{1176}=24.97 \mathrm{~N}-\mathrm{m} \\ P_1 &=\sqrt{3} V_{1} I_{1} \cos \phi\\ &=\sqrt{3} \times 115 \times 21.7 \times 0.804=3,450 \mathrm{~W}\\ \eta_{\text{Gross}} &=3,075 \times 100 / 3,450=89.5 \% \end{aligned}\]
The equivalent circuit of a 400 V, 3-phase induction motor with a star-connected winding has the following impedances per phase referred to the stator at standstill: \[\begin{aligned} \text{Stator:} ~& (0.4+j1)~\Omega\\ \text{Rotor:} ~&(0.6+j1)~\Omega \\ \text{Magnetizing branch:} ~&(10+j50)~\Omega. \end{aligned}\]
Find
maximum torque developed
slip at maximum torque and
pf at a slip of 5%.
Use approximate equivalent circuit.
\(P_2\) and hence \(T_m\) developed by rotor would be maximum when there is maximum transfer of power to the resistor \(R_{2}^{\prime} / s\)
It will happen when \(R_{2}^{\prime} / s\) equals the impedance looking back into the supply source. Hence,
\[\begin{aligned} \frac{R_{2}^{\prime}}{s_{m}} &=\sqrt{R_{1}^{2}+\left(X_{1}+X_{2}^{\prime}\right)^{2}} \\ \text { or } \quad s_{m} &=\frac{R_{2}^{\prime}}{\sqrt{R_{1}^{2}+\left(X_{1}+X_{2}^{\prime}\right)^{2}}}=\frac{0.6}{\sqrt{0.4^{2}+2^{2}}}=0.29 \text { or } 29 \% \end{aligned}\]
(i) Maximum value of gross torque developed by rotor \[T_{g \text { max }}=\frac{P_{\text {s, max }}}{2 \pi N_{s} / 60}=\frac{3 I_{2}^{\prime 2} R_{2}^{\prime} / s_{m}}{2 \pi N_{s} / 60} \mathrm{~N}-\mathrm{m}\]
\[\begin{aligned} I_{2}^{\prime} &=\frac{V_{1}}{\sqrt{\left(R_{1}+R_{2}\right)^{2}+\left(X_{1}+X_{2}^{\prime}\right)^{2}}}\\ &=\frac{400 / \sqrt{3}}{\sqrt{(0.4+0.6)^{2}+(1+1)^{2}}}=103.3 \mathrm{~A} \\ T_{g-m a x} &=\frac{3 \times 103.3^{2} \times 1 / 0.29}{2 \pi \times 1500 / 60}\\ &=351 \mathrm{~N}-\mathrm{m} \quad \text { ... assuming } N_{S}=1500 \mathrm{r} . \mathrm{pm} . \end{aligned}\]
(iii) The equivalent circuit for one phase for a slip of \(0.05\) as shown in Fig. (b). \[\begin{aligned} I_{2}^{\prime} &=231 /[(20+0.4)+j 2]=11.2-j 1.1 \\ I_{0} &=231 /(10+j 50)=0.89-j 4.4 \\ I_{1} &=I_{0}+I_{2}^{\prime}=12.09-j 5.5=13.28 \angle-24.4^{\circ} \\ \mathrm{p.f.}&=\cos 24.4^{\circ}=0.91 (\mathrm{lag}) \end{aligned}\]