\[\begin{aligned}
\text { output power }&=3986-150=3836 \mathrm{~W} \\
\text { stator core loss }&=50 \mathrm{~W} \\
\text { stator Cu loss } &=3 I_{1}^{2} R_{1}=3 \times
12.7^{2} \times 0.45=218 \mathrm{~W} \\
\text { Rotor Cu loss } & =3 I_{2}^{\prime 2}
R_{2}^{\prime}=s P_{2}=0.04 \times 4152=166 \mathrm{~W}\\
\text { Rotational losses }&=150 \mathrm{~W} \\
\text { Total loss }&=50+218+166+150=584 \mathrm{~W} \\
\eta&=3836 /(3836+584)=0.868 \text { or } 86.8 \%
\end{aligned}\]
Problem-2
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%
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:
Find
maximum torque developed
slip at maximum torque and
pf at a slip of 5%.
Use approximate equivalent circuit.
Solution-3
\(P_2\) 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,
\[\begin{aligned}
\frac{R_{2}^{\prime}}{s_{m}}
&=\sqrt{R_{1}^{2}+\left(X_{1}+X_{2}^{\prime}\right)^{2}} \\
\text { or } \quad s_{m}
&=\frac{R_{2}^{\prime}}{\sqrt{R_{1}^{2}+\left(X_{1}+X_{2}^{\prime}\right)^{2}}}=\frac{0.6}{\sqrt{0.4^{2}+2^{2}}}=0.29
\text { or } 29 \%
\end{aligned}\]
