3-Phase Induction Motors: Solved Problems

Problem-1

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 50 Hz. Calculate

The speed at which the magnetic field of the stator is rotating
The speed of the rotor when the slip is 0.04
The frequency of the rotor currents when the slip is 0.03
The frequency of the rotor currents at standstill

Solution-1

\[N_{s}=120 \mathrm{f} / P=120 \times 50 / 4=1500 \mathrm{rp} \mathrm{m}\]
Stator field revolves at synchronous speed, given by
rotor speed, \(N=N_{s}(1-s)=1500(1-0.04)=1440\) r.p.m.
\[f^{\prime}=s f=0.03 \times 50=1.5 \mathrm{r} . \mathrm{p} . \mathrm{s}=90 \mathrm{r} . \mathrm{p} . \mathrm{m}\]
frequency of rotor current,
\[s=1, \quad f^{\prime}=s f=1 \times f=f=50 \mathrm{~Hz}\]
At standstill,

Problem-2

A 3-phase induction motor having a star-connected rotor has an induced emf of 80 volts between slip rings at standstill on open circuit. The rotor has a resistance and reactance per phase of 1 \(\Omega\) and 4 \(\Omega\) respectively. Calculate current/phase and power factor when

slip-rings are short-circuited
slip-rings are connected to a star-connected rheostat of 3 \(\Omega\) per phase.

Solution-2

Standstill e.m.f./rotor phase \(=80 / \sqrt{3}=46.2 \mathrm{~V}\)
\[\begin{aligned} \text{Rotor impedance/phase} &=\sqrt{\left(1^{2}+4^{2}\right)}=4.12 \Omega \\ \pause \text{Rotor current/phase} & =46.2 / 4.12=11.2 \mathrm{~A} \\ \pause \text{Power factor} & =\cos \phi=1 / 4.12=0.243 \end{aligned}\]
slip rings are short-circuited:
As p.f. is low, the starting torque is also low.

\[\begin{aligned} \text{Rotor resistance/phase} & =3+1=4 \Omega\\ \pause \text { Rotor impedance/phase } &=\sqrt{\left(4^{2}+4^{2}\right)}=5.66 \Omega \\ \pause \therefore \text { Rotor current/phase } &=46.2 / 5.66=8.16 \mathrm{~A} \\ \pause \cos \phi & =4 / 5.66=0.707 . \end{aligned}\]
slip-rings connected to rheostat

\(\checkmark\) Hence, the starting torque is increased due to the improvement in the power factor. It will also be noted that improvement in p.f. is much more than the decrease in current due to increased impedance.

Problem-3

A 3-phase, 400-V, star-connected induction motor has a star-connected rotor with a stator to rotor turn ratio of 6.5. The rotor resistance and standstill reactance per phase are 0.05 \(\Omega\) and 0.25 \(\Omega\) respectively. What should be the value of external resistance per phase to be inserted in the rotor circuit to obtain maximum torque at starting and what will be the rotor starting current with this resistance?

Solution-3

Transformation ratio = \(\dfrac{\text{ Rotor turns/phase }}{\text{ Stator turns/phase }} = 1/6.5\)
Standstill rotor emf/phase \(E_2 = \dfrac{400}{\sqrt{3}} \times \dfrac{1}{6.5} = 35.5~\text{volts}\)
For max. starting torque \(R_2 = X_2 = 0.25~\Omega\)
external resistance/phase required \(= 0.25-0.005=0.2~\Omega\)
Rotor impedance/phase \(= \sqrt{0.25^2+0.25^2} = 0.3535~\Omega\)
Rotor current/phase \(I_2 = 35.5/0.3535 = 100~\mathrm{A (approx.)}\)

Problem-4

A 1100-V, 50-Hz delta connected induction motor has a star-connected slip-ring rotor with a phase transformation ratio of 3.8. The rotor resistance and standstill leakage reactance are 0.012 ohm and 0.25 ohm per phase respectively. Neglecting stator impedance and magnetizing current determine

The rotor current at start with slip-rings shorted
The rotor power factor at start with slip rings shorted
The rotor current at 4% slip with slip rings shorted
The rotor power factor at 4% slip with slip-rings shorted
The external rotor resistance per phase required to obtain a starting current of 100 A in the stator supply lines.

Solution-4

It should be noted that in a \(\Delta/ Y\) connection, primary phase voltage is the same as the line voltage.
The rotor phase voltage can be found by using the phase transformation ratio of \(K=1 / 3.8\)
Rotor phase voltage at standstill \(=1100 \times 1 / 3.8=289.5 \mathrm{~V}\)
Rotor impedence/phase \(=\sqrt{0.012^{2}+0.25^{2}}=0.2503 \Omega\)
Rotor phase current at start \(=289.5 / 0.2503=1157 \mathrm{~A}\)
p . f . \(=R_{2} / Z_{2}=0.012 / 0.2503=0.048 \text { lag }\)

\[\begin{aligned} X_{r} & =s X_{2}=0.04 \times 0.25=0.01 \Omega \\ \pause Z_{r}& =\sqrt{0.012^{2}+0.01^{2}}=0.0156 \Omega\\ \pause E_{r}& =s E_{2}=0.04 \times 289.5=11.58 \mathrm{~V} \\ \pause I_{2} & =11.58 / 0.0156=742.3 \mathrm{~A} \end{aligned}\]
at 4% slip
p.f. \(=0.01200 .0156=0.77\)
\[\begin{aligned} I_{2}&=I_{1} / K=100 \times 3.8=380 \mathrm{~A} \\ \pause &E_{2} \text { at standstill } =289.5 \mathrm{~V} \\ \pause Z_{2} & =289.5 / 380=0.7618 \Omega \\ \pause R_{2} & =\sqrt{Z_{2}^{2}-X_{2}^{2}}=\sqrt{0.7618^{2}-0.25^{2}}=0.7196 \Omega \end{aligned}\]
External resistance reqd \(/\) phase \(=0.7196-0.012=0.707 \Omega\)