# Three Phase Induction Machines

 Question 1
A three phase $415 \mathrm{~V}, 50 \mathrm{~Hz}, 6-pole, 960 RPM, 4 HP$ squirrel cage induction motor drives a constant torque load at rated speed operating from rated supply and delivering rated output. If the supply voltage and frequency are reduced by $20 \%$ the resultant speed of the motor in RPM (neglecting the stator leakage impednace and rotational losses) is ____ (Round off to the nearest integer)
 A 542 B 632 C 760 D 852
GATE EE 2023   Electrical Machines
Question 1 Explanation:
Given : Torque is constant.
We have, $\quad \tau \propto \frac{s V^{2}}{f} \quad ...(1)$
Synchronous speed, $N_{S}=\frac{120 f}{P}$
$=\frac{120 \times 50}{6}=1000 \mathrm{rpm}$

$\text { Slip, } S=\frac{1000-960}{1000}=0.04$

Now, from eqn. (1), we get
\begin{aligned} \frac{0.04 \times \mathrm{V}^{2}}{\mathrm{~F}} & =\frac{\mathrm{S}_{\text {new }} \times(0.8)^{2} \mathrm{~V}}{0.8 \mathrm{~F}} \\ \Rightarrow \quad \mathrm{S}_{\text {new }} & =0.05 \end{aligned}

Synchronous speed at reduced frequency i.e. $50 \times 0.8=40 \mathrm{~Hz}$.
$\mathrm{N}_{\mathrm{s}(\mathrm{new})}=\frac{120 \times 40}{6}=800 \mathrm{rpm}$
$\therefore$ Motor speed,

\begin{aligned} \mathrm{N}&=\left(1-\mathrm{S}_{\text {new }}\right) \mathrm{N}_{\mathrm{s} \text { (new) }} & =(1-0.05) \times 800 \\ & =760 \mathrm{rpm} \end{aligned}
 Question 2
A 3-phase, 415 V, 4-pole, 50 Hz induction motor draws 5 times the rated current at rated voltage at starting. It is required to bring down the starting current from the supply to 2 times of the rated current using a 3-phase autotransformer. If the magnetizing impedance of the induction motor and no load current of the autotransformer is neglected, then the transformation ratio of the autotransformer is given by _______. (round off to two decimal places).
 A 0.24 B 0.48 C 0.63 D 0.97
GATE EE 2022   Electrical Machines
Question 2 Explanation:
We have $I_L=x^2I_{sc}$
Here, $I_L=2I_f$ and $I_{sc}=5I_f$
From the above eqiations.
$2I_f=x^25I_f\Rightarrow x=\sqrt{\frac{2}{5}}=0.6324$

 Question 3
The frequencies of the stator and rotor currents flowing in a three-phase 8-pole induction motor are 40 Hz and 1 Hz, respectively. The motor speed, in rpm, is _______. (round off to nearest integer)
 A 254 B 365 C 542 D 585
GATE EE 2022   Electrical Machines
Question 3 Explanation:
We have, $f_r=sf_s$
Slip, $s=\frac{1}{40}=0.025$
Now, speed
$N=N_s(1-s)=\frac{120 \times 40}{8}(1-0.025)=585rpm$
 Question 4
An 8-pole, $50 \:Hz$, three-phase, slip-ring induction motor has an effective rotor resistance of $0.08\:\Omega$ per phase. Its speed at maximum torque is $650 \text{ RPM}$. The additional resistance per phase that must be inserted in the rotor to achieve maximum torque at start is _____________ $\Omega$. (Round off to 2 decimal places.) Neglect magnetizing current and stator leakage impedance. Consider equivalent circuit parameters referred to stator.
 A 0.25 B 0.44 C 0.52 D 0.68
GATE EE 2021   Electrical Machines
Question 4 Explanation:
\begin{aligned} N_{\max } &=650 \mathrm{rpm}, P=8,50 \mathrm{~Hz} \\ s_{m} &=\frac{750-650}{750}=0.1333 \\ s_{m} &=\frac{R_{2}}{X_{2}} \\ \therefore\qquad \qquad X_{2} &=\frac{R_{2}}{s_{m}}=\frac{0.08}{0.133}=0.601 \Omega \\ R_{2} &=0.08 \Omega, X_{2}=0.601 \Omega \end{aligned}
Condition for maximum $T_{\mathrm{s}}$
\begin{aligned} \Rightarrow \qquad \qquad R_{2} & =X_{2} \\ \therefore \qquad \qquad R_{2}+ R_{\text {ext }} & =X_{2} \\ \therefore \qquad \qquad R_{\text {ext }} & =0.601-0.08=0.521 \Omega \end{aligned}
 Question 5
The power input to a $\text{500 V, 50 Hz, 6-pole, 3-phase}$ induction motor running at $975 \text{ RPM}$ is $40 \text{ kW}$ . The total stator losses are $1 \text{ kW}$. If the total friction and windage losses are $2.025\text{~ kW}$, then the efficiency is _____________%.
 A 85 B 95 C 90 D 98
GATE EE 2021   Electrical Machines
Question 5 Explanation:
\begin{aligned} P_{i / p} &=40 \mathrm{~kW}, \quad \text { stator loss }=1 \mathrm{~kW}, \quad \mathrm{~F} \text { and } W=2.025 \mathrm{~kW} \\ \text { Stator } \mathrm{o} / \mathrm{p} &=40-1=39 \mathrm{~kW} \\ \qquad \text { Slip } &=\frac{1000-975}{1000}=0.025 \\ \text { Rotor } \mathrm{o} / \mathrm{p} &=\text { Rotor } \mathrm{i} / \mathrm{p} \times(1-\mathrm{s}) \\ &=39(1-0.025)=38.025 \mathrm{~kW} \\ \text { Motor } \mathrm{o} / \mathrm{p} &=38.025-2.025=36 \mathrm{~kW}\\ \eta&=\frac{\text { Motor output }}{\text { Motor input }}=\frac{36}{40} \times 100=90 \% \end{aligned}

There are 5 questions to complete.