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.