Question 1 |
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).
0.24 | |
0.48 | |
0.63 | |
0.97 |
Question 1 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
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 2 |
A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a
synchronous reactance of 1 ohm per phase with negligible armature resistance. The
shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in
the motor is 2 kW. The magnitude of the per phase excitation emf of the motor, in
volts, is __________. (round off to nearest integer).
258 | |
324 | |
128 | |
245 |
Question 2 Explanation:
Power input = 10 + 2 = 12 kW
\begin{aligned} P&=\sqrt{3}VI \cos\phi \\ I_a&=\frac{10 \times 10^3}{\sqrt{3} \times 400 \times 0.8}\\ &=21.65\angle 36.86^{\circ}A \end{aligned}
We have, Emf equation for motor
\begin{aligned} \bar{E_{ph}}&=\bar{V_{ph}}-\bar{I_{a}}\bar{Z_{s}}\\ &=\frac{400}{\sqrt{3}}-21.65\angle 36.83^{\circ}\times 1\angle 90^{\circ}\\ &=244.542\angle -4.062^{\circ}V \end{aligned}
\begin{aligned} P&=\sqrt{3}VI \cos\phi \\ I_a&=\frac{10 \times 10^3}{\sqrt{3} \times 400 \times 0.8}\\ &=21.65\angle 36.86^{\circ}A \end{aligned}
We have, Emf equation for motor
\begin{aligned} \bar{E_{ph}}&=\bar{V_{ph}}-\bar{I_{a}}\bar{Z_{s}}\\ &=\frac{400}{\sqrt{3}}-21.65\angle 36.83^{\circ}\times 1\angle 90^{\circ}\\ &=244.542\angle -4.062^{\circ}V \end{aligned}
Question 3 |
A 4-pole induction motor with inertia of 0.1 kg-m^2 drives a constant load torque of
2 Nm. The speed of the motor is increased linearly from 1000 rpm to 1500 rpm in
4 seconds as shown in the figure below. Neglect losses in the motor. The energy, in
joules, consumed by the motor during the speed change is ____________. (round
off to nearest integer)
1732 | |
2534 | |
1245 | |
3251 |
Question 3 Explanation:
We have
\begin{aligned} J\frac{d\omega }{dt}&=\tau _e-\tau _L\\ J\omega \frac{d\omega }{dt}&=P_e-P_L \;\;(\because P=\tau \omega )\\ \therefore P_e&=J\omega \frac{d\omega }{dt}+P_L\\ E&=\int P_edt\\ R&=J\int_{1000}^{1500}\omega d\omega +\int_{1000}^{1500}P_Ldt\\ &=\frac{J}{2}\left ( \frac{2 \pi}{60} \right )^2[1500^2-1000^2]+\frac{2 \pi \tau }{60}\int_{1000}^{1500}Ndt\\ &=\frac{0.1}{2}\left ( \frac{2 \pi}{60} \right )^2\times 125 \times 10^4+\frac{2 \pi}{60} \times 2\times [1000 \times 4+\frac{1}{2} \times 500 \times 4]\\ &=1732.586J \end{aligned}
\begin{aligned} J\frac{d\omega }{dt}&=\tau _e-\tau _L\\ J\omega \frac{d\omega }{dt}&=P_e-P_L \;\;(\because P=\tau \omega )\\ \therefore P_e&=J\omega \frac{d\omega }{dt}+P_L\\ E&=\int P_edt\\ R&=J\int_{1000}^{1500}\omega d\omega +\int_{1000}^{1500}P_Ldt\\ &=\frac{J}{2}\left ( \frac{2 \pi}{60} \right )^2[1500^2-1000^2]+\frac{2 \pi \tau }{60}\int_{1000}^{1500}Ndt\\ &=\frac{0.1}{2}\left ( \frac{2 \pi}{60} \right )^2\times 125 \times 10^4+\frac{2 \pi}{60} \times 2\times [1000 \times 4+\frac{1}{2} \times 500 \times 4]\\ &=1732.586J \end{aligned}
Question 4 |
A 280 V, separately excited DC motor with armature resistance of 1\Omega and constant
field excitation drives a load. The load torque is proportional to the speed. The
motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect
frictional losses in the motor. The speed, in rpm, at which the motor will run, if an
additional resistance of value 10\Omega is connected in series with the armature, is
_______. (round off to nearest integer)
482 | |
254 | |
365 | |
625 |
Question 4 Explanation:
Back emf,
E_{b1}=280-30 \times 1 =250V
When additional 10\Omega resistance connected in series with armature.
E_{b2}=280-(10+1) I_{a_2} =280-11I_{a_2}
Given: \tau \propto N
We know, \tau \propto \phi I_a \text{ (Here, }\phi \rightarrow constant)
So,
\begin{aligned} \tau &\propto I_a \\ \therefore \; N &\propto I_a \\ or \; \frac{N_2}{N_1}&=\frac{I_{a_2}}{I_{a_1}}\\ I_{a_2}&=N_2 \times \frac{30}{1000}\\ \therefore \; E_{b_2}&=280-\left ( 11 \times \frac{30}{1000} \right )N_2\\ &\text{We also have,}\\ E_b &\propto\phi \omega \\ \therefore \frac{E_{b_2}}{E_{b_1}}&=\frac{N_2}{N_1}\\ &\frac{280-\left ( 11 \times \frac{30}{1000} \right )N_2}{250}=\frac{N_2}{1000}\\ \Rightarrow N_2&=482.758rpm \end{aligned}
E_{b1}=280-30 \times 1 =250V
When additional 10\Omega resistance connected in series with armature.
E_{b2}=280-(10+1) I_{a_2} =280-11I_{a_2}
Given: \tau \propto N
We know, \tau \propto \phi I_a \text{ (Here, }\phi \rightarrow constant)
So,
\begin{aligned} \tau &\propto I_a \\ \therefore \; N &\propto I_a \\ or \; \frac{N_2}{N_1}&=\frac{I_{a_2}}{I_{a_1}}\\ I_{a_2}&=N_2 \times \frac{30}{1000}\\ \therefore \; E_{b_2}&=280-\left ( 11 \times \frac{30}{1000} \right )N_2\\ &\text{We also have,}\\ E_b &\propto\phi \omega \\ \therefore \frac{E_{b_2}}{E_{b_1}}&=\frac{N_2}{N_1}\\ &\frac{280-\left ( 11 \times \frac{30}{1000} \right )N_2}{250}=\frac{N_2}{1000}\\ \Rightarrow N_2&=482.758rpm \end{aligned}
Question 5 |
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)
254 | |
365 | |
542 | |
585 |
Question 5 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
Slip, s=\frac{1}{40}=0.025
Now, speed
N=N_s(1-s)=\frac{120 \times 40}{8}(1-0.025)=585rpm
Question 6 |
The type of single-phase induction motor, expected to have the maximum power
factor during steady state running condition, is
split phase (resistance start). | |
shaded pole. | |
capacitor start. | |
capacitor start, capacitor run. |
Question 7 |
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.
0.25 | |
0.44 | |
0.52 | |
0.68 |
Question 7 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}
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 8 |
A belt-driven \text{DC}
shunt generator running at 300\: \text{RPM}
delivers 100 \:\text{kW}
to a 200 \:V\: \text{DC}
grid. It continues to run as a motor when the belt breaks, taking 10 \:\text{kW}
from the \text{DC}
grid. The armature resistance is 0.025 \Omega, field resistance is 50\:\Omega, and brush drop is 2 \:V. Ignoring armature reaction, the speed of the motor is _____ \text{RPM}. (Round off to 2 decimal places.)
275.18 | |
254.12 | |
362.24 | |
148.44 |
Question 8 Explanation:
\begin{aligned} I_{L} &=\frac{100 \times 10^{3}}{200}=500 \mathrm{~A} \\ I_{S h} &=\frac{200}{50}=4 \mathrm{~A} \\ I_{a} &=504 \mathrm{~A} \\ E_{q} &=V+I_{a} R_{a}+\text { Brush drop } \\ &=200+504(0.025)+2 \mathrm{~V} \\ &=214.6 \mathrm{~V}\\ \text{In motoring case:} V I&=10 \mathrm{~kW}, V=200 \mathrm{~V}\\ \therefore \qquad \qquad I &=\frac{10000}{200}=50 A \\ I_{f} &=4 \mathrm{~A}, I_{a}=I_{L}-I_{f} \\ &=46 \mathrm{~A} \\ E_{b} &=V-I_{a} R_{a}-\text { Brush Drop } \\ &=200-46(0.025)-2=196.85 \mathrm{~V} \\ \frac{N_{m}}{N_{g}} &=\frac{E_{b}}{E_{g}} \\ N_{m} &=\frac{E_{b}}{E_{g}} \times N_{g} \\ \therefore \qquad \qquad N_{m} &=\frac{196.85}{214.6} \times 300=275.18 \mathrm{rpm} \end{aligned}
Question 9 |
In a single-phase transformer, the total iron loss is 2500\:W
at nominal voltage of 440\:V and frequency 50\:Hz. The total iron loss is 850\:W at 220\:V and 25\:Hz. Then, at nominal voltage and frequency, the hysteresis loss and eddy current loss respectively are
1600\:W and 900\:W
| |
900\:W and 1600\:W
| |
250\:W and 600\:W
| |
600\:W and 250\:W
|
Question 9 Explanation:
\begin{aligned} W_{i 1} &=2500 \mathrm{~W} \text { at } 440 \mathrm{~V}, 50 \mathrm{~Hz} \\ W_{i 2} &=850 \mathrm{~W} \text { at } 220 \mathrm{~V}, 25 \mathrm{~Hz} \\ W_{i 3} &=R_{e_{3}}+P_{H_{3}} \text { at } 440 \mathrm{~V}, 50 \mathrm{~Hz} \\ R_{\theta_{3}} &=?, \quad P_{h_{3}}=?, \quad \frac{v}{f}=\mathrm{constant} \\ \Rightarrow \qquad \qquad\quad 2500 &=A f+B f^{2} &\left\{\frac{400}{50}=\frac{220}{25}=\text { Constant }\right\} \\ \text { or, }\qquad \qquad \quad \frac{2500}{f} &=A+B f \\ \text{or},\qquad \qquad \frac{2500}{50}&=A+B(50) &\ldots(i)\\ \text{and}\qquad \qquad \frac{850}{25}&=\mathrm{A}+\mathrm{B}(25)&\ldots(ii) \end{aligned}
Solving (i) and (ii), we get
\begin{aligned} 25 B &=\frac{2500}{50}-\frac{850}{25}=\frac{2500-1700}{50} \\ &=\frac{800}{50}=16 \\ B &=\frac{16}{25} \\ \text{and from (i)},\qquad A &=50-\frac{16}{25} \times 50=50-32=18\\ \text{So, at} 50 \mathrm{~Hz}\\ P_{h}&=A f=18 \times 50=900 \mathrm{~W} \\ P_{e}&=B f=\left(\frac{16}{25}\right) \times(50)^{2}=1600 \mathrm{~W} \end{aligned}
Solving (i) and (ii), we get
\begin{aligned} 25 B &=\frac{2500}{50}-\frac{850}{25}=\frac{2500-1700}{50} \\ &=\frac{800}{50}=16 \\ B &=\frac{16}{25} \\ \text{and from (i)},\qquad A &=50-\frac{16}{25} \times 50=50-32=18\\ \text{So, at} 50 \mathrm{~Hz}\\ P_{h}&=A f=18 \times 50=900 \mathrm{~W} \\ P_{e}&=B f=\left(\frac{16}{25}\right) \times(50)^{2}=1600 \mathrm{~W} \end{aligned}
Question 10 |
An alternator with internal voltage of 1\angle \delta _{1}\text{p.u}
and synchronous reactance of \text{0.4 p.u} is connected by a transmission line of reactance \text{0.1 p.u} to a synchronous motor having synchronous reactance \text{0.35 p.u} and internal voltage of 0.85\angle \delta _{2}\text{p.u}.
If the real power supplied by the alternator is \text{0.866 p.u}, then \left( \delta _{1} -\delta _{2}\right) is _________ degrees. (Round off to 2 decimal places.)
(Machines are of non-salient type. Neglect resistances.)
If the real power supplied by the alternator is \text{0.866 p.u}, then \left( \delta _{1} -\delta _{2}\right) is _________ degrees. (Round off to 2 decimal places.)
(Machines are of non-salient type. Neglect resistances.)
60 | |
25.35 | |
68.6 | |
88 |
Question 10 Explanation:
Given real power in (p.u.), P=0.866
\begin{aligned} P &=\frac{E V \sin \left(\delta_{1}-\delta_{2}\right)}{X_{\mathrm{eq}}}=\left|\frac{1 \times 0.85}{0.4+0.1+0.35}\right| \sin \left(\delta_{1}-\delta_{2}\right) \\ \left(\delta_{1}-\delta_{2}\right) &=60^{\circ} \end{aligned}
There are 10 questions to complete.