Question 1 |
When the winding c-d of the singlephase, 50 \mathrm{Hz}, two winding transformer is supplied from an AC current source of frequency 50 \mathrm{~Hz}, the rated voltage of 200 \mathrm{~V} (rms), 50 \mathrm{~Hz}. is obtained at the open-circuited terminals a-b. The cross sectional area of the core 5000 \mathrm{~mm}^{2} and the average core length traversed by the mutual flux is 500 \mathrm{~mm}. The maximum allowable flux density in the core is B_{\max }=1 \mathrm{~Wb} / \mathrm{m}^{2} and the relative permeability of the core material is 5000. The leakage impedance of the winding a-b and winding c-d at 50 \mathrm{~Hz} are (5+j 100 \pi \times 0.16) \Omega and (11.5+j 100 \pi \times 0.36) \Omega, respectively. Considering the magnetizing characteristics to be linear and neglecting core loss, the self-inductance of the winding a-b in millihenry is ____ (Round off to 1 decimal place).
1256.3 | |
4152.4 | |
2218.4 | |
6523.8 |
Question 1 Explanation:
\begin{aligned}
\mathrm{A} & =5000 \mathrm{~mm}^{2} \\
\mathrm{I} & =500 \mathrm{~mm} \\
\mathrm{~B}_{\max } & =1 \mathrm{~Wb} / \mathrm{m}^{2} \\
\mathrm{~V}_{\mathrm{ab}} & =200 \mathrm{~V} \\
\sqrt{2} \pi \mathrm{fN}_{1} \phi_{\max } & =200 \\
\sqrt{2 \pi} \times 50 \times \mathrm{N}_{1} \times 1 & \times 5000 \times 10^{-6}=200 \quad[\because \phi=\beta \mathrm{A}] \\
\Rightarrow \quad \mathrm{N}_{1} & =181
\end{aligned}
Now, mutual inductance,
\begin{aligned} M&=\frac{N^{2}}{I / \mu A} \\ & =\frac{4 \pi \times 10^{-7} \times 5000 \times 5000 \times 10^{-6} \times(181)^{2}}{0.5} \\ & =2.05843 \mathrm{H} \end{aligned}
Therefore, self inductance of winding.
\begin{aligned} L & =\text { Mutual + leakage } \\ & =2.05843+0.16 \\ & =2.21843 \mathrm{H} \text { or } 2218.43 \mathrm{mH} \end{aligned}
Now, mutual inductance,
\begin{aligned} M&=\frac{N^{2}}{I / \mu A} \\ & =\frac{4 \pi \times 10^{-7} \times 5000 \times 5000 \times 10^{-6} \times(181)^{2}}{0.5} \\ & =2.05843 \mathrm{H} \end{aligned}
Therefore, self inductance of winding.
\begin{aligned} L & =\text { Mutual + leakage } \\ & =2.05843+0.16 \\ & =2.21843 \mathrm{H} \text { or } 2218.43 \mathrm{mH} \end{aligned}
Question 2 |
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)
542 | |
632 | |
760 | |
852 |
Question 2 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}
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 3 |
A three-phase synchronous motor with synchronous impedance of 0.1+j0.3 per unit per phase has a static stability limit of 2.5 per unit. The corresponding excitation voltage in per unit is ____ (Round off to 2 decimal places).
0.82 | |
1.59 | |
2.25 | |
3.64 |
Question 3 Explanation:
Developed power for motor,
P_{d e v}=\frac{V_{f}}{Z_{s}} \cos \left(\theta_{s}-\delta\right)-\frac{E_{F}^{2}}{2} \cos \theta_{s}
At max. power developed,
\delta=\theta_{\mathrm{S}}
and this is decide steady state stability of motor.
\therefore \mathrm{SSSL}, \mathrm{P}_{\text {dev }(max)}=\frac{V E_{F}}{\mathrm{Z}_{\mathrm{S}}}-\frac{\mathrm{E}_{\mathrm{F}}^{2}}{\mathrm{Z}_{\mathrm{S}}} \cos \theta_{\mathrm{s}}
where, V=1.0 \mathrm{pu}
\mathrm{Z}_{\mathrm{S}}=0.1+\mathrm{j} 0.3=0.316 \angle 71.6^{\circ} \mathrm{pu}
and \mathrm{SSSL}=2.5
\therefore \quad 2.5=\frac{1 \times E_{F}}{0.316}-\frac{E_{F}^{2}}{0.316} \cos 71.56^{\circ}
1.001 \mathrm{E}_{F}^{2}-3.16 \mathrm{E}_{F}+2.5=0
\Rightarrow \quad \mathrm{E}_{\mathrm{F}}=1.595 \mathrm{pu}
P_{d e v}=\frac{V_{f}}{Z_{s}} \cos \left(\theta_{s}-\delta\right)-\frac{E_{F}^{2}}{2} \cos \theta_{s}
At max. power developed,
\delta=\theta_{\mathrm{S}}
and this is decide steady state stability of motor.
\therefore \mathrm{SSSL}, \mathrm{P}_{\text {dev }(max)}=\frac{V E_{F}}{\mathrm{Z}_{\mathrm{S}}}-\frac{\mathrm{E}_{\mathrm{F}}^{2}}{\mathrm{Z}_{\mathrm{S}}} \cos \theta_{\mathrm{s}}
where, V=1.0 \mathrm{pu}
\mathrm{Z}_{\mathrm{S}}=0.1+\mathrm{j} 0.3=0.316 \angle 71.6^{\circ} \mathrm{pu}
and \mathrm{SSSL}=2.5
\therefore \quad 2.5=\frac{1 \times E_{F}}{0.316}-\frac{E_{F}^{2}}{0.316} \cos 71.56^{\circ}
1.001 \mathrm{E}_{F}^{2}-3.16 \mathrm{E}_{F}+2.5=0
\Rightarrow \quad \mathrm{E}_{\mathrm{F}}=1.595 \mathrm{pu}
Question 4 |
A separately excited DC motor rated 400 \mathrm{~V}, 15A, 1500 RPM drives a constant torque load at rated speed operating from 400 \mathrm{~V} DC supply drawing rated current. The armature resistance is 1.2 \Omega. If the supply voltage drops by 10 \% with field current unaltered then the resultant speed of the motor in RPM is ___ (Round off to the nearest integer).
1343 | |
2541 | |
1145 | |
1852 |
Question 4 Explanation:
Seperately excited motor :
Back emf, \mathrm{E}_{\mathrm{b}_{1}}=400-15 \times 1.2=382 \mathrm{~V}
Given: \quad \tau= constant
\because \quad \tau \propto \phi I_{a}
where \phi \rightarrow constant
\therefore \mathrm{I}_{\mathrm{a}} \rightarrow consant
If supply voltage is drops by 10 \%, then new value of supply voltage =0.9 \times 400=360 \mathrm{~V}
Now, back emf, \mathrm{E}_{\mathrm{b}_{2}}=360-15 \times 1.2=342 \mathrm{~V}
We have,
\begin{aligned} \mathrm{E}_{\mathrm{b}} & =\phi \omega_{\mathrm{m}} \\ \therefore \quad \frac{\mathrm{E}_{\mathrm{b}_{2}}}{\mathrm{E}_{\mathrm{b}_{1}}} & =\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \\ \Rightarrow \quad \frac{342}{382} & =\frac{\mathrm{N}_{2}}{1500} \\ \Rightarrow \quad \mathrm{N}_{2} & =342.93 \mathrm{rpm} \\ &\approx 1343 \mathrm{rpm} \end{aligned}
Back emf, \mathrm{E}_{\mathrm{b}_{1}}=400-15 \times 1.2=382 \mathrm{~V}
Given: \quad \tau= constant
\because \quad \tau \propto \phi I_{a}
where \phi \rightarrow constant
\therefore \mathrm{I}_{\mathrm{a}} \rightarrow consant
If supply voltage is drops by 10 \%, then new value of supply voltage =0.9 \times 400=360 \mathrm{~V}
Now, back emf, \mathrm{E}_{\mathrm{b}_{2}}=360-15 \times 1.2=342 \mathrm{~V}
We have,
\begin{aligned} \mathrm{E}_{\mathrm{b}} & =\phi \omega_{\mathrm{m}} \\ \therefore \quad \frac{\mathrm{E}_{\mathrm{b}_{2}}}{\mathrm{E}_{\mathrm{b}_{1}}} & =\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \\ \Rightarrow \quad \frac{342}{382} & =\frac{\mathrm{N}_{2}}{1500} \\ \Rightarrow \quad \mathrm{N}_{2} & =342.93 \mathrm{rpm} \\ &\approx 1343 \mathrm{rpm} \end{aligned}
Question 5 |
The four stator conductor \left(A, A^{\prime}, B\right. and \left.B^{\prime}\right) of a rotating machine are carrying DC currents of the same value, the directions of which are shown in figure (i). The rotor coils a-a^{\prime} and b-b^{\prime} are formed by connecting the back ends of conductor 'a' and 'a^{\prime}' and 'b' and 'b^{\prime}', respectively, as shown in figure (ii). The e.m.f. induced in coil a-a^{\prime} and coil b-b^{\prime} are denoted by E_{a-a^{\prime}} and E_{b-b^{\prime}}; respectively. If the rotor is rotated at uniform angular speed \omega \mathrm{rad} / \mathrm{s} in the clockwise direction then which of the following correctly describes the \mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}} and \mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}} ?
E_{a-a^{\prime}} and E_{b-b^{\prime}} have finite magnitudes and are in the same phase | |
E_{a-a^{\prime}} and E_{b-b^{\prime}} have finite magnitudes with \mathrm{E}_{\mathrm{b}-b^{\prime}} leading \mathrm{E}_{\mathrm{a-a}} | |
\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}} and \mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}} have finite magnitudes with E_{a-a^{\prime}} leading E_{b-b^{\prime}} | |
\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}=\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}=0 |
Question 5 Explanation:
Since, a-a' and b-b' are placed at GNA (Geometric Neutral Axis) therefore,
\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}=\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}=0
\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}=\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}=0
There are 5 questions to complete.