Question 1 |
A 20 MVA, 11.2 kV, 4-pole, 50 Hz alternator has an inertia constant of
15 MJ/MVA. If the input and output powers of the alternator are 15 MW and
10 MW, respectively, the angular acceleration in mechanical degree/s^2 is
__________. (round off to nearest integer)
25 | |
50 | |
75 | |
100 |
Question 1 Explanation:
We have, swing equation
\frac{2HS}{\omega _s}\frac{d^2\delta }{dt^2}=P_a
Put the values,
\begin{aligned} \frac{2 \times 15 \times 20}{360 \times 50}\frac{d^2\delta }{dt^2}&=15-10\\ \frac{d^2\delta }{dt^2}&=\frac{5 \times 360 \times 50}{2 \times 15 \times 20}\\ &=150 \text{ electric degree per} sec^2\\ \frac{d^2\delta }{dt^2}&=\alpha =\frac{2}{4} \times 150 \text{ Mech. degree/}sec^2\\ &=75\text{ Mech. degree/}sec^2 \end{aligned}
\frac{2HS}{\omega _s}\frac{d^2\delta }{dt^2}=P_a
Put the values,
\begin{aligned} \frac{2 \times 15 \times 20}{360 \times 50}\frac{d^2\delta }{dt^2}&=15-10\\ \frac{d^2\delta }{dt^2}&=\frac{5 \times 360 \times 50}{2 \times 15 \times 20}\\ &=150 \text{ electric degree per} sec^2\\ \frac{d^2\delta }{dt^2}&=\alpha =\frac{2}{4} \times 150 \text{ Mech. degree/}sec^2\\ &=75\text{ Mech. degree/}sec^2 \end{aligned}
Question 2 |
Two generating units rated for 250 MW and 400 MW have governor speed
regulations of 6% and 6.4%, respectively, from no load to full load. Both the
generating units are operating in parallel to share a load of 500 MW. Assuming free
governor action, the load shared in MW, by the 250 MW generating unit is
_________. (round off to nearest integer)
100 | |
150 | |
200 | |
250 |
Question 2 Explanation:
Let no-load frequency is 50 Hz.
Draw the curve :
From the curve,
\begin{aligned} \frac{50-f}{3}&=\frac{P_1}{250} \;\;\;..(1) \\ \frac{50-f}{3.2}&=\frac{P_2}{400} \;\;\;..(2) \end{aligned}
From eq. (1) & (2),
\begin{aligned} \frac{3P_1}{250}&= \frac{3.2P_2}{400} \\ 3P_1&=2P_2\;\;\;...(3) \end{aligned}
Given: P_1+P_2=500
From eq. (3),
P_1+1.5P_1=500
P_1=200MW
Draw the curve :
From the curve,
\begin{aligned} \frac{50-f}{3}&=\frac{P_1}{250} \;\;\;..(1) \\ \frac{50-f}{3.2}&=\frac{P_2}{400} \;\;\;..(2) \end{aligned}
From eq. (1) & (2),
\begin{aligned} \frac{3P_1}{250}&= \frac{3.2P_2}{400} \\ 3P_1&=2P_2\;\;\;...(3) \end{aligned}
Given: P_1+P_2=500
From eq. (3),
P_1+1.5P_1=500
P_1=200MW
Question 3 |
The fuel cost functions in rupees/hour for two 600 MW thermal power plants are
given by
Plant 1: \; C_1=350+6P_1+0.004P_1^2
Plant 2: \; C_2=450+aP_2+0.003P_2^2
where P_1 and P_2 are power generated by plant 1 and plant 2, respectively, in MW and a is constant. The incremental cost of power (\lambda ) is 8 rupees per MWh. The two thermal power plants together meet a total power demand of 550 MW. The optimal generation of plant 1 and plant 2 in MW, respectively, are
Plant 1: \; C_1=350+6P_1+0.004P_1^2
Plant 2: \; C_2=450+aP_2+0.003P_2^2
where P_1 and P_2 are power generated by plant 1 and plant 2, respectively, in MW and a is constant. The incremental cost of power (\lambda ) is 8 rupees per MWh. The two thermal power plants together meet a total power demand of 550 MW. The optimal generation of plant 1 and plant 2 in MW, respectively, are
200, 350 | |
250, 300 | |
325, 225 | |
350, 200 |
Question 3 Explanation:
\frac{dC_1}{dP_1}=IC_1=6+0.008P_1
\frac{dC_2}{dP_2}=IC_2=a+0.006P_2
For optimum generation,
IC_1=IC_2=\lambda
Therefore, 6+0.008P_1=8
P_1=250MW
Given, P_1+P_2=550MW
P_550-250=300MW
\frac{dC_2}{dP_2}=IC_2=a+0.006P_2
For optimum generation,
IC_1=IC_2=\lambda
Therefore, 6+0.008P_1=8
P_1=250MW
Given, P_1+P_2=550MW
P_550-250=300MW
Question 4 |
If only 5% of the supplied power to a cable reaches the output terminal, the power loss
in the cable, in decibels, is _________. (round off to nearest integer)
8 | |
13 | |
17 | |
15 |
Question 4 Explanation:
We have, the power loss (in decibel) in the cable
is given power loss = 95%
Power output as a % of power is 5%
P_L=10 \log \left ( \frac{95}{5} \right )=12.78
Power output as a % of power is 5%
P_L=10 \log \left ( \frac{95}{5} \right )=12.78
Question 5 |
The valid positive, negative and zero sequence impedances (in p.u.), respectively, for
a 220 kV, fully transposed three-phase transmission line, from the given choices are
1.1, 0.15 and 0.08 | |
0.15, 0.15 and 0.35 | |
0.2, 0.2 and 0.2 | |
0.1, 0.3 and 0.1 |
Question 5 Explanation:
We have,
X_0 \gt X_1=X_2
(for 3-\phi transposed transmission line)
X_0 \gt X_1=X_2
(for 3-\phi transposed transmission line)
Question 6 |
The geometric mean radius of a conductor, having four equal strands with each strand
of radius 'r', as shown in the figure below, is
4r | |
1.414r | |
2r | |
1.723r |
Question 6 Explanation:
Redraw the configuration:
\therefore \; GMR=(r' \times 2r\times 2r\times 2\sqrt{2}r)^{1/4}
Where, r'=0.7788r
Hence, GMR=1.723r
\therefore \; GMR=(r' \times 2r\times 2r\times 2\sqrt{2}r)^{1/4}
Where, r'=0.7788r
Hence, GMR=1.723r
Question 7 |
The most commonly used relay, for the protection of an alternator against loss of
excitation, is
offset Mho relay. | |
over current relay. | |
differential relay | |
Buchholz relay. |
Question 8 |
Suppose I_{A},\:I_{B} and I_{C} are a set of unbalanced current phasors in a three-phase system. The phase-B zero-sequence current I_{B0}=0.1\angle 0^{\circ} p.u. If phase-A current I_{A}=1.1\angle 0^{\circ} p.u and phase-C current I_{C}=\left ( 1\angle 120^{\circ}+0.1 \right ) p.u, then I_{B} in p.u is
1\angle 240^{\circ}-0.1\angle 0^{\circ} | |
1.1\angle 240^{\circ}-0.1\angle 0^{\circ} | |
1.1\angle -120^{\circ}+0.1\angle 0^{\circ} | |
1\angle -120^{\circ}+0.1\angle 0^{\circ} |
Question 8 Explanation:
\begin{aligned} I_{B O} &=\frac{1}{3}\left(I_{A}+I_{B}+I_{C}\right) \\ 0.1 &=\frac{1}{3}\left(1.1 \angle 0+I_{B}+1 \angle 120^{\circ}+0.1\right) \\ I_{B} &=0.3-1.1 \angle 0-0.1-1 \angle 120^{\circ} \\ I_{B} &=-0.9-1 \angle 120^{\circ} \\ I_{B} &=0.1+1 \angle 240^{\circ} \\ I_{B} &=1 \angle-120^{\circ}+0.1 \angle 0^{\circ} \end{aligned}
Question 9 |
A 3-Bus network is shown. Consider generators as ideal voltage sources. If rows 1,\:2 and 3 of the Y_{Bus} matrix correspond to Bus 1,\:2 and 3, respectively, then Y_{Bus} of the network is
\begin{bmatrix} -4j & j & j\\ j& -4j & j\\ j& j & -4j \end{bmatrix} | |
\begin{bmatrix} -4j & 2j & 2j\\ 2j& -4j & 2j\\ 2j& 2j & -4j \end{bmatrix} | |
\begin{bmatrix} - \frac{3}{4}j& \frac{1}{4}j & \frac{1}{4}j\\ \\ \frac{1}{4}j& -\frac{3}{4}j & \frac{1}{4}j\\ \\ \frac{1}{4}j& \frac{1}{4}j & -\frac{3}{4}j \end{bmatrix} | |
\begin{bmatrix} - \frac{1}{2}j& \frac{1}{4}j & \frac{1}{4}j\\ \\ \frac{1}{4}j& -\frac{1}{2}j & \frac{1}{4}j\\ \\ \frac{1}{4}j& \frac{1}{4}j & -\frac{1}{2}j \end{bmatrix} |
Question 9 Explanation:
\begin{aligned} V_{1} &=j 1 i_{1}+j 1\left(i_{1}+i_{2}+i_{3}\right) \\ V_{1} &=2 j i_{1}+j i_{2}+j i_{3} \\ \text{and}\qquad \qquad V_{2} &=j i_{1}+2 j i_{2}+j i_{3} \\ V_{3} &=j i_{1}+j i_{2}+2 j i_{3} \\ \therefore \qquad \left[\begin{array}{l} V_{1} \\ V_{2} \\ V_{3} \end{array}\right]&=\left[\begin{array}{ccc} 2 j & j & j \\ j & 2 j & j \\ j & j & 2 j \end{array}\right]\left[\begin{array}{l} i_{1} \\ i_{2} \\ i_{3} \end{array}\right]\\ Z_{\text {bus }}&=\left[\begin{array}{ccc} 2 j & j & j \\ j & 2 j & j \\ j & j & 2 j \end{array}\right] \\ Y_{\text {bus }}&=\left[\begin{array}{ccc} \frac{-3 j}{4} & \frac{j}{4} & \frac{j}{4} \\ \frac{j}{4} & \frac{-3 j}{4} & \frac{j}{4} \\ \frac{j}{4} & \frac{j}{4} & \frac{-3 j}{4} \end{array}\right] \end{aligned}
Question 10 |
In the figure shown, self-impedances of the two transmission lines are 1.5j\:p.u each, and Z_{m}=0.5j \:p.u is the mutual impedance. Bus voltages shown in the figure are in p.u. Given that \delta>0, the maximum steady-state real power that can be transferred in p.u from Bus-1 to Bus-2 is
\left | E \right |\left | V \right | | |
\frac{\left | E \right |\left | V \right |}{2} | |
2\left | E \right |\left | V \right | | |
\frac{3\left | E \right |\left | V \right |}{2} |
Question 10 Explanation:
\begin{aligned} L_{e q}&=\frac{L_{1} L_{2}-M^{2}}{\left(L_{1}+L_{2}-2 M\right)} \\ X_{\mathrm{eq}}&=\frac{1.5 \times 1.5-0.5^{2}}{1.5+1.5-2 \times 0.5}=1 \mathrm{p.u.} \end{aligned}
\begin{aligned} P_{\max }&=\frac{|E||V|}{1}\\ P_{\max }&=|E||V| \end{aligned}
There are 10 questions to complete.