Question 1 |
Bus 1 with voltage magnitude V_1 = 1.1 p.u. is sending reactive power Q_{12} towards bus
2 with voltage magnitude V_2 = 1 p.u. through a lossless transmission line of reactance
X. Keeping the voltage at bus 2 fixed at 1 p.u., magnitude of voltage at bus 1 is changed,
so that the reactive power Q_{12} sent from bus 1 is increased by 20%. Real power flow
through the line under both the conditions is zero. The new value of the voltage
magnitude, V_1, in p.u. (rounded off to 2 decimal places) at bus 1 is _______ .
0.118 | |
1.12 | |
1 | |
0.82 |
Question 1 Explanation:
With real power zero, load angle \delta =0
with initial values, V_{1}=1.1, \; \; \; V_{2}=1
Q_{12}=\frac{V_{1}^{2}}{X}-\frac{V_{1}V_{2}}{X}\sin \delta
\, \, =\frac{(1.1)^{2}}{X}-\frac{1.1\times 1}{X}\sin 0=\frac{0.11}{x}
With increased value of voltage,
new value of \; Q_{12}=1.2Q_{12},\; \; V_{2}=1
1.2\, Q_{12}=\frac{V_{1}^{2}}{X}-\frac{V_{1}\times 1}{X}=1.2\times \frac{0.11}{X}
V_{1}^{2}-V_{1}-0.132=0
V_{1}=1.12,\; \; -0.118
Hence the practical value in per unit,V_{1}=1.12 p.u.
with initial values, V_{1}=1.1, \; \; \; V_{2}=1
Q_{12}=\frac{V_{1}^{2}}{X}-\frac{V_{1}V_{2}}{X}\sin \delta
\, \, =\frac{(1.1)^{2}}{X}-\frac{1.1\times 1}{X}\sin 0=\frac{0.11}{x}
With increased value of voltage,
new value of \; Q_{12}=1.2Q_{12},\; \; V_{2}=1
1.2\, Q_{12}=\frac{V_{1}^{2}}{X}-\frac{V_{1}\times 1}{X}=1.2\times \frac{0.11}{X}
V_{1}^{2}-V_{1}-0.132=0
V_{1}=1.12,\; \; -0.118
Hence the practical value in per unit,V_{1}=1.12 p.u.
Question 2 |
Two buses, i and j, are connected with a transmission line of admittance Y, at the two
ends of which there are ideal transformers with turns ratios as shown. Bus admittance
matrix for the system is:
\begin{bmatrix} -t_it_jY & t_j^2 Y\\ t_i^2 Y & -t_it_jY \end{bmatrix} | |
\begin{bmatrix} t_it_jY & -t_j^2 Y\\ -t_i^2 Y & t_it_jY \end{bmatrix} | |
\begin{bmatrix} t_i^2 Y & -t_it_jY\\ -t_it_jY & t_j^2 Y \end{bmatrix} | |
\begin{bmatrix} t_it_jY & -(t_i-t_j)^2Y\\ -(t_i-t_j)^2Y & t_it_jY \end{bmatrix} |
Question 2 Explanation:
\begin{aligned} I&=Y(t_{i}V_{i}-V_{j}t_{j}) \\ I_{i}&=t_{i}I \\ &=t_{i}^{2}YV_{i}-t_{i}t_{j}YV_{j} \\ I_{j}&=-t_{j}I \\ &=-I_{i}t_{j}YV_{i}+t_{i}^{2}YV_{j} \\ \begin{bmatrix} I_{i}\\ I_{j} \end{bmatrix}&=\begin{bmatrix} t_{i}^{2}Y &-t_{i}t_{j}Y \\ -t_{i}t_{j}Y &t_{j}^{2}Y \end{bmatrix}\begin{bmatrix} V_{i}\\ v_{j} \end{bmatrix} \end{aligned}
Question 3 |
Out of the following options, the most relevant information needed to specify the real
power (P) at the PV buses in a load flow analysis is
solution of economic load dispatch | |
rated power output of the generator | |
rated voltage of the generator | |
base power of the generator |
Question 3 Explanation:
Most relevant information needed to specify P at PV buses is solution of economic load dispatch.
Question 4 |
A lossless transmission line with 0.2 pu reactance per phase uniformly distributed along
the length of the line, connecting a generator bus to a load bus, is protected up to 80%
of its length by a distance relay placed at the generator bus. The generator terminal
voltage is 1 pu. There is no generation at the load bus. The threshold pu current for
operation of the distance relay for a solid three phase-to-ground fault on the transmission
line is closest to:
1 | |
3.61 | |
5 | |
6.25 |
Question 4 Explanation:
I_{f}=\frac{1}{Z_{Th}}=\frac{1}{0.2}
=5 pu for 100% of line
Relay is operated for 80%
Z_{f}=0.8\, Z_{t}\Rightarrow 0.8\times 0.2=0.16\, p.u.
For 80% of line,
I_{f}=\frac{1}{0.16}=6.25\: p.u.
=5 pu for 100% of line
Relay is operated for 80%
Z_{f}=0.8\, Z_{t}\Rightarrow 0.8\times 0.2=0.16\, p.u.
For 80% of line,
I_{f}=\frac{1}{0.16}=6.25\: p.u.
Question 5 |
In a 132 kV system, the series inductance up to the point of circuit breaker locationis 50 mH. The shunt capacitanceat the circuit breaker terminal is 0.05 \mu F. The critical value of resistance in ohms required to be connected across the circuit breaker contacts which will give no transient oscillation is_____
100 | |
250 | |
500 | |
1000 |
Question 5 Explanation:
\begin{aligned} L&=50mH\\ C&=0.05\mu F\\ R_{cr}&=\frac{1}{2}\sqrt{\frac{L}{C}}\\ &=\frac{1}{2}\sqrt{\frac{50 \times 10^{-3}}{0.05 \times 10^{-6}}}\\ &=500\Omega \end{aligned}
Question 6 |
In the single machine infinite bus system shown below, the generator is delivering the real power of 0.8pu at 0.8 power factor lagging to the infinite bus. The power angle of the generator in degrees (round off to one decimal place) is _________
12.8 | |
28.4 | |
20.5 | |
32.6 |
Question 6 Explanation:
\begin{aligned} X &=0.25+0.2+0.4||0.4 \\ &=0.45+0.2=0.65pu \\ P&=V_{pu} \times I_{pu} \cos \phi \\ 0.8 &=1 \times I_{pu} \times 0.8 \\ I_{pu} &= 1pu\\ E &= V+jI_aX_s\\ &=1+1\angle -36.86^{\circ} \times j0.65 \\ &= 1.484 \angle 20.51^{\circ}pu\\ \delta &= 20.51^{\circ} \end{aligned}
Question 7 |
A 30 kV, 50 Hz, 50 MVA generator has the positive, negative, and zero sequence reactancesof 0.25 pu, 0.15 pu, and 0.05 pu, respectively. The neutral of the generator is grounded with a reactance so that the fault current for a bolted LG fault and that of a bolted three-phase fault at the generator terminal are equal. The value of grounding reactance in ohms (round off to one decimal place) is ______
2.2 | |
1.8 | |
3.6 | |
4.2 |
Question 7 Explanation:
\begin{aligned} X_1&=0.25 \; p.u.\\ X_2&=0.15 \; p.u.\\ X_0&=0.05 \; p.u.\\ I_{f(LG)}&=I_{f(3-\phi )}\\ \frac{3V_{pu}}{(X_1+X_2+X_0+3X_n)}&=\frac{V_{pu}}{X_1}\\ \frac{3 \times 1}{(0.25+0.15+0.05+3X_n)}&=\frac{1}{0.25}\\ \frac{3}{0.46+3X_n}&=\frac{1}{0.25}\\ \Rightarrow \;\; X_n&=0.1\; p.u.\\ X_n&=0.1 \times Z_B\\ &=0.1 \times \frac{30^2}{50}\\ &=1.8\Omega \end{aligned}
Question 8 |
A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per phase, and the capacitance is 0.01 \mu F/km per phase. The line is under open circuit condition at the receiving end and energized with 400 kV at the sending end, the receiving end line voltage in kV (round off to two decimal places) will be ___________.
418.85 | |
256.25 | |
458.45 | |
369.28 |
Question 8 Explanation:
\begin{aligned} V_s&= 400kV\\ l&=300km\\ L_1&=1mH/km/phase\\ C_1&=0.01 \mu F/km/phase\\ v&=\frac{1}{\sqrt{L_1C_1}}\\ &=\frac{1}{\sqrt{1 \times 10^{-3} \times 0.01 \times 10^{-6}}}\\ &=3.16 \times 10^5 km/sec\\ \beta '&=\frac{2 \pi f l}{v}\\ &=\frac{2 \pi \times 50 \times 300}{3.16 \times 10^5}=0.29\\ A&=1-\frac{\beta ^2}{2}\\ &=1-\frac{(0.29)^2}{2}=0.955\\ V_R&=\frac{V_S}{A}=\frac{400}{0.955}=418.85kV \end{aligned}
Question 9 |
The total impedance of the secondary winding, leads, and burden of a 5 A CT is 0.01 \Omega. If the fault current is 20 times the rated primary current of the CT, the VA output of the CT is ________
50 | |
100 | |
150 | |
200 |
Question 9 Explanation:
\begin{aligned} I_{sec}&=5 \times 20=100A \\ V &=I_{sec}R =100 \times 0.01\\ &=1V \\ \text{VA } &\text{output of the CT} \\ &= VI_{sec}=100 \times 1\\ &= 100VA \end{aligned}
Question 10 |
Five alternators each rated 5 MVA, 13.2 kV with 25% of reactance on its own base are connected in parallel to a busbar. The short-circuit level in MVA at the busbar is_________
50 | |
75 | |
100 | |
150 |
Question 10 Explanation:
Net reactance of parallel connection,
\begin{aligned} X&=\frac{0.25}{5}=0.05\;p.u.\\ I_{SC}&=\frac{1}{X}=\frac{1}{0.05}=20\; p.u.\\ \text{SC MVA}&=20 \times 5\\ &=100 \text{ MVA} \end{aligned}
\begin{aligned} X&=\frac{0.25}{5}=0.05\;p.u.\\ I_{SC}&=\frac{1}{X}=\frac{1}{0.05}=20\; p.u.\\ \text{SC MVA}&=20 \times 5\\ &=100 \text{ MVA} \end{aligned}
There are 10 questions to complete.