Question 1 |
A balanced delta connected load consisting of the series connection of one resistor (R=15 \Omega) and a capacitor (C=212.21 \mu \mathrm{F}) in each phase is connected to three-phase, 50 \mathrm{~Hz}, 415 \mathrm{~V} supply terminals through a line having an inductance of \mathrm{L}=31.83 \mathrm{mH} per phase, as shown in the figure. Considering the change in the supply terminal voltage with loading to be negligible, the magnitude of the voltage across the terminals V_{AB} in Volts is ____ (Round off to the nearest integer).
585 | |
632 | |
225 | |
415 |
Question 1 Explanation:
Given circuit :
where,
\begin{aligned} X_{L} & =2 \pi \times 50 \times 31.83 \times 10^{-3}=10 \Omega \\ R & =15 \Omega \\ X_{C} & =\frac{1}{2 \pi \times 50 \times 212.21 \times 10^{-6}}=15 \Omega \end{aligned}
Convert \Delta to star :
\begin{aligned} Z_{Y} & =\frac{Z_{\Delta}}{3} \\ & =\frac{15-j 15}{3}=5-j 5 \end{aligned}
Now, per phase equivalent circuit :
Using voltage division,
\begin{aligned} \mathrm{V}_{\mathrm{AN}} & =\frac{5-\mathrm{j} 5}{5-\mathrm{j} 5+10 \mathrm{j}} \times \frac{415 \angle 0^{\circ}}{\sqrt{3}} \\ & =\frac{415}{\sqrt{3}} \angle-90^{\circ} \mathrm{V} \\ \therefore \quad \quad\left|\mathrm{V}_{\mathrm{AB}}\right| & =\sqrt{3} \times \frac{415}{\sqrt{3}}=415 \mathrm{~V} \end{aligned}
where,
\begin{aligned} X_{L} & =2 \pi \times 50 \times 31.83 \times 10^{-3}=10 \Omega \\ R & =15 \Omega \\ X_{C} & =\frac{1}{2 \pi \times 50 \times 212.21 \times 10^{-6}}=15 \Omega \end{aligned}
Convert \Delta to star :
\begin{aligned} Z_{Y} & =\frac{Z_{\Delta}}{3} \\ & =\frac{15-j 15}{3}=5-j 5 \end{aligned}
Now, per phase equivalent circuit :
Using voltage division,
\begin{aligned} \mathrm{V}_{\mathrm{AN}} & =\frac{5-\mathrm{j} 5}{5-\mathrm{j} 5+10 \mathrm{j}} \times \frac{415 \angle 0^{\circ}}{\sqrt{3}} \\ & =\frac{415}{\sqrt{3}} \angle-90^{\circ} \mathrm{V} \\ \therefore \quad \quad\left|\mathrm{V}_{\mathrm{AB}}\right| & =\sqrt{3} \times \frac{415}{\sqrt{3}}=415 \mathrm{~V} \end{aligned}
Question 2 |
The two-bus power system shown in figure (i) has one alternator supplying a synchronous motor load through a Y-\Delta transformer. The positive, negative and zero-sequence diagrams of the system are shown in figures (ii), (iii) and (iv), respectively. All reactances in the sequence diagrams are in p.u. For a bolted line-to-line fault (fault impedance = zero) between phases 'b' and 'c' at bus 1, neglecting all pre-fault currents, the magnitude of the fault current (from phase 'b' to 'c') in p.u. is _____ (Round off to 2 decimal places).
7.22 | |
5.47 | |
6.98 | |
9.82 |
Question 2 Explanation:
From positive sequence network :
\begin{aligned} X_{1} & =\mathrm{j}(0.3 \| 0.2) \\ & =\mathrm{j} 0.12 \mathrm{pu} \end{aligned}
From negative sequence network :
\begin{aligned} \mathrm{X}_{2} & =\mathrm{j}(0.3 \| 0.2) \\ & =\mathrm{j} 0.12 \mathrm{pu} \end{aligned}
For \mathrm{L}-\mathrm{L} fault,
Fault current, I_{f}=\frac{\sqrt{3} E}{\left(X_{1}+X_{2}\right)} =\frac{\sqrt{3} \times 1}{0.12+0.12} =7.217 \mathrm{pu}
\begin{aligned} X_{1} & =\mathrm{j}(0.3 \| 0.2) \\ & =\mathrm{j} 0.12 \mathrm{pu} \end{aligned}
From negative sequence network :
\begin{aligned} \mathrm{X}_{2} & =\mathrm{j}(0.3 \| 0.2) \\ & =\mathrm{j} 0.12 \mathrm{pu} \end{aligned}
For \mathrm{L}-\mathrm{L} fault,
Fault current, I_{f}=\frac{\sqrt{3} E}{\left(X_{1}+X_{2}\right)} =\frac{\sqrt{3} \times 1}{0.12+0.12} =7.217 \mathrm{pu}
Question 3 |
The three-bus power system shown in the figure has one alternator connected to bus 2 which supplies 200 \mathrm{MW} and 40 MVAR power. Bus 3 is infinite bus having a voltage of magnitude \left|\mathrm{V}_{3}\right|=1.0 p.u. and angle of -15^{\circ}. A variable current source, |1| \angle \phi is connected at bus 1 and controlled such that the magnitude of the bus 1 voltage is maintained at 1.05 p.u. and the phase angle of the source current, \phi=\theta_{1} \pm \frac{\pi}{2}, where \theta_{1} is the phase angle of the bus 1 voltage. The three buses can be categorized for load flow analysis as
Bus 1 Slack bus Bus 2 P-|V| bus Bus 3 P-Q bus | |
Bus 1 P-|V| bus Bus 2 P-|V| bus Bus 3 Slack bus | |
Bus 1 P-Q bus Bus 2 P-Q bus Bus 3 Slack bus | |
Bus 1 P-|V| bus Bus 2 P-Q bus Bus 3 Slack bus |
Question 3 Explanation:
Concept : Classification of Bus system :
\begin{array}{|c|c|c|} \hline \\ \textbf{Bus Type} & \textbf{Specified} & \textbf{Unspecified} \\ &\textbf{Value} &\textbf{Value} \\ \hline \text{Slack Bus} & |\mathrm{V}|, \delta &\mathrm{P}, \mathrm{Q} \\ \text{PV Bus} &\mathrm{P},|\mathrm{V}| & \mathrm{Q}, \delta \\ \text{PQ (load) Bus} & \mathrm{P}, \mathrm{Q} & |\mathrm{V}|, \delta \\ \hline \end{array}
Now, from given power system network.
Bus-1 \Rightarrow PV bus
Bus-2 \Rightarrow \mathrm{PQ} (load) bus
Bus-3 \Rightarrow Slack bus
\begin{array}{|c|c|c|} \hline \\ \textbf{Bus Type} & \textbf{Specified} & \textbf{Unspecified} \\ &\textbf{Value} &\textbf{Value} \\ \hline \text{Slack Bus} & |\mathrm{V}|, \delta &\mathrm{P}, \mathrm{Q} \\ \text{PV Bus} &\mathrm{P},|\mathrm{V}| & \mathrm{Q}, \delta \\ \text{PQ (load) Bus} & \mathrm{P}, \mathrm{Q} & |\mathrm{V}|, \delta \\ \hline \end{array}
Now, from given power system network.
Bus-1 \Rightarrow PV bus
Bus-2 \Rightarrow \mathrm{PQ} (load) bus
Bus-3 \Rightarrow Slack bus
Question 4 |
A 50 \mathrm{~Hz}, 275 \mathrm{kV} line of length 400 \mathrm{~km} has the following parameters:
Resistance, R=0.035 \Omega / \mathrm{km};
Inductance, \mathrm{L}=1 \mathrm{mH} / \mathrm{km};
Capacitance, \mathrm{C}=0.01 \mu \mathrm{F} / \mathrm{km};
The line is represented by the nominal -\pi model. With the magnitudes of the sending end and the receiving end voltages of the line (denoted by V_{S} and V_{R}, respectively) maintained at 275 \mathrm{kV}, the phase angle difference (\theta) between V_{S} and V_{R} required for maximum possible active power to be delivered to the receiving end, in degree is ____ (Round off to 2 decimal places).
Resistance, R=0.035 \Omega / \mathrm{km};
Inductance, \mathrm{L}=1 \mathrm{mH} / \mathrm{km};
Capacitance, \mathrm{C}=0.01 \mu \mathrm{F} / \mathrm{km};
The line is represented by the nominal -\pi model. With the magnitudes of the sending end and the receiving end voltages of the line (denoted by V_{S} and V_{R}, respectively) maintained at 275 \mathrm{kV}, the phase angle difference (\theta) between V_{S} and V_{R} required for maximum possible active power to be delivered to the receiving end, in degree is ____ (Round off to 2 decimal places).
42.36 | |
64.88 | |
83.64 | |
98.25 |
Question 4 Explanation:
We have,
\mathrm{P}_{\mathrm{R}}=\frac{\mathrm{V}_{\mathrm{S}} \mathrm{V}_{\mathrm{R}}}{\mathrm{B}} \cos (\beta-\delta)-\frac{\mathrm{AV}_{\mathrm{R}}^{2}}{\mathrm{~B}} \cos (\beta-\alpha)
At max. power,
\delta=\beta
where, \beta= angle of T-parameter of B.
\pi-Model :
[T]=\begin{bmatrix} 1+\frac{yz}{2} &z \\ y\left ( 1+ \frac{yz}{4}\right )& 1+ \frac{yz}{2}\\ \end{bmatrix}=\begin{bmatrix} A & B\\ C&D \end{bmatrix}
\therefore \quad B=z
=(0.035 \times 400)+j(2 \pi \times 50 \times 10^{-3} \times 400) = 126.44\angle 83.643^{\circ}\Omega
\mathrm{P}_{\mathrm{R}}=\frac{\mathrm{V}_{\mathrm{S}} \mathrm{V}_{\mathrm{R}}}{\mathrm{B}} \cos (\beta-\delta)-\frac{\mathrm{AV}_{\mathrm{R}}^{2}}{\mathrm{~B}} \cos (\beta-\alpha)
At max. power,
\delta=\beta
where, \beta= angle of T-parameter of B.
\pi-Model :
[T]=\begin{bmatrix} 1+\frac{yz}{2} &z \\ y\left ( 1+ \frac{yz}{4}\right )& 1+ \frac{yz}{2}\\ \end{bmatrix}=\begin{bmatrix} A & B\\ C&D \end{bmatrix}
\therefore \quad B=z
=(0.035 \times 400)+j(2 \pi \times 50 \times 10^{-3} \times 400) = 126.44\angle 83.643^{\circ}\Omega
Question 5 |
The bus admittance \left(Y_{\text {bus }}\right) matrix of a 3-bus power system is given below.
\begin{bmatrix} -j15 &j10 &j5 \\ j10& -j13.5 &j4 \\ j5& j4 & -j8 \end{bmatrix}
Considering that there is no shunt inductor connected to any of the buses, which of the following can NOT be true?
\begin{bmatrix} -j15 &j10 &j5 \\ j10& -j13.5 &j4 \\ j5& j4 & -j8 \end{bmatrix}
Considering that there is no shunt inductor connected to any of the buses, which of the following can NOT be true?
Line charging capacitor of finite value is present in all three lines | |
Line charging capacitor of finite value is present in line 2-3 only | |
Line charging capacitor of finite value is present in line 2-3 only and shunt capacitor of finite value is present in bus 1 only | |
Line charging capacitor of finite value is present in line 2-3 only and shunt capacitor of finite value is present in bus 3 only |
Question 5 Explanation:
From Y_{\text {Bus }} matrix
\begin{aligned} & y_{10}=-j 15+j 10+j 5=0 \\ & y_{20}=j 10-j 13.5+j 4=j 0.5 \\ & y_{30}=j 5+j 4-j 8=j 1 \end{aligned}
Power system network :
Hence, option (A) and (C) will not be correct.
\begin{aligned} & y_{10}=-j 15+j 10+j 5=0 \\ & y_{20}=j 10-j 13.5+j 4=j 0.5 \\ & y_{30}=j 5+j 4-j 8=j 1 \end{aligned}
Power system network :
Hence, option (A) and (C) will not be correct.
There are 5 questions to complete.