Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
Consider a power system consisting of N number of buses. Buses in this power system are categorized into slack bus, PV buses and PQ buses for load flow study. The number of PQ buses is N_{L}. The balanced Newton-Raphson method is used to carry out load flow study in polar form. \text{H, S, M, and R} are sub-matrices of the Jacobian matrix J as shown below:
\begin{bmatrix} \Delta P\\ \Delta Q \end{bmatrix}=J\begin{bmatrix} \Delta \delta \\ \Delta V \end{bmatrix}, \text{where}\: J=\begin{bmatrix} H & S\\ M &R \end{bmatrix}
The dimension of the sub-matrix M is
\begin{bmatrix} \Delta P\\ \Delta Q \end{bmatrix}=J\begin{bmatrix} \Delta \delta \\ \Delta V \end{bmatrix}, \text{where}\: J=\begin{bmatrix} H & S\\ M &R \end{bmatrix}
The dimension of the sub-matrix M is
N_{L}\times \left ( N-1 \right ) | |
\left ( N-1 \right )\times \left ( N-1-N_{L} \right ) | |
N_{L}\times \left ( N-1+N_{L} \right ) | |
\left ( N-1 \right )\times \left ( N-1+N_{L} \right ) |
Question 4 Explanation:
\left[\begin{array}{l} \Delta P \\ \Delta Q \end{array}\right]=J\left[\begin{array}{l} \Delta \delta \\ \Delta V \end{array}\right] \text { where } J=\left[\begin{array}{ll} H & S \\ M & R \end{array}\right]
For size of M
Row = No. of unknown variables of Q=N_{L}
Column = No. of variable which has \delta=N_{L}+\left(N-1-N_{L}\right)
\begin{aligned} &=N-1\\ \text{So}, \quad\text{ size of }M&=N_{L} \times(N-1) \end{aligned}
For size of M
Row = No. of unknown variables of Q=N_{L}
Column = No. of variable which has \delta=N_{L}+\left(N-1-N_{L}\right)
\begin{aligned} &=N-1\\ \text{So}, \quad\text{ size of }M&=N_{L} \times(N-1) \end{aligned}
Question 5 |
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 5 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.
There are 5 questions to complete.