Demonstrative Video
Problem-1
The figure below shows the one line diagram of a simple four bus system. The table below gives line impedances identified by the buses on which these terminate. The shunt admittance at all the buses is assumed to be negligible.
Find YBUS , assuming that the line shown dotted is not connected.
What modifications need to be carried out in YBUS if the line shown dotted is connected.
Solution-1
\(R_{p.u.}\) and \(X_{p.u.}\) need to be converted to \(G_{p.u.}\) and \(B_{p.u.}\)
Impedance and Admittance: \[\begin{aligned} Z_{p.u.} & = R_{p.u.}+ j X_{p.u.} \\ Y_{p.u.} & = G_{p.u.} = j B_{p.u.} = \dfrac{1}{R_{p.u.}+ j X_{p.u.}} \end{aligned}\]
\[\begin{aligned} &\boxed{R_{p .u}+j X_{p u}} ~\quad \boxed{G_{p u}+j B_{p u}} \\ z_{1-2}~&0.05+j 0.15 \rightarrow 2.0-j 6.0 \\ z_{1-3}~&0.10+j 0.30 \rightarrow 1.0-j 3.0 \\ z_{2-3}~&0.15+j 0.45 \rightarrow 0.66-j 2.0 \\ z_{2-4}~&0.10+j 0.30 \rightarrow 1.0-j 3.0 \\ z_{3-4}~&0.05+j 0.15 \rightarrow 2.0-j 6.0 \end{aligned}\]
\[\begin{aligned}
&\boxed{R_{p .u}+j X_{p u}} ~\quad \boxed{G_{p u}+j B_{p
u}} \\
z_{1-2}~&0.05+j 0.15 \rightarrow 2.0-j 6.0 \\
z_{1-3}~&0.10+j 0.30 \rightarrow 1.0-j 3.0 \\
z_{2-3}~&0.15+j 0.45 \rightarrow 0.66-j 2.0 \\
z_{2-4}~&0.10+j 0.30 \rightarrow 1.0-j 3.0 \\
z_{3-4}~&0.05+j 0.15 \rightarrow 2.0-j 6.0
\end{aligned}\]
\[\begin{aligned} &Y_{Bus}=\left[\begin{array}{cccc} y_{13} & 0 & -y_{13} & 0 \\ 0 & y_{23}+y_{24} & -y_{23} & -y_{24} \\ -y_{13} & -y_{23} & y_{31}+y_{32}+y_{34} & -y_{34} \\ 0 & -y_{24} & -y_{34} & y_{43}+y_{42} \end{array}\right] \\ &Y_{Bus}=\left[\begin{array}{cccc} 1-j 3 & 0 & -1+j 3 & 0 \\ 0 & 1.66-j 5 & -0.66+j 2 & -1+j 3 \\ -1+j 3 & -0.66+j 2 & 3.66-j 11 & -2+j 6 \\ 0 & -1+j 3 & -2+j 6 & 3-j 9\\ \end{array}\right] \end{aligned}\]
Line added between Buses 1-2
\[\begin{aligned} &Y_{11,\text{new}}=Y_{11,\text{old}}+(2-j 6)=-3-j 9 \\ &Y_{12,\text{new}}=Y_{12,\text{old}}-(2-j 6)=-2+j 6=Y_{21,\text{new}} \\ &Y_{22,\text { new }}=Y_{22,\text{old}}+(2-j 6)=3.66-j 11 \end{aligned}\]
Modified Y-Bus: \[Y_{\text {Bus }}=\left[\begin{array}{cccc} 3-j 9 & -2+j 6 & -1+j 3 & 0 \\ -2+j 6 & 3.66-j 11 & -0.66+j 2 & -1+j 3 \\ -1+j 3 & -0.66+j 2 & 3.66-j 11 & -2+j 6 \\ 0 & -1+j 3 & -2+j 6 & 3-j 9 \end{array}\right]\]
Problem-2
Build \(\mathrm{Y}_{\mathrm{BUS}}\)
matrix using singular transformation for the power system given below.
The branch impedances of the lines are as follows: Self admittances of
Y- matrix \((\Omega)\) \[\begin{aligned}
\mathrm{Y}_{11} &=0.0094-\mathrm{j} 0.0134 &
\mathrm{Y}_{44} &=0.0154-\mathrm{j} 0.0231 \\
\mathrm{Y}_{22} &=0.0088-\mathrm{j} 0.0147 &
\mathrm{Y}_{55} &=0.0082-\mathrm{j} 0.0219 \\
\mathrm{Y}_{33} &=0.0186-\mathrm{j} 0.0082 &
\mathrm{Y}_{66} &=0.0176-\mathrm{j} 0.0294
\end{aligned}\]
Solution-2
Obtain Tree graph:
Note that a tree of a graph is not unique.
\[ [\hat{A}]=\begin{array}{c|ccccc} n \rightarrow & 0 & 1 & 2 & 3 & 4 \\ e \downarrow & & & & & \\ \hline 1 & 1 & 0 & 0 & 0 & -1 \\ 2 & 0 & -1 & 0 & 0 & 1 \\ 3 & 0 & 0 & -1 & 0 & 1 \\ 4 & 0 & 0 & 0 & -1 & 1 \\ 5 & 0 & 1 & -1 & 0 & 0 \\ 6 & 0 & 0 & -1 & 1 & 0 \end{array} \]
\(\Leftarrow\) Relating branches and nodes
Matrix rectangular \(\Rightarrow\) singular
\(1/-1\) if element oriented away/towards the node
\(0\) when no incidence
\[ [\hat{A}]=\begin{array}{c|ccccc}n\to&0&1&2&3&4\\\hline1&1&0&0&0&-1\\2&0&-1&0&0&1\\3&0&0&-1&0&1\\4&0&0&0&-1&1\\5&0&1&-1&0&0\\6&0&0&-1&1&0\end{array} \] Remove Reference (\(0^{th}\)) Node Column \[A=\left[\begin{array}{cccc} 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 1 \\ 0 & -1 & 0 & 1 \\ 0 & 0 & -1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & -1 & 1 & 0 \end{array}\right]\]
\[{\left[A^{\top}\right]=\left[\begin{array}{cccccc} 0 & -1 & 0 & 0 & 1 & 0 \\ 0 & 0 & -1 & 0 & -1 & -1 \\ 0 & 0 & 0 & -1 & 0 & 1 \\ -1 & 1 & 1 & 1 & 0 & 0 \end{array}\right]}\]
\[\begin{aligned} \mathrm{Y}_{11} &=0.0094-\mathrm{j} 0.0134 & \mathrm{Y}_{44} &=0.0154-\mathrm{j} 0.0231 \\ \mathrm{Y}_{22} &=0.0088-\mathrm{j} 0.0147 & \mathrm{Y}_{55} &=0.0082-\mathrm{j} 0.0219 \\ \mathrm{Y}_{33} &=0.0186-\mathrm{j} 0.0082 & \mathrm{Y}_{66} &=0.0176-\mathrm{j} 0.0294 \end{aligned}\]
\[Y_{B \cup S}=\left[A^{\top}\right] \times[Y] \times[A]\]
Problem-3
A transmission line represented in fig 1. A series capacitance is added to improve the performance of the transmission line as shown in fig 2. calculate the power transferred, if the capacitive reactance to inductive reactance ratio is
\(0.5\)
\(0.2\)
Solution-3
Power transferred to the load: \[P_{L1} = \dfrac{V_sV_r}{X_L}\sin\delta~\mathrm{Watts}\]
Adding a capacitor in series:
\[P_{L2} = \dfrac{V_sV_r}{X_L-X_C}\sin\delta~\mathrm{Watts}\]
Taking ratio \(P_{L2}/P_{L1}\), we get \[\dfrac{P_{L2}}{P_{L1}} = \dfrac{1}{1-\left(\dfrac{X_C}{X_L}\right)}\]
Given \(\frac{X_C}{X_L}=0.5\), we compute \[\begin{aligned} \dfrac{P_{L2}}{P_{L1}} & = \dfrac{1}{1-0.5} = 2\\ \Rightarrow ~P_{L2} & = 2P_{L1} \end{aligned}\] Power transferred will be 2-times the power transferred without series capacitance
Next, \(\dfrac{X_C}{X_L}=0.2\) \[\begin{aligned} \dfrac{P_{L2}}{P_{L1}} & = \dfrac{1}{1-0.2} = 1.25\\ \Rightarrow ~P_{L2} & = 1.25P_{L1} \end{aligned}\] Power transferred will be 1.25-times the power transferred without series capacitance