Solved Problems on Y-Bus Formation and Tuning of Transmission Lines

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.

  1. Find YBUS , assuming that the line shown dotted is not connected.

  2. What modifications need to be carried out in YBUS if the line shown dotted is connected.

image

Solution-1

  • \(R_{p.u.}\) and \(X_{p.u.}\) need to be converted to \(G_{p.u.}\) and \(B_{p.u.}\)

  • \[\begin{aligned} Z_{p.u.} & = R_{p.u.}+ j X_{p.u.} \\ Y_{p.u.} & = G_{p.u.} = j B_{p.u.} = \dfrac{1}{R_{p.u.}+ j X_{p.u.}} \end{aligned}\]
    Impedance and Admittance:
\[\begin{aligned} &\boxed{R_{p .u}+j X_{p u}} ~\quad \boxed{G_{p u}+j B_{p u}} \\ z_{1-2}~&0.05+j 0.15 \rightarrow 2.0-j 6.0 \\ z_{1-3}~&0.10+j 0.30 \rightarrow 1.0-j 3.0 \\ z_{2-3}~&0.15+j 0.45 \rightarrow 0.66-j 2.0 \\ z_{2-4}~&0.10+j 0.30 \rightarrow 1.0-j 3.0 \\ z_{3-4}~&0.05+j 0.15 \rightarrow 2.0-j 6.0 \end{aligned}\]
\[\begin{aligned} &\boxed{R_{p .u}+j X_{p u}} ~\quad \boxed{G_{p u}+j B_{p u}} \\ z_{1-2}~&0.05+j 0.15 \rightarrow 2.0-j 6.0 \\ z_{1-3}~&0.10+j 0.30 \rightarrow 1.0-j 3.0 \\ z_{2-3}~&0.15+j 0.45 \rightarrow 0.66-j 2.0 \\ z_{2-4}~&0.10+j 0.30 \rightarrow 1.0-j 3.0 \\ z_{3-4}~&0.05+j 0.15 \rightarrow 2.0-j 6.0 \end{aligned}\]
image
\[\begin{aligned} &Y_{Bus}=\left[\begin{array}{cccc} y_{13} & 0 & -y_{13} & 0 \\ 0 & y_{23}+y_{24} & -y_{23} & -y_{24} \\ -y_{13} & -y_{23} & y_{31}+y_{32}+y_{34} & -y_{34} \\ 0 & -y_{24} & -y_{34} & y_{43}+y_{42} \end{array}\right] \\ &Y_{Bus}=\left[\begin{array}{cccc} 1-j 3 & 0 & -1+j 3 & 0 \\ 0 & 1.66-j 5 & -0.66+j 2 & -1+j 3 \\ -1+j 3 & -0.66+j 2 & 3.66-j 11 & -2+j 6 \\ 0 & -1+j 3 & -2+j 6 & 3-j 9\\ \end{array}\right] \end{aligned}\]

  • Line added between Buses 1-2

\[\begin{aligned} &Y_{11,\text{new}}=Y_{11,\text{old}}+(2-j 6)=-3-j 9 \\ &Y_{12,\text{new}}=Y_{12,\text{old}}-(2-j 6)=-2+j 6=Y_{21,\text{new}} \\ &Y_{22,\text { new }}=Y_{22,\text{old}}+(2-j 6)=3.66-j 11 \end{aligned}\]
\[Y_{\text {Bus }}=\left[\begin{array}{cccc} 3-j 9 & -2+j 6 & -1+j 3 & 0 \\ -2+j 6 & 3.66-j 11 & -0.66+j 2 & -1+j 3 \\ -1+j 3 & -0.66+j 2 & 3.66-j 11 & -2+j 6 \\ 0 & -1+j 3 & -2+j 6 & 3-j 9 \end{array}\right]\]
Modified Y-Bus:

Problem-2

\[\begin{aligned} \mathrm{Y}_{11} &=0.0094-\mathrm{j} 0.0134 & \mathrm{Y}_{44} &=0.0154-\mathrm{j} 0.0231 \\ \mathrm{Y}_{22} &=0.0088-\mathrm{j} 0.0147 & \mathrm{Y}_{55} &=0.0082-\mathrm{j} 0.0219 \\ \mathrm{Y}_{33} &=0.0186-\mathrm{j} 0.0082 & \mathrm{Y}_{66} &=0.0176-\mathrm{j} 0.0294 \end{aligned}\]
image
matrix using singular transformation for the power system given below. The branch impedances of the lines are as follows: Self admittances of Y- matrix Build

Solution-2

  • Obtain Tree graph:

  • Note that a tree of a graph is not unique.

    • image
\[ [\hat{A}]=\begin{array}{c|ccccc} n \rightarrow & 0 & 1 & 2 & 3 & 4 \\ e \downarrow & & & & & \\ \hline 1 & 1 & 0 & 0 & 0 & -1 \\ 2 & 0 & -1 & 0 & 0 & 1 \\ 3 & 0 & 0 & -1 & 0 & 1 \\ 4 & 0 & 0 & 0 & -1 & 1 \\ 5 & 0 & 1 & -1 & 0 & 0 \\ 6 & 0 & 0 & -1 & 1 & 0 \end{array} \]

  • \(\Leftarrow\) Relating branches and nodes

  • Matrix rectangular \(\Rightarrow\) singular

  • \(1/-1\) if element oriented away/towards the node

  • \(0\) when no incidence

\[ [\hat{A}]=\begin{array}{c|ccccc}n\to&0&1&2&3&4\\\hline1&1&0&0&0&-1\\2&0&-1&0&0&1\\3&0&0&-1&0&1\\4&0&0&0&-1&1\\5&0&1&-1&0&0\\6&0&0&-1&1&0\end{array} \]
) Node Column Remove Reference (
\[{\left[A^{\top}\right]=\left[\begin{array}{cccccc} 0 & -1 & 0 & 0 & 1 & 0 \\ 0 & 0 & -1 & 0 & -1 & -1 \\ 0 & 0 & 0 & -1 & 0 & 1 \\ -1 & 1 & 1 & 1 & 0 & 0 \end{array}\right]}\]
\[\begin{aligned} \mathrm{Y}_{11} &=0.0094-\mathrm{j} 0.0134 & \mathrm{Y}_{44} &=0.0154-\mathrm{j} 0.0231 \\ \mathrm{Y}_{22} &=0.0088-\mathrm{j} 0.0147 & \mathrm{Y}_{55} &=0.0082-\mathrm{j} 0.0219 \\ \mathrm{Y}_{33} &=0.0186-\mathrm{j} 0.0082 & \mathrm{Y}_{66} &=0.0176-\mathrm{j} 0.0294 \end{aligned}\]
\[Y_{B \cup S}=\left[A^{\top}\right] \times[Y] \times[A]\]

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

  1. \(0.5\)

  2. \(0.2\)

image

Solution-3

  • \[P_{L1} = \dfrac{V_sV_r}{X_L}\sin\delta~\mathrm{Watts}\]
    Power transferred to the load:
  • Adding a capacitor in series:
    image
    \[P_{L2} = \dfrac{V_sV_r}{X_L-X_C}\sin\delta~\mathrm{Watts}\]
  • \[\dfrac{P_{L2}}{P_{L1}} = \dfrac{1}{1-\left(\dfrac{X_C}{X_L}\right)}\]
    , we get Taking ratio
  • \[\begin{aligned} \dfrac{P_{L2}}{P_{L1}} & = \dfrac{1}{1-0.5} = 2\\ \Rightarrow ~P_{L2} & = 2P_{L1} \end{aligned}\]
    , we compute Given
  • \[\begin{aligned} \dfrac{P_{L2}}{P_{L1}} & = \dfrac{1}{1-0.2} = 1.25\\ \Rightarrow ~P_{L2} & = 1.25P_{L1} \end{aligned}\]
    Next,