Admittance Bus Solved Problems
Demonstrative Video
Revision of Important Concept
The bus admittance matrix can be formed by inspection using the following guidelines.
The diagonal element \(Y_{i j}\) is given by sum of all the admittances connected to node-j.
The off-diagonal element \(Y_{j k}\) is given by negative of the sum of all the admittances connected between node-j and node-k.
Problem-1
For the network shown in the Figure,
Form the bus admittance matrix
Determine the reduced admittance matrix by eliminating node-4
All values are marked in p.u.
Solution-1
\[ \mathbf{Y}_{\text {bus }}=\left[\begin{array}{cccc} -(\mathrm{j} 0.5+\mathrm{j} 0.4+\mathrm{j} 0.4) & \mathrm{j} 0.5 & \mathrm{j} 0.4 & \mathrm{j} 0.4 \\ \mathrm{j} 0.5 & -(\mathrm{j} 0.5+\mathrm{j} 0.6) & \mathrm{j} 0.6 & 0 \\ \mathrm{j} 0.4 & \mathrm{j} 0.6 & -(\mathrm{j} 0.6+\mathrm{j} 0.5+\mathrm{j} 0.4) & \mathrm{j} 0.5 \\ \mathrm{j} 0.4 & 0 & \mathrm{j} 0.5 & -(\mathrm{j} 0.5+\mathrm{j} 0.4) \end{array}\right] \]
\[\mathbf{Y}_{\text {bus }}=\left[\begin{array}{cccc} -\mathrm{j} 1.3 & \mathrm{j} 0.5 & \mathrm{j} 0.4 & \mathrm{j} 0.4 \\ \mathrm{j} 0.5 & -\mathrm{j} 1.1 & \mathrm{j} 0.6 & 0 \\ \mathrm{j} 0.4 & \mathrm{j} 0.6 & -\mathrm{j} 1.5 & \mathrm{j} 0.5 \\ \mathrm{j} 0.4 & 0 & \mathrm{j} 0.5 & -\mathrm{j} 0.9 \end{array}\right]\]
The elements of new bus admittance matrix after eliminating the \(4^{\text {th }}\) row and \(4^{\text {th }}\) column is given by \[Y_{j k, n c w}=Y_{j k} \cdots \frac{Y_{j n} Y_{n k}}{Y_{n n}} ; \text { where } n=4 ; j=1,2,3 \text { and } k=1,2,3\]
The bus admittance matrix is symmetrical, \(\therefore Y_{k j, n e w}=Y_{j k, \text { new }}\)
\(\mathrm{Y}_{11, \text { new }}=\mathrm{Y}_{11}-\frac{\mathrm{Y}_{14} \mathrm{Y}_{41}}{\mathrm{Y}_{44}}=-\mathrm{j} 1.3-\frac{(\mathrm{j} 0.4)(\mathrm{j} 0.4)}{-\mathrm{j} 0.9}=-\mathrm{j} 1.12\)
\(\mathrm{Y}_{12, \mathrm{new}}=\mathrm{Y}_{12}-\frac{\mathrm{Y}_{14} \mathrm{Y}_{42}}{\mathrm{Y}_{44}}=\mathrm{j} 0.5-\frac{(\mathrm{j} 0.4 \times 0)}{-\mathrm{j} 0.9}=\mathrm{j} 0.5\)
\(\mathrm{Y}_{13, \mathrm{ncw}}=\mathrm{Y}_{13}-\frac{\mathrm{Y}_{14} \mathrm{Y}_{43}}{\mathrm{Y}_{44}}=\mathrm{j} 0.4-\frac{(\mathrm{j} 0.4)(\mathrm{j} 0.5)}{-\mathrm{j} 0.9}=\mathrm{j} 0.622\)
\(\mathrm{Y}_{21, \mathrm{new}}=\mathrm{Y}_{12, \text { new }}=\mathrm{j} 0.5\)
\(\mathrm{Y}_{22, \text { new }}=\mathrm{Y}_{22}-\frac{\mathrm{Y}_{24} \mathrm{Y}_{42}}{\mathrm{Y}_{44}}=-\mathrm{j} 1.1-\frac{(0)(0)}{-\mathrm{j} 0.9}=-\mathrm{j} 1.1\)
\(\mathrm{Y}_{23, \mathrm{new}}=\mathrm{Y}_{23}-\frac{\mathrm{Y}_{24} \mathrm{Y}_{43}}{\mathrm{Y}_{44}}=\mathrm{j} 0.6-\frac{(0)(\mathrm{j} 0.5)}{-\mathrm{j} 0.9}=\mathrm{j} 0.6\)
\(\mathrm{Y}_{31, \mathrm{new}}=\mathrm{Y}_{13, \mathrm{new}}=\mathrm{j} 0.622\)
\(\mathrm{Y}_{32, \mathrm{new}}=\mathrm{Y}_{23, \mathrm{new}}=\mathrm{j} 0.6\)
\(\mathrm{Y}_{33, \mathrm{new}}=\mathrm{Y}_{33}-\frac{\mathrm{Y}_{34} \mathrm{Y}_{43}}{\mathrm{Y}_{44}}=-\mathrm{j} 1.5-\frac{(\mathrm{j} 0.5)(\mathrm{j} 0.5)}{-\mathrm{j} 0.9}=-\mathrm{j} 1.222\)
The reduced bus admittance matrix after eliminating \(4^{\text {th }}\) row is shown below. \[\mathbf{Y}_{\text {bus }}=\left[\begin{array}{ccc} -\mathrm{j} 1.12 & \mathrm{j} 0.5 & \mathrm{j} 0.622 \\ \mathrm{j} 0.5 & -\mathrm{j} 1.1 & \mathrm{j} 0.6 \\ \mathrm{j} 0.622 & \mathrm{j} 0.6 & -\mathrm{j} 1.222 \end{array}\right]\]
Problem-2
Determine the bus admittance matrix of the system whose reactance diagram is shown in Fig. the currents and admittances are given in p.u.
Determine the reduced bus admittance matrix after eliminating node-3.
Solution-2
\[Y_{\text {bus }}=\left[\begin{array}{rrrr} -\mathrm{j} 2-\mathrm{j} 2-\mathrm{jl} & .0 & \mathrm{j} 2 & \mathrm{j} 1 \\ 0 & -\mathrm{j} 2-\mathrm{j} 4 & 0 & \mathrm{j} 2 \\ \mathrm{j} 2 & 0 & -\mathrm{j} 2-\mathrm{j} 2-\mathrm{j} 5 & \mathrm{j} 5 \\ \mathrm{j} 1 & \mathrm{j} 2 & \mathrm{j} 5 & -\mathrm{j} 1-\mathrm{j} 5-\mathrm{j} 2-\mathrm{j} 1 \end{array}\right]\] \[\therefore Y_{\text {bus }}=\left[\begin{array}{cccr} -\mathrm{j} 5 & 0 & \mathrm{j} 2 & \mathrm{j} 1 \\ 0 & -\mathrm{j} 6 & 0 & \mathrm{j} 2 \\ \mathrm{j} 2 & 0 & -\mathrm{j} 9 & \mathrm{j} 5 \\ \mathrm{j} 1 & \mathrm{j} 2 & \mathrm{j} 5 & -\mathrm{j} 9 \end{array}\right]\]
For eliminating node- 3 , \(Y_{bus}\) is rearranged by interchanging row-3 & row-4, and then interchanging column.3 & column - 4 .
After interchanging row \(-3 \&\) row -4 of \(Y_{\text {bus }}\) matrix we get, \[\mathbf{Y}_{\text {bus }}=\left[\begin{array}{rrrr} -\mathrm{j} 5 & 0 & \mathrm{j} 2 & \mathrm{j} 1 \\ 0 & -\mathrm{j} 6 & 0 & \mathrm{j} 2 \\ \mathrm{j} 1 & \mathrm{j} 2 & \mathrm{j} 5 & -\mathrm{j} 9 \\ \mathrm{j} 2 & 0 & -\mathrm{j} 9 & \mathrm{j} 5 \end{array}\right]\]
After interchanging column \(-3 \&\) column -4 of \(Y_{\text {bus }}\) matrix we get, \[\mathbf{Y}_{\text {bus }}=\left[\begin{array}{cccc} -\mathrm{j} 5 & 0 & \mathrm{j} 1 & \mathrm{j} 2 \\ 0 & -\mathrm{j} 6 & \mathrm{j} 2 & 0 \\ \mathrm{j} 1 & \mathrm{j} 2 & -\mathrm{j} 9 & \mathrm{j} 5 \\ \mathrm{j} 2 & 0 & \mathrm{j} 5 & -\mathrm{j} 9 \end{array}\right]\]
Now the last row & last column [i.e., \(4^{\text {th }}\) row and \(4^{\text {th }}\) column \(]\) of \(\mathrm{Y}_{\text {bus }}\) matrix can be eliminated.
The elements of \(Y_{bus,new}\) after eliminating \(4^{\text {th }}\) row and \(4^{\text {th }}\) column is given by
\[Y_{j k, n e w}=Y_{j k}-\frac{Y_{j n} Y_{n k}}{Y_{n n}} \text { where } n=4 ; j=1,2,3 \text { and } k=1,2,3\]
\(Y_{11, \text { new }}=Y_{11}-\dfrac{Y_{14} Y_{41}}{Y_{44}}=-j 5-\frac{(j 2)(j 2)}{-j 9}=-j 4.556\)
\(Y_{12, new}=Y_{12}-\dfrac{Y_{14} Y_{42}}{Y_{44}}=0-\frac{j 2 \times 0}{-j 9}=0\)
\(Y_{13, new}=Y_{13}-\dfrac{Y_{14} Y_{43}}{Y_{44}}=j 1-\frac{(j 2)(j 5)}{-j 9}=j 2.111\)
\(Y_{21, new}=Y_{12, \text { new }}=0\)
\(Y_{22 . \text { new }}=Y_{22}-\dfrac{Y_{24} Y_{42}}{Y_{44}}=-j 6-\frac{0 \times 0}{-j 9}=-j 6\)
\(Y_{23, new}=Y_{23}-\dfrac{Y_{24} Y_{43}}{Y_{44}}=j 2-\dfrac{0 \times j 5}{-j 9}=j 2\)
\(Y_{31, \text { new }}=Y_{13, new}=j 2.111\)
\(Y_{32, new}=Y_{23, \text { new }}=j 2\)
\(Y_{33, \text { new }}=Y_{33}-\dfrac{Y_{34} Y_{43}}{Y_{44}}=-j 9-\frac{j 5 \times j 5}{-j 9}=-j 6.222\)
The reduced bus admittance matrix after eliminating bus- 3 is given by, \[\mathbf{Y}_{\text {bus }}=\left[\begin{array}{ccc} -\mathrm{j} 4.556 & 0 & \mathrm{j} 2.111 \\ 0 & -\mathrm{j} 6 & \mathrm{j} 2 \\ \mathrm{j} 2.111 & \mathrm{j} 2 & -\mathrm{j} 6.222 \end{array}\right]\]