Electric Power Systems · System Analysis

Admittance Bus (Y-bus) Matrix

Solved Problems — Electric Power Systems

Dr. Mithun Mondal engineeringdevotion.com
Demonstrative Video

Admittance Bus Solved Problems

SECTION 01

Revision of Important Concept

The bus admittance matrix can be formed by inspection using the following guidelines.

  1. The diagonal element \(Y_{i j}\) is given by sum of all the admittances connected to node-j.

  2. 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.

SECTION 02

Problem-1

For the network shown in the Figure,

image

All values are marked in p.u.

SECTION 03

Solution-1

image
\[ \mathbf{Y}_{\text {bus }}=\left[\begin{array}{cccc} -(\mathrm{j} 0.5+\mathrm{j} 0.4+\mathrm{j} 0.4) & \mathrm{j} 0.5 & \mathrm{j} 0.4 & \mathrm{j} 0.4 \\ \mathrm{j} 0.5 & -(\mathrm{j} 0.5+\mathrm{j} 0.6) & \mathrm{j} 0.6 & 0 \\ \mathrm{j} 0.4 & \mathrm{j} 0.6 & -(\mathrm{j} 0.6+\mathrm{j} 0.5+\mathrm{j} 0.4) & \mathrm{j} 0.5 \\ \mathrm{j} 0.4 & 0 & \mathrm{j} 0.5 & -(\mathrm{j} 0.5+\mathrm{j} 0.4) \end{array}\right] \]
\[\mathbf{Y}_{\text {bus }}=\left[\begin{array}{cccc} -\mathrm{j} 1.3 & \mathrm{j} 0.5 & \mathrm{j} 0.4 & \mathrm{j} 0.4 \\ \mathrm{j} 0.5 & -\mathrm{j} 1.1 & \mathrm{j} 0.6 & 0 \\ \mathrm{j} 0.4 & \mathrm{j} 0.6 & -\mathrm{j} 1.5 & \mathrm{j} 0.5 \\ \mathrm{j} 0.4 & 0 & \mathrm{j} 0.5 & -\mathrm{j} 0.9 \end{array}\right]\]
  • \[Y_{j k, n c w}=Y_{j k} \cdots \frac{Y_{j n} Y_{n k}}{Y_{n n}} ; \text { where } n=4 ; j=1,2,3 \text { and } k=1,2,3\]
    column is given by row and The elements of new bus admittance matrix after eliminating the

  • The bus admittance matrix is symmetrical, \(\therefore Y_{k j, n e w}=Y_{j k, \text { new }}\)

  • \(\mathrm{Y}_{11, \text { new }}=\mathrm{Y}_{11}-\frac{\mathrm{Y}_{14} \mathrm{Y}_{41}}{\mathrm{Y}_{44}}=-\mathrm{j} 1.3-\frac{(\mathrm{j} 0.4)(\mathrm{j} 0.4)}{-\mathrm{j} 0.9}=-\mathrm{j} 1.12\)

  • \(\mathrm{Y}_{12, \mathrm{new}}=\mathrm{Y}_{12}-\frac{\mathrm{Y}_{14} \mathrm{Y}_{42}}{\mathrm{Y}_{44}}=\mathrm{j} 0.5-\frac{(\mathrm{j} 0.4 \times 0)}{-\mathrm{j} 0.9}=\mathrm{j} 0.5\)

  • \(\mathrm{Y}_{13, \mathrm{ncw}}=\mathrm{Y}_{13}-\frac{\mathrm{Y}_{14} \mathrm{Y}_{43}}{\mathrm{Y}_{44}}=\mathrm{j} 0.4-\frac{(\mathrm{j} 0.4)(\mathrm{j} 0.5)}{-\mathrm{j} 0.9}=\mathrm{j} 0.622\)

  • \(\mathrm{Y}_{21, \mathrm{new}}=\mathrm{Y}_{12, \text { new }}=\mathrm{j} 0.5\)

  • \(\mathrm{Y}_{22, \text { new }}=\mathrm{Y}_{22}-\frac{\mathrm{Y}_{24} \mathrm{Y}_{42}}{\mathrm{Y}_{44}}=-\mathrm{j} 1.1-\frac{(0)(0)}{-\mathrm{j} 0.9}=-\mathrm{j} 1.1\)

  • \(\mathrm{Y}_{23, \mathrm{new}}=\mathrm{Y}_{23}-\frac{\mathrm{Y}_{24} \mathrm{Y}_{43}}{\mathrm{Y}_{44}}=\mathrm{j} 0.6-\frac{(0)(\mathrm{j} 0.5)}{-\mathrm{j} 0.9}=\mathrm{j} 0.6\)

  • \(\mathrm{Y}_{31, \mathrm{new}}=\mathrm{Y}_{13, \mathrm{new}}=\mathrm{j} 0.622\)

  • \(\mathrm{Y}_{32, \mathrm{new}}=\mathrm{Y}_{23, \mathrm{new}}=\mathrm{j} 0.6\)

  • \(\mathrm{Y}_{33, \mathrm{new}}=\mathrm{Y}_{33}-\frac{\mathrm{Y}_{34} \mathrm{Y}_{43}}{\mathrm{Y}_{44}}=-\mathrm{j} 1.5-\frac{(\mathrm{j} 0.5)(\mathrm{j} 0.5)}{-\mathrm{j} 0.9}=-\mathrm{j} 1.222\)

\[\mathbf{Y}_{\text {bus }}=\left[\begin{array}{ccc} -\mathrm{j} 1.12 & \mathrm{j} 0.5 & \mathrm{j} 0.622 \\ \mathrm{j} 0.5 & -\mathrm{j} 1.1 & \mathrm{j} 0.6 \\ \mathrm{j} 0.622 & \mathrm{j} 0.6 & -\mathrm{j} 1.222 \end{array}\right]\]
row is shown below. The reduced bus admittance matrix after eliminating
SECTION 04

Problem-2

image
SECTION 05

Solution-2

\[Y_{\text {bus }}=\left[\begin{array}{rrrr} -\mathrm{j} 2-\mathrm{j} 2-\mathrm{jl} & .0 & \mathrm{j} 2 & \mathrm{j} 1 \\ 0 & -\mathrm{j} 2-\mathrm{j} 4 & 0 & \mathrm{j} 2 \\ \mathrm{j} 2 & 0 & -\mathrm{j} 2-\mathrm{j} 2-\mathrm{j} 5 & \mathrm{j} 5 \\ \mathrm{j} 1 & \mathrm{j} 2 & \mathrm{j} 5 & -\mathrm{j} 1-\mathrm{j} 5-\mathrm{j} 2-\mathrm{j} 1 \end{array}\right]\]

  • For eliminating node- 3 , \(Y_{bus}\) is rearranged by interchanging row-3 & row-4, and then interchanging column.3 & column - 4 .

  • \[\mathbf{Y}_{\text {bus }}=\left[\begin{array}{rrrr} -\mathrm{j} 5 & 0 & \mathrm{j} 2 & \mathrm{j} 1 \\ 0 & -\mathrm{j} 6 & 0 & \mathrm{j} 2 \\ \mathrm{j} 1 & \mathrm{j} 2 & \mathrm{j} 5 & -\mathrm{j} 9 \\ \mathrm{j} 2 & 0 & -\mathrm{j} 9 & \mathrm{j} 5 \end{array}\right]\]
    matrix we get, row -4 of After interchanging row
  • \[\mathbf{Y}_{\text {bus }}=\left[\begin{array}{cccc} -\mathrm{j} 5 & 0 & \mathrm{j} 1 & \mathrm{j} 2 \\ 0 & -\mathrm{j} 6 & \mathrm{j} 2 & 0 \\ \mathrm{j} 1 & \mathrm{j} 2 & -\mathrm{j} 9 & \mathrm{j} 5 \\ \mathrm{j} 2 & 0 & \mathrm{j} 5 & -\mathrm{j} 9 \end{array}\right]\]
    matrix we get, column -4 of After interchanging column

  • 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

\[Y_{j k, n e w}=Y_{j k}-\frac{Y_{j n} Y_{n k}}{Y_{n n}} \text { where } n=4 ; j=1,2,3 \text { and } k=1,2,3\]

  • \(Y_{11, \text { new }}=Y_{11}-\dfrac{Y_{14} Y_{41}}{Y_{44}}=-j 5-\frac{(j 2)(j 2)}{-j 9}=-j 4.556\)

  • \(Y_{12, new}=Y_{12}-\dfrac{Y_{14} Y_{42}}{Y_{44}}=0-\frac{j 2 \times 0}{-j 9}=0\)

  • \(Y_{13, new}=Y_{13}-\dfrac{Y_{14} Y_{43}}{Y_{44}}=j 1-\frac{(j 2)(j 5)}{-j 9}=j 2.111\)

  • \(Y_{21, new}=Y_{12, \text { new }}=0\)

  • \(Y_{22 . \text { new }}=Y_{22}-\dfrac{Y_{24} Y_{42}}{Y_{44}}=-j 6-\frac{0 \times 0}{-j 9}=-j 6\)

  • \(Y_{23, new}=Y_{23}-\dfrac{Y_{24} Y_{43}}{Y_{44}}=j 2-\dfrac{0 \times j 5}{-j 9}=j 2\)

  • \(Y_{31, \text { new }}=Y_{13, new}=j 2.111\)

  • \(Y_{32, new}=Y_{23, \text { new }}=j 2\)

  • \(Y_{33, \text { new }}=Y_{33}-\dfrac{Y_{34} Y_{43}}{Y_{44}}=-j 9-\frac{j 5 \times j 5}{-j 9}=-j 6.222\)

\[\mathbf{Y}_{\text {bus }}=\left[\begin{array}{ccc} -\mathrm{j} 4.556 & 0 & \mathrm{j} 2.111 \\ 0 & -\mathrm{j} 6 & \mathrm{j} 2 \\ \mathrm{j} 2.111 & \mathrm{j} 2 & -\mathrm{j} 6.222 \end{array}\right]\]
The reduced bus admittance matrix after eliminating bus- 3 is given by,