Question

Modify the matrices $Y_{11},Y_{12}$and $Y_{22}$determined in Problem 11.22 for (a) the case when circuit breakers B32 and B51 open to remove line$3-5$ ; and (b) the case when the load$P_{L3}+j{Q}_{L3}$ is removed.

Short answer

(a) The matrices $Y_{11},Y_{12}$and $Y_{22}$are

$Y_{11}=\left[\begin{array}{cccccc}-j35& -j20& j10& 0& 0& 0\\ j20& -j40& & j10& 0& 0\\ j10& & -j50& 0& j40& 0\\ 0& j10& 0& 2-j50.9& j40& 0\\ 0& 0& j40& j40& 1-j100.3& j20\\ 0& 0& 0& 0& j20& -j30\end{array}\right]$

Y11=-j35-j20j10000j20-j40j1000j103-j12000j1002-j50.9j400000j401-j100.3j200000j20-j30

$Y_{12}=\left[\begin{array}{ccc}j0.5& 0& 0\\ 0& j10& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& j10\end{array}\right]$

and $Y_{22}=\left[\begin{array}{ccc}-j5& 0& 0\\ 0& -j10& 0\\ 0& 0& -j10\end{array}\right]$

(b) The per unit admittance matrix $Y_{11},Y_{12}$and $Y_{22}$are,

$Y_{11}=\left[\begin{array}{cccccc}-j35& -j20& j10& 0& 0& 0\\ j20& -j40& & j10& 0& 0\\ j10& & -j50& 0& j40& 0\\ 0& j10& 0& 2-j50.9& j40& 0\\ 0& 0& j40& j40& 1-j100.3& j20\\ 0& 0& 0& 0& j20& -j30\end{array}\right]$

$Y_{12}=\left[\begin{array}{ccc}j0.5& 0& 0\\ 0& j10& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& j10\end{array}\right]$

and$Y_{22}=\left[\begin{array}{ccc}-j5& 0& 0\\ 0& -j10& 0\\ 0& 0& -j10\end{array}\right]$

Step-by-step solution

Write the given data from the question.

The number of the buses$N=6$

The number of the internal machine buses,$M=3$

The reactance between the bus 1 and 2,$X_{12}=j0.05\text{pu}$

The reactance between the bus 1 and 2,$X_{13}=j0.10\text{pu}$

The reactance between the bus 2 and 4,$X_{24}=j0.10\text{pu}$

The reactance between the bus 3 and 5,$X_{35}=j0.025\text{pu}$

The reactance between the bus 4 and 5,$X_{45}=j0.025\text{pu}$

The reactance between the bus 5 and 6,$X_{56}=j0.05\text{pu}$

Write the reactance of generators $G_1,G_2$and $G_3$.

$$\begin{gathered} {X\text{'}}_{d1}=j0.20 \\ {X\text{'}}_{d2}=j0.10 \\ {X\text{'}}_{d3}=j0.10 \end{gathered}$$

Write the inertial constant of generators$G_1,G_2$ and $G_3$.

$$\begin{gathered} H_1=25\text{pu} \\ {H}_2=6\text{pu} \\ {H}_3=3\text{pu} \end{gathered}$$

The real power on line 3,$P_{L3}=3.0\text{pu}$

The reactive power on line 3,$Q_{L3}=2.0\text{pu}$

The real power on line 4,$P_{L4}=2.0\text{pu}$

The reactive power on line 4,$Q_{L4}=0.9\text{pu}$

The real power on line 5,$P_{L5}=1.0\text{pu}$

The reactive power on line 4,$Q_{L5}=0.3\text{pu}$

 Step 2: Determine the equation to calculate the per-unit admittance matrices Y11,Y12and Y22.

The relationship between the admittance and the impedance is given as follows.

$Z=\frac{1}{Y}$ …… (1)

Here, Z is the impedance between the line and is the admittance between the line.

The admittance matrix to calculate the $Y_{22}$is given as follows.

$$\begin{gathered} Y_{22}=M\times M \\ {Y}_{22}=\left[\begin{array}{ccc}\frac{1}{j{X\text{'}}_{d1}}& 0& 0\\ 0& \frac{1}{j{X\text{'}}_{d2}}& 0\\ 0& 0& \frac{1}{j{X\text{'}}_{d3}}\end{array}\right] \end{gathered}$$.....(2)

The admittance matrix to calculate the$Y_{12}$ is given as follows.

$$\begin{gathered} Y_{12}=N\times M \\ {Y}_{12}=\left[\begin{array}{ccc}-\frac{1}{{X\text{'}}_{d1}}& 0& 0\\ 0& -\frac{1}{{X\text{'}}_{d2}}& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& -\frac{1}{{X\text{'}}_{d3}}\end{array}\right] \end{gathered}$$.....(3)

The equation to calculate the load admittance is given as follows.

$Y_{bus}=\frac{P_L-Q_L}{V^2}$....(4)

Calculate the per-unit admittance matrices  Y11,Y12and Y22when circuit breakers B32 and B51 open to remove line 3-5 .

Since the circuit breaker B32 and B51 circuit breaker are open, the reactance between the line 3 and 5 will not consider.

Draw the single line diagram as,

The admittance of the generator 1,

$$\begin{gathered} {Y\text{'}}_{d1}=-\frac{1}{j0.20} \\ {Y\text{'}}_{d1}=j5\mho \end{gathered}$$

The admittance of the generator 2,

$$\begin{gathered} {Y\text{'}}_{d2}=-\frac{1}{j0.10} \\ {Y\text{'}}_{d2}=j10\mho \end{gathered}$$

The admittance of the generator 3,

$$\begin{gathered} {Y\text{'}}_{d3}=-\frac{1}{j0.10} \\ {Y\text{'}}_{d3}=j10\mho \end{gathered}$$

Calculate the bus admittance matrix by including the load admittances and inverted generator impedances

Calculate the admittance between the line 1 and 2.

$$\begin{gathered} Y_{12}=-\frac{1}{j0.05} \\ {Y}_{12}=j20\mho \end{gathered}$$

Calculate the admittance between the line 1 and 3.

$$\begin{gathered} Y_{13}=Y_{31} \\ {Y}_{13}=-\frac{1}{j0.10} \\ {Y}_{13}=j10 \end{gathered}$$

Calculate the admittance of line 1.

$$\begin{gathered} Y_{11}=-Y_{12}-Y_{13}-{Y\text{'}}_{d1} \\ {Y}_{11}=-j20-j10-j5 \\ {Y}_{11}=-j35\mho \end{gathered}$$

Calculate the admittance between the line 4 and 2.

$$\begin{gathered} Y_{24}=Y_{42} \\ {Y}_{24}=-\frac{1}{j0.10} \\ {Y}_{24}=j10\mho \end{gathered}$$

Calculate the admittance of line 2.

$$\begin{gathered} Y_{22}=-Y_{12}-Y_{24}-{Y\text{'}}_{d2} \\ {Y}_{22}=-j20-j10-j10 \\ {Y}_{22}=-j40\mho \end{gathered}$$

Calculate the admittance between the line 3 and 5.

$$\begin{gathered} Y_{35}=Y_{53} \\ {Y}_{35}=0\text{pu} \end{gathered}$$

Calculate the admittance of line 3.

$$\begin{gathered} Y_{33}=-Y_{13}-Y_{35} \\ {Y}_{33}=-j10-0 \\ {Y}_{33}=-j10\mho \end{gathered}$$

Calculate the admittance between the line 4 and 5.

$$\begin{gathered} Y_{45}=Y_{54} \\ {Y}_{45}=-\frac{1}{j0.025} \\ {Y}_{45}=j40\mho \end{gathered}$$

Calculate the admittance of line 4.

$$\begin{gathered} Y_{44}=-Y_{42}-Y_{45} \\ {Y}_{44}=-j10-j40 \\ {Y}_{44}=-j50\mho \end{gathered}$$

Calculate the admittance between the line 5 and 6.

$$\begin{gathered} Y_{56}=Y_{65} \\ {Y}_{56}=-\frac{1}{j0.05} \\ {Y}_{56}=j20\mho \end{gathered}$$

Calculate the admittance of line 5.

$$\begin{gathered} Y_{55}=-Y_{35}-Y_{45}-Y_{65} \\ {Y}_{55}=-j40-j40-j20 \\ {Y}_{55}=-j100\mho \end{gathered}$$

Calculate the admittance of line 6.

$$\begin{gathered} Y_{66}=-Y_{65}-{Y\text{'}}_{d3} \\ {Y}_{66}=-j20-j10 \\ {Y}_{66}=-j30\mho \end{gathered}$$

Calculate the load admittance matrix at line 3,

$Y_{load3}=\frac{P_{L3}-Q_{L3}}{V_3^2}$

Substitute $3\text{pu}$for $P_{L3}$, 2.0 pu for $Q_{L3}$and 1pu for $V_3$into above equation.

$$\begin{gathered} Y_{load3}=\frac{3-j2}{1^2} \\ {Y}_{load3}=3-j2 \end{gathered}$$

Calculate the load admittance matrix at line 4,

$Y_{load4}=\frac{P_{L4}-Q_{L4}}{V_4^2}$

Substitute 2pu for $P_{L4}$, $0.9\text{pu}$for $Q_{L4}$and $1\text{pu}$for $V_4$into above equation.

$$\begin{gathered} Y_{load4}=\frac{2-j0.9}{1^2} \\ {Y}_{load4}=2-j0.9 \end{gathered}$$

Therefore, the matrix $Y_{11}$,

$Y_{11}=\left[\begin{array}{cccccc}-j35& -j20& j10& 0& 0& 0\\ j20& -j40& & j10& 0& 0\\ j10& & 3-j12& & 0& 0\\ 0& j10& 0& 2-j50.9& j40& 0\\ 0& 0& 0& j40& 1-j100.3& j20\\ 0& 0& 0& 0& j20& -j30\end{array}\right]$

Calculate the matrix $Y_{12}$.

Substitute $j0.20\text{pu}$for $X{\text{'}}_{d1}$,$j0.10\text{pu}$for $X{\text{'}}_{d2}$,and $j0.10\text{pu}$for $X{\text{'}}_{d3}$into equation (3).

$Y_{12}=\left[\begin{array}{ccc}-\frac{1}{j0.2}& 0& 0\\ 0& -\frac{1}{j0.1}& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& -\frac{1}{j0.1}\end{array}\right]$

$Y_{12}=\left[\begin{array}{ccc}j0.5& 0& 0\\ 0& j10& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& j10\end{array}\right]$

Calculate the matrix$Y_{22}$.

Substitute $j0.20\text{pu}$for$X{\text{'}}_{d1}$, $j0.10\text{pu}$for $X{\text{'}}_{d2}$,and $j0.10\text{pu}$for $X{\text{'}}_{d3}$into equation (2).

$Y_{12}=\left[\begin{array}{ccc}\frac{1}{j0.2}& 0& 0\\ 0& \frac{1}{j0.1}& 0\\ 0& 0& \frac{1}{j0.1}\end{array}\right]$

$Y_{22}=\left[\begin{array}{ccc}-j5& 0& 0\\ 0& -j10& 0\\ 0& 0& -j10\end{array}\right]$

Hence the matrices $Y_{11},Y_{12}$and $Y_{22}$are

$Y_{11}=\left[\begin{array}{cccccc}-j35& -j20& j10& 0& 0& 0\\ j20& -j40& & j10& 0& 0\\ j10& & -j50& 0& j40& 0\\ 0& j10& 0& 2-j50.9& j40& 0\\ 0& 0& j40& j40& 1-j100.3& j20\\ 0& 0& 0& 0& j20& -j30\end{array}\right]$

$Y_{12}=\left[\begin{array}{ccc}j0.5& 0& 0\\ 0& j10& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& j10\end{array}\right]$$Y_{22}=\left[\begin{array}{ccc}-j5& 0& 0\\ 0& -j10& 0\\ 0& 0& -j10\end{array}\right]$

Calculate the per-unit admittance matrices Y11,Y12and Y22when the load PL3+jQL3is removed.

Now the reactance between the bus 3 and 5 is also consider because it was removes for the part (1) only.

The admittance $Y_{33},Y_{35}$and $Y_{53}$value is changed.

Calculate the admittance of line 3.

$$\begin{gathered} Y_{33}=-Y_{13}-Y_{35} \\ {Y}_{33}=-j10-j40 \\ {Y}_{33}=-j50\mho \end{gathered}$$

Calculate the admittance between the line 3 and 5.

$$\begin{gathered} Y_{35}=Y_{53} \\ {Y}_{35}=-\frac{1}{j0.025} \\ {Y}_{35}=j40\mho \end{gathered}$$

Calculate the load admittance matrix at line 3,

$Y_{load3}=\frac{P_{L3}-Q_{L3}}{V_3^2}$

Substitute $0\text{pu}$for$P_{L3}$,$0\text{pu}$for $Q_{L3}$and$1\text{pu}$for $V_3$into above equation.

$$\begin{gathered} Y_{load3}=\frac{0-j0}{1^2} \\ {Y}_{load3}=0 \end{gathered}$$

Calculate the load admittance matrix at line 4,

$Y_{load4}=\frac{P_{L4}-Q_{L4}}{V_4^2}$

Substitute 2pu for $P_{L4}$ , $0.9\text{pu}$for $Q_{L4}$and $1\text{pu}$for$V_4$into above equation.

$$\begin{gathered} Y_{load4}=\frac{2-j0.9}{1^2} \\ {Y}_{load4}=2-j0.9 \end{gathered}$$

Calculate the load admittance matrix at line 5,

$Y_{load5}=\frac{P_{L5}-Q_{L5}}{V_5^2}$

Substitute 1pu for $P_{L5}$, $0.3\text{pu}$for $Q_{L5}$and 1pu for $V_5$into above equation.

$$\begin{gathered} Y_{load5}=\frac{1-j0.3}{1^2} \\ {Y}_{load5}=1-j0.3 \end{gathered}$$

The matrices $Y_{12}$and $Y_{22}$is same as 11.23 (a).

Therefore, the matrix ,$Y_{11}$

$Y_{11}=\left[\begin{array}{cccccc}-j35& -j20& j10& 0& 0& 0\\ j20& -j40& & j10& 0& 0\\ j10& & 3-j12& & 0& 0\\ 0& j10& 0& 2-j50.9& j40& 0\\ 0& 0& 0& j40& 1-j100.3& j20\\ 0& 0& 0& 0& j20& -j30\end{array}\right]$

$Y_{12}=\left[\begin{array}{ccc}j0.5& 0& 0\\ 0& j10& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& j10\end{array}\right]$

$Y_{22}=\left[\begin{array}{ccc}-j5& 0& 0\\ 0& -j10& 0\\ 0& 0& -j10\end{array}\right]$$Y_{22}=\left[\begin{array}{ccc}-j5& 0& 0\\ 0& -j10& 0\\ 0& 0& -j10\end{array}\right]$