Question

Repeat Problem 6.32 with the slack bus voltage changed to$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\angle }{\mathbf{30}}^{\mathbf{\circ }}$per unit.

Short answer

The voltage at bus 2 after three iterations is $V_2=0.963+0.115jper\mathit{}unit$.

Step-by-step solution

Given data

For a two-bus power system, $P-jQ=-0.7-(j0.4)$per unit load at bus 2, with the series impedance of $Z_L=0.05+j0.1$per unit with a fixed per unit voltage of $1.0\angle {30}^{\circ }$per unit.

General formula for Gauss-Seidel method

Write the general formula for Gauss-Seidel,

${\mathbf{X}}_{\mathbf{k}}\mathbf{(}\mathbf{i}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{A}}_{\mathbf{kk}}}\mathbf{\left[}{\mathbf{y}}_{\mathbf{k}}\mathbf{-}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{k}\mathbf{-}\mathbf{1}}{\mathbf{A}}_{\mathbf{kn}}{\mathbf{x}}_{\mathbf{n}}\mathbf{(}\mathbf{i}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{-}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{k}\mathbf{+}\mathbf{1}}^{\mathbf{N}}{\mathbf{A}}_{\mathbf{kn}}{\mathbf{x}}_{\mathbf{n}}\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{\right]}$ ......(1)

Here, $x_k(i+1)$is the k^th component of $x\left(\mathrm{i}\right)$; $x\left(\mathrm{i}\right)$is the i^thguess of the iteration; $A_{\mathrm{kk}}$

represents the coefficient of the k^th component of $x\left(\mathrm{i}\right)$; $y_k$is other variable of k^th component and N is the total number of buses.

Determine bus admittance.

Calculate the bus admittance,

$Y=\frac{1}{{\mathrm{Z}}_L}$

Substitute the given value of $Z_L$in the above equation,

$Y=\frac{1}{{\mathrm{Z}}_L}$

$=\frac{1}{0.05+j0.1}$

$=4-j8\mathrm{p}.\mathrm{u}.$

$=8.944\angle -63.{43}^{\circ }p\mathit{.}u\mathit{.}$

Matrix form of the above calculated admittance can be written as,

$$\begin{gathered} Y_{\mathrm{bus}}=\left[\begin{array}{cc}{Y}_{\mathit{11}}& Y_{12}\\ Y_{\mathit{21}}& Y_{\mathit{22}}\end{array}\right] \\ =\left[\begin{array}{cc}Y& -Y\\ -Y& Y\end{array}\right] \\ =\left[\begin{array}{cc}4-j8& -4+\mathrm{j}8\\ -4+\mathrm{j}8& 4-\mathrm{j}8\end{array}\right] \end{gathered}$$

Rewriting the matrix equation in Gauss-Seidel form

Rewrite the bus voltage equation in the general form of Gauss-Seidel,

$V_{\mathrm{k}}(\mathrm{i}+1)=\frac{1}{Y_{\mathrm{kk}}}\left[\frac{P_k-j{Q}_k}{V_k^{*}\mathit{\left(}I\mathit{\right)}}-\sum _{\mathrm{n}=1}^{\mathrm{k}-1}{Y}_{\mathrm{kn}}{V}_{\mathrm{n}}(\mathrm{i}+1)-\sum _{\mathrm{n}=\mathrm{k}+1}^{\mathrm{N}}{Y}_{\mathrm{kn}}{V}_{\mathrm{n}}\left(\mathrm{i}\right)\right]$

Here, $V_{\mathit{k}}$is the bus voltage of the k^th bus, P is the real power flowing on the k^th bus

and Q is the reactive power of the k^th bus,localid="1655286758775" $P_k-j{Q}_k=-0.7-(-j0.4)p\mathit{.}u\mathit{.}$

Solving the first iteration with i=0

For $i=0$,

Calculate the voltage on the second bus, i.e. $k=2,$

Substitute the value 0 for i and $\left[-0.7-(-j0.4)\right]$for $P_k\mathit{-}j{Q}_k$in equation (1)

Substitute the value $1\angle {30}^{\circ }$for $V_1$and $\left[-4+8j\right]$for $Y_{21}$from the bus admittance

matrix,

$$\begin{gathered} V_2\left(1\right)=\frac{1}{4-j8}\left[\frac{\left[-0.7-(-0.4j)\right]}{1\angle 0^{\circ }}-(-4+8j)(1\angle {30}^{\circ }\right] \\ =\frac{-0.7+0.4\mathrm{j}-(-4+8\mathrm{j})(0.866+j0.5)}{4-j8} \\ =\frac{-0.7+0.4j-(-3.464-2j+6.928j-4)}{4-\mathrm{j}8} \\ =\frac{-0.7+0.4\mathrm{j}-(-7.464+4.928j)}{4-\mathrm{j}8} \end{gathered}$$

Solve further as,

$V_2\left(1\right)=\frac{6.764+4.528j}{4-j8}$

Multiplying and dividing RHS by $(4-j8)$in the above equation,

$$\begin{gathered} V_2\left(1\right)=\frac{6.764+4.528j}{4-j8}\times \frac{4+j\mathit{8}}{4+\mathrm{j}\mathit{8}} \\ =\frac{27.05+18.112j+54.112j-36.224}{16-32j+32j+64} \\ =\frac{-9.174+72.224j}{80} \\ =-0.114+j0.9028p\mathit{.}u\mathit{.} \end{gathered}$$

In polar form the value of $V_2\left(1\right)$will look like,

$V_2\left(1\right)=0.910\angle -82.{803}^{\circ }p\mathit{.}u\mathit{.}$

Solving the second iteration withi=1 .

For $i=1$,

Calculate the voltage on the second bus, i.e.localid="1655287070930" $k=2$,

Substitute the value 1 for i and $\left[-0.7-(-j0.4)\right]$for $P_k\mathit{-}j{Q}_k$in equation (1)

localid="1655287477980" $$\begin{gathered} V_{\mathrm{k}}(\mathrm{i}+1)=\frac{1}{Y_{\mathrm{kk}}}\left[\frac{P_k-j{Q}_k}{V_k^{*}\mathit{\left(}I\mathit{\right)}}-\sum _{\mathrm{n}=1}^{\mathrm{k}-1}{Y}_{\mathrm{kn}}{V}_{\mathrm{n}}(\mathrm{i}+1)-\sum _{\mathrm{n}=\mathrm{k}+1}^{\mathrm{N}}{Y}_{\mathrm{kn}}{V}_{\mathrm{n}}\left(\mathrm{i}\right)\right] \\ {V}_2(1+1)=\frac{1}{Y_{22}}\left[\frac{P_2-j{Q}_2}{V_2^{*}\left(1\right)}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{2n}{V}_{\mathrm{n}}(0+1)-\sum _{\mathrm{n}=2+1}^{\mathrm{N}}{Y}_{2n}{V}_{\mathrm{n}}\left(1\right)\right] \\ {V}_2\left(2\right)=\frac{1}{4-j8}\left[\frac{-0.7-(-0.4j)}{0.910\angle -82.{803}^{\circ }}-\sum _{\mathrm{n}=1}^1{Y}_{21}{V}_1(1+1)-\sum _{\mathrm{n}=2+1}^2{Y}_{\mathrm{kn}}{V}_{\mathrm{n}}\left(1\right)\right] \end{gathered}$$

$=\frac{1}{4-j8}\left[\frac{-0.7+0.4j}{0.910\angle -82.{803}^{\circ }}-Y_{21}{V}_1\left(2\right)-0\right]$

Substitute the value localid="1655287077285" $1\angle {30}^{\circ }$for $V_1$and $\left[4-8j\right]$for $Y_{21}$from the bus admittance

matrix,

$$\begin{gathered} V_2\left(2\right)=\frac{1}{4-j8}\left[\frac{\left[-0.7+0.4j\right]}{(-0.114+j0.9028)}-(-4+8j)(1\angle {30}^{\circ })\right] \\ =\left[\begin{array}{c}\frac{1}{\left(4-j8\right)\left(-0.114+j0.9028\right)}\\ \left[(-0.7+0.4j)-(-4+8j)(0.866+j0.5)(-0.114+j0.9028)\right]\end{array}\right] \\ =\frac{\left[(-0.7+0.4j)-(-7.464+4.928j)(-0.114+j0.9028)\right]}{6.764} \end{gathered}$$

Further solving,

$$\begin{gathered} V_2\left(2\right)=\frac{\left[-0.7+0.4j)-(-7.464+4.928j\left)\right(-0.114+j0.9028\right]}{6.764} \\ =\frac{(-0.7+0.4j)-(0.85-6.73j-0.56j-4.44)}{6.764} \\ =\frac{(-0.7+0.4j)-(-3.59-7.29j)}{6.764} \\ ==\frac{-0.7+0.4j+3.59-7.29j}{6.764} \end{gathered}$$

Solve further as,

$$\begin{gathered} V_2\left(2\right)=\frac{2.89+7.69\mathrm{j}}{6.764} \\ =\frac{2.89+7.69\mathrm{j}}{6.764} \\ =0.427+1.13j\mathit{}p\mathit{.}u\mathit{.} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=1.208\angle 69.{30}^{\circ }p\mathit{.}u\mathit{.} \end{gathered}$$

Solving the third iteration with i=2.

For $i\mathit{=}\mathit{2}$

Calculate the voltage on the second bus, i.e. $k=2$,

Substitute the value 2 for i and $\left[-0.7-(-j0.4)\right]$for $P_k\mathit{-}j{Q}_k$in equation (1)

$$\begin{gathered} V_{\mathrm{k}}(\mathrm{i}+1)=\frac{1}{Y_{\mathrm{kk}}}\left[\frac{P_k-j{Q}_k}{V_k^{*}\mathit{\left(}i\mathit{\right)}}-\sum _{\mathrm{n}=1}^{\mathrm{k}-1}{Y}_{\mathrm{kn}}{V}_{\mathrm{n}}(\mathrm{i}+1)-\sum _{\mathrm{n}=\mathrm{k}+1}^{\mathrm{N}}{Y}_{\mathrm{kn}}{V}_{\mathrm{n}}\left(\mathrm{i}\right)\right] \\ {V}_2(2+1)=\frac{1}{Y_{22}}\left[\frac{P_2-j{Q}_2}{V_2^{*}\mathit{\left(}\mathit{2}\mathit{\right)}}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{2n}{V}_{\mathrm{n}}(2+1)-\sum _{\mathrm{n}=2+1}^{\mathrm{N}}{Y}_{2n}{V}_{\mathrm{n}}\left(2\right)\right] \\ {V}_2\left(3\right)=\frac{1}{4-j8}\left[\frac{-0.7-(-0.4j)}{1.208\angle 69.{30}^{\circ }}-\sum _{\mathrm{n}=1}^1{Y}_{21}{V}_1(2+1)-\sum _{\mathrm{n}=2+1}^2{Y}_{kn}{V}_{\mathrm{n}}\left(1\right)\right] \\ {V}_2\left(3\right)=\frac{1}{4-j8}\left[\frac{-0.7+0.4j}{1.208\angle 69.{30}^{\circ }}-Y_{21}{V}_1\left(3\right)-0\right] \end{gathered}$$

Substitute the value $1\angle {30}^{\circ }$for $V_1$and $\left[4-8j\right]$for $Y_{21}$from the bus admittance

matrix,

$V_2\left(3\right)=\frac{1}{4-j8}[\frac{\left[-0.7+0.4j\right]}{\left(0.427+1.13j\right)}-(-4-j8\left)\right(1\angle {30}^{\circ }\left)\right]$

$=\left[\begin{array}{c}\frac{1}{\left(4-j8\right)\left(0.427+1.13j\right)}\\ \left[(-0.7+0.4j)-(-4+8j)(0.866+j0.5)(-0.427+1.13j)\right]\end{array}\right]$

$=\left[\begin{array}{c}\frac{1}{1.708-3.416j+4.52j+9.04}\\ \left[(-0.7+0.4j)-(-3.464+6.928j-2j-4)(0.427+1.13j)\right]\end{array}\right]$

$=\frac{\left[(-0.7+0.4j)-(-7.464+4.928j)(0.427+1.13j)\right]}{10.748+1.104j}$

$$\begin{gathered} V_2\left(3\right)=\frac{\left[(-0.7+0.4j)-(-7.464+4.928j)(0.427+1.13j)\right]}{10.748+1.104j} \\ =\frac{(-0.7+0.4j)-(-3.18-8.434j)+(2.10j-5.568)}{10.748+1.104j} \\ =\frac{(-0.7+0.4j)-(-8.74-6.334j)}{10.748+1.104j} \\ =\frac{-0.7+0.4j+8.74-6.334j)}{10.748+1.104j} \end{gathered}$$

Further solving,

$$\begin{gathered} V_2\left(3\right)=\frac{8.04+6.734j}{10.748+1.104j} \\ =\frac{8.04+6.734j}{10.748+1.104j} \\ =\frac{10.448\angle 39.{951}^{\circ }}{10.805\angle 5.{865}^{\circ }} \\ =0.97\angle 6.{811}^{\circ }p\mathit{.}u\mathit{.} \end{gathered}$$

In rectangular form the value of $V_2\left(3\right)$will look like,

$V_2\left(3\right)=0.963+0.115jp.u.$

From the above iterations it can be seen that the value of second bus voltage after

3^rd iteration is achieved as $0.963+0.115jp.u.$