Question

For the three-bus system whose ${\mathbf{Y}}_{\mathbf{b}\mathbf{u}\mathbf{s}}$is given, calculate the second iteration value of ${\mathbf{V}}_{\mathbf{3}}$ using the Gauss-Seidel method. Assume bus 1 as the slack (with${\mathbf{V}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\angle }{\mathbf{0}}^{\mathbf{\circ }}$ ), and buses 2 and 3 are load buses with a per-unit load ${\mathbf{S}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{5}$ and . Use voltage guesses of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\angle }{\mathbf{0}}^{\mathbf{\circ }}$at both buses 2 and 3. The bus admittance matrix for a three-bus system is

${\mathbf{Y}}_{\mathbf{bus}}\mathbf{=}\mathbf{\left[}\begin{array}{ccc}\mathbf{-}\mathbf{j}\mathbf{10}& \mathbf{j}\mathbf{5}& \mathbf{j}\mathbf{5}\\ \mathbf{j}\mathbf{5}& \mathbf{-}\mathbf{j}\mathbf{10}& \mathbf{j}\mathbf{5}\\ \mathbf{j}\mathbf{5}& \mathbf{j}\mathbf{5}& \mathbf{-}\mathbf{j}\mathbf{10}\end{array}\mathbf{\right]}$

Short answer

The voltage at bus 3 after 2 iterations is achieved as$V_3\left(2\right)=1.115\angle 12.6^{\circ }\mathrm{per}\mathrm{unit}$.

Step-by-step solution

Given data

The bus admittance matrix for a three-bus system is

${\mathrm{Y}}_{\mathrm{bus}}=\left[\begin{array}{ccc}-\mathrm{j}10& \mathrm{j}5& \mathrm{j}5\\ \mathrm{j}5& -\mathrm{j}10& \mathrm{j}5\\ \mathrm{j}5& \mathrm{j}5& -\mathrm{j}10\end{array}\right]$

Voltage of bus 1 is ${\mathrm{V}}_1=1.0\angle 0^{\circ }$and load buses

$$\begin{gathered} {\mathrm{S}}_2=1+\mathrm{j}0.5\mathrm{per}\mathrm{unit} \\ ={\mathrm{P}}_2+j{Q}_2 \end{gathered}$$

Also,

$$\begin{gathered} {\mathrm{S}}_3=1.5+\mathrm{j}0.75\mathrm{per}\mathrm{unit} \\ ={\mathrm{P}}_3+j{Q}_3 \end{gathered}$$

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, $x_k(\mathrm{i}+1)$ is the k^th component of $x\left(\mathrm{i}\right)$; $x\left(\mathrm{i}\right)$ is the i^thguess of the iteration; $A_{KK}$ represents the coefficient of the k^th component of $\mathrm{x}\left(\mathrm{i}\right)$; $y_k$ is other variable of k^th component and N is the total number of buses.

Solving the first iteration by using Gauss-Seidel method

Rewrite the voltage equation in Gauss-Seidel form,

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

Calculate the voltage for bus 2 and consider voltages of bus 1 & 2 both to be $1\angle 0^{\circ }$.

For $i=0$, substitute the value $-j10$for $Y_{22}$ and $1+j0.5per\mathit{}unit$ for $P_2+j{Q}_2$

$V_2(0+1)=\frac{1}{Y_{22}}\left[\frac{P_2-j{Q}_2}{V_2^{*}\left(0\right)}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{2n}{V}_{\mathrm{n}}(0+1)-\sum _{\mathrm{n}=2+1}^{\mathrm{N}}{Y}_{2n}{V}_{\mathrm{n}}\left(0\right)\right]$

$V_2\left(1\right)=\frac{1}{-j10}\left[\frac{1-j0.5}{1\angle 0^{\circ }}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{2n}{V}_{\mathrm{n}}\left(1\right)-\sum _{\mathrm{n}=2+1}^3{Y}_{2n}{V}_{\mathrm{n}}\left(0\right)\right]$

$V_2\left(1\right)=\frac{1}{-j10}\left[\frac{1-j0.5}{1\angle 0^{\circ }}-Y_{21}{V}_1\left(1\right)-Y_{23}{V}_3\left(0\right)\right]$

$V_2\left(1\right)=\frac{1}{-j10}[\frac{1-j0.5}{1\angle 0^{\circ }}-Y_{21}1\angle 0^{\circ }-Y_{23}1\angle 0^{\circ }]$

Substitute the value$j5$for both$Y_{21}$and $Y_{\mathit{23}}$,

$V_2\left(1\right)=\frac{1}{-j10}[\frac{1-j0.5}{1\angle 0^{\circ }}-Y_{21}1\angle 0^{\circ }-Y_{23}1\angle 0^{\circ }]$

${\mathrm{V}}_2\left(1\right)=\frac{1}{-\mathrm{j}10}[\frac{1-\mathrm{j}0.5}{1\angle 0^{\circ }}-(\mathrm{j}5)1\angle 0^{\circ }-(\mathrm{j}5)1\angle 0^{\circ }]$

$V_2\left(1\right)=\frac{1}{-j10}[\frac{1-j0.5}{(1+j0)}-(j5\left)\right(1+j0)-(j5\left)\right(1+j0\left)\right]$

$V_2\left(1\right)=\frac{1}{-j10}[\frac{1-j0.5}{(1+j0)}-j10]$

Further Solving,

$V_2\left(1\right)=\left[\frac{1-j0.5-j10(1+j0)}{(1+j0)(-j10)}\right]$

$V_2\left(1\right)=\left[\frac{1-j0.5-j10}{(-j10)}\right]$

$$\begin{gathered} V_2\left(1\right)=\left[\frac{10.548\angle -84.56}{10\angle -90}\right] \\ {V}_2\left(1\right)=1.0548\angle 0.939\mathrm{per}\mathrm{unit} \end{gathered}$$

Voltage of the 2^ndbus in the polar form is,

$V_2\left(1\right)=1.055+0.017j\mathrm{per}\mathrm{unit}$

Solving the first iteration by using Gauss-Seidel method of bus 3(k=3)

Calculate the voltage for bus 3 and consider voltages of bus 1 & 3 both to be $1\angle 0^{\circ }$.

For $i=0$, substitute the value$-j10$ for$Y_{33}$ and $j5$ for $Y_{31}$ also $1.5-j0.75\mathit{}per\mathit{}unit$ for $P_3\mathit{-}\mathit{}j{Q}_3$in equation (1)

$V_3(0+1)=\frac{1}{Y_{33}}\left[\frac{P_3-j{Q}_3}{V_3^{*}\left(0\right)}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{3n}{V}_{\mathrm{n}}(0+1)-\sum _{\mathrm{n}=3+1}^{\mathrm{N}}{Y}_{3n}{V}_{\mathrm{n}}\left(0\right)\right]$

$V_3\left(1\right)=\frac{1}{-j10}\left[\frac{1.5-j0.75}{1\angle 0^{\circ }}-\sum _{\mathrm{n}=1}^1{Y}_{3n}{V}_{\mathrm{n}}\left(1\right)-\sum _{\mathrm{n}=3+1}^3{Y}_{3n}{V}_{\mathrm{n}}\left(0\right)\right]$

$V_3\left(1\right)=\frac{1}{-j10}\left[\frac{1.5-j0.75}{1\angle 0^{\circ }}-Y_{31}{V}_1\left(1\right)-Y_{31}{V}_1\left(0\right)\right]$

$V_3\left(1\right)=\frac{1}{-j10}\left[\frac{1.5-j0.75}{1\angle 0^{\circ }}-\left(j5\right)1\angle 0^{\circ }-\left(j5\right)(1.055+0.017j)\right]$

Further Solving,

$$\begin{gathered} V_3\left(1\right)=\frac{1}{-j10}\left[\frac{1.5-j0.75}{1\angle 0^{\circ }}-\left(j5\right)-\left(j5\right)(1.055+0.017j)\right] \\ {V}_3\left(1\right)=\frac{1}{-j10}\left[\frac{1.5-j0.75}{1\angle 0^{\circ }}-\left(j5\right)-(5.275j+0.085)\right] \\ {V}_3\left(1\right)=\left[\frac{1.5-j0.75-\left(j5\right)(1+j0)-(5.275j+0.085)(1+j0)}{(10\angle -{90}^{\circ })(1+j0)}\right] \\ {V}_3\left(1\right)=\left[\frac{1.5-j0.75-j5-5.275j-0.085)}{(10\angle -{90}^{\circ })}\right] \end{gathered}$$

The value of bus 3 voltage is as below,

$$\begin{gathered} V_3\left(1\right)=\left[\frac{1.415-j11.025}{(10\angle -{90}^{\circ })}\right] \\ {V}_3\left(1\right)=\left[\frac{11.115\angle 82.686}{(10\angle -{90}^{\circ })}\right] \end{gathered}$$

$V_3\left(1\right)=1.11\angle 0.918\mathrm{per}\mathrm{unit}$

The rectangular form of $V_3\left(1\right)$is

$V_3\left(1\right)=1.11+0.018j\mathrm{per}\mathrm{unit}$

Solving the second iteration by using Gauss-Seidel method of bus 2(k=2)

Calculate the voltage for bus 2 and consider voltage of bus 2 to be$1\angle 0^{\circ }$.

For $i=1$, substitute the value $-j10$ for $Y_{33}$ and $j5$ for $Y_{31}$ also $1.5-j0.75\mathit{}per\mathit{}unit$

for $P_3\mathit{-}\mathit{}j{Q}_3$in equation (1)

$$\begin{gathered} V_2(1+1)=\frac{1}{Y_{22}}\left[\frac{P_2-j{Q}_2}{V_2^{*}\left(1\right)}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{2n}{V}_{\mathrm{n}}(1+1)-\sum _{\mathrm{n}=2+1}^{\mathrm{N}}{Y}_{2n}{V}_{\mathrm{n}}\left(1\right)\right] \\ {V}_2\left(2\right)=\frac{1}{-j10}\left[\frac{1-j0.5}{1.0548\angle 0.939}-\sum _{\mathrm{n}=1}^1{Y}_{2n}{V}_{\mathrm{n}}\left(2\right)-\sum _{\mathrm{n}=2+1}^3{Y}_{2n}{V}_{\mathrm{n}}\left(1\right)\right] \\ {V}_2\left(2\right)=\frac{1}{-j10}\left[\frac{1-j0.5}{1.0548\angle 0.939}-Y_{21}{V}_1\left(2\right)-Y_{23}{V}_3\left(1\right)\right] \\ {V}_2\left(2\right)=\frac{1}{-j10}\left[\frac{1-j0.5}{1.0548\angle 0.939}-\left(j5\right)1\angle 0^{\circ }-\left(j5\right)(1.11\angle 0.918)\right] \end{gathered}$$

Further solving,

$V_2\left(2\right)=\frac{1}{-j10}\left[\frac{1-j0.5-j5(1.0548\angle 0.939)-(5.55\angle 4.59)(1.0548\angle 0.939)}{(1.0548\angle 0.939)}\right]$

$$\begin{gathered} V_2\left(2\right)=\frac{1}{-j10}\left[11.063\angle -79.{64}^{\circ }\right] \\ {V}_2\left(2\right)=1.1063\angle 10.{35}^{\circ }\mathrm{per}\mathrm{unit} \end{gathered}$$

Solving the second iteration by using Gauss-Seidel method of bus 3(k=3)

Calculate the voltage for bus 3 and consider voltage of bus 2 both to be$1\angle 0^{\circ }$.

For$i=1$, substitute the value$-j10$for$Y_{33}$ and $j5$ for $Y_{31}$ also $1.5-j0.75\mathit{}per\mathit{}unit$

for $P_3\mathit{-}\mathit{}j{Q}_3$in equation (1)

$V_3(1+1)=\frac{1}{Y_{33}}\left[\frac{P_3-j{Q}_3}{V_2^{*}\left(1\right)}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{2n}{V}_{\mathrm{n}}(1+1)-\sum _{\mathrm{n}=2+1}^{\mathrm{N}}{Y}_{2n}{V}_{\mathrm{n}}\left(1\right)\right]$

$V_3\left(2\right)=\frac{1}{-10j}\left[\frac{1.5-j0.75}{1.0548\angle 0.939}-\sum _{\mathrm{n}=1}^1{Y}_{2n}{V}_{\mathrm{n}}\left(2\right)-\sum _{\mathrm{n}=2+1}^3{Y}_{3n}{V}_{\mathrm{n}}\left(1\right)\right]$

$$\begin{gathered} V_3\left(2\right)=\frac{1}{-10j}\left[\frac{1.5-j0.75}{1.0548\angle 0.939}-Y_{21}{V}_1\left(2\right)-Y_{32}{V}_3\left(1\right)\right] \\ {V}_3\left(2\right)=\frac{1}{-j10}\left[\begin{array}{c}1.5-j0.75-\left(j5\right)(1\angle 0^{\circ })(1.0548\angle 0.939)\\ -\left(j5\right)(1.11\angle 0.918)(1.0548\angle 0.939)\end{array}\right] \end{gathered}$$

Solving the above equation,

$$\begin{gathered} V_3\left(2\right)=\frac{11.15\angle 77.{38}^{\circ }}{-10j} \\ {\mathrm{V}}_3\left(2\right)=1.115\angle 12.6^{\circ }\mathrm{per}\mathrm{unit} \end{gathered}$$

Thus, the voltage at bus 3 after 2 iterations is$1.115\angle 12.6^{\circ }\mathrm{per}\mathrm{unit}$.