Question

The bus admittance matrix of a three-bus power system is given by${\mathit{Y}}_{\mathbf{B}\mathbf{u}\mathbf{s}}\mathbf{=}\mathbf{-}\mathit{j}\left[\begin{array}{ccc}7& -2& -5\\ -2& 6& -4\\ -5& -4& 9\end{array}\right]\mathit{p}\mathit{e}\mathit{r}\mathit{u}\mathit{n}\mathit{i}\mathit{t}$

with${\mathit{V}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\angle }\mathbf{0}\mathbf{°}$ per unit;${\mathit{V}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}$ per unit; ${\mathit{P}}_{\mathbf{2}}\mathbf{=}\mathbf{60}\mathit{M}\mathit{W}$;${\mathit{P}}_{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{80}\mathit{M}\mathit{W}$; ${\mathit{Q}}_{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{60}\mathit{M}\mathit{V}\mathit{a}\mathit{r}$ (lagging) as a part of the power flow solution of the system. Find${\mathit{V}}_{\mathbf{2}}$ and ${\mathit{V}}_{\mathbf{3}}$within a tolerance of per unit by using the Gauss Seidel iteration method. Start with role="math" localid="1656307740606" ${\mathit{\delta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{V}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}$per unit, and${\mathit{\delta }}_{\mathbf{3}}\mathbf{=}\mathbf{0}$ .

Short answer

The voltage at bus 2 and 3 are $1.006\angle 4.29°\text{\hspace{0.33em}pu}$and$0.925\angle -3.57°\text{\hspace{0.33em}pu}$ .

Step-by-step solution

Write the given data from the question.

Total number of the buses,$N=3$

The admittance matrix,

$Y_{Bus}=-j\left[\begin{array}{ccc}7& -2& -5\\ -2& 6& -4\\ -5& -4& 9\end{array}\right]perunit$

Write the voltages of the buses.

$\begin{array}{l}{V}_1=1\angle 0°\text{\hspace{0.33em}pu}\\ V_2=1\angle 0°\text{\hspace{0.33em}pu}\\ V_3=1\angle 0°\text{\hspace{0.33em}pu}\end{array}$

Write the power on the buses.

$\begin{array}{c}{P}_2=60\text{\hspace{0.33em}MW}\\ P_3=-80\text{\hspace{0.33em}MW}\\ Q_3=-60\text{\hspace{0.33em}MVar}\end{array}$

Determine the equation to calculate the voltages V2and V3.

The equation to calculate the voltage on the bus is given as follows.

$V_{\mathrm{k}}(\mathrm{i}+1)=\frac{1}{{\mathrm{Y}}_{\mathrm{kk}}}[\frac{{\mathrm{P}}_{\mathrm{k}}-{\mathrm{jQ}}_{\mathrm{k}}}{{\mathrm{V}}_{\mathrm{k}}^{*}\left(\mathrm{i}\right)}-\sum _{\mathrm{n}=1}^{\mathrm{k}-1}{\mathrm{Y}}_{\mathrm{kn}}{\mathrm{V}}_{\mathrm{n}}\left(\mathrm{i}+1\right)-\sum _{\mathrm{n}=\mathrm{k}+1}^{\mathrm{N}}{\mathrm{Y}}_{\mathrm{kn}}{\mathrm{V}}_{\mathrm{n}}\left(\mathrm{i}\right)]$ …… (1)

Here,$\mathit{N}$is the total number of the buses, $\mathit{i}$is the number of the iteration and$\mathit{k}$is the bus number.

The equation to calculate the reactive power on bus$2$is given as follows.

$Q_2\left(i\right)=-lm\left[V_2^{*}\left(i\right)\sum _{k=1}^N{Y}_{2k}{V}_k\left(i\right)\right]$ …… (2)

Calculate the voltages v2andV3 . 

Iteration 1:

Calculate the reactive power on bus 2,

Substitute$3$for$N$, $2$for $K$, $0$for$i$into the equation (2).

$\begin{array}{l}{Q}_2\left(0\right)=-\text{Im}\left[V_2^{*}\left(0\right)\sum _{k=1}^3{Y}_{2k}{V}_k\right]\\ Q_2\left(0\right)=-\text{Im}\left[V_2^{*}\left(0\right)Y_{21}{V}_1\left(0\right)+V_2^{*}\left(0\right)Y_{22}{V}_2\left(0\right)+V_2^{*}\left(0\right)Y_{23}{V}_3\left(0\right)\right]\end{array}$

Substitute$2$$j$for $Y_{21}$,role="math" localid="1656308563640" $-6j$for$Y_{22}$, role="math" localid="1656308572810" $4j$for$Y_{23}$, $1\angle 0°$for$V_1\left(0\right)$, $1\angle 0°$for role="math" localid="1656308720343" $V_2\left(0\right)$and$1\angle 0°$for $V_3\left(0\right)$ into above equation.

$\begin{array}{l}{Q}_2\left(0\right)=-\text{Im}\left[{\left(1\angle 0°\right)}^{*}\left(-2j\right)\left(1\angle 0°\right)+{\left(1\angle 0°\right)}^{*}\left(6j\right)\left(1\angle 0°\right)+{\left(1\angle 0°\right)}^{*}\left(-4j\right)\left(1\angle 0°\right)\right]\\ Q_2\left(0\right)=-\text{Im}\left[\left(-2j\right)+\left(6j\right)+\left(-4j\right)\right]\\ Q_2\left(0\right)=-\text{Im}\left[6j-6j\right]\\ Q_2\left(0\right)=0\end{array}$

Calculate the voltage at bus 2,

Substitute$2$for$K$,and$3$for$N$into equation (1).

$V_2\left(1\right)=\frac{1}{Y_{22}}\left[\frac{P_2}{V_2^{*}\left(0\right)}-Y_{21}{V}_1\left(1\right)-Y_{23}{V}_3\left(0\right)\right]$

Substitute $-j6$for$Y_{22}$,$0.60\text{\hspace{0.33em}pu}$for$P_2$, $1\angle 0°\text{\hspace{0.33em}pu}$for$V_2^{*}$, $2j$for$Y_{21}$$4j$, for$Y_{23}$,$1\angle 0°\text{\hspace{0.33em}pu}$for$V_2$and $1\angle 0°\text{\hspace{0.33em}pu}$for$V_3$into above equation.

$\begin{array}{l}{V}_2\left(1\right)=\frac{1}{-j6}\left[\frac{0.6}{1}-j2\left(1\right)-4j\left(1\right)\right]\\ V_2\left(1\right)=\frac{1}{6\angle -90°}\left[0.6-6j\right]\\ V_2\left(1\right)=\frac{0.6-6j}{6\angle -90°}\\ V_2\left(1\right)=1.005\angle 5.71°\text{\hspace{0.33em}pu}\end{array}$

Calculate the voltage at bus 3.

Substitute$3$for$K$,and $3$for $N$into equation (1).

$V_3\left(1\right)=\frac{1}{Y_{33}}\left[\frac{P_3-j{Q}_3}{V_3^{*}\left(0\right)}-Y_{31}{V}_1\left(1\right)-Y_{32}{V}_2\left(1\right)\right]$

Substitute $-j9$for$Y_{33}$,$-0.80\text{\hspace{0.33em}pu}$for $P_3$,$j0.6\text{\hspace{0.33em}pu}$for $Q_3$,$1\angle 0°\text{\hspace{0.33em}pu}$for $V_1\left(1\right)$,$5j$for$Y_{31}$,$4j$for $Y_{32}$,$1.005\angle 5.71°\text{\hspace{0.33em}pu}$for $V_2\left(1\right)$into above equation.

$\begin{array}{l}{V}_3\left(1\right)=\frac{1}{-j9}\left[\frac{-0.8+j0.6}{1}-j5\left(1\right)-4j\left(1.005\angle 5.71°\right)\right]\\ V_3\left(1\right)=\frac{1}{9\angle -90°}\left[-0.8+j0.6-j5-4j\left(1.005\angle 5.71°\right)\right]\\ V_3\left(1\right)=\frac{-0.4-j8.4}{9\angle -90°}\\ V_3\left(1\right)=0.934\angle 2.73°\text{\hspace{0.33em}pu}\end{array}$

Calculate the error in the voltage of bus 2.

$\begin{array}{l}{V}_2\left(1\right)-V_2\left(0\right)=1.005-1\\ V_2\left(1\right)-V_2\left(0\right)=0.005\text{\hspace{0.33em}pu}\end{array}$

Calculate the error in the voltage of bus 2.

$\begin{array}{l}{V}_3\left(1\right)-V_3\left(0\right)=0.934-1\\ V_3\left(1\right)-V_3\left(0\right)=-0.066\text{\hspace{0.33em}pu}\end{array}$

Iteration 2:

Calculate the reactive power on bus 2,

Substitute $3$for$N$,$2$for $K$, $0$for$i$into the equation (2).

$\begin{array}{l}{Q}_2\left(1\right)=-\text{Im}\left[V_2^{*}\left(0\right)\sum _{k=1}^3{Y}_{2k}{V}_k\left(1\right)\right]\\ Q_2\left(1\right)=-\text{Im}\left[V_2^{*}\left(1\right)Y_{21}{V}_1\left(1\right)+V_2^{*}\left(1\right)Y_{22}{V}_2\left(1\right)+V_2^{*}\left(1\right)Y_{23}{V}_3\left(1\right)\right]\end{array}$

Substitute$2j$for$Y_{21}$,$-6j$for $Y_{22}$,$4j$for $Y_{23}$, $1\angle 0°$for$V_1\left(0\right)$,$1\angle 0°$for $V_2\left(0\right)$ and $1\angle 0°$for$V_3\left(0\right)$ into above equation.

$\begin{array}{l}{Q}_2\left(0\right)=-\text{Im}\left[\begin{array}{l}\left(1.005\angle -5.71°\right)\left(2\angle 90°\right)\left(1\angle 0°\right)\\ +\left(1.005\angle -5.71°\right)\left(6\angle -90°\right)\left(1.005\angle 5.71°\right)\\ +\left(1.005\angle -5.71°\right)\left(4\angle 90°\right)\left(0.934\angle 2.73°\right)\end{array}\right]\\ Q_2\left(0\right)=-\text{Im}\left[2.01\angle 84.29°+6.06\angle -90°+3.75\angle 81.56°\right]\\ Q_2\left(0\right)=-\text{Im}\left[0.828\angle -25.04°\right]\\ Q_2\left(0\right)=0.35\end{array}$

Calculate the voltage at bus 2,

Substitute$2$for$K$,and $3$for $N$into equation (1).

$V_2\left(2\right)=\frac{1}{Y_{22}}\left[\frac{P_2-j{Q}_2}{V_2^{*}\left(1\right)}-Y_{21}{V}_1\left(2\right)-Y_{23}{V}_3\left(1\right)\right]$

Substitute$-j6$for$Y_{22}$,$0.60\text{\hspace{0.33em}pu}$for $P_2$, $0.35\text{\hspace{0.33em}pu}$for$Q_2$,$1.005\angle 5.71°\text{\hspace{0.33em}pu}$for$V_2\left(1\right)$, $2j$for$Y_{21}$, $4j$for$Y_{23}$,$1\angle 0°\text{\hspace{0.33em}pu}$for $V_1\left(2\right)$and$0.934\angle -2.73°\text{\hspace{0.33em}pu}$for$V_3\left(1\right)$into above equation.

$\begin{array}{l}{V}_2\left(2\right)=\frac{1}{-j6}\left[\frac{0.6-j0.35}{1.005\angle -5.71}-j2\left(1\right)-4j\left(0.934\angle -2.73°\right)\right]\\ V_2\left(2\right)=\frac{1}{6\angle -90°}\left[0.628-0.287j-j2-0.177-j3.73\right]\\ V_2\left(2\right)=\frac{0.45-6.01}{6\angle -90°}\\ V_2\left(2\right)=1.006\angle 4.29°\text{\hspace{0.33em}pu}\end{array}$

Calculate the voltage at bus 3.

Substitute$3$for$K$,and 3 for$N$ into equation (1).

$V_3\left(2\right)=\frac{1}{Y_{33}}\left[\frac{P_3-j{Q}_3}{V_3^{*}\left(1\right)}-Y_{31}{V}_1\left(2\right)-Y_{32}{V}_2\left(2\right)\right]$

Substitute$-j9$ for $Y_{33}$, $-0.80\text{\hspace{0.33em}pu}$for$P_3$ ,$j0.6\text{\hspace{0.33em}pu}$ for $Q_3$, $1\angle 0°\text{\hspace{0.33em}pu}$for$V_1\left(2\right)$ , $5j$for$Y_{31}$ , $4j$for$Y_{32}$ ,$1.006\angle 4.29°\text{\hspace{0.33em}pu}$for $V_2\left(2\right)$,and$0.934\angle -2.73°$ for $V_3\left(1\right)$into above equation.

$\begin{array}{l}{V}_3\left(2\right)=\frac{1}{-j9}\left[\frac{-0.8+j0.6}{0.934\angle 2.73}-j5\left(1\angle 0°\right)-4j\left(1.006\angle 2.49°\right)\right]\\ V_3\left(2\right)=\frac{1}{9\angle -90°}\left[-0.52+j8.33\right]\\ V_3\left(2\right)=0.925\angle -3.57°\text{\hspace{0.33em}pu}\end{array}$

Calculate the error in the voltage of bus 2.

$\begin{array}{l}{V}_2\left(2\right)-V_2\left(1\right)=1.006-1.005\\ V_2\left(2\right)-V_2\left(1\right)=0.001\text{\hspace{0.33em}pu}\end{array}$

Calculate the error in the voltage of bus 2.

$\begin{array}{l}{V}_2\left(2\right)-V_2\left(1\right)=1.006-1.005\\ V_2\left(2\right)-V_2\left(1\right)=0.001\text{\hspace{0.33em}pu}\end{array}$

Since the error is less than$0.01\text{\hspace{0.33em}pu}$ .

Therefore, the voltage at bus 2 and 3 are$1.006\angle 4.29°\text{\hspace{0.33em}pu}$and $0.925\angle -3.57°\text{\hspace{0.33em}pu}$.