Question

The bus admittance matrix for the power system shown in Figure 6.24 is given by

${\mathbf{Y}}_{\mathbf{bus}}\mathbf{=}\mathbf{\left[}\begin{array}{cccc}\mathbf{3}\mathbf{-}\mathbf{9}\mathbf{j}& \mathbf{-}\mathbf{2}\mathbf{+}\mathbf{j}\mathbf{6}& \mathbf{-}\mathbf{1}\mathbf{+}\mathbf{j}\mathbf{3}& \mathbf{0}\\ \mathbf{-}\mathbf{2}\mathbf{+}\mathbf{j}\mathbf{6}& \mathbf{3}\mathbf{.}\mathbf{666}\mathbf{-}\mathbf{j}\mathbf{11}& \mathbf{-}\mathbf{0}\mathbf{.}\mathbf{666}\mathbf{+}\mathbf{2}& \mathbf{-}\mathbf{1}\mathbf{+}\mathbf{j}\mathbf{3}\\ \mathbf{-}\mathbf{1}\mathbf{+}\mathbf{j}\mathbf{3}& \mathbf{-}\mathbf{0}\mathbf{.}\mathbf{666}\mathbf{+}\mathbf{j}\mathbf{2}& \mathbf{3}\mathbf{.}\mathbf{666}\mathbf{-}\mathbf{j}\mathbf{11}& \mathbf{-}\mathbf{2}\mathbf{+}\mathbf{j}\mathbf{6}\\ \mathbf{0}& \mathbf{-}\mathbf{1}\mathbf{+}\mathbf{j}\mathbf{3}& \mathbf{-}\mathbf{2}\mathbf{+}\mathbf{j}\mathbf{6}& \mathbf{3}\mathbf{-}\mathbf{9}\mathbf{j}\end{array}\mathbf{\right]}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}\mathbf{}$

With the complex powers on load buses 2, 3, and 4 as shown in Figure 6.24,

determine the value for ${\mathbf{V}}_{\mathbf{2}}$that is produced by the first and second iterations of the Gauss-Seidel procedure. Choose the initial guess

${\mathbf{V}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{V}}_{\mathbf{3}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{V}}_{\mathbf{4}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\angle }\mathbf{0}\mathbf{°}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}\mathbf{}$.

Short answer

The value of ${\mathrm{V}}_2$after first iteration is $1.05\angle 1.93°perunit$and after second iteration is$1.04\angle 1.25°perunit$.

Step-by-step solution

 Step 1: Write the given data.

Initial guess for bus voltages are,

$$\begin{gathered} V_2\left(0\right)=1.0\angle 0°\mathrm{per}\mathrm{unit} \\ {V}_{\mathit{3}}\left(0\right)=1.0\angle 0°\mathrm{per}\mathrm{unit} \\ {V}_4\left(0\right)=1.0\angle 0°\mathrm{per}\mathrm{unit} \end{gathered}$$

The bus admittance matrix for the power system is,

${\mathrm{Y}}_{\mathrm{bus}}=\left[\begin{array}{cccc}3-9\mathrm{j}& -2+\mathrm{j}6& -1+\mathrm{j}3& 0\\ -2+\mathrm{j}6& 3.666-\mathrm{j}11& -0.666+2& -1+\mathrm{j}3\\ -1+\mathrm{j}3& -0.666+\mathrm{j}2& 3.666-\mathrm{j}11& -2+\mathrm{j}6\\ 0& -1+\mathrm{j}3& -2+\mathrm{j}6& 3-9\mathrm{j}\end{array}\right]\mathrm{per}\mathrm{unit}$

Write the general equation of Gauss-Seidel Method

Write the general formula for Gauss-Seidel,

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

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

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

Calculate the voltage for bus 2,

For $i=0$, substitute the value $\left(-2+j6\right)$for $Y_{21}$, $\left(-0.666+j2\right)$for &$\left(-1+j3\right)$for $Y_{24}$and $05-j0.2perunit$for $P_2-j{Q}_2$

$$\begin{gathered} V_2(0+1)=\frac{1}{{\mathrm{Y}}_{22}}\left[\frac{P_2-j{Q}_2}{V_2^{*}\left(0\right)}\begin{array}{c}-\sum _{\mathrm{n}=1}^{2-1}\end{array}{Y}_{2n}{V}_n(0+1)-\sum _{\mathrm{n}=2+1}^{\mathrm{N}}{Y}_{2n}{V}_{\mathrm{n}}\left(0\right)\right] \\ {V}_2\left(1\right)=\frac{1}{3.666-j11}\left[\frac{0.5-j0.2}{{(1+j0)}^{*}}\begin{array}{c}-\sum _{\mathrm{n}=1}^{2-1}\end{array}{Y}_{2n}{V}_n\left(1\right)-\sum _{\mathrm{n}=2+1}^4{Y}_{2n}{V}_{\mathrm{n}}\left(0\right)\right] \\ {V}_2\left(1\right)=\frac{1}{3.666-j11}[\frac{0.5-j0.2}{1+j0}\begin{array}{c}-Y_{21}{V}_1\left(1\right)\end{array}-Y_{23}{V}_3(0)-Y_{24}{V}_4(0\left)\right] \\ {V}_2\left(1\right)=\frac{1}{3.666-j11}\left[\begin{array}{c}\frac{0.5-j0.2}{1+j0}-(-2+j6)(1.04\angle 0°)-\\ (-0.666+j2)(1+j0)-(-1+j3)91\angle 0°)\end{array}\right] \end{gathered}$$

Further solving,

$$\begin{gathered} V_2\left(1\right)=\frac{4.246-j11.44}{3.666-j11} \\ =1.052+j0.035 \\ =1.05\angle 1.93°perunit \end{gathered}$$

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

Calculate the voltage for bus 3,

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

$$\begin{gathered} V_3(0+1)=\frac{1}{{\mathrm{Y}}_{33}}\left[\frac{P_3-j{Q}_3}{V_3^{*}\left(0\right)}\begin{array}{c}-\sum _{\mathrm{n}=1}^{3-1}\end{array}{Y}_{3n}{V}_n(0+1)-\sum _{\mathrm{n}=3+1}^{\mathrm{N}}{Y}_{3n}{V}_{\mathrm{n}}\left(0\right)\right] \\ {V}_3\left(1\right)=\frac{1}{3.666-j11}\left[\frac{-1+j0.5}{{(1+j0)}^{*}}\begin{array}{c}-\sum _{\mathrm{n}=1}^2\end{array}{Y}_{3n}{V}_n\left(1\right)-\sum _{\mathrm{n}=3+1}^4{Y}_{3n}{V}_{\mathrm{n}}\left(0\right)\right] \\ {V}_3\left(1\right)=\frac{1}{3.666-j11}[\frac{-1+j0.5}{1+j0}\begin{array}{c}-Y_{31}{V}_1\left(1\right)\end{array}-Y_{32}{V}_2(1)-Y_{34}{V}_4(0\left)\right] \\ {V}_3\left(1\right)=\frac{1}{3.666-j11}\left[\begin{array}{c}\frac{-1+j0.5}{1+j0}-(-1+j3)(1.04\angle 0°)-(-0.666+j2)(1.05\angle 1.93°)\\ -(-2+j6)(1\angle 0°)\end{array}\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} V_3\left(1\right)=\frac{2.81-j10.69}{3.666-j11} \\ =0.952-j0.062\mathrm{per}\mathrm{unit} \\ =0.95\angle -3.71°\mathrm{per}\mathrm{unit} \end{gathered}$$

Solving the first iteration by using Gauss-Seidel method of bus 4(k=4)

Calculate the voltage for bus 2 and consider voltage of bus 2 to be$1\angle 0°$.

For $i=0$, substitute the value localid="1655273673576" $0$for localid="1655273631182" $Y_{41}$, $-1+j3$for $Y_{42}$and $-2+j6$for $Y_{43}$also $0.3-j0.1perunit$for $P_4-j{Q}_4$in equation (2)

$$\begin{gathered} V_4(0+1)=\frac{1}{{\mathrm{Y}}_{44}}\left[\frac{P_4-j{Q}_4}{V_4^{*}\left(0\right)}\begin{array}{c}-\sum _{\mathrm{n}=1}^{3-1}\end{array}{Y}_{4n}{V}_n(0+1)-\sum _{\mathrm{n}=2+1}^{\mathrm{N}}{Y}_{2n}{V}_{\mathrm{n}}(1)\right] \\ {V}_4\left(1\right)=\frac{1}{-j10}\left[\frac{0.3+j0.1}{1-j0}\begin{array}{c}-\sum _{\mathrm{n}=1}^2\end{array}{Y}_{4n}{V}_n\left(1\right)-\sum _{\mathrm{n}=2+1}^3{Y}_{4n}{V}_{\mathrm{n}}\left(1\right)\right] \\ {V}_4\left(1\right)=\frac{1}{-j10}[0.965-j0.375\begin{array}{c}-Y_{41}{V}_1\left(1\right)\end{array}-Y_{42}{V}_2(1)-Y_{43}{V}_3(1\left)\right] \\ {V}_4\left(1\right)=\frac{1}{-j10}\left[\begin{array}{c}0.965-j0.375-\left(0\right)(1.04\angle 0°)-(-1+j3)(1.05\angle 1.93°)\\ -(-2+j6)(0.95\angle -371°)\end{array}\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} V_4\left(1\right)=\frac{2.98-j9.02}{3-j9} \\ =1.00-j0.0025 \\ =1.00\angle -0.144°\mathrm{per}\mathrm{unit} \end{gathered}$$

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

Calculate the voltage for bus 2,

For $i=1$, substitute the value$-2+j6$for$Y_{21}$,localid="1655274877994" $-0.666+j2$for$Y_{23}$and $-1+j3$for$Y_{24}$also$1.5-j0.75perunit$for $P_3-j{Q}_3$in equation (2)

$$\begin{gathered} V_2(1+1)=\frac{1}{{\mathrm{Y}}_{22}}\left[\frac{P_2-j{Q}_2}{V_2^{*}\left(1\right)}\begin{array}{c}-\sum _{\mathrm{n}=1}^{2-1}\end{array}{Y}_{2n}{V}_n(1+1)-\sum _{\mathrm{n}=2+1}^{\mathrm{N}}{Y}_{2n}{V}_{\mathrm{n}}\left(1\right)\right] \\ {V}_2\left(2\right)=\frac{1}{3.666-j11}\left[\frac{1-j0.5}{1.052-j0.035}\begin{array}{c}-\sum _{\mathrm{n}=1}^1\end{array}{Y}_{2n}{V}_n\left(2\right)-\sum _{\mathrm{n}=2+1}^4{Y}_{2n}{V}_{\mathrm{n}}\left(1\right)\right] \\ {V}_2\left(2\right)=\frac{1}{3.666-j11}[\frac{1-j0.5}{1.052-j0.035}\begin{array}{c}-Y_{21}{V}_1\left(2\right)\end{array}-Y_{23}{V}_3(1)-Y_{24}{V}_4(1\left)\right] \\ {V}_2\left(2\right)=\frac{1}{3.666-j11}\left[\begin{array}{c}\frac{1-j0.5}{1.052-j0.035}-(-2+j6)(1.04\angle 0°)-\\ (-0.666+j2)(0.95\angle -371°)-(-1+j3)\left(1\angle -0.144°\right)\end{array}\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} V_2\left(2\right)=\frac{4.062-j11.35}{3.666-j11} \\ {V}_2\left(2\right)=1.04+j0.023 \\ {V}_2\left(2\right)=1.04\angle 1.25°perunit \end{gathered}$$

Thus, the value of 2^nd bus voltage after second iteration becomes,

$V_2\left(2\right)=1.04\angle 1.25°\mathrm{per}\mathrm{unit}$.