Question

Repeat Problem 6.34 except assume the bus 1 (slack bus) voltage of ${\mathit{V}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{05}\mathbf{\angle }\mathbf{0}\mathbf{°}$.

Short answer

The second iteration value of third bus is${\mathit{\text{V}}}_3\left(2\right)=1.156\angle 11.85°\mathrm{V}$.

Step-by-step solution

Given data

The bus admittance matrix for a three-bus system is

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

Voltage of bus 1 is$V_1\left(0\right)=1.05\angle 0°$, $V_2\left(0\right)=1.0\angle 0^{°}$and$V_3\left(0\right)=1.0\angle 0°$also load buses are as follows:

Write the expression for the$S_2$.

$$\begin{gathered} S_{\mathit{2}}=1+j0.5\mathrm{per}\mathrm{unit} \\ =P_2+j{Q}_2 \end{gathered}$$

Write the expression for the$S_3$.

$$\begin{gathered} S_3=1.5+j0.75\mathrm{per}\mathrm{unit} \\ =P_3+j{Q}_3 \end{gathered}$$

General formula for Gauss-Seidel method.

Write the general formula for Gauss-Seidel.

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

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,

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

Calculate the voltage for bus 2,

For $i=0$.

Substitute the value $-j10$for$Y_{22}$and $1+j0.5perunit$for $P_2+j{Q}_2$in equation (1).

$$\begin{gathered} V_2(0+1)=\frac{1}{Y_{22}}\left[\frac{P_2-j{Q}_2}{V_2^{*}\left(0\right)}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{2n}{V}_n(0+1)-\sum _{\mathrm{n}=2+1}^{\mathrm{N}}{Y}_{2n}{V}_{\mathrm{n}}\left(0\right)\right] \\ {V}_2\left(1\right)=\frac{1}{-j10}\left[\frac{1-j0.5}{{(1+j0)}^{*}}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{2n}{V}_n\left(1\right)-\sum _{\mathrm{n}=2+1}^3{Y}_{2n}{V}_{\mathrm{n}}\left(0\right)\right] \\ {V}_2\left(1\right)=\frac{1}{-j10}\left[\frac{1-j0.5}{(1-j0)}-Y_{21}{V}_1\left(1\right)-Y_{23}{V}_3\left(0\right)\right] \\ {V}_2\left(1\right)=\frac{1}{-j10}\left[\frac{1-j0.5}{(1-j0)}-\left(j5\right)(1.05\angle 0°)-\left(j5\right)(1+j0)\right] \end{gathered}$$

Further solving.

$$\begin{gathered} V_2\left(1\right)=\frac{1}{-j10}\left[1-j0.5-(j5.25)-\left(j5\right)\right] \\ {V}_2\left(1\right)=1.075+j0.1perunit \end{gathered}$$

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$1\angle 0^o$.

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)

$$\begin{gathered} V_3(0+1)=\frac{1}{Y_{33}}\left[\frac{P_3-j{Q}_3}{V_3^{*}\left(0\right)}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{3n}{V}_n(0+1)-\sum _{\mathrm{n}=3+1}^{\mathrm{N}}{Y}_{3n}{V}_{\mathrm{n}}\left(0\right)\right] \\ {V}_3\left(1\right)=\frac{1}{-j10}\left[\frac{1.5-j0.75}{{(1+j0)}^{*}}-\sum _{\mathrm{n}=1}^1{Y}_{3n}{V}_n\left(1\right)-\sum _{\mathrm{n}=3+1}^3{Y}_{3n}{V}_{\mathrm{n}}\left(0\right)\right] \\ {V}_3\left(1\right)=\frac{1}{-j10}\left[\frac{1.5-j0.75}{(1-j0)}-Y_{31}{V}_1\left(1\right)-Y_{31}{V}_1\left(0\right)\right] \\ {V}_3\left(1\right)=\frac{1}{-j10}\left[\frac{1.5-j0.75}{(1-j0)}-\left(j5\right)1.05\angle 0°-\left(j5\right)(1.075+0.1j)\right] \end{gathered}$$

Further solving,

$$\begin{gathered} V_3\left(1\right)=\frac{1}{-j10}\left[1.5-j0.75-j5.25+0.5-j5.375\right] \\ =1.1375+j0.2perunit \end{gathered}$$

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$1\angle 0^o$.

For$i=1$, 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 (2)

$$\begin{gathered} V_2(1+1)=\frac{1}{Y_{22}}\left[\frac{P_2-j{Q}_2}{V_2^{*}\left(1\right)}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{2n}{V}_n(1+1)-\sum _{\mathrm{n}=2+1}^{\mathrm{N}}{Y}_{2n}{V}_{\mathrm{n}}\left(1\right)\right] \\ {V}_2\left(2\right)=\frac{1}{-j10}\left[\frac{1-j0.5}{{(1.075+j0.1)}^{*}}-\sum _{\mathrm{n}=1}^1{Y}_{2n}{V}_n\left(2\right)-\sum _{\mathrm{n}=2+1}^3{Y}_{2n}{V}_{\mathrm{n}}\left(1\right)\right] \\ {V}_2\left(2\right)=\frac{1}{-j10}[0.965-j0.375-Y_{21}{V}_1\left(2\right)-Y_{23}{V}_3\left(1\right)] \\ {V}_2\left(2\right)=\frac{1}{-j10}[0.965-j0.375-j5.25+1-j5.68] \end{gathered}$$

Deriving the final result,

$$\begin{gathered} V_2\left(2\right)=1.13125+j01965 \\ =1.148\angle 9.85°\mathrm{per}\mathrm{unit} \end{gathered}$$

Solving for the second iteration.

Solve for the second iteration by gauss seidel method for bus 3.

$$\begin{gathered} V_3(1+1)=\frac{1}{Y_{33}}\left[\frac{P_3-j{Q}_3}{V_2^{*}\left(1\right)}-\sum _{\mathrm{n}=1}^{2-1}{Y}_{2n}{V}_n(1+1)-\sum _{\mathrm{n}=2+1}^{\mathrm{N}}{Y}_{2n}{V}_{\mathrm{n}}\left(1\right)\right] \\ {V}_3\left(2\right)=\frac{1}{-10j}\left[\frac{1.5-j0.75}{(1.1375+j0.2)}-\sum _{\mathrm{n}=1}^1{Y}_{2n}{V}_n\left(2\right)-\sum _{\mathrm{n}=2+1}^3{Y}_{3n}{V}_{\mathrm{n}}\left(1\right)\right] \\ {V}_3\left(2\right)=\frac{1}{-10j}[1.392-j0.415-Y_{21}{V}_1\left(2\right)-Y_{32}{V}_3\left(1\right)] \\ {V}_3\left(2\right)=\frac{1}{-10j}[1.392-j0.415-(j5\left)\right(1.05+j0)-(j5\left)\right(1.13+j0.19] \end{gathered}$$

Solving further,

$\begin{array}{l}{V}_3\left(2\right)=\frac{1}{-10j}\left[1.392-j0.415-j5.25+0.9825-j5.656\right]\\ V_3\left(2\right)=1.132+j0.2374\mathrm{per}\mathrm{unit}\\ V_3\left(2\right)=1.156\angle 11.85°\mathrm{V}\end{array}$

Thus, the value of the second iteration for bus 3 is $1.156\angle 11.85°\mathrm{V}$.