Question

For a two-bus power system, a$\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{4}$per unit load at bus 2 is supplied by a generator at bus 1 through a transmission line with series impedance of $\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{1}$per unit. With bus 1 as the slack bus with a fixed per-unit voltage of$\mathbf{1}\mathbf{\angle }{\mathbf{0}}^{\circ }$, use the Gauss-Seidel method to calculate the voltage at bus 2 after three iterations.

Short answer

The voltage at bus 2 after three iteration is$0.9168\angle -3.{126}^{\circ }\mathrm{pu}$

Step-by-step solution

Write the given data by the question.

Transmission line impedance,$Z_L=0.05+j0.1\mathrm{pu}$

Voltage on bus 1,$V_1=1\angle 0^{\circ }$

Load on bus 2,$S_2=0.7+j0.4$

Write the given data by the question.

The equation to calculate the diagonal element of the$Y_{\mathit{bus}}$matrix of the system

is given as follows.

${\mathbf{Y}}_{\mathbf{kk}}$= Sum of the admittance connected to bus$\mathit{k}$.

The equation to calculate the non-diagonal element of the matrix of the system is given as follows.

${\mathbf{Y}}_{\mathbf{kn}}$= (-sum of admittance connected between bus $\mathit{k}$and$\mathit{n}$)

The equation to calculate the admittance of the transmission line is given as follows.

$\mathbf{Y}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Z}}_{\mathbf{L}}}$

The equation to calculate the voltage at bus using the Gauss-Seidel method is given as follows.

${\mathbf{V}}_{\mathbf{k}}\mathbf{(}\mathbf{i}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Y}}_{\mathbf{kk}}}\left[\frac{P_{2-}\mathit{j}{\mathit{Q}}_{\mathit{k}}}{V_k^{*}}-\sum _{n=1}^{k-1}{Y}_{\mathrm{kn}}{V}_n(i+1)-\sum _{n=k+1}^N{Y}_{\mathrm{kn}}{V}_n\left(i\right)\right]$

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

Calculate the voltage at bus 2.

Calculate the diagonal elements of admittance matrix.

Substitute$0.05+j0.1\mathrm{pu}$for $Z_{\mathit{L}}$into equation (1).

$$\begin{gathered} Y_{\mathit{11}}=Y_{22} \\ {Y}_{11}=\frac{1}{0.05+j0.1} \\ {Y}_{11}=4-j8 \\ {Y}_{11}=8.944\angle -63.{43}^{\circ }\mathrm{pu} \end{gathered}$$

Calculate the non-diagonal element of admittance matrix.

$$\begin{gathered} Y_{\mathit{1}2}=Y_{21} \\ {Y}_{12}=\frac{1}{0.05+j0.1} \\ {Y}_{12}=4-j8 \\ \mathrm{The}\mathrm{admittance}\mathrm{matrix}\mathrm{is}{Y}_{\mathit{Bus}}\left[\begin{array}{cc}4-j\mathrm{8}& \mathrm{-}\mathrm{4}\mathrm{+}j\mathrm{8}\\ \mathrm{-}\mathrm{4}\mathrm{+}j\mathrm{8}& \mathrm{}\mathrm{4}\mathrm{-}j\mathrm{8}\end{array}\right] \end{gathered}$$

Calculate voltage of bus 2 after first iteration.

Substitute $2$for$k$, for $i$into equation (2).

$$\begin{gathered} V_{\mathit{2}}(0+1)=\frac{1}{Y_{\mathit{22}}}\left[\begin{array}{cc}\frac{P_{\mathit{2}}\mathit{-}j{Q}_2}{V_2^{*}\left(0\right)}& Y_{21}{V}_1(0+1)\end{array}\right] \\ {\mathrm{V}}_{\mathit{2}}\left(1\right)=\frac{1}{{\mathrm{Y}}_{\mathit{22}}}\left[\begin{array}{cc}\frac{{\mathrm{P}}_{\mathit{2}}\mathit{-}{\mathrm{jQ}}_2}{{\mathrm{V}}_2^{*}\left(0\right)}-& {\mathrm{Y}}_{21}{V}_1\left(1\right)\end{array}\right] \end{gathered}$$

Substitute $4-j8\mathrm{pu}$for$Y_{22}$,$-0.7$for$P_{\mathit{2}}$,$-j0.4$for$Q_2$,$-4+j8\mathrm{pu}$for $Y_{21}$and $1\angle 0^{\circ }$for $V_1$ into above equation.

$$\begin{gathered} V_2\left(1\right)=\frac{1}{4-j8}\left[\begin{array}{cc}\frac{-0.7-(-j0.4)}{1\angle 0^{\circ }}-& (-4+j8(1)\end{array}\right] \\ {V}_2\left(1\right)=\frac{1}{4-\mathrm{j}8}\left[\begin{array}{cc}\frac{-0.7+\mathrm{j}0.4}{1\angle 0^{\circ }}+& 4-\mathrm{j}8\end{array}\right] \\ {V}_2\left(1\right)=\frac{-0.7+j0.4+4-j8}{4-j8} \\ {V}_2\left(1\right)=0.925-j0.05\mathrm{pu} \end{gathered}$$

Convert into polar form.

$V_2\left(1\right)=0.926\angle -3.{09}^{\circ }\mathrm{pu}$

Calculate the voltage at bus 2 after second iteration.

Substitute 2 for$k$, $1$for $i$into equation (4).

$$\begin{gathered} V_2(1+1)=\frac{1}{Y_{22}}\left[\begin{array}{cc}\frac{P_2-j{Q}_2}{V_2^{*}\left(0\right)}-& {\mathrm{Y}}_{21}{\mathrm{V}}_1(1+1)\end{array}\right] \\ {\mathrm{V}}_2\left(2\right)=\frac{1}{{\mathrm{Y}}_{22}}\left[\begin{array}{cc}\frac{{\mathrm{P}}_2-{\mathrm{jQ}}_2}{{\mathrm{V}}_2^{*}\left(1\right)}-& {\mathrm{Y}}_{21}{\mathrm{V}}_1\left(2\right)\end{array}\right] \end{gathered}$$

Substitute $4-j8\mathrm{pu}$for$Y_{22}$,$-0.7$for$P_2$,$-j0.4$for$Q_2$,$-4+j8\mathrm{pu}$for $Y_{21}$and$0.926\angle -3.{09}^{\circ }\mathrm{pu}$for$V_2^{*}\mathit{}\left(1\right)$into above equation.

role="math" localid="1655198496415" $$\begin{gathered} V_2\left(2\right)=\frac{1}{4-j8}\left[\begin{array}{cc}\frac{-0.7-(-j0.4)}{{(0.926\angle -3.{09}^{\circ })}^{*}}-(-4+j8)& \left(1\right)\end{array}\right] \\ {\mathrm{V}}_2\left(2\right)=\frac{1}{4-\mathrm{j}8}\left[\begin{array}{cc}\frac{-0.7-(-\mathrm{j}0.4)}{(0.926\angle -3.{09}^{\circ }}+& 4-j8\end{array}\right] \\ {V}_2\left(2\right)=\frac{3.268-j7.528}{4-j8} \\ {\mathrm{V}}_2\left(2\right)=0.9162-j0.05\mathrm{pu} \end{gathered}$$

Convert into polar form.

$V_2\left(2\right)=0.918\angle -3.{095}^{\circ }\mathrm{pu}$

Calculate the voltage at bus 2 after third iteration.

Substitute $2$for$k$,$2$for $i$into equation (4).

$$\begin{gathered} V_2(2+1)=\frac{1}{Y_{22}}\left[\begin{array}{cc}\frac{P_2-j{Q}_2}{V_2^{*}\left(0\right)}-Y_{21}{V}_1& (2+1)\end{array}\right] \\ {\mathrm{V}}_2\left(3\right)=\frac{1}{{\mathrm{Y}}_{22}}\left[\begin{array}{cc}\frac{{\mathrm{P}}_2-{\mathrm{jQ}}_2}{{\mathrm{V}}_2^{*}\left(2\right)}-{\mathrm{Y}}_{21}{\mathrm{V}}_1& \left(3\right)\end{array}\right] \end{gathered}$$

Substitute $4-j8\mathrm{pu}$for$Y_{22}$,$-0.7$for$P_2$, $-j0.4$for$Q_2$,$-4+j8\mathrm{pu}$for $Y_{21}$and$0.918\angle -3.{095}^{\circ }\mathrm{pu}$forole="math" localid="1655198257204" $V_2\left(2\right)$ into above equation.

$$\begin{gathered} V_2\left(3\right)=\frac{1}{4-j8}\left[\begin{array}{cc}\frac{-0.7-(-j0.4)}{{(0.918\angle -3.{095}^{\circ })}^{*}}-(-4+j8)& \left(1\right)\end{array}\right] \\ {\mathrm{V}}_2\left(3\right)=\frac{1}{4-\mathrm{j}8}\left[\begin{array}{cc}\frac{-0.7-(-\mathrm{j}0.4)}{(0.918\angle -3.{095}^{\circ }}+& 4-j8\end{array}\right] \\ {V}_2\left(3\right)=\frac{3.262-j7.52}{4-j8} \\ {\mathrm{V}}_2\left(3\right)=0.915-j0.05\mathrm{pu} \end{gathered}$$

Convert into polar form.

$V_2\left(2\right)=0.9168\angle -3.{126}^{\circ }\mathrm{pu}$