Question

Use the Gauss-Seidel method to solve the following equations that

contain terms that are often found in power flow equations.

$$\begin{gathered} \mathit{x}\mathbf{=}\left(1/\left(-20j\right)\right)\mathbf{*}\left[\left(-1+0.5j\right)/\left(x_1\right)*-\left(j10\right)*x_2-\left(j10\right)\right] \\ \mathit{x}\mathbf{=}\left(1/\left(-20j\right)\right)\mathbf{*}\left[\left(-3+j\right)/\left(x_2\right)*-\left(j10\right)*x_1-\left(j10\right)\right] \end{gathered}$$

Use an initial estimate of ${\mathbf{x}}_{\mathbf{1}}\left(0\right)\mathbf{=}\mathbf{1}$and${\mathbf{x}}_{\mathbf{2}}\left(0\right)\mathbf{=}\mathbf{1}$, a stopping of$\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$

Short answer

The values of $x_1$and $x_2$achieved by solving the given power flow equations using

Gauss-Seidel method are:

$$\begin{gathered} x_1=0.9405-0.1373j \\ {x}_2=0.889-0.213j \end{gathered}$$

Step-by-step solution

Write the given data.

The two power flow equations are:

$$\begin{gathered} x=(1/(-20\mathrm{j}))*[(-1+0.5\mathrm{j})/\left({\mathrm{x}}_1\right)*-\left(\mathrm{j}10\right)*{\mathrm{x}}_2-\left(\mathrm{j}10\right)] \\ x=(1/(-20\mathrm{j}))*[(-3+\mathrm{j})/\left({\mathrm{x}}_2\right)*-\left(\mathrm{j}10\right)*{\mathrm{x}}_1-\left(\mathrm{j}10\right)] \end{gathered}$$

Initial values for the above equations are: $x_1\left(0\right)=1$and $x_2\left(0\right)=1$, and a tolerance

value of $\epsilon =0.05$.

General formula of Gauss-Seidel Method

Write the general formula for the Gauss-Seidel method,

${\mathbf{x}}_{\mathbf{k}}\left(i+1\right)\mathbf{=}\frac{\mathbf{1}}{{\mathbf{A}}_{\mathbf{KK}}}\left[y_k-\sum _{n=1}^{k-1}{A}_{\mathrm{kn}}{x}_n\left(i+1\right)-\sum _{n=k+1}^n{A}_{\mathrm{kn}}{x}_n\left(i\right)\right]$

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.

Write the given power flow equations in general form.

Write the general form of power flow equation.

$$\begin{gathered} x_1\left(i+1\right)=\frac{1}{-20j}\left[\frac{-1+0.5j}{{\left[x_1\left(i\right)\right]}^{*}}-j10\times x_2\left(i\right)-j10\right]......\left(1\right) \\ {x}_1\left(i+1\right)=\frac{1}{-20j}\left[\frac{-3+j}{{\left[x_2\left(i\right)\right]}^{*}}-j10\times x_1\left(i+1\right)-j10\right]......\left(2\right) \end{gathered}$$

And the equation of tolerance will look like,

$\left|\frac{x_k\left(i+1\right)-x_k\left(i\right)}{x_k\left(i\right)}\right|<\epsilon$

First iteration, determine the values of x1 and x2 for  i = 10

Substituting the value 0 for i in equations (1) and (2)

$$\begin{gathered} x_1\left(1\right)=\frac{1}{-20j}\left[\frac{-1+0.5j}{{\left[x_1\left(0\right)\right]}^{*}}-j10\times x_2\left(0\right)-j10\right] \\ {x}_2\left(1\right)=\frac{1}{-20j}\left[\frac{-3+j}{{\left[x_2\left(0\right)\right]}^{*}}-j10\times x_1\left(1\right)-j10\right] \end{gathered}$$

Now substituting the initial values in the above equation.

$$\begin{gathered} x_1\left(1\right)=\frac{1}{-20j}\left[\frac{-1+0.5j}{{\left[1\right]}^{*}}-j10\times 1-j10\right] \\ {x}_2\left(1\right)=\frac{1}{-20j}\left[\frac{-3+j}{{\left[1\right]}^{*}}-j10\times x_1\left(1\right)-j10\right] \end{gathered}$$

Further solving the equations,

$$\begin{gathered} x_1\left(1\right)=\frac{1}{-20j}\left[-1+0.5j-j10-j10\right] \\ =\frac{-1-19.5j}{-20j} \\ =\left(0.05j\right)\left[-1-19.5j\right] \\ =0.975-0.05j \end{gathered}$$

$$\begin{gathered} x_2\left(1\right)=\frac{1}{-20j}\left[-3+j-j-j10\left[0.0975-0.05j\right]-j10\right] \\ =\frac{1}{-20j}\left[-3+j-9.75j-0.5-10j\right] \\ =\left(0.05j\right)\left[-3-18.75j\right] \\ =0.9375-0.175j \end{gathered}$$

Calculating tolerance level,

$$\begin{gathered} \left|\frac{x_1\left(0+1\right)-x_1\left(0\right)}{x_1\left(0\right)}\right|<\epsilon \\ \left|\frac{x_1\left(1\right)-x_1\left(0\right)}{x_1\left(0\right)}\right|<\epsilon \\ \left|\frac{0.975-0.05j-1}{1}\right|<\epsilon \\ \left|-0.025-0.05j\right|<\epsilon \end{gathered}$$

The tolerance level is 0.025 as of now.

Second iteration, determine the values of x1 and x2 for i=1

Substituting the value 1 for i in equations (1) and (2)

$$\begin{gathered} x_1\left(1+1\right)=\frac{1}{-20j}\left[\frac{-1+0.5j}{{\left[x_1\left(1\right)\right]}^{*}}-j10\times x_2\left(1\right)-j10\right] \\ {x}_1\left(2\right)=\frac{1}{-20j}\left[\frac{-1+0.5j}{{\left[x_1\left(1\right)\right]}^{*}}-j10\times x_2\left(1\right)-j10\right]......\left(3\right) \end{gathered}$$

Also,

$$\begin{gathered} x_2\left(1+1\right)=\frac{1}{-20j}\left[\frac{-3+j}{{\left[x_2\left(1\right)\right]}^{*}}-j10\times x_1\left(1+1\right)-j10\right] \\ {x}_2\left(2\right)=\frac{1}{-20j}\left[\frac{-3+j}{{\left[x_2\left(1\right)\right]}^{*}}-j10\times x_1\left(2\right)-j10\right].......\left(4\right) \end{gathered}$$

Substituting the value of in the equation (3).

$$\begin{gathered} x_1\left(2\right)=\frac{1}{-20j}\left[\frac{-1+0.5j}{{\left[x_1\left(1\right)\right]}^{*}}-j10\times x_2\left(1\right)-j10\right] \\ =\frac{1}{-20j}\left[\frac{-1+0.5j}{{\left[0.975-0.05j\right]}^{*}}-j10\left[0.9375-0.175j\right]-j10\right] \\ =\left(0.05j\right)\left[\frac{-1+0.5j}{{\left[0.975-0.05j\right]}^{*}}-j10\left[0.9375-0.175j\right]-j10\right] \\ =\left(0.05j\right)\left[-0.996+j0.564-j9.375-1.75-j10\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} x_1\left(2\right)=\left(0.05j\right)\left[-2.746-j18.811\right] \\ =0.9405-j0.137 \end{gathered}$$

Substituting the value of$x_2\left(1\right)$in the equation (4)

$$\begin{gathered} x_1\left(2\right)=\frac{1}{-20j}\left[\frac{-3+j}{{\left[x_2\left(1\right)\right]}^{*}}-j10\times x_1\left(2\right)-j10\right] \\ =\frac{1}{-20j}\left[\frac{-3+j}{{\left[0.9375-0.175j\right]}^{*}}-j10\left[0.09405-j0.1373\right]-j10\right] \\ =\frac{1}{-20j}\left[\frac{-3+j}{\left[0.9375-0.175j\right]}-j10\left[0.09405-j0.1373\right]-j10\right] \\ =\left(0.05j\right)\left[-2.89+j1.608-j19.405-1.373\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} x_1\left(2\right)=\left(0.05j\right)\left[-4.236-j17.797\right] \\ =0.889-j0.213 \end{gathered}$$

Calculating the tolerance value,

$$\begin{gathered} \left|\frac{x_1\left(1+1\right)-x_1\left(1\right)}{x_1\left(1\right)}\right|<\epsilon \\ \left|\frac{x_1\left(2\right)-x_1\left(1\right)}{x_1\left(1\right)}\right|<\epsilon \\ \left|\frac{\left(0.9405-j0.1373\right)-\left(0.975-0.05j\right)}{\left(0.975-0.05j\right)}\right|<\epsilon \\ \left|\frac{0.9405-j0.1373-0.975-0.05j}{\left(0.975-0.05j\right)}\right|<\epsilon \end{gathered}$$

Further calculating the tolerance value

$$\begin{gathered} \left|\frac{-0.345-j0.0873}{\left(0.975-0.05j\right)}\right|<\epsilon \\ \left|-0.0307-j0.092\right|<\epsilon \end{gathered}$$

Looking at the value of tolerance level, $\epsilon$it is has reached 0.03 further more

iterations will tend to exceed the value beyond 0.05. Thus, the values will be

$x_1=0.9405-0.1373jand{x}_2=0.889-0.213j$