Question

Equipment ratings and per-unit reactances for the system shown in Figure 9.19 are given as follows:

Synchronous generators:

Transformers:

$\begin{array}{l}\mathbf{}\mathbf{T}\mathbf{1}\mathbf{}\mathbf{}\mathbf{}\mathbf{100}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{25}\mathbf{/}\mathbf{230}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_{\mathbf{1}}\mathbf{=}{\mathbf{X}}_{\mathbf{2}}\mathbf{=}{\mathbf{X}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\\ \mathbf{T}\mathbf{1}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{100}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{/}\mathbf{230}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_{\mathbf{1}}\mathbf{=}{\mathbf{X}}_{\mathbf{2}}\mathbf{=}{\mathbf{X}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\end{array}$

Transmission lines:

Using a$\mathbf{100}\mathbf{MVA}\mathbf{,}\mathbf{230}\mathbf{kV}$ base for the transmission lines, draw the perunit sequence networks and reduce them to their Thevenin equivalents, “looking in” at bus 3. Neglect$\mathbf{\Delta }\mathbf{-}\mathbf{Y}$ phase shifts. Compute the fault currents for a bolted three-phase fault at bus 3.

Short answer

The three-phase fault current are$\left[\begin{array}{c}5.73\angle -90°\\ 5.73\angle -150°\\ 5.73\angle -30°\end{array}\right]\text{pu}$ .

Step-by-step solution

Write the given data from the question:

Write the rating of the generators.

The power rating of generator $G1$is $100\mathrm{MVA}$, the voltage rating is $25\text{\hspace{0.17em}kV}$, the reactance $X_1$and $X_2$are equal to $0.20\text{\hspace{0.17em}per\hspace{0.17em}unit}$ and the value of $X_0$is $0.05\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of generator $G2$is $100MVA$, the voltage rating is $13.8\mathrm{kV}$, the reactance $X_1$and $X_2$are equal to $0.20\text{\hspace{0.17em}per\hspace{0.17em}unit}$ and the value of $X_0$is $0.05\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

Write the rating of transformers.

The power rating of transformer$T1$is$100MVA$, the voltage rating is$25/230\mathrm{kV}$, and the value of${\mathrm{X}}_1,{\mathrm{X}}_2$ and$X_0$is$0.05\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of transformer $T2$is $100MVA$, the voltage rating is $13.8/230\mathrm{kV}$, and the value of $X_1,X_2$and $X_0$is $0.05\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

Write the rating for the line.

The power rating of line $TL12$ is $100MVA$, the voltage rating is $230kV$, and the value of $X_1,X_2$is $0.1\text{\hspace{0.17em}per\hspace{0.17em}unit}$ and $X_0$ is $0.3\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of line $TL13$is $100\text{\hspace{0.17em}MVA}$, the voltage rating is $230kV$, and the value of ${\mathrm{X}}_1,{\mathrm{X}}_2$ is $0.1\text{\hspace{0.17em}per\hspace{0.17em}unit}$ and $X_0$is $0.3\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of line$TL23$ is$100\text{\hspace{0.17em}MVA}$ , the voltage rating is $230kV$, and the value of$X_1,X_2$ is$0.1\text{\hspace{0.17em}per\hspace{0.17em}unit}$ and$X_0$ is$0.3\text{\hspace{0.17em}per\hspace{0.17em}unit}$ .

Determine the formulas to calculate the fault current for a bolted three phase fault at bus 3.

The equation to calculate the phase current is given as follows.

$\left[\begin{array}{c}{\mathrm{I}}_{\mathrm{a}}\\ {\mathrm{I}}_{\mathrm{b}}\\ {\mathrm{I}}_{\mathrm{c}}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathrm{a}}^2& \mathrm{a}\\ 1& \mathrm{a}& {\mathrm{a}}^2\end{array}\right]\left[\begin{array}{c}{\mathrm{I}}_0\\ {\mathrm{I}}_1\\ {\mathrm{I}}_2\end{array}\right]$ …… (1)

Here ${\mathrm{I}}_{\mathrm{a}},{\mathrm{I}}_{\mathrm{b}}$ and${\mathbf{I}}_{\mathbf{C}}$ are the phase current,$I_0,I_1$ and${\mathit{I}}_{\mathbf{2}}$ are the sequence current.

Draw the per unit sequence network and calculate the fault current.

Draw the positive sequence network,

Convert the delta 123 into star.

$\begin{array}{l}{X}_1=\frac{j0.1\times j0.1}{j0.1+j0.1+j0.1}\\ X_1=j0.033\text{\hspace{0.17em}pu}\end{array}$

Draw the positive sequence circuit as,

Calculate the equivalent impedance of positive sequence network

$\begin{array}{c}{Z}_1=[(j0.25+j0.033)\parallel (j0.25+j0.033)]+j0.033\\ Z_1=j0.283\parallel j0.283+j0.033\\ Z_1=j0.1415+j0.033\\ Z_1=j0.1745\text{\hspace{0.17em}pu}\end{array}$

Draw the negative sequence network.

The equivalent impedance for negative sequence network is same as the positive sequence network.

$Z_2=-j0.1745\text{\hspace{0.17em}pu}$

The generators are grounded with the neutral of$j0.03\text{\hspace{0.17em}pu}$ reactance. The neutral reactance is $j0.09\text{\hspace{0.17em}pu}$.

Draw the zero-sequence network.

Convert the delta 123 into star.

$\begin{array}{l}{X}_1=\frac{j0.3\times j0.3}{j0.3+j0.3+j0.3}\\ X_1=j0.1\text{\hspace{0.17em}pu}\end{array}$

Again, draw the zero-sequence network.

Calculate the equivalent impedance of zero-sequence network between the terminal 3 and ground.

$\begin{array}{l}{Z}_0=[(j0.09+j0.05+j0.05+j0.1)\parallel (j0.05+j0.1)]+j0.1\\ Z_0=\left(j0.29\right)\parallel j0.15+j0.1\\ Z_0=j0.099+j0.1\\ Z_0=j0.199\text{\hspace{0.17em}pu}\end{array}$

Draw the equivalent circuit for zero sequence network when the fault is occurred at bus 3.

The zero-sequence fault current is zero.

Draw the equivalent circuit for positive sequence network when the fault is occurred at bus 3.

Calculate the positive sequence fault current.

$\begin{array}{l}{I}_1=\frac{1\angle 0°}{j0.1745}\\ I_1=5.73\angle -90°\text{\hspace{0.17em}pu}\end{array}$

Draw the equivalent negative sequence network.

The negative sequence fault current is zero.

Calculate the fault current in three phases.

Hence the three-phase fault current are$\left[\begin{array}{c}5.73\angle -90°\\ 5.73\angle -150°\\ 5.73\angle -30°\end{array}\right]\text{\hspace{0.17em}pu}$ .