Question

For the system of Problem 9.13, compute the fault current for the following faults at bus 3: (a) a single line-to-ground fault through a fault impedance${\mathbf{Z}}_F=\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{1}$ per unit, (b) a line-to-line fault through a fault impedance${\mathbf{Z}}_F=\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{1}$ per unit, (c) a double line-to-ground fault through a common fault impedance to ground${\mathbf{Z}}_F=\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{1}$ per unit.

Short answer

(a)

The value of sub-transient fault current from single line-to-ground fault is $\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}1.92\angle -90°\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$.

(b)

The value of sub-transient fault current from line-to-line fault is $\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}0\\ 2.2\angle 180°\\ 2.2\angle 0°\end{array}\right]\text{\hspace{0.33em}pu}$.

(c)

The value of sub-transient fault current from double line-to-ground fault is $\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}0\\ 2.61\angle 15.9°\\ 2.61\angle 164.1°\end{array}\right]\text{\hspace{0.33em}pu}$.

The value of fault current in the neutral is $I_n=1.43\angle 90°\text{\hspace{0.33em}pu}$.

Step-by-step solution

Write the given data from the question.

As reference Problem 9.13

The fault impedance in per unit is ${\text{Z}}_{\text{F}}=\text{j0.1\hspace{0.33em}perunit}$.

Theline-to-line fault through fault impedance ${\text{Z}}_{\text{F}}=\text{j0.1\hspace{0.33em}perunit}$.

The common line-to-ground fault impedance is ${\text{Z}}_{\text{F}}=\text{j0.1\hspace{0.33em}perunit}$.

Determine the formula of line-to-line fault and line-to-ground fault.

Write the formula of sub-transient fault current from line-to-ground fault.

$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathbf{a}}^2& a\\ 1& a& {\mathbf{a}}^2\end{array}\right]\left[\begin{array}{c}{\mathbf{I}}_0\\ {\mathbf{I}}_1\\ {\mathbf{I}}_2\end{array}\right]$ …… (1)

Here ${\mathbf{I}}_0$,${\mathbf{I}}_1$, and ${\mathbf{I}}_2$are phase sequence currents.

Write the formula of neutral fault current for a double line-to-ground fault.

${I^{\text{'}\text{'}}}_n=\mathbf{3}{\mathbf{I}}_0$ …… (2)

Here,${\mathbf{I}}_0$ is positive sequence current.

(a) Determine the sub-transient fault current and fault voltages from a bolted single line to ground fault.

Reference as a Problem 9.13 from the text book.

Draw a circuit diagram of per unit positive-sequence network.

Draw a circuit diagram of positive-sequence network for a single line-to-ground fault occurs at bus 3.

Calculate the equivalent impedance of the positive sequence impedance.

$\begin{array}{c}{Z}_1=\left(j0.25\parallel j0.15\right)+j0.125+\left(j0.25\parallel j0.25\right)\\ =\frac{\left(j0.25\right)\left(j0.15\right)}{\left(j0.25\right)\left(j0.15\right)}+j0.125+\frac{\left(j0.25\right)\left(j0.25\right)}{\left(j0.25\right)+\left(j0.25\right)}\\ =j0.344\text{\hspace{0.17em}pu}\end{array}$

Draw a circuit diagram in per unit of negative sequence network

The per-unit negative sequence impedance of fault at bus 3 is similar to positive sequence impedance.

${\text{Z}}_{\text{2}}=j0.344\text{\hspace{0.33em}pu}$

The two generators are grounded through a common neutral reactance of $\mathrm{j}0.08333\mathrm{\Omega }$are three times the neutral reactance of$\mathrm{j}0.25\mathrm{\Omega }$is add in the zero-sequence network

Draw a circuit diagram of zero-sequence network.

Draw a circuit diagram of zero-sequence network when fault happen at bus 3.

Calculate the per unit zero sequence impedance.

$\begin{array}{c}{Z}_0=\left(j0.4\parallel j0.35\right)+\left(j0.1\parallel j0.7125\right)\\ =\frac{\left(j0.4\right)\left(j0.35\right)}{\left(j0.4\right)+\left(j0.35\right)}+j0.3+\frac{\left(j0.1\right)\left(j0.7125\right)}{\left(j0.1\right)+\left(j0.7125\right)}\\ =j0.187+j0.3+j0.0877\\ =j0.575\text{\hspace{0.17em}pu\hspace{0.33em}}\end{array}$

A single line-to-ground fault occurs in a system through fault impedance at $Z_{\text{F}}$.

Determine the fault current in the sequence components.

$\begin{array}{c}{I}_0=I_1=I_2\\ =\frac{E_{\text{g}}}{\left({\text{Z}}_0+{\text{Z}}_1+{\text{Z}}_2+3Z_{\text{F}}\right)}\\ =\frac{1}{j\left(0.575+0.344+0.344+\left(0.1\right)\times 3\right)}\\ =-j0.64\text{\hspace{0.33em}pu}\end{array}$

Determine the value of sub-transient fault current from single line-to-ground fault.

Substitute$-j0.64\text{\hspace{0.33em}pu}$for$I_0$into equation (1).

$\begin{array}{c}\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}3I_0\\ \left(1+a^2+a\right)I_0\\ \left(1+a^2+a\right)I_0\end{array}\right]\\ =\left[\begin{array}{c}1.92\angle -90°\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Therefore, the sub-transient fault current from single line-to-ground fault$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]$in pu is $\left[\begin{array}{c}1.92\angle -90°\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$.

(b) Determine the sub-transient fault current and fault voltages from line to line fault.

The fault current in a zero-sequence component for line to line fault is zero.

Determine the fault current in the positive and negative sequence domain.

$\begin{array}{c}{I}_1=-I_2\\ =\frac{E_{\text{g}}}{j\left({\text{Z}}_{\text{1}}+{\text{Z}}_{\text{2}}+Z_{\text{F}}\right)}\\ =\frac{1}{j\left(0.344+0.344+0.1\right)}\\ =-j1.27\text{\hspace{0.33em}pu}\end{array}$

Determine the sub-transient fault current from line to line fault in the system.

Substitute$-j1.27$ for $I_1$and$j1.27$ for$I_2$ into equation (1)

$\begin{array}{c}\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ -j1.27\\ j1.27\end{array}\right]\\ =\left[\begin{array}{c}0-j1.27+j1.27\\ 0+\left(-j1.27\right)\left(1\angle -120\right)+\left(j1.27\right)\left(1\angle 120\right)\\ 0+\left(-j1.27\right)\left(1\angle 120\right)+\left(j1.27\right)\left(1\angle -120\right)\end{array}\right]\\ =\left[\begin{array}{c}0\\ 2.2\angle 180°\\ 2.2\angle 0°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Hence, the sub-transient fault currents in phase of a system$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]$ is $\left[\begin{array}{c}0\\ 2.2\angle 180°\\ 2.2\angle 0°\end{array}\right]\text{\hspace{0.33em}pu}$

(c) Determine the sub-transient fault current and fault voltages from double line to ground fault.

Determine the positive-sequence fault current component for a double line-to-ground fault.

$I_1=\frac{E_R}{j\left(Z_1+\left(\frac{Z_2\left(Z_0+3Z_{\text{F}}\right)}{Z_2+Z_0+3Z_{\text{F}}}\right)\right)}$

Here, $E_R$is required voltage, $Z_1,Z_2\&Z_0$is sequence impedance, $Z_F$is fault impedance.

Substitute 1 for $E_R$, 0.344 for $Z_1$, 0.344 for $Z_2$, 0.575 for$Z_0$ and 0.3 for $Z_F$into above equation.

$\begin{array}{c}{I}_1=\frac{1}{j\left(0.344+\left(\frac{\left(0.344\right)\left(0.575+0.3\right)}{0.344+0.575+0.3}\right)\right)}\\ =-j1.69\text{\hspace{0.33em}pu}\end{array}$

Determine the negative-sequence fault current component for a double line-to-ground fault.

$\begin{array}{c}{I}_2=\left(-I_1\right)\frac{Z_0+3Z_{\text{F}}}{Z_2+Z_0+3Z_{\text{F}}}\\ =\left(j1.69\text{\hspace{0.33em}pu}\right)\left(\frac{j0.575+j0.3}{j0.344+j0.575+j0.3}\right)\\ =j1.213\text{\hspace{0.33em}pu}\end{array}$

Determine the zero-sequence fault current component for a double line-to-ground fault.

$\begin{array}{c}{I}_0=\left(-I_1\right)\frac{Z_2}{Z_2+Z_0+3Z_{\text{F}}}\\ =\left(j1.69\text{\hspace{0.33em}pu}\right)\left(\frac{j0.344}{j0.344+j0.575+j0.3}\right)\\ =j0.477\text{\hspace{0.17em}pu}\end{array}$

Determine the sub-transient fault current from double line to ground fault in the system.

Substitute$j0.477$ for $I_0$, $j1.213$for$I_1$ and$-j1.69$ for$I_2$ into equation (1)

$\begin{array}{c}\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}j0.477\\ j1.213\\ -j1.69\end{array}\right]\\ =\left[\begin{array}{c}j0.477+j1.213-j1.69\\ j0.477+\left(-j1.213\right)\left(1\angle -120\right)+\left(-j1.69\right)\left(1\angle 120\right)\\ j0.477+\left(-j1.213\right)\left(1\angle 120\right)+\left(-j1.69\right)\left(1\angle -120\right)\end{array}\right]\\ =\left[\begin{array}{c}0\\ 2.61\angle 15.9°\\ 2.61\angle 164.1°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Hence, the sub-transient fault currents from double line-to-ground fault of a system $\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]$is $\left[\begin{array}{c}0\\ 2.61\angle 15.9°\\ 2.61\angle 164.1°\end{array}\right]\text{\hspace{0.33em}pu}$.

Determine the double line-to-ground fault current in the neutral path.

Substitute $j0.477$for $I_0$into equation (2).

$\begin{array}{c}{I}_n=3\left(j0.477\right)\\ =1.43\angle 90°\text{\hspace{0.33em}pu}\end{array}$

Therefore the neutral fault current for a double line-to-ground fault is $I_n=1.43\angle 90°\text{\hspace{0.33em}pu}$.