Question

Repeat Problem 9.38 for a bolted double line-to-ground fault at bus 1.

Short answer

The fault current and voltages are$\left[\begin{array}{c}0\\ 8.605\angle 147°\\ 8.605\angle 33°\end{array}\right]\text{\hspace{0.33em}pu}$ and $\left[\begin{array}{c}0.750\\ 0.4733\angle 217.6°\\ 0.4733\angle 142.4°\end{array}\right]\text{\hspace{0.33em}pu}$respectively.

Step-by-step solution

Write the given data from the question.

The pre fault voltage for each bus in the positive sequence network$V_F=1\angle 0°\text{\hspace{0.33em}pu}$,

Write the zero-sequence impedance matrix,

$Z_{bus0}=j\left[\begin{array}{ccc}0.10& 0& 0\\ 0& 0.20& 0\\ 0& 0& 0.10\end{array}\right]perunit$

Write the positive and negative-sequence impedance matrix,

$\begin{array}{l}{Z}_{bus1}=Z_{bus2}\\ Z_{bus1}=j\left[\begin{array}{ccc}0.12& 0.08& 0.04\\ 0.08& 0.12& 0.06\\ 0.04& 0.06& 0.08\end{array}\right]perunit\end{array}$

Determine the equations to calculate the per-unit fault current and per-unit voltage at bus 2 for a bolted line to line fault at bus 1. 

The equation to calculate the positive sequence fault current is given as follows.

${\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{1}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{F}}}{{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{1}}\mathbf{+}{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{2}}||Z_{11-0}}$ …… (1)

The equation to calculate the negative sequence fault current is given as follows.

${\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{2}}\mathbf{=}\mathbf{-}{\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{1}}\left(\frac{Z_{11-0}}{Z_{11-0}+Z_{11-2}}\right)$ …… (2)

The equation to calculate the zero-sequence fault current is given as follows.

${\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{2}}\mathbf{=}\mathbf{-}{\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{1}}\left(\frac{Z_{11-0}}{Z_{11-0}+Z_{11-2}}\right)$ …… (3)

The equation to calculate the voltage at bus (from 9.5.9 of textbook equation) is given as follows.

localid="1656754669186" $\left[\begin{array}{c}{\mathrm{V}}_{2-0}\\ {\mathrm{V}}_{2-1}\\ {\mathrm{V}}_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ {\mathrm{V}}_{\mathrm{F}}\\ 0\end{array}\right]-\left[\begin{array}{ccc}{\mathrm{Z}}_{2-0}& 0& 0\\ 0& {\mathrm{Z}}_{2-1}& 0\\ 0& 0& {\mathrm{Z}}_{2-2}\end{array}\right]\left[\begin{array}{c}{\mathrm{I}}_{2-0}\\ {\mathrm{I}}_{2-1}\\ {\mathrm{I}}_{2-2}\end{array}\right]$ …… (4)

The equation to calculate the fault voltage is given as,

localid="1656754660009" $\left[\begin{array}{c}{\mathrm{V}}_{2\mathrm{ag}}\\ {\mathrm{V}}_{2\mathrm{bg}}\\ {\mathrm{V}}_{2\mathrm{cg}}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathrm{a}}^2& \mathrm{a}\\ 1& \mathrm{a}& {\mathrm{a}}^2\end{array}\right]\left[\begin{array}{c}{\mathrm{V}}_{2-0}\\ {\mathrm{V}}_{2-1}\\ {\mathrm{V}}_{2-3}\end{array}\right]$ …… (5)

The equation to calculate the fault current is given as,

localid="1656754704820" $\left[\begin{array}{c}{\mathrm{I}}_{1\mathrm{a}}\\ {\mathrm{I}}_{1\mathrm{b}}\\ {\mathrm{I}}_{1\mathrm{c}}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathrm{a}}^2& \mathrm{a}\\ 1& \mathrm{a}& {\mathrm{a}}^2\end{array}\right]\left[\begin{array}{c}{\mathrm{I}}_{1-0}\\ {\mathrm{I}}_{1-1}\\ {\mathrm{I}}_{1-2}\end{array}\right]$…… (6)

Calculate the per-unit fault current and per-unit voltage at bus 2 for a bolted line to line fault at bus 1. 

Calculate the positive sequence component of fault current.

Substitute $1\angle 0°\text{\hspace{0.33em}pu}$for$V_F$, $j0.12\text{\hspace{0.33em}pu}$for$Z_{11-1}$, $j0.12\text{\hspace{0.33em}pu}$for$Z_{11-2}$,and$j0.10\text{\hspace{0.33em}pu}$for $Z_{11-0}$into equation (1).

$\begin{array}{l}{I}_{1-1}=\frac{1\angle 0°}{j\left(0.12+j0.12||j0.10\right)}\\ I_{1-1}=\frac{1\angle 0°}{j0.1745}\\ I_{1-1}=-j5.729\text{\hspace{0.33em}pu}\end{array}$

Calculate the negative sequence component of fault current.

Substitute$-j5.729\text{\hspace{0.33em}pu}$for $I_{1-1}$, $j0.12\text{\hspace{0.33em}pu}$for$Z_{11-2}$, and$j0.10\text{\hspace{0.33em}pu}$for$Z_{11-0}$into equation (2).

$\begin{array}{l}{I}_2=-\left(-j5.729\text{\hspace{0.33em}pu}\right)\times \frac{j0.10}{j0.10+j0.12}\\ I_2=j5.729\times \frac{j0.10}{j0.22}\\ I_2=j2.604\text{\hspace{0.33em}pu}\end{array}$

Calculate the zero-sequence component of fault current.

Substitute $-j5.729\text{\hspace{0.33em}pu}$for$I_{1-1}$,$j0.12\text{\hspace{0.33em}pu}$for $Z_{11-2}$and $j0.10\text{\hspace{0.33em}pu}$for $Z_0$into equation (3).

$\begin{array}{l}{I}_0=-\left(-j5.729\text{\hspace{0.33em}pu}\right)\times \frac{j0.12}{j0.1+j0.12}\\ I_0=j5.729\text{\hspace{0.33em}pu}\times \frac{j0.12}{j0.22}\\ I_0=j3.125\text{\hspace{0.33em}pu}\end{array}$

Calculate the voltages at bus 2.

Substitute$j3.125\text{\hspace{0.33em}pu}$ for $I_{1-0}$,$-j5.729\text{\hspace{0.33em}pu}$for $I_{1-1}$,$j2.604\text{\hspace{0.33em}pu}$for $I_{1-2}$, $0\text{\hspace{0.33em}pu}$for$Z_{2-0}$,$j0.08\text{\hspace{0.33em}pu}$for$Z_{2-1}$and$j0.08\text{\hspace{0.33em}pu}$for $Z_{2-2}$into equation (4).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{ccc}0& 0& 0\\ 0& j0.08& 0\\ 0& 0& j0.08\end{array}\right]\left[\begin{array}{c}j3.125\text{\hspace{0.33em}pu}\\ -j5.729\text{\hspace{0.33em}pu}\\ j2.604\text{\hspace{0.33em}pu}\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 0.4583\\ -0.2083\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 0.5417\\ 0.2083\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault voltage.

Substitute$\left[\begin{array}{c}0\\ 0.5417\\ 0.2083\end{array}\right]\text{\hspace{0.33em}pu}$ for $\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]$into equation (5).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ 0.5417\\ 0.2083\end{array}\right]\\ \left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{c}0.750\\ 0.4733\angle 217.6°\\ 0.4733\angle 142.4°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault current.

Substitute$j3.125\text{\hspace{0.33em}pu}$ for$I_{1-0}$,$-j5.729\text{\hspace{0.33em}pu}$for$I_{1-1}$ ,$j2.604\text{\hspace{0.33em}pu}$for $I_{1-2}$,$I_{1-2}$into equation (6).

$\begin{array}{l}\left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}j3.125\text{\hspace{0.33em}pu}\\ -j5.729\text{\hspace{0.33em}pu}\\ j2.604\text{\hspace{0.33em}pu}\end{array}\right]\\ \left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{c}0\\ 8.605\angle 147°\\ 8.605\angle 33°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Therefore, the fault current and voltages are$\left[\begin{array}{c}0\\ 8.605\angle 147°\\ 8.605\angle 33°\end{array}\right]\text{\hspace{0.33em}pu}$and$\left[\begin{array}{c}0.750\\ 0.4733\angle 217.6°\\ 0.4733\angle 142.4°\end{array}\right]\text{\hspace{0.33em}pu}$respectively.