Question

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

Short answer

The fault current and voltages are $\left[\begin{array}{c}-j8.824\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$and $\left[\begin{array}{c}0.5294\angle 0°\\ 0.9056\angle 253°\\ 0.9056\angle 107°\end{array}\right]\text{\hspace{0.33em}pu}$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 single line to ground fault at bus 1.

The equation to calculate the sequence fault current for bolted single line to ground fault is given as follows.

$\begin{array}{l}{\mathbf{I}}_{\mathbf{1}\mathbf{-}\mathbf{1}}\mathbf{=}{\mathbf{I}}_{\mathbf{1}\mathbf{-}\mathbf{0}}\\ {\mathbf{I}}_{\mathbf{1}\mathbf{-}\mathbf{1}}\mathbf{=}{\mathbf{I}}_{\mathbf{1}\mathbf{-}\mathbf{2}}\\ {\mathbf{I}}_{\mathbf{1}\mathbf{-}\mathbf{1}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{F}}}{{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{0}}\mathbf{+}{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{2}}\mathbf{+}{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{2}}}\end{array}$ …… (1)

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

$\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]\mathbf{=}\left[\begin{array}{c}0\\ V_F\\ 0\end{array}\right]\mathbf{-}\left[\begin{array}{ccc}{Z}_{2-0}& 0& 0\\ 0& Z_{2-1}& 0\\ 0& 0& Z_{2-2}\end{array}\right]\left[\begin{array}{c}{I}_{2-0}\\ I_{2-1}\\ I_{2-2}\end{array}\right]$ …… (2)

The equation to calculate the fault voltage is given as,

role="math" localid="1655959724582" $\mathbf{\left[}\begin{array}{c}{\mathbf{V}}_{\mathbf{2}\mathbf{ag}}\\ {\mathbf{V}}_{\mathbf{2}\mathbf{bg}}\\ {\mathbf{V}}_{\mathbf{2}\mathbf{cg}}\end{array}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{c}{\mathbf{V}}_{\mathbf{2}\mathbf{-}\mathbf{0}}\\ {\mathbf{V}}_{\mathbf{2}\mathbf{-}\mathbf{1}}\\ {\mathbf{V}}_{\mathbf{2}\mathbf{-}\mathbf{3}}\end{array}\mathbf{\right]}$ …… (3)

The equation to calculate the fault current is given as,

$\left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{I}_{1-0}\\ I_{1-1}\\ I_{1-2}\end{array}\right]$ …… (4)

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

Calculate thesequence fault current for bolted single line to ground fault

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

(1).

$\begin{array}{l}{I}_{1-1}=\frac{1\angle 0°}{j0.10+j0.12+j0.12}\\ I_{1-1}=\frac{1\angle 0°}{j0.34}\\ I_{1-1}=-j2.941\text{\hspace{0.33em}pu}\end{array}$

Calculate the voltages at bus 2.

Substitute $-j2.941\text{\hspace{0.33em}pu}$for $I_{1-0},I_{1-1},I_{1-2}$, 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 (2).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{ccc}0& 0& 0\\ 0& j0.08& 0\\ 0& 0& j0.08\end{array}\right]\left[\begin{array}{c}-j2.941\\ -j2.941\\ -j2.941\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 0.2353\\ 0.2353\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 0.7647\\ -0.2353\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault voltage.

Substitute$\left[\begin{array}{c}0\\ 0.7647\\ -0.2353\end{array}\right]\text{\hspace{0.33em}pu}$ for$\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]$into equation (3).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ 0.7647\\ -0.2353\end{array}\right]\\ \left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{c}0.5294\angle 0°\\ 0.9056\angle 253°\\ 0.9056\angle 107°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault current.

Substitute$-j2.941\text{\hspace{0.33em}pu}$ for ${\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{0}}\mathbf{,}{\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{1}}\mathbf{,}{\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{2}}$into equation (4).

$\begin{array}{l}\left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}-j2.941\\ -j2.941\\ -j2.941\end{array}\right]\\ \left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{c}-j8.824\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Therefore, the fault current and voltages are$\left[\begin{array}{c}-j8.824\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$and$\left[\begin{array}{c}0.5294\angle 0°\\ 0.9056\angle 253°\\ 0.9056\angle 107°\end{array}\right]\text{\hspace{0.33em}pu}$respectively.