Question

Repeat Problem 9.43 for a bolted single line-to-ground fault at bus 1.

Short answer

The fault current and voltages are $\left[\begin{array}{c}-j6\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$and localid="1656416729820" $\left[\begin{array}{c}0.6\angle 0°\\ 0.916\angle 250.89°\\ 0.916\angle 109.10°\end{array}\right]\text{\hspace{0.33em}pu}$respectively.

Step-by-step solution

Write the given data from the question.

Write the zero, positive and negative sequence matrices for two bus three phase system.

$\begin{array}{l}{\mathrm{Z}}_{\mathrm{bus}0}=\mathrm{j}\left[\begin{array}{cc}0.10& 0\\ 0& 0.10\end{array}\right]\mathrm{per}\mathrm{unit}\\ {\mathrm{Z}}_{\mathrm{bus}1}={\mathrm{Z}}_{\mathrm{bus}2}\\ {\mathrm{Z}}_{\mathrm{bus}2}=\mathrm{j}\left[\begin{array}{cc}0.20& 0.10\\ 0.10& 0.30\end{array}\right]\mathrm{per}\mathrm{unit}\end{array}$

The pre fault voltage at each bus in positive sequence network $V_F=1.03\text{\hspace{0.33em}pu}$.

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 2 (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,

$\left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-3}\end{array}\right]$ …… (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 the sequence fault current for bolted single line to ground fault

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

$\begin{array}{l}{I}_{1-1}=\frac{1\angle 0°}{j0.10+j0.2+j0.2}\\ I_{1-1}=\frac{1\angle 0°}{j0.5}\\ I_{1-1}=-j2\text{\hspace{0.33em}pu}\end{array}$

Calculate the voltages at bus 2.

Substitute$-j2\text{\hspace{0.33em}pu}$ for $I_{1-0},I_{1-1},I_{1-2}$, 0 pu for $Z_{2-0}$, $j0.1\text{\hspace{0.33em}pu}$for$Z_{2-1}$ and$j0.1\text{\hspace{0.33em}pu}$ for$Z_{2-2}$ into equation (2).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{ccc}0& 0& 0\\ 0& j0.1& 0\\ 0& 0& j0.1\end{array}\right]\left[\begin{array}{c}-j2\\ -j2\\ -j2\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 0.2\\ 0.2\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 0.8\\ -0.2\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault voltage.

Substitute$\left[\begin{array}{c}0\\ 0.8\\ -0.2\end{array}\right]\text{\hspace{0.33em}pu}$for$\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]$ into equation (3).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ 0.8\\ -0.2\end{array}\right]\\ \left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{c}0.6\angle 0°\\ 0.916\angle 250.89°\\ 0.916\angle 109.10°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault current.

Substitute$-j2\text{\hspace{0.33em}pu}$ for$I_{1-0},I_{1-1},I_{1-2}$ into equation (4).

$\begin{array}{l}\left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}-j2\\ -j2\\ -j2\end{array}\right]\\ \left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{c}-j6\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Therefore, the fault current and voltages are$\left[\begin{array}{c}-j6\\ 0\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$ and $\left[\begin{array}{c}0.6\angle 0°\\ 0.916\angle 250.89°\\ 0.916\angle 109.10°\end{array}\right]\text{\hspace{0.33em}pu}$respectively.