Question

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

Short answer

The fault current and voltages are $\left[\begin{array}{c}0\\ 1.32\angle 250.89°\\ 1.32\angle 109.10°\end{array}\right]\text{\hspace{0.33em}pu}$and $\left[\begin{array}{c}0.750\\ 0.5728\angle 229.106°\\ 0.5728\angle 130.89°\end{array}\right]\text{\hspace{0.33em}pu}$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,

$\begin{array}{l}{Z}_{bus0}=j\left[\begin{array}{cc}0.10& 0\\ 0& 0.10\end{array}\right]\mathrm{per}\mathrm{unit}\\ Z_{bus1}=Z_{bus2}\\ Z_{bus1}=j\left[\begin{array}{cc}0.20& 0.10\\ 0.10& 0.30\end{array}\right]\mathrm{per}\mathrm{unit}\end{array}$

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 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.

localid="1656400693336" ${\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{2}}\mathbf{=}\mathbf{-}{\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{0}}}{{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{0}}\mathbf{+}{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{2}}}\mathbf{\right)}$ …… (2)

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

localid="1656400697791" ${\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{0}}\mathbf{=}\mathbf{-}{\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{0}}\mathbf{+}{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{2}}}\mathbf{\right)}$ …… (3)

The equation to calculate the voltage at bus 2 (from 9.5.9 of textbook equation) is given as follows.

localid="1656400702393" $\mathbf{\left[}\begin{array}{c}{\mathbf{V}}_{\mathbf{2}\mathbf{-}\mathbf{0}}\\ {\mathbf{V}}_{\mathbf{2}\mathbf{-}\mathbf{1}}\\ {\mathbf{V}}_{\mathbf{2}\mathbf{-}\mathbf{2}}\end{array}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{0}\\ {\mathbf{V}}_{\mathbf{F}}\\ \mathbf{0}\end{array}\mathbf{\right]}\mathbf{-}\mathbf{\left[}\begin{array}{ccc}{\mathbf{Z}}_{\mathbf{2}\mathbf{-}\mathbf{0}}& \mathbf{0}& \mathbf{0}\\ \mathbf{0}& {\mathbf{Z}}_{\mathbf{2}\mathbf{-}\mathbf{1}}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& {\mathbf{Z}}_{\mathbf{2}\mathbf{-}\mathbf{2}}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{c}{\mathbf{I}}_{\mathbf{2}\mathbf{-}\mathbf{0}}\\ {\mathbf{I}}_{\mathbf{2}\mathbf{-}\mathbf{1}}\\ {\mathbf{I}}_{\mathbf{2}\mathbf{-}\mathbf{2}}\end{array}\mathbf{\right]}$ …… (4)

The equation to calculate the fault voltage is given as,

localid="1656400707143" $\mathbf{\left[}\begin{array}{c}{\mathbf{V}}_{\mathbf{2}\mathbf{a}\mathbf{g}}\\ {\mathbf{V}}_{\mathbf{2}\mathbf{b}\mathbf{g}}\\ {\mathbf{V}}_{\mathbf{2}\mathbf{c}\mathbf{g}}\end{array}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{c}{\mathbf{V}}_{\mathbf{2}\mathbf{-}\mathbf{0}}\\ {\mathbf{V}}_{\mathbf{2}\mathbf{-}\mathbf{1}}\\ {\mathbf{V}}_{\mathbf{2}\mathbf{-}\mathbf{3}}\end{array}\mathbf{\right]}$ …… (5)

The equation to calculate the fault current is given as,

localid="1656400715027" $\mathbf{\left[}\begin{array}{c}{\mathbf{I}}_{\mathbf{1}\mathbf{a}}\\ {\mathbf{I}}_{\mathbf{1}\mathbf{b}}\\ {\mathbf{I}}_{\mathbf{1}\mathbf{c}}\end{array}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& {\mathbf{a}}^{\mathbf{2}}& \mathbf{a}\\ \mathbf{1}& \mathbf{a}& {\mathbf{a}}^{\mathbf{2}}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{c}{\mathbf{I}}_{\mathbf{1}\mathbf{-}\mathbf{0}}\\ {\mathbf{I}}_{\mathbf{1}\mathbf{-}\mathbf{1}}\\ {\mathbf{I}}_{\mathbf{1}\mathbf{-}\mathbf{2}}\end{array}\mathbf{\right]}$ …… (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 $V_F$, $j0.2\text{\hspace{0.33em}pu}$for $Z_{11-1}$, $j0.2\text{\hspace{0.33em}pu}$for $Z_{11-2}$,and $j0.10\text{\hspace{0.33em}pu}$for $Z_{11-0}$into equation (1).

$\begin{array}{l}{I}_{1-1}=\frac{1\angle 0°}{j\left(0.2+j0.2||j0.10\right)}\\ I_{1-1}=\frac{1\angle 0°}{j0.267}\\ I_{1-1}=-j3.75\text{\hspace{0.33em}pu}\end{array}$

Calculate the negative sequence component of fault current.

Substitute $-j3.75\text{\hspace{0.33em}pu}$for $I_{1-1}$,$j0.2\text{\hspace{0.33em}pu}$for $Z_{11-2}$, and$j0.10\text{\hspace{0.33em}pu}$for$Z_{11-0}$into equation (2).

$\begin{array}{l}{I}_{1-2}=-\left(-j3.75\text{\hspace{0.33em}pu}\right)\times \frac{j0.10}{j0.10+j0.2}\\ I_{1-2}=j5.729\times \frac{j0.10}{j0.30}\\ I_{1-2}=j1.25\text{\hspace{0.33em}pu}\end{array}$

Calculate the zero-sequence component of fault current.

Substitute $-j3.75\text{\hspace{0.33em}pu}$for $I_{1-1}$, $j0.2\text{\hspace{0.33em}pu}$for $Z_{11-2}$and $j0.10\text{\hspace{0.33em}pu}$for$Z_0$into equation (3).

$\begin{array}{l}{I}_{1-0}=-\left(-j3.75\text{\hspace{0.33em}pu}\right)\times \frac{j0.2}{j0.1+j0.2}\\ I_{1-0}=j5.729\text{\hspace{0.33em}pu}\times \frac{j0.2}{j0.30}\\ I_{1-0}=j2.5\text{\hspace{0.33em}pu}\end{array}$

Calculate the voltages at bus 2.

Substitute$j2.5\text{\hspace{0.33em}pu}$for $I_{1-0}$, $-j3.75\text{\hspace{0.33em}pu}$for $I_{1-1}$, $j1.25\text{\hspace{0.33em}pu}$for $I_{1-2}$, $0\text{\hspace{0.33em}pu}$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 (4).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{ccc}0& 0& 0\\ 0& j0.1& 0\\ 0& 0& j0.1\end{array}\right]\left[\begin{array}{c}j2.5\text{\hspace{0.33em}pu}\\ -j3.75\text{\hspace{0.33em}pu}\\ j1.25\text{\hspace{0.33em}pu}\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 0.375\\ -0.125\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 0.625\\ 0.125\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault voltage.

Substitute $\left[\begin{array}{c}0\\ 0.625\\ 0.125\end{array}\right]\text{\hspace{0.33em}pu}$for $\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]$into equation (5).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ 0.625\\ 0.125\end{array}\right]\text{\hspace{0.33em}pu}\\ \left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{c}0.750\\ 0.5728\angle 229.106°\\ 0.5728\angle 130.89°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault current.

Substitute $j2.5\text{\hspace{0.33em}pu}$for $I_{1-0}$, $-j3.75\text{\hspace{0.33em}pu}$for $I_{1-1}$, $j1.25\text{\hspace{0.33em}pu}$for $I_{1-2}$into equation (6).

$\begin{array}{l}\left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}j2.5\text{\hspace{0.33em}pu}\\ -j3.75\text{\hspace{0.33em}pu}\\ j1.25\text{\hspace{0.33em}pu}\end{array}\right]\\ \left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{c}0\\ 1.32\angle 250.89°\\ 1.32\angle 109.10°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Therefore, the fault current and voltages are $\left[\begin{array}{c}0\\ 1.32\angle 250.89°\\ 1.32\angle 109.10°\end{array}\right]\text{\hspace{0.33em}pu}$and $\left[\begin{array}{c}0.750\\ 0.5728\angle 229.106°\\ 0.5728\angle 130.89°\end{array}\right]\text{\hspace{0.33em}pu}$respectively.