Question

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

Short answer

The fault current and voltages are$\left[\begin{array}{c}0\\ 7.217\angle 180°\\ 7.217\angle 0°\end{array}\right]\text{\hspace{0.33em}pu}$ and $\left[\begin{array}{c}1\\ 0.5774\angle 210°\\ 0.5774\angle 107°\end{array}\right]\text{\hspace{0.33em}pu}$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 line to line fault at bus 1. 

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

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

$\left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-3}\end{array}\right]$ …… (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 line to line fault at bus 1.

Calculate the sequence fault current for bolted line to line fault

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

$\begin{array}{l}{I}_{1-1}=-I_{1-2}\\ I_{1-1}=\frac{1\angle 0°}{j0.12+j0.12}\\ I_{1-1}=\frac{1\angle 0°}{j0.24}\\ I_{1-1}=-j4.167\text{\hspace{0.33em}pu}\end{array}$

Calculate the voltages at bus 2.

Substitute$0\text{\hspace{0.33em}pu}$for$I_{1-0}$,$-j4.167\text{\hspace{0.33em}pu}$for$I_{1-1}$ ,$j4.167\text{\hspace{0.33em}pu}$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 (2).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{ccc}0& 0& 0\\ 0& j0.08& 0\\ 0& 0& j0.08\end{array}\right]\left[\begin{array}{c}0\\ -j4.167\text{\hspace{0.33em}pu}\\ j4.167\text{\hspace{0.33em}pu}\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 0.3333\\ -0.3333\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 0.6667\\ 0.3333\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault voltage.

Substitute$\left[\begin{array}{c}0\\ 0.6667\\ 0.3333\end{array}\right]\text{\hspace{0.33em}pu}$for$\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]$ into equation (3).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ 0.6667\\ 0.3333\end{array}\right]\\ \left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{c}1\\ 0.5774\angle 210°\\ 0.5774\angle 107°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault current.

Substitute$0\text{\hspace{0.33em}pu}$for$I_{1-0}$,$-j4.167\text{\hspace{0.33em}pu}$for $I_{1-1}$,$j4.167\text{\hspace{0.33em}pu}$for$I_{1-2}$ into equation (4).

$\begin{array}{l}\left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ -j4.167\text{\hspace{0.33em}pu}\\ j4.167\text{\hspace{0.33em}pu}\end{array}\right]\\ \left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{c}0\\ 7.217\angle 180°\\ 7.217\angle 0°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Therefore, the fault current and voltages are$\left[\begin{array}{c}0\\ 7.217\angle 180°\\ 7.217\angle 0°\end{array}\right]\text{\hspace{0.33em}pu}$and$\left[\begin{array}{c}1\\ 0.5774\angle 210°\\ 0.5774\angle 107°\end{array}\right]\text{\hspace{0.33em}pu}$respectively.