Question

The zero-, positive-, and negative-sequence bus impedance matrices for a three-bus three-phase power system are

$\begin{array}{l}{\mathbf{Z}}_{\mathbf{b}\mathbf{u}\mathbf{s}\mathbf{0}}\mathbf{=}\mathbf{j}\left[\begin{array}{ccc}0.10& 0& 0\\ 0& 0.20& 0\\ 0& 0& 0.10\end{array}\right]\mathbf{p}\mathbf{e}\mathbf{r}\mathbf{u}\mathbf{n}\mathbf{i}\mathbf{t}\\ {\mathbf{Z}}_{\mathbf{b}\mathbf{u}\mathbf{s}\mathbf{1}}\mathbf{=}{\mathbf{Z}}_{\mathbf{b}\mathbf{u}\mathbf{s}\mathbf{2}}\mathbf{=}\mathbf{j}\left[\begin{array}{ccc}0.12& 0.08& 0.04\\ 0.08& 0.12& 0.06\\ 0.04& 0.06& 0.08\end{array}\right]\mathbf{p}\mathbf{e}\mathbf{r}\mathbf{u}\mathbf{n}\mathbf{i}\mathbf{t}\end{array}$

Determine the per-unit fault current and per-unit voltage at bus 2 for a bolted three-phase fault at bus 1. The pre fault voltage is$1.0$ per unit.

Short answer

The fault current and voltages are $\left[\begin{array}{c}8.33\angle -90°\\ 8.33\angle 150°\\ 8.33\angle 30°\end{array}\right]\text{\hspace{0.33em}pu}$and $\left[\begin{array}{c}0.333\angle 0°\\ 0.333\angle 240°\\ 0.333\angle 120°\end{array}\right]\text{\hspace{0.33em}pu}$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 three-phase fault at bus 1.

For three phase faults, the zero and negative sequence currents are zero.

$\begin{array}{l}{\mathbf{I}}_{\mathbf{1}\mathbf{-}\mathbf{0}}\mathbf{=}{\mathbf{I}}_{\mathbf{1}\mathbf{-}\mathbf{2}}\\ {\mathbf{I}}_{\mathbf{1}\mathbf{-}\mathbf{0}}\mathbf{=}\mathbf{0}\end{array}$

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

${\mathit{I}}_{\mathbf{1}\mathbf{-}\mathbf{1}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{F}}}{{\mathbf{Z}}_{\mathbf{11}\mathbf{-}\mathbf{1}}}$ …… (1)

Here${\mathit{Z}}_{\mathbf{11}\mathbf{-}\mathbf{1}}$is the positive sequence impedance.

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

localid="1656754573551" $\left[\begin{array}{c}{\mathrm{V}}_{2-0}\\ {\mathrm{V}}_{2-1}\\ {\mathrm{V}}_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ {\mathrm{V}}_{\mathrm{F}}\\ 0\end{array}\right]-\left[\begin{array}{ccc}{\mathrm{Z}}_{2-0}& 0& 0\\ 0& {\mathrm{Z}}_{2-1}& 0\\ 0& 0& {\mathrm{Z}}_{2-2}\end{array}\right]\left[\begin{array}{c}{\mathrm{I}}_{2-0}\\ {\mathrm{I}}_{2-1}\\ {\mathrm{I}}_{2-2}\end{array}\right]$ …… (2)

The equation to calculate the fault voltage is given as,

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

The equation to calculate the fault current is given as,

localid="1656754588554" $\left[\begin{array}{c}{\mathrm{I}}_{1\mathrm{a}}\\ {\mathrm{I}}_{1\mathrm{b}}\\ {\mathrm{I}}_{1\mathrm{c}}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathrm{a}}^2& \mathrm{a}\\ 1& \mathrm{a}& {\mathrm{a}}^2\end{array}\right]\left[\begin{array}{c}{\mathrm{I}}_{1-0}\\ {\mathrm{I}}_{1-1}\\ {\mathrm{I}}_{1-2}\end{array}\right]$ …… (4)

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

Calculate the positive sequence fault current

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

$\begin{array}{l}{I}_{1-1}=\frac{1\angle 0°}{j0.12}\\ I_{1-1}=-j8.33\text{\hspace{0.33em}pu}\end{array}$

Calculate the voltages at bus 2.

Substitute$0\text{\hspace{0.33em}pu}$for $I_{1-0}$, $-j8.33\text{\hspace{0.33em}pu}$for$I_{1-1}$, $0\text{\hspace{0.33em}pu}$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).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{ccc}0& 0& 0\\ 0& j0.08& 0\\ 0& 0& j0.08\end{array}\right]\left[\begin{array}{c}0\\ -j8.33\\ 0\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 0.666\\ 0\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 0.333\\ 0\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault voltage.

Substitute$\left[\begin{array}{c}0\\ 0.333\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$ for $\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]$into equation (3).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ 0.333\\ 0\end{array}\right]\\ \left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{c}0.333\angle 0°\\ 0.333\angle 240°\\ 0.333\angle 120°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault current.

Substitute$0\text{\hspace{0.33em}pu}$ for$I_{1-0}$ ,$-j8.33\text{\hspace{0.33em}pu}$ for$I_{1-1}$ and$0\text{\hspace{0.33em}pu}$ for $I_{1-2}$into equation (4).

$\begin{array}{l}\left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ -j8.33\\ 0\end{array}\right]\\ \left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{c}8.33\angle -90°\\ 8.33\angle 150°\\ 8.33\angle 30°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Hence the fault current and voltages are $\left[\begin{array}{c}8.33\angle -90°\\ 8.33\angle 150°\\ 8.33\angle 30°\end{array}\right]\text{\hspace{0.33em}pu}$and $\left[\begin{array}{c}0.333\angle 0°\\ 0.333\angle 240°\\ 0.333\angle 120°\end{array}\right]\text{\hspace{0.33em}pu}$respectively.