Question

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

$\begin{array}{c}{\mathbf{Z}}_{\mathbf{bus}\mathbf{0}}\mathbf{=}\mathbf{j}\left[\begin{array}{ccc}0.10& 0& 0\\ 0& 0.20& 0\\ 0& 0& 0.10\end{array}\right]\mathbf{per}\mathbf{unit}\\ {\mathbf{Z}}_{\mathbf{bus}\mathbf{1}}\mathbf{=}{\mathbf{Z}}_{\mathbf{bus}\mathbf{2}}\mathbf{=}\mathbf{j}\left[\begin{array}{ccc}0.12& 0.08& 0.04\\ 0.08& 0.12& 0.06\\ 0.04& 0.06& 0.08\end{array}\right]\mathbf{per}\mathbf{unit}\end{array}$

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$\mathbf{1}\mathbf{.}\mathbf{0}$ per unit.

Short answer

The fault current and voltages are$\left[\begin{array}{c}8.33\angle -90°\\ 8.33\angle 150°\\ 8.33\angle 30°\end{array}\right]\text{\hspace{0.17em}pu}$ and$\left[\begin{array}{c}0.333\angle 0°\\ 0.333\angle 240°\\ 0.333\angle 120°\end{array}\right]\text{\hspace{0.17em}pu}$ respectively.

Step-by-step solution

Write the given data from the question.

The pre fault voltage for each bus in the positive sequence network, ${\mathrm{V}}_{\mathrm{F}}=1\angle 0°\text{\hspace{0.17em}pu}$

Write the zero-sequence impedance matrix,

${\mathrm{Z}}_{\mathrm{bus}0}=\mathrm{j}\left[\begin{array}{ccc}0.10& 0& 0\\ 0& 0.20& 0\\ 0& 0& 0.10\end{array}\right]\mathrm{per}\mathrm{unit}$

Write the positive and negative-sequence impedance matrix,

$\begin{array}{l}{\mathrm{Z}}_{\mathrm{bus}1}={\mathrm{Z}}_{\mathrm{bus}2}\\ {\mathrm{Z}}_{\mathrm{bus}1}=\mathrm{j}\left[\begin{array}{ccc}0.12& 0.08& 0.04\\ 0.08& 0.12& 0.06\\ 0.04& 0.06& 0.08\end{array}\right]\mathrm{per}\mathrm{unit}\end{array}$

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.

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

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

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

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

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

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

Calculate the positive sequence fault current

Substitute $1\angle 0°\text{\hspace{0.17em}pu}$for $V_f$ and $j0.12\text{\hspace{0.17em}pu}$for $Z_{11}$into the equation (1).

$\begin{array}{l}{I}_{1-1}=\frac{1\angle 0°}{j0.12}\\ I_{1-1}=-j8.33\text{\hspace{0.17em}pu}\end{array}$

Calculate the voltages at bus 2.

Substitute $0\text{\hspace{0.17em}pu}$for $I_{1-0}$, $-j8.33\text{\hspace{0.17em}pu}$for $I_{1-1}$, $0\text{\hspace{0.17em}pu}$ for $I_{1-2}$, $0\text{\hspace{0.17em}pu}$for $Z_{2-0}$, $j0.08\text{\hspace{0.17em}pu}$for $Z_{2-1}$and $j0.08\text{\hspace{0.17em}pu}$for $Z_{2-2}$into equation (2).

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

Calculate the fault voltage.

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

$\begin{array}{l}\left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ 0.333\\ 0\end{array}\right]\\ \left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{c}0.333\angle 0°\\ 0.333\angle 240°\\ 0.333\angle 120°\end{array}\right]\text{\hspace{0.17em}pu}\end{array}$

Calculate the fault current.

Substitute $0\text{\hspace{0.17em}pu}$ for $I_{1-0}$, $-j8.33\text{\hspace{0.17em}pu}$for $I_{1-1}$and $0\text{\hspace{0.17em}pu}$ for $I_{1-2}$into equation (4).

$\begin{array}{l}\left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ -j8.33\\ 0\end{array}\right]\\ \left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{c}8.33\angle -90°\\ 8.33\angle 150°\\ 8.33\angle 30°\end{array}\right]\text{\hspace{0.17em}pu}\end{array}$

Hence the fault current and voltages are$\left[\begin{array}{c}8.33\angle -90°\\ 8.33\angle 150°\\ 8.33\angle 30°\end{array}\right]\text{\hspace{0.17em}pu}$ and $\left[\begin{array}{c}0.333\angle 0°\\ 0.333\angle 240°\\ 0.333\angle 120°\end{array}\right]\text{\hspace{0.17em}pu}$respectively.