Question

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

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

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 1.03 per unit.

Short answer

The fault current and voltages are$\left[\begin{array}{c}5\angle -90°\\ 5\angle 150°\\ 5\angle 30°\end{array}\right]\text{\hspace{0.33em}pu}$ and $\left[\begin{array}{c}0.5\angle 0°\\ 0.5\angle 240°\\ 0.5\angle 120°\end{array}\right]\text{\hspace{0.33em}pu}$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 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 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 three-phase fault at bus 1.

Calculate the positive sequence fault current

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

$\begin{array}{l}{I}_{1-1}=\frac{1\angle 0°}{j0.2}\\ I_{1-1}=-j5\text{\hspace{0.33em}pu}\end{array}$

Calculate the voltages at bus 2.

Substitute $0\text{\hspace{0.33em}pu}$for $I_{1-0}$, $-j5\text{\hspace{0.33em}pu}$for $I_{1-1}$,0 pu for $I_{1-2}$, 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).

$\begin{array}{c}\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{ccc}0& 0& 0\\ 0& j0.1& 0\\ 0& 0& j0.1\end{array}\right]\left[\begin{array}{c}0\\ -j5\\ 0\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 0.5\\ 0\end{array}\right]\\ \left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]=\left[\begin{array}{c}0\\ 0.5\\ 0\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault voltage.

Substitute $\left[\begin{array}{c}0\\ 0.5\\ 0\end{array}\right]\text{\hspace{0.33em}pu}$for $\left[\begin{array}{c}{V}_{2-0}\\ V_{2-1}\\ V_{2-2}\end{array}\right]$into equation (3).

$\begin{array}{l}\left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ 0.5\\ 0\end{array}\right]\\ \left[\begin{array}{c}{V}_{2ag}\\ V_{2bg}\\ V_{2cg}\end{array}\right]=\left[\begin{array}{c}0.5\angle 0°\\ 0.5\angle 240°\\ 0.5\angle 120°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the fault current.

Substitute 0 pu for $I_{1-0}$,$-j5\text{\hspace{0.33em}pu}$for$I_{1-1}$and 0 pu for $I_{1-2}$into equation (4).

$\begin{array}{l}\left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ -j5\\ 0\end{array}\right]\\ \left[\begin{array}{c}{I}_{1a}\\ I_{1b}\\ I_{1c}\end{array}\right]=\left[\begin{array}{c}5\angle -90°\\ 5\angle 150°\\ 5\angle 30°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Therefore, the fault current and voltages are $\left[\begin{array}{c}5\angle -90°\\ 5\angle 150°\\ 5\angle 30°\end{array}\right]\text{\hspace{0.33em}pu}$and $\left[\begin{array}{c}0.5\angle 0°\\ 0.5\angle 240°\\ 0.5\angle 120°\end{array}\right]\text{\hspace{0.33em}pu}$respectively.