Question

Using the bus impedance matrices determined in Problem 9.51, verify the fault currents for the faults given in Problems 9.10, 9.24, 9.25, 9.26, and 9.27.

Short answer

(a)

The value of sub-transient fault current in the system${I^{\text{'}}}_a$ is $-j0.609\text{\hspace{0.33em}pu}$.

(b)

The value of sub-transient fault current in the system${I^{\text{'}}}_a$ is $-j0.483\text{\hspace{0.33em}pu}$.

(c)

The value of sub-transient fault current in the system${I^{\text{'}}}_a$ is $-j0.483\text{\hspace{0.33em}pu}$.

(d)

The value of sub-transient fault current in phases of a system$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]$ is $\left[\begin{array}{c}0\\ -2.113\angle 180°\\ -2.113\angle 0°\end{array}\right]\text{\hspace{0.33em}p.u.}$

(e)

The value of sub-transient fault current in phases of a system$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]$ is$\left[\begin{array}{c}0.125\angle -90\\ 0.642\angle 157.5°\\ 0.642\angle 22.48°\end{array}\right]\text{\hspace{0.33em}p.u.}$

Step-by-step solution

Write the given data from the question.

As reference Problem 9.13

The fault impedance in per unit is ${\text{Z}}_{\text{F}}=\text{j0.1\hspace{0.33em}perunit}$.

Thevalue of base MVA is 100 MVA.

The value of voltage in the zone of generator G2 is 15 kV.

Determine the formula of line-to-line fault and line-to-ground fault.

Write the formula of sub-transient fault current in the system.

${I^{\text{'}\text{'}}}_a={\mathbf{I}}_1$ …… (1)

Here,${\mathbf{I}}_1$ are phase sequence currents.

Write the formula of sub-transient fault current in the system.

${I^{\text{'}\text{'}}}_a=\mathbf{3}{\mathbf{{\rm I}}}_0$ …… (2)

Here,${\mathbf{{\rm I}}}_0$ are phase sequence currents.

Write the formula of sub-transient fault current in the system.

${I^{\text{'}\text{'}}}_a=\mathbf{3}{\mathbf{{\rm I}}}_0$ …… (2)

Here,${\mathbf{{\rm I}}}_0$ are phase sequence currents.

Write the formula of sub-transient fault current in the system.

$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathbf{a}}^2& a\\ 1& a& {\mathbf{a}}^2\end{array}\right]\left[\begin{array}{c}{\mathbf{I}}_0\\ {\mathbf{I}}_1\\ {\mathbf{I}}_2\end{array}\right]$ …… (3)

Here $\begin{array}{c}{\mathbf{I}}_0\end{array}$,${\mathbf{I}}_1$ androle="math" localid="1656425516200" $\mathbf{I}{\mathbf{\text{'}}}_2$ are phase sequence currents.

(a) Determine the sub-transient fault current in the system.

Determine the per-unit positive sequence reactance for generator 1.

$\begin{array}{c}{X}_1={X^{\text{'}\text{'}}}_d\\ =0.2{\left(\frac{12}{10}\right)}^2\left(\frac{100}{50}\right)\\ =0.576\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per-unit negative sequence for generator 1.

$X_2=0.576\text{\hspace{0.17em}p.u.}$

Determine the per-unit zero sequence reactance for generator 1.

$\begin{array}{c}{X}_0=0.1{\left(\frac{12}{10}\right)}^2\left(\frac{100}{50}\right)\\ =0.288\text{\hspace{0.17em}p.u.}\end{array}$

Determine the per-unit positive sequence reactance for generator 2.

$\begin{array}{c}{X}_1={X^{\text{'}\text{'}}}_d\\ =0.2\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per-unit negative sequence for generator 2.

$X_2=0.23\text{\hspace{0.17em}p.u.}$

T2 Determine the per-unit zero sequence reactance for generator 2.

$X_0=0.1\text{\hspace{0.33em}p.u.}$

Determine the per unit reactance for transformer T1.

$\begin{array}{c}{X}_{\text{T1}}=0.10\left(\frac{100}{50}\right)\\ =0.2\text{\hspace{0.17em}p.u.}\end{array}$

Determine the per unit reactance for transformer T2.

$X_{\text{T2}}=0.10\text{\hspace{0.33em}p.u.}$

Determine the base impedance value.

$\begin{array}{c}{Z}_{\text{base}}=\frac{{\left(138\times {10}^3\right)}^2}{1000\times {10}^6}\\ =190.44\mathrm{\Omega }\end{array}$

Determine the per unit positive and negative quantities for lines.

$\begin{array}{c}{X}_1=X_2\\ =\left(\frac{40}{190.44}\right)\\ =0.210\text{\hspace{0.17em}p.u.}\end{array}$

Determine the per unit zero sequence reactance for line.

$\begin{array}{c}{X}_0=\left(\frac{100}{190.44}\right)\\ =0.5251\text{\hspace{0.33em}p.u.}\end{array}$

Draw the positive-sequence network of the system.

For a system with positive-sequence impedance, write the expression for the bus impedance matrix.

$\begin{array}{c}{Z}_{bus}=\frac{1}{{\text{Y}}_{\text{bus}}}\\ ={{\text{Y}}_{\text{bus}}}^{-1}\end{array}$

Determine the${\text{Y}}_{\text{bus}}$matrix.

$Y_{\text{bus}}=j\left[\begin{array}{ccccc}0.776& -0.2& 0& 0& 0\\ -0.2& 0.41& -0.21& 0& 0\\ 0& -0.21& 0.42& -0.21& 0\\ 0& 0& -0.21& 0.31& -0.1\\ 0& 0& 0& -0.1& 0.3\end{array}\right]\text{\hspace{0.33em}p.u.}$

Determine the bus impedance matrix for positive-sequence impedance.

$\begin{array}{c}{Z}_{\text{bus}}=-j{\left[\begin{array}{ccccc}0.776& -0.2& 0& 0& 0\\ -0.2& 0.41& -0.21& 0& 0\\ 0& -0.21& 0.42& -0.21& 0\\ 0& 0& -0.21& 0.31& -0.1\\ 0& 0& 0& -0.1& 0.3\end{array}\right]}^{-1}\\ =-j\left[\begin{array}{ccccc}1.6397& 1.362& 1.0975& 0.8331& 0.2777\\ 1.362& 5.2846& 4.2584& 3.2323& 1.0774\\ 1.0975& 4.2584& 7.2688& 5.5173& 1.8391\\ 0.8331& 3.2323& 5.5173& 7.8023& 2.6008\\ 0.2777& 1.0774& 1.8391& 2.6008& 4.2003\end{array}\right]\text{\hspace{0.33em}p.u.}\end{array}$

Draw a circuit diagram of per-unit negative sequence network.

Determine the $Y_{\text{bus}}$.

$Y_{\text{bus}}=j\left[\begin{array}{ccccc}0.776& -0.2& 0& 0& 0\\ -0.2& 0.41& -0.21& 0& 0\\ 0& -0.21& 0.42& -0.21& 0\\ 0& 0& -0.21& 0.31& -0.1\\ 0& 0& 0& -0.1& 0.33\end{array}\right]\text{\hspace{0.33em}p.u.}$

Determine the bus impedance matrix for positive-sequence impedance.

$\begin{array}{c}{Z}_{\text{bus}}=-j{\left[\begin{array}{ccccc}0.776& -0.2& 0& 0& 0\\ -0.2& 0.41& -0.21& 0& 0\\ 0& -0.21& 0.42& -0.21& 0\\ 0& 0& -0.21& 0.31& -0.1\\ 0& 0& 0& -0.1& 0.33\end{array}\right]}^{-1}\\ =-j\left[\begin{array}{ccccc}1.6376& 1.354& 1.0839& 0.8138& 0.2466\\ 1.354& 5.2536& 4.2056& 3.1576& 0.9569\\ 1.0839& 4.2056& 7.1787& 5.3899& 1.6333\\ 0.8138& 3.1576& 5.3899& 7.6221& 2.3097\\ 0.2466& 0.9569& 1.6333& 2.3097& 3.7302\end{array}\right]\text{\hspace{0.33em}p.u.}\end{array}$

Determine the diagram for zero-sequence impedance.

Determine the$Y_{\text{bus}}$matrix.

$Y_{\text{bus}}=j\left[\begin{array}{ccccc}0.488& -0.2& 0& 0& 0\\ -0.2& 0.7251& -0.5251& 0& 0\\ 0& -0.5251& 1.0502& 0.6261& 0\\ 0& 0& -0.5251& 0.6261& 0\\ 0& 0& 0& 0& 0.1\end{array}\right]\text{\hspace{0.33em}p.u.}$

Determine the bus impedance matrix for positive-sequence impedance.

$\begin{array}{c}{Z}_{\text{bus}}=-j\left[\begin{array}{ccccc}0.488& -0.2& 0& 0& 0\\ -0.2& 0.7251& -0.5251& 0& 0\\ 0& -0.5251& 1.0502& -0.5251& 0\\ 0& 0& -0.5251& 0.6261& 0\\ 0& 0& 0& 0& 0.1\end{array}\right]\\ =-j\left[\begin{array}{ccccc}2.9287& 2.146& 1.8479& 1.5498& 0\\ 2.146& 5.2363& 4.509& 3.7816& 0\\ 1.8479& 4.509& 5.5225& 4.6316& 0\\ 1.5498& 3.7816& 4.6316& 5.4817& 0\\ 0& 0& 0& 0& 10\end{array}\right]\text{\hspace{0.33em}p.u.}\end{array}$

Determine the positive-sequence impedance of $Y_{\text{bus}}$.

$\begin{array}{c}{Y}_{\text{bus}}=Z_{11}\\ =1.6397\text{\hspace{0.33em}p.u.}\end{array}$

Determine the negative-sequence impedance of $Y_{\text{bus}}$.

$\begin{array}{c}{Y}_{\text{bus}}=Z_{11}\\ =1.6376\end{array}$

Determine the zero-sequence impedance of $Y_{\text{bus}}$.

$\begin{array}{c}{Y}_{\text{bus}}=Z_{11}\\ =2.9287\end{array}$

When a 3-phase fault occurs in a system:

Determine the fault current in a positive-sequence component.

$\begin{array}{c}{I}_1=\frac{E_{\text{g}}}{Z_1}\\ =\frac{1}{j1.6397}\\ =-j0.609\text{\hspace{0.33em}pu}\end{array}$

Since the fault is symmetrical, the zero and negative sequence currents are zero.

$I_0=I_2=0$

Determine the sub-transient fault current in the system.

Substitute$-j0.609\text{\hspace{0.33em}pu}$for$I_1$into equation (1).

$\begin{array}{c}{I^{\text{'}\text{'}}}_a=I_1\\ =-j0.609\text{\hspace{0.17em}pu}\end{array}$

Therefore, the value of sub-transient fault current in the system${I^{\text{'}}}_a$is $-j0.609\text{\hspace{0.33em}pu}$.

(b) Determine the sub-transient fault current in the system.

When a single line-to-ground fault occurs in a system:

Positive, negative, and zero sequence fault currents are all the same for a single line to ground fault.

Determine the fault current in the sequence components.

$\begin{array}{c}{I}_0=\frac{E_g}{\left(Z_0+Z_1+Z_2\right)}\\ =\frac{1}{j\left(2.9287+1.6397+1.6376\right)}\\ =\frac{1}{j6.206}\\ =-j0.161\text{\hspace{0.33em}pu}\end{array}$

Determine the sub-transient fault current in the system.

Substitute$-j0.161$ for$I_0$ into equation (2).

$\begin{array}{c}{I^{\text{'}\text{'}}}_a=3I_0\\ =3\left(-j0.161\right)\\ =-j0.483\text{\hspace{0.33em}pu}\end{array}$

(c) Determine the sub-transient fault current in the system.

When a single line-to-ground fault through fault impedance occurs in a system:

Positive, negative, and zero sequence fault currents are all the same for a single line to ground fault.

Determine the fault current in the sequence components.

$\begin{array}{c}{I}_0=\frac{E_g}{\left(Z_0+Z_1+Z_2+3Z_{\text{F}}\right)}\\ =\frac{1}{j\left(2.9287+1.6397+1.6376+3\left(0.05\right)\right)}\\ =\frac{1}{j6.356}\\ =-j0.157\text{\hspace{0.33em}pu}\end{array}$

Determine the sub-transient fault current in the system.

Substitute$-j0.157$ for$I_0$ into equation (2).

$\begin{array}{c}{I^{\text{'}\text{'}}}_a=3I_0\\ =3\left(-j0.157\right)\\ =-j0.472\text{\hspace{0.33em}pu}\end{array}$

(d) Determine the sub-transient fault current in the system.

When a line-to-line fault occurs in a system.

For line-line line fault.

$\begin{array}{l}{I}_1=-I_2\\ I_0=0\end{array}$

Determine the fault current in a positive-sequence component:

$\begin{array}{c}{I}_1=\frac{E_{\text{g}}}{j\left(Z_1+Z_2\right)}\\ =\frac{1}{j\left(1.6397+1.6376\right)}\\ =\frac{1}{j3.2773}\\ =-j0.305\text{\hspace{0.33em}pu}\end{array}$

Determine the fault current in a positive-sequence component.

$\begin{array}{c}{I}_2=-I_1\\ =j0.305\text{\hspace{0.17em}p.u.}\end{array}$

Determine the fault current in a zero-sequence component.

$I_0=0$

Determine the sub-transient fault current in phases of a system.

Substitute 0 for $I_0$,$-j0.305\text{\hspace{0.17em}pu}$ for $I_1$and$j0.305\text{\hspace{0.33em}pu}$ for$I_2$ into equation (3).

$\begin{array}{c}\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ -j0.305\text{\hspace{0.33em}pu}\\ j0.305\text{\hspace{0.33em}pu}\end{array}\right]\\ =\left[\begin{array}{c}0\\ -2.113\angle 180°\\ -2.113\angle 0°\end{array}\right]\text{\hspace{0.33em}p.u.}\end{array}$

Therefore, the value of sub-transient fault current$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]$ in phases of a system is $\left[\begin{array}{c}0\\ -2.113\angle 180°\\ -2.113\angle 0°\end{array}\right]\text{\hspace{0.33em}p.u.}$

(e) Determine the sub-transient fault current in phases of a system.

When bolted double line-to-ground fault occurs in a system:

Determine the fault current in a positive-sequence component.

$\begin{array}{c}{I}_1=\frac{E_{\text{g}}}{j\left(X_1+\left(X_2\parallel \left(X_0+3X_0\right)\right)\right)}\\ =\frac{1}{j\left(1.6397+\left(\frac{\left(1.6376\right)\left(0.73\right)}{1.6376+0.73}\right)\right)}\\ =\frac{1}{j2.144}\\ =-j0.466\text{\hspace{0.17em}pu}\end{array}$

Determine the fault current in a negative-sequence component.

$\begin{array}{c}{I}_2=\frac{Z_0}{Z_0+Z_1+3Z_0}\left(-I_1\right)\\ =\left(\frac{2.9287}{2.9287+1.6397+1.6376}\right)\left(j0.466\right)\\ =\left(0.4719\right)\left(j0.466\right)\\ =j0.219\text{\hspace{0.33em}pu}\end{array}$

Determine the fault current in a sequence component.

$\begin{array}{c}{I}_0=\frac{Z_2}{Z_0+Z_1+3Z_2}\left(-I_1\right)\\ =\left(\frac{1.6376}{2.9287+1.6397+1.6376}\right)\left(j0.466\right)\\ =\left(0.2638\right)\left(j0.466\right)\\ =j0.122\text{\hspace{0.33em}pu}\end{array}$

Determine the sub –transient fault currents in phases of a system.

$\begin{array}{c}\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}j0.122\\ -j0.466\\ j0.219\end{array}\right]\\ =\left[\begin{array}{c}j0.122-j0.466+j0.219\\ j0.122-j0.466a^2+j0.219a\\ j0.122-j0.466a+j0.219a^2\end{array}\right]\\ =\left[\begin{array}{c}0.1256\angle -90\\ 0.642\angle 157.5°\\ 0.642\angle 22.48°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Therefore, the value of sub-transient fault current$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]$ in phases of a system is $\left[\begin{array}{c}0.125\angle -90\\ 0.642\angle 157.5°\\ 0.642\angle 22.48°\end{array}\right]\text{\hspace{0.33em}p.u.}$