Question

Using the bus impedance matrices determined in Problem 9.47, verify the fault currents for the faults given in Problems 9.4(b), 9.4(c), and 9.19 (a through d).

Short answer

(a)The value of fault current in the phase sequence component$I_0$,data-custom-editor="chemistry" $I_1$and$I_2$is$0$,$-j7.89\text{\hspace{0.33em}p.u}$and $0$.

(b) The value of subtransient fault current of single-line-to ground fault is$-j9.146\text{\hspace{0.33em}p.u}$ .

(c) The value of subtransient fault current of single-line-to ground fault is$-j9.146\text{\hspace{0.33em}p.u}$ .

(d) The value of subtransient fault current of a single line-to-ground fault through fault impedance,$-j5.87\text{\hspace{0.33em}p.u}$.

(e) The value of subtransient fault current of a bolted line-to-line fault${I^{\text{'}\text{'}}}_{\text{a}}$,${I^{\text{'}\text{'}}}_{\text{b}}$and${I^{\text{'}\text{'}}}_{\text{c}}$is$0$, $6.84\angle 180°\text{\hspace{0.33em}p.u}$,$6.84\angle 0°\text{\hspace{0.33em}p.u}$.

(f) The value of subtransient fault current of bolted double line-to-ground fault$-j10.86\text{\hspace{0.33em}p.u}$,.

Step-by-step solution

Write the given data from the question.

As reference of problem 9.47.

Refer to the Figure 9.17 from the textbook.

Assume the following data for a single-line diagram of a three-phase power system.

The value of base MVA is$\text{1000\hspace{0.33em}MVA}$.

The value of voltage base for generator side is$\text{15\hspace{0.33em}kV}$.

The neutral reactance is $\text{0.05\hspace{0.17em}per\hspace{0.33em}unit}$.

Assume that the arc impedance,$Z_{\text{F}}=15+j0\Omega$ .

The base impedance of the network, $Z_{\text{base}}=250\Omega$.

Determine the formula of sequence impedance bus matrix and subtransient fault current.

Write the formula of fault current in the positive-sequence component.

${\mathbf{I}}_1=\frac{{\mathbf{E}}_g}{{\mathbf{Z}}_{11-1}}$ …… (1)

Here,role="math" localid="1656483053732" ${\mathbf{E}}_g$is generated voltage and${\mathbf{Z}}_{\mathbf{11}-\mathbf{1}}$is line to line impedance.

Write the formula of subtransient fault current in the system.

role="math" localid="1656483175386" ${I^{\text{'}\text{'}}}_a=\mathbf{3}{\mathbf{I}}_0$ …… (2)

Here, role="math" localid="1656483229073" ${\mathbf{I}}_0$is zero sequence component.

Write the formula of subtransient fault current in the phase of the system.

${I^{\text{'}\text{'}}}_a=\mathbf{0}$ …… (3)

Here,${I^{\text{'}\text{'}}}_a$is phase of the system

Write the formula of subtransient fault current in the phase of the system.

${I^{\text{'}\text{'}}}_b=-\mathbf{j}\sqrt{3}{\mathbf{I}}_1$ …… (4)

Here${\mathbf{I}}_1$, is positive sequence component.

Write the formula of subtransient fault current in the phase of the system.

${I^{\text{'}\text{'}}}_c={I^{\text{'}\text{'}}}_b$ …… (5)

Here,${I^{\text{'}\text{'}}}_b$is subtransient fault current in the phase of the system

(a)Determine thefault current in the phase sequence component.

As reference of problem 9.4 (b).

Determine the base impedance of transmission line.

$\begin{array}{c}{Z}_{\text{base}}=\frac{{\left(500\times {10}^3\right)}^2}{1000\times {10}^6}\\ =250\Omega \end{array}$

Determine the per unit quantities for lines.

Determine the per unit zero quantity for line 1-2.

$\begin{array}{c}{X}_{12}=\frac{150\Omega }{250\Omega }\\ =0.6\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per unit zero quantities for line 1-3 and 2-3.

$\begin{array}{c}{X}_{13}=X_{23}\\ =\frac{100\Omega }{250\Omega }\\ =0.4\text{\hspace{0.33em}p.u}\end{array}$

Determine the per unit positive and negative quantities for line 1-2.

$\begin{array}{c}{X}_{12}=\frac{50\Omega }{250\Omega }\\ =0.2\text{\hspace{0.33em}p.u.}\end{array}$

Determine the per unit positive and negative quantities for line 1-3 and 2-3.

$\begin{array}{c}{X}_{13}=X_{23}\\ =\frac{40\Omega }{250\Omega }\\ =0.16\text{\hspace{0.33em}p.u.}\end{array}$

The circuit's zero-sequence bus admittance matrix is shown below.

$Y_{\text{bus0}}=j\left[\begin{array}{ccc}14.167& -1.67& -2.5\\ -1.67& 14.167& -2.5\\ -2.5& -2.5& 13.34\end{array}\right]$

Determine the bus impedance, $Z_{\text{bus0}}$using the following expression.

$Z_{\text{bus0}}=\frac{1}{Y_{\text{bus0}}}$

Here,$Y_{\text{bus0}}$is zero-sequence bus admittance matrix.

Substitute$\left[\begin{array}{ccc}14.167& -1.67& -2.5\\ -1.67& 14.167& -2.5\\ -2.5& -2.5& 13.34\end{array}\right]$for$Y_{\text{bus0}}$into above equation.

$\begin{array}{c}{Z}_{\text{bus0}}=\frac{1}{\left[\begin{array}{ccc}14.167& -1.67& -2.5\\ -1.67& 14.167& -2.5\\ -2.5& -2.5& 13.34\end{array}\right]}\\ =j\left[\begin{array}{ccc}0.0748& 0.0117& 0.0162\\ 0.0117& 0.0748& 0.0162\\ 0.0162& 0.0162& 0.081\end{array}\right]\end{array}$

Therefore, the value of zero-sequence bus impedance matrix is,

$Z_{\text{bus0}}=j\left[\begin{array}{ccc}0.0748& 0.0117& 0.0162\\ 0.0117& 0.0748& 0.0162\\ 0.0162& 0.0162& 0.081\end{array}\right]$

Draw the circuit diagram of per unit positive-sequence single line diagram.

The circuit's positive-sequence bus admittance matrix is shown below.

$Y_{\text{bus1}}=j\left[\begin{array}{ccc}14.8214& -5& -6.25\\ -5& 14.5834& -6.25\\ -6.25& -6.25& 16.2037\end{array}\right]$

Determine the bus impedance,$Z_{\text{bus1}}$using the following expression.

$Z_{\text{bus1}}=\frac{1}{Y_{\text{bus1}}}$

Here, $Y_{\text{bus1}}$is positive-sequence admittance matrix.

Substitute$\left[\begin{array}{ccc}14.8214& -5& -6.25\\ -5& 14.5834& -6.25\\ -6.25& -6.25& 16.2037\end{array}\right]$for$Y_{\text{bus1}}$into above equation.

$\begin{array}{c}{Z}_{\text{bus1}}=\frac{1}{\left[\begin{array}{ccc}14.8214& -5& -6.25\\ -5& 14.5834& -6.25\\ -6.25& -6.25& 16.2037\end{array}\right]}\\ =j\left[\begin{array}{ccc}0.1266& 0.0771& 0.0786\\ 0.0771& 0.1291& 0.0795\\ 0.0786& 0.0795& 0.1227\end{array}\right]\end{array}$

Therefore, the value of positive-sequence bus impedance matrix is,

$Z_{\text{bus1}}=j\left[\begin{array}{ccc}0.1266& 0.0771& 0.0786\\ 0.0771& 0.1291& 0.0795\\ 0.0786& 0.0795& 0.1227\end{array}\right]$

Draw the circuit diagram of per unit negative-sequence single line diagram.

The circuit's negative-sequence bus admittance matrix is shown below.

$Y_{\text{bus2}}=j\left[\begin{array}{ccc}14.8214& -5& -6.25\\ -5& 14.5834& -6.25\\ -6.25& -6.25& 16.2037\end{array}\right]$

Determine the bus impedance, $Z_{\text{bus2}}$using the following expression.

$Z_{\text{bus2}}=\frac{1}{Y_{\text{bus2}}}$

Here, $Y_{\text{bus2}}$is negative-sequence bus admittance matrix.

Substitute$\left[\begin{array}{ccc}14.8214& -5& -6.25\\ -5& 14.5834& -6.25\\ -6.25& -6.25& 16.2037\end{array}\right]$for$Y_{\text{bus2}}$into above equation.

$\begin{array}{c}{Z}_{\text{bus2}}=\frac{1}{\left[\begin{array}{ccc}14.8214& -5& -6.25\\ -5& 14.5834& -6.25\\ -6.25& -6.25& 16.2037\end{array}\right]}\\ =j\left[\begin{array}{ccc}0.1266& 0.0771& 0.0786\\ 0.0771& 0.1291& 0.0795\\ 0.0786& 0.0795& 0.1227\end{array}\right]\end{array}$

Therefore, the value of negative-sequence bus impedance matrix is,

$Z_{\text{bus2}}=j\left[\begin{array}{ccc}0.1266& 0.0771& 0.0786\\ 0.0771& 0.1291& 0.0795\\ 0.0786& 0.0795& 0.1227\end{array}\right]$

Assume that a $3$-phase fault occurs in the system with a pre-fault voltage of.$\text{1.0perunit}$

Determine the fault current in the positive-sequence component.

Substitute$1$for${\text{E}}_{\text{g}}$and$j0.1266$for$Z_{11-1}$into equation (1).

$\begin{array}{c}{I}_1=\frac{1\text{\hspace{0.33em}V}}{j0.1266}\\ =-j7.89\text{\hspace{0.33em}p.u}\end{array}$

Determine the fault currents in the negative-and zero-sequence component are zero.

$\begin{array}{c}{I}_0=I_2\\ =0\end{array}$

(b) Determine the subtransient fault current of single-line-to ground fault.

As reference of problem 9.4 (c).

Assume that a single line-to ground fault occurs in the system with a pre-fault voltage of$\text{1.0perunit}$

Determine the fault current in the sequence components.

$\begin{array}{c}{I}_0=I_1=I_2\\ =\frac{E_{\text{g}}}{Z_{11-1}+Z_{11-1}+Z_{11-2}}\\ =\frac{1}{j\left(0.0748+0.1266+0.1266\right)}\\ =-j3.0487\text{\hspace{0.33em}p.u}\end{array}$

Determine the subtransient fault current in the system.

Substitute$-j3.0487$for$I_0$into equation (2).

$\begin{array}{c}{I^{\text{'}\text{'}}}_{\text{a}}=3\left(-j3.0487\right)\\ =-j9.146\text{\hspace{0.33em}p.u}\end{array}$

(c) Determine the subtransient fault current of single-line-to ground fault.

As reference of problem 9.19 (a).

Assume that a single line-to ground fault occurs in the system with a pre-fault voltage of$\text{1.0perunit}$.

Determine the fault current in the sequence components.

$\begin{array}{c}{I}_0=I_1=I_2\\ =\frac{E_{\text{g}}}{Z_{11-1}+Z_{11-1}+Z_{11-2}}\\ =\frac{1}{j\left(0.0748+0.1266+0.1266\right)}\\ =-j3.0487\text{\hspace{0.33em}p.u}\end{array}$

Determine the subtransient fault current in the system.

Substitute$-j3.0487$for$I_0$into equation (2).

$\begin{array}{l}B=3\left(-j^3.0487\right)\\ =-j9-1460.u\end{array}$

(d) Determine the subtransient fault current of a single line-to-ground fault through fault impedance.

As reference of problem 9.19 (b).

Consider that a system with a pre fault voltage of$\text{1.0perunit}$experiences a single line-to-ground fault through fault impedance.

Determine the per unit fault impedance of the system.

$\begin{array}{c}{Z}_{\text{F}}\left(\text{pu}\right)=\frac{Z_{\text{F}}}{Z_{\text{base}}}\\ =\frac{15\Omega }{250\Omega }\\ =0.06\text{\hspace{0.33em}p.u}\end{array}$

Determine the fault current in the sequence components.

$\begin{array}{c}{I}_0=I_1=I_2\\ =\frac{E_{\text{g}}}{Z_{11-0}+Z_{11-1}+Z_{11-2}+3Z_{\text{F}}\left(\text{pu}\right)}\\ =\frac{1}{j\left[0.0748+0.1266+0.1266+3\left(0.06\right)\right]}\\ =-j1.956\text{\hspace{0.33em}p.u}\end{array}$

Determine the subtransient fault current in the system.

Substitute$-j1.956$for$I_0$into equation (3).

$\begin{array}{c}{I^{\text{'}\text{'}}}_{\text{a}}=3\left(-j1.956\right)\\ =-j5.87\text{\hspace{0.33em}p.u}\end{array}$

(e) Determine the subtransient fault current of a bolted line-to-line fault.

As reference of problem 9.19 (c).

Assume that the system experiences a bolted line-to-line fault with a pre-fault voltage of$\text{1.0perunit}$.

Determine the fault current in the positive-sequence component.

$\begin{array}{c}{I}_1=-I_2\\ =\frac{E_{\text{g}}}{j\left(Z_{11-1}+Z_{11-2}\right)}\\ =\frac{1}{j\left(0.1266+0.1266\right)}\\ =-j3.95\text{\hspace{0.33em}p.u}\end{array}$

The fault current in the zero-sequence component,$I_0=0$.

Determine the subtransient fault currents in the phases of the system.

${I^{\text{'}\text{'}}}_{\text{a}}=0$

Determine the subtransient fault currents in the phases of the system.

Substitute$-j3.95$for$I_1$into equation (4).

$\begin{array}{c}{I^{\text{'}\text{'}}}_{\text{b}}=-j\sqrt{3}\left(-j3.95\right)\\ =6.84\angle 180°\text{\hspace{0.33em}p.u}\end{array}$

Determine the subtransient fault currents in the phases of the system.

${I^{\text{'}\text{'}}}_{\text{c}}=6.84\angle 0°\text{\hspace{0.33em}p.u}$

(f) Determine the subtransient fault current of bolted double line-to-ground fault.

As reference of problem 9.19 (d).

Determine the bolted double line-to-ground fault occurs in the system.

Determine the fault current in the positive-sequence component

$\begin{array}{c}{I}_1=\frac{E_{\text{g}}}{j\left[X_1+X_2{\text{\hspace{0.17em}parallel\hspace{0.33em}with\hspace{0.33em}X}}_{\text{0}}\right]}\\ =\frac{1\text{\hspace{0.33em}V}}{j\left[0.1266+\frac{\left(0.1266\right)\left(0.0748\right)}{\left(0.1266\right)+\left(0.0748\right)}\right]}\\ =\frac{1\text{\hspace{0.33em}V}}{j0.1736}\\ =-j5.76\text{\hspace{0.33em}p.u}\end{array}$

Determine the fault current in the zero-sequence component.

$\begin{array}{c}{I}_0=I_1\left(\frac{Z_{11-2}}{Z_{11-0}+Z_{11-1}}\right)\\ =\left(-j5.76\right)\left(\frac{j0.1266}{j0.1266+j0.0748}\right)\\ =\left(-j5.76\right)\left(0.6285\right)\\ =-j3.62\text{\hspace{0.17em}p.u}\end{array}$

Determine the subtransient fault current.

Substitute $-j3.62$for $I_0$into equation (2).

$\begin{array}{c}{I^{\text{'}\text{'}}}_{\text{a}}=3\left(-j3.62\right)\\ =-j10.86\text{\hspace{0.33em}p.u}\end{array}$

Therefore, the value of subtransient fault current are computed and verify is $-j10.86\text{\hspace{0.33em}p.u}$.