Question

(a) Compute the 3 x 3 per-unit zero-, positive-, and negative-sequence bus impedance matrices for the power system given in Problem 9.1. Use a base of 1000 MVA and 765 kV in the zone of line 1-2.

(b) Using the bus impedance matrices determined in Problem 9.42, verify the fault currents for the faults given in Problems 9.3, 9.14, 9.15, 9.16, and 9.17.

Short answer

(a)

The value of 3 x 3 per unit zero-sequence bus impedance matrix, $Z_{\text{bus0}}$for the system is $j\left[\begin{array}{ccc}0.0614& 0.0192& 0.0237\\ 0.0192& 0.0614& 0.0237\\ 0.0237& 0.0237& 0.0639\end{array}\right]$.

The value of 3 x 3 per unit positive-sequence bus impedance matrix,$Z_{\text{bus1}}$for the system is $j\left[\begin{array}{ccc}0.095& 0.0714& 0.0688\\ 0.0714& 0.0959& 0.0692\\ 0.0688& 0.0692& 0.0852\end{array}\right]$.

The value of 3 x 3 per unit negative-sequence bus impedance matrix,$Z_{\text{bus2}}$for the system is $j\left[\begin{array}{ccc}0.0973& 0.0738& 0.0717\\ 0.0738& 0.0983& 0.0722\\ 0.0717& 0.0722& 0.0888\end{array}\right]$.

(b)

The value of sub-transient fault current in the system $I_0$is $0$.

The value of sub-transient fault current in the system${I^{\text{'}\text{'}}}_a$is $-j11.825\text{\hspace{0.17em}p.u}$.

The value of sub-transient fault current in the system${I^{\text{'}\text{'}}}_a$is$-j7.431\text{\hspace{0.33em}p.u.}$

The value of sub-transient fault current in the phases of the system $\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}0\\ -9\\ 9\angle 0°\text{\hspace{0.33em}p.u}\end{array}\right]$.

The value of sub-transient fault current in the system$I_0$is $-j13.73\text{\hspace{0.33em}p.u}$.

Step-by-step solution

Write the given data from the question.

As reference Problem 9.1

Thevalue of base MVA is 1000 MVA.

The value of fault impedance$Z_{\text{F}}$is $30+j0\mathrm{\Omega }$.

The value of base impedance of the network$Z_{\text{base}}$is $585.225\mathrm{\Omega }$.

The value of voltage in the zone of line 1-2 is 765 kV.

Determine the formula of 3 x 3 per-unit zero-, positive-, and negative-sequence bus impedance matrices for the power system and sub-transient fault current in the system.

Write the formula of per unit-zero sequence bus impedance matrix.

${\mathbf{Z}}_{\mathrm{bus}0}=\frac{1}{{\mathbf{Y}}_{\mathrm{bus}0}}$ …… (1)

Here,${\mathbf{Y}}_{\mathrm{bus}0}$ is zero-sequence bus admittance matrix.

Write the formula of per unit-positive sequence bus impedance matrix.

${\mathbf{Z}}_{\mathrm{bus}1}=\frac{1}{{\mathbf{Y}}_{\mathrm{bus}1}}$ …… (2)

Here,${\mathbf{Y}}_{\mathrm{bus}1}$ is positive-sequence bus admittance matrix.

Write the formula of per unit-negative sequence bus impedance matrix.

${\mathbf{Z}}_{\mathrm{bus}2}=\frac{1}{{\mathbf{Y}}_{\mathrm{bus}2}}$ …… (3)

Here,${\mathbf{Y}}_{\mathrm{bus}2}$ is negative-sequence bus admittance matrix.

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

${\mathbf{I}}_0={\mathbf{I}}_2$ …… (4)

Here,${\mathbf{I}}_2$ are phase sequence currents.

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

${I^{\text{'}\text{'}}}_a=\mathbf{3}{\mathbf{{\rm I}}}_0$ …… (5)

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

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

$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathbf{a}}^2& a\\ 1& a& {\mathbf{a}}^2\end{array}\right]\left[\begin{array}{c}{\mathbf{I}}_0\\ {\mathbf{I}}_1\\ {\mathbf{I}}_2\end{array}\right]$ …… (6)

Here, ${\mathbf{I}}_0$,${\mathbf{I}}_1$ and${I^{\text{'}}}_2$ are phase sequence currents.

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

See the textbook's figure 9.17, which is a single-line diagram of a three-phase power system.

Determine the base impedance of transmission line.

$\begin{array}{c}{Z}_{\text{base}}=\frac{{\left(765\times {10}^3\right)}^2}{1000\times {10}^6}\\ =\frac{585225\times {10}^6}{1000\times {10}^6}\\ =585.225\mathrm{\Omega }\end{array}$

Determine the per unit quantities for lines.

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

$\begin{array}{c}{\mathrm{X}}_{12}=\frac{150\mathrm{\Omega }}{585.225\mathrm{\Omega }}\\ =0.256\text{\hspace{0.17em}p.u.}\end{array}$

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

$\begin{array}{c}{\mathrm{X}}_{13}={\mathrm{X}}_{23}\\ =\frac{100\mathrm{\Omega }}{585.225\mathrm{\Omega }}\\ =0.17\text{\hspace{0.33em}p.u}\end{array}$

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

$\begin{array}{c}{X}_{12}=\frac{50}{585.225\mathrm{\Omega }}\\ =0.085\text{\hspace{0.33em}p.u.}\end{array}$

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

$\begin{array}{c}{X}_{13}=X_{23}\\ =\frac{40\mathrm{\Omega }}{585.225\mathrm{\Omega }}\\ =0.068\text{\hspace{0.17em}p.u.}\end{array}$

Draw a circuit diagram of per unit zero-sequence single line diagrams.

Determine the zero-sequence bus admittance matrix for the circuit.

$Y_{\text{bus0}}=j\left[\begin{array}{ccc}-19.78& 3.90625& 5.88\\ 3.90625& -19.78& 5.88\\ 5.88& 5.88& -20\end{array}\right]$

Now, determine the bus impedance matrix, $Z_{\text{bus0}}$.

Substitute $\left[\begin{array}{ccc}-19.78& 3.90625& 5.88\\ 3.90625& -19.78& 5.88\\ 5.88& 5.88& -20\end{array}\right]$for $Y_{\text{bus0}}$into equation (1).

$\begin{array}{c}{Z}_{\text{bus0}}=\frac{1}{j\left[\begin{array}{ccc}-19.78& 3.90625& 5.88\\ 3.90625& -19.78& 5.88\\ 5.88& 5.88& -20\end{array}\right]}\\ =j\left[\begin{array}{ccc}0.0614& 0.0192& 0.0237\\ 0.0192& 0.0614& 0.0237\\ 0.0237& 0.0237& 0.0639\end{array}\right]\end{array}$

Therefore,the value of 3 x 3 per unit zero-sequence bus impedance matrix,$Z_{\text{bus0}}$for the system is $j\left[\begin{array}{ccc}0.0614& 0.0192& 0.0237\\ 0.0192& 0.0614& 0.0237\\ 0.0237& 0.0237& 0.0639\end{array}\right]$.

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

Determine the positive-sequence bus admittance matrix for the circuit.

$Y_{\text{bus1}}=j\left[\begin{array}{ccc}-30.042& 11.7647& 14.7058\\ 11.7647& -29.8039& 14.7058\\ 14.7058& 14.7058& -35.5544\end{array}\right]$

Now, determine the bus impedance matrix, $Z_{\text{bus1}}$.

Substitute $\left[\begin{array}{ccc}-30.042& 11.7647& 14.7058\\ 11.7647& -29.8039& 14.7058\\ 14.7058& 14.7058& -35.5544\end{array}\right]$for $Y_{\text{bus1}}$into equation (2).

$\begin{array}{c}{Z}_{\text{bus1}}=\frac{1}{j\left[\begin{array}{ccc}-30.042& 11.7647& 14.7058\\ 11.7647& -29.8039& 14.7058\\ 14.7058& 14.7058& -35.5544\end{array}\right]}\\ =j\left[\begin{array}{ccc}0.095& 0.0714& 0.0688\\ 0.0714& 0.0959& 0.0692\\ 0.0688& 0.0692& 0.0852\end{array}\right]\end{array}$

Therefore, the value of 3 x 3 per unit zero-sequence bus impedance matrix, $Z_{\text{bus1}}$for the system is $j\left[\begin{array}{ccc}0.095& 0.0714& 0.0688\\ 0.0714& 0.0959& 0.0692\\ 0.0688& 0.0692& 0.0852\end{array}\right]$

Determine the positive-sequence bus admittance matrix for the circuit.

$Y_{\text{bus2}}=j\left[\begin{array}{ccc}-30.042& 11.7647& 14.7058\\ 11.7647& -29.8039& 14.7058\\ 14.7058& 14.7058& -35.5544\end{array}\right]$

Now, determine the bus impedance matrix,$Z_{\text{bus2}}$.

Substitute $\left[\begin{array}{ccc}-30.042& 11.7647& 14.7058\\ 11.7647& -29.8039& 14.7058\\ 14.7058& 14.7058& -35.5544\end{array}\right]$for$Y_{\text{bus2}}$into equation (3).

$\begin{array}{c}{Z}_{\text{bus2}}=\frac{1}{j\left[\begin{array}{ccc}-30.042& 11.7647& 14.7058\\ 11.7647& -29.8039& 14.7058\\ 14.7058& 14.7058& -35.5544\end{array}\right]}\\ =j\left[\begin{array}{ccc}0.0973& 0.0738& 0.0717\\ 0.0738& 0.0983& 0.0722\\ 0.0717& 0.0722& 0.0888\end{array}\right]\end{array}$

Therefore, the value of 3 x 3 per unit zero-sequence bus impedance matrix,$Z_{\text{bus2}}$for the system is$j\left[\begin{array}{ccc}0.0973& 0.0738& 0.0717\\ 0.0738& 0.0983& 0.0722\\ 0.0717& 0.0722& 0.0888\end{array}\right]$.

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

Assume that the system experiences a 3-phase fault with a prefault voltage of 1.0 per unit.

Determine the fault current in the positive sequence components.

$\begin{array}{c}{I}_1=\frac{E_R}{Z_{11-1}}\\ =\frac{1\text{\hspace{0.33em}V}}{j0.095\Omega }\\ =-j10.53\text{\hspace{0.33em}p.u}\end{array}$

Determine the sub-transient fault current in the system.

Substitute 0 for$I_0$into equation (4).

$I_0=0$

Assume that the system experiences a single line-to-ground fault with a prefault voltage of 1.0 per unit.

Determine the fault current in the sequence components.

$\begin{array}{c}{I}_0=I_1\\ =I_2\\ =\frac{E_{\text{g}}}{Z_{11-0}+Z_{11-1}+Z_{11-2}}\\ =\frac{1\text{\hspace{0.33em}V}}{j\left(0.0614+0.095+0.0973\right)}\end{array}$

Solve further as,

role="math" localid="1656414216442" $I_0=-j3.942\text{\hspace{0.33em}p.u}$

Determine the sub-transient fault current in the system.

Substitute$-j3.942$for$I_0$into equation (5).

$\begin{array}{c}{I^{\text{'}\text{'}}}_a=3\left(-j3.942\right)\\ =-j11.825\text{\hspace{0.33em}p.u}\end{array}$

Consider a system with a pre-fault voltage of 1.0 per unit that experiences a single line-to-ground fault through fault impedance.

Determine the per unit fault impedance of the system.

$\begin{array}{c}{Z}_{\text{F}\left(\text{Pu}\right)}=\frac{Z_{\text{F}}}{Z_{\text{base}}}\\ =\frac{30\mathrm{\Omega }}{585.225\mathrm{\Omega }}\\ =0.05\text{\hspace{0.33em}p.u}\end{array}$

Determine the fault current in the sequence components.

$\begin{array}{c}{I}_0=I_1\\ =I_2\\ =\frac{E_{\text{g}}}{Z_{11-0}+Z_{11-1}+Z_{11-2}+3Z_{\text{F}}\left(\text{pu}\right)}\\ =\frac{1}{j\left[0.0614+0.095+0.0973+3\left(0.05\right)\right]}\end{array}$

Solve further as,

$I_0=-j2.477\text{\hspace{0.33em}p.u}$

Determine the sub transient fault current in the system.

Substitute$-j2.477$for$I_0$into equation (5).

$\begin{array}{c}{I^{\text{'}\text{'}}}_a=3\left(-j2.477\right)\\ =-j7.431\text{\hspace{0.33em}p.u}\end{array}$

Consider a system with a pre-fault voltage of 1.0 per unit that experiences a bolted line-to-line fault through fault impedance.

Determine the fault current in the positive-sequence component.

$\begin{array}{c}{I}_1=-I_2\\ =\frac{E_{\text{g}}}{Z_{11-1}+Z_{11-2}}\\ =\frac{1}{j\left(0.095+0.0973\right)}\\ =-j5.2\text{\hspace{0.33em}p.u}\end{array}$

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

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

Substitute 0 for ${I^{\text{'}\text{'}}}_a$, -9pu for${I^{\text{'}\text{'}}}_{\text{b}}$ and$9\angle 0°\text{\hspace{0.33em}p.u}$ for${I^{\text{'}\text{'}}}_c$ into equation (6).

$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ -j5.2\text{\hspace{0.33em}p.u}\\ j5.2\text{\hspace{0.17em}p.u}°\end{array}\right]$

Assume that a bolted double line-to-ground fault occurs in the system.

Determine the fault current in a positive-sequence component.

$\begin{array}{c}{I}_1=\frac{E_{\text{g}}}{\left[X_1+X_2\text{\hspace{0.33em}parallel\hspace{0.33em}with\hspace{0.33em}}{X}_0\right]}\\ =\frac{1\text{\hspace{0.33em}V}}{j\left[0.095+\frac{\left(0.0973\right)\left(0.0614\right)}{\left(0.0973\right)+\left(0.0614\right)}\right]}\\ =\frac{1\text{\hspace{0.33em}V}}{j0.1326}\\ =-j7.5389\text{\hspace{0.33em}p.u}\end{array}$

Determine the fault current in the zero-sequence component.

$\begin{array}{c}{I}_0=I_1\left(\frac{Z_{11-2}}{Z_{11-0}+Z_{11-1}}\right)\\ =\left(-j7.5389\right)\left(\frac{j0.095}{j0.0614+j0.095}\right)\\ =\left(-j7.5389\right)\left(0.607\right)\\ =-j4.579\text{\hspace{0.33em}p.u}\end{array}$

Determine the sub-transient fault current.

Substitute$-j4.579$ for$I_0$ into equation (5).

${I^{\text{'}\text{'}}}_a=-j13.73\text{\hspace{0.17em}p.u}$

Therefore, the bus impedance matrices are obtained, the fault current are determine and verified.