Question

The system shown in Figure 9.28 is the same as in Problem 9.48 except that the transformers are now $\mathbf{Y}–\mathbf{Y}$connected and solidly grounded on both sides. (a) Determine the bus impedance matrix for each of the three sequence networks. (b) Assume the system to be operating at nominal system voltage without prefault currents when a bolted single-line-toground fault occurs on phase A at bus$\mathbf{3}$. Compute the fault current, the current out of phase C of machine$\mathbf{2}$ during the fault, and the line-toground voltages at the terminals of machine $\mathbf{2}$during the fault.

Short answer

(a) The value of zero sequence bus impedance matrix is .

${\overline{\sout{Z}}}_{\text{bus0}}=\left[\begin{array}{cccc}j0.1553& j0.1407& j0.0493& j0.0347\\ j0.01407& j0.1999& j0.0701& j0.0493\\ j0.0493& j0.0701& j0.1999& j0.1407\\ j0.0347& j0.0493& j0.1407& j0.1553\end{array}\right]$

(b)

The value of fault current, the current out of phase C of machine $2$during the fault is .

The value of line-to-ground voltage of machine$2$ during the fault${\overline{V}}_{\text{ya}}$ ,${\overline{V}}_{\text{yb}}$ and ${\overline{V}}_{\text{yc}}$is $3.346\angle 0°\text{\hspace{0.33em}kV}$, $11.763\angle -121.8\text{\hspace{0.33em}kV}$and$11.763\angle 121.8°\text{\hspace{0.33em}kV}$ .

Step-by-step solution

Write the given data from the question.

Assume that the transformers are now$\text{Y–Y}$ connected and solidly grounded on both sides.

Determine the formula ofzero sequence bus impedance matrix fault current, the current out of phase C of machine2 during the fault and line-to-ground voltage of machine2 during the fault.

Write the formula of zero sequence bus impedance matrix fault current.

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

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

Write the formula of fault current, the current out of phase C of machine$2$.

${\overline{\mathbf{I}}}_c={\overline{\mathbf{I}}}_{a0}+\mathbf{a}{\overline{\mathbf{I}}}_{a1}+{\mathbf{a}}^2{\overline{\mathbf{I}}}_{a2}$ …… (2)

Here${\overline{\mathbf{I}}}_{a0}$, $\mathbf{a}{\overline{\mathbf{I}}}_{a1}$, and Write the formula of ${\mathrm{a}}^2{\overline{\mathrm{I}}}_{\mathrm{a}2}$are symmetrical components.

Write the formula of line-to-ground voltage of machine$2$.

$\left[\begin{array}{c}{\overline{V}}_{4a}\\ {\overline{V}}_{4b}\\ {\overline{V}}_{4c}\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{\overline{V}}_{4a0}\\ {\overline{V}}_{4a1}\\ {\overline{V}}_{4a2}\end{array}\right]$

Here,${\overline{\mathbf{V}}}_{4a0}$, ${\overline{\mathbf{V}}}_{4a1}$and ${\overline{\mathbf{V}}}_{4a2}$are phase a sequence voltages at bus (4), terminals of machine .

(a)Determine thevalue ofzero sequence bus impedance matrix.

The ${\overline{\sout{Z}}}_{\text{bus1}}$and${\overline{\sout{Z}}}_{\text{bus2}}$are identical to those in answer to issue 9.51. The zero-sequence network is modified as indicated below since the transformer is firmly grounded on both sides:

Draw the circuit diagram of zero-sequence network.

The following diagram illustrates the series connection of the Thevenin equivalents of the sequence networks for the single line-to-ground fault:

Determine the value of zero sequence bus impedance matrix fault current.

Substitute$\left[\begin{array}{cccc}j0.1553& j0.1407& j0.0493& j0.0347\\ j0.01407& j0.1999& j0.0701& j0.0493\\ j0.0493& j0.0701& j0.1999& j0.1407\\ j0.0347& j0.0493& j0.1407& j0.1553\end{array}\right]$for$Y_{\text{bus0}}$into equation (1).

${\overline{\sout{Z}}}_{\text{bus0}}=\left[\begin{array}{cccc}j0.1553& j0.1407& j0.0493& j0.0347\\ j0.01407& j0.1999& j0.0701& j0.0493\\ j0.0493& j0.0701& j0.1999& j0.1407\\ j0.0347& j0.0493& j0.1407& j0.1553\end{array}\right]$

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

${\overline{\sout{Z}}}_{\text{bus0}}=\left[\begin{array}{cccc}j0.1553& j0.1407& j0.0493& j0.0347\\ j0.01407& j0.1999& j0.0701& j0.0493\\ j0.0493& j0.0701& j0.1999& j0.1407\\ j0.0347& j0.0493& j0.1407& j0.1553\end{array}\right]$

(b) Determine the fault current, the current out of phase C of machine and line-to-ground voltage of machine during the fault.

Draw the circuit diagram of phase sequence of fault current network.

Determine the fault current in phase sequence.

$\begin{array}{c}{\overline{I}}_{fA0}={\overline{I}}_{fA1}={\overline{I}}_{fA2}\\ =\frac{1\angle 0°}{j\left(0.1696+0.1696+0.1999\right)}\\ =-j1.8549\end{array}$

${\overline{I}}_{fA}=3{\overline{I}}_{fA0}=-j5.5648=931\angle 270°\text{\hspace{0.33em}A}$

Therefore, the base current in HV transmission line is

$\frac{100,000}{\sqrt{3}\times 345}=167.35\text{\hspace{0.17em}A}$

Determine the phase- zero sequence voltage at bus (4), terminals of machine$2$.

$\begin{array}{c}{\overline{V}}_{\text{4a0}}=-{\sout{\overline{Z}}}_{430}{\overline{I}}_{\text{fA0}}\\ =-\left(j0.1407\right)\left(-j1.8549\right)\\ =-0.2610\end{array}$

Determine the phase- positive sequence voltage at bus (4), terminals of machine.$2$

$\begin{array}{c}{\overline{V}}_{\text{4A1}}=1-\left(j0.1211\right)\left(-j1.8549\right)\\ =0.7754\end{array}$

Determine the phase- positive sequence voltage at bus (4), terminals of machine.$2$

$\begin{array}{c}{\overline{V}}_{\text{4a2}}=-\left(j0.1211\right)\left(-j1.8549\right)\\ =-0.2246\end{array}$

Note: The HV and LV circuits of the Y-Y linked transformer are indicated by the subscripts and a, respectively. Phase shifting is not involved.

Determine the value of line-to-ground voltage of machine $2$during the fault.

Substitute $-0.2610$for${\overline{V}}_{\text{4a0}}$, $0.7754$for ${\overline{V}}_{\text{4a1}}$and $-0.2246$for${\overline{V}}_{\text{4a2}}$into equation (3).

$\begin{array}{c}\left[\begin{array}{c}{\overline{V}}_{4\text{a}}\\ {\overline{V}}_{\text{4b}}\\ {\overline{V}}_{\text{4c}}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}-0.2610\\ 0.7754\\ -0.2246\end{array}\right]\\ =\left[\begin{array}{c}0.2898+j0\\ -0.5364-j0.866\\ -0.5364+j0.866\end{array}\right]\\ =\left[\begin{array}{c}0.2898\angle 0°\\ 1.0187\angle -121.8°\\ 1.0187\angle 121.8°\end{array}\right]\end{array}$

Multiplied by$20/\sqrt{3}$in above equation.

$\left[\begin{array}{c}{\overline{V}}_{4\text{a}}\\ {\overline{V}}_{\text{4b}}\\ {\overline{V}}_{\text{4c}}\end{array}\right]=\left[\begin{array}{c}3.346\angle 0°\text{\hspace{0.33em}kV}\\ 11.763\angle -121.8\text{\hspace{0.33em}kV}\\ 11.763\angle 121.8°\text{\hspace{0.33em}kV}\end{array}\right]$

Determine the symmetrical components of phase- zero sequence current.

$\begin{array}{c}{\overline{I}}_{\text{a0}}=-\frac{{\overline{V}}_{\text{4a0}}}{j{X}_0}\\ =\frac{0.2610}{j0.04}\\ =-j6.525\end{array}$

Determine the symmetrical components of phase- positive sequence current.

$\begin{array}{c}{\overline{I}}_{\text{a1}}=\frac{{\overline{V}}_f-{\overline{V}}_{4\text{a1}}}{j{X}^{\text{'}\text{'}}}\\ =\frac{1.0-0.7754}{j0.2}\\ =-j1.123\end{array}$

Determine the symmetrical components of phase- negative sequence current.

$\begin{array}{c}{\overline{I}}_{\text{a2}}=-\frac{{\overline{V}}_{\text{4a2}}}{j{X}_2}\\ =\frac{0.2246}{j0.2}\\ =-j1.123\end{array}$

Determine the fault current, the current out of phase C of machine$2$.

Substitute $-j6.525$for${\overline{I}}_{\text{a0}}$, $-j1.123$for ${\overline{I}}_{\text{a1}}$and$-j1.123$ for ${\overline{I}}_{\text{a2}}$into equation (2).

$\begin{array}{c}{\overline{I}}_{\text{c}}=-j6.525+a\left(-j1.123\right)+a^2\left(-j1.123\right)\\ =-j5.402\end{array}$

Determine the base current in the machine circuit.

$\begin{array}{c}{I}_B=\frac{100\times {10}^3}{\sqrt{3}\left(20\right)}\\ =2886.751\text{\hspace{0.33em}A}\end{array}$