Question

For a double line-to-ground fault through a fault impedance ${\mathit{Z}}_{\mathbf{F}}$, the sequence networks are to be connected ________, at the fault terminal; additionally, ________ is to be included in series with the zero-sequence network.

Short answer

The sequence network should be connected in the parallel and impedance should be connected in the series with zero sequence network.

Step-by-step solution

Write the condition of double line ground fault.

Let assume $I_a,I_b,I_c$are the phase current, $I_0,I_1,I_2$are the sequence current, $V_o,V_1,V_2$are the sequence voltage, $V_a,V_b,V_c$are the phase voltages and $V_{ag},V_{bg},V_{cg}$are the line to ground voltages.

Let assume the double line to fault occurs from the phase b to c through the impedance $Z_F$.

The fault condition of the double line to ground fault the current become zero.

$I_a=0$ …… (1)

The Line to ground voltages of fault line becomes equal.

${\mathbf{V}}_{\mathrm{bg}}\mathbf{=}{\mathbf{V}}_{\mathrm{cg}}$ …… (2)

${\mathbf{V}}_{\mathrm{bg}}\mathbf{=}{\mathbf{Z}}_F\left({\mathbf{I}}_b\mathbf{+}{\mathbf{I}}_c\right)$ …… (3)

Derive to find the connection of the sequence network, and impedance is to be included in series with zero sequence network.

The fault condition in sequence domain.

$\begin{array}{c}{I}_0+I_1+I_2=I_a\\ I_0+I_1+I_2=0\\ I_0=-\left(I_1+I_2\right)\end{array}$ ……. (4)

From equation (2).

$\begin{array}{c}{V}_{bg}=V_{cg}\\ V_0+a{V}_1+a^2{V}_2=V_0+a^2{V}_1+a{V}_2\\ a{V}_1+a^2{V}_2=a^2{V}_1+a{V}_2\\ a{V}_1+a^2{V}_1=a{V}_2-a^2{V}_2\end{array}$

Solve further as,

$\begin{array}{c}\left(a-a^2\right)V_1=\left(a-a^2\right)V_2\\ V_1=V_2\end{array}$ …… (5)

From equation (3),

$\begin{array}{c}{V}_{bg}=Z_F\left(I_b+I_c\right)\\ V_0+a^2{V}_1+a{V}_2=Z_F\left(I_0+a^2{I}_1+a{I}_2+I_0+a{I}_1+a^2{I}_2\right)\\ V_0+a^2{V}_1+a{V}_1=Z_F\left(2I_0+\left(a^2+a\right)I_1+\left(a+a^2\right)I_2\right)\end{array}$

Substitute$V_1$ for$V_2$ into above equation. [From equation (5)]

$V_0+\left(a^2+a\right)V_1=Z_F\left(2I_0+\left(a^2+a\right)I_1+\left(a+a^2\right)I_2\right)$

Substitute -1 for $a+a^2$into above equation.

$\begin{array}{c}{V}_0+\left(-1\right)V_1=Z_F\left(2I_0+\left(-1\right)I_1+\left(-1\right)I_2\right)\\ V_0-V_1=Z_F\left(2I_0-I_1-I_2\right)\end{array}$

Substitute$I_0$ for$-I_1-I_2$ into above equation. [From equation (4)]

$\begin{array}{c}{V}_0-V_1=Z_F\left(2I_0+I_0\right)\\ V_0-V_1=Z_F\left(3I_0\right)\\ V_0-V_1=3Z_F{I}_0\end{array}$

To satisfy the above equations, the sequence network should be connected in the series. The impedance $3Z_F$should be connected in the series with zero sequence network.