Question

Consider Figures 8.13 and 8.14 of the text with reference to a Y-connected synchronous generator (grounded through neutral impedance ${\mathbf{Z}}_n$) operating at no load. For a line-to-ground fault occurring on phase of the generator, list the constraints on the currents and voltages in the phase domain, transform those into the sequence domain, and then obtain a sequence-network representation. Also, find the expression for the fault current in phase.

Short answer

The expression for the fault current in phase $a$,$I_a$ is $\frac{3V_{\text{F}}}{Z_0+Z_1+Z_2+\left(3Z_{\text{F}}\right)}$.

The required sequence network interconnection in Figure 2.

Step-by-step solution

Write the given data from the question.

Assume a line-to-ground fault from phase of the generator to groundwith reference to a Y-connected synchronous generator (grounded through neutral impedance $Z_n$) operating at no load.

Consider fault present at phase $a$.

Draw a three-phase bus system.

Figure 1

Draw the sequence circuit of interconnected network.

Figure 2

Determine the formula of the fault current in phase.

Write the formula of fault current in phase-.

${\mathbf{I}}_a={\mathbf{{\rm I}}}_0\mathbf{+}{\mathbf{I}}_1\mathbf{+}{\mathbf{I}}_2$ …… (1)

Here, ${\mathbf{{\rm I}}}_0\mathbf{,}\mathbf{}{\mathbf{I}}_1$ and${\mathbf{I}}_2$ are zero, positive and negative sequence current.

(a) Determine the expression of fault current in phase.

Determine the value of current in phase domain for single line-to-ground fault from figure1.

$\begin{array}{c}{I}_b=I_c\\ =0\end{array}$

Determine the value of voltage in phase domain for single line-to-ground fault from figure1.

$V_{ag}=Z_{\text{F}}{I}_a$ …… (2)

Here, $Z_{\text{F}}$is fault impedance and $I_a$is phase current.

Determine the expression for sequence currents.

role="math" localid="1656160355486" $\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]$

Here, $I_a$, $I_b$and $I_c$are the phase currents.

The value of$a$and$a^2$is,

$\begin{array}{l}a=1\angle 120°\\ a^2=1\angle -120°\end{array}$

Substitute $0$for $I_b$, $0$for $I_c$into above equation.

Determine the expression for sequence voltages.

$\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]$

Here, $V_a$,$V_b$and $V_c$are the phase voltages.

Determine the expression for phase- $a$voltage.

$V_a=V_0+V_1+V_2$

Substitute $I_0+I_1+I_2$ for role="math" localid="1656160842493" $I_a$and $V_0+V_1+V_2$for $V_a$into equation (2).

$V_0+V_1+V_2=Z_{\text{F}}(I_0+I_1+I_2)$

Determine the value of current in sequence domain for single line-to-ground fault from figure 2.

$\begin{array}{c}{I}_0=I_1\\ =I_2\end{array}$

Determine the value of voltage in sequence domain for single line-to-ground fault from figure 2.

$V_0+V_1+V_2=\left(3Z_{\text{F}}\right)I_1$

Determine the zero sequence current expression from figure 2.

$I_0=\frac{V_{\text{F}}}{Z_0+Z_1+Z_2+\left(3Z_{\text{F}}\right)}$

Here, $V_F$is fault voltage,$Z_0$is positive sequence component,$Z_1$is positive sequence component and$Z_2$is negative sequence component and$Z_{\text{F}}$is fault impedance.

Determine the expression of fault current in phase $a$.

Therefore, all sequence currents are same.

$\begin{array}{c}{I}_0=I_1\\ =I_2\end{array}$

Substitute$I_0$for$I_1$and $I_0$for$I_2$into equation (1).

…… (3)

Substitute $\frac{V_{\text{F}}}{Z_0+Z_1+Z_2+\left(3Z_{\text{F}}\right)}$for $I_0$into equation (3)

$I_a=\frac{3V_{\text{F}}}{Z_0+Z_1+\left(3Z_{\text{F}}\right)}$

Hence, the expression for the fault current in phase$a$$I_a$ is $\frac{3V_{\text{F}}}{Z_0+Z_1+Z_2+\left(3Z_{\text{F}}\right)}$.