Question

Reconsider the synchronous generator of Problem 8.36. Obtain sequence network representations for the following fault conditions.

(a) A short-circuit between phases$\mathbf{b}$ and $\mathbf{c}$.

(b) A double line-to-ground fault with phase’s$\mathbf{b}$ and$\mathbf{c}$ grounded.

Short answer

1. The fault conditions of a short-circuit between phases $b$and$c$are

$I_b=\frac{(-j\sqrt{3})V_{\text{F}}}{Z_1+Z_2+Z_{\text{F}}}$and $I_c=\frac{\left(j\sqrt{3}\right)V_{\text{F}}}{Z_1+Z_2+Z_{\text{F}}}$.

The sequence network representation is shown below:

(b) The fault conditions of a double line-to-ground fault with phase and grounded are

The sequence network representation is shown below:

Step-by-step solution

Write the given data from the question.

Assume a line-to-line fault from phase$b$ to $c$.

Consider fault present at phase$b$ to $c$.

Draw a three-phase bus system.

Figure 1

Draw the sequence circuit of interconnected network.

Figure 2

Determine the formula of the fault conditions of a short-circuit between phases b and c and fault conditions of a double line-to-ground fault with phase b and c grounded.

Write the formula of fault current in phase-.

${\mathbf{I}}_b=({\mathbf{I}}_0\mathbf{+}{\mathbf{a}}^2{\mathbf{I}}_1\mathbf{+}{\mathbf{aI}}_2)$ …… (1)

Here, localid="1656162920401" ${\mathit{I}}_{\mathbf{0}}$, ${\mathit{I}}_{\mathbf{1}}$and ${\mathit{I}}_{\mathbf{2}}$are zero, positive and negative sequence current and$\mathbf{a}$is identities.

Write the formula of fault current in phase-.

$\begin{array}{l}{\mathbf{I}}_c=-{\mathbf{I}}_b\\ {\mathbf{I}}_c=-({\mathbf{I}}_0\mathbf{+}{\mathbf{a}}^2{\mathbf{I}}_1\mathbf{+}{\mathbf{aI}}_2)\end{array}$ …… (2)

Here,${\mathit{I}}_{\mathbf{0}}$, ${\mathit{I}}_{\mathbf{1}}$and ${\mathit{I}}_{\mathbf{2}}$are zero, positive and negative sequence current and $\mathbf{a}$is identities.

Write the formula of fault conditions in phase $b$ domain in double line-to-ground fault.

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

Here,${\mathbf{Z}}_F$is fault impedance and${\mathbf{I}}_b$are fault current at phase$\mathit{b}$and${\mathit{I}}_{\mathbf{C}}$are fault current at phase.

Write the formula of fault conditions in phase domain in double line-to-ground fault.

${\mathbf{V}}_{\mathrm{cg}}={\mathbf{V}}_{\mathrm{bg}}$

localid="1656166370688" ${\mathbf{V}}_{\mathrm{cg}}={\mathbf{{\rm Z}}}_F({\mathbf{I}}_b\mathbf{+}{\mathbf{I}}_c)$ …(4)

Here, ${\mathbf{Z}}_F$is fault impedance and ${\mathbf{I}}_b$are fault current at phase$b$and $I_C$are fault current at phase $C$.

(a) Determine the fault conditions of a short-circuit between phases b and c.

Write the equations of line-to-line fault in phase domain.

$I_a=0$

Here,$I_a$line current in phase domain.

Write the equations of line-to-ground fault in phase domain.

$V_{ag}=Z_{\text{F}}{I}_a$ …… (5)

Here, $Z_F$is fault impedance and $I_a$is line current in phase domain.

Determine the equations for zero, positive and negative sequence currents.

$\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]$ …… (6)

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

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

Determine the equations for zero, positive and negative 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.

Now substitute $0$for $-I_a$and $-I_b$for $I_c$into equations (6).

Substitute $I_0+a^2{I}_1+a{I}_2$for $I_b$, $V_0+a^2{V}_1+a{V}_2$for $V_b$and $V_0+a{V}_1+a^2{V}_2$for $V_c$into equation (5).

…… (8)

Now substitute$0$for$I_0$and$-I_1$for$I_2$into equation (8).

…… (9)

Determine the fault conditions in sequence component from line to line fault from equation (7) and (9).

$\begin{array}{c}{I}_1=-I_2\\ I_0=0\\ V_1-V_2=Z_{\text{F}}{I}_1\end{array}$

Hence, fault current are connected parallel with all positive, negative and zero sequence component with fault impedance,$Z_{\text{F}}$as shown in figure 2.

Write the equation for fault current from the figure 2.

$\begin{array}{c}{I}_1=-I_2\\ =\frac{V_{\text{F}}}{Z_1+Z_2+Z_{\text{F}}}\end{array}$

Write the expression for current in phase- $b$.

Substitute $0$for $I_0$and $-I_1$for $I_2$into equation (1).

Write the expression for phase- role="math" localid="1656165005140" $a$current.

Write the expression for phase- $b$current.

$I_b=\frac{(-j\sqrt{3})V_{\text{F}}}{Z_1+Z_2+Z_{\text{F}}}$

Write the expression for fault current in phase- $c$.

Substitute $\frac{(-j\sqrt{3})V_{\text{F}}}{Z_1+Z_2+Z_{\text{F}}}$for $I_b$into equation (2).

$I_c=\frac{\left(j\sqrt{3}\right)V_{\text{F}}}{Z_1+Z_2+Z_{\text{F}}}$

Therefore, the faults current in phase-$c$ represent in terms of sequence component from line to line fault and voltages are obtained.

(b) Determine the fault conditions of a double line-to-ground fault with phase b and c grounded.

Determine the double line-to-ground fault with phase $b$and $c$grounded. If fault present at phase $b$and$c$when impedance$Z_F$is also obtained at phase and.

Draw a bus system of three phases.

Figure 3

Draw the sequence circuit of interconnected network.

Figure 4

Write the equation for the fault conditions in phase domain in double line-to-ground fault.

$I_a=0$ …… (10)

Here,$I_a$line current in phase domain.

Transforming equation (10) into sequence domain

$I_0+I_1+I_2=0$ …… (11)

Here,$I_0$are zero sequence components, $I_1$is positive sequence component and $I_2$is negative sequence component.

Substitute $V_0+a{V}_1+a^2{V}_2$for $V_{cg}$and $V_0+a^2{V}_1+a{V}_2$for $V_{bg}$into equation (4)

…..(12)

Substitute$V_0+a^2{V}_1+a{V}_2$for $V_{bg}$,$I_0+a^2{I}_1+a{I}_2$for$I_b$and$I_0+a{I}_1+a^2{I}_2$for$I_c$into equation (3).

As further solve as

$V_0-V_1=\left(3Z_{\text{F}}\right)I_0$ …… (13)

Summarize the equation (11), (12) and (13) in terms of Fault conditions in sequence domain from double line to ground fault.

$\begin{array}{c}{I}_0+I_1+I_2=0\\ V_2=V_1\\ V_0-V_1=\left(3Z_{\text{F}}\right)I_0\end{array}$

Determine the positive sequence current equation from figure 4.

Determine the negative sequence current equation from figure 4.

$I_2=\frac{(Z_0+3Z_{\text{F}})}{Z_2+(Z_0+3Z_{\text{F}})}(-I_1)$

Determine the zero sequence current equation from figure 4.

$I_0=\frac{Z_2}{Z_2+(Z_0+3Z_{\text{F}})}(-I_1)$

Therefore, the sequence components are found and the fault conditions and currents of double line to ground fault are found.