Question

Repeat Problem 9.21 for a bolted double line-to-ground fault.

Short answer

The value of sub transient fault current is$\left[\begin{array}{c}\begin{array}{c}0\\ 3.975\angle 144.3^{\circ }\end{array}\\ 3.975\angle 35.{57}^{\circ }\end{array}\right]perunit$and$\left[\begin{array}{c}\begin{array}{c}0\\ 4.590\angle 144.3^{\circ }\end{array}\\ 3.590\angle 35.{57}^{\circ }\end{array}\right]kA$.

The value of contributions to the fault current from each line is$\left[\begin{array}{c}\begin{array}{c}0.686\angle -{90}^{\circ }\\ 2.342\angle {158}^{\circ }\end{array}\\ 2.342\angle {22}^{\circ }\end{array}\right]kA$.

The value of contributions to the fault current from each transformer is$\left[\begin{array}{c}\begin{array}{c}0.6864\angle {90}^{\circ }\\ 2.376\angle 131.1^{\circ }\end{array}\\ 2.376\angle 48.9^{\circ }\end{array}\right]kA$.

Step-by-step solution

Write the given data from the question.

Consider the impedance of the circuit are $Z_0$, $Z_1$, and $Z_2$.

Consider the $I_0$Is the current flowing through impedance$Z_0$.

Consider the $I_1$Is the current flowing through impedance$Z_1$.

Consider the $I_2$Is the current flowing through impedance$Z_2$.

Consider the $I_2$Is the current flowing through impedance$Z_2$.

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

Consider the $V_F$is the pre-fault voltage at each bus in the positive-sequence network.

Determine the formula of per unit reactance of generator, motor and neutral reactor.

Write the formula of sub transient fault current.

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

Here, $I_a$is a phase current, $I_1$is the current flowing through impedance$Z_1$and $I_2$is the current flowing through impedance$Z_2$.

Write the formula of contributions to the fault current from each line.

$\left[\begin{array}{c}\begin{array}{c}{I}_{G1-a}\\ I_{G2-b}\end{array}\\ I_{G3-c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}\begin{array}{c}{I}_{L-0}\\ I_{L-1}\end{array}\\ I_{L-2}\end{array}\right]$ …… (2)

Here, $I_{L-0}$is the fault current by zero sequence line, $I_{L-1}$is fault current by positive sequence line and $I_{L-2}$is fault current by negative sequence line.

Write the formula of contributions to the fault current from each transformer.

$\left[\begin{array}{c}\begin{array}{c}{I}_{T1-a}\\ I_{T1-b}\end{array}\\ I_{T1-c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}\begin{array}{c}{I}_{T1-0}\\ I_{T1-1}\end{array}\\ I_{T1-2}\end{array}\right]$ …… (3)

Here, $I_{T1-0}$is the fault current by zero sequence transformers$T1$, $I_{T1-1}$is the fault current by positive sequence transformer$T1$, $I_{T1-2}$is fault current by negative sequence transformer$T1$.

 Determine the sub transient fault current, contributions to the fault current from each line and contributions to the fault current from each transformer.

Draw a bolted single line-to-ground fault at the fault bus 1 selected the value of impedance and $V_F$.

Consider the expression for positive sequence current through each impedance.

$I_1=\frac{V_F}{Z_1+Z_2+Z_0}$

Here, $V_F$is fault voltage, $Z_1$is positive sequence impedance, $Z_2$is negative sequence impedance and $Z_0$is zero sequence impedance.

Substitute $1\angle 0^{\circ }$for$V_F$, j.267 for$Z_1$, j.27 for $Z_2$and j0.1919 for $Z_0$into above equation.

$$\begin{gathered} I_1=\frac{1\angle 0^{\circ }}{j.267+j.27j0.1919} \\ =-j2.637perunit \end{gathered}$$

Determine the negative sequence current through each impedance.

$I_2=-I_1\left(\frac{Z_0}{Z_0+Z_2}\right)$

Here, $I_1$is positive sequence current, $Z_0$is zero sequence impedance and $Z_2$is negative sequence impedance.

Substitute j2.637 for$I_1$, .1919 for$Z_0$
, .27 for $Z_2$into above equation.

$$\begin{gathered} I_2=j2.637\left(\frac{.1919}{.1919+.27}\right) \\ =j1.096perunit \end{gathered}$$

Determine the zero sequence current through each impedance.

$I_0=-I_1\left(\frac{Z_2}{Z_0+Z_2}\right)$

Here, $I_1$is positive sequence current, $Z_0$is zero sequence impedance and $Z_2$is negative sequence impedance.

Substitute j2.637 for$I_1$, .27 for $Z_2$and .1919 for$Z_0$into above equation.

$$\begin{gathered} I_0=j2.637\left(\frac{.27}{.27+.1919}\right) \\ =j1.541perunit \end{gathered}$$

Determine the sub transient fault current.

Substitute -j0.4847 for$I_0$, -j0.4847 for $I_1$and -j0.4847 for $I_2$into equation (1).

$$\begin{gathered} \left[\begin{array}{c}\begin{array}{c}{I}_a\\ I_b\end{array}\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}\begin{array}{c}j1.541\\ -j2.637\end{array}\\ +j1.096\end{array}\right] \\ =\left[\begin{array}{c}\begin{array}{c}0\\ 3.975\angle 144.3^{\circ }\end{array}\\ 3.975\angle 35.{57}^{\circ }\end{array}\right]perunit \end{gathered}$$

Change in kA.

$\left[\begin{array}{c}\begin{array}{c}{I}_a\\ I_b\end{array}\\ I_c\end{array}\right]=\left[\begin{array}{c}\begin{array}{c}0\\ 4.590\angle 144.3^{\circ }\end{array}\\ 3.590\angle 35.{57}^{\circ }\end{array}\right]kA$

The value of sub transient fault current is$\left[\begin{array}{c}\begin{array}{c}0\\ 3.975\angle 144.3^{\circ }\end{array}\\ 3.975\angle 35.{57}^{\circ }\end{array}\right]perunit$and$\left[\begin{array}{c}\begin{array}{c}0\\ 4.590\angle 144.3^{\circ }\end{array}\\ 3.590\angle 35.{57}^{\circ }\end{array}\right]kA$.

Determine the contributions of zero sequence fault current.

Draw the circuit diagram of zero sequence network.

Determine the currents from Transformer as.

$$\begin{gathered} I_{T-0}=j1.541\left(\frac{.9581}{.24+.9581}\right) \\ =-j1.232perunit \end{gathered}$$

Determine the currents from line as.

$$\begin{gathered} I_{L-0}=j1.541\left(\frac{.24}{.24+.9581}\right) \\ =j0.3087perunit \end{gathered}$$

Determine the contributions of positive sequence fault current.

Draw the circuit diagram of positive sequence network.

Determine the currents from Transformer as.

$$\begin{gathered} I_{T-1}=j2.637\left(\frac{.4583}{.64+.4583}\right) \\ =-j1.100perunit \end{gathered}$$

Determine the currents from line as.

$$\begin{gathered} I_{L-1}=-j2.637\left(\frac{.64}{.64+.4583}\right) \\ =-j1.537perunit \end{gathered}$$

Determine the contributions of negative sequence fault current.

Draw the circuit diagram of negative sequence network.

Determine the currents from Transformer as.

$$\begin{gathered} I_{T-2}=j1.096\left(\frac{.4671}{.64+.4671}\right) \\ =-j0.4624perunit \end{gathered}$$

Determine the currents from line as.

$$\begin{gathered} I_{L-1}=-j1.96\left(\frac{.64}{.64+.4671}\right) \\ =j0.6336perunit \end{gathered}$$

Determine the contribution to the fault current from each transformer.

Substitute j1.232 for$I_{T-0}$, -j1.10 for $I_{T-1}$and j0.4624 for $I_{T-2}$into equation (3).

$$\begin{gathered} \left[\begin{array}{c}\begin{array}{c}{I}_{T1-a}\\ I_{T1-b}\end{array}\\ I_{T1-c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}\begin{array}{c}j1.232\\ -j1.10\end{array}\\ j0.4624\end{array}\right] \\ =\left[\begin{array}{c}\begin{array}{c}j0.5944\\ 2.058\angle 131.1^{\circ }\end{array}\\ 2.058\angle 48.9^{\circ }\end{array}\right]perunit \end{gathered}$$

Change in kA.

$\left[\begin{array}{c}\begin{array}{c}{I}_{T1-a}\\ I_{T1-b}\end{array}\\ I_{T1-c}\end{array}\right]=\left[\begin{array}{c}\begin{array}{c}0.6864\angle {90}^{\circ }\\ 2.376\angle 131.1^{\circ }\end{array}\\ 2.376\angle 48.9^{\circ }\end{array}\right]kA$.

Therefore, the value of contributions to the fault current from each transformer is$\left[\begin{array}{c}\begin{array}{c}0.6864\angle {90}^{\circ }\\ 2.376\angle 131.1^{\circ }\end{array}\\ 2.376\angle 48.9^{\circ }\end{array}\right]kA$.

Determine the contribution to the fault current from each line.

Substitute j.3087 for$I_{L-0}$, -j1.537 for $I_{L-1}$and j0.6336 for $I_{L-2}$into above equation (2).

$$\begin{gathered} \left[\begin{array}{c}\begin{array}{c}{I}_{L-a}\\ I_{L-b}\end{array}\\ I_{L-c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}\begin{array}{c}j.3087\\ -j1.537\end{array}\\ j0.6336\end{array}\right] \\ =\left[\begin{array}{c}\begin{array}{c}-j0.594\\ 2.028\angle {158}^{\circ }\end{array}\\ 2.028\angle {22}^{\circ }\end{array}\right]perunit \end{gathered}$$

Change in.

$\left[\begin{array}{c}\begin{array}{c}{I}_{L-a}\\ I_{L-b}\end{array}\\ I_{L-c}\end{array}\right]=\left[\begin{array}{c}\begin{array}{c}0.686\angle -{90}^{\circ }\\ 2.342\angle {158}^{\circ }\end{array}\\ 2.342\angle {22}^{\circ }\end{array}\right]kA$

The value of contributions to the fault current from each line is$\left[\begin{array}{c}\begin{array}{c}0.686\angle -{90}^{\circ }\\ 2.342\angle {158}^{\circ }\end{array}\\ 2.342\angle {22}^{\circ }\end{array}\right]kA$.