Question

Determine the subtransient fault current in per-unit and in kA, as well as contributions to the fault current from each line, transformer, and generator connected to the fault bus for a bolted single line-to-ground fault at the fault bus selected in Problem 9.9.

Short answer

The value of sub transient fault current is$\left[\begin{array}{c}\begin{array}{c}-j1.454\\ 0\end{array}\\ 0\end{array}\right]perunit$and$\left[\begin{array}{c}\begin{array}{c}-j8.396\\ 0\end{array}\\ 0\end{array}\right]kA$.

The value of contributions to the fault current from each generator is$\left[\begin{array}{c}\begin{array}{c}3.463\angle -{90}^{\circ }\\ 1.731\angle 89.4^{\circ }\end{array}\\ 1.731\angle 90.6^{\circ }\end{array}\right]kA$.

The value of contributions to the fault current from each transformer is$\left[\begin{array}{c}\begin{array}{c}4.932\angle -{90}^{\circ }\\ 1.731\angle -90.6^{\circ }\end{array}\\ 1.731\angle -89.4^{\circ }\end{array}\right]kA$.

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 generator.

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

Here, $I_{G1-0}$is the fault current by zero sequence generator G1, $I_{G1-1}$is fault current by positive sequence generator G1 and $I_{G1-2}$is fault current by negative sequence generator G1.

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 the fault current by negative sequence transformer T1.

 Determine the sub transient fault current, contributions to the fault current from each generator 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 $V_F$.

Determine the sequence current through each impedance.

$$\begin{gathered} I_0=I_1 \\ {I}_1=I_2 \\ {I}_2=\frac{V_F}{Z_0+Z_1+Z_2} \end{gathered}$$

Substitute the value and solve as,

$$\begin{gathered} I_2=\frac{1\angle 0^{\circ }}{j1.3502+j0.354+j0.3586} \\ =-j0.4847perunit \end{gathered}$$

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).

$$\begin{gathered} \left[\begin{array}{c}\begin{array}{c}{I}_a\\ I_b\end{array}\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}\begin{array}{c}-j0.4847\\ -j0.4847\end{array}\\ -j0.4847\end{array}\right] \\ =\left[\begin{array}{c}\begin{array}{c}-j1.454\\ 0\end{array}\\ 0\end{array}\right]perunit \end{gathered}$$

Rewrite as change in kA.

$\left[\begin{array}{c}\begin{array}{c}{I}_a\\ I_b\end{array}\\ I_c\end{array}\right]=\left[\begin{array}{c}\begin{array}{c}-j8.396\\ 0\end{array}\\ 0\end{array}\right]kA$

Determine the contribution of fault current by zero sequence for generator G1 as:$I_{G1-0}=0$

Determine the contribution of fault current by zero sequence for transformer T1 as:$I_{T1-0}=-j0.4847perunit$

Draw the circuit diagram of positive sequence network.

Determine the currents from Generator G1 as.

$$\begin{gathered} I_{G1-1}=-j0.4847\left(\frac{0.92}{0.92+.576}\right) \\ =-j0.2981perunit \end{gathered}$$

Determine the currents from Transformer T1 as.

$$\begin{gathered} I_{T1-1}=-j0.4847\left(\frac{.576}{.92+.576}\right) \\ =-j0.1866perunit \end{gathered}$$

Draw the circuit diagram of negative sequence network.

Determine the current of Generator G1.

$$\begin{gathered} I_{G1-2}=-j0.4847\left(\frac{0.95}{0.95+.576}\right) \\ =-j0.3017perunit \end{gathered}$$

Determine the current of Transformer T1.

$$\begin{gathered} I_{T1-2}=-j0.4847\left(\frac{.576}{.95+.576}\right) \\ =-j0.1830perunit \end{gathered}$$

Determine the contribution to the fault current from each generator.

Substitute 0 for $I_{G1-0}$, -j0.2981 for $I_{G1-1}$and -j0.3017 for $I_{G1-2}$into equation (2).

$$\begin{gathered} \left[\begin{array}{c}\begin{array}{c}{I}_{G1-a}\\ I_{G1-b}\end{array}\\ I_{G1-c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}\begin{array}{c}0\\ -j0.2981\end{array}\\ -j0.3017\end{array}\right] \\ =\left[\begin{array}{c}\begin{array}{c}-j.5998\\ 0.2999\angle 89.4^{\circ }\end{array}\\ 0.2999\angle {90}^{\circ }\end{array}\right]perunit \end{gathered}$$

Change in kA.

$\left[\begin{array}{c}\begin{array}{c}{I}_{G1-a}\\ I_{G1-b}\end{array}\\ I_{G1-c}\end{array}\right]=\left[\begin{array}{c}\begin{array}{c}3.463\angle -{90}^{\circ }\\ 1.731\angle 89.4^{\circ }\end{array}\\ 1.731\angle 90.6^{\circ }\end{array}\right]kA$.

Therefore, the value of contributions to the fault current from each generator is$\left[\begin{array}{c}\begin{array}{c}3.463\angle -{90}^{\circ }\\ 1.731\angle 89.4^{\circ }\end{array}\\ 1.731\angle 90.6^{\circ }\end{array}\right]kA$.

Determine the contribution to the fault current from each transformer.

Substitute -j0.4847 for $I_{T1-0}$, -j0.1866 for $I_{T1-1}$and -j0.1830 for $I_{T1-2}$into equation (3).

$$\begin{gathered} \left[\begin{array}{c}\begin{array}{c}{I}_{T1-a}\\ I_{T1-b}\end{array}\\ I_{T1-c}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}\begin{array}{c}-j0.4847\\ -j0.1866\end{array}\\ -j0.1830\end{array}\right] \\ =\left[\begin{array}{c}\begin{array}{c}-j.8543\\ 0.2999\angle -90.6^{\circ }\end{array}\\ 0.2999\angle -89.4^{\circ }\end{array}\right]perunit \end{gathered}$$

Rewrite as change in kA.

$\left[\begin{array}{c}\begin{array}{c}{I}_{T1-a}\\ I_{T1-b}\end{array}\\ I_{T1-c}\end{array}\right]=\left[\begin{array}{c}\begin{array}{c}4.932\angle -{90}^{\circ }\\ 1.731\angle -90.6^{\circ }\end{array}\\ 1.731\angle -89.4^{\circ }\end{array}\right]kA$.

Therefore, the value of contributions to the fault current from each transformer is$\left[\begin{array}{c}\begin{array}{c}4.932\angle -{90}^{\circ }\\ 1.731\angle -90.6^{\circ }\end{array}\\ 1.731\angle -89.4^{\circ }\end{array}\right]kA$.