L-L fault currents and voltages, 30 MVA generator

Data: 30 MVA, 11 kV, $Z_1 = Z_2 = j0.2$, $Z_0 = j0.05$ pu. Unloaded, $E_a = 1\angle0^\circ$ pu. Fault between phases b and c (no ground).

$I_{base} = \dfrac{30\times10^6}{\sqrt3\times11\times10^3} = 1574.6$ A; $V_{base,ph} = 11/\sqrt3 = 6.351$ kV.

Fault conditions

$I_a = 0$, $I_b = -I_c$, $V_b = V_c$ → $I_{a0} = 0$, $I_{a2} = -I_{a1}$, $V_{a1} = V_{a2}$. Positive and negative networks in parallel; zero network not involved.

Sequence currents

$$I_{a1} = -I_{a2} = \frac{E_a}{Z_1+Z_2} = \frac{1}{j0.4} = -j2.5\ \text{pu}, \quad I_{a0} = 0$$

Line currents and fault current

$$\begin{aligned} I_a &= 0 \\ I_b &= (a^2 - a)I_{a1} = (-j\sqrt3)(-j2.5) = -4.330\ \text{pu} \\ I_c &= -I_b = 4.330\ \text{pu} \\ |I_f| &= 4.330 \times 1574.6 = 6818\ \text{A} \end{aligned}$$

Line-to-neutral voltages

$$\begin{aligned} V_{a1} = V_{a2} &= E_a - Z_1I_{a1} = 1 - (j0.2)(-j2.5) = 0.5\ \text{pu}, \quad V_{a0} = 0 \\ V_a &= V_{a1}+V_{a2} = 1.0\ \text{pu} = 6.351\ \text{kV} \\ V_b &= (a^2 + a)V_{a1} = -V_{a1} = -0.5\ \text{pu} = 3.175\ \text{kV}\ (\angle180^\circ) \\ V_c &= V_b = -0.5\ \text{pu} = 3.175\ \text{kV}\ (\angle180^\circ) \end{aligned}$$

Results

The line voltage between the faulted phases is zero, and $V_{ab} = V_{ac} = 1.5$ pu $= 9.53$ kV.