SLG fault on 20 MVA generator

Data: 20 MVA, 13.8 kV, $Z_1 = j0.20$ pu, $Z_2 = j0.30$ pu. The zero-sequence impedance is not given; assume $Z_0 = j0.10$ pu (typical value for such a generator, about half of $Z_1$) and a solidly grounded neutral. Unloaded, $E_a = 1\angle0^\circ$ pu.

Base current: $I_{base} = \dfrac{20\times10^6}{\sqrt3 \times 13.8\times10^3} = 836.7$ A; base phase voltage $= 13.8/\sqrt3 = 7.967$ kV.

Fault current (phase a to ground)

Sequence networks in series:

$$\begin{aligned} I_{a1} = I_{a2} = I_{a0} &= \frac{E_a}{Z_1+Z_2+Z_0} = \frac{1}{j0.6} = -j1.667\ \text{pu} \\ I_f &= 3I_{a1} = -j5.0\ \text{pu} \\ |I_f| &= 5 \times 836.7 = 4184\ \text{A} \end{aligned}$$

Sequence voltages

$$\begin{aligned} V_{a1} &= 1 - (j0.2)(-j1.667) = 0.6667 \\ V_{a2} &= -(j0.3)(-j1.667) = -0.5 \\ V_{a0} &= -(j0.1)(-j1.667) = -0.1667 \end{aligned}$$

Phase voltages

$$\begin{aligned} V_a &= V_{a0}+V_{a1}+V_{a2} = 0 \\ V_b &= V_{a0} + a^2V_{a1} + aV_{a2} = 1.0408\angle{-103.90^\circ}\ \text{pu} \\ V_c &= V_{a0} + aV_{a1} + a^2V_{a2} = 1.0408\angle{103.90^\circ}\ \text{pu} \end{aligned}$$

Line-to-line voltages

$$\begin{aligned} V_{ab} &= V_a - V_b = 1.0408\angle{76.10^\circ}\ \text{pu} = 1.0408 \times 7.967 = 8.29\ \text{kV} \\ V_{bc} &= V_b - V_c = 2.0207\angle{-90^\circ}\ \text{pu} = 16.10\ \text{kV} \\ V_{ca} &= V_c - V_a = 1.0408\angle{103.90^\circ}\ \text{pu} = 8.29\ \text{kV} \end{aligned}$$

Answer (with $Z_0 = j0.1$ pu): $I_f = 5$ pu $= 4.18$ kA; $V_{ab} = V_{ca} = 8.29$ kV, $V_{bc} = 16.10$ kV.

General form for any $Z_0$: $I_f = \dfrac{3}{|Z_1+Z_2+Z_0|}$ pu, so with the data given the fault current is $3/(0.5 + X_0)$ pu. $V_{bc}$ rises above the rated 13.8 kV because $Z_2 > Z_1$.