L-L fault line voltages, 25 MVA alternator

For a line-to-line fault at the generator terminals, $I_{a1} = -I_{a2} = E_a/[j(X_1+X_2)]$ and $I_{a0} = 0$, and for a bolted fault $V_{a1} = V_{a2}$; the phase voltages follow from these.

Data: $X_1 = 0.25$, $X_2 = 0.35$, $X_0 = 0.1$ pu; $E_a = 1\angle0^\circ$ pu (no-load prefault); fault between phases b and c, $Z_f = 0$.

Sequence currents

$$I_{a1} = -I_{a2} = \frac{E_a}{j(X_1 + X_2)} = \frac{1}{j0.60} = -j1.667\ \text{pu}, \qquad I_{a0} = 0$$

Sequence voltages

$$\begin{aligned} V_{a1} &= E_a - jX_1I_{a1} = 1 - j0.25(-j1.667) = 1 - 0.4167 = 0.5833\ \text{pu} \\ V_{a2} &= -jX_2I_{a2} = -j0.35(j1.667) = 0.5833\ \text{pu} \\ V_{a0} &= 0 \end{aligned}$$

Phase voltages

$$\begin{aligned} V_a &= V_{a1} + V_{a2} = 1.1667\angle0^\circ\ \text{pu} \\ V_b &= a^2V_{a1} + aV_{a2} = (a^2 + a)(0.5833) = -0.5833\ \text{pu} \\ V_c &= aV_{a1} + a^2V_{a2} = -0.5833\ \text{pu} \end{aligned}$$

Line voltages

$$\begin{aligned} V_{ab} &= V_a - V_b = 1.1667 + 0.5833 = 1.75\angle0^\circ\ \text{pu} \\ V_{bc} &= V_b - V_c = 0 \\ V_{ca} &= V_c - V_a = -1.75 = 1.75\angle180^\circ\ \text{pu} \end{aligned}$$

These are in per unit of the base phase voltage $13.2/\sqrt3 = 7.621$ kV:

$$|V_{ab}| = |V_{ca}| = 1.75 \times 7.621 = 13.34\ \text{kV}, \qquad V_{bc} = 0$$

(Fault current for reference: $I_b = -j\sqrt3 I_{a1} = 2.887$ pu $= 2.887 \times 1093.5 = 3157$ A.)

Answer: $V_{ab} = 13.34\angle0^\circ$ kV, $V_{bc} = 0$, $V_{ca} = 13.34\angle180^\circ$ kV.