L-G fault and 3-phase fault, 30 MVA generator

Data: 30 MVA, 11 kV, $Z_1 = Z_2 = j0.2$, $Z_0 = j0.05$ pu, solidly grounded, unloaded, so $E_a = 1\angle0^\circ$ pu.

Base current: $I_{base} = \dfrac{30\times10^6}{\sqrt3 \times 11\times10^3} = 1574.6$ A; base phase voltage $= 11/\sqrt3 = 6.351$ kV.

(i) L-G fault on phase a

Sequence networks in series:

$$\begin{aligned} I_{a1} = I_{a2} = I_{a0} &= \frac{E_a}{Z_1+Z_2+Z_0} = \frac{1}{j0.45} = -j2.222\ \text{pu} \\ I_f = I_a &= 3I_{a1} = -j6.667\ \text{pu} \\ |I_f| &= 6.667 \times 1574.6 = 10\,497\ \text{A} \end{aligned}$$

Sequence voltages

$$\begin{aligned} V_{a1} &= E_a - Z_1 I_{a1} = 1 - (j0.2)(-j2.222) = 0.5556 \\ V_{a2} &= -Z_2 I_{a2} = -0.4444 \\ V_{a0} &= -Z_0 I_{a0} = -0.1111 \end{aligned}$$

Phase voltages ($V = V_{a0} + a^2 V_{a1} + aV_{a2}$ for phase b, etc.)

$$\begin{aligned} V_a &= 0.5556 - 0.4444 - 0.1111 = 0 \\ V_b &= V_{a0} + a^2V_{a1} + aV_{a2} = 0.8819\angle{-100.89^\circ}\ \text{pu} \\ V_c &= V_{a0} + aV_{a1} + a^2V_{a2} = 0.8819\angle{100.89^\circ}\ \text{pu} \end{aligned}$$

Line-to-line voltages

$$\begin{aligned} V_{ab} &= V_a - V_b = 0.8819\angle{79.11^\circ}\ \text{pu} \Rightarrow 0.8819 \times 6.351 = 5.60\ \text{kV} \\ V_{bc} &= V_b - V_c = 1.7321\angle{-90^\circ}\ \text{pu} \Rightarrow 1.7321 \times 6.351 = 11.0\ \text{kV} \\ V_{ca} &= V_c - V_a = 0.8819\angle{100.89^\circ}\ \text{pu} \Rightarrow 5.60\ \text{kV} \end{aligned}$$

(Line voltages are converted with the phase base 6.351 kV because the pu values are on a phase-voltage base.)

(ii) Three-phase fault

$$I_f = \frac{E_a}{Z_1} = \frac{1}{j0.2} = -j5\ \text{pu} \Rightarrow 5 \times 1574.6 = 7873\ \text{A}$$

Results

The L-G current is larger than the 3-phase current because $Z_0$ (0.05) is much less than $Z_1$.