DLG fault on 30 MVA generator

Data: 30 MVA, 11 kV, $Z_1 = Z_2 = j0.2$, $Z_0 = j0.05$ pu, solidly grounded, unloaded ($E_a = 1\angle0^\circ$). Fault: phases b and c to ground.

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

Fault conditions

$I_a = 0$, $V_b = V_c = 0$, giving $V_{a0} = V_{a1} = V_{a2}$: the three sequence networks are in parallel.

Sequence currents

$$\begin{aligned} Z_2 \| Z_0 &= \frac{(j0.2)(j0.05)}{j0.25} = j0.04 \\ I_{a1} &= \frac{E_a}{Z_1 + Z_2\|Z_0} = \frac{1}{j0.24} = -j4.1667\ \text{pu} \\ I_{a2} &= -I_{a1}\frac{Z_0}{Z_2+Z_0} = j4.1667 \times 0.2 = j0.8333\ \text{pu} \\ I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0} = j4.1667 \times 0.8 = j3.3333\ \text{pu} \end{aligned}$$

Line currents

$$\begin{aligned} I_a &= I_{a0} + I_{a1} + I_{a2} = 0 \\ I_b &= I_{a0} + a^2I_{a1} + aI_{a2} = 6.614\angle{130.89^\circ}\ \text{pu} \Rightarrow 6.614 \times 1574.6 = 10\,415\ \text{A} \\ I_c &= I_{a0} + aI_{a1} + a^2I_{a2} = 6.614\angle{49.11^\circ}\ \text{pu} \Rightarrow 10\,415\ \text{A} \end{aligned}$$

Fault current (to ground)

$$I_f = I_b + I_c = 3I_{a0} = j10.0\ \text{pu} \Rightarrow 10 \times 1574.6 = 15\,746\ \text{A}$$

Line-to-neutral voltages

$$\begin{aligned} V_{a1} = V_{a2} = V_{a0} &= E_a - Z_1I_{a1} = 1 - (j0.2)(-j4.1667) = 0.1667\ \text{pu} \\ V_a &= 3V_{a1} = 0.5\ \text{pu} = 0.5 \times 6.351 = 3.175\ \text{kV} \\ V_b &= V_c = 0 \end{aligned}$$

Results