L-G fault at bus 3, parallel lines

Base: 100 MVA; 11 kV on generator side, 220 kV on line side. All data are already on this base. $X_n = 0.06$ pu for each generator. Pre-fault voltage $E = 1\angle0^\circ$ pu, system unloaded.

Single line diagram

 G1        T1                 L1            T2        G2
 (~)--(1)--)(--(2)====================(3)--)(--(4)--(~)
  |       Yg Yg     ====== L2 =======      Yg Yg     |
 Xn                                                  Xn
 |                                                   |
gnd                      F (L-G at bus 3)           gnd

Per-unit reactances

Positive sequence network

 E1-j0.15-(1)-j0.1-(2)-j0.05-(3)-j0.1-(4)-j0.15-E2
                            |
                            F

$$Z_1 = j(0.15+0.1+0.05)\ \|\ j(0.1+0.15) = j0.30\ \|\ j0.25 = j0.1364\ \text{pu}$$

Negative sequence network

Same as positive, without the sources:

$$Z_2 = j0.1364\ \text{pu}$$

Zero sequence network

Both transformers are Yg–Yg, so zero-sequence current passes through them; generator neutrals are grounded through $X_n$, which appears as $3X_n$.

 ref-j0.23-(1)-j0.1-(2)-j0.05-(3)-j0.1-(4)-j0.23-ref
                             |
                             F

$$Z_0 = j(0.23+0.1+0.05)\ \|\ j(0.1+0.23) = j0.38\ \|\ j0.33 = j0.1766\ \text{pu}$$

Fault current (L-G at bus 3, $Z_f = 0$)

Sequence networks in series:

$$\begin{aligned} I_{a1} = I_{a2} = I_{a0} &= \frac{E}{Z_1+Z_2+Z_0} = \frac{1}{j(0.1364+0.1364+0.1766)} = \frac{1}{j0.4494} = -j2.2255\ \text{pu} \\ I_f &= 3I_{a0} = -j6.676\ \text{pu} \end{aligned}$$

Base current at 220 kV:

$$I_{base} = \frac{100\times10^6}{\sqrt3 \times 220\times10^3} = 262.43\ \text{A}$$ $$I_f = 6.676 \times 262.43 = 1752\ \text{A}$$

Phase currents at the fault

• $I_a = I_f = 6.676\angle{-90^\circ}$ pu $= 1.752$ kA
• $I_b = I_c = 0$ (healthy phases, unloaded system)

Fault MVA (for reference)

Fault MVA $= |I_f| \times$ base MVA (with $V_f = 1$) $= 6.676 \times 100 = 667.6$ MVA.

Answer: L-G fault current at bus 3 ≈ 6.68 pu ≈ 1.75 kA.

Note: $Z_0$ (0.1766) is larger than $Z_1$ (0.1364) mainly because of the neutral reactances, so here the L-G current (6.68 pu) is less than the three-phase fault current ($1/0.1364 = 7.33$ pu).