LLG fault sequence currents at bus 3

Approach: find the Thevenin impedances at bus 3 from the bus impedance matrices of the positive and zero sequence networks ($Z_2 = Z_1$ since all $X_1 = X_2$), then connect the networks for an LLG fault. Pre-fault voltage $V_f = 1\angle0^\circ$ pu, no load; $Z_f = j0.1$ pu is the impedance from the faulted phases' common point to ground.

Positive sequence network

Generator + transformer branches to reference: bus 1: $j(0.15+0.1) = j0.25$; bus 2: $j0.25$. Lines: 1-2 $j0.12$, 1-3 $j0.15$, 2-3 $j0.25$.

   E1          E2
   |           |
  j0.25       j0.25
   |           |
  (1)--j0.12--(2)
    \         /
   j0.15   j0.25
      \     /
       (3) F

Forming $Y_{bus}$ and inverting gives

$$Z_{bus}^{(1)} = j\begin{bmatrix} 0.1445 & 0.1055 & 0.1299 \\ 0.1055 & 0.1445 & 0.1201 \\ 0.1299 & 0.1201 & 0.2200 \end{bmatrix}$$

So $Z_1 = Z_2 = Z_{33}^{(1)} = j0.2200$ pu.

Zero sequence network

• G1 (Yg) + T1 (Yg/Yg): zero-seq path passes, bus 1 to reference $j(0.05+0.1) = j0.15$
• G2 behind T2 (Δ on generator side, Yg on bus side): bus 2 to reference through $j0.1$ only; G2 is cut off
• Lines: 1-2 $j0.3$, 1-3 $j0.35$, 2-3 $j0.71$

$$Z_{bus}^{(0)} = j\begin{bmatrix} 0.1035 & 0.0310 & 0.0796 \\ 0.0310 & 0.0793 & 0.0470 \\ 0.0796 & 0.0470 & 0.3032 \end{bmatrix}$$

So $Z_0 = j0.3032$ pu.

Sequence currents for LLG fault (phases b, c to ground through $Z_f$)

The negative network is in parallel with the zero network plus $3Z_f$:

$$\begin{aligned} Z_0 + 3Z_f &= j0.3032 + j0.3 = j0.6032 \\ I_{a1} &= \frac{V_f}{Z_1 + \dfrac{Z_2(Z_0+3Z_f)}{Z_2 + Z_0 + 3Z_f}} = \frac{1}{j0.2200 + j\dfrac{0.2200 \times 0.6032}{0.8232}} = \frac{1}{j0.3812} = -j2.6236\ \text{pu} \\ I_{a2} &= -I_{a1}\frac{Z_0+3Z_f}{Z_2+Z_0+3Z_f} = j2.6236 \times \frac{0.6032}{0.8232} = j1.9225\ \text{pu} \\ I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0+3Z_f} = j2.6236 \times \frac{0.2200}{0.8232} = j0.7011\ \text{pu} \end{aligned}$$

Check: $I_{a1} + I_{a2} + I_{a0} = -j2.6236 + j1.9225 + j0.7011 = 0 = I_a$.

Phase and fault currents

$$\begin{aligned} I_b &= I_{a0} + a^2 I_{a1} + a I_{a2} = 4.075\angle{165.05^\circ}\ \text{pu} \\ I_c &= I_{a0} + a I_{a1} + a^2 I_{a2} = 4.075\angle{14.95^\circ}\ \text{pu} \\ I_f &= I_b + I_c = 3I_{a0} = j2.103\ \text{pu} \end{aligned}$$

Answer: $I_{a1} = -j2.624$ pu, $I_{a2} = j1.923$ pu, $I_{a0} = j0.701$ pu; fault (ground) current $= 2.103$ pu; $|I_b| = |I_c| = 4.075$ pu.