L-L fault at point B

Assumptions: base 20 MVA; base voltages 6 kV (generator side), 66 kV (line), 3.3 kV (motor side). The L-L fault needs only the positive and negative sequence networks, so zero-sequence data and the neutral impedances are not used. Pre-fault voltage $V_f = 1$ pu, no-load.

Per-unit reactances (20 MVA base)

$$\begin{aligned} X_{T1} &= 0.04 \times \frac{20}{50} = 0.016\ \text{pu} \\ Z_{base,66} &= \frac{66^2}{20} = 217.8\ \Omega,\quad X_{line} = \frac{30}{217.8} = 0.1377\ \text{pu} \\ X_{T2} &= 0.04 \times \frac{20}{40} = 0.02\ \text{pu} \\ \text{Generator: } X_1 &= 0.10,\ X_2 = 0.05;\quad \text{Machine at D: } X_1 = 0.10,\ X_2 = 0.05 \end{aligned}$$

Sequence networks seen from B

Positive:
 E-j0.1-A-j0.016-B-j0.1377-C-j0.02-D-j0.1-E
                |
           (fault point)
Negative:  same, with j0.05 for both machines, no emf

Thevenin impedances at B:

$$\begin{aligned} Z_1 &= j(0.10+0.016)\ \|\ j(0.1377+0.02+0.10) = j0.116\ \|\ j0.2577 = j0.0800\ \text{pu} \\ Z_2 &= j(0.05+0.016)\ \|\ j(0.1377+0.02+0.05) = j0.066\ \|\ j0.2077 = j0.0501\ \text{pu} \end{aligned}$$

L-L fault current (phases b and c)

$$\begin{aligned} I_{a1} &= -I_{a2} = \frac{V_f}{Z_1+Z_2} = \frac{1}{j0.1301} = -j7.687\ \text{pu} \\ |I_f| = |I_b| = |I_c| &= \sqrt3\,|I_{a1}| = \frac{\sqrt3}{0.1301} = 13.31\ \text{pu} \end{aligned}$$

Base current at B (66 kV):

$$I_{base} = \frac{20 \times 10^6}{\sqrt3 \times 66 \times 10^3} = 174.95\ \text{A}$$ $$I_f = 13.31 \times 174.95 = 2329.5\ \text{A}$$

Answer: L-L fault current at B ≈ 13.31 pu ≈ 2.33 kA (in phases b and c, opposite in direction).