Fault level at end of two 200 km lines

A bolted three-phase fault is balanced, so only the positive-sequence reactance diagram is needed. Base: 100 MVA, 11 kV on the generator side and 132 kV on the line side. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation). As described, both transformers feed a common 132 kV bus, and two parallel 200 km lines run from that bus to F.

Per-unit reactances

$X_{new} = X_{old}\dfrac{S_{new}}{S_{old}}$ (voltage bases match the ratings):

Line: $X_L = 0.2 \times 200 = 40\ \Omega$ per line.

$$\begin{aligned} Z_{base} &= \frac{132^2}{100} = 174.24\ \Omega \\ X_L &= \frac{40}{174.24} = 0.2296\ \text{pu} \end{aligned}$$

Reactance diagram

 E=1 --[G1 j0.15]--[T1 j0.10]--+
                               |   +--[L1]--+
                            132 kV |        |-- F
                               |   +--[L2]--+
 E=1 --[G2 j0.20]--[T2 j0.16]--+

Thevenin reactance at F

$$\begin{aligned} X_{G1T1} &= 0.15 + 0.10 = 0.25 \\ X_{G2T2} &= 0.20 + 0.16 = 0.36 \\ X_{gen} &= \frac{0.25 \times 0.36}{0.25 + 0.36} = 0.1475 \\ X_{lines} &= \frac{0.2296}{2} = 0.1148 \\ X_{th} &= 0.1475 + 0.1148 = 0.2623\ \text{pu} \end{aligned}$$

Fault current and fault MVA

$$\begin{aligned} I_f &= \frac{1}{X_{th}} = \frac{1}{0.2623} = 3.8121\ \text{pu} \\ I_{base} &= \frac{100 \times 10^6}{\sqrt{3} \times 132 \times 10^3} = 437.39\ \text{A} \\ I_f &= 3.8121 \times 437.39 = 1667.3\ \text{A} \\ \text{Fault MVA} &= \frac{S_{base}}{X_{th}} = \frac{100}{0.2623} = 381.21\ \text{MVA} \end{aligned}$$

Answer: total fault current $\approx 1.667$ kA (at 132 kV), fault level $\approx 381.2$ MVA.