Fault MVA at bus 5 with inductive lines

Base: 100 MVA, 11 kV (generator side), 132 kV (line side), $f = 50$ Hz. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation).

Per-unit reactances

• G1, G2: $X'' = 0.20$ pu each (already on 100 MVA, 11 kV).
• T1, T2: $X = 0.05$ pu each.
• So from each generator EMF to its 132 kV bus: $0.20 + 0.05 = 0.25$ pu.

Lines ($X = 2\pi fL$):

$$\begin{aligned} Z_{base} &= \frac{132^2}{100} = 174.24\ \Omega \\ X_{L1} &= 2\pi(50)(1.2 \times 10^{-3} \times 100) = 37.699\ \Omega = \frac{37.699}{174.24} = 0.2164\ \text{pu} \\ X_{L2} = X_{L3} &= 2\pi(50)(1.0 \times 10^{-3} \times 50) = 15.708\ \Omega = 0.0902\ \text{pu} \end{aligned}$$

Reactance diagram

  E=1                         E=1
   |                           |
 j0.25 (G1+T1)               j0.25 (G2+T2)
   |                           |
 bus3 ------- L1 j0.216 ------- bus4
    \                         /
  L3 j0.090            L2 j0.090
       \                   /
            bus 5 (F)

Thevenin reactance at bus 5

Both generator EMFs are equal (1 pu), and the two halves of the network are identical (0.25 pu to each source, $X_{L2} = X_{L3}$). So buses 3 and 4 are always at the same potential and line L1 carries no current; it can be removed. The two paths (0.25 + 0.0902) are then in parallel:

$$\begin{aligned} X_{th} &= \frac{0.25 + 0.0902}{2} = 0.1701\ \text{pu} \end{aligned}$$

(The same value is obtained as $Z_{55}$ of the bus impedance matrix.)

Fault current and fault MVA

$$\begin{aligned} I_f &= \frac{1}{0.1701} = 5.8797\ \text{pu} \\ I_{base} &= \frac{100 \times 10^6}{\sqrt{3}(132 \times 10^3)} = 437.39\ \text{A} \\ I_f &= 5.8797 \times 437.39 = 2571.7\ \text{A} \\ \text{Fault MVA} &= \frac{100}{0.1701} = 587.97\ \text{MVA} \end{aligned}$$

Answer: fault MVA $\approx 588.0$ MVA and fault current $\approx 2.572$ kA at 132 kV.