Short circuit MVA of feeder circuit breaker

The short-circuit MVA that the feeder circuit breaker must interrupt is the fault level at bus 3, since the fault is just beyond the breaker and the feeder impedance up to F is negligible. Base: 100 MVA, 11 kV / 132 kV. Pre-fault $V_3 = 1$ pu and pre-fault current is zero, so all source EMFs are 1 pu.

Per-unit reactances (100 MVA base)

• G1, G2: 0.10 pu each; T1, T2: 0.05 pu each, so each source branch is $0.10 + 0.05 = 0.15$ pu.
• Lines 1-2, 1-3, 2-3: 0.10 pu each.

  E=1                 E=1
   |                   |
 j0.15               j0.15
   |                   |
 bus1 ---- j0.1 ---- bus2
     \               /
    j0.1          j0.1
        \         /
          bus 3 --CB-- F

Thevenin reactance at bus 3

The network is symmetrical about bus 3: buses 1 and 2 are always at the same potential, so line 1-2 carries no current. Each path is $0.15 + 0.10 = 0.25$ pu, and the two are in parallel:

$$X_{th} = \frac{0.25 \times 0.25}{0.25 + 0.25} = 0.125\ \text{pu}$$

Short-circuit MVA of the breaker

$$\begin{aligned} I_f &= \frac{V_3}{X_{th}} = \frac{1}{0.125} = 8\ \text{pu} \\ \text{SC MVA} &= \frac{S_{base}}{X_{th}} = \frac{100}{0.125} = 800\ \text{MVA} \\ I_f &= 8 \times \frac{100 \times 10^6}{\sqrt{3} \times 132 \times 10^3} = 8 \times 437.39 = 3499.1\ \text{A} \end{aligned}$$

Answer: the circuit breaker should be rated for a short-circuit capacity of 800 MVA, with a fault current of about 3.50 kA at 132 kV. (For the 11/110 kV variant the per-unit values and the 800 MVA result are the same; only the current in amperes changes, to 4.20 kA at 110 kV.)