Specifying breaker on 132/66 kV transformer

Breaker X is on the 66 kV side, between the transformer and the load bus. It must carry the full-load current of the transformer continuously, and it must interrupt the fault current that flows through it for a fault just beyond it (on the load bus).

Assumptions: the source is an infinite bus (zero impedance), the load bus has no other source, and the fault current is limited only by the transformer impedance (14% on a 50 MVA base).

i) Rated continuous current

I_rated = S / (√3 × V)
        = 50 × 10⁶ / (√3 × 66 × 10³)
        = 437.39 A

So breaker X must have a continuous current rating of at least 437.4 A; the next standard rating, 630 A, is chosen (IEC R10 series).

ii) Breaking capacity

For a three-phase fault on the load bus (just after X), the only impedance is the transformer reactance:

Fault MVA = Base MVA / Z(pu)
          = 50 / 0.14
          = 357.14 MVA

Fault current at 66 kV:
I_f = 357.14 × 10⁶ / (√3 × 66 × 10³)
    = 3124.2 A  ≈ 3.12 kA

A fault on the transformer side of X is fed directly from the infinite bus and is cleared by breaker Y, so X only has to break the current fed through the transformer.

Making current (peak, first cycle) = 2.55 × 3.124 = 7.97 kA.

Answer: rated continuous current = 437.4 A (choose the 630 A standard); breaking capacity = 357.14 MVA (about 3.12 kA at 66 kV). In practice a standard 72.5 kV, 630 A breaker with a rated short-circuit breaking current of 12.5 kA (the lowest standard value commonly offered at this voltage, well above the required 3.12 kA) would be specified.