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Transformer short circuit fault calculation clarification

  1. Nov 5, 2008 #1

    I'm trying to perform a short circuit fault calculation on this transformer and load I have when a 3 phase bolted short (worse case) occurs.

    The transformer specs are below:
    Z(transformer)=5%, Z(cable)= .00306 + j.0027, Delta/Star 6.6kV/400V

    I can understand calculating the short circuit available current at the load:
    Ifault(load) = Vphase/Z(transformer)
    where Z(transformer) is the impedance of one winding, and the voltage across it would be 400/1.73 ~=230V
    Ifault(load) = 230/.05
    = 4.6kA

    Now calculating the fault down at the load is where I get a little confused:
    The example I'm following states that the load current fault for a 3 phase bolted short is:
    Ifault(load) = Vphase/(Z(transformer) + Z(cable))

    My only guess/understanding of why they use Vphase is that Vphase represents the voltage drop from the neutral (mid point) of the WYE transformer to the end of 1 phase? I've drawn a diagram to help clarify what I mean, its hosted at:

    Can someone please explain to me why they use Vphase in the load fault calculation?
  2. jcsd
  3. Nov 6, 2008 #2
    If = Vph / Ztr(ohm)

    Ztr(ohm) = %Z / Zb

    Zb = V2/S

    %Z : Transformer per unit impedance
    V : Transformer Rated voltage (V)
    S : Transformer Rated power (VA) :smile:

  4. Nov 6, 2008 #3
    Thanks for your reply. My question was however clarification on why they use phase voltage when they're calculating at the load end, total fault current that the transformer can deliver.
  5. Nov 12, 2008 #4
    Hi! I would say:

    They use the phase voltage because this fits with the winding impedance, and the load is in star connection as well. This gives you the current in a secondary winding.

    Within this logic, you could equally well compute everything with line voltages and equivalent delta connections. It would be consistent, but less direct and less usable.

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