3-phase and 1-phase short circuit calculations

Discussion in 'General Electronics Chat' started by shespuzzling, Dec 14, 2010.

  1. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009
    88
    0
    1. Can somebody please explain why the 3-phase short circuit current calculation is:
    I_{SC}=I_{FL})/Z_{p.u.}
    I can't make any sense of this, and the only way it does make sense to me is if I assume that in the symmetrical component analysis, the voltage in the postivie sequence is equal to 1.0 p.u., but I don't think that is always the case so I'm not comfortable with the above generalization (unless I'm missing something). Here's my reasoning for the above:

    I_{a(p.u.)}=I_{pos.seq(p.u.)}=V_{p.u.}/Z_{p.u.}

    I_{SC}=I_{p.u.}*I_{FL}

    I_{SC}=(V_{p.u.}*I_{FL})/Z_{p.u.}

    2. During a single phase to ground fault, why do we say that the currents in phase B and C are 0? I can't understand why that would be the case, if only one phase is grounded. If the system is connected to a load, and one phase is grounded, the other two phases are still complete circuits, so why won't there be current?

    Thanks in advance....I can't find answers to these questions anywhere.
     
  2. subtech

    Senior Member

    Nov 21, 2006
    123
    4
    1. The first equation infers a permanent short at the terminals of the transformer. This is important (mandatory actually) for the protection engineer to know for several reasons.

    First, the conductors that are connected to the transformer terminals must be sized correctly in order to carry the available short circuit current for a period of time sufficient that a protective device (fuse or circuit breaker) can operate to disconnect the source from the offending transformer and de-energize it.
    Conductors that are to be connected to transformer terminals will have a known "fuse" rating. This means that the conductor will carry a specified amount of current for a specified amount of time without melting and separating. Often, the time period is 1 second.


    Second, the engineer must take into account the amount of short circuit current available if he is to design a rigid bus. Often solid copper bars (sometimes aluminum) are used as conductors in the stead of wires and these bars must be braced and supported adequately if they are to physically survive the tremendous physical stresses that are placed on them when a fault (short) occurs downstream of such a bus.


    There are other aspects to consider, but these are prominent.

    2. The unfaulted phases are often neglected in fault calculation situations because their influence on on the faulted section is minimal. Calculation of fault values is simplified greatly if these values are ignored. This is an accepted practice in most instances for the protection engineer.
     
    Last edited: Dec 14, 2010
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