DETERMINATION OF ELECTRICAL FAULT CURRENT

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A short circuit is an abnormal condition of the supply system, caused by a demage or "short-circuiting" of the normal insulation of the system. The task of a Short Circuit Protection Device is to bring the effects of this faulty condition under control and reduce the demages which it may cause.

For the appropriate selection of the protection device, the expected short-circuit current at the point of installation must be known. If the actual short-circuit current is higher than the switching capacity of the protective device, a reliable interruption of the short-circuit currrent is not fully assured. Extensive damage and service interruption could be the result.

The Value of the short-circuit current depends primarily on:

  • The capacity of the supply source (ratings of transformer or generator)
  • The Type of supply system
  • The distance from the supply system
  • The size of the conductors and the different devices lying between the fault and the supply and
  • The impedance of the fault.
  • The usual faults encounter are illustrated as below : -
    Image Short Circuit Between All 3 Line Conductors
    Isc (3 Phase) = V(phase)/(1.732 x Z) Amp

    where
    Isc = Fault Current
    V(phase) = Voltage between pahases
    Z = Impedence of each coil (phase) of generator or transformer
    Image Short Circuit Between 2 Line Conductors
    Isc (2 Phase) = V(phase)/(2 x Z) Amp

    where
    Isc = Fault Current
    V(phase) = Voltage between pahases
    Z = Impedence of each coil (phase) of generator or transformer
    Image Short Circuit Between Line And Neutral Conductor
    Isc (1 Phase) = V(phase)/(1.732 x Z) Amp

    where
    Isc = Fault Current
    V(phase) = Voltage between pahases
    Z = Impedence of single coil (phase) of generator or transformer
    Impedance of Transformer and Generator
    The actual Impedance of transformers, generators and other supply sources are not specified in ohmic value in most cases. They are usually indicated as percent (%) and they can be read out from the name plate. Otherwise it is generally assumed to be 5%.

    Calculation of fault current in real world

  • Assuming a transformer with 1000KVA at 415V and 50Hz
  • Current rating of transformer = 1000KVA x 1000 / (1.732 x 415V)
  • = 1,391 Ampere
  • Maximum theoritical(permissible) impedance (2 x Z) of transformer = 415V / 1,391A
  • = 0.2983465 Ohm
  • Assuming a 5% impedance = 5% x 0.2983465 Ohm = 0.014917 Ohm
  • The expected 2 line conductor fault current = 415V / 0.014917 Ohm = 27,820 Ampere
  • Nevertheless, it is unusual to calculate fault current as in above manner
  • An alternative way of calculation shall be Isc= Current rating x 100/(percent impedance)
  • In the same example, the fault current = 1,391A x 100/5(%) = 27,820 Ampere
  • As such, the minimum KA rating of MCCB or ACB shall be >28KA
  •