Table of Contents

Estimating Fault Current Availability

Short-circuit current available in bolted-fault conditions. The system fault current is the sum of all connected source fault currents.

Utility Transformers

For utility transformers, a worst-case estimate of available short-circuit current in a three-phase fault can be determined by dividing the transformer ampere rating by the impedance of the transformer.1) This assumes an infinite utility bus with unlimited current; as this is rarely if ever the case, actual fault current will be limited by available utility bus current.

Three-Phase Fault

I ~ = ~ {kVA * 1000}/{sqrt{3} * V} I3p ~ = ~ I / {Z1}

I3p ~=~ {kVA * 1000}/{sqrt{3} * V * Z1}


For example, a 630kVA/400VAC, 5.76% impedance transformer would be capable of delivering approximately 15,800 amperes of fault current into a three-phase fault circuit.

I ~ = ~ {630kVA * 1000}/{sqrt{3} * 400} ~ = ~ 910 ~ Amps

I3p ~ = ~ 910 / 0.0576 ~ = ~ 15,833 ~ Amps


Phase to Phase Fault

Ipp ~=~ I / {(Z1 + Z2)}

Ipp ~=~ {kVA * 1000} / {V * (Z1 + Z2)}

Ipp ~=~ {630 * 1000} / {400 * (0.0576 + 0.0576)} ~=~ 13,671 ~Amps


Phase to Neutral Fault

Ipn ~=~ {3 * I}/ {(Z0 + Z1 + Z2)}

Ipn ~=~ {3*VA} / {sqrt{3} * V * (Z0 + Z1 + Z2)}

Ipn ~=~ {3 * (630 * 1000)}/{sqrt{3} * 400 * (0.0576 + 0.0576 + 0.04896)} ~=~ 16,618 ~ Amps



Voltage: V = VLL
sqrt{3} 1.732
Positive Sequence Impedance: Z1 ~=~ {Transformer Impedance %}/100
Negative Sequence Impedance: Z2 ~=~ Z1
Zero Sequence Impedance: Z0 ~=~ 0.85 * Z12)

Generators

The three-phase bolted fault current from a generator is roughly the rated ampacity of the generator, divided by the subtransient reactance of the machine at that rating. Subtransient reactance is noted on the manufacturer's performance data sheet for that generator, using the “submittal data” on the data sheet, not the implemented or de-rated ampacity of the installation. The Cummins Power Application Manual on Gensets contains calculations for converting that submittal data to another rating in the section on calculating fault currents.3)

Three-Phase Bolted Fault

For example, with a 200kW/480vAC machine with 12% subtransient reactance (X“d), the available fault current at the generator terminals is approximately 2,500 Amps. If this generator is paralleled with two other identical machines, the total available from all generator sets is the sum of the available current from each machine, or 7,500 Amps.


I ~ = ~ {kW * 1000}/{1.732 * V * PF}

I ~ = ~ {200kW * 1000}/{1.732 * 480V *0.8PF} ~ = ~ 300 ~ Amps


I3p ~ = ~ I / {X prime prime d}

I3p ~ = ~ 300 / 0.12 ~ = ~ 2,500 ~ Amps


Single Phase - Neutral Bolted Fault

A single phase-neutral bolted fault may have higher short-circuit current than a Line-Line fault on wye-connected generators with 2/3 pitch windings. The product of the three phase currents divided by the sum of the Subtransient Reactance (X”d), Negative Sequence Reactance (X2) and the Zero Sequence Reactance (X0) gives the available phase-neutral fault current.4)

For a generator with full load current at 200 kW, 480 volts, 60 Hz of 300 amps:

  • Subtransient Reactance (X“d) = 0.110 pu
  • Negative Sequence Reactance (X2) = 0.110 pu
  • Zero Sequence Reactance (X0) = 0.019 pu
    • Note: “pu” = per unit, otherwise considered as percent.


Ipn ~=~ {3 * I} / {X prime prime d + X2 + X0}



Ipn ~=~ {3 * 300Amps}/(0.110 + 0.110 + 0.019) ~ = ~ 3,766 Amps


References

1)
https://www.jcalc.net/transformer-short-circuit-fault-current-calculator
2)
For core-type transformers with delta-star-earth connections.
3)
Application Manual – Liquid Cooled Generator Sets T030f, Rev May 2010, pg 5-36 / p113
4)
Caterpillar EDS 70.4, Generator Winding Pitch and Harmonics, pg. 6