Calculate single phase fault from 3-phase fault using calculator

[Tainted]

Suppose I have available fault from utility 80000A (208V 3phase symmetrical)

I want to use this fc2 calculator by EATON they have both options, single phase and 3-phase: http://faultcurrentcalculator.agili...ver=1.5&noCache=1706038356709#language-select


If I use this calculator and input 30 feet of 3-phase 250KCMIL copper in non magnetic conduit. The available fault current would be 38560 amps at the end of the run.

If there are single phase loads downstream of 250KCMIL wiring mentioned above, how would I use this calculator for that?

Would I have to start with 38560 amps at "single phase" even though this value was calculated L-L-L?

 

[Carultch]

Suppose I have available fault from utility 80000A (208V 3phase symmetrical)

I want to use this fc2 calculator by EATON they have both options, single phase and 3-phase: http://faultcurrentcalculator.agili...ver=1.5&noCache=1706038356709#language-select


If I use this calculator and input 30 feet of 3-phase 250KCMIL copper in non magnetic conduit. The available fault current would be 38560 amps at the end of the run.

If there are single phase loads downstream of 250KCMIL wiring mentioned above, how would I use this calculator for that?

Would I have to start with 38560 amps at "single phase" even though this value was calculated L-L-L?

Fault current is point-to-point, and is indifferent to the details of the distribution on the load side of any given point.

You started with your utility fault current, you then account for the impedance of the service conductors, to your main panelboard, to find the fault current at the main panelboard. This then becomes the starting point, for the fault current calculation to subpanels, and then along the branch circuit to each load.

If you derive a single phase load or single phase subpanel from a 3-phase main panel, you follow the 3-phase formulas for the service conductors, and then follow the single phase formulas for the single phase feeder from the 3-phase main panel.

Essentially, what you are calculating, is the maximum possible short circuit current at any given point. A textbook case of an ideal voltage source would have infinite fault current, since it'd always produce the same voltage for any load. A real voltage source has an inherent source impedance that reduces this. That impedance, is what reduces the fault current to a finite value. The more impedance you accumulate, as you travel through more conductor length / more transformer impedances, the lower your fault current will diminish.