Available fault current @ single phase tr. connected to 3-phase grid

[rotee]

How do I model calculation of phase to ground fault @ single phase distribution transformer (secondary side, 120,240) Using 3-phase transformer model?

 

[dpc]

This is more complicated than you might think, because of the split winding. You can find references with approximate values. Usually the transformer impedance is so high with small distribution transformers, that the source impedance can be ignored. Getting good transformer reactance and X/R data can be difficult.

Beeman's Industrial Power System Handbook covers this calculation, IIRC. The best reference I have found if you actually want to calculate this is "A Practical Guide to Short Circuit Calculations" by Conrad St. Pierre.

 

[rotee]

Thanks DPC,
Do those two books model it as three phase tr. and than fault current in this 3-phase model tr. would be equal fault current @ my 1-phase tr. ???

I know how to model ph-ph fault @1-ph tr using 3-ph model.
(3-ph model: 3x z, 3x kVA, delta connection secondary, and phase to phase fault, standard 3ph calculations)

 

[magoo2]

I've got a copy of an old GE paper on the subject. The calculation method is done in per unit on the transformer kVA base and is a single-phase calculation. The three-phase source is simply represented as a source impedance. Results are available for phase-to-phase and phase-to-neutral faults (full winding-240 V and half-winding 120 V faults).

This doesn't however work from a three-phase transformer model, so I'm not sure you'd have any interest in it.

 

[dpc]

Check this: thread238-186381

The link I gave for GE publication GET-3550F is still good. Download that and print it out for future reference.

 

[waross]

You entitled the thread
" Available fault current @ single phase tr. connected to 3-phase grid"
Available fault current is the symmetrical component of a fault current. Breakers and switchgear are rated in available fault current. The equipment will withstand the asymmetric currents associated with the rated level of available fault current.
Available fault current is calculated in the first instance by assuming an infinite supply and dividing the full load current by the percent impedance of the transformer. If the desired switchgear is rated for this magnitude of available fault current no problem. If the current exceeds the rating of the desired switchgear the next step is to consider the impedance of the feeders and the reduction of available fault current at the switchgear.
If you
The asymmetric fault current may be double the magnitude of the available fault current. You must know or be able to derive the X/R ratio of the transformer and also the available current from the source if less than infinite, if your application requires the asymmetric fault current.
respectfully