Transformer kVA and impedance - short-circuit calculation

[dinos]

I need to model a 4160-480/277V, 1500/2000kVA transformer (Class AA/FA) which has 6.63% impedance.

Does anyone know if the proper calculation of short-circuit current at the secondary of the transformer is supposed to be based on the self-cooled rating (1500kVA, 1804FLA) or is it supposed to be based on the forced air rating (2000kVA, 2406FLA)?

Using an infinite bus at the primary the difference in calculated fault current is considerable (27210A self-cooled kVA basis, and 36290A forced air basis).

 

[jim dungar]

I need to model a 4160-480/277V, 1500/2000kVA transformer (Class AA/FA) which has 6.63% impedance.

Does anyone know if the proper calculation of short-circuit current at the secondary of the transformer is supposed to be based on the self-cooled rating (1500kVA, 1804FLA) or is it supposed to be based on the forced air rating (2000kVA, 2406FLA)?

Using an infinite bus at the primary the difference in calculated fault current is considerable (27210A self-cooled kVA basis, and 36290A forced air basis).

The standard %Z is based on the standard KVA.
Fan cooling changes the way heat is handled not the construction of the transformer.

 

[kenaslan]

Use the higher value. There is no guarantee that the forced cooling fans will be operational during a short circuit.

 

[jim dungar]

Use the higher value. There is no guarantee that the forced cooling fans will be operational during a short circuit.

The Nameplate %Z is based on the Nameplate self-cooled kVA. To use any other combination is bad engineering.
Short circuit calculations do not care how much load current is flowing.

Large MVA size transformers may actually list a %Z for each increased MVA option, but this is not a normal practice for the majority of transformers.

Short circuit software can easily make the %Z adjustments from self-cooled to higher kVA output as long as you provide the correct base %Z and base kVA.