Short circuit fault condition KA

[kev2632]

Hi,
I'm just learning and I've got what is probably a stupid question here but I'd appreciate the help and a simple basic explanation of this.

Why is the KA rating fault current of a low voltage 440v switchboard higher than the high voltage 6.6KV ?

Thanks

 

[waross]

The ASCC, Available Short Circuit Current is the rated full load current divided by the transformer percent impedance voltage.
You do the math:
A 1000 KVA transformer; what is the rated secondary current at 440Volts and at 6.6kVolts?

 

[kev2632]

The ASCC, Available Short Circuit Current is the rated full load current divided by the transformer percent impedance voltage.
You do the math:
A 1000 KVA transformer; what is the rated secondary current at 440Volts and at 6.6kVolts?

That's what I'm unsure on please if you could help/explain?

Thanks

 

[wcaseyharman]

1000000/(440*sqrt(3))=1,312.16
If the transformer is a 5% impedance:
1312.16/.05=26,243.2

1000000/(6600*sqrt(3))=87.477
87.477/.05=1,749.54

 

[waross]

Note that the actual, first cycle, fault current may approach 2.8 times the calculated Available Short Circuit current.
The exact value depends on the X:R ratio of the transformer and the point-on-wave of switching.
When switchgear is rated for ASCC, this is taken into account.