3 Individual Xfmrs SC Calculation

[farmaped]

When calculating 3-phase max available short circuit current at transformer secondary, considering 3 individual transformers.
This is best asked with a simple example:
Primary Service: 4160V 3P 3W
Three (3) individual cans each: 100kVA; 2400:277; Z=2%; Connected D-Y.
Secondary Service: 480V 3P 4W
.
Max Avail Fault Amps= (300kVA)/(480*SQRT(3)*(.02*2))=9,021 Amps
OR
Max Avail Fault Amps=(100kVA)/(277V*.02)=18,050 Amps
.
Which is correct? Thanks

 

[david luchini]

When calculating 3-phase max available short circuit current at transformer secondary, considering 3 individual transformers.
This is best asked with a simple example:
Primary Service: 4160V 3P 3W
Three (3) individual cans each: 100kVA; 2400:277; Z=2%; Connected D-Y.
Secondary Service: 480V 3P 4W
.
Max Avail Fault Amps= (300kVA)/(480*SQRT(3)*(.02*2))=9,021 Amps
OR
Max Avail Fault Amps=(100kVA)/(277V*.02)=18,050 Amps
.
Which is correct? Thanks

Your suggested setup won't give you 480V 3P 4W on the secondary, it will give you 832V. You will need (3) 4160:277 transformers connected D-Y to get 480V 3P 4W on the secondary.

I think either...

Max Avail Fault Amps= (300kVA)/(480*SQRT(3)*(.02))=18,042 Amps
OR
Max Avail Fault Amps=(100kVA)/(277V*.02)=18,050 Amps

would be correct.

 

[ron]

Take the FLA output and divide by 2%, so 18kA

 

[farmaped]

Thank you both, I was overthinking it thinking secondary Amps would have to go through both windings and thus require 2 x %Z for the calcs.

 

[winnie]

When you connect transformer secondaries in series, their absolute (not percent) impedance adds up, but so does their EMF. _Approximately_ this is a wash, and you can say that a transformer bank has about the same percent impedance as the transformers that make up the bank.

The big error in this approximation is that it ignores the effect of phase angle on impedance.

-Jon