ElecPwrPEOct11
Well-known member
Hi, I'm hoping this is a quick one to explain.
This problem has a 6.6kV utility service with source impedance of 0.94%. It is connected to a transformer serving a building at 480V. The transformer has 4.25% impedance and rated current of 1,720A. It asks for Isc at the transformer's secondary.
The problem solution simply takes:
Isc = FLA / Zeq
Zeq = 0.0094 + 0.0425= 0.0519 pu
FLA = given = 1720A
Isc = 33,141A
This is a fairly simple pu problem but I don't understand why you don't need to convert voltage bases for the utility source impedance. My solution was to find a new Zsource and then use FLA/ Zeq.
It seems like the utility voltage should matter in this sort of problem, but the solution doesn't include it at all. Can anyone explain why? Thanks for any last minute help!
This problem has a 6.6kV utility service with source impedance of 0.94%. It is connected to a transformer serving a building at 480V. The transformer has 4.25% impedance and rated current of 1,720A. It asks for Isc at the transformer's secondary.
The problem solution simply takes:
Isc = FLA / Zeq
Zeq = 0.0094 + 0.0425= 0.0519 pu
FLA = given = 1720A
Isc = 33,141A
This is a fairly simple pu problem but I don't understand why you don't need to convert voltage bases for the utility source impedance. My solution was to find a new Zsource and then use FLA/ Zeq.
It seems like the utility voltage should matter in this sort of problem, but the solution doesn't include it at all. Can anyone explain why? Thanks for any last minute help!