calculating single phase fault
calculating single phase fault
(OP)
I'm trying to reflect a fault level through a single phase transformer winding. I know I can use the %Z and the rated current to get the maximum unlimited secondary fault current. But If I know the actual level on the primary, how do I relect it thru to the secondary? I cannot just use the turns ratio cause I'm looking at current....or can I?
Data:
12,470V pri, 240/120V (center-tap), 15Kva, 1.3%Z
Available on the 3ph line (steady state)
bolt=1653A, L-L=1494A, L-N=2107A, X/R=2.1
Data:
12,470V pri, 240/120V (center-tap), 15Kva, 1.3%Z
Available on the 3ph line (steady state)
bolt=1653A, L-L=1494A, L-N=2107A, X/R=2.1






RE: calculating single phase fault
If connected line to neutral, do the same for a line to ground fault.
RE: calculating single phase fault
He is trying to model the 1ph transformer using a 3ph transformer. Then running the 3ph fault results to then convert back to 1ph results.....thus he keeps the system in one (3ph) file. So he was trying the different 3ph configurations and getting results, but did not know what was correct. As I like to calculate by hand, he asked me and I drew a blank....again thanks for the jog.
After all this, he gave up on his model.
RE: calculating single phase fault
-Calculate the fault on the 120V (or 240V) side of the dist transformer.
-Reflect this faulted transformer's seen impedance to primary side. For a solid fault, this would be 1.3%
-Convert this impedance value to system base.
-Then do a manual single-phase to ground fault analysis, using the impedance value you found before as fault impedance (any fault on the secondary of the single phase trf would be a single phase trf for the system)
-You can use EDSA to convert amp values of short circuits to impedance values.
Hope it helps
RE: calculating single phase fault
This should come out to same value if you follow what stevenal said. (Around 4600A for 240V L-L fault or half of that for 120V L-N fault)