I need to model utility two single phase Delta-Delta transformers; one transformer has 75kVA, Z=1.3%, and the other has 50kVA, Z=1.2%.
These two single phase transformers are configured as open delta configuration to take 3 phase 13.2kV LL and step it down into delta secondary 240/120V
as shown in the

I'm working with SKM to calculate Arc flash IE's on the buses fed off these utility transformers. I need to model everything as 3 phase.

I plan to have two cases;
1- MAX SC - Model a 3-phase (Delta-Delta) 75kVA transformer with its open delta rating of 43.3kva (75x57.7%) and Z=1.3%, with 10% tolerance.
2- MIN SC - Model a 3-phase (Delta-Delta) 50kVA transformer with its open delta rating of 28.9.3kva (50x57.7%) and Z=1.2%, with 10% tolerance.

Am I on the right track to end up with conservative Arc flash results? Any other suggestion? Thanks in advance for your help.

I do not know of a accurate way to model that. 1584 is based on a single bolted fault per XFMR. I would ignore the XFMR and only use min/max fault currents - good luck.

I just found this on Mike Holt's forum. It was from "Bob" who is a senior member and Auburn University fan.


You will need to get the %Z for the transformers as has been stated.
One of the most popular uses of open delta is to supply a small three phase load and a large single phase load. Typically, a small transformer is installed alongside an existing 120/240 volt lighting transformer.
Calculate the available fault current for the large transformer. This will be the available fault current on "A" phase.
Calculate the available fault current on the small transformer. This will be the available fault current on "C" phase.
Add the currents vectorially. This will be the available fault current on "B" phase.
Assume a 50 kva %Z 2.5 and a 15 kva %Z 2.5 both 240 volt.
50 kva fault 208 amps/0.025 = 8333 amps Ia
15 kva fault 62.5 amps/0.025 = 2500 amps Ic
Ia + Ib + Ic = 0
Ib@240 = -Ia @0 - Ic@120
Ib = -8333 - (-2500x0.5 + J2500x0.866)
Ib = -8333 - (-1250 + J2165)

If you include the primary impedance the fault current wll be lower.
Ib = -8333 + 1250 - J2165.
Ib = sqrt(-7083� - J2165�)
Ib = 7406 amps.

Robert, your calculation is correct but what do you put into you software? What about the calculation when the available fault current is given? The difference between infinite bus and a week system can be huge, I recently saw the difference being 8 times.

It is difficult to determine the fault from open delta and high leg transformers without a lot of vector math. Best guess and many scenarios is along with engineering judgement is sometime the only way to go.