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

[email protected] = -Ia @0 -

[email protected]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.