3-Phase fault current of closed delta transformer

[dk101010]

We have a padmount closed delta transformer with mid tapped on the largest leg to provide single phase(120/240volts). According to the manufacturer, the largest leg is rated at 150kVA, and the other two are rated at 100kVA. The %Z of this transformer is 3.3, How do I calculate the three phase fault current when one of the leg is larger than the other two?

 

[waross]

For a zero ohms fault:
From the KVA per winding, calculate the full load current per winding. Divide by the percent impedance voltage to determine the available fault current of each winding (symetrical fault current).
You can use 1.73 times the available fault current to determine the current of the phase fed by the two 100 KVA windings. Use vector addition to get the current of the other two phases.
As the percent impedance is often used to determine the interupting rating of breakers and switches, the figure of 3.3% may be only for the winding with the highest available fault current. This is the available fault current for selecting the interupting rating of switches and breakers.
If this is for protective relaying or an arc flash study, you must consider the X/R ratio of each winding to get the actual fault current (asymetrical current).
This may be different for the 150 KVA winding and the 100 KVA windings.
For a fault with impedance the problem is a little more involved.
If you care to look up a link to the data sheet for this transformer, we may be able to help a little more.
respectfully