Short Circuit Calcs - Open Delta

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bronco

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Utility is providing three phase fault data at 12kv side of two pot open delta. Any suggestions on how to model this for fault calculations? The main service into the building is 240 volt, 3 phase, 4 wire. Not sure how to get from 3 phase fault data, through two single phase transformers and back to 3 phase 240 v.

Thanks for any help.
 
My first guess is to calculate for the 120/240v 1ph section, and figure the high leg will be lower.
 
Why do you need the fault current?
To simply select SCCR and AIC ratings of devices, or to conduct a full blown short circuit and/or arc flash study?
 
This can't be modeled in Etap because a single phase transformer can't be connected to a 3 phase bus. Believe me I've tried. Even sent my diagrams to OTI. If I assume the 3 phase fault current goes through one transformer (ignoring it is single phase or 3 phase), I get a fault current of 2ka. Just not sure that's the correct way to model it.
 
bronco said:
If I assume the 3 phase fault current goes through one transformer (ignoring it is single phase or 3 phase), I get a fault current of 2ka. Just not sure that's the correct way to model it.
Click to expand...
That is the wrong way.

For simplistic calculations of short circuit, you can assume it is a closed delta system made up of 3 identical transformers (3 phase KVA and %Z = (3) x 1 phase KVA and %Z).
 
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.
 
Well, I tried a 3 phase 75 kva at 3%Z and ended up with a higher short circuit current (5.9ka) on the secondary side of the transformer than the primary side. I also put in 43.3 kva (86.6%) and ended up 3.4 ka. I started with 5.43ka on the primary side.

Still doesn't seem right. Any suggestions?
 
bronco said:
Well, I tried a 3 phase 75 kva at 3%Z and ended up with a higher short circuit current (5.9ka) on the secondary side of the transformer than the primary side. I also put in 43.3 kva (86.6%) and ended up 3.4 ka. I started with 5.43ka on the primary side.

Still doesn't seem right. Any suggestions?
Click to expand...
It is not at all unusual that the fault current on the secondary of a step-up transformer is higher than the value on the primary.

Effectively your primary fault value is about 113MVA and on your 75KVA secondary it is 2.45MVA, so the transformer has made a difference.
 
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