Calculating available fault current for open delta setup

[sw_ross]

How would you calculate the available fault current for a 3-phase 240 volt open delta?
This is in a small commercial building in a residential neighborhood. The poco adds a smaller xfmr by the neighborhood 120/240 xfmr to provide the C phase just for this one building.

Thanks

 

[ptonsparky]

First, ask the the POCO for details. They should be able to tell you what it is at their transformers. If nothing else you need to start with transformer size.

 

[sw_ross]

First, ask the the POCO for details. They should be able to tell you what it is at their transformers. If nothing else you need to start with transformer size.

Yeah I have a call in to the poco for info, still waiting for a call back.

Of the 2 xfrmr's there's a kva rating on one (37.5), the other one the labeling is weathered off. They're both similar in size (physical size) and have been in place for many years.

I'll see what info I get from poco. If I don't get info I'll play around with the numbers and see if I can come up with a reasonable number.

 

[electrofelon]

For the electrical theory side of things, there was a thread on this last month:

https://forums.mikeholt.com/showthread.php?t=197011

 

[sw_ross]

Info from poco on open delta

Info from poco on open delta

The poco finally got back to me with the specs on the xfrmr's for my open delta setup.
My panel is 3-phase 240v with high-leg

One xfrmr is 75 kva with impedance of 1.6%
Second xfrmr is 37.5 kva with impedance of 2.6%

How the heck do I calculate an AFC number for my new panel?!

Thanks!

 

[topgone]

The poco finally got back to me with the specs on the xfrmr's for my open delta setup.
My panel is 3-phase 240v with high-leg

One xfrmr is 75 kva with impedance of 1.6%
Second xfrmr is 37.5 kva with impedance of 2.6%

How the heck do I calculate an AFC number for my new panel?!

Thanks!

Take the higher rated transformer and compute for the open-delta short-circuit available. Most POCOs publish maximum service equipment short-circuit available current. For pole-top pigs, my files here tells me it's 24,344 Amps!

 

[Julius Right]

It could be like this ?

 

[sw_ross]

It could be like this ? View attachment 22911

Yes,
I don't know if the 75kva is the main xfrmr (where the neutral is derived) and the 35 kva is just added for the C phase?, but that would be my assumption, although I could be wrong.

 

[Julius Right]

I think you are right. By reviewing the opening words:
"The poco adds a smaller xfmr by the neighborhood 120/240 xfmr to provide the C phase just for this one building".
it can be deduced that the new transformer without a 120 V socket is the smaller transformer. My mistake!:ashamed1:

 

[kwired]

Yes,
I don't know if the 75kva is the main xfrmr (where the neutral is derived) and the 35 kva is just added for the C phase?, but that would be my assumption, although I could be wrong.

Unless you have some rare situation where the C phase has a majority of the load on it, the C phase transformer is always the smaller one.

 

[Julius Right]

From Cooper-Bussmann Short Circuit Current Calculations the short-circuit current for the used half of secondary winding is 1.5 times the short-circuit current of entire transformer. Then the short-circuit reactance has to be 1.5 times less.
The 37.5 transformer impedance Zsmall=0.24^2/37.5*1000*2.6%=0.039936 ohm
The 75 kVA half impedance will be Zhalf=0.24^2/75*1000*1.6%/1.5=0.008192 ohm
For 208 V circuit Z208=Zsmall+Zhalf=0.048128 ohm
Then the expected short-circuit current will be 208/0.048128=4321.81 A

 

[kwired]

How would you calculate the available fault current for a 3-phase 240 volt open delta?
This is in a small commercial building in a residential neighborhood. The poco adds a smaller xfmr by the neighborhood 120/240 xfmr to provide the C phase just for this one building.

Thanks

If you are only interested in the maximum available fault current, most of the time it will be the fault current associated with line to line on the 120/240 portion. If you have a very short feed from the transformer it may be line to neutral fault current - still likely to be on the 120/240 portion though. High leg to neutral will have high leg transformer impedance plus half the other transformer impedance to limit fault current.

 

[Julius Right]

From Cooper-Bussmann Short Circuit Current Calculations the short-circuit current for the used half of secondary winding is 1.5 times the short-circuit current of entire transformer. Then the short-circuit reactance has to be 1.5 times less.
The 37.5 transformer impedance Zsmall=0.24^2/37.5*1000*2.6%=0.039936 ohm
The 75 kVA half impedance will be Zhalf=0.24^2/75*1000*1.6%/1.5=0.008192 ohm
For 208 V circuit Z208=Zsmall+Zhalf=0.048128 ohm
Then the expected short-circuit current will be 208/0.048128=4321.81 A

"Then the short-circuit reactance has to be 1.5 times less."
It has to be :
Then the short-circuit impedance has to be 1.5 times less.:weeping: