Did I do this correctly?
- Thread starter Ponchik
- Start date
Elect117
Senior Member
- Location
- California
- Occupation
- Engineer E.E. P.E.
You calculated the load based on 220.84(C) per unit?
I did.
So if my calculated load came out to 39,500 per unit, I multiply it by (4) then take 45% based on the 220.84(B) and size the service conductors accordingly?
39.5kw x 4 = 158kw
158kw x .45 = 71.1kw <<<<<< I size my service conductors according to this.
Am I making sense? or I am off base?
Elect117
Senior Member
- Location
- California
- Occupation
- Engineer E.E. P.E.
* This is a minimum service size as required by the NEC *
I think you are on the right track.
Just make sure you ran through all the items in 220.84. For example, is there electric cooking equipment? Every unit have it's own air conditioner? etc.
That is the section for the optional method. You can still use the standard method and add up each unit.
So if your math is correct, you would get 71.1kw/240 = 296.3A, 300A MCB, 350kcmil copper, and you can reduce the neutral based on 220.61 / 310.12(D).
I would suggest providing a larger service based on the premise layout and any design choices that could imply future solar or car charging. It would be a waste of effort if someone else has to come back and replace the service in a couple of years.
I think you are on the right track.
Just make sure you ran through all the items in 220.84. For example, is there electric cooking equipment? Every unit have it's own air conditioner? etc.
That is the section for the optional method. You can still use the standard method and add up each unit.
So if your math is correct, you would get 71.1kw/240 = 296.3A, 300A MCB, 350kcmil copper, and you can reduce the neutral based on 220.61 / 310.12(D).
I would suggest providing a larger service based on the premise layout and any design choices that could imply future solar or car charging. It would be a waste of effort if someone else has to come back and replace the service in a couple of years.