Newbie simple (yea right) question for 3-phase

[ggunn]

The three phase resultant is exactly the second result, 72.9A.
That would be appropriate for feeder and OCPD determination, and should be the same number you get using the power formula.
There is no question, IMHO, about how the NEC says to do that calculation in the case of actual loads.
But for sources, the specific language seems to rule out doing it the same way. If 690 had you add the power and then determine the current, it would give your result. But it seems to require "adding" the currents instead without saying just how that should be done.

(21000)(1.25)/(208)(sqrt 3) = 72.86. 80A breaker. (There is no 75A IIRC, but my code book is at work.)

Good. That's how I have been doing it.

 

[Smart $]

Well, I think it affects anything which uses the term 'inverter output circuit current', which is to say that it affects both 705.12(D) and 705.60(B) and perhaps some other stuff.

Well that's sort of why I prefer this elsewhere. It only affects stipulations that use the term sum without any indication it can be vector sum... and the only place that I know of that does that is 705.12(D)(2). To try and put the revision in as you suggest will require it to be put in wherever 'inverter output circuit current' is covered.

To me, this part seems appropriate for a short Informational Note.

I think I'm going to change my submission as first posted to an informational note... and we'll see what happens.

 

[Penwar]

Sheesh

Sheesh

I never thought that my Newbie Simple thread would turn into such a workup. Actually glad that it did in that it made it more evident to me that there is more than one way to skin this cat.

 

[Smart $]

I never thought that my Newbie Simple thread would turn into such a workup. Actually glad that it did in that it made it more evident to me that there is more than one way to skin this cat.

Sometimes it's the simple questions that turn into the longest threads on this forum... :happyyes: