if i have a 400 amp 3 phase service and i want to parallel the feeders from the 400 amp disconect to a 400 amp panel, what would be the min. size condutors i could use? I would also like to know how you came to this with 2002 nec ref..ect. thanks

310-16 pair of 3/0 copper @ 75c.

that is the same way i have thought however, i was told something different by a electric inspector today...He told me to go to 310.16
and find the wire size for 400 amps...which is 600mcm. then he said the total cir mils of my parallel feeders had to equal the cir mil of the 600 mcm...which means (2) sets of 300mcm..i need to show him this is wrong and why it is wrong...please help! thanks

First thing I would do is use his own logic against him. With your code book open to 310-16......"Well if that is the case ,sir. How would you apply this method to a 1200 amp service?"

thanks...you guys are the best! i mean it.

One of the main reasons to use parallel conductors is to reduce over-all cost. Point this little number out to him as well. MCM size is not proportional to the amount of amperage it can carry. Take a look at the 75 degree column and compare 1500 kcmil to 1000 kcmil 625 amps as opposed to 545 amps. Gosh that extra 500 mcm bought me a whole 80 amps of extra capacity.

Are all the conductors in the same raceway? This will impact on the ampacity of the system. I can't agree with any of the answer unless I know more details

Oh sure, George, muddy it up with derating
My answer is per standard paralleling using separate raceways and following requirements of 310.4

Had the inspector used this "logic" to derate he would still be giving the wrong information.

[This message has been edited by kentvw (edited 12-15-2004).]

Okay kentvw- you made me get the book out. I used the situation you discribe with 2 raceways. Used THHN insulation in a damp or dry location. Turns out that if you are required to count the neutral as you would for a wye, you are still okay with 3/0. 225a x .8 x 2 = 360 Round up to 400 per 240.4(B). If you install all the conductors in the same race way your in trouble 225a x .7 x 2 = 315 and round up to 350. I've used the code reference Table 310.15(B)(2)(a) The other inspectors approach was incorrect IMHO and I've never heard of his approach used by any instructor out there.

George, Just because I have done something a certain way for years does not necessarily make it right.
Could you then comment on 310.15(4)(a) and also 310.15 (4) (c)
An electricians call? an engineers call?

Must admit that I did not considered derating in my earlier posts.

[This message has been edited by kentvw (edited 12-16-2004).]

Okay here's my understanding of the following:
(Quote)Could you then comment on 310.15(4)(a) and also 310.15 (4) (c)
An electricians call? an engineers call? (Unquote)

(a) is talking about a Delta system where the common (neutral) conductor only carries the unbalance current between the two phase conductors and the two phase conductors are 180 degrees out of phase with each other so there could be and probably is some cancellation taking place thus reducing the current flow in the common conductor.

(b) is talking about a Wye system where the two phase conductors are only 120 degrees out of phase so we don't have the same amount of cancellation as we had with the Delta system. Code panel says we need to count this as a current carring conductor.

(c) is talking again about the Wye and addressing not only the phase issue but the harmonics issue due to the non linear load placing extra duty on the conductor.

As to who's call this is? Everyone's.

The inspectors approach would be entirely appropriate if all of the conductors were bundled together in one place. If, working table 310.16 and the derating rules and the like you get something different, I posit that this is simply an artifact of the way 310.16 simplifies a complex situation.

The ampacity as determined by 310.16 is not an absolute number set by physical law; it is an approximate number set by making conservative estimates of things like thermal insulation and air flow and all the various factors which influence ampacity. I presume that this is the reason for the 'under engineering supervision' note.

The ampacity of a conductor is set by several factors; the heat generated by the conductor, the thermal compatibility of the insulation system, the heat carried away by the surroundings, and any other heat being produced in the surroundings.

If you run copper at the same current density (same amps per circular mil), then you will get the same heat production per unit of copper. Doesn't matter if you put 230A in a #4/0 or 115A in each of two #0 conductors; for the same length you will produce the same total heat. Bundle the two #0s together and the factors affecting ampacity will be the same as the #4/0.

It is only because the #0 conductors can be spaced apart than the total ampacity of 2 #0 conductors is greater than that of a single #4/0; you can get better heat dissipation.

-Jon

All the conductors were running through a 12" nipple, so derating was not a factor.315(b)(2)(a). This inspector got mad at my written response and sent back a note "refer to 90.4".(authority having jurisdiction bla bla blaaa).....that was pretty weak if you ask me.

He also wanted me to supply my ground rod with a 1/0 thhn. I had already brought a 1/0 to building steel. I quoted 250.53(E) (supplemental grounding electrodeshall not be required to be larger than #6 copper).
He didn't like that one much either.

All said and done there is a piece of 1/0 going to the ground rod and i added (1) more set of 3/0 thhn to make him happy about the whole cir mil idea. Had to change all my lugs to make that one work . What can you do? Thanks again for all your help.

"If you run copper at the same current density (same amps per circular mil), then you will get the same heat production per unit of copper."
When you parallel those conductors you end up with a lot more surface area to dissipate that heat and you increase the skin effect.

PIP,
I think the inspector is confusing this with the requirements of the International Residenial Code, Table E3503.1