ECN Forum Forums Code Related Discussion NEC & other Code issues sizing of parallel feeders

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).]