The example in the original post is dealing with equipment over 100A, so we would use 110.14(C)(1)(b). As such we would use

**EITHER** a

75?C rated conductor (eg, THWN) **OR** a 90?C rated conductor (eg, THWN-2) **PROVIDED **ampacity of such conductors does not exceed the 75?C ampacity of the conductor size used. Applying this to the previous example, 2 sets of 400mcm Cu (THWN-2) would have an ampacity of 670A, not 760A.[/QUOTE]

As I stated before, the step where I used 2 sets of 400kcmil at their 760A combined ampacity rating was for verifying compliance with 215.2(A)(1), which has no stipulation that 110.14(C) requirements must be combined in this verification. If you have any gripe whatsoever with my steps, based on what you are trying to point out, it would be with my first step verifying compliance with 110.14(C).

I think this is fairly clear. From the original example, Noncontinuous load (250A) plus 125% of the continuous load (400A*1.25) = 750A. The smallest feeder conductor size shall have an allowable ampacity without any adjustment of correction factors of not less than 750A. As the terminations for the circuit are rated 75?C, 110.14(C) requires me to use the ampacity based on the 75?C rating, even if I am using a 90?C rated conductor. The smallest conductor that gives me an ampacity (from Table 310.15(B)(16)) that is not less than 750A is 2 sets of 500mcm Cu (380A*2 = 760A.)

This just takes us back to the questioning the predominance of 215.2(A)(1) second sentence over first sentence. Where is it stipulated???

Having established our minimum conductor size as 2 sets of 500mcm Cu (maybe someday I'll stop using mcm and start using kcmil...) we need to address the first sentence in 215.2(A)(1). Our load current is 650A. If there are no correction factors (for ambient temperature) or adjustment factors (for number of ccc's in the raceway) then we already know that our feeder ampacity (760A) is large enough to supply the load (650A.) Let's imagine that the feeder was sharing a raceway with another feeder so that there were 6 ccc's, and I was planning to use 500mcm THWN because I had it readily available. The adjusted ampacity of 2 sets of 500mcm THWN would be 608A. This is clearly too small to supply the 650A load. I could look at using 2 sets of 600mcm THWN which would give me an adjusted ampacity of 672A...large enough to supply the load. **OR** using general statement from 110.14(C), I could look at using 2 sets of 500mcm THWN-2. 110.14(C) permits me to use the higher temperature rating for ampacity adjustment, so...430A*2*0.8=688A. This is also large enough to supply the load. Either 2 sets of 600mcm THWN or 2 sets of 500mcm THWN-2 would be acceptable for this example.

I'm agreeable to the result(s) you've established here, but I notice you use the calculated load of 650A as a basis of determination. Why use it here and not for 110.14(C)???

Here is another example of selecting the conductors applying both 215.2 and 110.14(C) together...

http://www.mikeholt.com/news/archive/html/master/conductor.htm

FWIW, I've searched the internet for articles relating to this matter. The ones that deal with continuous loading do follow the method you are using. I've not found any that support my method conclusively. However, I found a bevy of articles that to my mind seem to completely avoid the issue...