Re: 110.14(C) question
Andrew,
"There is more than one way to skin a cat." Whatever methodical steps you take to select the conductors for the given conditions is fine as long as the correct results occur. The post which I used to illustrate what works best for me was, in retrospect, not clearly enough presented. At the risk of repetition, my suggested approach is as follows:
DO THE TERMINATIONS [110.14(c)] LAST!!!
Step 1. Calculate minimum conductor ampacity for the calculated load per 210.19(A)(1). This value must be not less than 100% of the non-continuous plus 125% of the continuous load. Call this MCLA, for minimum conductor load amps.
Step 2. Adjust MCLA for (a) ambient temperature by DIVIDING BY the appropriate correction factor in Table 310.16 and (b) for more than three current-carrying conductors by DIVIDING BY the appropriate adjustment factor from Table 310.15(B)(2)(a).
Step 3. From the appropriate column in Table 310.16 select a conductor size whose allowable ampacity in the table is NOT LESS THAN calculated in Step 2.
Step 4. Lastly, check to make sure that the terminations will not be exposed- from MCLA- to more current than is allowed by Table 310.16. This is easily done visually by looking at the values stated in the table. If MCLA exceeds the maximum allowable, go up one or more size conductors until you satisfy the termination temperature restrictions.
ILLUSTRATION FROM HANDBOOK 210.20.
1. MCLA = 31.25.
2. 31.25 divided by 0.8 = 39.06.
3. Minimum 90C copper conductor size is 10AWG.
4. 10AWG won't work because it will cause the 60C terminal to reach its limit at 30 amps, but MCLA (31.25) is greater than 30 amps. Go up one size to 8AWG which will not cause the terminal to reach its maximum allowable 60C rating until 40 amps- well above MCLA of 31.25.
ILLUSTRATION FROM HANDBOOK 220.10.
1. MCLA = 375.
2. 375 divided by 0.8 = 469.
3. Minimum 90C copper conductor size is 600kcmil.
4. 600kcmil will work, because it won't cause a 75C terminal to reach its limit until 420 amps- a value higher than the MCLA of 375.
In Step 2, for the examples given, no information was given that it was necessary to do an additional adjustment for ambient temperature. Had such been the case, then the answer in that Step 2 would need to have been increased by dividing by the appropriate correction factor from Table 310.16. Steps 3 and 4, then would be conducted based on the higher value calculated thereby. For example, if the ambient temperature was given to be 100 deg F, the values in Step 2 in the above ILLUSTRATIONS would need to have been divided by .91 before proceeding to Step 3.
End result: Whatever works!