210.19 .... "the minimum
branch-circuit conductor size, before the application of any
adjustment or correction factors, shall have an allowable ampacity
not less than the noncontinuous load plus 125 percent
of the continuous load."
I think this means we multiply 39 amps continuous load by 125% and get 48.8 amps...this is the load that we will now be working with. We do not go back to 39 amps.
After having posted the correct method to determine the minimum size conductor, and assuming you have read the entire thread, I guess all there is left to say is, I respectfully disagree with your approach.
I, like Smart$, will also have to respectfully disagree with your approach.
You do not fully post 210.19(A)(1)...
"Branch circuit conductors shall have an
ampacity not less than the
maximum load to be served. Where a branch circuit supplies continuous load or any combination of continuous and non-continuous loads, the
minimum branch-circuit
conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the non-continuous load plus 125% of the continuous load."
The first sentence says that the conductors must have an ampacity not less than the maximum load to be served. The maximum load is 39A. The circuit will never see more than 39A load.
The second sentence tells you minimum allowable conductors size. 39*1.25=48.75. The
minimum conductor size is #8AWG (which has an ampacity of 50.)
#6 THHN has a starting ampacity of 75. Its adjusted ampacity is 75*0.91*0.7=47.78.
The adjusted ampacity of 47.78 is large enough to serve the load (39A) and the #6 is not smaller than the minimum conductor size, so #6 THHN would be acceptable.
Example D3(a) in Annex D gives a good example of this.