Avoiding plans or engineering with Voltage Drop calcs like this is always fun. Many engineers have trouble with some VD calcs, unless using a verified modeling tool.
Forum member "kingpb" used to share his corporate-purchased, modeling-systems results on this forum, which some tried to emulate with spreadsheets. The problem here breaks my spreadsheet, without some agreement on neutral current thru each section of the circuit.
I find it entertaining as well. It keeps the old cobwebs cleaned out.
Modeling tools verified by whom? We have found many errors in canned software. It always pays to make your own verification. I wrote my own software as well as the software for the company where I worked. I guess since they paid me you could say it was corporate-purchased.
Jon beat me to the point, but here's a similar way neutral current could affect wire size, if AC power factor = 1.
Assuming all lights are working on a 3-wire 120/240v ckt, one obvious place neutral current wont equal zero (cancel between phases) is End Of Line, 2.5A, 1600 ft total, shown on Smart'$ diagram.
A #8cu neutral is under 3% VD when plugging in 2.5A, 800ft for calculators that need 1-way distance.
Using Smart'$ diagram, size phase B by checking max VD at each load.
Wire between panel & Load1
320 ft. 1-way with 12.5A from all fixture loads = #6cu < 3%
Wire between panel & Load2
640 ft. 1-way with 10.0A from 4 fixture loads = #4cu < 3%
Wire between panel & Load3
960 ft. 1-way with 07.5A from 3 fixture loads = #4cu ~ 3%
Wire between panel & Load4
1280 ft. 1-way with 05.0A from 2 fixture loads = #4cu < 3%
Wire between panel & Load5
1600 ft. 1-way with 02.5A from 1 fixture load = #6cu < 3%
If this method is Ok for < 3% VD, then the smaller wire works.
#8 neutral,
#4 hot on phase B,
#6 hot on Phase A.
The best way would be to model the system impedances and loads.
I have written a program for this and also have some canned software for such things, but where is the fun in that? Besides, my program is not graphically driven and is a little more cumbersome to set up the data than in Excel. It was written back in the Fortran/Basic days. While I have modified it from time to time, I just have not felt like converting it to take full advantage of Visual. The exercises I do here are strictly for my enjoyment.
I like to take the chance to walk back through the calcs. Using canned software and not walking back through the calcs sometimes can come up and bite you because you might forget some of the under-lying assumptions that were made for the program.
Now consider this exercise for fun:
I made a quick and dirty modification of my spreadsheet. To keep from re-writing the whole thing, I made 3 calcs: one for phase A, one for phase B and one for the neutral, with the correct amps in each line section. I assumed the loads were simple so that equal currents from A & B will cancel at the neutral. This means that 1/2 of the neutral line segments have no current and the other 1/2 have 2.5 amps.
So, my thought was that the normal single-phase calc assumes that the neutral is the same as the phase and uses a multiplier of "2" (3-phase calcs use a 1.732 multiplier). So if I take 1/2 of the % drop from the A or B calc, and add it to 1/2 the % drop from the neutral calc, I would get the total drop. I may add that this is approximate for this circuit, but much closer than using no shared neutral.
See if you agree with this logic or have I over-thunk it?
Anyway, using Smart $'s lengths, allowing for about a 3% drop, with <1.5% on the neutral, the end results for the short run, long run, and neutral are: 100% pf: #3|#2|#8 = 1.90%|1.65%|1.13% => short=3.03% , long=2.77% .... 90% pf: #3|#2|#8 = 1.87%|1.65%|1.05% => short=2.92% , long=2.70%
Using equal sized wire yields: 100% pf with #4 = 2.36%|2.68%|0.45% => short=2.81% , long=3.13% .... 90% pf with #4 = 2.29%|2.60%|0.45% => short=2.72% , long=3.03%
You can compare this with the conservative (non-shared neutral) results of: 100% pf: short run with #2 = 2.74%, long run with #1 = 2.60% .... 90% pf: short run with #2 = 2.75%, long run with #1 = 2.69%
This means a #2 or #1 in the conservative case vs a #4 for the more precise case (but maybe not so smart...more later): 1.6 times the cmil in one run and 2 times the cmil in the other run for an average of 1.8 times. That is pretty close to doubling the wire size so the contractor was correct along that line of thinking.
I was wrong in thinking the neutral current would be reduced by a fraction when it is actually reduced down to one fixture load for 1/2 the segments and zero for the other.
Now for the rub: I don't know that a total size reduction is such a good idea. Your worst case scenario would be losing one leg (or all the lights on one leg) and the neutral would then carry full current from the other leg and you would have a full 2-wire voltage drop (the conservative case).
I don't think you can necessarily consider single phase loads as "balancing out" the neutral. Otherwise we could say we will balance our panels and use a very small neutral. Three-phase loads can be excluded from the neutral of course.
Using ramsy's conductors yields:
100% pf: #6|#4|#8 = 3.73%|2.68%|1.13% => short=4.86% , long=3.81% .... 90% pf: #6|#4|#8 = 3.53%|2.60%|1.05% => short=4.58% , long=3.64%
In conclusion, you could reduce the wire sizes for volt drop with the shared neutral, but it may not be the best idea if you consider that you might lose some of the loads on one phase.
A final point: You could just use 240 volt fixtures and you could just use #8 all the way with no neutral.