Further to my last post, which was perhaps a bit unclear given the time of the morning and the rush I was in to get my son to college…..
From the OPs original post I’m assuming you’re thinking that the neutral might neeed to be sized for the sum of the three loads – 92+78+78=248 amps.
The answer to that is no. Because it’s three phase, the currents are going to be 120 degrees out of phase with each other. So, if L1 is at it’s peak, L2 and L3 will be 120 degrees before and 120 degrees after L1 in their cycle so will in fact be reducing the overall current in the neutral by a proportion of each of their maximum currents. For arguements sake, lets say that is 39 amps each (( 78*Cos(60) )then your actual neutral load is 92-78 = 14 amps.
A fully balanced 3 phase load will have no neutral current. If L1, L2 and L3 are all 78 amps then the neutral current is (78 – 2*(78Cos60)) which is 78-(2*39) which is zero.
So, sizing of the conductors is based on the highest phase conductor; in this case 92 amps. Neutral is sized by the regs to be the same size as the line conductor. Where harmonics are present there may be a factor in sizing to be applied to derate the cable (up to 0.86 of the cable rating I believe from a very quick look at the regs). So by increasing the cable size, the current rating goes up to cover the extra harmonic induced currents in the neutral.
Hopefully this makes sense and I’m not talking rubbish (it wouldn’t be the first time).
Rich.