In calculating your voltage drop going from a #8 to a #2awg for a 46 amp load I came out with about a 500' one way wire run this would give you a resistance path for fault current of .6055 ohms for the #8/#2 awg path, at 208 volts and an infinite buss ahead of this circuit would only provide 343 amps on a bolted fault, from a trip curve from Square D this would put it into the 2-9 second range on time to trip, well out of the instantaneous range, anything above 5 seconds is considered insulation damaging at this current level, add the impedance of the wire path and transformer ahead of this breaker and your well within the range that would easily damage the #8 EGC, I don't think trying to get the code to allow smaller EGC's in this case would be the right answer, the real reason why the NEC requires when you up size the ungrounded conductors for VD you have to up size the EGC proportionately because the added resistance will lower the current to the point the breaker can't protect the wire anymore.
Appreciate the effort you put into that analysis... but only being provided with basic results and not actual calculations with the variables and constants used, I'm going to surmise there is fault in your premise. Don't be despondent about it, for it is the same fault in premise I see in a lot of Vd calculations... and I'll reiterate I am only surmising fault at this point.
The fault centers on an assumed conductor temperature. Consider that when you determine the circuit will have insufficient voltage for the load using the minimum size circuit conductors required by Code: 46A load on #6 w/#10 EGC. Because of more than 3 conductors are in the raceway, the minimum 'normal' ampacity is upped by a factor of 143% (the inverse of derating 70%). The temperature of the other conductors has no bearing on this increase... it is only assumed that they will operate at the same temperature.. so let's revise the scenario, to make it easier to understand, to not more than 3 conductors in raceway, so we don't have to derate... and go with: 46A load on #8 w/#10 EGC, ungrounded upsized to #4... because of voltage drop using #8...
Now, say you do the voltage drop calculation using the IEEE exact voltage drop formula (link
). Note the value for R
is from Chapter 9, Table 9. Proceed to that table and tell me at what temperature that value of R
is given at....?
Next, tell me how much current has to be carried by the circuit conductor for it to be at that temperature [it's okay to make assumptions; hint: Table 310.15(B)(16) 75?C column
]. Now tell me if the conductor will be at that temperature while conducting 46A of current. If not, why not? Without calculating or having to provide empirical evidence, make an assumption and tell me will it be higher or lower?
If you've arrive at the same conclusion I always do, you'll be starting to get the gist of why I surmise fault in the premise.
Let me give you a simpler approach. If our concern is voltage drop, draw up an 'on-paper' circuit with 3 resistors in series across a voltage supply. Let the middle resistor represent the load, and the two resistors to each side represent the voltage drop of the conductors. The load resistor's value is (E-3%E)/I, where I = 46A, which would make the 'wire' resistors 1.5%E/I. These values represent the limit no matter what size, ampacity, or length of conductors you use. You can also determine I?R values, and these will also be the limit.
When you use a larger, longer conductor for voltage drop reasons, you will be spreading the same or less I?R loss over a greater distance, which result in a lower point temperature of the circuit conductors. In comparison, the temperature of the #10 EGC will be: a) higher, b) the same, or c) lower?