Sizing equipment grounding conductors for voltage drop (and other increases). NEC section 250.122(B) 

Thank you!

can u xplain this a little bit different.>>/? i dnt know if i get what you were saying

@@trope5105 Go to en.wikipedia.org/wiki/American_wire_gauge, to get your data. Pick a size, and look for the area in kcmil. I'll use #6 gauge as an example. The area is 26.3 kcmil. Now pick a size that is 3 gauge sizes larger than it. In other words, subtract 3 from the gauge size number above, and look up the corresponding kcmil. For my example, that would be #3 gauge, which is 52.6 kcmil. Compare the two kcmil numbers. Notice that 52.6 kcmil is twice the area of 26.3 kcmil? That's what I mean by "every 3 steps on the AWG scale corresponds to a doubling in kcmil." The unit of kcmil is a unit of cross-sectional area. I'll leave it to you to look up its definition. Try it with other examples, and see if you get the same pattern. Subtracting and adding on the AWG scale, corresponds to multiplying and dividing the kcmil respectively. This is what I mean by "incremental change in AWG is the same thing as multiplicative change in cross-sectional area".

@@trope5105 Now let's unpack what I mean by negative logarithmic scale. Take the data from the American Wire Gauge table, and paste it into a spreadsheet. Keep only the first column, and the kcmil column. Replace "the aughts" as follows. This is what numbers they would be, if the pattern continued with numbers, instead of renaming them as aught values. 1/0 becomes 0 2/0 becomes -1 3/0 becomes -2 4/0 becomes -3 Make a graph with the kcmil on the horizontal, and the AWG size on the vertical. Compare this with the shape of the graph of the logarithmic function. You'll notice that it looks just like a logarithmic function, except it is negated. Now set the horizontal axis to logarithmic scale. Instead of seeing linear increments between tickmarks, you'll see powers of ten. 1, 10, 100, etc, and on the opposite side, 0.1, 0.01, 0.001 etc (instead of the negative numbers). You also see the logarithmic curve has become a straight line. This shows you that the logarithmic curve, really is a logarithmic function. You've simply compressed the scale on the kcmil, in a format called a logarithmic scale.

That is debatable. In the 2017 it would not have worked be a use it specifically mentioned the load. Now it doesn't give a starting point (an increase from what??), but one could argue that if the conductors exceed the minimum required for the load they have been increased. This is one of the many problems with this rule.

May I use the 10 as hot, 8's as neutral/ground?

TAKE EVERYTHING IM ABOUT TO SAY WITH A GRAIN OF SALT. THIS IS NOT ELECTRICAL ADVICE PROCEED WITH CAUTION. So first we need to calculate the voltage dropoff of your system, assuming you're running it in copper. The formula 2kil/cma applies here. I chose the k value of 12.9 for copper. With that being calculated. The volts dropped according to my calculations is about 6.2 volts. So if we plug that into ohm's law that tells us that what might come in as 80 amps on one side gets dropped to 75.866 amps on the other. So if I check the ampacity chart located at 310.16 we have a few different categories depending on wire type. If I assume you're in the 90 degree category then we take the number 95 for for the #4 wire's ampacity. However we need to adjust this because it is a constant load. So if I remember correctly we need to multiply this number by .8 because it is a consistent load and that number is 76 amps, which is the ampacity of #4 wire. Which is just slightly higher than the 75.866 amps that would be possible from that OCPD and therefore to code? There could be other rules that I'm just unaware of. Also, I don't know about the rules of a connection of those sorts in a panel, but that's what my math got me to haha, good luck. REMEMBER THIS IS NOT ADVICE, I AM JUST GUESSING IF I WAS IN YOUR SITUATION, HOW ID CALCULATE IT. PLEASE CONSULT A PROFESSIONAL.

You'd have to do the math.

Since the EMT is a different metal entirely than copper, it isn't a 1-for-1 match of cross sectional area to a wire EGC. The Steel Tube Institute has data applicable to the limitations on using conduit as the EGC, in terms of what maximum circuit size they recommend for each trade size, and what maximum length you can effectively count upon the conduit to be the EGC.

We are speaking two different languages here. The EGC is never sized using 250.66. Can you elaborate?

@@RyanJacksonElectrical sorry, I’m talking about the electrode grounding conductor

@@poolsdoc9178 Voltage drop doesn't matter for that, just use 250.66. Don't forget to read (A), (B), and (C) though.

@@RyanJacksonElectrical THANKS! I’ll check it out.

250.122(F) for parallel raceways.

@@RyanJacksonElectrical thank you.

Yes. The argument has long been that increasing the size of the ungrounded conductor yields higher fault current and, therefore, could damage the EGC. So it isn't just a voltage drop issue. Although I'm not convinced, personally.

@@RyanJacksonElectrical yeah...it makes no sense seeing as the code specifically states there is NO need to increase EGC for conductor fill or temperature adjustments. Then the EGC is good enough. This code needs to be changed....otherwise I will omit the EGC and use the conduit and be compliant.

Not sure, sorry.

There's rounding errors in the way the table is transcribed. Some numbers round the logarithmic formula for KCMIL up, some numbers round down. If you use the logarithmic formula for AWG instead of the look-up tables, you'll find that the 402% number really "should be" 400%. Until you cross over to sizes above #4/0, where the AWG scale is no longer used, the same number of gauge size increments of the ungrounded wire size, is the same number of gauge size increments of the EGC wire size. You aren't required to let rounding errors govern a design decision on this issue.

10.380 x 4.02 = 41,740.

But if your starting value is 10380 and your final value is 41740 then that’s an increase of about 302% 🤔 (41740-10380)/10380 X 100 = 302.12%

@@jlugolok If you increase 10,380 by 100 percent, what do you get?

@@RyanJacksonElectrical Normally I would very confidently say 20,760 but now I’m not too sure 😂

Ah I think I know what happened, 100% increase of 10,380 is 20,760 because you are increasing 10,380 by 10,380. So if we go back to the Initial problem, you would be increasing 10,380 by 41,740 which gives you 52,107.6. If you use 41,740 you would not be taking into consideration that your starting number was 10 awg (10,380). Or am I way off here? Thank you for taking the time to respond by the way.