2023 NEC 250.122(B) Edge cases.

[SterTheDer]

I am a new EIT(1yr)(Just passed my PE today) in consulting and am trying to understand 250.122(B). I made a calculator in Excel to help with this, and in applying it to a recent project I encountered a scenario that I am not 100% sure how to handle. I have a workaround that I believe addresses this and want to pick brains to see if I'm on the right track.

Scenario: I have a 400A breaker feeding a 208 3phase load at 220ft away. Without regarding voltage drop, the minimum conductor size is 600kcmil per 310.16
Now, upsizing for voltage drop we must increase the size to 750kcmil. This represents a 25% increase in wire size and thus would proportionally increase EGC from #3 to #2.
So far so good.
However, we want to avoid using cables larger than 600kcmil (Something about the difficulty of installation I am told) and thus we are going to run parallel conduits.
With (2) runs of cable, I can size for voltage drop sufficiently with #4/0 wire. The tricky part comes when you try to determine what the EGC should be. Looking at kcmil, we went from 600kcmil to 423kcmil, for a decrease in wire area. NEC 250.122(F)(2)(a) "...The Equipment grounding conductor in each multiconductor cable shall be sized in accordance with 250.122...", so we know it cannot be smaller than #3. However, since our kcmil decreased it is not immediately clear how to proceed.

My workaround goes as follows:
Since we are going from a single cable run to a dual cable run, our baseline should be based upon a dual cable run, pre-voltage drop.
(2) runs of #3/0 are sufficient for a 400A load per 310.16. So our base is (2) sets of #3/0 with #3 EGC.
Upsizing for voltage drop to (2) runs of #4/0 gives us an increase of 26%, dictating the need to upsize our EGC to #2.
This seems to fit the spirit of the rule and provides similar results to the original scenario.

Am I on the right track, or is there something I am missing?

 

[infinity]

Since we are going from a single cable run to a dual cable run, our baseline should be based upon a dual cable run, pre-voltage drop.
(2) runs of #3/0 are sufficient for a 400A load per 310.16. So our base is (2) sets of #3/0 with #3 EGC.
Upsizing for voltage drop to (2) runs of #4/0 gives us an increase of 26%, dictating the need to upsize our EGC to #2.
This seems to fit the spirit of the rule and provides similar results to the original scenario.

This is correct with the exception of your 26% increase number. How did you determine that the increased from parallel #3/0 to parallel #4/0 was 26%? I didn't run the numbers but off the top of my head if the original minimum size EGC was #3 and you increased it by 26% it would be larger than #2. You should be comparing the circular mils area of the conductors.

Welcome to the Forum.

 

[wwhitney]

This is correct with the exception of your 26% increase number. How did you determine that the increased from parallel #3/0 to parallel #4/0 was 26%?

AWG is a geometric progression, and it's very close to true that jumping three sizes doubles the area. So the ratio of areas of adjoining sizes is close to cube root of 2 = 1.26.

Thus every jump of one AWG size increases the area by 26%. And if you increase the ungrounded conductors by one AWG, you need to increase the EGC by one AWG. Keeps it simple.

Cheers, Wayne

 

[infinity]

Yes, you're correct. As I said earlier I didn't have a chance to actually look at the numbers but in my head 26% seemed like a high number. Looking at Chapter 9 Table 8 I agree with the OP.

 

[Tulsa Electrician]

I ran the numbers on the voltage drop.
208 three phase
400 amp load
220'
3% ?

I get 2 sets 3/0 cu
6.24/(400*.0000395*1.732)
228.02 feet
3/0 in steel conduit .079 table 9
I know that is not what he ask.
Just trying to see how he got 4/0

Maybe doing 2% for feeder to a panel for 5% over all feeder/ branch.
If so than I get larger than 4/0.

Maybe using a power factor in the equation?
Or did I just mess up the math.

 

[electrofelon]

I am a new EIT(1yr)(Just passed my PE today)

I ran the numbers on the voltage drop.
208 three phase
400 amp load
220'
3% ?

I get 2 sets 3/0 cu
6.24/(400*.0000395*1.732)
228.02 feet
3/0 in steel conduit .079 table 9
I know that is not what he ask.
Just trying to see how he got 4/0

Maybe doing 2% for feeder to a panel for 5% over all feeder/ branch.
If so than I get larger than 4/0.

Maybe using a power factor in the equation?
Or did I just mess up the math.

Ster, I also know this wasn't your question, but as an EIT I would like to offer some advice for the future. I agree with Tulsa your voltage drop calcs seem excessively conservative, unless you have some specific reason for doing so, it is going to cost your client excessive money and waste resources to use the full equipment / feeder rating for a VD calc. You should be using a more realistic number for your current which is going to depend on the situation of course, and everyone probably has their own number. Personally I would probably use around 66%, and I can pretty much never see needing to go above 80%. Your allowable percentage could be bumped up to, again depending on the load profile and how long it's likely to sit up at its Max. Us electricians have been in the field a long time know how lightly loaded most equipment is, and how conservative the NEC calcs are.

 

[SterTheDer]

Thank you all for the responses!
Here's some additional background:
This circuit is from a 1200A Main Distribution Panel to a 400A Panelboard via direct burred HDPE conduit. This 400A panel feeds a handful of 125A RV pedestals.
I am using 208V 3Ph, 0.8 Power Factor, Copper conductor, Nonmetallic conduit, 220Ft, 400A.

Table 9 gives X to be 0.039 ohms to Neutral per kft and R to be 0.023 ohms to neutral per kft.
Voltage drop is then calculated to be I*Z.
Z equivalent is R*PF+X(sin(arccos(PF)) =(0.0184+0.0234) = 0.0418ohms per kft.
Divide by 1000, multiply by 220ft, Z = 0.009196 ohms.
400A *1.73 * 0.009196 = 6.36V dropped for a single 600kcmil.
6.36/208 = 3.059% >3%.

Running it again for the parallel conduit scenario:

3/0 Copper in PVC, at 0.8 PF gives us Z = (0.042*0.8 + 0.077*sin(arccos(0.8))) = 0.079ohms per kft
Divide by 1000, multiply by 220ft, Z = 0.0175 ohms.
(400A / 2) *1.73* 0.0154 =6.07 dropped for dual 3/0.
6.07/208= 2.918% <3%. (Oh hey, look at that, this doesn't need upsizing actually. My excel sheet is using a lookup table and reports slightly higher values than my hand calculation(It was reporting this at 3%). Time to make the case to my manager that I should spend a week re-doing the entire spreadsheet ;p Only half kidding)

4/0 Copper in PVC, at 0.8 PF gives us Z = (0.041*0.8 + 0.062*sin(arccos(0.8))) = 0.07ohms per kft
Divide by 1000, multiply by 220ft, Z = 0.0154 ohms.
(400A / 2) *1.73* 0.0154 =5.32V dropped for dual 4/0.
5.32/208= 2.56% <3%.

That is how I decided upon dual 4/0.
Now for the upsize calculation:

Compared to:

Ampacity of Overcurrent Device in Circuit:	400​	Amps	Copper GND
Base wire size:	3/0	AWG/kCMIL	3	AWG/kCMIL
Base Number of parallel runs	2	Runs	2​	Runs
Upsized Current Carrying Conductor:	4/0	AWG/kCMIL	2	AWG/kCMIL	Upsized EGC
Number of parallel runs	2​	Runs	2​	Runs
Base wire as CMIL*QTY	335,600	CMIL	52,620	CMIL	NEC minimum EGC
Upsized wires,* QTY Wires as CMIL	423,200	CMIL	66,360	CMIL	Upsized EGC
How much was wire upsized?	1.26​	multiplier	1.26​	multiplier

Regarding why the 3% number specifically, that's what we've been using for branches. I agree, this is quite conservative, given that 400A is the breaker 100% load, so a more realistic load would be 80% (As that is how we'd size the breaker, 125% of continuous load) or 320A.
Assuming 0.8 Power factor is also conservative, given that these will be RV pedestals at a convention center in texas so the majority of the load will be AC units. So being perfectly reasonable, we could likely get away with a single 400kcmil conductor, and have 1.9ish percent voltage drop if we used 320A at unity power factor, relying upon the fact that if that PF degraded to .8 we'd still be chill on the Vdrop, and if the Ampacity was exceeded, our 400A breaker would pop (as it would anyway even with larger wire).
So I agree, its excessively conservative. In this case, not my circus not my monkeys. And this client has had us re-route and re-calculate these numbers 3-4 times now. (There's 160 RV spots and 34 distribution panels, so you can guess how long it takes to re-calculate every time they move stuff around)
I am assisting another PE in this case, and am using their prescribed ampacity and power factor. Client is a few step away from gold-plating everything, we're using NEMA 4x for everything.

Learning lots, thanks for the feedback and warm welcome!

 

[Tulsa Electrician]

Wow that was pretty cool.
Thank you for doing that.

Almost forgot.
Congratulations on you PE

 

[texie]

Congrats on the PE! While I'm not one, I know this is great milestone. Sounds like you will do well based on your post.

 

[tortuga]

Regarding why the 3% number specifically, that's what we've been using for branches.

Check the IECC code for Iowa to be sure there is not a specific minimum voltage drop requirement in your energy code.
I think it was first added to the energy code in 2013/2014, in California its California title 24 energy code 130.5 (c) Voltage Drop. in most other states its in 2018 IECC C405.9.
In CA, OR and WA its "Feeder conductors and branch circuits combined to be sized for a maximum of 5% voltage drop total".