Ampacity Adjustment

[Alwayslearningelec]

When adjusting amapacity due to # CCC in raceway or ambient temp you don't use 125% even if it's a continious load?

 

[augie47]

The derated ampacity must be equal or greater than the continuous load required ampacity
If you have a 40 amp continuous load you need a conductor with a 50 amp ampacity after derating

For example if you had a 40 amp continuous load and 8 #6 THWN-2 in a conduit you could use the 75 amp 90 deg for derating to 52 amps and be o.,k. based on a 50 amp load.

 

[david luchini]

When adjusting amapacity due to # CCC in raceway or ambient temp you don't use 125% even if it's a continious load?

You are correct. The 125% for continuous load is not used for adjustment or correction factors.

 

[wwhitney]

The derated ampacity must be equal or greater than the continuous load required ampacity
If you have a 40 amp continuous load you need a conductor with a 50 amp ampacity after derating

These statements are not true.

The unadjusted ampacity at the termination temperature rating must be greater than or equal to 125% of the continuous load (plus 100% of the non-continuous load). The derated ampacity needs only to be equal or greater to 100% of the load, both continuous and non-continuous.

One way to think of this is that the 125% continuous use multiplier is only necessary for terminations (where you don't need to derate). That's because a wire's ampacity is already a continuous rating (see the definition).

So if you have a 40A continuous load, you need a conductor with 41A after derating (so that you can protect it at the required 50A per 240.4(B)), and 50A at the 60C or 75C unadjusted ampacity (depending on termination rating).

For example if you had a 40 amp continuous load and 8 #6 THWN-2 in a conduit you could use the 75 amp 90 deg for derating to 52 amps and be o.,k. based on a 50 amp load.

The ampacity after derating only has to be 41A. So in your example, you could further have a 50C ambient temperature (0.82 correction factor for 90C rated insulation) for a final ampacity of 75A * 0.7 CCCs * 0.82 temperature = 43A.

Cheers, Wayne

 

[augie47]

If he has a 40 amp continuous load the ampacity of the conductor would need to be 50 amps.

210.19:.........not less than the larger of 210.19(A)(1)(a) or (b). (a) Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the minimum branch-circuit conductor size shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load. (b) The minimum branch-circuit conductor size shall have an allowable ampacity not less than the maximum load to be served after the application of any adjustment or correction factors.

 

[david luchini]

If he has a 40 amp continuous load the ampacity of the conductor would need to be 50 amps.

210.19:.........not less than the larger of 210.19(A)(1)(a) or (b). (a) Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the minimum branch-circuit conductor size shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load. (b) The minimum branch-circuit conductor size shall have an allowable ampacity not less than the maximum load to be served after the application of any adjustment or correction factors.

The ampacity needs to be 40A (the maximum load to be served) after the application of any adjustment or correction factors. That's part (b) in what you posted.

 

[wwhitney]

If he has a 40 amp continuous load the ampacity of the conductor would need to be 50 amps.

Not in the Article 100 sense. Let me make explicit the implicit meaning of 210.19 for you:

210.19:.........not less than the larger of 210.19(A)(1)(a) or (b). (a) Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the minimum branch-circuit conductor size shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load before application of any adjustment or correction factors (b) The minimum branch-circuit conductor size shall have an allowable ampacity not less than the maximum load to be served after the application of any adjustment or correction factors.

The first bolded text is not spelled out in the NEC, but it is implied by the second bolded text. [This section's text could be improved.]

Cheers, Wayne

 

[wwhitney]

[This section's text could be improved.]

The 2011 NEC's version of 210.19(A)(1) was clearer:

2011 NEC 210.19(A)(1) General. Branch-circuit conductors shall have an ampacity not less than the maximum load to be served. Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the minimum branch-circuit conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load.

The text I bolded was removed in the 2014 NEC edit, but to my understanding there was no accompanying intention to change the rule.

Cheers, Wayne

 

[NoahsArc]

2020's 210.19/310.14/110.14 is a mess, and the whole aspect of wiring things really needs to be straightened out in the book given how central it is.

There is no excuse for how poorly structured this all is in 2020. Even knowing what to do (do I now?), I'm getting tripped up on how it's stated.
(a) says 125+100 in accordance with 310.14 (which directs to 310.15) and (b) says 100+100 in accordance with 310.15...
Either way, terminations also need to be considered separately based on their temperature ratings, per 110.14.

It doesn't explicitly state you don't derate, and 310.14A3 directs you to 310.15 for derating. So... Is this not to code in 2020 now?

 

[NoahsArc]

Not in the Article 100 sense.......

If this is 2023, then great, it's fixed. 2020, which we're on now, is horrid.
Either way, the bouncing around through different articles for something so central is fairly bad top-level design.

 

[wwhitney]

If this is 2023, then great, it's fixed.

No, it's not. If you read what I wrote, the first bolded text in that quote is my insertion; that bolded text is not in the NEC. But it is implied by the second bolded text.

Cheers, Wayne

 

[kwired]

My understanding is the reasoning for 125% for continuous loads was always for termination ratings of standard overcurrent devices, and the reason why you don't need to apply that 125% for 100% rated devices.

What this boils down to for most common situations we encounter is the terminals are only rated 75C but the conductor insulation is usually rated 90C. You need minumum conductor ampacity based on 125% for continuous loads for the sake of the overcurrent device rating.

That will leave you fine for adjustments in the raceway as is until you have ambient or number of conductors to cause an impact, but since you likely have 90C conductors to begin with you have a little more room for such adjustments before you need to think about increasing conductor size.

12 AWG THHN/THWN would still have 75C ampacity of 25 amps, but usually is limited to 20 amps max overcurrent protection per the small conductor rule in 240.4(D) so a continuous load of over 17 amps would be over 20 amps after adding the 25% and would need to be larger conductor for the sake of termination temperature.

Insulation temperature is a separate issue and that same 12 AWG conductor is good for 30 amps before you start making adjustments.

If your adjusted ampacity ends up requiring a larger conductor then it will naturally be larger than the minimum size needed if adjustments never applied so that part no longer really matters, other than you could run a smaller conductor part way and a larger conductor through the portion of the circuit where adjustments do apply.

 

[wwhitney]

My understanding is the reasoning for 125% for continuous loads was always for termination ratings of standard overcurrent devices, and the reason why you don't need to apply that 125% for 100% rated devices.

Yes, and this is why you apply the 125% continuous factor for the termination check, which uses the termination temperature tabular ampacity without any adjustment or correction. And when doing the "run of the wire" check, which uses the insulation temperature ampacity with adjustment and correction, no 125% continuous factor is required.

Cheers, Wayne

 

[NoahsArc]

No, it's not. If you read what I wrote, the first bolded text in that quote is my insertion; that bolded text is not in the NEC. But it is implied by the second bolded text.

Cheers, Wayne

Alright, so I still don't understand (per the strict reading of the 2020 at least) how "in accordance with 310.14" means "before the application of any adjustment or correction factors", when 310.14(A)(3)'s informational note tells you that (A)(3) more or less requires 310.15's deratings. I know INs aren't enforceable, but I'd never just wave that clear of one away anywhere else in the book; it's basically saying "hey you! we mean go to the next section right now!"

Am I just reading this badly, or is it worded badly? I'm open to it being an issue solely on my end, that's been the case before. I'm not surprised others struggle with it too.

To me, this is clear as mud how it's worded in the book, and it feels like one of those multiple choice questions that's almost impossible to answer due to how badly it's constructed.

 

[wwhitney]

This came up on the previous thread on this topic, but it is worth repeating. See Example D3(a) in the Annex, the section "Ungrounded Conductors". It goes through the two checks, which it calls "(1) terminations, and (2) conditions of use throughout the raceway run."

On the second check, no 125% continuous use factor is used. The check uses the "actual load VA" rather than the "total load VA" figure that includes a 125% continuous use factor. [Not the best terminology in my opinion.]

Cheers, Wayne

 

[wwhitney]

Alright, so I still don't understand (per the strict reading of the 2020 at least) how "in accordance with 310.14" means "before the application of any adjustment or correction factors", when 310.14(A)(3)'s informational note tells you that (A)(3) more or less requires 310.15's deratings.

Yeah, again, it's terrible wording. If you read 210.19(A)(1)(a) in isolation, the language obviously would mean that you do apply correction and adjustment factors.

But you have to read 210.19(A)(1)(a) in context and compare the language to 210.19(A)(1)(b). Part (a) refers to 310.14; part (b) refers to 310.15, so something is intended by this difference. And part (b) says "after the application of any adjustment or correction factors," while part (a) does not have that language.

So in comparison, the upshot is that for (a) you are not required to apply adjustment and correction factors. Just terribly worded.

Cheers, Wayne

 

[wwhitney]

OK, here it is straight from the CMP's mouth, although I don't know why the CMP can't write the code section to properly reflect what they say in their committee statement.

Some background: the 2020 NEC added the references to 310.14 in 210.19(A)(1)(a) and 310.15 in 210.19(A)(1)(b). In the first draft report, they both actually referred to 310.15(B). Which was incorrect, as 310.15(B) includes ampacity adjustment and corrections, and so would have required adjustment and correction for both (a) and (b), rendering section (b) moot.

In response, Don submitted Public Comment 652-NFPA 70-2018 on the First Draft report to change (b) to say "The minimum branch-circuit conductor size shall have an a table ampacity not less than the maximum load to be served." This was actually backwards, as (a) is supposed to use the table ampacity at the termination temperature, while (b) is supposed to use the ampacity after adjustment and correction.

So the CMP rejected the Public Comment, with the statement (emphasis mine): "The existing language provides the necessary clarity. Section 210.19 (A)(1)(a) does not use the correction and adjustment factors. Incorporating this change would imply that the adjustment and correction factors would be used when calculating continuous and non continuous loads mentioned in 210.19(A)(1)(a)."

As to both sections referring to 310.15(B), this was changed for the Second Draft Report to the final 2020 NEC language by Second Correlating Revision No. 13-NFPA 70-2019. The committee statement on this change is uninformative: "The Correlating Committee revised the language in 210.19(A)(1)(a) from 310.15(B) to 310.14 for correlation. The Correlating Committee also revised 210.19(A)(1)(b) from 310.15(B) to 310.15 for correlation."

Cheers, Wayne

 

[Dennis Alwon]

You have me confused also.

a.) I have a 25 amp load that would be 31.25 after the 125% for continuous load was added.

b.) I have the same 25 amp load but I have an adjustment factor of 70% because of number of conductors in a conduit. 25/.7 = 35.7 amps

This means I need a conductor good for 35.7 amps

Is this what you all are saying.--- nec 2020

 

[wwhitney]

You have me confused also.

You basically have it correct, but let me elaborate on your answers:

a.) I have a 25 amp load that would be 31.25 after the 125% for continuous load was added.

If you have a 25A continuous load, then 210.19(A)(1)(a) (and the similar sections for feeders and SECs) says that at the terminations you need an unadjusted and uncorrected ampacity of at least 31.25A. That would be using the table column for the termination temperature, typically 75C, occasionally 60C.

b.) I have the same 25 amp load but I have an adjustment factor of 70% because of number of conductors in a conduit. 25/.7 = 35.7 amps
This means I need a conductor good for 35.7 amps

If you have a 25A load, continuous or non-continuous, doesn't matter, then 210.19(A)(1)(b) (and the similar sections for feeders and SECs) says that you need a conductor with a final ampacity of 25A. With a 0.7 adjustment factor and no temperature correction, that means you can use the table column for the conductor insulation temperature, typically the 90C column, and look for a conductor that is at least 35.7A in that column.

Putting those two together for a 25A continuous load, 75C termination temperature, and 90C insulation temperature, then 210.19(A)(1) (and the similar sections for feeders and SECs) says that you need a conductor whose 75C column entry is at least 31.25A, and whose 90C column entry is at least 35.7A.

However, for a continuous 25A load, you also need a 35A OCPD. So 240.4 also requires that you have a final ampacity of at least 30.5A so you can apply 240.4(B). That means that you also need a conductor whose 90C column entry is at least 30.5 / 0.7 = 43.6A. This requirement is stricter than 210.19(A)(1)(b), so the final answer is a conductor whose 75C column entry is at least 31.25A, and whose 90C column entry is at least 43.6A.

This procedure never requires doing the computation 25A * 125% / 0.7 = 44.6A. That figure is irrelevant.

Cheers, Wayne

 

[NoahsArc]

So greatest of:
Termination sizing (usually 75c) @125% no AF
Wire sizing (usually 90c) @100% x AF
OCPD sizing (next up) @125% -> resulting ocpd wire size minimum (lower size + 0.5a) x AF
?

I do wonder if we couldn't simplify some of the code sometimes, especially when the committee itself gets tripped up on its own proposed verbiage changes...