210.19(A)

P

Pinnie

Senior Member
Location
Ohio
Occupation
Commercial Electrician
Struggling with this code rule. You have to use the greater of either the (100% non-continuous+ 125% continuous load) or (the ampacity after applying 310.15).

(1) is the load ampacity being raised due to conditions

(2) is wire ampacity being raised or lowered due to conditions

Say you have a 27 amp continuous load

(1) = 34amps
Got it

(2) 10awg is 35 at 75C but then you run into 240.4(d). So go up to 8awg.

It seems (2) is determined based on (1), which is what I always thought anyway. Where does the “greater of the two” come in?
 
(1) means take 125% of the continuous load plus 100% of the non-continuous load and compare that to the 310.16 table entry for the column corresponding to the termination temperature rating.

(2) means take 100% of the load and compare that to the 310.16 - 310.20 table entry based on the insulation temperature rating, adjusted and corrected for number of CCCs and ambient temperature.

Now for a 27A continuous load, you will need a minimum 27*125% = 34A breaker (next standard size is 35A). So Article 240 already tells you #8 is your smallest allowable size (unless this is a motor load or some other 240.4(G) application). Even at 60C passes test (1), 40A > 34A. So test (2) will only require you to upsize if the adjustment and correction is very significant. With 90C insulation, #8 has a table ampacity of 55A, and so your adjustment and correction factors together would have to be less than 0.5.

Cheers, Wayne
 
wwhitney said:
(1) means take 125% of the continuous load plus 100% of the non-continuous load and compare that to the 310.16 table entry for the column corresponding to the termination temperature rating.

(2) means take 100% of the load and compare that to the 310.16 - 310.20 table entry based on the insulation temperature rating, adjusted and corrected for number of CCCs and ambient temperature.

Now for a 27A continuous load, you will need a minimum 27*125% = 34A breaker (next standard size is 35A). So Article 240 already tells you #8 is your smallest allowable size (unless this is a motor load or some other 240.4(G) application). Even at 60C passes test (1), 40A > 34A. So test (2) will only require you to upsize if the adjustment and correction is very significant. With 90C insulation, #8 has a table ampacity of 55A, and so your adjustment and correction factors together would have to be less than 0.5.

Cheers, Wayne
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So for option 1 you don’t derate to find the answer, you just 125% continuous + 100 noncontiuous.

And option 2 you don’t have to take into account continuous or not you just derate?

(i know “derate” is old hat)
 
Pinnie said:
So for option 1 you don’t derate to find the answer, you just 125% continuous + 100 noncontiuous.

And option 2 you don’t have to take into account continuous or not you just derate?

(i know “derate” is old hat)
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(1) is for ensuring the conductor is large enough for the load. Continuous loads can increase the amount of heat a conductor needs to endure. So you multiply the load amps by 125% and add 100% of the noncontinuous.

(2) is to ensure that the environment the wire is in does not put more heat on the conductor than it can handle. This only relates to the load because the wire needs to withstand number 1 as well. So you take the insulation rating (60°, 75°, 90° or whatever) amapacity and derate it based on the enviromental conditions. A common condition is the number of current carrying conductors. Another is where ran strapped to the roof.

At the end of the day, both (1) and (2) are to be taken into consideration. But (1) helps your select your wire size for the load and (2) provides reasons to increase the conductor size based on environmental concerns.
 
What confuses me is in option 2 you can change the adjustment and correction factors by choosing different wire or raceway. Or by choosing a different route through different temperature environments. So I get option 1. 34a. Easy. Option 2 this is my thought process: “well what would my 8awgs derate to if I use this raceway here that has 4 CCCs already. 310.15(C)(1) would bring the ampacity down to 80% of 55a. So 44a. Okay that’s more than I need. So let’s find that smallest wire that can achieve 27 amps at 80% ampacity. Go to 310.16. Go to the next size down. 10 AWG is 40a at 90C. 40 x .8 = 32a. Okay so does that mean my first answer is bigger and now I need a wire that can do 34 amps?”

Where is my misstep?
 
Am I wrong in thinking you size the wire based on the result of option 1 or option 2? Like they will both provide different ampacities and that I’ll size my wire to the bigger of the two?
 
Pinnie said:
Am I wrong in thinking you size the wire based on the result of option 1 or option 2? Like they will both provide different ampacities and that I’ll size my wire to the bigger of the two?
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Each of option 1 and option 2 will give you different minimum wire sizes. Then you must pick a wire size at least as large as the larger of the two answers (*). Only option 2 is actually computing ampacity (any usages of the word "ampacity" in (1) in the code are misuses). Option 1 is calculating what we could call the "termination size".

Cheers, Wayne

(*) With continuous loads, sometimes you have to go larger because of 210.20 and 240.4. E.g. a 100A continuous load requires a 125A breaker, and then 240.4(B) says the wire ampacity (option 2) must be at least 111A, which is more than the 100A that (2) requires.
 
wwhitney said:
Each of option 1 and option 2 will give you different minimum wire sizes. Then you must pick a wire size at least as large as the larger of the two answers (*). Only option 2 is actually computing ampacity (any usages of the word "ampacity" in (1) in the code are misuses). Option 1 is calculating what we could call the "termination size".

Cheers, Wayne

(*) With continuous loads, sometimes you have to go larger because of 210.20 and 240.4. E.g. a 100A continuous load requires a 125A breaker, and then 240.4(B) says the wire ampacity (option 2) must be at least 111A, which is more than the 100A that (2) requires.
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The way it’s worded is really tripping me up. Man the code is hard to learn I feel dumb. I don’t know which words they are using loosely and ones they actually hold to their definitions in this code rule.

“(A) General. Branch circuit conductor shall have an ampacity, not less than the larger of the following and comply with a 110.14 (C) for equipment terminations:”

This implies both of the following options will provide an ampacity that you size a conductor to.

“(1) where a branch circuit supplies, continuous loads or any combination of continuous and noncontinuous loads the minimum branch circuit conductor size shall have ampacity not less than the noncontinuous load plus 125 percent of the continuous load in accordance with 310.14”

This is self explanatory. The result is an ampacity minimum that the conductor needs to withstand.

“(2) the minimum branch circuit conductor shall have an ampacity not less than the maximum load to be served after the application of any adjustment or correction factors with 310.15”

This just does not jive with my brain. The maximum load to be served is just the load. So this literally says the conductor shall have an ampacity of the load after derating. So the load is what you size the wire to. That’s it. 30a load, 30 amp wire. I have to derate no matter what. Option 1 will always be greater based on this wording.
 
Pinnie said:
What confuses me is in option 2 you can change the adjustment and correction factors by choosing different wire or raceway. Or by choosing a different route through different temperature environments. So I get option 1. 34a. Easy. Option 2 this is my thought process: “well what would my 8awgs derate to if I use this raceway here that has 4 CCCs already. 310.15(C)(1) would bring the ampacity down to 80% of 55a. So 44a. Okay that’s more than I need. So let’s find that smallest wire that can achieve 27 amps at 80% ampacity. Go to 310.16. Go to the next size down. 10 AWG is 40a at 90C. 40 x .8 = 32a. Okay so does that mean my first answer is bigger and now I need a wire that can do 34 amps?”

Where is my misstep?
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Your first step is correct. You would use #8. You would then just verify that #8 is good for the environment and do any adjustments required to the selected #8. You would not start over. You start with one to get a conductor size and verify that size still works for the physical environment you are installing that conductor in.

You would not try to do them separately and pick the smallest.

They both have to work. You can not have 30 current carry conductors and use #8. Then your environment is forcing you to increase the wire size to get to the required ampacity of 34A.
 
Elect117 said:
the required ampacity of 34A.
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For a 27A continuous load, 210.19(A) tells us the minimum (actual) ampacity is 27A. While 210.20 tells us the minimum breaker size is 34A, so a 35A breaker. For which 240.4(B) tells us the minimum ampacity is 31A.

So the required ampacity is 31A. The only thing that needs to be at least 34A is the termination check, which is not actually an ampacity.

Cheers, Wayne
 
Pinnie said:
I don’t know which words they are using loosely and ones they actually hold to their definitions in this code rule.
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That's a definite flaw in the code language, so it's no wonder you find it confusing, it is confusing.

Pinnie said:
Option 1 will always be greater based on this wording.
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In terms of Amps, yes, Option 1 will always be greater. But in terms of how you convert Amps to a physical wire size, the procedure is different for Option 1 and Option 2. Option 1 does not require derating, while Option 2 does. [This is a flaw in the language, it is only implicitly communicated by mentioning derating in the text of Option 2, while omitting any mention of it in the text of Option 1.] So even though the Amp value in Option 2 will not exceed Option 1, the physical wire size required by Option 2 may exceed that required by Option 1, due to derating.

Cheers, Wayne
 
wwhitney said:
For a 27A continuous load, 210.19(A) tells us the minimum (actual) ampacity is 27A. While 210.20 tells us the minimum breaker size is 34A, so a 35A breaker. For which 240.4(B) tells us the minimum ampacity is 31A.

So the required ampacity is 31A. The only thing that needs to be at least 34A is the termination check, which is not actually an ampacity.

Cheers, Wayne
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I have always included it. After reading it back a couple of times, I realize you are right on the money. I was using the calculated load with continuous ratings when comparing to adjustment factors. Which I guess raises a question to me about why the impact of adjustment factors is separate from the impact of continuous use. I wonder if it has to do with the failure of the conductor's insulation in conduit due to the lack of de-rating and separately the failure of protecting the conductor due to the nuisance overloading of a continuously ran load?
 
Elect117 said:
Which I guess raises a question to me about why the impact of adjustment factors is separate from the impact of continuous use.
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Ampacity is a continuous rating. So for the wire itself, there is no need to use a 125% factor.

But equipment is more varied, and it may not be the case that equipment that requires a continuous current has been designed to allow its supply conductors to be sized at only 100% of the required current. If the equipment is tested with such sized supply conductors, it may be that the terminal temperature exceeds 90C. In which case the equipment requires a larger wire size at the termination to reduce the terminal temperature.

So the default is to size the wire at the termination at 125% of the continuous current; if the equipment has been successfully tested with wires sized at 100% of the continuous current, it will be so labeled, and the exception allows us to omit the 125% factor at that termination.

Cheers, Wayne
 
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I don’t understand why derating is brought into it at all. If I size using option 1 and then run the conductors in an environment with a temperature higher that 86F I have to derate. It’s a separate issue. We should size the wire ampacity based off the load, then if the conditions require derating, that ampacity needs satisfied after any derating. Why is it so complicated.
 
Pinnie said:
I don’t understand why derating is brought into it at all.
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Imagine we always install a junction box butted up against the equipment termination compartment. In that case we could size the wire going from the junction box to the equipment just based on option 1 (at least 210.19(A)-wise, there's still 210.20 and 240.4). There's no derating required because there's no raceway involved. And we could size the wire going from the junction box to the junction box at the other end (through a raceway or cable) just based on option 2 (again 210.19(A)-wise). Here derating is involved because of the raceway/cable.

But if we don't want to have all those junction boxes and all those splices, and just want to use a single piece of wire from termination to termination, then we need to pick our wire size to be the larger of the two sizes above.

Cheers, Wayne
 
So is option one the termination rating “ampacity” and option 2 is the “middle of the wire ampacity like in 215.2(A)(1) Ex.2?
 
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