Sizing Conductors

[Charles R. Miller]

Sorry I get them both and got it wrong.

That's okay Dennis, I was just kidding. Besides, it was another way to shamelessly plug the magazine and my article.

 

[Smart $]

...

I don't care if they accept my wording; I just want the references to be clear and understandable. I would like to know other thoughts on this.

IMO, to help with clarity, remove the word allowable from all ampacity statements. This word modifies the defined term ampacity in an undefined manner.

If we have to adjust and correct a table value for the condiition(s) of use, then the resulting value of that adjustment and correction would be the allowable ampacity, not the value listed in the table.

If only the table value should be considered, then the combined term allowable ampacity should not be used.

Clarity would come from using wording such as "the ampacity value in the appropriate table for conductor type and condition of use, or as determined per engineering supervision per 310.15(C) or 310.60(D) as applicable".

 

[Dennis Alwon]

I just went thru 230,42 and the CMP seems to be saying that service conductors should have the 125% plus any derating. Their response seems off on this.

 

[Smart $]

To satisfy 110.14(C) and 210.19(A)(1),

Section 110.14(C) is only for terminations. Section 210.19(A)(1) provides the wording for continuous loads. After multiplying the load by 125 percent, select a conductor from the column as determined by 110.14(C). This is also the way the calculation is performed in Example D3(a). You could select a conductor from the 90?C column (using this example), but then you will have to turn around and select the conductor from the 75?C column when determining the minimum size conductor at the overcurrrent device termination.

110.14(C) says nothing about continuous loads factored to 125%, nor does it mention any of the sections associated with 125% factoring.

To satisfy 110.14(C) for a 75?C termination limitation, and using your example of 39A continuous load, we only have to choose the smallest size from the 75?C column of 310.15(B)(16) having an ampacity value not less than the maximum load of 39A. We do not have to factor the 39A to 125% and base the minimum size to 48A. In this case, it just happens to be an 8AWG copper conductor either way. But if we were using aluminum, 8AWG would also suffice... but not if you use 125% continuous load factoring. That would raise the minimum size for an aluminum conductor to 6AWG.

 

[kwired]

I mentioned earlier that I do not like the wording in 210.19(A)(1). It is obvious by the replies to this one question that this section is difficult to understand. Although I have a great job writing and teaching the NEC, I don't like the Code being difficult to understand. Because of this, I submitted a proposal to change the wording of this section. For this section, my proposal was accepted in principal and I like the change that was made. Here is the draft copy of this section:

210.19(A)(1) General. Branch-circuit conductors shall have an ampacity not less than the maximum load to be served. Conductors shall be sized to carry not less than the larger of (a) or (b). [ROP 2?131]
(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. [ROP 2?131]
(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. [ROP 2?131]

I think the new wording is easy to understand and is what the Code is trying to say. Although it was accepted at the proposal stage, it may be rejected in the comment stage. This is what the Technical Correlating Committee (TCC) said about the proposal:

TCC Action: It was the action of the Correlating Committee that this proposal be reconsidered and correlated with the actions taken on Proposals 2-201 and 2-202. The Correlating Committee directs that the panel clarify whether the 125 percent is applied before or after the correction factors for consistency. The Correlating Committee also directs that this proposal be submitted to Code-Making Panel 6 for comment. This action will be considered as a public comment.

I submitted a similar proposals for feeder conductors in 215.2(A)(1) and, like the one for branch-circuit conductors, it was accepted in principal. I also submitted one for service conductors in 230.42 and it was rejected. Here is the response from the committee:

Panel Statement: The current text is clear that conditions of use must be considered in addition to continuous loading.

Even their statement is not clear. Do they mean to calculate the conditions of use along with the continuous loads or perform the calculations separate?

I don't care if they accept my wording; I just want the references to be clear and understandable. I would like to know other thoughts on this.

Your proposals are exactly what I have have been taught and have always understood is the way it is done, and states it in a way that is easier to understand than what has been in print for years.

I was taught when selecting a conductor size the part a of your proposal is when we are selecting a conductor based on the termination temperature rating. You will never select a conductor smaller than what you come up with part a. Part b of your proposal is selecting a conductor based on the insulation temperature rating, the temp away from the termination is not necessarily going to be same and in fact will be higher in cases where there is multiple conductors in a raceway, crossing a high ambient area, etc. plus conductor insulation typically is 90^oC where terminal temperature is only 60^oC or 75^oC. There are cases where low ambient will actually allow increased ampacity instead of decreased. If result of b allows for smaller conductor than a you must still have a conductor large enough to satisify terminal temperature rating. If you were sizing a 75^oC conductor to land on 75^oC terminals there really wouldn't be two results to pick from, first step you would select a 75^oC, second step you would see if it still has sufficient ampacity after deration adjustments, but it will never be possible for it to be smaller than it was in the first step.

 

[Charles R. Miller]

110.14(C) says nothing about continuous loads factored to 125%, nor does it mention any of the sections associated with 125% factoring.

To satisfy 110.14(C) for a 75?C termination limitation, and using your example of 39A continuous load, we only have to choose the smallest size from the 75?C column of 310.15(B)(16) having an ampacity value not less than the maximum load of 39A. We do not have to factor the 39A to 125% and base the minimum size to 48A. In this case, it just happens to be an 8AWG copper conductor either way. But if we were using aluminum, 8AWG would also suffice... but not if you use 125% continuous load factoring. That would raise the minimum size for an aluminum conductor to 6AWG.

I mentioned that the overcurrent device is a factor when selecting a conductor. Multiplying the continuous loads by 125 percent and then selecting a conductor from the 75?C column is a shortcut. This shortcut is for the provision in 210.20(A).

210.20 Overcurrent Protection. Branch-circuit conductors and equipment shall be protected by overcurrent protective devices that have a rating or setting that complies with 210.20(A) through (D).
(A) Continuous and Noncontinuous Loads. Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the rating of the overcurrent device shall not be less than the noncontinuous load plus 125 percent of the continuous load.

If we are sizing aluminum conductors (as you mentioned), 8 AWG conductors would not be permitted. We can select an 8 AWG aluminum conductor because it has an allowable ampacity of 40 amperes in the 75?C column. But we must also consider 210.20(A) and 240.4. In accordance with 210.20(A), the overcurrent device must not be less than 125 percent of the continuous load. The rating of the fuse or breaker must be at least 49 amperes (39 ? 125% = 48.75 = 49). Since 49 is not a standard rating, we must use a 50-ampere fuse or breaker. Looking at the conductor we selected, it only has an allowable ampacity (out of the 75?C column) of 40 amperes. Because the 8 AWG aluminum conductor has an ampacity of 40 amperes, it must be protected with a 40 ampere overcurrent device. Therefore we must select a 6 AWG conductor, which has an ampacity of 50 amperes.

 

[kwired]

I mentioned that the overcurrent device is a factor when selecting a conductor. Multiplying the continuous loads by 125 percent and then selecting a conductor from the 75?C column is a shortcut. This shortcut is for the provision in 210.20(A).

210.20 Overcurrent Protection. Branch-circuit conductors and equipment shall be protected by overcurrent protective devices that have a rating or setting that complies with 210.20(A) through (D).
(A) Continuous and Noncontinuous Loads. Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the rating of the overcurrent device shall not be less than the noncontinuous load plus 125 percent of the continuous load.

If we are sizing aluminum conductors (as you mentioned), 8 AWG conductors would not be permitted. We can select an 8 AWG aluminum conductor because it has an allowable ampacity of 40 amperes in the 75?C column. But we must also consider 210.20(A) and 240.4. In accordance with 210.20(A), the overcurrent device must not be less than 125 percent of the continuous load. The rating of the fuse or breaker must be at least 49 amperes (39 ? 125% = 48.75 = 49). Since 49 is not a standard rating, we must use a 50-ampere fuse or breaker. Looking at the conductor we selected, it only has an allowable ampacity (out of the 75?C column) of 40 amperes. Because the 8 AWG aluminum conductor has an ampacity of 40 amperes, it must be protected with a 40 ampere overcurrent device. Therefore we must select a 6 AWG conductor, which has an ampacity of 50 amperes.

Overcurrent device has nothing to do with selecting minimum size of conductor needed (under 800 amps anyway).

You have a load you want to supply, the first thing you need to figure out is what size conductor is needed, then you look at overcurrent protection. A conductor with a derated ampacity must be protected according to that derated ampacity not to the values in the 310.15 ampacity tables. (next standard size higher is permitted up to 800 amps)

 

[Smart $]

I mentioned that the overcurrent device is a factor when selecting a conductor. Multiplying the continuous loads by 125 percent and then selecting a conductor from the 75?C column is a shortcut. This shortcut is for the provision in 210.20(A).

I understand it is a shortcut... and I also forgot to mention in my previous post that I'm aware this is how it is done in the D3(a) Example. It is mislabeled as pertaining to 210.19(A)(1), when it pertains to how 210.20 Overcurrent Protection (OCP) may affect the minimum conductor size determination. Nonetheless, it is not a proper application of the requirements.

First, kwired is correct regarding the ampacity value to which the OCP rule applies, namely the "derated" value, not the table value in the 75?C column. We are determining protection of conductors, not device terminals. Additionaly, there is no requirement that the allowable ampacity [i.e. derated ampacity] for 210.20(A) be determined using Table 310.15(B)(16) values. Not all installations, wire types, and conditions of use are determined using that table's values.

Kwired's is somewhat correct on OCP rating having nothing to do with minimum wire size... but that is not entirely correct. Permitting a higher standard rating for under 800 ampere OCP does, in many instances, provide some leeway... but not all. In summation, there are no shortcuts.

 

[Smart $]

I mentioned that the overcurrent device is a factor when selecting a conductor. Multiplying the continuous loads by 125 percent and then selecting a conductor from the 75?C column is a shortcut. This shortcut is for the provision in 210.20(A).

...

I believe I have found the root of misconception. 110.14(C)(1)(b)(2) states:

Conductors with higher temperature ratings, provided
the ampacity of such conductors does not exceed the
75?C (167?F) ampacity of the conductor size used, or
up to their ampacity if the equipment is listed and identified
for use with such conductors

Misconception results from red-highlighted text... but the green-highlighted text is a viable exception. However, note this is regarding "Termination provisions of equipment for circuits rated over 100 amperes, or marked for conductors larger than 1 AWG". Example D3(a) falls into this category, and nothing was stated regarding the equipment listing and identification for the purpose of using conductors up to their ampacity. Such forces using the default 75?C-size table ampacity. However, it should be noted that there are instances where the "derated" conductor ampacity is less than the 75?C-column value. That alone requires checking for the lesser of the two, rather than automatically assuming the table value to be the lower of the two.

However... the example presented for this thread is not subject to that requirement... and we are assuming the equipment provisions fall under:

110.14(C)(1)(a)(3) Conductors with higher temperature ratings if the equipment is listed and identified for use with such conductors.

Anyway, I rewrote my earlier method to include aluminum conductors and check for OCP affecting the minimum size...

110.14(C)(1)Termination Limitations
39A load
75?C terminal ratings
minimum 8AWG Cu conductor required (50A; 10AWG is 35A)
minimum 8AWG Al conductor required (40A; 10AWG is 30A)

210.19(A)(1) Conductors — Minimum Ampacity and Size
39A continuous load
39A ? 125% = 48.8A
8AWG minimum 90?C THHN Cu (55A; 10AWG is 40A)
6AWG minimum 90?C THHN Al (55A; 8AWG is 45A)

310.15(B)(2)&(3) Ambient Temperature Correction and Adjustment Factors
39A load
40?C ambient temperature correction factor for THHN is 0.91
8 current-carrying conductors in raceway adjustment factor is 70% (0.7)
39A ? 0.91 ? 0.7 = 61.2A minimum table ampacity required
6AWG minimum 90?C THHN Cu (75A; 8AWG is 55A)
4AWG minimum 90?C THHN Al (75A; 6AWG is 55A)

210.20 Overcurrent Protection: (A) Continuous and Noncontinuous Loads & (B) Conductor Protection.
OCP rating ≥ 125% continuous load + (100%) non-continuous load
39A continuous load
39A ? 125% = 48.8A minimum OCP rating required
Minimum OCPD standard rating ≥ 48.8A is 50A [240.4(B), 240.6(A)]
Next lower standard rating is 45A [240.6(A)]
Adjusted and corrected conductor ampacity must be ≥ 46A [240.4(B)]

6AWG is the largest of the required minimum sizes for 90?C THHN Cu
75A ? 0.91 ? 0.7 = 47.8A
Meets above OCP conditions: YES

4AWG is the largest of the required minimum sizes for 90?C THHN Al
75A ? 0.91 ? 0.7 = 47.8A
Meets above OCP conditions: YES

 

[kenaslan]

I need a drink after reading all the posts. The code writer must have been an atty in a former life.

 

[ggunn]

I need a drink after reading all the posts. The code writer must have been an atty in a former life.

In a related story...

I remember reading a story years ago about a lawyer new to an area where people started calling his office asking for "Fatty". He had had his phone number listed with his last name and first initial with an indication of his profession, and when he looked himself up in the phone book (remember phone books?) he saw what was happening. Whoever had entered his listing had left a crucial space out of it so it read "Jones, Fatty".

 

[A-1Sparky]

What if this load were noncontinuous? Wouldn't you apply the same correction factors (.91 and .7) to the 39 amp load? And end up with the same calculations and same conductor size (#6)? So, why use the 39 amp load instead of the 39x1.25 (48.75) for calculations for a continuous load? To me, that doesn't make sense. Maybe I'm just not looking at it the right way.

 

[Dennis Alwon]

What if this load were noncontinuous? Wouldn't you apply the same correction factors (.91 and .7) to the 39 amp load? And end up with the same calculations and same conductor size (#6)? So, why use the 39 amp load instead of the 39x1.25 (48.75) for calculations for a continuous load? To me, that doesn't make sense. Maybe I'm just not looking at it the right way.

The reason is that you don't need to derate twice. If the continuous load is covered by the other de-rating factors then their is no issue. The correction factor for temp and wire fill automatically covers the 1.25 needed for continuous load or vice versa.

 

[kwired]

What if this load were noncontinuous? Wouldn't you apply the same correction factors (.91 and .7) to the 39 amp load? And end up with the same calculations and same conductor size (#6)? So, why use the 39 amp load instead of the 39x1.25 (48.75) for calculations for a continuous load? To me, that doesn't make sense. Maybe I'm just not looking at it the right way.

The 125% factor for continuous loads is for termination temperature and not for conductor insulation temperature. The conductor insulation can handle 100% load at its temperature rating. The overcurrent device sinks heat into the conductor - they are designed to do this, and that is why 125% of continuous load is required to be used for selecting ampacity. See the exceptions that say a breaker designed for loading to 100% can have conductors selected at 100% of continuous load. You will not find this type of breaker in typical miniature breakers installed in "load centers" so most people most of the time need to just assume they are dealing with a non-100% continuous rated breaker. Common fuses for general purposes are the same way.

 

[A-1Sparky]

The reason is that you don't need to derate twice. If the continuous load is covered by the other de-rating factors then their is no issue. The correction factor for temp and wire fill automatically covers the 1.25 needed for continuous load or vice versa.

Thanks for the clarification, Dennis and kwired. I agree that the wording needs to be changed in the NEC to lessen the confusion.

 

[Smart $]

What if this load were noncontinuous? Wouldn't you apply the same correction factors (.91 and .7) to the 39 amp load? And end up with the same calculations and same conductor size (#6)? So, why use the 39 amp load instead of the 39x1.25 (48.75) for calculations for a continuous load? To me, that doesn't make sense. Maybe I'm just not looking at it the right way.

Perhaps because the determination of ratings is, shall we say, interactive over several aspects of an installation. The following is a rewrite of Post #49 changed to 39A noncontinuous load (changes indicated with red text). Take special note of the OCP rating determination.

110.14(C)(1)Termination Limitations
39A load
75?C terminal ratings
minimum 8AWG Cu conductor required (50A; 10AWG is 35A)
minimum 8AWG Al conductor required (40A; 10AWG is 30A)

210.19(A)(1) Conductors ? Minimum Ampacity and Size
39A noncontinuous load
39A ? 100% = 39A
10AWG minimum 90?C THHN Cu (40A; 12AWG is 30A)
8AWG minimum 90?C THHN Al (45A; 10AWG is 35A)

310.15(B)(2)&(3) Ambient Temperature Correction and Adjustment Factors
39A load
40?C ambient temperature correction factor for THHN is 0.91
8 current-carrying conductors in raceway adjustment factor is 70% (0.7)
39A ? 0.91 ? 0.7 = 61.2A minimum table ampacity required
6AWG minimum 90?C THHN Cu (75A; 8AWG is 55A)
4AWG minimum 90?C THHN Al (75A; 6AWG is 55A)

210.20 Overcurrent Protection: (A) Continuous and Noncontinuous Loads & (B) Conductor Protection.
OCP rating ≥ 125% continuous load + (100%) noncontinuous load
39A noncontinuous load
39A minimum OCP rating required
Minimum OCPD standard rating ≥ 39A is 40A [240.4(B), 240.6(A)]
Next lower standard rating will not affect determination for a noncontinuous only load
Adjusted and corrected conductor ampacity must be ≥ 39A [210.19(A)(1)]

6AWG is the largest of the required minimum sizes for 90?C THHN Cu
75A ? 0.91 ? 0.7 = 47.8A
Meets above OCP conditions: YES

4AWG is the largest of the required minimum sizes for 90?C THHN Al
75A ? 0.91 ? 0.7 = 47.8A
Meets above OCP conditions: YES

45A and 50A OCPD rating also acceptable

 

[mlnk]

So 210.19 should say: "separate from the application of any adjustment..." instead of "before application of any adjustment..." Think of: turn off power before you get electrocuted. Not: sift the flour before adding the egg.
But i still like the idea of derating conductors: 75 ampacity * .7 = 52.5 ampacity (derated)
instead of taking a load of say, 20 amps dividing by .7 to get 28.6 which is the load that you should pretend you have in order to correct for conditions of use. Even if you get the same answer