Overcurrent Protection Conductor Bundling Clarification

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eeRyanC

Member
Location
Seattle, WA, USA
Hello,

I am having a difficult time with an example provided with the '08 NEC handbook under article 210.20(A) (I do not have a previous example of the handbook to check how long this example has been around). The example states to determine the min. size overcurrent protective device and min. conductor size for the following ckt.: 25A continuous load, 60 deg. C overcurrent device terminal rating, THWN conductors, and 4 current, carrying cu. conductors in a raceway.

I don't understand the methodology in the solution where the bundled conductors are accounted for. In the example, the calculated load of the circuit is adjusted by the .8 derating factor to get 25A/.8 = a conductor ampacity of 31.25A. My understanding was that the bundling derating had to be applied to the conductor ampacity (not limited by termination rating) rather than the calculated load. Doesn't the derating have to be figured into the conductor ampacity in order to protect the conductors against overcurrent protection per 240-4? In this particular example, the resultant conductor size is still correct because the THWN insulators are rated at 75 deg which yields 50*.8 = 40A which is greater than the 35A breaker. However, if type TW or UF insulators were used, we would get 40*.8 = 32A which would not be adequate. Is this correct?

-Ryan
 

chris kennedy

Senior Member
Location
Miami Fla.
Occupation
60 yr old tool twisting electrician
In this particular example, the resultant conductor size is still correct because the THWN insulators are rated at 75 deg which yields 50*.8 = 40A which is greater than the 35A breaker. However, if type TW or UF insulators were used, we would get 40*.8 = 32A which would not be adequate. Is this correct?

-Ryan

You should start derating at 90?C for both THWN and UF. Where UF cable is installed as nonmetallic-sheathed cable, the ampacity of Type UF cable is determined according to 334.80, as stated in 340.10(4).
334.80 Ampacity.
The ampacity of Types NM, NMC, and NMS cable shall be determined in accordance with 310.15. The ampacity shall be in accordance with the 60?C (140?F) conductor temperature rating. The 90?C (194?F) rating shall be permitted to be used for ampacity derating purposes, provided the final derated ampacity does not exceed that for a 60?C (140?F) rated conductor.

110.14(C) Temperature Limitations. The temperature rating associated with the ampacity of a conductor shall be selected and coordinated so as not to exceed the lowest temperature rating of any connected termination, conductor, or device. Conductors with temperature ratings higher than specified for terminations shall be permitted to be used for ampacity adjustment, correction, or both.
 

chris kennedy

Senior Member
Location
Miami Fla.
Occupation
60 yr old tool twisting electrician
I am having a hard time following you..........

Here is the example.


Calculation Example
Determine the minimum-size overcurrent protective device and the minimum conductor size for the following circuit:
? 25 amperes of continuous load
? 60?C overcurrent device terminal rating
? Type THWN conductors
? Four current-carrying copper conductors in a raceway
Solution
STEP 1.
Determine the size of the overcurrent protective device (OCPD). Referring to 210.20(A), 125 percent of 25 amperes is 31.25 amperes. Thus, the minimum standard-size overcurrent device, according to 240.6(A), is 35 amperes.
STEP 2.
Determine the minimum conductor size. The ampacity of the conductor must not be less than 125 percent of the 25-ampere continuous load, which results in 31.25 amperes. The conductor must have an allowable ampacity of not less than 31.25 amperes before any adjustment or correction factors are applied.
Because of the 60?C rating of the overcurrent device terminal, it is necessary to choose a conductor based on the ampacities in the 60?C column of Table 310.16. The calculated load must not exceed the conductor ampacity. Therefore, an 8 AWG conductor with a 60?C allowable ampacity of 40 amperes is the minimum size permitted. Conductors with a higher allowable ampacity based on their insulation temperature rating may be used, but only at a 60?C allowable ampacity.
STEP 3.
Because there are four current-carrying conductors in the raceway, Table 310.15(B)(2)(a) applies. First, calculate the ampacity of the conductor using the ampacity value calculated above:

Conductor Ampacity=Calculated load?Percent adjustment factor per T310.15(B)(2)(a)

25A?0.8=31.25A

The conductor ampacity determined in Step 2 is not the calculated load that is used in this step. It is a calculation that is used to determine a minimum conductor ampacity and circular mil area to ensure sufficient dissipation of heat at the terminals of overcurrent protective devices supplying continuous loads. The calculated load is 25 amperes. The phrase ?before the application of any adjustment or correction factors? has the effect of not requiring that the conductor be subjected to a ?double-derating? where continuous loads are supplied by a conductor that is also subject to ampacity adjustment, because there are more than three current-carrying conductors in a cable or raceway or is subject to ampacity correction due to the installation being made where the ambient temperature exceeds 86?C.
 

eeRyanC

Member
Location
Seattle, WA, USA
My confusion is based on this: The example applies the 80% conductor bundling derating factor to the calculated circuit load. However, 310.15(2)(a) seems to indicate that the conductor bundling derating is to be applied to the circuit ampacity (not calculated load) in order to protect the conductors from overheating.

The example provides the following formula:
Conductor Ampacity = [ Calculated Load ] / [ Percent Adjustment Factor from Table 310.15(B)(2)(a) ]
My understanding based on 310.15(2)(a) was to be applied to the "Rated Circuit Load" as opposed to the "Calculated Load".
 

eeRyanC

Member
Location
Seattle, WA, USA
Chris - First of all, thanks for posting the example. Regarding your note with respect where to start derating types THWN and UF conductors, I think we are getting away from my question, but I will address them. Table 310.13(A) shows THWN rated for 75 deg. and shows UF rated for 60 deg. I am not really concerned with specific exceptions to the temperature rating for type UF conductors since I was only using them as an example of a conductor rated at 60 deg.

Thanks!
 

infinity

Moderator
Staff member
Location
New Jersey
Occupation
Journeyman Electrician
I agree THWN is only rated for 75 degrees so any adjustment would start from there.
 

erickench

Senior Member
Location
Brooklyn, NY
Ryan, it is because the ampacity is not given that you have to go in reverse and figure out what it is. If the load is 25A then you must divide it by .8 which comes up with 31.25A. The ampacity would have to be equal to or greater than this number. If you had been given a certain wire size, insulation, temperature, etc. then you would apply the derating factor to the ampacity as determined from NEC table 310.16. Lets suppose you were told to use say no. 8 THHW at normal temperature. The ampacity as determined from table 310.16 is 50A. You would then multiply this number by the derating factor .8 and come up with 40A. This is the maximum amount of current allowed. But because it is so much higher than 31.25A it's a bit of an overkill in terms of conductor sizing. The next lower size would be No. 10 rated at 35A.
 
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eeRyanC

Member
Location
Seattle, WA, USA
Erickench, I think we're getting on the right track. I agree that the solution attempts to work backwards. However, I do not agree with working backwards from the calculated load (as opposed to the circuit's rated load). Article 310.15(B)(2)(a) supplies a derating for conductor ampacity. If I have a 120V, 20A branch circuit with a calculated discontinuous load of 500W, working backwards from the calculated load will never force me to upsize my conductors (500/.35 = 1429W < 2400) when in fact a bundle of more than 6 current-carrying #12 THWN conductors would need to be up-sized (25*.70 = 17.5A < 20A). I hope this illustrates my confusion. Is this correct?

Thanks!
 

erickench

Senior Member
Location
Brooklyn, NY
I don't know what happened to my reply. Anyhow, the circuits rated load is the same as the protection which is determined after the conductor ampacity is calculated. This method is used only if the ampacity and protection is not known. Unless the NEC clearly specifies a certain branch circuit rating like for small appliance circuits which are 20A you would use this method. I would think the method is more applicable in the sizing of feeders since branch circuits tend to be standardized in some cases. For instance General Purpose Circuits are either 15A or 20A. You would have to use conductor sizes #14 and #12 respectively.
 

eeRyanC

Member
Location
Seattle, WA, USA
Eric - Not sure I follow you. Seems to me that in most cases (including branch circuits and feeders) the first thing that I want to size is the OCPD based on my calculated load. That gives me my rated circuit amps which allows me to size my conductors accordingly. Yes, you could start by sizing your conductors to meet your calculated load, select your OCPD, and then upsize your conductors as necessary, but why make life difficult?
 

benaround

Senior Member
Location
Arizona
Hello,

I am having a difficult time with an example provided with the '08 NEC handbook under article 210.20(A) (I do not have a previous example of the handbook to check how long this example has been around). The example states to determine the min. size overcurrent protective device and min. conductor size for the following ckt.: 25A continuous load, 60 deg. C overcurrent device terminal rating, THWN conductors, and 4 current, carrying cu. conductors in a raceway.

I don't understand the methodology in the solution where the bundled conductors are accounted for. In the example, the calculated load of the circuit is adjusted by the .8 derating factor to get 25A/.8 = a conductor ampacity of 31.25A. My understanding was that the bundling derating had to be applied to the conductor ampacity (not limited by termination rating) rather than the calculated load. Doesn't the derating have to be figured into the conductor ampacity in order to protect the conductors against overcurrent protection per 240-4? In this particular example, the resultant conductor size is still correct because the THWN insulators are rated at 75 deg which yields 50*.8 = 40A which is greater than the 35A breaker. However, if type TW or UF insulators were used, we would get 40*.8 = 32A which would not be adequate. Is this correct?

-Ryan


Ryan, In the example, 25a x 1.25 for continious load = 31.25a that is not a "bundling" derating.

Now we need OCPD for 31.25a and that is a 35a OCPD, OK, so far ?

Now conductor size for 31.25a with 4 conductors in a raceway = derate by 80% or 31.25/.8

and that is 39.06a . Look in the 60deg. column to find #8 AWG is rated for 40amps.

I hope that this helps.
 

eeRyanC

Member
Location
Seattle, WA, USA
I hate to drag this out, but I can't resist.
Ryan, In the example, 25a x 1.25 for continious load = 31.25a that is not a "bundling" derating.
Now we need OCPD for 31.25a and that is a 35a OCPD, OK, so far ?
Yep.
Now conductor size for 31.25a with 4 conductors in a raceway = derate by 80% or 31.25/.8 and that is 39.06a .
You lost me here. In order to protect the conductors, you need to work backwards from the OCPD rated current.
35/.8 = 43.75a
Look in the 60deg. column to find #8 AWG is rated for 40amps.
Lost me again. We are dealing with THWN conductors. I would check the 60 deg. column when checking the terminations, but we are checking the conductor runs here (hence the bundling derating). Therefore, I would Look in the 75deg. column to find #8 AWG is rated for 50amps which is clearly satisfactory.
Again, I don't have any problem with the example's result - it is the methodology I disagree with. Of course the whole reason for this post was because I wanted to check my understanding.
 

infinity

Moderator
Staff member
Location
New Jersey
Occupation
Journeyman Electrician
I agree with Benaround's calculations.

31.25 amps is the required conductor ampacity before derating. Since you have 4 CCC's you'll need to derate to 80% from the value listed in table 310.16 for THWN. One way of doing that is to divide you minimum conductor ampacity by 80% or multiply it by 125%. Either way you'll end up with needing a conductor rated for 39 amps. #8 THWN is listed as 50 amps but you're limited to it's 60 degree rating due to the terminal that it's attached to. This give you a #8 THWN with an ampacity of 40 amps. You could do the calculation in reverse to check the answer.
 

benaround

Senior Member
Location
Arizona
You lost me here. In order to protect the conductors, you need to work backwards from the OCPD rated current.
35/.8 = 43.75a

Lost me again. We are dealing with THWN conductors. I would check the 60 deg. column when checking the terminations, but we are checking the conductor runs here (hence the bundling derating). Therefore, I would Look in the 75deg. column to find #8 AWG is rated for 50amps which is clearly satisfactory.
Again, I don't have any problem with the example's result - it is the methodology I disagree with. Of course the whole reason for this post was because I wanted to check my understanding.

OK, We already know that 35a OCPD will be used, we need to determine the minimum

conductor size, and we know that the 60deg. column is required no matter what the

insulation (thwn) is rated for. You say, 75deg. to find #8 rated for 50a, but, you can only

use it at the 60deg. column rating, are you with me ??

Also, we are not protecting the conductors from the OCPD, we are protecting them from

the amount of current they are rated for, hence 31.25/.8=39.06a, a conductor rated over

39.06a is the minimum size conductor allowed, have I got you thinking ???
 

eeRyanC

Member
Location
Seattle, WA, USA
OK, We already know that 35a OCPD will be used, we need to determine the minimum

conductor size, and we know that the 60deg. column is required no matter what the

insulation (thwn) is rated for. You say, 75deg. to find #8 rated for 50a, but, you can only

use it at the 60deg. column rating, are you with me ??

Also, we are not protecting the conductors from the OCPD, we are protecting them from

the amount of current they are rated for, hence 31.25/.8=39.06a, a conductor rated over

39.06a is the minimum size conductor allowed,

I think I know what the disconnect is between our lines of thought. It looks to me like you are approaching your conductor ampacity sizing as one step, where as I break it into two separate steps.

At least we agree on the first step!
1. Size OCPD.
(25amp continuous load)*(1.25 cont. load factor) = 31.25amp
Next breaker size up = 35amp OCPD
2. Size conductors for 60 deg. terminal rating
31.25amp load --> at least #8 THWN required @ 60deg. [Note that I didn't apply the bundling derating when checking the terminal amps]
3. Check that conductors are protected by OCPD
(35amp rated circuit load) / (.8 conductor bundling derating) = 43.75amp load --> at least #8 THWN required @75deg.

It sounds to me like you are combining step 2 & 3, which may work for you on this problem, but won't work in all cases. The good news, your terminations are going to be ok, since you are overly protecting them by applying the bundling derating. The bad news is that if instead of using the 75deg. THWN, we use a conductor rated at 60deg., it looks to me like you are going to undersize your conductors and possibly start a fire - here's why:

Using your method, nothing will change since you were already limiting yourself to the 60deg. ampacity --> you select #8 conductors.
However, if my circuit load approaches the OCPD limit (i.e. equipment malfunctions, another device is added to circuit in remodel, etc) the effective continuous circuit amps will be as high as 43.75amps. Those #8 conductors are gonna be smokin!

have I got you thinking ???

You sure do.

Happy Thanksgiving folks!
 

infinity

Moderator
Staff member
Location
New Jersey
Occupation
Journeyman Electrician
I think I know what the disconnect is between our lines of thought. It looks to me like you are approaching your conductor ampacity sizing as one step, where as I break it into two separate steps.

At least we agree on the first step!
1. Size OCPD.
(25amp continuous load)*(1.25 cont. load factor) = 31.25amp
Next breaker size up = 35amp OCPD
2. Size conductors for 60 deg. terminal rating
31.25amp load --> at least #8 THWN required @ 60deg. [Note that I didn't apply the bundling derating when checking the terminal amps]
3. Check that conductors are protected by OCPD
(35amp rated circuit load) / (.8 conductor bundling derating) = 43.75amp load --> at least #8 THWN required @75deg.

It sounds to me like you are combining step 2 & 3, which may work for you on this problem, but won't work in all cases. The good news, your terminations are going to be ok, since you are overly protecting them by applying the bundling derating. The bad news is that if instead of using the 75deg. THWN, we use a conductor rated at 60deg., it looks to me like you are going to undersize your conductors and possibly start a fire - here's why:

Using your method, nothing will change since you were already limiting yourself to the 60deg. ampacity --> you select #8 conductors.
However, if my circuit load approaches the OCPD limit (i.e. equipment malfunctions, another device is added to circuit in remodel, etc) the effective continuous circuit amps will be as high as 43.75amps. Those #8 conductors are gonna be smokin!



You sure do.

Happy Thanksgiving folks!


Why are you using circuit load of 35 amps in your calculation? You said that the circuit load is 31.25 amps in step 1.


You need a conductor that is good for 39.06 amps (31.25*1.25) after derating. #8 THWN is 50 amps at 75 degrees. 50 * 80% = 40 amps. #8 THWN is OK.
 

erickench

Senior Member
Location
Brooklyn, NY
Infinity, are you sure you have to apply the 125% rule? 25 amps is the actual load. The purpose of dividing it by .8 is to increase the size of the conductor ampacity. If you look at the reciprocal for .8 it becomes 1.25 or 125%. Now what if there were 15 conductors in the raceway. The adjustment factor would be .5(50%). The reciprocal of that is 2. Are you going to multiply this number by 1.25(125%) or is 1/.8=1.25 already in accordance with the minimum 125% rule?
 

infinity

Moderator
Staff member
Location
New Jersey
Occupation
Journeyman Electrician
Infinity, are you sure you have to apply the 125% rule? 25 amps is the actual load. The purpose of dividing it by .8 is to increase the size of the conductor ampacity. If you look at the reciprocal for .8 it becomes 1.25 or 125%. Now what if there were 15 conductors in the raceway. The adjustment factor would be .5(50%). The reciprocal of that is 2. Are you going to multiply this number by 1.25(125%) or is 1/.8=1.25 already in accordance with the minimum 125% rule?



There are two factors that need to be applied here:
1- The 125% factor since this is a continuous load
2- The 80% derating factor for 4 CCC's in the raceway.

There are many ways mathematically to do this. Benaround had it right in post #13.
 
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