Wire Derating Table 310.15( C )( 1 )

[rbauerff]

I am considering a modification to the values in table 310.15( C )( 1 ), Adjustment Factors for more than 3 Current Carrying Conductors. This is for a specific application only. In this forum I am looking for feedback to point out any flaws in my thinking.

What is the maximum number of # 12 awg, cu, conductors that I can run in a raceway with each ungrounded conductor protected at 20 amps.

Using table 310.16, 90 deg C column, #12 awg, THWN-2 wire is good for 30 amps. Using table 10.15( C )( 1 ) I can use 9 wires (30 amps x 70 percent = 21 amps.

Table 310.15( C )( 1 ), is based on each wire being fully loaded. This is mentioned in the handbook notes following this table. However, in my application all conductors have been selected for 80 percent load.

If I have 9 #12 awg, 20 amp circuits loaded at 20 amps, and my wire resistance is 2 ohms per thousand feet from table 9 in the annexes, this arrangement will generate 7200 watts per thousand feet. 9 x 20A x 20A x 2.0 ohms/1000 feet.

If my wire is only loaded to 80 percent, the amount of heat generated per wire is 16A x 16A x 2 ohms/1000 feet = 512 watts.

The number of 80 percent wires needed to generate 7200 watts is 7200/512 =14.

My conclusion is that I can run 14 #12awg wires in a common raceway.

Issues that I have not considered but could effect my conclusion, increasing the number of wires.
1. I have not considered an ambient temperature adjustment.
2. the 2 ohm/1000 feet value I believe is the worst case condition and the actual value may be less.
3. If going from 9 to 14 wires requires a larger conduit this will make my calculation more conservative.

Thank you for your comments

 

[LarryFine]

My conclusion is that I can run 14 #12awg wires in a common raceway.

As long as you use 15a OCP.

 

[charlie b]

The rule is not based on current values, but rather on the number of wires that meet the description of current-carrying. It does not matter how heavily any wire is loaded or in what size conduit they reside. My belief is that the rule was written that way because any "do this except when that or if the wires are such and such" language would be impossible to write or to enforce.

 

[rbauerff]

The rule is not based on current values, but rather on the number of wires that meet the description of current-carrying. It does not matter how heavily any wire is loaded or in what size conduit they reside. My belief is that the rule was written that way because any "do this except when that or if the wires are such and such" language would be impossible to write or to enforce.

I agree with your statement regarding the rule. The NEC does allow me, as an engineer to do a thermal balance calculation to determine the number of wires that can be run together for my specific situation. However, I don't have the information needed to to do a thermal balance calculation and it would be a bad idea to even try it. The code has determined that the heat generated from the 9 wires will be dispersed into the environment without damage to the wire. What I have done is compared the heat generated by wires loaded to 100 percent with wires loaded to 80 percent and assumed that the heat lost to the environment does not change. Since I have never seen this type of analysis done before, my concern is that I may be missing something.

 

[rbauerff]

I agree with your statement regarding the rule. The NEC does allow me, as an engineer to do a thermal balance calculation to determine the number of wires that can be run together for my specific situation. However, I don't have the information needed to to do a thermal balance calculation and it would be a bad idea to even try it. The code has determined that the heat generated from the 9 wires will be dispersed into the environment without damage to the wire. What I have done is compared the heat generated by wires loaded to 100 percent with wires loaded to 80 percent and assumed that the heat lost to the environment does not change. Since I have never seen this type of analysis done before, my concern is that I may be missing something.

Table B.2(11) in Informative Annex B allows 24 wires with a load diversity of 50 percent.

 

[Jpflex]

I am considering a modification to the values in table 310.15( C )( 1 ), Adjustment Factors for more than 3 Current Carrying Conductors. This is for a specific application only. In this forum I am looking for feedback to point out any flaws in my thinking.

What is the maximum number of # 12 awg, cu, conductors that I can run in a raceway with each ungrounded conductor protected at 20 amps.

Using table 310.16, 90 deg C column, #12 awg, THWN-2 wire is good for 30 amps. Using table 10.15( C )( 1 ) I can use 9 wires (30 amps x 70 percent = 21 amps.

Table 310.15( C )( 1 ), is based on each wire being fully loaded. This is mentioned in the handbook notes following this table. However, in my application all conductors have been selected for 80 percent load.

If I have 9 #12 awg, 20 amp circuits loaded at 20 amps, and my wire resistance is 2 ohms per thousand feet from table 9 in the annexes, this arrangement will generate 7200 watts per thousand feet. 9 x 20A x 20A x 2.0 ohms/1000 feet.

If my wire is only loaded to 80 percent, the amount of heat generated per wire is 16A x 16A x 2 ohms/1000 feet = 512 watts.

The number of 80 percent wires needed to generate 7200 watts is 7200/512 =14.

My conclusion is that I can run 14 #12awg wires in a common raceway.

Issues that I have not considered but could effect my conclusion, increasing the number of wires.
1. I have not considered an ambient temperature adjustment.
2. the 2 ohm/1000 feet value I believe is the worst case condition and the actual value may be less.
3. If going from 9 to 14 wires requires a larger conduit this will make my calculation more conservative.

Thank you for your comments

You didn’t specify what size raceway you are using in the beginning of your example. I don’t have my code book so I’ll have to go mostly off of memory. So I think you are referring to 1/2” inch EMT for nine 12 AWG conductor fill max?

When trying to determine the heat just from the voltage drop there are two formulas I remember but I have always got different answers depending on which formula I use and I don’t understand why?

VD formula 1: 20 i (amperes) x 1.62 (ohms for 1,000 feet at 12 AWG) = 32.4 volts dropped?

Or

VD formula 2: 2 x K(12.9 copper) x i (20 amperes assume load plus voltage drop total) x D 1,000 ft (distance of 1 run) / CM 6,530 (12 AWG circular mill) = 79.01 voltage dropped?

If i use the voltage drop in the first formula we get 32.4 E x 20 i ampere = 648 watts included in the total 20 I ampere loaded circuit, this means that the load plus this voltage drop resistive 1 phase load is 20 amperes total


Second formula 79.01 voltage drop x 20 i amperes = 1,580 watts from voltage drop resistive load at 1,00 feet

I do not know why either formula does not yield equal consistent results when trying to figure out the extra wattage generated on one wire with 20 amperes at 1,000 feet?

 

[mikeames]

What I have done is compared the heat generated by wires loaded to 100 percent with wires loaded to 80 percent and assumed that the heat lost to the environment does not change. Since I have never seen this type of analysis done before, my concern is that I may be missing something.

My guess is that would be non-linear in a confined space such as a round raceway. When you add conductors under load they may be exponentially impacted. The heat generated is linear, but the heat escaping is def not.

 

[don_resqcapt19]

Table B.2(11) in Informative Annex B allows 24 wires with a load diversity of 50 percent.

That was in Article 310 for a long time, but was moved to the Annex when the code making panel could not agree on a definition of what 50% diversity actually means.

 

[charlie b]

The NEC does allow me, as an engineer to do a thermal balance calculation to determine the number of wires that can be run together for my specific situation.

Speaking as an engineer, I would not recommend doing it, even if it were allowed. But please tell me the code article that led you to infer that your method is allowed.

 

[charlie b]

That was in Article 310 for a long time, but was moved to the Annex when the code making panel could not agree on a definition of what 50% diversity actually means.

I believe there are some members of our industry who consider the phrase, "50% load diversity" to be synonymous with, "50% Diversity Factor." I am not among them. That phrase is well defined in EE text books. In my opinion, that definition does not fit the context of the cited Annex Table.

 

[don_resqcapt19]

I believe there are some members of our industry who consider the phrase, "50% load diversity" to be synonymous with, "50% Diversity Factor." I am not among them. That phrase is well defined in EE text books. In my opinion, that definition does not fit the context of the cited Annex Table.

All I know is that the CMP could not agree on a definition for the NEC and because of that, they moved that ampacity adjustment table to the Annex.

 

[Jpflex]

You didn’t specify what size raceway you are using in the beginning of your example. I don’t have my code book so I’ll have to go mostly off of memory. So I think you are referring to 1/2” inch EMT for nine 12 AWG conductor fill max?

When trying to determine the heat just from the voltage drop there are two formulas I remember but I have always got different answers depending on which formula I use and I don’t understand why?

VD formula 1: 20 i (amperes) x 1.62 (ohms for 1,000 feet at 12 AWG) = 32.4 volts dropped?

Or

VD formula 2: 2 x K(12.9 copper) x i (20 amperes assume load plus voltage drop total) x D 1,000 ft (distance of 1 run) / CM 6,530 (12 AWG circular mill) = 79.01 voltage dropped?

If i use the voltage drop in the first formula we get 32.4 E x 20 i ampere = 648 watts included in the total 20 I ampere loaded circuit, this means that the load plus this voltage drop resistive 1 phase load is 20 amperes total


Second formula 79.01 voltage drop x 20 i amperes = 1,580 watts from voltage drop resistive load at 1,00 feet

I do not know why either formula does not yield equal consistent results when trying to figure out the extra wattage generated on one wire with 20 amperes at 1,000 feet?

Correction first equation 2 x length

 

[charlie b]

Speaking as an engineer, I would not recommend doing it, even if it were allowed. But please tell me the code article that led you to infer that your method is allowed.

I was hoping to get an answer to this question. Rbauerff, can you chime in?

 

[kwired]

I am considering a modification to the values in table 310.15( C )( 1 ), Adjustment Factors for more than 3 Current Carrying Conductors. This is for a specific application only. In this forum I am looking for feedback to point out any flaws in my thinking.

What is the maximum number of # 12 awg, cu, conductors that I can run in a raceway with each ungrounded conductor protected at 20 amps.

Using table 310.16, 90 deg C column, #12 awg, THWN-2 wire is good for 30 amps. Using table 10.15( C )( 1 ) I can use 9 wires (30 amps x 70 percent = 21 amps.

Table 310.15( C )( 1 ), is based on each wire being fully loaded. This is mentioned in the handbook notes following this table. However, in my application all conductors have been selected for 80 percent load.

If I have 9 #12 awg, 20 amp circuits loaded at 20 amps, and my wire resistance is 2 ohms per thousand feet from table 9 in the annexes, this arrangement will generate 7200 watts per thousand feet. 9 x 20A x 20A x 2.0 ohms/1000 feet.

If my wire is only loaded to 80 percent, the amount of heat generated per wire is 16A x 16A x 2 ohms/1000 feet = 512 watts.

The number of 80 percent wires needed to generate 7200 watts is 7200/512 =14.

My conclusion is that I can run 14 #12awg wires in a common raceway.

Issues that I have not considered but could effect my conclusion, increasing the number of wires.
1. I have not considered an ambient temperature adjustment.
2. the 2 ohm/1000 feet value I believe is the worst case condition and the actual value may be less.
3. If going from 9 to 14 wires requires a larger conduit this will make my calculation more conservative.

Thank you for your comments

You already can make ampacity adjustments for fixed loads based on actual current draw or maximum operating current of the load.

General purpose receptacle outlet circuits is where you must assume a possible 20 amps on a 20 amp overcurrent device

Connect several fixed 10 amp loads to 20 amp breakers and place them all in one raceway you can use that 10 amps for load when making ampacity adjustments which would allow many more conductors in the raceway (if large enough raceway) than if you were simply feeding receptacle outlets on 20 amp circuits. Actually would allow unlimited number as the 12AWG @90C conductor adjusted to 35% for 41 or more conductors is still above your 10 amp load per conductor.

BTW why the 9 x 20A x 20A x 2.0 ohms/1000 feet?

shouldn't it be 9 x 20 x 2.0 / 1000?

 

[wwhitney]

Connect several fixed 10 amp loads to 20 amp breakers and place them all in one raceway you can use that 10 amps for load when making ampacity adjustments

No, using 20A breakers means your final ampacity needs to be at least 16A so you can apply 240.4(B). If you use 15A breakers, then you could let me the ampacity get as low as your 10A load rating. Except under the 2023 NEC, which if IIRC added 10A as a standard size, in which case you'd need 11A ampacity to apply 240.4(B).

Cheers, Wayne

 

[kwired]

No, using 20A breakers means your final ampacity needs to be at least 16A so you can apply 240.4(B). If you use 15A breakers, then you could let me the ampacity get as low as your 10A load rating. Except under the 2023 NEC, which if IIRC added 10A as a standard size, in which case you'd need 11A ampacity to apply 240.4(B).

Cheers, Wayne

I can agree with that. But not the 16 amp figure, more like anything over 15 for minimum final conductor ampacity can still have 20 amp overcurrent protection.

And could still have basically unlimited number of 12 AWG in a raceway adjusted to 35% which is 10.5 amps which is more than 10 and protect them at 15 amps.

OP was basing the 80% most likely off continuous load requirements for minimum conductor ampacity, which technically does not adjust the ampacity of a conductor but rather increases the minimum ampacity, before adjustments, to 125% of the load. In that case no changes to current wording is needed. If you must have a minimum ampacity of 16 and are over 9 CCC's with 12 AWG conductor the adjusted ampacity is less than 16, you will need a larger conductor in that case. But even if he was not considering continuous loads, his situation still defined the minimum ampacity as being 16 so still in the same boat.