NEC Chapter 9 Note 7

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tryingEE

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Chapter 9 note 7 in the NEC is kind of confusing in my opinion. From an example it shows the quantity of wires that will fit within the size. This is done by taking the conduits fill area divided by the conductors area to find the number of conductors able to fit. For example, 1.25" RMC at 40% fill is 0.610. The area of a #10 AWG is 0.0211. These divided is equal to 28.91, which given that the decimal is over .8, you can fit 29 conductors within it.

I am trying to apply this math a different way, and I want to see if it "maths" out. My problem is:
I have (4) 600 KCMIL conductors and a #3 AWG grounding conductor. The total area from these conductors is 4.005 (4*0.9729+0.1134 (these values derived from Chapter 9 Table 5 THW section). Then, moving to Chapter 9 Table 4, the 40% fill for a 3" conduit is 3.538. Shucks, it is too small. However, if I divide the 3.538 by 4.005, I get 0.8834. Since this is above .8, am I allowed to use the 3" conduit? Or should I play it safe an use a 3.5" conduit because its permittance is 4.618 which gives ample space?
 
You cannot use 3" conduit. I am not sure what your last few sentences have to do with the calculation. Use 3.5" and welcome to the forum
 
Dennis Alwon said:
You cannot use 3" conduit. I am not sure what your last few sentences have to do with the calculation. Use 3.5" and welcome to the forum
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Essentially, rather than dividing conduit 40% area by conductor area to get the number of wires allowed in that conduit, I am dividing the 40% fill area by the total area of all the conductors I plan on putting into the conduit (like thinking my 4 CCC and grounding wire is one singular mega conductor). Typically, that value should be greater than 1, because you shouldn't go over the 40% area allowed in the conduit. However, if I get a value above 0.8, could I still use that conduit based on Chapter 9 note 7? I italicized "shouldn't" because if you can't go over that area allowed, why is note 7 a thing?
I respect your opinion and believe you have more merit and knowledge on the NEC than I wish I could obtain, I am only questioning again because I have been checking my calculations with this calculator: https://www.southwire.com/calculator-conduit
It has been agreeing with my math, try it with the parameters of my question, you will get 3" conduit if you use THHN.
 
tryingEE said:
Chapter 9 note 7 in the NEC is kind of confusing in my opinion. From an example it shows the quantity of wires that will fit within the size. This is done by taking the conduits fill area divided by the conductors area to find the number of conductors able to fit. For example, 1.25" RMC at 40% fill is 0.610. The area of a #10 AWG is 0.0211. These divided is equal to 28.91, which given that the decimal is over .8, you can fit 29 conductors within it.

I am trying to apply this math a different way, and I want to see if it "maths" out. My problem is:
I have (4) 600 KCMIL conductors and a #3 AWG grounding conductor. The total area from these conductors is 4.005 (4*0.9729+0.1134 (these values derived from Chapter 9 Table 5 THW section). Then, moving to Chapter 9 Table 4, the 40% fill for a 3" conduit is 3.538. Shucks, it is too small. However, if I divide the 3.538 by 4.005, I get 0.8834. Since this is above .8, am I allowed to use the 3" conduit? Or should I play it safe an use a 3.5" conduit because its permittance is 4.618 which gives ample space?
Click to expand...
Note 7 does not apply because you have two different size conductors in the raceway.
(7)
When calculating the maximum number of conductors or cables permitted in a conduit or tubing, all of the same size (total cross-sectional area including insulation), the next higher whole number shall be used to determine the maximum number of conductors permitted when the calculation results in a decimal greater than or equal to 0.8. When calculating the size for conduit or tubing permitted for a single conductor, one conductor shall be permitted when the calculation results in a decimal greater than or equal to 0.8.
Click to expand...
 
don_resqcapt19 said:
Note 7 does not apply because you have two different size conductors in the raceway.
Click to expand...
Ahhhh I see, I need to be more thorough with my reading. Thank you Don! And thank you to Dennis as well! I guess my question still stands if my math thought process would still work if they were the same size.
 
@tryingEE I have a suggestion and a question. Why are you using THW. Everything today is thwn-2 insulation which is significantly smaller than Thw.
Not sure what you are wiring but it may be possible to downsize the neutral.
 
Dennis Alwon said:
@tryingEE I have a suggestion and a question. Why are you using THW. Everything today is thwn-2 insulation which is significantly smaller than Thw.
Not sure what you are wiring but it may be possible to downsize the neutral.
Click to expand...
I had no reason behind THW other than it was in the 75 degrees column. These conductors are feeders for panelboards. Would thwn-2 be the better option in this application?
 
tryingEE said:
I had no reason behind THW other than it was in the 75 degrees column. These conductors are feeders for panelboards. Would thwn-2 be the better option in this application?
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Yes, no one really uses THW anymore. I'm not even sure if it's even readily available.
 
tryingEE said:
I had no reason behind THW other than it was in the 75 degrees column. These conductors are feeders for panelboards. Would thwn-2 be the better option in this application?
Click to expand...
The 75° rating is really only important for calculating ampacity and conductor size. Use currently available insulation for new raceway fill. My electrical supply house stopped stocking THW some 30+ years ago. I started specifying THHN 45 years ago.
 
This part of Note 7 strikes me as a bit unexpected: "When calculating the size for conduit or tubing permitted for a single conductor, one conductor shall be permitted when the calculation results in a decimal greater than or equal to 0.8."

Table 1 says that for 1 conductor, the allowable fill is 53%. But Note 7 always applies, so the allowable fill is really 66.25%.

Cheers, Wayne
 
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