Feeder Calculations and Derating

[winterovereverything]

Howdy Folks,

First time poster here with a feeder calculation question. I'm nearly certain on the right path but, I just want to double check my work here. The short of it is this: I have feeders that, following the principles of 215.2(a) or (b) have boiled down to 19.2 amps (for 215.2(a)) for the 125% continuous load calculation and 23.8 amps for (215.2(b)) after adjustment and correction factors. Here's my train of thought: Feeder conductors would be of THHN and so 12 AWG, would be the appropriate wire size.

I derated off the 90 degree column, and the ampacity of the wire under the 60 degree and 90 degree is sufficient for the original load to be supported without adjustment factors (16.56) amps. Am I wrong? Should I size up to 10 AWG to support the adjusted load of 23.8 amps? I understand the application of 110.14(c) as well as when it is permitted to apply adjustment and correction factors as seen in 310.15(a), and that the conductor's ampacity is ultimately in regards to the terminations (which in this case would be 60 degrees) but I want to make sure that I'm following the right path where that ampacity under the 60 degree column is only to support the actual load and not the adjusted or corrected load.

Thanks.

 

[Dennis Alwon]

I don't understand your train of thought. 19.2 amps x 125% = 24 amps. Are you using nm cable? Your conductor must be rated for 24 amps minimum after and de-rating, etc

 

[winterovereverything]

I don't understand your train of thought. 19.2 amps x 125% = 24 amps. Are you using nm cable? Your conductor must be rated for 24 amps minimum after and de-rating, etc

Hey Dennis,

Let me clarify. That 19.2 Amps is the total load for one of the units being supported by the feeder and has already factored in the continuous load as 215.2 (A) (1) calls for. Just didn't include the calculation.

 

[Dennis Alwon]

So this is a branch circuit not a feeder and the calculated load is 19.2 after 125% was added to it. Then #12 is fine. And yes, if the conductor is rated 90C then you can de-rate from the 90C column but generally, the overcurrent protective device must not be greater than 60C for nm cable and 75 for the rest.

 

[winterovereverything]

These are feeders. It's just an imagined project I've assigned myself for a small area with a few receptacles and lights (just wanted to make sure that I'm on the right track). So, in summary, although part b of 215.2 calls for derating off the 90 degree column, and those derating calculations call for 23 amps, I can still use 12 awg since its 90 degree ampacity is 30 amps and since it's 60 degree ampacity can support the original load without correction or adjustment. Is that a correct line of thinking?

 

[Dennis Alwon]

These are feeders. It's just an imagined project I've assigned myself for a small area with a few receptacles and lights (just wanted to make sure that I'm on the right track). So, in summary, although part b of 215.2 calls for derating off the 90 degree column, and those derating calculations call for 23 amps, I can still use 12 awg since its 90 degree ampacity is 30 amps and since it's 60 degree ampacity can support the original load without correction or adjustment. Is that a correct line of thinking?

Your are confusing me. If your load is 19 amps after the 125% then how are we getting to 23 amps. If you have other de-rating issues then your conductor must be rated for 23 amps minimum.

I guess the 23.8 amps was what confused me. I assume you have other de-rating factors in here so that is where 23.8 comes in.

#10 wire is needed

 

[Dennis Alwon]

Are you installing an overcurrent protective device at the end of this #10 wire? then it is a feeder otherwise it is a branch circuit. Only knowing part of the job makes it hard to respond correctly. I think or hope I have it correct now. haha

 

[winterovereverything]

Hi Dennis, These conductors have an overcurrent device at the end of their run so they would be feeders. I’m sorry for being so vague.

215.2 calls for feeder conductors to be sized by either a) 125% of the continuous load or b) an ampacity for the maximum load to be served after adjustment or correction factors. 19.2 is the given ampacity after the continuous load has been implemented and so that would require a 12 AWG wire in compliance with 110.14.

When solving for b) I applied my derating factors to the 90 degree column of 310.16 and got a required ampacity of 23.8 amps. Now under the 90 degree, 12 AWG is worth 30 amps but in compliance with 110.14, the actual ampacity is worth 20 amps. Can’t I still use the 12 AWG wire since it’s 60 degree ampacity is sufficient to carry the non- adjusted load and since the 90 degree column is sufficient for the adjusted load? Is that a correct line of thinking?

 

[Dennis Alwon]

Hi Dennis, These conductors have an overcurrent device at the end of their run so they would be feeders. I’m sorry for being so vague.

215.2 calls for feeder conductors to be sized by either a) 125% of the continuous load or b) an ampacity for the maximum load to be served after adjustment or correction factors. 19.2 is the given ampacity after the continuous load has been implemented and so that would require a 12 AWG wire in compliance with 110.14.

When solving for b) I applied my derating factors to the 90 degree column of 310.16 and got a required ampacity of 23.8 amps. Now under the 90 degree, 12 AWG is worth 30 amps but in compliance with 110.14, the actual ampacity is worth 20 amps. Can’t I still use the 12 AWG wire since it’s 60 degree ampacity is sufficient to carry the non- adjusted load and since the 90 degree column is sufficient for the adjusted load? Is that a correct line of thinking?

The 90 degree column is only good for the de-rating purposes. if you have more de-rating to do and you end up being above 20 amps then you cannot use 90 degree column for any reason. #10 wire must be used so you cannot use number 12 in this situation

 

[winterovereverything]

Dennis thanks a ton! I had remembered at one point someone telling me otherwise, and that I could so long as the derated ampacity and the non-derated ampacity satisfy the maximum load to be served, but I heard wrong. You’re the best!

 

[david luchini]

Dennis thanks a ton! I had remembered at one point someone telling me otherwise, and that I could so long as the derated ampacity and the non-derated ampacity satisfy the maximum load to be served, but I heard wrong. You’re the best!

If I'm reading your situation correctly, #12 should be fine

 

[winterovereverything]

How so? See I’m partial to the argument on both sides of the coin and I think this stems from a vast interpretation of 110.14(c) that everyone seems to have. My original idea is that so long as that a derated conductor can sufficiently support the load, that the conductor’s ampacity of the given load doesn’t exceed that of the 60 degree column, and the maximum load before derating does not exceed that of the 60 degree column, that it is appropriate to use. However Dennis does bring up the point that a 90 degree conductor is only for derating.

 

[david luchini]

How so? See I’m partial to the argument on both sides of the coin and I think this stems from a vast interpretation of 110.14(c) that everyone seems to have. My original idea is that so long as that a derated conductor can sufficiently support the load, and that the conductor’s ampacity of the given load doesn’t exceed that of the 60 degree column, that it is appropriate to use. However Dennis does bring up the point that a 90 degree conductor is only for derating.

What are you derating for? Temperature? Number of current carrying conductors?

 

[winterovereverything]

Temperature

 

[david luchini]

Temperature

What temperature?

 

[winterovereverything]

Sorry about the late response. I was on a call. Temp is 66 degrees celsius.

 

[wwhitney]

I have feeders that, following the principles of 215.2(a) or (b) have boiled down to 19.2 amps (for 215.2(a)) for the 125% continuous load calculation and 23.8 amps for (215.2(b)) after adjustment and correction factors. . . . I derated off the 90 degree column, and the ampacity of the wire under the 60 degree and 90 degree is sufficient for the original load to be supported without adjustment factors (16.56) amps. . . .

Let me back-calculate: the load is 6A non-continuous and 10.56A continuous. That gives your total load of 16.56A, and your 215.2(A) figure of 19.2 A = 6 + 1.25*10.56.

215.3 says your OCPD must be at least 19.2A, so you select a 20A OCPD.

Your temperature derating factor is 16.56/23.8 = 0.696.

So 215.2(a) [didn't confirm the section numbers, going off the OP] says that you need a conductor whose table ampacity at the termination temperature is at least 19.2A. #12 AWG works. 215.2(b) says that the table ampacity at the conductor insulation temperature must be at least 23.8A. #12AWG works as long as the insulation temperature is at least 75C.

Lastly we have to check 240.4(B) to be sure the conductor has sufficient ampacity (after derating) to be protected at 20A. But as your total load is 16.56A and we've already ensured that the ampacity is sufficient for the load, it's sufficient for a 20A OCPD under 240.4(B). [An example where this check could control would be a continuous load of 19A, which requires a 25A OCPD; if the conductor ampacity (after derating) came out to 19.5A, that would pass the previous checks, but not this check.]

Cheers, Wayne

 

[wwhitney]

Your are confusing me. If your load is 19 amps after the 125% then how are we getting to 23 amps. If you have other de-rating issues then your conductor must be rated for 23 amps minimum.

I guess the 23.8 amps was what confused me. I assume you have other de-rating factors in here so that is where 23.8 comes in.

#10 wire is needed

Dennis you are overlooking that there are two separate checks to be done, and we don't need to apply the 125% continuous use factor and the derating factors at the same time. So #12 turns out to be OK.

For the first check (which Example D3(a) in the Annex calls the "termination" check), we use the termination temperature ampacity column, with no derating, and with a 125% continuous use factor. For the second check (which Example D3(a) in the Annex calls the "conditions of use through the raceway run" check), we use the insulation temperature ampacity column, with derating, and without any 125% continuous use factor.

Cheers, Wayne

 

[david luchini]

Sorry about the late response. I was on a call. Temp is 66 degrees celsius.

66 degree celsius would require a 90deg conductor with an unadjusted ampacity of at least 26.3, not 23.8.

#12 THHN at 66 deg C would have an ampacity of 30A * 0.63 = 18.9A. This is large enough for your 16.56A load, so #12 would be acceptable.

 

[winterovereverything]

Let me back-calculate: the load is 6A non-continuous and 10.56A continuous. That gives your total load of 16.56A, and your 215.2(A) figure of 19.2 A = 6 + 1.25*10.56.

215.3 says your OCPD must be at least 19.2A, so you select a 20A OCPD.

Your temperature derating factor is 16.56/23.8 = 0.696.

So 215.2(a) [didn't confirm the section numbers, going off the OP] says that you need a conductor whose table ampacity at the termination temperature is at least 19.2A. #12 AWG works. 215.2(b) says that the table ampacity at the conductor insulation temperature must be at least 23.8A. #12AWG works as long as the insulation temperature is at least 75C.

Lastly we have to check 240.4(B) to be sure the conductor has sufficient ampacity (after derating) to be protected at 20A. But as your total load is 16.56A and we've already ensured that the ampacity is sufficient for the load, it's sufficient for a 20A OCPD under 240.4(B). [An example where this check could control would be a continuous load of 19A, which requires a 25A OCPD; if the conductor ampacity (after derating) came out to 19.5A, that would pass the previous checks, but not this check.]

Cheers, Wayne

Wayne I think you answered my question here, albeit inadvertently. So even though my conductor at 60 degrees is rated at 20 amps, it still remains a viable option since its ampacity can suffice the maximum total load of my feeders. There's been a lot of talk around the shop, that the ampacities under the 60 degree column need to support, not the maximum total load of my feeders, but the derated load of my feeders (23.8 amps).