Double Derating 210.20a

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met123

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I'm still having trouble with the "double derating" In the NEC 2008 Handbook at 210.20a it gives as example of a 25A continuous load with 4 Current Carrying Conductors (CCC). It says the calculated load is 25A and to use this with the derating from 310.15b2a this being 25A/.80 = 31.25 to size the conductor to avoid double derating. But I would normally use 25A X 1.25 = 31.25A for a continuous load to size the conductor which would bring me to the same value without derating for more than 3 CCC. The engineer that I'm working with said it makes no sense to him either in that when you have 4-6 CCC there is no affect on the calculations as there would be with just multiplying by 1.25 for a continuous load. How would you interpret this?
Also, at the end of the example it says "where the ambient temperature exceeds 86C" I believe this should be 86F.
 
480sparky

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0.8 and 1.25 are reciprocals, so either method (divide by 0.8 or multiply by 1.25) works. Enter either number into a calculator, and push the 1/x key, and the other number is shown.

Since I don't have the Handbook, I can't look at the example you're referencing. But if you're wondering about derating for both 4+ CCCs and ambient temp., both must be calculated.
 
charlie b

charlie b

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I would have to see the handbook, to see how the example is worded, in order to form an opinion on its methodology. A key question I would have is whether they are stating that the current is 25 amps, and that the load is continuous, or whether they are stating that the calculated load (including any adjustments for continuous loads) is 25 amps.

But I will say that applying a 1.25 factor to a continuous load is not a "derating" action. It is defining the amount of load for which the conductor must be sized. When you apply an 80% factor because there are more than 3 CCCs in a raceway, you are changing the rated ampacity of the conductor. It was rated for one value, and now, because of the conditions of use, is has a lower rating. That is a derating action.
 
S'mise

S'mise

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Like the others said don't confuse derating. A 25 amp load continuously used is calculated at 125% or, 25x1.25=31.25 If 4-6 ccc the wire rating is derated by 80% then look up the ampacity of wire (310.16) and multiply it times .8 if temp is above 86F (yes your right it's F) then also derate for temperature. (correction factors on bottom of page)
Bottom line you are looking for an ampacity over 31.25 Amps. But if the wire type has a temp rating higher than required (60 or 75C) then you dont have to derate if the ampacity is at least the values in the lower colum on the left.

Sorry, I fear I confused you more.
 
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S'mise

S'mise

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met123 said:
Guess my question then is, what is "double derating"?
Click to expand...

I did not read the handbook but i assume they are talking about...

Derating #1 is for more than 3 ccc
derating #2 is for temp (double derating)

EG. If you use Thhn wire which is rated for 90C you can use the higher ampacity in the 90C colum as long as it's not higher than the amount in the 75C colum. Another words, you only have to derate for "more than 3 ccc" and also for temperature.
 
Mr. Bill

Mr. Bill

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From this example I think #8 AWG is needed. Either at 60C or 75C.
40A (@ 60C) * 80% (per Table 310.15(B)(2)(a)) = 32A ampacity
25A continuous load * 125% (210.19(A)(1)) = 31.25A min ampacity
The derated #8 AWG has an ampacity not less than 31.25A.
Or am I looking at this wrong? I know technically it's not double derating but that's how many think of it. Continuous current * 125% * 125% = normal conductor ampacity required. It's just a short hand calculation method. Technically it is wrong but is easier and the result is the same.

(I'm still working from the 2005 NEC so my section #s may be a bit off)
 
erickench

erickench

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I know what you're saying. I've wondered about that myself. It's like your multiplying the same number twice by 125%. You take the continuous load and multiply it by 1.5625 which is the square of 1.25. That's just the way the NEC is written though I do question it. Make a proposal.
 
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inspector141

inspector141

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I try not to use the term 'double derate' in my calculation class. Because some will imply that your derating the same thing twice, when in fact each derating action has nothing to do with each other, but have everything to do with each other as far as ampacity of the conductor is concerned.
That is, derating a conductor(s) because of 4ccc in a raceway, adds to the heat of the circuit conductors. Derating a conductor(s) because of an ambient temp above 86 degrees F, adds to the heat of the circuit conductors.
One action does not cancel the other action. Heat is heat. They are additive in the derating of the conductor.
 
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met123

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double derating

double derating

The question came up because I was applying the 125% for continuous load and then using that value to apply the 4-6 ccc (divide by .8) to calculated the needed ampacity. I was told that this is double derating and was refered to the example in 210.20.a. Was told to use 100% of load and then apply the 4-6 ccc which would bring me back to the exact amount I would have by applying the 125% only and not derating for 4-6 ccc. This just does not make sense to me.
 
inspector141

inspector141

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The 125% is for the continous load and has nothing to do with derating the conductors. One is for calculating the circuit amps, but derating occurs for the ampacity of the conductor. Once derated, the ampacity of the conductor must be equal to or greater than the calculated load of the circuit or feeder.
 
augie47

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all of a sudden I feel stupid (fitting :) ).
If you have 4 cc conductors in a conduit, you derate by a percentage value of 80% of the conductor ampacity after it's been adjusted for ambient, correct ? 4 #12 THHN is a 113? F ambient would have an ampacity of 20.88 amps (30 x .87 x ,8) correct ?

Does this not mean that the maximum continuous load would be 16.7 amps ?
 
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SegDog

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factor = multiply = percentage

factor = multiply = percentage

Hello Met,

As a math tutor, I immediately noticed that dividing by .80 is not the same as factor .80 . Both column headings use the terms factor and percentage, and are multipliers.

I have fits when I read about continuous loads, especially when it's combined with the word plus non-continuous.

The continuous load times 1.25 is the base amperage for the calculation prior to adjustments. It's just coincidental that 4-6 CCC comes out the same.

Best of
 
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SegDog

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Philadelphia
scanned Handbook section

scanned Handbook section

Hello Met,

I just read the scanned section and agree that it's confusing.

Apparently, calculated load is not calculated load when it comes to continuous loads alone that require adjustment, de-rating, or above rating.

Reminds me of one of the government agencies that I worked for. The regulations actually strangled logic.

"hype and chump-change" - Nikpour
 
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