Derating of 4-6 Current Carrying Conductors in conduit

[met123]

How to derate 4-6 current carrying conductors (CCC) in conduit, 310.15B2a.
This is how I've done it in the past. Say for a 125A load and 4 each THHN CCC.

Conductor Size:
125A X 1.25 continuous load = 156.25A
156.25A/.8 (310.15B2a) = 195.3A from table 310.16 column 90C THHN s/b 2/0 conductors. Some might argue s/b 3/0 because of the .3A over 195A. But knowing 100% of load is not continuous I would go with the 2/0. Check for 75C terminal rating 156.25A < 175A so 2/0 will work.

I'm being told that this is double derating per NEC 210.20a. To size conductors should use actual load of 125A/.8 (310.15B2a) = 156.25A not the continuous load calculated after multipying by 1.25 or the 195.3A for derating. Which would select THHN 1/0 from 90C column. This in effect removes the derating from 310.15b2a for 4-6 CCC and makes no sense to me because this is the way the conductor would be selected with 3 or less CCC.

 

[Smart $]

How to derate 4-6 current carrying conductors (CCC) in conduit, 310.15B2a.
This is how I've done it in the past. Say for a 125A load and 4 each THHN CCC.

Conductor Size:
125A X 1.25 continuous load = 156.25A
156.25A/.8 (310.15B2a) = 195.3A from table 310.16 column 90C THHN s/b 2/0 conductors. Some might argue s/b 3/0 because of the .3A over 195A. But knowing 100% of load is not continuous I would go with the 2/0. Check for 75C terminal rating 156.25A < 175A so 2/0 will work.

I'm being told that this is double derating per NEC 210.20a. To size conductors should use actual load of 125A/.8 (310.15B2a) = 156.25A not the continuous load calculated after multipying by 1.25 or the 195.3A for derating. Which would select THHN 1/0 from 90C column. This in effect removes the derating from 310.15b2a for 4-6 CCC and makes no sense to me because this is the way the conductor would be selected with 3 or less CCC.

Your method is acceptable for determination of conductor size. Just remember that your first result is the minimum required ampacity of the conductor.

 

[charlie b]

I don?t think you are looking at the situation correctly. Even if you have the right answer (I didn?t check your figures), it is not ?right? if you did not use the right method.

There are really two very, very different things going on here. You have combined them into a single thing, and I fear that will occasionally give you the wrong answer. Here is how I suggest you do it.

Step 1: Figure out the load. That is, figure out how many amps will be flowing in the conductors, and therefore what minimum ampacity you need those conductors to have.

Step 2: Find a conductor size that has that minimum ampacity.

In your case, I do not know why you are starting with 125 amps. That is a good, round number, and therefore I suspect it is not the result of a load calculation. When you perform a load calculation, the distinction between continuous and non-continuous loads are part of the process. What I mean is that, by the time you reach your final answer, you will have already added the extra 25% for the continuous portions of the loads. So you don?t start with 125 amps, and multiply it by 125%. You take 100% of the non-continuous loads, and add 125% of the continuous loads, and that is the current your selected conductors must be able to handle.

Thus, if you want to do an example using 125 amps as the load, then I have to ask whether this figure is intended to have already incorporated the extra 25% for continuous loads. If not, then you have to tell us how much of the 125 amps is to be treated as continuous.

Only after this is settled do you have a load value that can be used as the criterion for selecting the conductor size.

Now on to Step 2. The bottom line here is that you don?t apply the derating factor to the load. You apply it to the tabulated values of conductor ampacity. A 2/0 is good for 195 amps at 90C. Derate that by 80% because of having four current carrying conductors, and the ampacity of the 2/0 becomes 156. You then compare that value of ampacity with the required value you obtained in Step 1.

Welcome to the forum.

 

[met123]

double derating

double derating

The 125A load is an example. What I'm saying is given a 125A load do we use the 125A or the 125AX1.25 assuming 100% continous load for the derating of 4 CCC.

 

[infinity]

The 125A load is an example. What I'm saying is given a 125A load do we use the 125A or the 125AX1.25 assuming 100% continous load for the derating of 4 CCC.

I would approach this in two steps:

First find the minimum conductor ampacity, in this case 125 amps * 125%= 156.25 amps.

Now find a conductor that is equal or greater in ampacity after you've applied your derating adjustments. It's that simple.

 

[charlie b]

What I'm saying is given a 125A load do we use the 125A or the 125AX1.25 assuming 100% continous load for the derating of 4 CCC.

What I am saying is that "given a 125 amp load" is not enough information. So you tell me:
(1) Is the value of "125 amps" the amount of connected load, and is all of it continous, or
(2) Is the value of "125 amps" the amount of connected load, and is only a part of it continuous, or
(3) Does the value "125 amps" represent a calculated load amount, with that calculation having already accounted for an extra 25% for the portion that is continuous?

These are the three different ways I could interpret a question that begins with "the load is 125 amps." Which of these would you like to solve?

 

[crossman gary]

Man...

This NEC stuff is hard!:wink:

 

[charlie b]

Let me complete the three examples I give above.

(1) If all the 125 amp load is continuous, then you must have a conductor with an ampacity of 156.3 amps. The conductor you select must have an ampacity, before derating, of 195.3 amps, in order for its derated ampacity to be at least as high as 156.3 amps. Given that the tabulated values are in whole numbers, and that 195.3 ?rounds down? to 195, I would accept a 2/0 for this application.

(2) If the connected load is 125 amps, and if (as an example) we say that 80 amps of the load is continuous (so that 45 amps is non-continuous), then the ?calculated load? would be 100% of 45 plus 125% of 80, or 145 amps. The conductor you select must have an ampacity, before derating, of 181.3 amps, in order for its derated ampacity to be at least as high as 145 amps. I would select a 2/0 for this application.

(3) Now let us consider a total calculated load of 125 amps. One way you can get there is if there were 100 amps of non-continuous load and 20 amps of continuous load. That is, 100% of 100 plus 125% of 20 gives you 125 amps. The conductor you select must have an ampacity, before derating, of 156.3 amps, in order for its derated ampacity to be at least as high as 125 amps. I would select a 1/0 for this application.

Does this help?

 

[Smart $]

I would approach this in two steps:

First find the minimum conductor ampacity, in this case 125 amps * 125%= 156.25 amps.

Now find a conductor that is equal or greater in ampacity after you've applied your derating adjustments. It's that simple.

That's what he is doing... only a little different than most, but in fact the better way IMO.

After determining the minimum required circuit ampacity (MRCA), others generally choose a conductor then derate its ampacity (DCA), compare to the MRCA with the logic test: DCA<MRCA = go bigger, DCA>MRCA=OK, but can I go smaller... DCA_smaller>MRCA= OK, if not go back, etc... plus you have to derate the selected conductor for each size you choose. Why go through all that... Simply divide your MRCA by the derating factor and you automatically know the minimum to-be-derated ampacity of the conductor you need.

 

[infinity]

Sounds to me like this just an example of a load of 125 amps that is considered a continuous load.

 

[infinity]

That's what he is doing... only a little different than most, but in fact the better way IMO.

After determining the minimum required circuit ampacity (MRCA), others generally choose a conductor then derate its ampacity (DCA), compare to the MRCA with the logic test: DCA<MRCA = go bigger, DCA>MRCA=OK, but can I go smaller... DCA_smaller>MRCA= OK, if not go back, etc... plus you have to derate the selected conductor for each size you choose. Why go through all that... Simply divide your MRCA by the derating factor and you automatically know the minimum to-be-derated ampacity of the conductor you need.

Yes, you're correct there are mathematical ways to short cut to the correct sized conductor. I was just trying to keep it as simple as possible.

 

[crossman gary]

I bet the calculators are crunching trying to find the odd example where that method misses by a conductor size.

 

[Smart $]

Let me complete the three examples I give above.

(1) If all the 125 amp load is continuous, then you must have a conductor with an ampacity of 156.3 amps. The conductor you select must have an ampacity, before derating, of 195.3 amps, in order for its derated ampacity to be at least as high as 156.3 amps. Given that the tabulated values are in whole numbers, and that 195.3 ?rounds down? to 195, I would accept a 2/0 for this application.

(2) If the connected load is 125 amps, and if (as an example) we say that 80 amps of the load is continuous (so that 45 amps is non-continuous), then the ?calculated load? would be 100% of 45 plus 125% of 80, or 145 amps. The conductor you select must have an ampacity, before derating, of 181.3 amps, in order for its derated ampacity to be at least as high as 145 amps. I would select a 2/0 for this application.

(3) Now let us consider a total calculated load of 125 amps. One way you can get there is if there were 100 amps of non-continuous load and 20 amps of continuous load. That is, 100% of 100 plus 125% of 20 gives you 125 amps. The conductor you select must have an ampacity, before derating, of 156.3 amps, in order for its derated ampacity to be at least as high as 125 amps. I would select a 1/0 for this application.

Does this help?

charlie,

Seems to me you are doing the exact same thing he is doing without realizing it because you're not writing it out in math form. I've highlighted where you divide by the derating factor....

 

[Smart $]

I bet the calculators are crunching trying to find the odd example where that method misses by a conductor size.

I wish them the best of luck because there not going to find it, unless they resort to pinning it on rounding error(s).

 

[charlie b]

Seems to me you are doing the exact same thing he is doing without realizing it because you're not writing it out in math form.

I knew that; I did realize that. I don't object to doing the math backwards. But I believe it to be important that we realize that the derating factor does not alter the load, and is not applied to the load. The derating factor changes the ampacity of the conductors. That is the "Step 2" to which I refered earlier. That does not mean that we must start with the tables, and multiply the derating factor to values in the tables until we get an answer that is big enough. Sure, dividing the load by the derating factor gets us the result quicker. But it is still worth taking it back to basics, to make sure the concept is clear, before we start using mathematical shortcuts.

That said, my intent in post #8 I was to complete my explanation that just telling us a load value does not give us enough information, that we also need to know whether, and how, the continuous load is accounted for.

 

[glene77is]

Charlie B

You wrote
"...make sure the concept is clear,
before we start using mathematical shortcuts."

Excellent advice!