derating

[MR. S]

OK, what size copper THHN feeder conductors are needed for an 85 amp continuous load and a 105 amp non-continuous load, with 4 current carrying conductors and run in a room with a 34 degree celcius ambient temperature.
We are comming up with a 250 Kcmil?

 

[Volta]

I agree, assuming the conductors have a 90 deg C insulation rating, and will operate at 75 deg C. The final ampacity would be 222.7.

 

[MR. S]

How did you get a 250 Kcmil from the 75 degree column with a 222.7 ampacity?

 

[david luchini]

How did you get a 250 Kcmil from the 75 degree column with a 222.7 ampacity?

He didn't. He got 250 mcm from the 90 degree column... You said THHN.

A agree with 250mcm as well, the ampacity is 222.7. #4/0 (the next smaller size) THHN would have an ampacity of 200, and would not be properly protected by the minimum required 225A OCPD.

 

[Volta]

True. As you suggested 250 kcmil, I started with that with the 90 deg c column which allows for 290 amps.

Then to account for the 4 CC conductors, that was derated 80%, which brings the capacity down to 232 amps.

To that we would find the temperature requires another derating of 96%. So 232 x .96 = 222.7.

That is not higher than the 75 deg C column for that size wire, which is 255 amps, so is allowed to be considered the final current-carrying capacity.

Your continuous load of 85 amps would have 25% added, and taken as 106.25. To which you are adding 105 amps, for a minimum required ampacity of 211.25. So the wire you selected is sufficient, and as shown by David, also the smallest allowed.

Now 220.5(B) does allow for rounding, so we can call it 11 amps to spare.

 

[augie47]

add my agreement, but, with a question. FOUR current carrying conductors. ? Bit unusual. Do you have that big a non-linear load ?

 

[kwired]

add my agreement, but, with a question. FOUR current carrying conductors. ? Bit unusual. Do you have that big a non-linear load ?

Two single phase circuits in one raceway, one of them being the feeder in question?

 

[kwired]

Here we go again, but not with a motor circuit this time.

215.2(A)(1) minimum size and rating - not more than 600 volts - general -

Feeder conductors shall have an ampacity not less than required to supply the load as calculated in Parts III, IV, and V of Article 220. The minimum feeder-circuit conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load.

85 continuous x 1.25 = 106.25
105 non continuous

106.25 + 105 = 211.25 This is minimum size conductor required before adjustment factors and will be at 75C unless terminations are otherwise marked. This is the minimum size required by the termination temperature rating, the insulation temperature rating comes next. You make ampacity adjustments for insulation rating at 100 percent of load, but also at 90C (unless using other than 90C conductors) Again, the 125% factor for continuous loads is for the termination temp and is before ampacity adjustments.

85 cont + 105 non cont (both at 100%) = 190. Adjustments for insulation temp @ 90C will be based on a load of 190 amps.

I like to multiply the load by the reciprocal of the adjustment factors to come up with a minimum ampacity required instead of selecting a conductor, derating it then if it is too small doing it all over again with another size.

We have a factor of 1.25 (which is 1/.80) for an adjustment for number of conductors in the raceway.
We also have a factor of 1.04 (which is 1/.96) for an adjustment for ambient temperature.

190 x 1.25 x 1.04 = 247 amps is our minimum ampacity needed for a 90C conductor after accounting for needed adjustments.
247 @ 90C needs a 4/0 conductor which is good for up to 260 amps, and is not greater than the minimum size needed for termination temp @ 75C so we do not need an increase to the minimum size needed before adjustments.

4/0 90C conductor is the minimum size conductor needed for this application.

215.3 overcurrent protection -

...the rating of the overcurrent device shall not be less than the noncontinuous load plus 125 percent of the continuous load.

85 continuous X 1.25 = 106.25
105 non continuous

106.25 + 105 = 211.25

A 225 amp overcurrent device satisfies 215.3.

240.4(B) Overcurrent Devices Rated 800 Amperes or Less.

The next higher standard overcurrent device rating (above the ampacity of the conductors being protected) shall be permitted to be used, provided all of the following conditions are met:

(1) The conductors being protected are not part of a branch circuit supplying more than one receptacle for cord-and-plug-connected portable loads.
(2) The ampacity of the conductors does not correspond with the standard ampere rating of a fuse or a circuit breaker without overload trip adjustments above its rating (but that shall be permitted to have other trip or rating adjustments).
(3) The next higher standard rating selected does not exceed 800 amperes.

A 225 amp overcurrent device satisfies 240.4(B).

 

[david luchini]

Here we go again, but not with a motor circuit this time.

Yes, here we go again.

4/0 90C conductor is the minimum size conductor needed for this application.

I'm afraid a 4/0 90C is incorrect. Yes, 4/0 would be the correct minimum conductor size before the application of any adjustment and correction factors (per 215.2(A)(1)). And, Yes, 4/0 90C would have adequate ampacity for the load after adjustment/correction (also per 215.2(A)(1)).

BUT, the 4/0 90C would not be properly protected per 240.4. The ampacity of the 4/0 90C for the given conditions would be 260*0.8*0.96 = 199.68 ==> 200A. 200A is a standard OCPD rating, so the 4/0 conductor cannot be protected at the next size up (225A) using 240.4(B).

 

[kwired]

Yes, here we go again.



I'm afraid a 4/0 90C is incorrect. Yes, 4/0 would be the correct minimum conductor size before the application of any adjustment and correction factors (per 215.2(A)(1)). And, Yes, 4/0 90C would have adequate ampacity for the load after adjustment/correction (also per 215.2(A)(1)).

BUT, the 4/0 90C would not be properly protected per 240.4. The ampacity of the 4/0 90C for the given conditions would be 260*0.8*0.96 = 199.68 ==> 200A. 200A is a standard OCPD rating, so the 4/0 conductor cannot be protected at the next size up (225A) using 240.4(B).

I see what you are getting at. I am totally stumped at this point and need to do some research and thinking. The method I described when coming up with this ampacity I have used for some time now, works pretty well, and was taught to me by a guy kind of like Mike Holt, but not as well known. He teaches code and has published materials with the same kind of details and even drawings like Mike puts out in his materials that he gives out in his classes that outline the entire process I used and put a lot of time into producing them. I may just have to read through what I have again and see if I come up with something relating to this problem or maybe I missed something while doing this problem.

While at it I may just have to look at our motor circuit from another thread and see what I come up with related to the problem I had there also.

I do have a book from this guy that was based on 2011 NEC that I received when attending one of his classes. My calculation in this thread is a summarized version of what he goes through in the book, and was entirely from memory also, so maybe I missed something.

 

[david luchini]

I see what you are getting at. I am totally stumped at this point and need to do some research and thinking. The method I described when coming up with this ampacity I have used for some time now, works pretty well, and was taught to me by a guy kind of like Mike Holt, but not as well known. He teaches code and has published materials with the same kind of details and even drawings like Mike puts out in his materials that he gives out in his classes that outline the entire process I used and put a lot of time into producing them. I may just have to read through what I have again and see if I come up with something relating to this problem or maybe I missed something while doing this problem.

You're method is technically correct, especially for both parts of 215.2(A)(1) and 215.3. You just didn't follow all the way through when applying 240.4.

I like to multiply the load by the reciprocal of the adjustment factors to come up with a minimum ampacity required instead of selecting a conductor, derating it then if it is too small doing it all over again with another size.

I think using this method is what leads you astray. Not that it is wrong, but again, it didn't lead you to the correct answer regarding 240.4. Perhaps the way to do it would be to apply the reciprocals like you did to get the minimum ampacity, and then double check by applying the adjustment factors to the starting conductor ampacity based on the result. For instance, 190*1.25*1.04=247 needed starting ampacity, which would be at least a 4/0 90C --- Check: a 4/0 90C has a starting ampacity of 260. 260*0.8*0.96=200. A 200 ampcacity conductor cannot be protected by the 225A min ocpd required by 215.3.

The way I approached it was, I know I would need a final adjusted ampacity of not less than 190A to supply the load, but I also know that I would have a 225A min OCPD. 240.4(B) allows protection at the next size up, so I would have to have a final adjusted ampacity of not less than 201A to comply with both 215.2(A)(1) and 240.4. So, 201/0.8/0.96 = 261.7 would be my starting ampacity. 4/0 90C is too small, so 250 mcm 90C would be the minimum allowed.

 

[ggunn]

A 200 ampcacity conductor cannot be protected by the 225A min ocpd required by 215.3.

But a 201A ampacity conductor could, right?

 

[david luchini]

But a 201A ampacity conductor could, right?

Yes. :happyyes:

 

[kwired]

You're method is technically correct, especially for both parts of 215.2(A)(1) and 215.3. You just didn't follow all the way through when applying 240.4.



I think using this method is what leads you astray. Not that it is wrong, but again, it didn't lead you to the correct answer regarding 240.4. Perhaps the way to do it would be to apply the reciprocals like you did to get the minimum ampacity, and then double check by applying the adjustment factors to the starting conductor ampacity based on the result. For instance, 190*1.25*1.04=247 needed starting ampacity, which would be at least a 4/0 90C --- Check: a 4/0 90C has a starting ampacity of 260. 260*0.8*0.96=200. A 200 ampcacity conductor cannot be protected by the 225A min ocpd required by 215.3.

The way I approached it was, I know I would need a final adjusted ampacity of not less than 190A to supply the load, but I also know that I would have a 225A min OCPD. 240.4(B) allows protection at the next size up, so I would have to have a final adjusted ampacity of not less than 201A to comply with both 215.2(A)(1) and 240.4. So, 201/0.8/0.96 = 261.7 would be my starting ampacity. 4/0 90C is too small, so 250 mcm 90C would be the minimum allowed.

I agree with everything you said here, just don't recall ever having this problem pointed out and not sure just how often it maybe has been a problem. Maybe there is another step to the process I am missing, maybe there is a flaw in this guys method and it needs another step. When I get the chance I will review his material and see what I find. Maybe he has some logical explanation as to why what I came up with is right, but I agree with you for now that 240.4(B) is in violation with the way I did it.

 

[david luchini]

Your continuous load of 85 amps would have 25% added, and taken as 106.25. To which you are adding 105 amps, for a minimum required ampacity of 211.25. So the wire you selected is sufficient, and as shown by David, also the smallest allowed.

Now 220.5(B) does allow for rounding, so we can call it 11 amps to spare.

I hate to beat a dead horse, but I'm going to do it anyway...


The minimum required ampacity to carry the load is 190, not 211.25. The selected 250mcm would have 33 amps to spare.

 

[david luchini]

When I get the chance I will review his material and see what I find.

Please let us know what you find. Thanks.

 

[cpinetree]

This calculator makes things quick and easy.:happyyes:
http://www.electrician2.com/calculators/wireocpd_ver_1.html

(211.25 amps total load matches post #5 + #8)

 

[Smart $]

...
I think using this method is what leads you astray. Not that it is wrong, but again, it didn't lead you to the correct answer regarding 240.4. ...

It will if applied correctly. What I do is figure actual load and minimum OCPD first...

85 + 105 = 190A actual load
85 ? 125% + 105 = 211A minimum OCPD

Less than 800A permits protection with next larger standard OCPD at 225A>211A [240.4(B)]

Minimum conductor ampacity protected by 225A OCPD is 201A (and greater than actual load)...

For THHN:

201A ? 80% ? 0.96 = 262A minimum,

4/0 rated at 260A too small, so minimum 250kcmil required.

 

[david luchini]

Minimum conductor ampacity protected by 225A OCPD is 201A (and greater than actual load)...

For THHN:

201A ? 80% ? 0.96 = 262A minimum,

4/0 rated at 260A too small, so minimum 250kcmil required.

It's like deja vu all over again...:lol:

The way I approached it was, I know I would need a final adjusted ampacity of not less than 190A to supply the load, but I also know that I would have a 225A min OCPD. 240.4(B) allows protection at the next size up, so I would have to have a final adjusted ampacity of not less than 201A to comply with both 215.2(A)(1) and 240.4. So, 201/0.8/0.96 = 261.7 would be my starting ampacity. 4/0 90C is too small, so 250 mcm 90C would be the minimum allowed.

 

[Smart $]

It's like deja vu all over again...:lol:

Vodoo who do what? ...

Mine's more effective, at the very least in the use of white space