90C conductor terminated on 75C question.

[Paintsniffer69]

To my understanding, I thought you could not use 90C conductor on 75C terminations. How is the answer to the problem below correct? My temperature correction factor for 75C cable @ 50C is 0.75. Bundling is 0.80. MCA without adjustment factors is 190A. Applying adjustment factors gets me to (190)/((0.75 x 0.80)) = 316.67 A. I would select choice D of 500kcmil @ 380A (75C).

Problem statement:

THWN-2 Copper conductor is routed from an electrical room to a process area in a cable tray while being in contact with 5 other conductors. The ambient temperature is recorded as 30ºC but in certain areas through which this tray is routed, temperature can increase up to 50°C. This conductor will terminate on load terminals that are rated as 75ºC. This conductor will be used to supply 80A continuous load and 90A non-continuous load. Determine the minimum conductor size requirement according to NEC®.

Answer choices:
A. 2/0
B. 4/0
C. 250kcmil (correct answer per solution key)
D. 500kcmil

Answer explanation:

Even though load terminals are 75ºC rated, since THWN-2 is a 90ºC rated copper conductor, we can apply derating factors for temperature and bundling on its 90ºC ampacity instead of 75ºC ampacity rating.

According to Table 310.15(B)(2)(a), 50ºC temperature derating factor for 90ºC cable is 0.82.

According to Table 310.15(B)(3)(a), 6 conductor bundling derating factor is 0.80.

Total connected load that needs to be supplied = 80A x 1.25 + 90A = 190A (125% for continuous load and 100% for non-continuous).

Min conductor ampacity required = 190A / (0.80 × 0.82) = 289.6A

According to Table 310.15(B)(16), 250kcmil THW-2 copper conductor has 90ºC rating of 290A.

 

[Dennis Alwon]

Look at 110.14(C), You deal with the weakest link. 90C wire can be used but we cannot use the 90C rating except for de-rating. If every connection and piece of equipment was also rated 90C then we could use 90C at the 90C rating.

You are basically never going to see where everything is rated 90C. In general everything new is rated 75C somewhere on the circuit so the final overcurrent protective device after de-rating cannot be higher than the 75C column

 

[Paintsniffer69]

Look at 110.14(C), You deal with the weakest link. 90C wire can be used but we cannot use the 90C rating except for de-rating. If every connection and piece of equipment was also rated 90C then we could use 90C at the 90C rating.

You are basically never going to see where everything is rated 90C. In general everything new is rated 75C somewhere on the circuit so the final overcurrent protective device after de-rating cannot be higher than the 75C column

"(C) temperature limitations. The temperature rating associated with the ampacity of a conductor shall be selected and coordinated so as not to exceed the lowest temperature rating of any connected termination, conductor, or device. Conductors with temperature ratings higher than specified for terminations shall be permitted to be used for ampacity adjustment, correction, or both."

Thanks, Dennis. That has been my understanding as well.

To provide more context to my question, it is being asked in an academic sense, not a practical sense. I'm beginning to study for the PE Power exam, and I got the question I posted wrong.

Isn't there one additional step the solution omitted? Don't you need to backtrack from the adjusted 90C conductor ampacity to verify the equivalent 75C conductor ampacity or something?

 

[Dennis Alwon]

So your temperature correction factor is based on wire insulation.

Since you are using the value of 90C wire-- #2 Thwn-2 then obviously the final step will entail a 90C wire.

Work backwards

190 amps is need

250kcm at 90 C = 290 amps then de-rate

290 x.8 x.82 = 190.24

250 kcm at 75C= 255 amps so our ocpd is not larger than 255 amps. In fact a 200 overcurrent protective device is all that is needed

 

[wwhitney]

The answer and its methodology are incorrect. There are 2 checks:

1) Termination check, no ampacity adjustment and correction. This includes the 125% continuous use factor and the termination temperature limit. So you need a conductor that at the 75C table rating is good for 190A.

2) As Example D3(a) puts it "conditions of use throughout the raceway run." Now the ampacity is only required to be 170A, and you can start at the 90C insulation temperature rating, but you must apply any adjustment or correction factors. So you need a conductor whose table 90C rating is good for 170A / (0.80 * 0.82) = 259A.

That means the minimum Cu conductor size is in fact 4/0, which has a 90C rating of 260A and a 75C rating of 230A.

Cheers, Wayne

 

[jaggedben]

Wayne, why is the 125% factor on the continuous load not required to be part of the conditions of use calculation?

 

[wwhitney]

Wayne, why is the 125% factor on the continuous load not required to be part of the conditions of use calculation?

Say we are dealing with a feeder, then 215.2 provides the guidance. [For a branch circuit, 210.19 has equivalent language.]

The short answer:

2020 NEC 215.2(A)(1)(b) [which is 215.2(A)(2) in the 2023 NEC] tells us that when applying ampacity adjustment and correction factors, the minimum feeder ampacity is just the sum of the loads, no 125% for continuous loads. And 110.14(C) says that since we are applying the adjustment and correction factors, we can use the 90C ampacity of the conductors.

Elaboration:

2020 NEC 215.2(A)(1)(a) [which is 215.2(A)(1) in the 2023 NEC] tells us, without mentioning ampacity adjustment and correction, that the "ampacity" needs to be at least 100% of the noncontinuous loads plus 125% of the continuous loads. The NEC errs in using the word ampacity here, as that term always includes "conditions of use", but the contrast with 215.2(A)(1)(b) is clear. So the intention in 215.2(A)(1)(a) is that the check be made without ampacity adjustment and correction. As such, 110.14(C) tells us we are limited to the termination temperature ampacity

Further points in favor of that interpretation of 215.2(A)(1)(a): if ampacity adjustment and correction were to be used in 215.2(A)(1)(a), then 215.2(A)(1)(b) would be moot, as the required ampacity in 215.2(A)(1)(a) would always be at least as large as in 215.2(A)(1)(b) and would be calculated the same way. Also, 690.8(B), for the case of PV conductors, is providing the same sizing guidance with clearer language: 690.8(B)(1) is "Without Adjustment and Correction Factors" and requires a 125% continuous use factor, while 690.8(B)(2) is "With Adjustment and Correction Factors" and does not require a continuous use factor.

If you think about it, the two checks are for terminations vs "throughout the raceway run". At terminations, ampacity adjustment for conductor count does not apply, as the termination is typically in an enclosure with much more volume per conductor. Apparently ampacity correction for temperature does not apply, I guess because the equipment will have its own temperature limitations that control (such as the 40C rated ambient temperature of a breaker). And lastly, because the 125% continuous use factor exists only due to the limitations of breakers, it only applies at terminations, one of which will be at a breaker (and there are exceptions for the 125% continuous factor in 215.2(A)(1)(a) for various cases where there is no breaker).

Cheers, Wayne

 

[jim dungar]

And lastly, because the 125% continuous use factor exists only due to the limitations of breakers, it only applies at terminations, one of which will be at a breaker (and there are exceptions for the 125% continuous factor in 215.2(A)(1)(a) for various cases where there is no breaker).

The 125% factor applies to both breakers and fusible switches, which are both tested per UL standards with the OCPD in open air but are installed in enclosures in the field.

 

[Paintsniffer69]

The answer and its methodology are incorrect. There are 2 checks:

1) Termination check, no ampacity adjustment and correction. This includes the 125% continuous use factor and the termination temperature limit. So you need a conductor that at the 75C table rating is good for 190A.

2) As Example D3(a) puts it "conditions of use throughout the raceway run." Now the ampacity is only required to be 170A, and you can start at the 90C insulation temperature rating, but you must apply any adjustment or correction factors. So you need a conductor whose table 90C rating is good for 170A / (0.80 * 0.82) = 259A.

That means the minimum Cu conductor size is in fact 4/0, which has a 90C rating of 260A and a 75C rating of 230A.

Cheers, Wayne

Your comment makes sense. I see your reply, #7, later down in the thread, and I agree with your interpretation. The wording in 215.2(A)(1) implies the conductor size is based on the worst-case load scenario.

Re-writing out what you said to confirm my comprehension.

Max[A, B] = The load used to size the conductor

A = Conductor size per 215.2(A)(1)(a), accounts for noncontinuous loads @ 100% and continuous loads @ 125%

B = Conductor size per 215.2(A)(1)(b), accounts for adjustment or temperature correction factors and neglects the continuous load factor for continuous loads

A = 1.25 * 80A + 1.00 * 90A = 190A

B = 80A + 90A = 170A.... (170A) / (0.80 * 0.82)) = 259A,

Max[190A, 259A] = 259A. Select 4/0AWG CU @ 90C (We can use the 90C column since we are considering bundling and or correction factors).

At first glance, it seems our 90C wire of 4/0AWG CU (260A) exceeds the 75C rating of 4/0AWG CU (230A). However, the 90C conductor current-carrying capability is restricted due to the number of conductors in the raceway and the increased ambient temperature. The actual ampacity of the 90C conductor is 260A * 0.80 * 0.82 = 170.56A, which is greater than the load (necessary to be larger than) and less than the 75C rating of 4/0AWG CU (230A) letting us know we won't exceed the 75C termination ratings.

We can conclude the conductor selection of 90C 4/0AWG CU (260A) is sufficient given the problem statement's parameters.

 

[Paintsniffer69]

Say we are dealing with a feeder, then 215.2 provides the guidance. [For a branch circuit, 210.19 has equivalent language.]

The short answer:

2020 NEC 215.2(A)(1)(b) [which is 215.2(A)(2) in the 2023 NEC] tells us that when applying ampacity adjustment and correction factors, the minimum feeder ampacity is just the sum of the loads, no 125% for continuous loads. And 110.14(C) says that since we are applying the adjustment and correction factors, we can use the 90C ampacity of the conductors.

Elaboration:

2020 NEC 215.2(A)(1)(a) [which is 215.2(A)(1) in the 2023 NEC] tells us, without mentioning ampacity adjustment and correction, that the "ampacity" needs to be at least 100% of the noncontinuous loads plus 125% of the continuous loads. The NEC errs in using the word ampacity here, as that term always includes "conditions of use", but the contrast with 215.2(A)(1)(b) is clear. So the intention in 215.2(A)(1)(a) is that the check be made without ampacity adjustment and correction. As such, 110.14(C) tells us we are limited to the termination temperature ampacity

Further points in favor of that interpretation of 215.2(A)(1)(a): if ampacity adjustment and correction were to be used in 215.2(A)(1)(a), then 215.2(A)(1)(b) would be moot, as the required ampacity in 215.2(A)(1)(a) would always be at least as large as in 215.2(A)(1)(b) and would be calculated the same way. Also, 690.8(B), for the case of PV conductors, is providing the same sizing guidance with clearer language: 690.8(B)(1) is "Without Adjustment and Correction Factors" and requires a 125% continuous use factor, while 690.8(B)(2) is "With Adjustment and Correction Factors" and does not require a continuous use factor.

If you think about it, the two checks are for terminations vs "throughout the raceway run". At terminations, ampacity adjustment for conductor count does not apply, as the termination is typically in an enclosure with much more volume per conductor. Apparently ampacity correction for temperature does not apply, I guess because the equipment will have its own temperature limitations that control (such as the 40C rated ambient temperature of a breaker). And lastly, because the 125% continuous use factor exists only due to the limitations of breakers, it only applies at terminations, one of which will be at a breaker (and there are exceptions for the 125% continuous factor in 215.2(A)(1)(a) for various cases where there is no breaker).

Cheers, Wayne

Thank you for the detailed reply this helps a lot. By the way, the current version of the PE Power Exam utilizes NEC 2017 so all of my preparation has been based on that. 215.2(A)(1) in 2020 echoes what is said in 2017, except with the added clarification of "... and shall comply with 110.14(C)." And 215.2(A)(1)(b) 2020 has ".. after the application of any adjustment or correction factors in accordance with 310.14."

It'd be nice if NFPA would spell it out for us in the next release of the code... one can dream.

 

[wwhitney]

A = 1.25 * 80A + 1.00 * 90A = 190A

B = 80A + 90A = 170A.... (170A) / (0.80 * 0.82)) = 259A,

Max[190A, 259A] = 259A. Select 4/0AWG CU @ 90C (We can use the 90C column since we are considering bundling and or correction factors).

Not quite. The table lookup for (A) is limited to the 75C column, while the lookup in (B) can use the 90C column. So you need to do the table lookups for conductor size separately, and then just take the maximum conductor size.

At first glance, it seems our 90C wire of 4/0AWG CU (260A) exceeds the 75C rating of 4/0AWG CU (230A).

The wording of 110.14(C) might make you think that's a possible concern, but it's just bad wording. That comparison doesn't matter. All 110.14(C) means is that in check (A) you are limited to the termination temperature rating, while check (B) can use the insulation temperature rating.

Cheers, Wayne

 

[Paintsniffer69]

Not quite. The table lookup for (A) is limited to the 75C column, while the lookup in (B) can use the 90C column. So you need to do the table lookups for conductor size separately, and then just take the maximum conductor size.

You're right. After the MCA calculation for item A, my comment should read "Select 3/0AWG CU @ 75C (200A). Use of 90C is not permitted." Now our Max[A, B] is Max[3/0AWG CU @75C, 4/0AWG CU @ 90C].

The wording of 110.14(C) might make you think that's a possible concern, but it's just bad wording. That comparison doesn't matter. All 110.14(C) means is that in check (A) you are limited to the termination temperature rating, while check (B) can use the insulation temperature rating.

I added that bit when I read the last paragraph in comment #3 of the this thread: https://forums.mikeholt.com/threads/conductor-ampacity-derating-90-degree-column.99044/.

 

[Dennis Alwon]

Not quite. The table lookup for (A) is limited to the 75C column, while the lookup in (B) can use the 90C column. So you need to do the table lookups for conductor size separately, and then just take the maximum conductor size.


The wording of 110.14(C) might make you think that's a possible concern, but it's just bad wording. That comparison doesn't matter. All 110.14(C) means is that in check (A) you are limited to the termination temperature rating, while check (B) can use the insulation temperature rating.

Cheers, Wayne

@wwhitney -- thanks Wayne, I think this was brought up years ago and I believe Smart$ was involved in it. I can't remember but one of the writers to an electrical magazine wrote a calculation question here and I, of course, made the same mistake as I did here. Smart, of course, as smart as he was got it correct. I believe the 2014 NEC made changes to 215.2(A)(1) to clarify what you were saying. I finally see it....

 

[Dennis Alwon]

I will guarantee there would be a lot of exams where the answer giver is exactly like Paintsniffer got. And test takers will answer it as I did which is correct on the test but incorrect per NEC...hahaha

 

[ggunn]

Wayne, why is the 125% factor on the continuous load not required to be part of the conditions of use calculation?

The short answer is that if you have used the 75 degree column to size the wire so that it won't get over 75 degrees in continuous use, and the wire sized using the 90 degree ampacity derate for conditions of use is not smaller that that, you have already taken it into account. But you already know that, don't you? I didn't read everything in the thread.

 

[wwhitney]

The short answer is that if you have used the 75 degree column to size the wire so that it won't get over 75 degrees in continuous use, and the wire sized using the 90 degree ampacity derate for conditions of use is not smaller that that, you have already taken it into account.

Right, but the point was that when doing the wire sizing "using the 90 degree ampacity derate for conditions of use," you don't need to use a 125% factor for continuous use. If you do use a 125% factor there, you may be unnecessarily oversizing the wire.

Cheers, Wayne

 

[jaggedben]

No I didn't already know that. I was under the impression that the 125% factor for continuousness was an across-the-board circuit current requirement. At least, until recently.

Also the interesting case is where the 90C wire in conditions of use would have to be larger than the termination check if the 125% factor is required.

I've not had time to fully review Wayne's answer to see if I agree that's what the code says. But it sounds reasonable to me.

 

[ggunn]

Right, but the point was that when doing the wire sizing "using the 90 degree ampacity derate for conditions of use," you don't need to use a 125% factor for continuous use. If you do use a 125% factor there, you may be unnecessarily oversizing the wire.

If you have 75 degree terminations, you must ensure that the wire will not get hotter than 75 degrees. If the conductors are in continuous use, that means that 125% of the current cannot exceed the ampacity in the 75 degree column. The conditions of use calculation is completely separate, and most of the time the minimum wire size from the COU calculation is smaller than the CU calculation yields (in my world, anyway), so the CU calculation drives the conductor size since both CU and COU must be satisfied.

 

[wwhitney]

If you have 75 degree terminations, you must ensure that the wire will not get hotter than 75 degrees. If the conductors are in continuous use, that means that 125% of the current cannot exceed the ampacity in the 75 degree column.

It is correct that for a continuous load, the NEC requires that 125% of the load not exceed the unadjusted termination ampacity.

However, the "that means" portion of your statement is not correct. To keep the conductor temperature at the termination from exceeding 75C, the continuous current could be up to the full 75C ampacity. See the definition of ampacity--it is already a continuous rating. As such, the NEC is being overly conservative here.

The only reason the 125% continuous use factor exists in the NEC is to account for the thermal effects of enclosing possibly multiple OCPD within an enclosure, the resulting heat buildup, and the possibility of nuisance tripping of an OCPD at or below its rating with continuous loading. Then because of the need for the conductor to still be protected by the resulting upsized OCPD (240.4), that has implications for conductor sizing. The inclusion of the 125% factor in the conductor sizing rules (215.2 and 210.19) has the effect of ruling out the use of the "next size up" rule 240.4(B), at least for installations controlled by the termination ampacity check.

Cheers, Wayne

P.S. I'm not sure why your post was a reply to mine--your post doesn't contradict or amend anything of mine you quoted. Namely that you don't need to use a 125% continuous use factor when doing the conditions of use ampacity check.

 

[wwhitney]

The answer and its methodology are incorrect. There are 2 checks:

OK, my statement above requires correction. The problem answer is correct (250 kcmil is the minimum size) if we assume no 100% rated OCPD, but not for the reason given in the answer key in the OP.

There is a 3rd check required (one I've overlooked before, and managed to do so here, even though I was aware of it):

3) 240.4 requires that the ampacity (which mean the "conditions of use" result, check 2 in my first post) be sufficient for the minimum OCPD size. 125% of continuous load plus non-continuous load = 190A, so the minimum OCPD size is 200A (unless the OCPD is 100% rated, then it could be 175A, but 100% rated 175A OCPD is rare or non-existent). As the next lower OCPD size is 175A, 240.4(B) would make the minimum ampacity 176A. 4/0 Cu is too small, as under the conditions of use its ampacity is only 171A. So 250 kcmil Cu is in fact the correct size.

That also means that to my point of not needing to use the 125% continuous use factor in the "conditions of use check," while it's correct on the face of it, check (3) often has the result of requiring something similar. Which was maybe what ggunn was trying to tell me.

Now if the ambient temperature in the problem were 47C instead of 50C, then we could use the formula for temperature correction to calculate a sharper temperature correction factor, so that the ampacity of 4/0 Cu is in fact 276A, and then that would be the minimum size.

Cheers, Wayne