Code Treatment of Adjustment Factor for Higher Conductor Temp Rating

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bjhuffine

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I'm trying to wrap my mind around this concept so that I can ensure full understanding. The scenario is based on a calculation example in the 2011 Handbook. But because I've been studying the NEC 2008 code book for some time now, any code references below are from the 2008 edition. Also, please bear with me here, this scenario is somewhat complex:

OCPD terminations rated at 75 C

125A noncontinuous + 200A continuous loads

Therefore, per 215.3: OCPD=125 + 1.25*200 = 375A... per 240.6 OCPD = 400A

3-phase, 4-wire feeder with nonlinear load. therefore, per310.15(B)(4)(c), will be treated as current-carrying, therefore all four in the same raceway will be subject to the adjustment factor applied via 310.15(B)(2)(a), or 80%.

per 215.2(A)(1), Min Feeder Conductor Ampacity = 125 + 1.25 * 200 = 375A

So let's assume a 600kcmil conductor @ 90C (rated 475A) is chosen. This means:

600kcmil @ 90C Ampacity = 475A * 80% = 380A which can be protected by the 400A OCPD so we're good here.

However, 110.14 states that when a higher temp rated conductor is terminated on a 75C termination, the conductor ampacity used must be at the lower temperature (ie 75C). So the question is, how does this ampacity get treated?

Option 1:

My original assumption is that the adjustment factor is designed to derate the conductor ampacity as a result of the number of current-carrying conductors in the same race, which doesn't change. So the 80% still applies:

600kcmil @ 75C = 420A * 80% = 336A

This also means that the load current cannot exceed this value and taking into account that the continous load adjustment of 125% is based on the additional heat considerations when continuous, I'll neglect this factor as heat is already taken into account when using this factor as compared to the 90C insulation. Therefore:

Load Current = 125 + 200 = 325A < 336A so we're good.

If I hadn't done this, then it wouldn't have validated, in other words:

125 + 1.25 * 200 = 375A > 336A conductor ampacity at 75C, therefore selection is not good.

Alternatively I could've said:

Option 2:

But since the NEC doesn't seem to clear on this, I wondered if it could've been treated with the 75C requirement from 114 without adjusting the ampacity:

600kcmil @ 75C = 420A

which means that the min feeder ampacity requirement of 375A (using the 125% adjustment for continuous load) is met by the 420A ampacity.

This doesn't seem right to me because we are neglecting the 80% adjustment for multiple current-carrying conductors in the same raceway.



I know from the example calculation, they state that the 600kcmil @ 90C is a fit for this application and that the 75C value should be compared to the load current. What they don't say is how. And it doesn't exactly seem clear to me in the code itself which is best. Any help would be appreciated.
 
Option 3:
? Min Feeder Conductor Ampacity = 125 + 1.25 * 200 = 375A
? 600kcmil @ 90C Ampacity = 475A * 80% = 380A.
? 380 is higher than the minimum required of 375.
? You are done.
All that 110.14 does to this situation is make you verify that you don’t use the conductor’s “calculated ampacity,” in this case 380, if that value is higher than the same conductor’s “tabulated ampacity at 75C,” in this case 420.
 
Okay, I read the IAEI article from jumper and charlie b's comment. Again, I really appreciate you helping me clarify this. I definitely want to ensure I have a firm grasp.

So the rules are as follows:

  1. Regardless, the required ampacity (125 + 1.25*200=375A) must be met. So where any mention of load is of concern it is referencing the required ampacity based on conductor capabilities to sufficiently carry the load without overheating?
  2. 110.14 states "Unless the equipment is listed and marked otherwise, conductor ampacities used in determining equipment termination providions shall be based on Table 310.16 as appropriately modified by 310.15(B)(6)". This essentially states that other tables can be used for conductor ampacities with exception of any conductor ampacity at termiantion. In other words when terminating a conductor, T310.16 is the only table allowed.
  3. Since a 90C conductor is being considered for 75C terminals, it gets special consideration where it is the smaller of:
    • The 600kcmil @90C with Adjustment Factors applied (475A * 80% = 380A)
    • The 600kcmil @75C without requiring the adjustment factors (420A)

So the smaller of the 90C (calculated) or 75C (tabulated) is the 90C @ 380A. This ensures a less likelyhood of overheating the equipment to which it terminates. And since 380A>the required 375A, we're good.

I think I've been in a mode of considering the conductor as either a 75C or 90C conductor. The fact is that it is a 90C conductor to which we want to ensure that the appropriate ampacity (calculated vs tabulated) is used. I assume then that if the tabulated (75C) was lower, it would be used as the ampacity of the conductor without adjustments for the 90C conductor that is physically installed. In other words a margin of safety.

Does that sound right?
 
I fear that I am having some difficulty following your description. Let me put it in other terms, and see if this helps.

First of all, we need to keep in mind the basic concept that a chain is no stronger than its weakest link. Regarding conductor ampacities, the conductor?s insulation system may be able to handle a great deal of heat, but the device to which the conductor is connected, and specifically the termination points, may not be able to handle as much heat. Thus, the terminations are usually the weakest link, in that they are usually rated for only 75C, whereas the conductor?s insulation system may be rated for 90C. That is why we get to use the 90C values in table 310.16, when we apply adjustment or correction factors, and why we cannot use a final calculated ampacity that is higher than the value shown in the 75C column.
 
The end of the wire is not the middle of it.
The derating process is for more than 3 current carrying conductors IN A CONDUIT (and BTW conductors that only carry the unbalanced load do not count). The middle of the conductor.
The termination rule applies to the end of the conductor. That end is not in a conduit.
So actually you can use 500MCM for your example. 500MC ampacity is 380 at 75 degress, no deration required, and next higher size overload protection can be used.

bjhuffine said:
I'm trying to wrap my mind around this concept so that I can ensure full understanding. The scenario is based on a calculation example in the 2011 Handbook. But because I've been studying the NEC 2008 code book for some time now, any code references below are from the 2008 edition. Also, please bear with me here, this scenario is somewhat complex:

OCPD terminations rated at 75 C

125A noncontinuous + 200A continuous loads

Therefore, per 215.3: OCPD=125 + 1.25*200 = 375A... per 240.6 OCPD = 400A

3-phase, 4-wire feeder with nonlinear load. therefore, per310.15(B)(4)(c), will be treated as current-carrying, therefore all four in the same raceway will be subject to the adjustment factor applied via 310.15(B)(2)(a), or 80%.

per 215.2(A)(1), Min Feeder Conductor Ampacity = 125 + 1.25 * 200 = 375A

So let's assume a 600kcmil conductor @ 90C (rated 475A) is chosen. This means:

600kcmil @ 90C Ampacity = 475A * 80% = 380A which can be protected by the 400A OCPD so we're good here.

However, 110.14 states that when a higher temp rated conductor is terminated on a 75C termination, the conductor ampacity used must be at the lower temperature (ie 75C). So the question is, how does this ampacity get treated?

Option 1:

My original assumption is that the adjustment factor is designed to derate the conductor ampacity as a result of the number of current-carrying conductors in the same race, which doesn't change. So the 80% still applies:

600kcmil @ 75C = 420A * 80% = 336A

This also means that the load current cannot exceed this value and taking into account that the continous load adjustment of 125% is based on the additional heat considerations when continuous, I'll neglect this factor as heat is already taken into account when using this factor as compared to the 90C insulation. Therefore:

Load Current = 125 + 200 = 325A < 336A so we're good.

If I hadn't done this, then it wouldn't have validated, in other words:

125 + 1.25 * 200 = 375A > 336A conductor ampacity at 75C, therefore selection is not good.

Alternatively I could've said:

Option 2:

But since the NEC doesn't seem to clear on this, I wondered if it could've been treated with the 75C requirement from 114 without adjusting the ampacity:

600kcmil @ 75C = 420A

which means that the min feeder ampacity requirement of 375A (using the 125% adjustment for continuous load) is met by the 420A ampacity.

This doesn't seem right to me because we are neglecting the 80% adjustment for multiple current-carrying conductors in the same raceway.



I know from the example calculation, they state that the 600kcmil @ 90C is a fit for this application and that the 75C value should be compared to the load current. What they don't say is how. And it doesn't exactly seem clear to me in the code itself which is best. Any help would be appreciated.
Click to expand...
 
Long day at work, so I'm just now able to read this. Charlie B... yes, that is the direction I was going. I just figured if I could accurately bullet the major points of consideration then hopefully I'm on the right track. Though I hadn't quite thought of it as the weakest link before. That's a really good analogy.

To add to that, I also recently read where manufacturers may use a 90C termination on a piece of equipment, but that there really aren't any 90C equipment. In other words, the temperature rating requirement applies for the whole piece of distribution/service equipment, not just the lugs to which the conductor lands. Since there's really not 90C equipment (or if there is, there are very few), more importance is placed on using the 75C for over 100A and 60C for under.

Strife... thanks for your comment. I hear you on the fact that the adjustment factors applies to the "middle" of the conductors instead of the point of termination which is not in the raceway but in an enclosure instead. That's a great point that can help one be in the right frame of mind when reading this. The derating factor is designed to be applied due to the heat generated by multiple conductors in a raceway. This means that the ampacity for a 90C conductor would be derated at the 90C column value for the "middle" (as long as it is lower than the tabulated 75C column value that is necessary to check because of the termination).

But I don't think a 500kcmil with 75C insulation would work because in this scenario it would be run as one of four current-carrying conductors in the same raceway, each the same size. Therefore it too would have to be derated per 310.15(B)(2)(a) at the 80% adjustment factor which would lower the ampacity to 380 * 80% = 304A which cannot be protected by the minimum 400A OCPD. 304A conductor ampacity is also less than the 375A minimum ampacity allowed for the load in this case which means the "middle" of the conductor could potentially overheat. At least that's the way I understand it. In fact, the NEC 2011 handbook to which this scenario was derived first attempted to use a 500kcmil conductor and explained why it couldn't be used (though most of its explanation was related to the min OCPD requirement). My original problem was related to a conductor with a higher insulation value. I was confused about why the higher temp rating could be derated but not at the 75C value. I now understand.

Thank you all!
 
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