Feeder Calculations and Derating

[david luchini]

Wayne I think you answered my question here, albeit inadvertently. So even though my conductor at 60 degrees is rated at 20 amps, it still remains a viable option since its ampacity can suffice the maximum total load of my feeders. There's been a lot of talk around the shop, that the ampacities under the 60 degree column need to support, not the maximum total load of my feeders, but the derated load of my feeders (23.8 amps).

I think you have the concept backwards. The "load" isn't derated, the ampacity of the conductor is derated.

You need a conductor that has an ampacity of at least 16.56A after the application of any adjustment or correction factors.

 

[wwhitney]

Wayne I think you answered my question here, albeit inadvertently. So even though my conductor at 60 degrees is rated at 20 amps, it still remains a viable option since its ampacity can suffice the maximum total load of my feeders. There's been a lot of talk around the shop, that the ampacities under the 60 degree column need to support, not the maximum total load of my feeders, but the derated load of my feeders (23.8 amps).

I would say that your #12 conductors with 90C insulation have an ampacity of 30A * derating, so 30*sqrt(.4) = 19A where they are in a 66C ambient. The terminations would have an ampacity of 20A or 25A depending on whether they are 60C or 75C rated. And 240.4(D) would limit the OCPD to 20A for applications not subject to 240.4(G).

Note also that for 30C ambient, I would therefore say that the #12 Cu 90C conductors have an ampacity of 30A. But that won't be the controlling factor, the termination ampacity or 240.4(D) would be. [Except for special cases, like a motor load and a run of 90C #12 Cu between separate enclosures containing only 90C rated wire connectors. Then you can use the #12 Cu at 30A.]

Cheers, Wayne

 

[winterovereverything]

I think you have the concept backwards. The "load" isn't derated, the ampacity of the conductor is derated.

You need a conductor that has an ampacity of at least 16.56A after the application of any adjustment or correction factors.

You're right, maybe this is just a mix up on my behalf here. And that ampacity would be selected from the 60 degree column, correct?

 

[wwhitney]

You're right, maybe this is just a mix up on my behalf here. And that ampacity would be selected from the 60 degree column, correct?

No, conductor ampacity is determined from the column corresponding to conductor insulation temperature (and is subject to ampacity correction and adjustment, and must be at least the total load without any continuous use factors, and must be sufficient under 240.4 for the OCPD size selected).

Termination ampacity is determined from the column corresponding to termination temperature (and is not subject to ampacity correction and adjustment, and must be at least 125% of the continuous load plus the non-continuous load unless using a 100% rated OCPD).

Cheers, Wayne

 

[winterovereverything]

No, conductor ampacity is determined from the column corresponding to conductor insulation temperature (and is subject to ampacity correction and adjustment, and must be at least the total load without any continuous use factors, and must be sufficient under 240.4 for the OCPD size selected).

Termination ampacity is determined from the column corresponding to termination temperature (and is not subject to ampacity correction and adjustment, and must be at least 125% of the continuous load plus the non-continuous load unless using a 100% rated OCPD).

Cheers, Wayne

Got it. Thank you guys for the clarification and for helping a very insane man grow a little more sane since this question was presented to me Also my apologies for the vague wording in any of these posts. Been a while since I got some decent sleep and I think all that time has finally managed to rout me.

 

[Dennis Alwon]

Dennis you are overlooking that there are two separate checks to be done, and we don't need to apply the 125% continuous use factor and the derating factors at the same time. So #12 turns out to be OK.

For the first check (which Example D3(a) in the Annex calls the "termination" check), we use the termination temperature ampacity column, with no derating, and with a 125% continuous use factor. For the second check (which Example D3(a) in the Annex calls the "conditions of use through the raceway run" check), we use the insulation temperature ampacity column, with derating, and without any 125% continuous use factor.

Cheers, Wayne

I keep forgetting that once you use the 125% for continuous loads then you don't need to double de-rate unless the other de-rating factors are higher than the 125%.

At least that is how I learned it

 

[Dennis Alwon]

@david luchini why are you using .63 from the 40C table instead of .58 from the 30C table

 

[wwhitney]

@david luchini why are you using .63 from the 40C table instead of .58 from the 30C table

The 30C table has entries that are every 5C, so for each entry you have the worst case factor if the temperature were actually rounded up to the next multiple of 5C. For 66C that makes the effect significant as you'd actually be derating for 70C ambient.

The table entries are themselves coming from a formula, which just says that the temperature rise is proportional to the power dissipated, which itself is proportional to the current squared. That means the allowable current is proportional to the square root of the allowable temperature rise.

In other words, whiles the table ampacities for the 90C column are based on the a 60C temperature rise (90C - 30C ambient), with 66C ambient we can only permit a 24C temperature rise (90C - 66C). That means we are allowed 24/60 = 40% of our normal temperature rise. And therefore the allowable ampacity will be sqrt(40%) = 63.2% of our normal ampacity. If you use this procedure for multiples of 5C, you will generate all the values in the table.

Either that or he just looked at the wrong table, which by coincidence gives the same value. : - )

Cheers, Wayne

 

[wwhitney]

I keep forgetting that once you use the 125% for continuous loads then you don't need to double de-rate unless the other de-rating factors are higher than the 125%.

It's not quite that simple, as the 125% factor (without ampacity correction or adjustment) get used with the termination temperature ampacity, while the normal derating (ampacity correction and adjustment) gets used with the conductor insulation temperature. So it really is two separate checks.

Cheers, Wayne

 

[Dennis Alwon]

The 30C table has entries that are every 5C, so for each entry you have the worst case factor if the temperature were actually rounded up to the next multiple of 5C. For 66C that makes the effect significant as you'd actually be derating for 70C ambient.

The table entries are themselves coming from a formula, which just says that the temperature rise is proportional to the power dissipated, which itself is proportional to the current squared. That means the allowable current is proportional to the square root of the allowable temperature rise.

In other words, whiles the table ampacities for the 90C column are based on the a 60C temperature rise (90C - 30C ambient), with 66C ambient we can only permit a 24C temperature rise (90C - 66C). That means we are allowed 24/60 = 40% of our normal temperature rise. And therefore the allowable ampacity will be sqrt(40%) = 63.2% of our normal ampacity. If you use this procedure for multiples of 5C, you will generate all the values in the table.

Either that or he just looked at the wrong table, which by coincidence gives the same value. : - )

Cheers, Wayne

Wayne, the NEC does not expect us to use that much physics or engineering when we try to calculate wire size. IMO, those tables are there for our use and although I still don't quite get the difference between the tables it is rare that the 40C table is used. Sorry I wish I could grasp all the info you posted.

 

[wwhitney]

Wayne, the NEC does not expect us to use that much physics or engineering when we try to calculate wire size.

No, that's why the tables are there. But the formula I explained is also there as Equation 310.15(B). Use whichever you like. If you use the table, you'll get 0.58, which is the exact value for 70C. If you use the formula, you get 0.63, which is the exact value for 66C.

It's really not that complicated once you get used to it. The temperature correction factor is just "sqrt(actual temperature headroom/standard temperature headroom)". Where temperature headroom is just the conductor insulation temperature less the ambient temperature, and "standard" means using the ambient values listed on the source ampacity table, 30C usually.

Cheers, Wayne

 

[david luchini]

@david luchini why are you using .63 from the 40C table instead of .58 from the 30C table

Because I've been a bit under the weather and read the wrong table. Thanks for catching that.

 

[Dennis Alwon]

Because I've been a bit under the weather and read the wrong table. Thanks for catching that.

I thought that might be the situation but I wanted to check to make sure I wasn't missing something. I hope you feel better

 

[wwhitney]

Because I've been a bit under the weather and read the wrong table. Thanks for catching that.

Ah, but you coincidentally got a more accurate value, rather than rounding the ambient up to 70C as using the tables would do.

Since (90 - 70) / (90 - 40) = 40% = (90 - 66) / (90 - 30).

Cheers, Wayne

 

[ramsy]

Wayne, the NEC does not expect us to use that much physics or engineering when we try to calculate wire size. IMO, those tables are there for our use and although I still don't quite get the difference between the tables it is rare that the 40C table is used. Sorry I wish I could grasp all the info you posted.

Your brain was trained to use Ohm's law to solve for Volts, Ohms, or Watts. You didn't need NFPA tables to break out the increments for each volt, ohm, or watt.

Why not solve for temperature corrected ampacity with a formula:

Per 310.15(B): I'-I*Sqrt(Tc-T'a/Tc-Ta)

I′ = ampacity corrected for ambient temperature
I = ampacity shown in the tables
Tc = temperature rating of conductor (°C)
Ta′ = new ambient temperature (°C)
Ta = ambient temperature used in the table (°C)

-----
Or solve for conductor temperature:
-----

T2 = (Ambient + Temp.Rise)
T2 = T1 + (TR-T1) * Load^2 / Imax^2

Load = Circuit Amps
T1 = Ambient Temperature
T2 = Estimated Conductor Temperature
TR = Conductor Max Temperature Rating
Imax = Max Amps at TR, derated for ccc's

 

[chorty55]

Your brain was trained to use Ohm's law to solve for Volts, Ohms, or Watts. You didn't need NFPA tables to break out the increments for each volt, ohm, or watt.

Why not solve for temperature corrected ampacity with a formula:

Per 310.15(B): I'-I*Sqrt(Tc-T'a/Tc-Ta)

I′ = ampacity corrected for ambient temperature
I = ampacity shown in the tables
Tc = temperature rating of conductor (°C)
Ta′ = new ambient temperature (°C)
Ta = ambient temperature used in the table (°C)

-----
Or solve for conductor temperature:
-----

T2 = (Ambient + Temp.Rise)
T2 = T1 + (TR-T1) * Load^2 / Imax^2

Load = Circuit Amps
T1 = Ambient Temperature
T2 = Estimated Conductor Temperature
TR = Conductor Max Temperature Rating
Imax = Max Amps at TR, derated for ccc's

I'm not following that formula.

 

[letgomywago]

What are the 2 loads you have? And why do you need a feeder for them? That'll help us with our answers. Also wiring method you wish to use since you'll have a sub panel If it's a true feeder?

 

[ramsy]

I'm not following that formula.

T2 = (Ambient + Temp.Rise)
T2 = T1 + (TR-T1) * Load^2 / Imax^2

Load = Circuit Amps
T1 = Ambient Temperature
T2 = Estimated Conductor Temperature
TR = Conductor Max Temperature Rating
Imax = Max Amps at TR, derated for ccc's
---
It solves table 310.16 for ambient temperatures other than 30C.

For example try #12 THWN-2 with 20A load @ 40C ambient, a common solar array primary circuit.

T2 = 40 + (90-40) * 20^2 / 30^2
T2 = 62.2C

Not allowed with NM cables, combiners, or disconnects limited to 60C.