Annex D Example D3(a)

[david luchini]

Is it not to account for the amount of heat generated for overloads and additional ampacity required to prevent thermal damage to the insulation?

Sorry I missed this while reading the thread on my phone.

The load could potentially be 136A for overload.

The additional 25% doesn't have anything to do with overload. I has to do with continuous load. The additional 25% creates a larger termination size to help dissipate the heat from the portion of the load that is continuous.

In Example D3(a), the load is 119A. 51A of the load will be non-continuous and 68A of the load will be continuous. But together, the maximum load will be 119A. According to the load calculation, the highest current the feeder conductors will see is 119A, not 136A.

So the 2/0 conductor has an ampacity of 131A and will carry a load of 119A. It's operating temperature will not exceed it's insulation rating. And the 150A c/b will carry a load of 119A. It will not exceed it's rating. And the 2/0 lugs on the c/b can carry 175A without exceeding its temperature rating, but will only be carrying 119A. So the terminations will not exceed it's temperature rating either.

 

[xptpcrewx]

Sorry I missed this while reading the thread on my phone.

No Problem.

The additional 25% doesn't have anything to do with overload. I has to do with continuous load. The additional 25% creates a larger termination size to help dissipate the heat from the portion of the load that is continuous.

Ok. This clarifies the last assumption I was making. Do you have a reference I could read up on about the purpose of the additional 25% and how its related to the mass of the termination? Thanks again!

 

[david luchini]

Do you have a reference I could read up on about the purpose of the additional 25% and how its related to the mass of the termination? Thanks again!

I've got nothing. I would suggest the interweb, but I just saw an online article from the International Association of Electrical Inspectors that seems to be confused on the issue.

 

[xptpcrewx]

I've got nothing. I would suggest the interweb, but I just saw an online article from the International Association of Electrical Inspectors that seems to be confused on the issue.

Interesting. So we are applying 125% and there may not be a well known understanding for what the exact purpose is? :happysad:

 

[david luchini]

Interesting. So we are applying 125% and there may not be a well known understanding for what the exact purpose is? :happysad:

I think the understanding of the purpose is well known. I think people make mistakes in their explanations.

For example, in the IAEI article I mentioned "100% vs 80%: Choosing the right OCPD solution" (I'll let you look it up rather than posting a link),

you can see that they say that "load current" = Noncontinuous Load Amps + 1.25*Continuous Load Amps. This is obviously incorrect.

 

[xptpcrewx]

I think the understanding of the purpose is well known. I think people make mistakes in their explanations.

For example, in the IAEI article I mentioned "100% vs 80%: Choosing the right OCPD solution" (I'll let you look it up rather than posting a link),

you can see that they say that "load current" = Noncontinuous Load Amps + 1.25*Continuous Load Amps. This is obviously incorrect.

Unfortunately its not obvious to me at all. For something to be obvious I would need a standard to reference; i.e. IEEE, ANSI, UL, ASTM etc.

Anybody?

 

[david luchini]

Unfortunately its not obvious to me at all. For something to be obvious I would need a standard to reference; i.e. IEEE, ANSI, UL, ASTM etc.

Anybody?

https://www.schneider-electric.com/...20.1573179392.1551128226-555531687.1548692531

Maybe this explanation will help

 

[xptpcrewx]

https://www.schneider-electric.com/...20.1573179392.1551128226-555531687.1548692531

Maybe this explanation will help

A little but not much. This is about non-100% circuit breakers having to be de-rated due to being enclosed. This doesn't specifically apply to conductors.

 

[tortuga]

Unfortunately its not obvious to me at all. For something to be obvious I would need a standard to reference; i.e. IEEE, ANSI, UL, ASTM etc.

Anybody?

Ahh you said anybody.
By definition the ampacity of a conductor is the load it can carry continuously.
This is not really a engineering explanation but by art 100 Definition

Ampacity. The maximum current, in amperes, that a conduc-
tor can carry continuously under the conditions of use without
exceeding its temperature rating. (CMP-6)

That definition was added first in the 1965 NEC and was changed to include the wording "continuously" in the 84 code.
That was the same year they ditched the Rosch method of calculating ampacity and adopted the Neher-McGrath method.
The Rosch method of calculating ampacity was published in "Electrical Engineering" March 1938 in a paper entitled " The Current-Carrying Capacity or Rubber-Insulated Conductors". The Neher-McGrath method was first developed for high voltage, I could post some more info about it if you want.

 

[xptpcrewx]

Ahh you said anybody.
By definition the ampacity of a conductor is the load it can carry continuously.
This is not really a engineering explanation but by art 100 Definition

That definition was added first in the 1965 NEC and was changed to include the wording "continuously" in the 84 code.
That was the same year they ditched the Rosch method of calculating ampacity and adopted the Neher-McGrath method.
The Rosch method of calculating ampacity was published in "Electrical Engineering" March 1938 in a paper entitled " The Current-Carrying Capacity or Rubber-Insulated Conductors". The Neher-McGrath method was first developed for high voltage, I could post some more info about it if you want.

Tortuga, We are getting warmer!

If you look at the Neher-McGrath formula I is the "ampacity" which, as you have stated, is the load a conductor can carry "continuously", so why are we taking 125% of I for continuous loads if I is already accounts for this? :?

 

[tortuga]

Tortuga, We are getting warmer!

If you look at the Neher-McGrath formula I is the "ampacity" which, as you have stated, is the load a conductor can carry "continuously", so why are we taking 125% of I for continuous loads if I is already accounts for this? :?

From the UL Molded Case Circuit Breakers Marking and Application Guide page 13, 2016 Edition

38.
100 Percent Continuous Rated — Unless otherwise marked for continuous use at 100
percent of its current rating, a circuit breaker is intended for use at no more than 80 percent of its
rated current where in normal operation the load will continue for three hours or more. A breaker
with a frame size of 250 A or more, or a multi-pole breaker of any current rating greater than 250 V,
may be marked to indicate it is suitable for continuous use at 100 percent of its current rating. The
marking is “Suitable for continuous operation at 100 percent of rating only if used in a circuit breaker
enclosure Type ____or in a cubicle space______by_____ by _____ inches” or an equivalent
statement. This type of breaker may also be marked to indicate it is to be used with wire sized for a
75°C conductor with 90°C insulation and used with 90°C wire connectors.

https://www.ul.com/wp-content/uploads/2014/09/CircuitBreaker_MG.pdf

 

[david luchini]

If you look at the Neher-McGrath formula I is the "ampacity" which, as you have stated, is the load a conductor can carry "continuously", so why are we taking 125% of I for continuous loads if I is already accounts for this? :?

I think you've already answered this.

Do you have a reference I could read up on about the purpose of the additional 25% and how its related to the mass of the termination

The load in the example D3(a) is 119A. A #1awg conductor would carry a 119A load, but the Code makes you use a minimum #1/0awg conductor, which forces the terminations to be larger.

But the conductor only has to have an ampacity to supply the load. The additional 25% really doesn't relate to the conductor sizing, other than as a means to increase the termination sizing.

 

[xptpcrewx]

I think you've already answered this.

Not sure I follow what you mean here...

 

[xptpcrewx]

One other question. Under the feeder neutral conductor section of this example, it states that 210.11(B) does not apply to these buildings. I’m looking at 210.11(B) and I don’t see anything that excludes industrial buildings. Am I missing something?