Using 100% rated breakers in a service

[johnsciolino]

I'm looking for some straight answers why you would use a 100% rated Main Breaker in a service.
There's an example in the Annex of the code book that calculates the following:
Continuous load 56,600va
NonContinuous load 42,400va
25% of continuous 14,200va
Total 113,200va
113,200/(480*1.732)= 136amp
It says to select the next size breaker of 150amps or use a 125amp 100% rated breaker.
Does the feeder come into play when using this exception? 240.42 A(1) Exception No. 2


This is from a friend of mine, i think we are talking about the same thing but it seems he is selecting the 100% rating based on the conductor size.
He is saying I can terminate four sets of 600mcm in a 2000amp breaker with 80% rating or a 1600amp breaker at 100% rating. I can't find any examples of this in the code book or other references to the 100% breaker rating. This is for the service feeder to the main.

So, for the purposes of this discussion, let’s assume the load has been calculated per 230.42 (2) and this has resulted in a size of 1,560 amps. My view is as follows:

1. The service entrance conductors must be rated to carry not less than 1,560 amps.
2. The main breaker must be rated for a current not less than 1600 amps since it is the closet size breaker setting above the calculated load (it could be higher but that’s irrelevant)
3. Based on the capacities listed in 310.15, the most efficient combination of conductors which meet this criteria would be for sets of 4-#600kcmiil since this equals 1600A. Incidentally, four sets of 500kcmil would not comply.
4. I have the option of terminating these conductors in breaker which is listed at 100% of it’s trip setting.
5. I have the option of terminating these conductors in a 2,000A breaker which is rated at 80% trip setting.

The way the Annex explains it, I think the conductor is a separate calculation from selecting the 100% breaker, see attached (I'm using this example from 2011 book but its also explained in 2017), see attached. It seems he is working backwards and sizing the main to the feeder.
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In a recent plan, we have a 1000amp Main Breaker Service panel
Load calc is 270,200va or 750.6amps
Feeder shown is 3 sets of 4-600mcm copper in a 3-1/2" PVC pipe underground 475a x 3=1425amp 1425 X .8=1140amps (assuming the neutral is non-linear)
75 degree column says 420a X 3=1260amps (this is what the breaker lugs are good for)
I am being told that this Main Breaker needs to be rated 100% by the reasoning above. I believe the wire is a separate issue from selecting a 100% rated breaker.
Technically, I think we could use 3 sets of 4-500mcm on a standard 1000a breaker based on the load calculation and the wire size since the calculation is taking into account continuous and non-continuous loads. The way the code reads, technically, I think we could use an 600amp 100% rated main in lieu of a 1000amp, the wire would be part of that assembly and calculated on the continuous and non-continuous load only (750.6 amps). Based on the calculation, I think you could also use a standard 800amp Main breaker with the same wire size. Am I wrong in this thinking?

 

[johnsciolino]

I can't find any examples of this in the code book or other references to the 100% breaker rating other than 230.42(A). 230.31 is for size and rating of conductors, 230.42 is service entrance conductors. I think where some get confused is 215.3 is for feeder OCP but feeders are not service conductors by definition.

 

[david luchini]

He is saying I can terminate four sets of 600mcm in a 2000amp breaker with 80% rating

This would be a violation of 230.90(A).

In a recent plan, we have a 1000amp Main Breaker Service panel
Load calc is 270,200va or 750.6amps
Feeder shown is 3 sets of 4-600mcm copper in a 3-1/2" PVC pipe underground 475a x 3=1425amp 1425 X .8=1140amps (assuming the neutral is non-linear)
75 degree column says 420a X 3=1260amps (this is what the breaker lugs are good for)
I am being told that this Main Breaker needs to be rated 100% by the reasoning above. I believe the wire is a separate issue from selecting a 100% rated breaker.
Technically, I think we could use 3 sets of 4-500mcm on a standard 1000a breaker based on the load calculation and the wire size since the calculation is taking into account continuous and non-continuous loads. The way the code reads, technically, I think we could use an 600amp 100% rated main in lieu of a 1000amp, the wire would be part of that assembly and calculated on the continuous and non-continuous load only (750.6 amps). Based on the calculation, I think you could also use a standard 800amp Main breaker with the same wire size. Am I wrong in this thinking?

You don't say how much of the load is continuous and how much is non-continuous, but the adjustment factor for 9 current carrying conductors is 0.7 not 0.8. Plus, you might have trouble fitting twelve 600mcm into a 3.5" conduit.

 

[d0nut]

He is saying I can terminate four sets of 600mcm in a 2000amp breaker with 80% rating or a 1600amp breaker at 100% rating.

This is not correct. 4 sets of 600s is 1680A using the 75 deg column. You can terminate that on a standard 1600A breaker or a 100% rated 1600A breaker. You cannot protect these conductors with a 2000A circuit breaker.

The 100% rated breaker lets you calculate the load based on 100% of the continuous and noncontinuous loads. A standard breaker requires 100% of the noncontinuous loads plus 125% of the continuous loads. Once you have appropriately calculated the load, you need to choose conductors that will support the load and an overcurrent device that will protect those conductors.

The first example you gave is calculated correctly.
Based on that example, if you used a standard breaker your load calc would be 136A. You would need 136A of conductor and you could protect it with a 150A circuit breaker.
If you used a continuous breaker, your load calc would be 119A. You would need 119A of conductor and you could protect it with a 125A circuit breaker.

 

[johnsciolino]

I realize the violation of 230.90(A) and overlooked the .7 factor for the parallel conductors (310.15(B)(3)(a) but that doesn't answer the question of the original post and reasoning given in red for using 100% breakers.
Or are you trying to say 475a x 3=1425amp, 1425 X .7=997.5amps is below the rating of the 1000amp main so 100% breaker has to be used?
This really doesn't follow the Annex example.
Out of the 270,200va, only 15,200 of it is a continuous load, remainder of 255,000 is non-continuous load.

 

[david luchini]

I realize the violation of 230.90(A) and overlooked the .7 factor for the parallel conductors (310.15(B)(3)(a) but that doesn't answer the question of the original post and reasoning given in red for using 100% breakers.
Or are you trying to say 475a x 3=1425amp, 1425 X .7=997.5amps is below the rating of the 1000amp main so 100% breaker has to be used?
This really doesn't follow the Annex example.
Out of the 270,200va, only 15,200 of it is a continuous load, remainder of 255,000 is non-continuous load.

For a 270,200va load (15,200cont/255,000non-cont.), you would need an 800A minimum OCPD. Using 100% rated does nothing for you, it would still be 800A minimum.

If the OCPD is 1000A, and the ampacity of the conductors is 997.5A, then you are violating 230.90(A). Using a 100% rated OCPD doesn't change that.
The conductors are too small.

As for the part in red...You could use a 1600A, 100% rated OCPD with 4 sets of 600mcm. If you use a 2000A, standard OCPD, you would need to increase to 5 sets of 600mcm.

 

[electrofelon]

Here is an article that may be worth a read:

https://library.e.abb.com/public/dfd81f0d5f1f4f09b89a42b4b67d1fc4/61914333_circuit_breaker_ratings3-5.pdf

I only skimmed your example, but typically one would use a 100% breaker where it results in a cost savings on wire, and (potentially) keeps you from having to go up to a larger frame sized breaker.

 

[johnsciolino]

If the OCPD is 1000A, and the ampacity of the conductors is 997.5A, then you are violating 230.90(A). Using a 100% rated OCPD doesn't change that.
The conductors are too small.

Let's circle back on this one, if I have 3 sets of parallel feeders in 3 separate conduits and maintain spacing per 310.15(B)(3)(a), why would I count 9 conductors for de-rating? I would only apply this if I couldn't maintain spacing for more than 24" in length the way I read it.

 

[david luchini]

Let's circle back on this one, if I have 3 sets of parallel feeders in 3 separate conduits and maintain spacing per 310.15(B)(3)(a), why would I count 9 conductors for de-rating? I would only apply this if I couldn't maintain spacing for more than 24" in length the way I read it.

I don't see where "spacing" would have anything to do with it. If you have 3 sets of parallel feeders in 3 separate conduits, you wouldn't count 9 conductors for de-rating. Your original post talked about putting parallel conductors in a single conduit.

 

[johnsciolino]

No, parallel conductors in separate conduits.
The adjustment factors seems like a broad statement when it comes to parallel, there's several factors but I don't see how you would apply this to parallels in separate conduits. The reduction for 8 parallels would be 45%, I have never seen this applied in my 40 yrs in this trade and have several sets of drawings not in compliance if this were the case.

(3) Adjustment Factors.
(a) More than Three Current-Carrying Conductors. Where the
number of current-carrying conductors in a raceway or cable
exceeds three, or where single conductors or multiconductor
cables are installed without maintaining spacing for a continuous
length longer than 600 mm (24 in.) and are not installed
in raceways, the allowable ampacity of each conductor shall be
reduced as shown in Table 310.15(B)(3)(a). Each current carrying
conductor of a paralleled set of conductors shall be
counted as a current-carrying conductor.

If I applied this to the attached example, the parallel shown would be derated to 1923amps (475 x 9= 4275amp x .45=1923amps). I think in this case, the EE might of used the 75 degree column and derated it for 4 conductors (3024amps).

And what would be the purpose of this note?
(3) Adjustment factors shall not apply to underground
conductors entering or leaving an outdoor trench if those
conductors have physical protection in the form of rigid metal
conduit, intermediate metal conduit, rigid polyvinyl chloride
conduit (PVC), or reinforced thermosetting resin conduit
(RTRC) having a length not exceeding 3.05 m (10 ft), and if
the number of conductors does not exceed four.

 

[david luchini]

No, parallel conductors in separate conduits.
The adjustment factors seems like a broad statement when it comes to parallel, there's several factors but I don't see how you would apply this to parallels in separate conduits. The reduction for 8 parallels would be 45%, I have never seen this applied in my 40 yrs in this trade and have several sets of drawings not in compliance if this were the case.

(3) Adjustment Factors.
(a) More than Three Current-Carrying Conductors. Where the
number of current-carrying conductors in a raceway or cable
exceeds three, or where single conductors or multiconductor
cables are installed without maintaining spacing for a continuous
length longer than 600 mm (24 in.) and are not installed
in raceways, the allowable ampacity of each conductor shall be
reduced as shown in Table 310.15(B)(3)(a). Each current carrying
conductor of a paralleled set of conductors shall be
counted as a current-carrying conductor.

If I applied this to the attached example, the parallel shown would be derated to 1923amps (475 x 9= 4275amp x .45=1923amps). I think in this case, the EE might of used the 75 degree column and derated it for 4 conductors (3024amps).

And what would be the purpose of this note?
(3) Adjustment factors shall not apply to underground
conductors entering or leaving an outdoor trench if those
conductors have physical protection in the form of rigid metal
conduit, intermediate metal conduit, rigid polyvinyl chloride
conduit (PVC), or reinforced thermosetting resin conduit
(RTRC) having a length not exceeding 3.05 m (10 ft), and if
the number of conductors does not exceed four.

In the attached example, there is NO derating.

9 sets of 600mcm...
9*340A = 3060A.

There is not more than 3 current-carrying conductors in each raceway, so no need for adjustment factors.