Min size conductor and 250.122(B)

[david luchini]

What is the minimum size cable I need? To solve that I would do following:

40*1.25 = 50A

Now I do 50/.8=62.5A

So would I be correct to have #6 as my minimum size in the above example or not?

#6 is correct, but your calculation is not.

Your breaker size is 50A, so you need a conductor with an ampacity of greater than 45 in order to protect the conductor with the "next size up" rule. 45A/0.8=56.25A, so your conductor ampacity before the application of the adjustment factor must be greater than 56.25.

You can use the 90deg rating for the adjustment. #8 has an ampacity of 55 at 90deg, so it is too small. #6 would be adequate.

 

[hhsting]

#6 is correct, but your calculation is not.

Your breaker size is 50A, so you need a conductor with an ampacity of greater than 45 in order to protect the conductor with the "next size up" rule. 45A/0.8=56.25A, so your conductor ampacity before the application of the adjustment factor must be greater than 56.25.

You can use the 90deg rating for the adjustment. #8 has an ampacity of 55 at 90deg, so it is too small. #6 would be adequate.

I know the load is 40A and its continuous and cable would be 4 cc then

1. 40*1.25= 50A

2. 40/.8= 50A

Both are same so I need minimum 50A cable.

#8 would do it at 75C

 

[wwhitney]

I know the load is 40A and its continuous and cable would be 4 cc then

1. 40*1.25= 50A

2. 40/.8= 50A

Both are same so I need minimum 50A cable.

#8 would do it at 75C

David's point is that we have considered so far these sections: 210.19 (branch circuit conductor minimum ampacity), 210.20 (branch circuit overcurrent protection), and 110.14(C) (termination temperature limitations). But we have neglected 240.4 ("conductors . . . shall be protected against overcurrent in accordance with their ampacities").

So in your example above, the #8s with 90C insulation in a conduit with 4 CCCs have an ampacity there of 55A * 0.8 =44A. Then 240.4 tells us that the OCPD for those conductors must be no more than 45A (using 240.4(B), as 45A is a standard breaker size per 240.6(A). As a 45A breaker is too small per 210.20, this configuration doesn't work. The #8s are too small, so we have to bump up to #6, and verify that works (which it does).

Cheers, Wayne

 

[david luchini]

#8 would do it at 75C

#8 at 75C adjusted for 4 ccc's is 50*0.8=40A, which cannot be protected by a 50A c/b.
#8 at 90C adjusted fo4 4 ccc's is 55*0.8=44A, which cannot be protected by a 50A c/b.

#6 at 75C adjusted for 4 ccc's is 65*0.8=52A, which can be protected by a 50A c/b. You would need a #6 min. conductor.

 

[wwhitney]

So to summarize all of the above, I propose the following conductor sizing algorithm, which aims to be complete. Please point out any errors or omissions:

(1) Start with the demand as continuous X + non-continuous Y, and a choice of wire type with corresponding insulation temperature.
(2) Determine the minimum breaker size per 210.20 or 210.3; call it B. That's the smallest size in Table 240.6(A) that exceeds 125% X + Y, except in the case of 100% rated breakers.
(3) 240.4 then provides a minimum ampacity based on B; call that M. [E.g. if B = 50A, M = 45.5A]
(4) For the terminations, check the table column for the termination temperature (no adjustment or correction), and find the smallest wire size would table entry is at least max(M, 125% X + Y)
(5) Elsewhere, consider at each point the number of CCCs and the ambient temperature. Each of those gives an adjustment/correction factor; multiply them at each point. Call the lowest combined factor W; it controls wire sizing away from the terminations.
(5.1) If the insulation temperature is different for wet and dry, do step (5) separately for wet locations and for dry locations.
(6) Find the smallest size wire whose table entry for the insulation temperature column is at least max(M, X+Y) / W.
(6.1) If the insulation temperature is different for wet and dry, do step (6) separately for wet locations and for dry locations, and take the larger size.
(7) Choose the larger of (4) or (6).

Cheers, Wayne

 

[don_resqcapt19]

#8 at 75C adjusted for 4 ccc's is 50*0.8=40A, which cannot be protected by a 50A c/b.
#8 at 90C adjusted fo4 4 ccc's is 55*0.8=44A, which cannot be protected by a 50A c/b.

#6 at 75C adjusted for 4 ccc's is 65*0.8=52A, which can be protected by a 50A c/b. You would need a #6 min. conductor.

You need to apply 210.19(A) for the minimum permitted size.
8 AWG with a 75°C rating satisfies both 210.19(A)(1)(a) and (A)(1)(b).

In (A)(1)(a) you do not apply the adjustments and/or corrections.
In (A)(1)(b) you do not apply the 125% continuous load factor.
Both cases require a 50 amp conductor and assuming everything is suitable for 75°C ampacities, an 8 AWG copper conductor is the minimum permitted by the code for this load.

Note that the language in this section and its equivalent in 215 is very poor. I have submitted PIs for a couple of cycles to make it clear that (A)(1)(a) refers to the ampacity directly from the ampacity table without any adjustments or corrections. The panel comment is that is already clear in the existing code language.

 

[hhsting]

You need to apply 210.19(A) for the minimum permitted size.
8 AWG with a 75°C rating satisfies both 210.19(A)(1)(a) and (A)(1)(b).

In (A)(1)(a) you do not apply the adjustments and/or corrections.
In (A)(1)(b) you do not apply the 125% continuous load factor.
Both cases require a 50 amp conductor and assuming everything is suitable for 75°C ampacities, an 8 AWG copper conductor is the minimum permitted by the code for this load.

Note that the language in this section and its equivalent in 215 is very poor. I have submitted PIs for a couple of cycles to make it clear that (A)(1)(a) refers to the ampacity directly from the ampacity table without any adjustments or corrections. The panel comment is that is already clear in the existing code language.

So 240.4 does not apply at all? Post#24 doesnt apply?

 

[wwhitney]

You need to apply 210.19(A) for the minimum permitted size.

210.19(A) is a requirement, but it's not the only requirement. 240.4 still applies, as I forgot until David pointed it out.

Cheers, Wayne

 

[david luchini]

So 240.4 does not apply at all? Post#24 doesnt apply?

Yes, 240.4 applies. You cannot protect a conductor with an ampacity of 44 (#8, 90deg, 4ccc's)with a 50A c/b. So you would need at a #6 or larger conductor.

 

[david luchini]

Note that the language in this section and its equivalent in 215 is very poor. I have submitted PIs for a couple of cycles to make it clear that (A)(1)(a) refers to the ampacity directly from the ampacity table without any adjustments or corrections. The panel comment is that is already clear in the existing code language.

They never should have changed it in the first place. It was fine the way it was.

 

[hhsting]

Yes, 240.4 applies. You cannot protect a conductor with an ampacity of 44 (#8, 90deg, 4ccc's)with a 50A c/b. So you would need at a #6 or larger conductor.

Well #8 awg cable you have 50A at 75C. 50*0.8=40A due to 4 CC.

So now my cable has ampacity of 40A instead of 50A.

However the load is continuous so new cable ampacity based off of #8 is 40x1.25=50A and next standard size is 50A according to 240.4

I dont see where #8 is not sufficient? You are forgetting this is continuous load

 

[Dsg319]

Well #8 awg cable you have 50A at 75C. 50*0.8=40A due to 4 CC.

So now my cable has ampacity of 40A instead of 50A.

However the load is continuous so new cable ampacity based off of #8 is 40x1.25=50A and next standard size is 50A according to 240.4

I dont see where #8 is not sufficient? You are forgetting this is continuous load

#8 is not sufficient. You have a 40amp continues load therefor the Reasoning for 50 OCPD. Although your #8 conductor is still usable at 40amps after being derated, you cannot protect a 40amp conductor with a 50amp OCPD. As the next size up rule would not apply in your situation being your conductor ampacity after adjustment can be protected with the same size OCPD.

 

[hhsting]

#8 is not sufficient. You have a 40amp continues load therefor the Reasoning for 50 OCPD. Although your #8 conductor is still usable at 40amps after being derated, you cannot protect a 40amp conductor with a 50amp OCPD. As the next size up rule would not apply in your situation being your conductor ampacity after adjustment can be protected with the same size OCPD.

But the load is continuous so even after derating due to 4 CC you have to apply 125% that brings to 50A. No next size up rule is used. where in NEC it says not to apply 125% due to continuous load after applying derating due to 4 CC?


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[Dsg319]

But the load is continuous so even after derating due to 4 CC you have to apply 125% that brings to 50A. No next size up rule is used. where in NEC it says not to apply 125% due to continuous load after applying derating due to 4 CC?


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Pretty sure after adjustments conductors ampacity only needs be 100% of load to be served(40amps). But you still cannot protect a conductor after adjustments with a larger OCPD.Don’t have my book handy someone else chime in.

Sill yet your #8 is only good for 40ampere OCPD after adjustments.

 

[augie47]

I thought david's post #29 summed it up:

Yes, 240.4 applies. You cannot protect a conductor with an ampacity of 44 (#8, 90deg, 4ccc's)with a 50A c/b. So you would need at a #6 or larger conductor.

 

[hhsting]

Pretty sure after adjustments conductors ampacity only needs be 100% of load to be served(40amps). But you still cannot protect a conductor after adjustments with a larger OCPD.Don’t have my book handy someone else chime in.

Sill yet your #8 is only good for 40ampere OCPD after adjustments.

Load of 40A 208V single phase is EV charger. NEC 2017 section 625.41 says to size overcurrent protection per 125%. That section does not say anything about not doing it after derating 4 CC. I just dont see in code where it says not to apply 125% after derating 4 CC? I.e just take 100% of ampacity after derating

 

[david luchini]

Well #8 awg cable you have 50A at 75C. 50*0.8=40A due to 4 CC.

So now my cable has ampacity of 40A instead of 50A.

However the load is continuous so new cable ampacity based off of #8 is 40x1.25=50A and next standard size is 50A according to 240.4

I dont see where #8 is not sufficient? You are forgetting this is continuous load

The part in red is where you went wrong. The ampacity of the cable is 40A due to 4 ccc's.

The #8 has sufficient ampacity for the 40A load, but cannot be protected by a 40A c/b per 240.4.

 

[hhsting]

The part in red is where you went wrong. The ampacity of the cable is 40A due to 4 ccc's.

The #8 has sufficient ampacity for the 40A load, but cannot be protected by a 40A c/b per 240.4.

What about 625.41 sizing breaker says do 125%? 40A load is EV chargers.

 

[david luchini]

where in NEC it says not to apply 125% due to continuous load after applying derating due to 4 CC?

210.19(A)

What about 625.41 sizing breaker says do 125%? 40A load is EV chargers.

That's the part that requires you to have a 50A c/b. That's why you can't use your #8 conductor with an ampacity of 40.

 

[hhsting]

210.19(A)

That's the part that requires you to have a 50A c/b. That's why you can't use your #8 conductor with an ampacity of 40.

210.19(A)(1) says to pick larger of the two but in my case they are the same or 125% is larger?