followup question for closed post ampacity-adjustment.2580477

[newbee_one]

wwhitney replied to the above thread on April 17, 2024 (framed around finding the appropriate conductor size for a 25A continuous load across 8 conductors in a conduit):

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You have me confused also.

You basically have it correct, but let me elaborate on your answers:

a.) I have a 25 amp load that would be 31.25 after the 125% for continuous load was added.

If you have a 25A continuous load, then 210.19(A)(1)(a) (and the similar sections for feeders and SECs) says that at the terminations you need an unadjusted and uncorrected ampacity of at least 31.25A. That would be using the table column for the termination temperature, typically 75C, occasionally 60C.

b.) I have the same 25 amp load but I have an adjustment factor of 70% because of number of conductors in a conduit. 25/.7 = 35.7 amps
This means I need a conductor good for 35.7 amps

If you have a 25A load, continuous or non-continuous, doesn't matter, then 210.19(A)(1)(b) (and the similar sections for feeders and SECs) says that you need a conductor with a final ampacity of 25A. With a 0.7 adjustment factor and no temperature correction, that means you can use the table column for the conductor insulation temperature, typically the 90C column, and look for a conductor that is at least 35.7A in that column.

Putting those two together for a 25A continuous load, 75C termination temperature, and 90C insulation temperature, then 210.19(A)(1) (and the similar sections for feeders and SECs) says that you need a conductor whose 75C column entry is at least 31.25A, and whose 90C column entry is at least 35.7A.

However, for a continuous 25A load, you also need a 35A OCPD. So 240.4 also requires that you have a final ampacity of at least 30.5A so you can apply 240.4(B). That means that you also need a conductor whose 90C column entry is at least 30.5 / 0.7 = 43.6A. This requirement is stricter than 210.19(A)(1)(b), so the final answer is a conductor whose 75C column entry is at least 31.25A, and whose 90C column entry is at least 43.6A.

This procedure never requires doing the computation 25A * 125% / 0.7 = 44.6A. That figure is irrelevant.

Cheers, Wayne

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Regarding the (in the blue text area above):

First sentence: Where did the 35A OCPD come from? Is it simply the upgrade of the minimum 240.4(b) amperage (30.5) to the next standard rating?

Second sentence: I interpreted the use of 30.5A as the minimum load needed before 240.4(b) could be used (assumedly to avoid the use 240.4(d) small conductors clause).

Third sentence: I'm curious how the 30.5 / 0.7 = 43.6 calculation would change if the continuous load was 35A not 25A. Would the 240.4(b) calculation then be 35 / 0.7 = 50A? Giving a conductor whose 75C column entry is at least (35 * 1.25 = 43.75A), and whose 90C column entry is at least 50A.

thanks in advance for any help.

 

[Dennis Alwon]

Here is the original thread https://forums.mikeholt.com/threads/ampacity-adjustment.2580477/\ I don't have time to get into this right now so if someone wants to read it

 

[wwhitney]

First sentence: Where did the 35A OCPD come from? Is it simply the upgrade of the minimum 240.4(b) amperage (30.5) to the next standard rating?

For a branch circuit, 210.20. For a feeder, 215.3. Etc.

Second sentence: I interpreted the use of 30.5A as the minimum load needed before 240.4(b) could be used (assumedly to avoid the use 240.4(d) small conductors clause).

It's to be able to use 240.4(B) in lieu of the basic requirement at the beginning of 240.4. If the load does not fall under 240.4(G), then 240.4(D) would still apply, so the minimum 35A breaker would require a minimum #8 conductor (assuming the use of copper).

Third sentence: I'm curious how the 30.5 / 0.7 = 43.6 calculation would change if the continuous load was 35A not 25A. Would the 240.4(b) calculation then be 35 / 0.7 = 50A? Giving a conductor whose 75C column entry is at least (35 * 1.25 = 43.75A), and whose 90C column entry is at least 50A.

If you have a 35A continuous load, and your derating (ampacity adjustment for number of CCCs times ampacity correction for ambient temperature) comes out to 0.7, then your minimum breaker size would be 35*1.25 = 44A, or 45A in practice (a standard size in 240.6(A)). Your minimum 60C or 75C (depending on termination rating) ampacity column entry would be 44A. And with 90C rated insulation, your minimum 90C ampacity column entry would be 40.5/0.7 = 58A, in order to use 240.4(B) (40A is the next smallest standard size). 40.5 exceeds the 210.19 / 215.2 requirement of at least 35, so that is moot.

Cheers, Wayne

 

[newbee_one]

@wwhitney - thanks for that. I'm going to dig in deeper to figure out where the 30.5A (for the 25 amp example) and 40.5A come from for the x / 0.7 90C ampacity look ups. I get the 0.7 from the 310.15(c)(1) table. I'm struggling with the 30.5 and 40.5 derivations. I'll get there - thanks again!

 

[wwhitney]

@wwhitney - thanks for that. I'm going to dig in deeper to figure out where the 30.5A (for the 25 amp example) and 40.5A come from for the x / 0.7 90C ampacity look ups.

It's just that 30.5A rounds up to 31A, and as a 35A OCPD is required, that's the smallest ampacity that may be protected by a 35A OCPD under 240.4(B). If the ampacity is only 30A, then 30A is the largest size OCPD you can use.

Cheers, Wayne

 

[newbee_one]

That one I got. I'm not sure where the 30.5 came from. why not 31, 32 33.75? For 25A where did 30.5 come from?

 

[wwhitney]

That one I got. I'm not sure where the 30.5 came from. why not 31, 32 33.75? For 25A where did 30.5 come from?

A 25A continuous load requires a 35A breaker. 210.19 only requires a 25A ampacity conductor for a 25A continuous load, but 240.4 requires a minimum 31A ampacity conductor for a 35A breaker. So you need a 31A ampacity conductor.

If we want to extend the above to one more significant figure, 31A becomes 30.5A because of the principle expressed in 220.5(B). But if you want to stick with whole numbers of amps, 31A is the minimum.

Cheers, Wayne