Looking for clarification on understanding 110.14 (C)(1)(a) and (b).

Rfloyd100

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Location
Montana
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Journeyman electrician
I've been working my way through the Understanding The NEC vol. 1. I'm having a hard time understanding when and what C° column to use. I have copied and pasted the practice questions below. I've read through 110.14 and am still just not grasping it. I'd really appreciate some feed back. Thanks!!

Question #1: What size conductors rated 90°C are required between terminals rated 90°C for a circuit supplying a 44A continuous. load?

(a) 10 AWG

(b) 8 AWG

(c) 6 AWG

(d) 4 AWG

Question #2:
Question: What size XHHW-2 copper conductor can be used to interconnect 90°C rated power distribution blocks protected by a 400A overcurrent protective device serving a 320A continuous load?


(a) 250 kcmil (b) 300 kcmil (c) 350 kcmil (d) 400 kcmil


Side note: in case anyone has the Mike Holt understanding national electrical code volume one based on the 2020 NEC, and wants to physically see the questio, these questions are on page 77 and page 127.
 
1) What ever size that has an ampacity of 125% of 44 amps in the 90°C column. See 215.2(A)(1) if the conductors are feeder conductors or 210.19(A)(1) if they are branch circuit conductors.

2) again we have a continuous load so we need a conductor ampacity of 1.25 x 320 or 400 amps. This question is taking advantage of Exception #2 to 215.2(A)(1) so you need a conductor with an ampacity of at least 400 amps in the 90°C column.
 
I am in agreement with don's answer on #1 and have to admit I don't fully understand the Exception he referenced in question #2 but, assuming he is correct on the 400 amps, you dont have that as an option.
Hopefully don can explain in a bit more detail for both of us.
 
Do you understand what temperature rating you are to use? If so then you cannot use the 90C rating except for de-rating purpose so you must use the 75C column
 
On #2 it has to do with the conductor between the 90 c blocks and the load.
I am in agreement with don's answer on #1 and have to admit I don't fully understand the Exception he referenced in question #2 but, assuming he is correct on the 400 amps, you dont have that as an option.
Hopefully don can explain in a bit more detail for both of us.
In this case it can be sized to the total load of 320 based on exception #2.
The way I understand it, the conductor is based off total load without the 25% increase for the continues or the total of continues and non continues.

The question says 320 amp load using the exception. Since it is between the 90c blocks I will choose a xwwh-2 90c cu conductor rated no less than 320 amps.
 
The question says 320 amp load using the exception. Since it is between the 90c blocks I will choose a xwwh-2 90c cu conductor rated no less than 320 amps.
Other than motor loads and the like, a conductor with 320A ampacity can't be protected at more than 350A. Since the OCPD is specified at 400A, the minimum conductor ampacity is 351A.

Cheers, Wayne
 
I am in agreement with don's answer on #1 and have to admit I don't fully understand the Exception he referenced in question #2 but, assuming he is correct on the 400 amps, you dont have that as an option.
Hopefully don can explain in a bit more detail for both of us.
I did not do #2 correctly. The others are correct that we only need a conductor with an ampacity as the language in the exception does not include the 125% of the continuous load like the parent text does. So we need a conductor rated at least 320 amps for the load, but since we have a 400 amp OCPD, we need to bump that up to a conductor with an ampacity of at least 400 amps in the 90°C column. It does not change the answer that I originally gave, it just changes the path to get to the answer.
 
Do you understand what temperature rating you are to use? If so then you cannot use the 90C rating except for de-rating purpose so you must use the 75C column
This. If you are using 90 degree wire and 75 degree terminals, without any deratings the 90 degree column tells you how much current will heat the wire to 90 degrees and the 75 degree column tells you how much current will heat the wire to 75 degrees, so you have to use the 75 degree column to protect the terminals.
 
So we need a conductor rated at least 320 amps for the load, but since we have a 400 amp OCPD, we need to bump that up to a conductor with an ampacity of at least 400 amps in the 90°C column.
240.4(B) permits a conductor with an ampacity of at least 351A to be protected at 400A. So the answer is 400 kcmil, the largest choice provided. None of the provided choices has a 90C ampacity of at least 400A.

Cheers, Wayne
 
I see where I went wrong. I blew right past 400 amp OCPD. Based on the question alone using the round up rule. I see what the correct answer should be.
Thank you.
 
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