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Old 10-18-2009, 03:46 PM   #1
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Default Derating wire

How do you derate wires? my jman explained it so fast and didn't give me much info. just said look at the table 310.15(b)(2)(a) for percentage and number of current carrying conductors and look at the 310.16 and past that he didn't explain much, could someone give me a explanation of how its done? cause i would like to understand this more then just look at these tables explanation. Thanks

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Old 10-18-2009, 07:23 PM   #2
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Conductors carry current. Current causes heat. Too much current (as a result of too many conductors) creates too much heat. Too much heat causes insulation to break down. Failed insulation causes fires, shorts, shock hazards and other nasty stuff.

By reducing the amount of current a given conductor is allowed to carry, heat is reduced and all the nasty stuff is (theoretically) avoided.

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Old 10-18-2009, 09:17 PM   #3
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Thanks, so how do i use the table? i understand that w/o doing anything with derate you only use 80% of the breaker. so when you have x number of Current carrying conductors do you take that number off the table and multiply it by the voltage(what ever it may be) and does it affect breaker size or do you use the same breaker size as you would per the table?
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Old 10-18-2009, 09:21 PM   #4
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in the CEC, and i dont imagine the NEC being any different, you multiply it against the original max current allowed in the conductor. say 14awg @15 amps max......multiply the 80% against it, now you can only run 12 amps thru it, then if you have any other derating factor, you multiply against that, so if you keep going, you can end up being able run very little current through it, so you may have to switch up to a larger conductor size.

does this help any?
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Old 10-18-2009, 09:27 PM   #5
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Thanks, so how do i use the table? i understand that w/o doing anything with derate you only use 80% of the breaker. so when you have x number of Current carrying conductors do you take that number off the table and multiply it by the voltage(what ever it may be) and does it affect breaker size or do you use the same breaker size as you would per the table?
You only 'use 80% of the breaker' if you have a continuous load.

As for the table, you multiply a conductor's ampacity from Tables 310.16-310.19 by the values in the table.

If you have 6 CCCs, you multiply the ampacity from T310.16-19 by 0.8 (80%). If you have 11, you multiply by 0.5 (50%).
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Old 10-18-2009, 09:27 PM   #6
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80% only if continuous load.
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Old 10-18-2009, 09:34 PM   #7
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ok that makes more sense, thanks, so 80% is for continuous loads only, such as recepticals and anything thats on constantly. and when you derate with current carrying conduction you just take the percentage and multiply it by the amps to get my overcurrent protection. how does ambient temp play it? is that the same and i take the lower of the two and how do you figure out the temp?

sorry for lots of questions on it, but i wanna get this down and not have a confusion explanation by my jman.
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Old 10-18-2009, 09:37 PM   #8
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Both would apply. So if you had 20 ccc in a 200 degree attic, you'd have major probs. A continuous load is a load that is on for 3 hours or more at a time, like lighting in an office etc...
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Old 10-18-2009, 09:40 PM   #9
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here is were the 90c column can come into play.You use the full ampere rating of the wire ,disregaurd the asterisk as it only applies to the breaker.
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Old 10-18-2009, 09:42 PM   #10
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so if you have 6 ccc's of #12 thhn , the amapacity would be 24 amps.
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Old 10-18-2009, 10:01 PM   #11
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so if you have 6 ccc's of #12 thhn , the amapacity would be 24 amps.
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Old 10-18-2009, 10:11 PM   #12
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where in the NEC08 does it state 80% for continuous loads??
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Old 10-18-2009, 10:24 PM   #13
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where in the nec08 does it state 80% for continuous loads??

210.19(a)(1)
210.20(a)
215.2(a)(1)
215.3
230.42(a)(1)
230.42(a)(2)
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Old 10-18-2009, 10:37 PM   #14
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ok looked at those articles and it states 125% am i missing something here?
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Old 10-18-2009, 10:55 PM   #15
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ok looked at those articles and it states 125% am i missing something here?

125% is the reciprocal of 80%

It's nothing but math. Six of one, half dozen of the other.
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Old 10-18-2009, 11:20 PM   #16
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ok that makes sense now, i was taking it as a literal translation of 125%. i got it down now. Thanks guys
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Old 10-20-2009, 10:21 PM   #17
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Also, you have to count your neutrals as current carrying conductors when they are NOT part of a MULTIWIRE branch circuit (dedicated neutrals are 'ccc's). When your neutral belongs to a 'boat' or multiwire branch circuit, single phase or 3 phase, you don't have to count it. The derating rules also apply to bundling romex and there are slightly different derating rules that apply to bundling MC but I'm not gonna' get out the code book right now.

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