Wire in conduit derating
snguyen4 Registered Users Posts: 7 ✭
I'm new to this forum and I've been reading/searching but can't find the answer I'm looking for. I live in Garden Grove, CA and the inspector wants me to submit a derating with my plans. I was able to figure out the derating for the wire size and breaker for the main panel but I don't know how to derate it for for the cable run on my roof in conduit. I have about 50 feet of 4 (2+ & 2-) #10 AWG THWN 90C and 1 #8 AWG ground in 3/4 EMT conduit running on my roof from one end of the panel to the inverter. How would I show the derating of the run for the inspector?
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Here is a program that implements the NEC calculations for wire in conduit (as I understand):
There are two deratings that you need to account for. One is the derating for the number of current carrying conductors in a raceway, which is Table 310.15(B)(3)(a) in the NEC (you have 4 CCC's, which derates the ampacity of your conductors to 80%) and the other is the derating for temperature of rooftop conduit (distance from rooftop), which is Table 310.15(B)(3)(c). Tip: always put your rooftop conduit up on those 4" blocks rather than laying it directly on the roof; that drops your adder from 60 to 30 degrees F.
Is there any way you can help me put what you said in some kind of formula or table that I can use to submit for a permit? I understand what it's asking for but putting it on paper and showing it is another thing. I will have 1 pos or neg (does it make any different if you run the pos or neg from the end of the string to the front J-box to run to the inverter?) in 3/4EMT that runs about 40 feet to the j-box and from there I will have 4 #10THWN and 1 #8 ground in 3/4EMT running about 30 feet from the j-box to the inverter. I'm going to mount the EMT on the rails which will be 3" off the roof. Attachment not found.Attachment not found.
OK I see one or 2 things.
1) you are using a pair of 30 amp breakers in the main panel, that is over the max 40 amps you can have for back feed without derating the panel
2) Some AHJ will only want a single connect to the main panel, they require a AC combiner to a single AC main panel breaker.
A 4000 watt inverter should output less than 20 amps.
His 4kW inverters are rated at 17 amps...
17 amps * 1.25 NEC = 21.25 amp branch circuit
So, I believe he cannot get away with 20 amp breakers.
It is possible to use a smaller array, but with the NEC using 1.25*1.25=1.56 "safety factor" for solar--You have to have a much smaller array to get under the 20 amp branch circuit on the 240 VAC side.
I was also think of using a 150amp main breaker on my soon to be install 200amp main panel because the inspector said that I need at least a 150amp main. So if I run 2x25amp or 2x30amp it should still be ok right?
Solar-Dave... I hope you're not right about the AC combiner to a single AC main panel breaker because that's just more work and $$$
I'm still working on the derating (giving me a headache right now) so I have not submit anything yet. Hopefully they will not make me use a AC combiner box.
Thank you guys for all the help so far
All I can tell you is what my AHj did. A AC combiner and a single disconnect labeled for the AC solar source before the panel connection for the fire dept. The utility required a separate meter socket and meter for the solar so they could get the real annual solar output, I assume for t he REC or Federal compliance for Alternative energy installs.
@ggunn- How does this look? Is this the derating that you had in mind? Am I even close?
Calculate maximum Circuit Current [690.8(A)(1)]
Imax = Isc 7.63 x 1.25 = 9.54A x 1.25 = 11.9A
Calculate the minimum overcurrent protective device (OCPD) [690.8(B)(1)(a)]
OCPD = Imax 11.9 x 1.25 = 14.9A x 1.25 = 18.62A → 20A [690.9(C)]
Module maximum fuse rating is 15A so minimum and maximum are the same.
Calculate minimum conductor size without conditions of use [690.8(B)(2)(a)]
Minimum conductor ampacity = Imax 11.9 x 1.25 = 14.9A → 14 AWG (minimum bldg. wire)
Calculate minimum conductor size base on Imax with conditions of use [690.8(B)(2)(b)]
Conduit fill 4-6 adjustment factor → 0.8 according to Table 310.15(B)(3)(a)
Sunlit conduit temperature adder above .5” to 3.5” → 22°C according to Table 310.15(B)(3)(c)
Ambient temperature adjustment factor → 22°C + 39°C (LA ambient temperature) = 61°C → 0.58
Minimum conductor ampacity = Imax 11.9 divided by conduit fill adj factor divided by 0.58 temp adj factor = 25.64A → 12 AWG
15 Amp Overcurrent protection: 690.8(B)(2)(c)
12 AWG → ampacity = 30amps x 0.8 x 0.58 = 13.92 amps (fails because of 690.0(C))
10 AWG → ampacity = 40amps x 0.8 x 0.58 = 18.56 amps (okay to use)
I would use 4" blocks to get the conduit at least 3.5" above the roof. It makes a big difference in your temperature adder.
If I have room I will try to get it to 4" but right now I'm just trying to get something together to submit to the City. Does my calculation look like something the City may be looking for?
I can't speak for your city, but what AHJ's generally want is assurance that you've taken into account all the variables which could cause a fire and/or damage to their equipment. I have not checked all your numbers.