Table 310.15(B)(6) for Parallel Service

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Proposal 6-74 in the 95ROP was to prohibit the use of the reduced wire sizes in parallel. Panel 6 rejected the Proposal with this statement: "Conductors 1/0 and larger are permitted to be paralleled by section 310-4.
 
It is a single service lateral to a quadplex then it feeds to 4 seperate brk-meters-panels. When you look at 250.66 it talks about CM if you go parallel, but that is not mentioned in this table and believe it was put in for simplicity in addition to residential demand. I just need to make sure there isn't a code floating out there that says he can run parallel outside sect 310
 
It gets worse :mad: I used 2008 310.4(B)(2) which says it must have the same size in circular mil area. Well 4/0 is 211,600 x 2 = 423200 CM, but the table is saying it must be 600000 CM (600kCMIL) for 400A so this really kills it. So now I agree that you can use parallel cables for 310.15(B)(6) but you have to make sure the CM are equal and not the current ratings. You can still derate for bundling and I read somewhere that this table takes into consideration the worse case ambient temperature for a residential service which is why a temp derating is not included.

Thanks for everyone's response it really helped.
 
don_resqcapt19 said:
Proposal 6-74 in the 95ROP was to prohibit the use of the reduced wire sizes in parallel. Panel 6 rejected the Proposal with this statement: "Conductors 1/0 and larger are permitted to be paralleled by section 310-4.
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I have to say, it is sometimes maddening when they answer a question not asked and ignore the direct problem: Table 310.15(B)(6) does not offer a parallel conductor option.
 
Bjenks, to find the ampacity of parallel conductors you add the ampacity of the individual conductors not the circular mils. You add the circular mils of each conductor to find the grounding electrode and main bonding jumper size. The service you have described is NEC compliant with parallel 4/0 AL conductors if they are installed in separate raceways.
 
George Stolz said:
I have to say, it is sometimes maddening when they answer a question not asked and ignore the direct problem: Table 310.15(B)(6) does not offer a parallel conductor option.
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But T310.16 does? If so, where?
 
curt swartz said:
Bjenks, to find the ampacity of parallel conductors you add the ampacity of the individual conductors not the circular mils. You add the circular mils of each conductor to find the grounding electrode and main bonding jumper size. The service you have described is NEC compliant with parallel 4/0 AL conductors if they are installed in separate raceways.
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Oops your right, I jumped the gun on that one. 310.4(B)(3) is saying the parallel conductors must be exactly the same for matching impedence not current rating.

I sat down and read each line of 310.15(B)(6) very carefully and then read the comments from the handbook for 2008. Couple of points to clarify it all.

1) This table only applies when it is carrying 100% of the dwelling units load. So you can't use the table for a quadplex's service lateral only from the tap to the individual panel.
2) In the comments it says "For the 2008 code, the panel clarified that this permission to use the table applies only to conductors carrying 100 percent of the dwelling unit's diversified load. Provided a single set of 3-wire, single-phase...", so that should mean no parallel cables.
3) this is my own thoughts but if we could do parallel feeds by taking the load and dividing by 2 and then selecting a rated cable, then this table could be good for up to 800A in a dwelling unit. However when you look at the table it has a header of "Service or Feeder Rating" while 310.16 is "Allowable Ampacities of Insulated Conductors..."

So after going all over the place I am sticking with my first thought, that you can't parallel cables for table 310.15(B)(6) and a new one for me is you can't use it unless it is carrying the full current of the dwelling unit.
 
Bjenks said:
table 310.15(B)(6) and a new one for me is you can't use it unless it is carrying the full current of the dwelling unit.
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That one caught me by suprise too.

In the past, I've used a combo 200A Meter Main to supply a 200A MLO panel inside the home.
I've always used 4/0 al for the feeder between the panels per 310.15(B)(6).

Sometimes, I would install a couple of 2 pole breakers in the Meter Main to supply the outside A/C compressors, if they were handy.
Under the new rule, I don't think that this is allowed.
Since the 4/0 feeder wouldn't be carrying 100% of the power to the dwelling...I can't use 310.15(B)(6).

The ironic thing is that I could use 310.15(B)(6)..... power all the circuits (A/C compressors included) from the MLO panel... and be in compliance, even though it would add the additional compressor load to the 4/0 al feeder.

So...in a nut shell...and in my humble opinion...if you plan on using 310.15(B)(6), there's no need to buy a Meter/Main with additional breaker space,
unless you're planning on powering a out building or other remote structure from the "free" breaker spaces.

If you sized your Main Power Feeder from 310.15(B)(6), the moment that you place a breaker in the Meter/Main to power a house circuit, you are technically non-compliant.

Just my opinion, and interested in hearing yours
steve
 
You are correct, however, you would still be able to use it for the service lateral as long as all the loads on the main disco serves a single dwelling unit. Also I believe the code also says that you do not have to make the feeder larger than the service just because you are using another table for the feeder sizing. So I think you would be OK to do it the same way as long as the compressor is for the dwelling unit.
 
480sparky said:
But T310.16 does? If so, where?
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Obviously, it doesn't - but it also doesn't provide a narrow window of opportunity to make use of it.

Table 310.15(B)(6) is very specific about what it allows and when, and parallel conductors are a provision not covered in the allowances. So, if we don't meet the specific criteria provided, we have to revert to the SOP of using 310.16 any way we see fit (unless of course we run afoul of what they intend for 310.4 to say ;) ).
 
Table 310.15(B)(6) does not mention, or make use of in any way, the word "ampacity." It is absolutely not saying that a 1/0 copper has an ampacity of 175 amps under the limited installation conditions it addresses. It says you can use a 1/0, if the service rating is 175 amps. But that is not the same thing. You cannot take two 1/0 in parallel, and add a 175 ampacity to a 175 ampacity to get a total ampacity of 350. There are no such words in the table or in the instructions that describe how to use the table. Your only option, as far as this table is concerned, is to look up your required service rating, and see what size conductors that table allows.
 
charlie b said:
Table 310.15(B)(6) does not mention, or make use of in any way, the word "ampacity." It is absolutely not saying that a 1/0 copper has an ampacity of 175 amps under the limited installation conditions it addresses. It says you can use a 1/0, if the service rating is 175 amps. But that is not the same thing. You cannot take two 1/0 in parallel, and add a 175 ampacity to a 175 ampacity to get a total ampacity of 350. There are no such words in the table or in the instructions that describe how to use the table. Your only option, as far as this table is concerned, is to look up your required service rating, and see what size conductors that table allows.
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That is so obvious now that you pointed it out. Thanks.
 
If indeed the service has a 400 amp main, the 4/0 al is paralled in seperate raceways, at 180 amps each, 360 amps total, this would comply with 240.4(B) Devices rated at 800 amps or less. Therefore 310.15 is not needed and the installation is compliant????
 
jamesoftn said:
If indeed the service has a 400 amp main, the 4/0 al is paralled in seperate raceways, at 180 amps each, 360 amps total, this would comply with 240.4(B). . . .
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Yes if, AND ONLY IF, the calculated load is below 360 amps. You must always start with the calculated load. That is Step One. For Step Two, you find a conductor that has that ampacity. Finally, as Step Three, you pick an overcurrent device to protect that conductor. But if the calculated load is between 360 and 400, then a pair of 4/0 aluminum does not have the ampacity to carry that load, and you can't use that for the service conductors.

What Table 310.15(B)(6) does to the three step process I outline above is to replace Step Two with a step that does not use ampacity as its criteria.
 
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