If you have 2 4” parallel conduits going from a transformer to a tap-can to feed a 100amp, 200amp and a 400amp meters. It calls for 3 sets of 4 #350. But like I said only 2 conduits. How would I go about derating. I’m pretty sure I have too. Thanks. P.s. I cannot add another conduit considering the pad for the transformer is already poured.
Parallel conductors
- Thread starter Rriley88
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I think you need another conduit to meet the rules for parallel conductors in the same pipe. Are the conduits between the transformer and the cabinet owned by the utility? They don't have to follow NEC rules. You could try going down a size and reducing the neutrals and going with two sets of four in each pipe.
LarryFine
Master Electrician Electric Contractor Richmond VA
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Welcome to the forum.
Could you use one conduit for the 100a and 200a, and the other one for the 400a, and do no paralleling?
That is what I was thinking. We cannot add another pipe as the slab for the transfomer and sand is already poured. I was thinking doing 2 sets of 500kcmil
I am thinking 2 sets of 500
I am thinking 2 sets of un
What size wire would I run for that? Just 350?
Julius Right
Senior Member
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- Electrical Engineer Power Station Physical Design Retired
If 6+(2 neutrals) single core cables 600 V 350 mcm copper it run in 4" RMC conduit and the over insulation diameter is 0.87" area 0.6 in.^2 [8*0.6=4.8 in^2].
According to NEC Tab.4 Article 344 for more than 2 cables 5.153 in.^2 for RMC 4".
If the neutral is not loaded then 6 single core cables in a conduit -derating factor 0.8 the permissible ampacity will be 280 A.
So, for 100 A and 200 A connections 2 sets of 4*350 kcml copper it seems good.
On the other hand 350 kcml copper ampacity it is only 350 A [according to Table 310.15(B)(16) for 350 kcml 90oC 350 A].
If we take into consideration NEC 110.14 Electrical Connections , and the terminals are not rated for 90oC, you have to employ the ampacity for 75oC [only 310 A] In this case only 600 kcml is required.
According to NEC Tab.4 Article 344 for more than 2 cables 5.153 in.^2 for RMC 4".
If the neutral is not loaded then 6 single core cables in a conduit -derating factor 0.8 the permissible ampacity will be 280 A.
So, for 100 A and 200 A connections 2 sets of 4*350 kcml copper it seems good.
On the other hand 350 kcml copper ampacity it is only 350 A [according to Table 310.15(B)(16) for 350 kcml 90oC 350 A].
If we take into consideration NEC 110.14 Electrical Connections , and the terminals are not rated for 90oC, you have to employ the ampacity for 75oC [only 310 A] In this case only 600 kcml is required.
Julius Right
Senior Member
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- Electrical Engineer Power Station Physical Design Retired
However, for 100 A and 200A part, even if you have to go to 75oC and 8 loaded cables it will be still 0.7*310=217 A.
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With 2-4" empty conduits there are probably at least ten different options. As Augie asked what is the calculated load and how did you come up with 500 kcmil?
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I would be looking at one set of 350s to feed the 100 and 200 amp services in one raceway and one set of 500s in the other to feed the 400 amp service....subject to change if I had the actual load calculations.
As Rob said, there are a lot of combinations that will work with the two conduits and they don't all require parallel sets. The use of parallel sets may reduce the amount of copper required, but the splices and taps may eat of that cost savings. It may be possible to run 3 sets of conductors, one for each of the 3 services. Two sets in one 4" conduit and one set in the other.
As Rob said, there are a lot of combinations that will work with the two conduits and they don't all require parallel sets. The use of parallel sets may reduce the amount of copper required, but the splices and taps may eat of that cost savings. It may be possible to run 3 sets of conductors, one for each of the 3 services. Two sets in one 4" conduit and one set in the other.
What do you mean by calculated load exactly? I have a 1 line diagram that may help. I’m not quiet understanding what you’re asking.
Yes it is a 120/208 3 phase system
Don’t you have to have the same amount and same size of conductors in each raceway for parallel runs though? You can’t put 4 conductors in 1 pipe and 8 in the other can you?
Can you split them up like that? 8 in 1 conduit and 4 in the other?
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You could if you were able to do something like what Larry and Doug are suggesting because it would be different feeds to different loads.
It's only parallel conductors if they are connected at both ends.
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It’s the result of using Article 220. Use the calculated load to determine conductor size and OCPD, based on75 Dec C ampacity
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- New Jersey
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- Journeyman Electrician
Let's assume that the person who came up with the 3 sets of 350's calculated the load. You didn't say copper or aluminum but let's assume aluminum. That's 3*250=750 amps. All you need to do is find the equivalent of 750 amps using only 2 sets of conductors. Parallel 750 aluminum would work (385*2=770 amps) or parallel 500 copper (380*2=760 amps). There are other possible combinations including using 4 smaller conductor sets, two sets in each raceway.
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