Parallel Conductors

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jonny1982

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If I have parallel 250's in one conduit, and it is 90 degree THHN Copper, is the amp rating 580 (290 plus 290) or is it 430 (250,000 circular mills plus 250,000 circular mills equals 500,000 circular mills which is 500 wire and amp rating of 430?

Thanks
 
iwire said:
You add amps not sizes.
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Because, among other things, in the really large sizes the AC resistance (not just the reactive component of complex impedance) is is higher than naively expected because of skin effect (uneven current distribution over the cross section of the wire, with higher current near the surface.)
 
GoldDigger said:
Because, among other things, in the really large sizes the AC resistance (not just the reactive component of complex impedance) is is higher than naively expected because of skin effect (uneven current distribution over the cross section of the wire, with higher current near the surface.)
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People love to say 'skin effect' but is it really a factor at 60 hz?


I think the biggest factor is that a big fat conductor can't shed the heat as well as two smaller ones.
 
iwire said:
People love to say 'skin effect' but is it really a factor at 60 hz?


I think the biggest factor is that a big fat conductor can't shed the heat as well as two smaller ones.
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Since I am not talking about rated ampacity but the actual DC resistance versus AC resistance numbers which can be found in tables in the NEC, I think that skin effect is significant at 60Hz as long as the wires are large enough.
Note also that the scale depth of the skin effect will result in a higher percentage change in resistance the higher the conductivity of the wire is. (I think. I have not run the equations.....)

As for heat dissipation, it is true that the ratio of electrically conductive area (cross section) increases as the square of the diameter while the surface area increases as the first power of the diameter.
 
jonny1982 said:
So it would be a base amp a city of 580 then?
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You can use the 90C for de-rating so if you have 6 current carrying conductor in one conduit then you have 580 amps times 80%-- T. 310.15(B)(3)(a)- which is 464 amps. If you have 8 current carrying conductor then you multiply by 70%==406 amps
 
iwire said:
People love to say 'skin effect' but is it really a factor at 60 hz?
Click to expand...

ISTR someone writing here a while ago that at 60Hz, the skin effect is about 8.5mm (~5/16"), so for most of what we'd be using, it doesn't matter. I've always assumed this is why you don't see AC bus bars thicker than about 1/2-3/4". Never really thought about whether the SE applies to the individual conductor or the overall assembly of bare bundled strands in a cable (e.g. 4/0), but assumed the latter, since Litz wire is made with insulated strands.

http://electronics.stackexchange.co...fect-a-factor-for-low-current-ac-applications has some info about it.
 
jonny1982 said:
If I have parallel 250's in one conduit, and it is 90 degree THHN Copper, is the amp rating 580 (290 plus 290) or is it 430 (250,000 circular mills plus 250,000 circular mills equals 500,000 circular mills which is 500 wire and amp rating of 430?

Thanks
Click to expand...

250 kcmil THHN has a terminal ampacity at 75C of 255A, and a conductor ampacity at 90C of 290A.

When you parallel two sets of 250 kcmil, you add up the terminal ampacities. Then you add up the conductor ampacities, and apply conditions of use factors. Paralleling two sets within the same conduit means that you have a de-facto 0.8 derate factor on the conductor ampacity (assuming neutral doesn't count). You might have others as well.

2 parallel sets of 250 kcmil THHN copper in the same conduit has a total terminal ampacity of 510 Amps. Assuming 30C ambient, it has a total conductor ampacity of 464A. You need to take the lesser of the two, when claiming the ampacity of this feeder, which is 464A.


The reason that you do not add up the KCMIL when calculating parallel feeder amps, is the surface area to volume ratio. Heat is generated in the body of the wire, and must be dissipated through the surface of the wire. The more surface you get, the more it can dissipate heat. That is why you can use less metal to carry the same current, with more parallel sets.

When calculating a parallel feeder set's ability to curtail voltage drop, parallel 250 kcmil will perform approximately just as well as 500 kcmil for the same situation. For AC, the smaller wires comprising the same total KCMIL feeder will perform slightly better, due to effects that are not present in DC.
 
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