What gauge wire is proper for a 200 foot distance

I'm getting ready to run power to my barn from my house. It's about a 200 foot run. I'm planning on running a 220V 60A in two 30A legs from my house to be split into four 15A breakers in the barn. My question is this... Is 8 gauge wire sufficient over this distance or should I go with 6 gauge? Also, can the ground wire be of a lighter gauge than the two hots and neutral? I'm planning on using 4 single wires through conduit rather than direct burial.
Thanks,
Doug
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Your description is contradictory. "220V 60A" and "two 30A legs" are not the same thing. If you have a double-pole 30A breaker, you have 30A at 220V, not 60A; conversely, if you have 60A at 220V, then you have two 60A legs, not 30A.

Well, that depends entirely on whether you're running 60A at 220V, or 30A at 220V. If the latter, you're fine with 8ga. If the former, you need 6ga at a minimum: 8ga does not meet Code for 60A, period.

Minimum size for the equipment grounding conductor is 10ga, whether at 30A or 60A.
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Doug Miller (alphageek at milmac dot com)
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Can't answer it with the info you provided. It depends on how much current is actually going to be used in your three circuits, how evenly it will be split between the legs, and even how long the circuits will be. Pardoxically, 30a/0a actually requires larger wire than 30a/30a; since the former is 120v and the latter is 240v.
Figure that out, and plug it into http://www.electrician.com/vd_calculator.html I would have to look up the acceptable VD (3%?) but I expect someone will tell you that.
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Utter nonsense from Toller, as usual. The ampacity of any given wire is the same under the Code, regardless of whether it's at 240V or 120V.
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Toller has a point. The ampacity of the wire indeed is the same whether it's 120V or 240V, secondly, the total voltage drop of the 30A is the same whether it's 240V or 120V, _but_, the percentage at 120V is double that of 240V.
Ie: say the voltage drop at 30A thru the cable is 3V. That's 3% of 120V and 1.5% of 240V. You have to size the wire so the percentage droop is permissible at 120V.
Thus, when you're computing allowable voltage percentage droop of a feeder circuit, you have to compute the percentage of the voltage drop at 120V, not 240V.
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(Chris Lewis) wrote:

Well, yes, but to make the blanket statement that 30A at 120V requires larger wire than 30A at 240V -- as he did -- is in fact utter nonsense.
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On Mon, 07 Nov 2005 02:12:47 GMT, snipped-for-privacy@milmac.com (Doug Miller) wrote:

This really all depends on how the circuit is loaded. If he loads up one phase and nothing on the other you have to look at this as a 120v circuit. If he balances the A and B phase then it is a 240v circuit and the neutral has only the unbalanced load.
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Yes, I know that. But Toller wasn't talking specifically about the neutral. He may well have been *thinking* about the neutral, but that isn't what he wrote. He made the general statement that 30A at 120V "requires" larger wire than 30A at 240V. And, as a general statement, that's nonsense.
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He was talking about 120V, then of _course_ he was talking about the neutral.
Or, do you get 120V from a hot and something _other_ than a neutral?
Sheesh.
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On Mon, 07 Nov 2005 03:11:25 -0000, snipped-for-privacy@nortelnetworks.com (Chris Lewis) wrote:

In the center tapped system the US has, you also get 120v from 2 balanced loads across 240v, which is 50% more efficient than a single 120v load. The neutral only carries the unbalanced load.
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We have exactly the same center tapped system in Canada. I know exactly how it works, unbalanced loads and all.
Sure, with a perfectly balanced load, you can disconnect the neutral and still have 120V on both 120V circuits.
But you'd better be good at switching things on and off in perfect synchronization...
We all know what happens to 120V circuits if you lose the neutral, don't we?
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(Chris Lewis) wrote:

You missed the point rather badly, I'm afraid. Refer to the statement to which I was objecting: the claim that circuit conductors had to be larger for 120V than for 240V. This is of course nonsense. Lewis is pointing out that, in a 240V system, the neutral carries only the unbalanced load (and thus, by implication, can safely be smaller than the ungrounded conductors). While true, that's not particularly relevant to the statement I was disputing: the blanket assertion that wires "have" to be larger at 120V than 240V. That just isn't so. The size of the UNgrounded conductors is the same.
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You seem to be entirely missing the point - as in being totally unaware of NEC-permissible voltage drop (which is a percentage) in long circuits. 200' is a long circuit.
The correct wire size for a circuit is dependent both on the ampacity of the wire AND wire _length_.
I said _nothing_ about reducing the size of the neutral, because in a 240/120V circuit, you must assume that at times the load will be 100% unbalanced.
_Think_: NEC permits a certain V=IR drop in a circuit in terms of a percentage of the circuit voltage. A fully unbalanced load (120v) will have the same V=IR drop as a fully balanced load (240V), but at 120V, the percentage drop is _doubled_.
It's entirely possible for a given 120V circuit at, say, 30A requiring _larger_ wire than a 240V circuit at 30A due to V=IR losses stated as a percentage being too high.
As a corrollary, a 120V circuit with a given wire size at 30A can go only half as far as a 240V circuit at 30A can go.
While a short 30A circuit can be carried quite easily on #10, at 200 feet, while the ampacity of #10 is still sufficient, the V=IR losses are too high. At 200', to select the correct wire for a 240V/120V 30A circuit, you have to consult the NEC tables to find the right wire size. Or compute yourself, calculating the voltage drop of two lengths of whatever size wire, as a percentage of _120V_, not _240V_, and stay under the NEC maximum % drop.
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On Mon, 07 Nov 2005 14:31:25 -0000, snipped-for-privacy@nortelnetworks.com (Chris Lewis) wrote:

The NEC only mentions voltage drop in a fine print note. (AKA "gee whiz" info) It is not part of the enforceable code. You are correct that voitage drop is based on amps and when taken as a percentage of cercuit voltage "x" volts dropped will be a lower percentage of 240v than 120v.
For the original poster, #8 copper should be fine with his anticipated 30a max possible load (two pairs of 15a breakers). Load diversity would generally hold it way under that. His efficiency will be better if he tries to balance that load. Max VD would be 9.336v @30a which is only ~4% of 240v. If his load balance is worst case he will be losiong that all on one side and that would be a bad DESIGN decision.
He wil bump up against a strange article of the code and his ground wire will be required to be #8 also. That means he has to pull wire in pipe since 8ga cable will usually have 10ga EGC. You can buy full size EGC cable, but not at the BORG.
250.122(B) Increased in Size. Where ungrounded conductors are increased in size, equipment grounding conductors, where installed, shall be increased in size proportionately according to circular mil area of the ungrounded conductors.
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He didn't make a blanket statement like that. Read within the context of the thread: when voltage drops are taken into account, a fully unbalanced 120V load _does_ require larger wire (to maintain a given voltage drop percentage) than a balanced 240V load.
Of _course_, likely the nearest size wire that maintains given voltage percentage at 240 is likely large enough to maintain acceptable percentage at 120 anyway.
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Based on a 60 amp load you need 4 awg or 2 awg AL. Lower the load and the wire sizes can get smaller. Yes the ground can be smaller. Might be time to do a load calculation to be sure your not putting money in the ground for nothing.
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How do you figure that? 6ga copper is adequate for 60A. And as I noted in my initial response to him, there's some question whether he has 60A, or 30A.
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Doug Miller (alphageek at milmac dot com)
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