Electrical Forum - Neutral between phases

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Subject - Neutral between phases
David Hyatt	You can use one neutral to supply 2 hots in single phase service, one neutral to 3 hots in three phase. I always in single phase use one neutral to serve phase A and phase B, but would it make a difference if one neutral feed two circuits on the same phase (N-A-A) Or (N-B-B)? Just wondering if it makes a difference.
JimmyDee	No you can't do that because the neutral will be carrying the load of both hots and really wouldn't be called a neutral in that situation. Jim
Electricman	David if youre talking about general circuitry in a residence where nm cable is being used then there is a chance of overloading the neutral. Say you have a circuit that draws 10 amps and another that draws 8 amps , if you use the same phase to power these circuits you would add the amps to calculate neutral load N A A = 18 amps. But if these circuits where on different phases you would subtract to calculate N A B = 2 amps.So if these circuits where on a 15 amp breaker utilizing the first configuration would be a neutral overload.
iwire	We bat this around for fun at another forum. An example. Four 15 amp circuits in pipe. Four, phase A hot conductors 14 AWG One, Neutral 6 AWG. Can we do this if we want? Is it NEC compliant? How about this one. Again four 15 amp circuits in pipe. Two 14 AWG hots on phase A. Two 14 AWG hots on Phase B. One 10 AWG neutral. Will this work without overload? Is it NEC compliant? Bob
Ryan_J	Question number 1 doesn't comply after derating 80%. a #4 nuetral would comply though. Question number 2 doesn't comply after derating either. If this is single phase, we need a #8 nuetral. If it is three phase, we need #4 nuetral. Anybody else? Assuming single phase
Ryan_J	Well come on bob....how did I do? :D
David Hyatt	Ya-ll are messing with my head now. You are saying put as many circuits as you want as long as you up size the neutral?
Ryan_J	Yep, pretty much :D There really isn't anything in the code that says you can't. I'm not saying that its a neat thing to do or that anyone should do it, but it would be legal.
iwire	quote: Originally posted by Ryan_J Well come on bob....how did I do? :D You did to good, I hate a smart guy.......only kidding. I had never thought of or seen anyone else bring up the deratting issue. I would have to agree with your corrections. quote: Ya-ll are messing with my head now. You are saying put as many circuits as you want as long as you up size the neutral? Sure, isn't that what happens at a panel? I can't see any advantage in running branch circuits this way but IMO it is not a violation.
lctrc789	I have seen some good ones on this before. You are right on a single phase if one of the legs is pulling 24 amps and the other leg is pulling 16 apms the neutral between the secondary lines balance the load 24-16 =8 amps. But if you have two lines on the same leg (say A phase) one pulling 24 amps and the other pulling 2 amps then (with one neutral) then you have a problem. You will not pull the difference betwen phases, but together you will in this case have a neutral pulling 26 amps. Three phase transformer neutral currents are very tricky when you apply single phase loads on them, here we use two transformers in small businesses to get 3 phase loads, an open delta connection. This is harder to figure then an easy three pahse load, because in this matter the transformers is only 87% of the rated power of the two together. This calcualtion is a little different because of the neutral can pull more and it is like this. TN= N1 + (N2-N3) So say that line 1 has a current flow of 100 amps line two has a current flow of 75 amps, and line three has a current flow of 75 amps. the amount of neutral current would be 100+(75-50) 100+25 Neutral current flow would be 125 amps.
Ryan_J	Pat I disagree with your calculation. The formula for determining three phase nuetral current is: Sqaure root of: (A squared + B squared + C squared) - (AB)+(BC)+(CA)= Total nuetral current. In your example, (100 sqaured + 75 squared + 75 squared)-(10075)+(7575)+(75*75)=.... (10,000+5625+5625)-(7500+5625+7500)=..... 21,250-20,625=625.... Sqaure root of 625=25 25 amps total nuetral current. In your example, there would have to be some 27th order harmonic currents to get the nuetral current that high.
Pierre Belarge	There are a couple of things the average electrician can get from this post. 1. If you are not sure how to calculate the grounded conductor current, use the usual method of a full sized grounded conductor with the phase leg(s). 2. Yes you can use and 'oversized' grounded conductor with different phase conductors - a pretty neat trick when you know how to apply it.