amperage on a neutural

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roger said:
Of the current measured on the ungrounded conductors, how much is flowing in L-L loads and how much is L-N?

Roger
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Mathematically it makes no difference which loads are L-L or L-N unless you are calculating the neutral current based on the individual powerfactors for each load.

The sum of the currents entering a node is zero. It doesn't matter if they are just passing through (so to speak) or getting returned on the neutral. (The circuit is a "black box".)

The imbalance in the currents is the result of having multiple powerfactors per load, more specifically, per phase. The simplest solution is a 47 degree phase shift on the higher leg, but there could be many combinations of both leading and lagging powerfactors that give the same result.
 
Rick Christopherson said:
Mathematically it makes no difference which loads are L-L or L-N unless you are calculating the neutral current based on the individual powerfactors for each load.
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Rick, we are not trying to calculate anything, the OP is questioning the neutral current he is reading, take L-L loads out of the equation and we will have a better starting point.

It would be nice to know if the OP did verify zero sequencing though.



Roger
 
Cold Fusion said:
Roger -
I'm must be short on coffee today. What does that mean?

cf
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Make sure the sum off all conductors equal zero by putting a clamp around all of them. If a clamp on meter reads current when placed around all the conductors at once I would be looking into SG-1's (post number 8) suspicion.

Roger
 
roger said:
Rick, we are not trying to calculate anything, the OP is questioning the neutral current he is reading, take L-L loads out of the equation and we will have a better starting point.
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Whether you realize it or not, you are trying to calculate something. By subtracting out the L-L loads, all you are doing is, well, subtraction. The core problem remains the same, but you simply reduce the magnitude of the line currents. It actually makes the problem more complicated than it needs to be, given the information provided. As I already said, it is applicable only if you know the powerfactor of individual loads.
roger said:
It would be nice to know if the OP did verify zero sequencing though.
....
If a clamp on meter reads current when placed around all the conductors at once I would be looking into SG-1's (post number 8) suspicion.
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That situation (a super-node) is applicable only if there is ground-neutral bonding at the panel in question.

If that is in fact the case, then all you have done is forgotten one path into the node. However the existence of a super-node shared among the neighbors is far less common than electricians make it out to be. The reason why it gets presented so often is because it is the easy answer to a situation that those individuals do not fully understand. The easiest way to identify a super-node is to measure the current in the main ground before and after cutting the main breaker/disconnect. If there remains current in the ground wire, then you have a super-node. However, in nearly all cases of single-family residences, the ground current drops to zero when the main is open. When that happens, then you know absolutely that there is not a super-node situation.
 
Rick, you are looking at this as an engineer and that is the problem.

An electrician walking into a house will have a few meters and is looking for a reason for the problem, they ordinarily will not be given time to analyze every circuit so simplifying the task and coming up with a reasonable / approximate answer is the best way to approach it and eliminating the L-L loads is a start to this conclusion

Now, if you are on call and available when the OP or our service man needs you, we can take trouble shooting to a different level. ;)

Roger
 
I'm prone to brain farts, but here's the OP

I measured the amperage on a neutural that is part of a 3 wire 100 amp panel, 120/240 volt. One hot carried 21 amps, second leg carried 38amps and the neutural carried 26 amps. Does this sound normal for the nuetural to carry so much current?
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If every single load was L-N,( and wouldn't that be the worst possible situation for current on the neutral) it would not have that much current on the neutral.

If there were ANY L-L loads there would be less on the neutral yet.

So, I'm missing what the L-L loads have to do with it.
It's 4 amps too much for the worst possible situation, with those magnitudes of current on each phase..


Like I said, I often miss the obvious, but I can't think of a way that the neutral could NORMALLY carry that much current under the load conditions he described.

I would say it is NOT normal for the OP's neutral to carry that much current.
 
realolman said:
I would say it is NOT normal for the OP's neutral to carry that much current.
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I agree and SG-1 has suggested a possible solution.

Neutral2.gif




Roger
 
Well now that I look at it that does seem like alot of amps on the neutral.
38-21=17A. Where are the additional 9 amps coming from? Non-linear load perhaps?
 
I guess I just didn't see where the OP asked anything beyond whether there was too much amperage on the neutral.

That's a nice, very understandable diagram, and quite possible. Maybe probable, even .:)
 
roger said:
Rick, you are looking at this as an engineer and that is the problem.
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No Roger, I am not looking at this as an engineer. It is basic circuit analysis that any electrician should know, unless of course they stopped teaching electricity to electricians.
roger said:
...they ordinarily will not be given time to analyze every circuit so simplifying the task and coming up with a reasonable / approximate answer is the best way to approach it and eliminating the L-L loads is a start to this conclusion
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So you're suggesting that 3 circuit paths is more simplified than 2 circuit paths? :-? Is this kind of like the new math?

In one breath (of the same sentence) you state that they don't have time to analyze every circuit, and yet you suggest they subtract out the L-L currents from the measured feed currents. So how are they supposed to do that without analyzing all of those circuits? (I certainly hope you wouldn't suggest they start flipping breakers on active loads just to suit their curiosity.)

What you are not understanding is that figuring out the problem is EXACTLY the same regardless whether there are any L-L loads or not. It makes absolutely no difference to figuring it out. By trying to eliminate the L-L current, you are actually making the problem more complicated, not less complicated.
 
Rick, you've been putting that reply together for over an hour, I will have to conceed that I have made this complicated. :grin:

Roger
 
It is not so much that I supect an open neutral somewhere as much as I think the first step should be to determine where the extra neutral current is originating. The numbers do not match simple Single or 3-Phase calculations.

Is the extra 10 Amps being generated by the premisis wiring & loads or is it part of somebody elses load that is just passing through ?
 
If there were actual L-L loads the unbalance obtained by subtracting would be more than the measured value of the neutral and not less. That's what's so strange about this problem.:-?
 
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