substation guy
Member
how do you calculate neutral current on an unbalanced wye connected load i thought you just subtract the three amp readings from each leg think im wrong though anyone help me out here?
neutral currents
- Thread starter substation guy
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- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
Okay here's another look.
- Location
- New Jersey
- Occupation
- Journeyman Electrician
Okay here's another look.
Three different formulas which one is right?
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
Mine, of course. Charlie's is really the same perhaps but I think there is a parenthesis missing to make it accurate.
Howard Burger
Senior Member
- Location
- N of Anchrage, AK
OK, guys. Thanks for the opportunity to demonstrate my lack of theory...
If on balanced 3 ph loads to resistive loads, like say 3 circuits at 10 amps each to incandescent lights, why does the neutral notcarry any return current (and please, don't say 'because the formula says the answer is zero'). Taking this to it's illogical extreme, if I run three circuits to these lights, and the neutral has no return current, why should I run a neutral at all?
Reason I ask, an apprentice and I were discussing non-linear loads and the importance of upsizing neutral, and I got into the 'when you don't count the neutral as a ccc' and this question came up. Thanks.
hb
If on balanced 3 ph loads to resistive loads, like say 3 circuits at 10 amps each to incandescent lights, why does the neutral notcarry any return current (and please, don't say 'because the formula says the answer is zero'). Taking this to it's illogical extreme, if I run three circuits to these lights, and the neutral has no return current, why should I run a neutral at all?
Reason I ask, an apprentice and I were discussing non-linear loads and the importance of upsizing neutral, and I got into the 'when you don't count the neutral as a ccc' and this question came up. Thanks.
hb
- Location
- New Jersey
- Occupation
- Journeyman Electrician
In essence you've created a 3 phase, 10 amp WYE load. The current will flow through all three paths created by this connection so no neutral return path is needed.
Howard Burger
Senior Member
- Location
- N of Anchrage, AK
I'm sorry, I don't understand. 'So no neutral return path is needed'. But I have to run a neutal with each hot leg to make it work.
In theory, if all three loads were equal, that is true.
only if all three phases are not equaly loaded. The neutral carries only the unbalanced load (theory).
Howard Burger
Senior Member
- Location
- N of Anchrage, AK
And I think, there is where I'm letting it go. 'In theory' it should be 'A'. In reality, you'd better do 'B' or it won't work. When I tell the new guys why it's important not to open the neutral on an mwbc I explain that if a neutral is opened that the laws of electricity change from paralell to series, and different rules apply, because that's the way electricity is made. I'ts a God thing. Don't worry about it, just do it right. (Sigh) As Pierre said in another post somewhere, there's so much to learn ....
You don't have to run a neutral to make it work (i.e. balanced or equal loads on all three lines), but line to neutral loads can be switched on and off individually, or loads can fail such as incandescent lamps thereby unbalancing the mwbc. If you switch one or two off, the neutral is necessary to make the second and/or third work properly.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
No "perhaps" about it. Gus had an error or two in the brackets and parenthesis stuff. But yours and mine are mathematically the same, and I posted first. So there! :grin:
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Howard, the answer to your question is not easy to explain, without being there in person, with a pencil and paper handy. I will try, though.
It starts by noting that ?amps? is not the same as ?amps? Confused, yet? What I mean is that the amps you measure on each of the three phases is different from the others in an important way. They are out of phase from each other by 120 degrees. That is, at the moment in time that Phase A current is at its maximum positive value, the Phase B current is at a smaller value, is negative, has past its point of being maximum negative, and is getting closer to being zero. At the same moment, the Phase B current is also negative, and is approaching its maximum negative value. The next phase that will reach a maximum positive value will be Phase B, and then Phase C, and then Phase A again. The simple way to say this is that they are not the same value at any moment in time, so you can?t just add the three and come up with a total value.
In a practical sense, for a perfectly balanced system, the current that leaves the source on the Phase A conductor will return to the source on the Phase B and C wires, and none will flow in the neutral wire.
It starts by noting that ?amps? is not the same as ?amps? Confused, yet? What I mean is that the amps you measure on each of the three phases is different from the others in an important way. They are out of phase from each other by 120 degrees. That is, at the moment in time that Phase A current is at its maximum positive value, the Phase B current is at a smaller value, is negative, has past its point of being maximum negative, and is getting closer to being zero. At the same moment, the Phase B current is also negative, and is approaching its maximum negative value. The next phase that will reach a maximum positive value will be Phase B, and then Phase C, and then Phase A again. The simple way to say this is that they are not the same value at any moment in time, so you can?t just add the three and come up with a total value.
In a practical sense, for a perfectly balanced system, the current that leaves the source on the Phase A conductor will return to the source on the Phase B and C wires, and none will flow in the neutral wire.
We have been here before. The formula given works only for the specific case of the same power factor for each phase.
In general, it doesn't.
But for the electrician level calculation and the "close enough" factor, it works fine.
Roger
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