Calculating Neutral Load Question?

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Besoeker

Senior Member
Location
UK
I am not sure why that is a shame.

They teach electricians what is expected of electricians, they teach engineers what is expected of them.
So it is expected of electricians that they will use a formula they have been taught that will give them the wrong answer every time except for one very specific set of circumstances?

Whilst I respect you, I cannot agree that is a reasonable thing to do. Better to give them the tools to do it correctly or not at all.
 

iwire

Moderator
Staff member
Location
Massachusetts
So it is expected of electricians that they will use a formula they have been taught that will give them the wrong answer every time except for one very specific set of circumstances?

The subject of what is required of an electrician is a tough one as it is very different from place to place, even job to job.

A great many, I would say the large majority of US electricians will never have a need to do a neutral current calculation other than during a test.

Those that are in positions where that is required will quickly learn that there is more to it than what we were taught.

Whilst I respect you, I cannot agree that is a reasonable thing to do. Better to give them the tools to do it correctly or not at all.

That is cool, but would put many EEs out of work as most are not taught about the NEC.
 

Besoeker

Senior Member
Location
UK
The subject of what is required of an electrician is a tough one as it is very different from place to place, even job to job.
Totally agree. Whilst there would be overlap, I would expect a maintenance electrician in say, a paper mill, to have different skills to one wiring domestic residences.

A great many, I would say the large majority of US electricians will never have a need to do a neutral current calculation other than during a test.
Probably the same here.

Those that are in positions where that is required will quickly learn that there is more to it than what we were taught.
Maybe this is where we politely disagree.
"More to it" suggests that it is a starting point. A simple approach. It's one that has been cited on this forum a number of times at least as far back as 2008.

Problem is that for most situations it will produce an incorrect answer and sometimes a very incorrect answer.
Given that, it's worse than useless and it shouldn't be taught at all in my opinion.

That is cool, but would put many EEs out of work as most are not taught about the NEC.
Would that be a bad thing?
:D
 

fmtjfw

Senior Member
If I use a true RMS ammeter will it show the real current in a conductor? I assume that it samples the waveform and sums it.

In terms of "calculating" neutral currents when there are varying power factors (different between phases), I assume that you need to calculate the capacitance, inductance, and resistance. I assume the inductance is not so easy to calculate, I suppose motors have different inductance based on loading.

In terms of "calculating" harmonics, I assume you can sometimes get harmonic content from device cut sheets and from standards.

In terms of tests these factors are seldom included.


So what's a curious electrician to do?
 

Besoeker

Senior Member
Location
UK
If I use a true RMS ammeter will it show the real current in a conductor? I assume that it samples the waveform and sums it.
TBH, I don't know how but your assumption seems reasonable. If it's a non-linear waveform, I often record it on a storage scope, drop it into a spreadsheet, and do the calcs based on that.

In terms of "calculating" neutral currents when there are varying power factors (different between phases), I assume that you need to calculate the capacitance, inductance, and resistance. I assume the inductance is not so easy to calculate, I suppose motors have different inductance based on loading.
For induction motors, the most ubiquitous type of three phase motor, there is an approximately fixed magnetising (quadrature) component at no load resulting in low power factor. As loading increases the mostly active component increases resulting in higher power factor.

In terms of "calculating" harmonics, I assume you can sometimes get harmonic content from device cut sheets and from standards.
Some manufacturers provide that information. Typically on mainstream variable speed drives they do and some provide calculation tools.
But in my experience the culprits are things like computers and televisions. Not that any single one contributes a lot. Just that there are huge numbers of them.

So what's a curious electrician to do?
Ask and continue asking.
Reflect on the answers.
And ask again.
 

infinity

Moderator
Staff member
Location
New Jersey
Occupation
Journeyman Electrician
Since Kam is an electrician studying for an exam (and I'll assume not a PE exam) a simple explanation is all that's required. Anybody got one? :)
 

Kam1

Member
Location
Chicago, IL
Since Kam is an electrician studying for an exam (and I'll assume not a PE exam) a simple explanation is all that's required. Anybody got one? :)

I appreciate everyone helping out. This is a question asked on the Chicago Electricians Exam. Just to clarify, the question was dealing with a single phase, 3 wire system. So, if I am not mistaken the neutral load will carry 0A?
 

Besoeker

Senior Member
Location
UK
I appreciate everyone helping out. This is a question asked on the Chicago Electricians Exam. Just to clarify, the question was dealing with a single phase, 3 wire system. So, if I am not mistaken the neutral load will carry 0A?
I expect that was the answer expected.
 

mivey

Senior Member
Problem is that for most situations it will produce an incorrect answer and sometimes a very incorrect answer.
Given that, it's worse than useless and it shouldn't be taught at all in my opinion.
I agree. It is not even worthy to be called a "rule of thumb". It is for academia at best and I would just call it an interesting observation, not something that is useful as a tool.

We might not all be good at complex math but we should at least be able to use a straightedge and a compass. I have even used sections of pipe or range rods, etc on the ground and a tape measure to demonstrate the vector relationships. It really is not that much work to make the correct calculation. FWIW, if I am working in AutoCAD, I will sketch the vectors for a quick solution and for the fun of it. For most of this stuff I just use a complex-math calculator or Excel (unless I can just directly read the current).

If an EC can calculate box fill, feeder loads, bend conduit etc. it really is just child's play. IMO, there are other things the EC deals with that are much more complicated.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Since Kam is an electrician studying for an exam (and I'll assume not a PE exam) a simple explanation is all that's required. Anybody got one? :)
I will try. Let?s use, as an example, a 3-wire circuit that serves some heating appliance with two heating elements. Let ?the first element? be powered from the Phase A wire and the neutral (call them the black wire and the white wire). Let ?the second element? be powered from the Phase B wire and the neutral (call them the red wire and the same white wire). Let the two elements be controlled separately.


  • If only the first element has been turned on, then during one half cycle current will flow from the panel, along the black wire, through the first element, and back to the panel on the white wire. During the other half cycle, current will flow from the panel, along the white wire, through the first element (?backwards?), and back to the panel on the black wire. This is an unbalanced loading situation.

  • If only the second element has been turned on, then during one half cycle current will flow from the panel, along the red wire, through the second element, and back to the panel on the white wire. During the other half cycle, current will flow from the panel, along the white wire, through the second element (?backwards?), and back to the panel on the red wire. This is another unbalanced loading situation.

  • If both elements have been turned on, then during one half cycle current will flow from the panel, along the black wire, through the first element, then through the second element (?backwards?), and back to the panel on the red wire. During the other half cycle, current will flow from the panel, along the red wire, through the second element, then through the first element (?backwards?), and back to the panel on the black wire. This is a balanced loading situation, and you can see that the neutral current is zero.
 

Kam1

Member
Location
Chicago, IL
Thank You everyone!!

Thank You everyone!!


I will try. Let?s use, as an example, a 3-wire circuit that serves some heating appliance with two heating elements. Let ?the first element? be powered from the Phase A wire and the neutral (call them the black wire and the white wire). Let ?the second element? be powered from the Phase B wire and the neutral (call them the red wire and the same white wire). Let the two elements be controlled separately.


  • If only the first element has been turned on, then during one half cycle current will flow from the panel, along the black wire, through the first element, and back to the panel on the white wire. During the other half cycle, current will flow from the panel, along the white wire, through the first element (?backwards?), and back to the panel on the black wire. This is an unbalanced loading situation.
  • If only the second element has been turned on, then during one half cycle current will flow from the panel, along the red wire, through the second element, and back to the panel on the white wire. During the other half cycle, current will flow from the panel, along the white wire, through the second element (?backwards?), and back to the panel on the red wire. This is another unbalanced loading situation.
  • If both elements have been turned on, then during one half cycle current will flow from the panel, along the black wire, through the first element, then through the second element (?backwards?), and back to the panel on the red wire. During the other half cycle, current will flow from the panel, along the red wire, through the second element, then through the first element (?backwards?), and back to the panel on the black wire. This is a balanced loading situation, and you can see that the neutral current is zero.


I am happy to say that I have passed my exam! Thanks to everyone who helped me understand concepts and questions I had about the code. I hope everyone the best of luck.
 

jumper

Senior Member
I am happy to say that I have passed my exam! Thanks to everyone who helped me understand concepts and questions I had about the code. I hope everyone the best of luck.

excellent.jpg
 
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