neutral currents

[Smart $]

But for the electrician level calculation and the "close enough" factor, it works fine.

Roger

In the field, or should I say using field measurements it can deviate from the calculation quite a bit. I've tried using the "basic" formula when trying to determine which neutral belonged to which line conductors at panelboards, where several mwbc come into the enclosure in one conduit and are not grouped. It is not as easy as you might think, for the basic calculation result can be way off when power factors differ.

 

[Howard Burger]

Smart $, thanks for a response that I can use next time I get into this discussion at work.

And Charlie, thank you. I can see in my mind's eye the phases pulsing (at 60 hz) and cancelling out at the source in accordance with the formula. Part of my understanding delimma is having always run a grounded conductor for return current, so I choke at the idea of the phase cycling through (or into) the filament of a bulb such that the bulb would actucally produce light. But I'm a little closer to understanding it, now.

hb

 

[Besoeker]

I believe that the OP originally indicated three equal resistive loads - why would they not have the same power factor ?

OP

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?

Unbalanced indicates neither equal nor resistive.

 

[Besoeker]

It is not as easy as you might think, for the basic calculation result can be way off when power factors differ.

Totally agree.
The formula quoted can give quite misleading results.

 

[glene77is]

Inifinity,
I think you are referring to the grouping of the terms,
under the radical sign.
The "+" and "-" symbols are correct, in both cases.

More importantly, the Power Factor is not figured in.
In another thread, we had 118 Amps, 125 Amps, and 128 Amps
which should yield a Neutral Current of 8.89 Amps in a Balanced System..
But, the REAL Neutral Current was 60 Amps!

Big Harmonic Power Factor!

I bet you know all this stuff, you just need another cup of coffee.

 

[jghrist]

Just to confuse things further, for the formula to work, not only do the power factors have to be equal in all phases, but ?-? loads cannot be included. For example, say you have a single resistive 10A load connected from A to B, with no neutral. The current in ?A will be 10A. The current in ?B will be 10A. The current in ?C will be 0A.

A?+B?+C?-A?B-B?C-C?A = 100+100+0-100-0-0 = 100
sqrt(100)=10

But there is no neutral, so there cannot be any current in the neutral.

 

[Besoeker]

Inifinity,
I think you are referring to the grouping of the terms,
under the radical sign.
The "+" and "-" symbols are correct, in both cases.

More importantly, the Power Factor is not figured in.
In another thread, we had 118 Amps, 125 Amps, and 128 Amps
which should yield a Neutral Current of 8.89 Amps in a Balanced System..
But, the REAL Neutral Current was 60 Amps!

Big Harmonic Power Factor!

Displacement rather than harmonic power factor.
If the currents are 118A, 125A, and 128A in A,B, and C respectively with PFs of 0.872 lag, 1.0 and 1.0, the resulting neutral current as a result of Ia being displaced is 60A. The currents add vectorially.

On the other hand, harmonic currents generally add arithmetically with others of the same order.