4W Neutral Current

[That Man]

I can't quite tell if I'm having a senior moment, or if I've stumbled on something I haven't considered before, though I suspect it's the first one. I'm calculating voltage drops on a 3 phase 4 wire system where all the loads are single phase LN.

At the the end of the circuit there is a 3A load on phase A to N, so A current is 3A, N current is 3A.
Second to last load is 3A phase B to N, so B current is 3A, N Current is...
My guess is a phase shift of 60 degrees with an amplitude of 1.73x 3 = 5.2A.
First load is 3A phase A to N, so A current is 3A, N current is...
My guess is zero, as this N would cancel out the 2 other N's, so current at the N terminal from the source panel is zero.

Does this sound correct?

 

[That Man]

A photo that tells the story better

A photo that tells the story better

Here's a photo that maybe explains it better.

 

[That Man]

Not sure how I messed that up. Try this photo instead:

 

[ptonsparky]

I’m not smart enough to follow your reasoning, but your neutral current will be zero

 

[That Man]

So, I found a circuit simulator online. Turns out I wasn't quite there. Here's a screenshot:



This tells me that only phase shifts occur, and there are no additive effects. So my diagram would look more like this:



So, from the source, 0A to the first load, 3A to the second load, 3A to the third load.
The phase angle is 0 degrees between the first and second load,
60 degrees between the second and third load,
and from the N to the second load is 30 degrees.

Interesting. I think I'm done, but I'll be monitoring this thread if anyone has any other comments.

 

[LarryFine]

So, from the source, 0A to the first load, 3A to the second load, 3A to the third load.

This is correct. Start with all three 3a loads, zero neutral current.

Now, reduce one load by 1a. What happens to the neutral current?

It rises by that same one amp. Now do the same with 2a or with 3a.

 

[Tony S]

I_n=√((I_a²+I_b²+I_c²)-((I_axI_b)+(I_axI_c)+(I_bxI_c)))

 

[ptonsparky]

I_n=√((I_a²+I_b²+I_c²)-((I_axI_b)+(I_axI_c)+(I_bxI_c)))

View attachment 23444

Now that's familiar and is in my spreadsheet for all those formula I am supposed to know, but don't.

 

[david luchini]

I_n=√((I_a²+I_b²+I_c²)-((I_axI_b)+(I_axI_c)+(I_bxI_c)))

This formula works when the loads have the same power factor, but not when they have different power factors.

 

[gar]

190730-0903 EDT

That Man:

You need to take the phasor (vector) sum of the currents.

.

 

[Tony S]

This formula works when the loads have the same power factor, but not when they have different power factors.

I should have added that.

Get harmonics involved and it goes completely out of the window.

 

[synchro]

Deleted duplicate of post below.

 

[synchro]

190730-0903 EDT

That Man:

You need to take the phasor (vector) sum of the currents.

.

Which in this case is easy to do:

View attachment 23446

You can see that the vector summation of phasors A and B has the same magnitude as them (and therefore an equal RMS current), and also the phasor sum is +- 60 degrees from phasors A and B respectively. This is because they form an equilteral triangle with equal sides and all 60 degree angles.

You can also see that the phasor sum is 180 degrees from phasor C and so they all cancel when C is added in.

If you were interested in the vector difference between A and B (for example if they were wye phase voltages and you wanted the line voltage) then the resulting magnitude would be sqrt (3) times that of one phase.

 

[That Man]

Thanks for the follow-ups everyone. I was trying to develop a simplification of 3 phase 4 wire unbalanced voltage drop that avoids the vector arithmatic, and I think I've got it. I don't think I can take it any further without introducing vectors, and at that point, I would use the phasor diagram method, as it will handle differing power factors. Phasors don't fit neatly on a drawing though, when you have a couple dozen legs to calculate. Here's what I've come up with: