Calculation of neutral current

[gar]

100817-1505 EST

LAMO:

Professor Bill Dow would tell you to figure it out for yourself.

Do the third harmonics flow in the neutral? I would answer yes.

Then I suggest your real question is do they cancel? Draw waveforms and see what happens.

.

 

[LMAO]

100817-1505 EST

LAMO:

Professor Bill Dow would tell you to figure it out for yourself.

Do the third harmonics flow in the neutral? I would answer yes.

Then I suggest your real question is do they cancel? Draw waveforms and see what happens.

.

may I ask why 3rd harmonics flow in neutral? and not in phases?

 

[LarryFine]

may I ask why 3rd harmonics flow in neutral? and not in phases?

They flow in the phases, but only add (as in 'in phase') in the neutral.

 

[Smart $]

may I ask why 3rd harmonics flow in neutral? and not in phases?

As Larry said... but it sounded like you knew this when you asked your first question...

If you have a balanced 3 phase, wye, 4 wire and you have 10A rms 3rd harmonic in each phase, would these harmonics flow through neutral? And why?

If each phase (line) has 10A of 3rd harmonic current, there would be 30A on the neutral.

 

[LMAO]

As Larry said... but it sounded like you knew this when you asked your first question...


If each phase (line) has 10A of 3rd harmonic current, there would be 30A on the neutral.

OK, what I was really asking was "why third harmonics add up in neutral, as oppose to canceling each other out".
tell me if my reasoning is correct:
for phases a, b and c:
i(a)=i*sin(wt+0);
i(b)=i*sin(wt+120);
i(a)=i*sin(wt+240);
for the third harmonics, we have i(a)=(3wt+0) and i(b)=(3wt+3*120) and i(c)=(3wt+3*240).
That's why they add up, the phase shifts are all multiples of 360.
Is my reasoning right?

 

[LarryFine]

They add up because the positive peaks of the third harmonics occur simultaneouly, as do the negative peaks.

 

[jghrist]

OK, what I was really asking was "why third harmonics add up in neutral, as oppose to canceling each other out".
tell me if my reasoning is correct:
for phases a, b and c:
i(a)=i*sin(wt+0);
i(b)=i*sin(wt+120);
i(a)=i*sin(wt+240);
for the third harmonics, we have i(a)=(3wt+0) and i(b)=(3wt+3*120) and i(c)=(3wt+3*240).
That's why they add up, the phase shifts are all multiples of 360.
Is my reasoning right?

Yes, if you add sin(), e.g. i3(b)=i3*sin(3wt+3*120)

 

[ronaldrc]

Hello, my two cents as usual a picture

I down loaded the free Mathematica Player Demo by Wolfram Research,Inc.

The first graph is the default graph all Rs or current are set for 1.0 and magneta or red line equals zero current.

The second graphs is currents set as close as I could get them with the sliders to 50,40 and 40 amps. which where 50,39,39 amps. and with no Pf.

The third graph is currents set as close as I could get them with the sliders to 50,40 and 40 amps. which where 50,39,39 amps. PFs. as close to .707,.866 and .707 as I could them with the sliders
which where .7, .86 and .7 these PFs are lag,lag and lead as to the OP. post.


The forth graph is currents set as close as I could get them with the sliders to 50,40 and 40 amps. which where 50,39,39 amps. PFs. as close to .707,.866 and .707 as I could them with the sliders
which where .7, .86 and .7 these PFs are lead,lead and lag backwards as to the OP. post.


Click here to see Graphs



Ronald

 

[ronaldrc]

Hello, my two cents as usual a picture

I down loaded the free Mathematica Player Demo by Wolfram Research,Inc.

The first graph is the default graph all Rs or current are set for 1.0 and magneta or red line equals zero current.

The second graphs is currents set as close as I could get them with the sliders to 50,40 and 40 amps. which where 50,39,39 amps. and with no Pf.

The third graph is currents set as close as I could get them with the sliders to 50,40 and 40 amps. which where 50,39,39 amps. PFs. as close to .707,.866 and .707 as I could them with the sliders
which where .7, .86 and .7 these PFs are lag,lag and lead as to the OP. post.


The forth graph is currents set as close as I could get them with the sliders to 50,40 and 40 amps. which where 50,39,39 amps. PFs. as close to .707,.866 and .707 as I could them with the sliders
which where .7, .86 and .7 these PFs are lead,lead and lag backwards as to the OP. post.


Click here to see Graphs

Hey everybody

Sorry about that seems like I ran everyone off from this thread.

But if anyone is interested in the graphics above you can download the interactive player here.

http://www.wolfram.com/products/player/

Did anyone examine the pictures I might have gotten the lead and lag currents backwards
if so sparkys answer was wrong?

Ronald

 

[Lady Engineer]

Hey everybody

Sorry about that seems like I ran everyone off from this thread.

But if anyone is interested in the graphics above you can download the interactive player here.

http://www.wolfram.com/products/player/

Ronald

No you didn't run anyone off, Ronald. We just had really in depth conversation about this topic, I even added my two cents, and someone thought maybe my method was erroneous, so I threw my hat into the ring. However, everyone's input should always be welcome!

 

[david luchini]

Did anyone examine the pictures I might have gotten the lead and lag currents backwards

Unfortunately, the Wolfram Mathmetica Player doesn't provide accurate graphs.

If you look at your graph with phase A: R=.5, Xc=0, Xl=0 you will see wave have a magnitude of 0.5, as expected.

But if you look at the next graph where Phase A is: R=0.5, Xc=0.7, Xl=0 you see that the magnitude is still 0.5, when the value should be 0.86.

 

[Pham Van Hung]

Unfortunately, the Wolfram Mathmetica Player doesn't provide accurate graphs.

If you look at your graph with phase A: R=.5, Xc=0, Xl=0 you will see wave have a magnitude of 0.5, as expected.

But if you look at the next graph where Phase A is: R=0.5, Xc=0.7, Xl=0 you see that the magnitude is still 0.5, when the value should be 0.86.

The first field indicates current not resistance. So if you adjust to 0.5 to phase A, that means 0.5A and the phase A voltage on the graphic is 1V. Therefore, R will be 2 Ohms.
When you add Xc = 0.7, then z=Sqrt(2*2 + 0.7*0.7) =2.118962 Ohms.
As the voltage on phase A is still = 1V, so the current will be 1 divided by 2.118962 and actually you get 0.471929A.
This value looks like 0.5A when you see it on the graphs. That is where you get confusion.

However, the Player is just an approximation as mentioned somewhere in this post.

 

[david luchini]

The first field indicates current not resistance. So if you adjust to 0.5 to phase A, that means 0.5A and the phase A voltage on the graphic is 1V. Therefore, R will be 2 Ohms.
When you add Xc = 0.7, then z=Sqrt(2*2 + 0.7*0.7) =2.118962 Ohms.
As the voltage on phase A is still = 1V, so the current will be 1 divided by 2.118962 and actually you get 0.471929A.
This value looks like 0.5A when you see it on the graphs. That is where you get confusion.

I don't believe this is correct. The simulator says to vary the "loads" of a simple power system. Resistance, Capacitive Reactance and Inductive Reactance are "loads." Current is not load.

For instance, start with Phase A, R=1, and all other values set to zero. You will see phase A plotted as a sine wave with with magnitude of 1. Resistance would be 1 ohm under the scenario that you suggest. If you then add a reactance of Xl=1.75, then the impedance of the circuit should be z=sqrt(1*1 + 1.75*1.75)=2.016 Ohms. So current should be 0.49614 Amps, but you will see that the magnitude of Phase A hasn't changed. It is still 1.

For some reason, the Wolfram simulator does not properly calculated the magnitude of the current properly using the circuit impedance of Z=R+jX. It uses only R in figuring magnitude, but uses X in determining phase shift.

Edit: I see what you're saying. It incorrectly figures the current as you adjust the resistance. Basically, it doesn't model anything right.

 

[ronaldrc]

Hello

I think the wolfram simulator gives a good aproximation you can not adjust it to the fine point you need, 39 was as close as I could get to 40.

I believe since this is a neutral current simulator it would stand to reason the R would
equal current.

I'm not sure which is which on leading and lagging current XC or XL?

I beleive sparky and Rattus are right and the current is somewhere in the range of 56 amps. but according to this simulator it is around 62 amps. but like I said I couldn't get the sliders to adjust very close, I don't know how much that would affect the reading?

I do know if you turn any R all the way up it reads 1.75 I assume current.

If you have one R slider set to 1.75 and everything else at zero the current equals 1.75 on the R and the neutral as it should.

If you slide any two R sliders to 1.75 and every thing else to zero the currents equal 1.75 on the R and neutral as it should.

If you slide all three Rs at any setting as long as they are equal the neutral current equals zero as it should.

All the sliders when slid left or right for the PFs seem to hold their shape I beleive it is correct.

I know the designer has put a lot of work into this project and deserves a bit more respect that.

Ronald

 

[rattus]

Ronald,

Although this program is somewhat educational, I find that the best way to visualize this problem is with a phasor diagram. It is easy enough then to compute In in a spreadsheet.

Bear in mind though, that there are two valid results unless the phase sequence is stated.

 

[david luchini]

I believe since this is a neutral current simulator it would stand to reason the R would
equal current.

I can't think of any situation that for a constant voltage and a power factor of 1.0, that current would be proportional to the Resistance. In fact, the current should be inversely proportional to the resistance. For example, if I=1 and R=1, then for I=0.5, R should equal 2.

I'm not sure which is which on leading and lagging current XC or XL?

Your examples have reversed the leading and lagging power factors, but it really doesn't change the general point. But Xc should provide a leading power factor, and Xl should provide a lagging power factor.

All the sliders when slid left or right for the PFs seem to hold their shape I beleive it is correct.

This is the problem with the simulator, the sliders for "PF" do not hold the proper shapes when moved left and right. For example, if you set phases B and C to zero, and worked on only phase A with the setting of R=0.5 & Xc=0.7 you would expect a value (with your method) of 0.5A leading by 45 degrees. But if you plug that into the simulator and look at the vector plot, you will see that the current leads the voltage by approximately 55 degrees. And if you plug in R=0.5, Xc=0.866 you would expect the vector plot to show 0.5 leading by 30 degrees, but the simulator shows it leading by 60 degrees.

The simulator does not accurately model current/power factor, so the graphs in your example are not accurate. For instance, if you look at your third example graph, you will see that phase B & C currents are on top of each other. But for voltage angles of -30, -150 (for BC, ie, 120 degrees apart) if one current leads by 45 deg. and one lags by 30 deg, the two phase currents should be 45 degrees apart. But they are not.

I know the designer has put a lot of work into this project and deserves a bit more respect that.

I'm sure the designer put alot of work into the project, and it provides a nice graphical representation IF you can decipher how to "work" the inputs to provide accurate results. But as it is designed, it does not provide accurate results. This is not disrespectful of the designer, it is a statement of fact that must be understood if the simulator is to provide accurate results.

 

[ronaldrc]

Rattus and David

I've been on the internet all day, I believe I went plum to the end of it.My comp. is just crawling now.

Rattus that shows you how much I know I thought that was a phasor graph and the other one was the vector. I'm too old and uninterested in it now to try to learn the math for phasors and vectors.

Someone brought the wolfram simulator up and I down loaded it just out of curiosity.
Like Dave said it gives a good idea of what lagging and leading currents do.

But we couldn't use it for a dependable way to figure neutral current.Great graphics and
close to the real thing though.

Ronald

 

[david luchini]

But we couldn't use it for a dependable way to figure neutral current.Great graphics and
close to the real thing though.

Ronald, just FYI, if you plug in:

Phase A: R=0.5, Xl=0.5 (50A, .707Pf lagging)
Phase C: R=0.4, Xl=0.2309 (40A, .866Pf lagging)
Phase B: R=0.4, Xc=0.4 (40A, .707Pf leading)

You will get an accurate model of the values in the original post (assuming an ABC phase rotation.) You will note that in the inputs you have to reverse phase B and C, because the way it plots the values is Phase A, followed by Phase C, followed by Phase B.

 

[ronaldrc]

Dave

You're right I hadn't even caught that B and C are reversed I will reverse my numbers tomorrow and see what I come up with. I never could get the number R 40 though it always went from 39 to 41. Can you get the 40?

Good night:Ronald

 

[david luchini]

Dave

You're right I hadn't even caught that B and C are reversed I will reverse my numbers tomorrow and see what I come up with. I never could get the number R 40 though it always went from 39 to 41. Can you get the 40?

Good night:Ronald

Yes, the thing to do is expand the input selection by clicking on the "plus" sign to the right of the slider. That will give you a box where you can type in an exact value. Give it a try.