Kirchoff's law

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winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Smart $ said:
Umm, the formula is for a 120? 3? 4W voltage system. I believe it to be impossible to have three equal currents at those angles on 4W mwbc of such a system.

It would be difficult, but not impossible, to set up loads that give the currents and phase angles that Rattus suggested.

I could, for example, set up 3 line to neutral loads. One is resistive, one has a significant capacitive component (leading power factor), and the third has significant inductive component (lagging power factor). Put enough effort into being intentionally unbalanced, and you could get all three legs nearly in phase.

In reality, with reasonable loading, one expects similar power factor on each of the legs on a feeder. The phase angles will all be approximately (but not exactly) 120 degrees apart, and thus an equation that depends upon 120 degree phase angles will be a good approximation.

However, consider the following situation: A panel that feeds only resistive loads, say heaters in a process. You have 30A of line-neutral loads on phase A, 10A of line-neutral loads on phase B, 10A of line-neutral loads on phase C, and 20A of line-line loads from phase B to phase C. What is the current on the feeder neutral?

-Jon
 

rattus

Senior Member
Possible, but unlikely:

Possible, but unlikely:

winnie said:
It would be difficult, but not impossible, to set up loads that give the currents and phase angles that Rattus suggested.

I could, for example, set up 3 line to neutral loads. One is resistive, one has a significant capacitive component (leading power factor), and the third has significant inductive component (lagging power factor). Put enough effort into being intentionally unbalanced, and you could get all three legs nearly in phase.

In reality, with reasonable loading, one expects similar power factor on each of the legs on a feeder. The phase angles will all be approximately (but not exactly) 120 degrees apart, and thus an equation that depends upon 120 degree phase angles will be a good approximation.

However, consider the following situation: A panel that feeds only resistive loads, say heaters in a process. You have 30A of line-neutral loads on phase A, 10A of line-neutral loads on phase B, 10A of line-neutral loads on phase C, and 20A of line-line loads from phase B to phase C. What is the current on the feeder neutral?

-Jon

Winnie, although such a set of loads is unlikely, we agree it is possible, and this example was provided only to demonstrate a case where the neutral current formula does not apply.

I also agree that the expectation that the load current separation is approximately 120 degrees is reasonable, therefore the algebraic formula provides a good approximation.

I am too sleepy to compute In this early in the day.
 

rattus

Senior Member
Right!

Right!

K2500 said:
I was just going for hypothetical, I realize that I'm in over my head, but was just trying to get a feel for the math. I have'nt learned a thing yet without getting my hands dirty.

By the way, my new answer is 15.52Arms...

15.5A is correct. This formula can be evaluated more easily with a spreadsheet or a programmable calculator. If the separation between load currents is close to 120 degrees, then the algebraic formula is a reasonable approximation.

The derivation of the formula is rather messy, so we won't worry with that just yet.
 

rattus

Senior Member
charlie b said:
The algebraic formula does not take angles into account. It uses the RMS values of currents in the three phases, to obtain the RMS value of current in the neutral. So a set of currents at 0, +90, and -90 degrees will not sum to zero, using that formula.

I have a copy of a proof of the validity of that formula. Actually, it is not the entire proof. It is a set of instructions on how to construct the proof. It tells you what to multiply, and which Trig Identities to apply, but it does not show you the math. I received it as a Word file in an email from another member of this Forum, who had said he would send it to anyone who asked for it.

It has been a while since I performed any detailed analysis of unsymmetrical loading, and I fear I don?t have the time now to review my old textbooks. What I recall, and it may well be wrong, is that the mechanism by which you obtain an imbalanced loading condition is going to drag the angles between the three phase currents away from the balanced 0, 120, 240 pattern. That does not depend on the power factor of the loads either.

No question that the formula is valid for 120 degree separation.

For other angles I would use vector math to compute magnitude and phase of In. However, I think this would rarely be necessary. Still it is easy enough to do.

I have both solutions in a spreadsheet, and the value of In comes out practically the same in either case.
 

K2500

Senior Member
Location
Texas
rattus said:
15.5A is correct. This formula can be evaluated more easily with a spreadsheet or a programmable calculator. If the separation between load currents is close to 120 degrees, then the algebraic formula is a reasonable approximation.

I was hoping that it turned out correct, as far a spreadsheet, I would not know how to apply one.

I have started to reseach power factor so that I might better understand some of these relations. It seems I have a lot to learn.

rattus said:
The derivation of the formula is rather messy, so we won't worry with that just yet.

That will be just fine with me:) .
 

Smart $

Esteemed Member
Location
Ohio
rattus said:
Smart, you need to explain this one a little better. I cannot see why In has to be zero for the case of 120 degree separation between load currents. If I am missing something, please educate me.
It is my belief that any current on the neutral shifts load currents away from the supply phase angle. The image below depicts how I see voltage drop in such a scenario. Granted the shift in the phase angles may well be minimal... but your response to charlie stated "load currents are separated by exactly 120 degrees"... where, according to my belief, that only occurs when there is no balancing current "returning" on the neutral.

Notexactly120degrees.gif
 

Smart $

Esteemed Member
Location
Ohio
winnie said:
It would be difficult, but not impossible, to set up loads that give the currents and phase angles that Rattus suggested.
OK, I'll concede on that point. It was quite late in the awake cycle of my day and a state of delirium had already set in. I literally do not know what I was thinking at that point :wink:

...

However, consider the following situation: A panel that feeds only resistive loads, say heaters in a process. You have 30A of line-neutral loads on phase A, 10A of line-neutral loads on phase B, 10A of line-neutral loads on phase C, and 20A of line-line loads from phase B to phase C. What is the current on the feeder neutral?
Without knowing power factors and circuit wiring details a high-precision figure cannot be calculated. Taking the conventional approach, the current would be calculated as:
√(30?+10?+10?-30?10-10?10-30?10) = 20A​
Are you implying "we" should calculate or determine the neutral current some other way?
 

rattus

Senior Member
I still don't get it:

I still don't get it:

Smart, I don't see your reasoning at all. Just evaluate the OP's problem with the algebraic formula and with phasor math. The result is,

In = 15.5A @ -63

It is just a matter of Kirchoff's current law.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Smart $ said:
Taking the conventional approach, the current would be calculated as:
?(30?+10?+10?-30?10-10?10-30?10) = 20A​
Are you implying "we" should calculate or determine the neutral current some other way?

No, I agree with the results that you get.

However I note that you did not run the calculation on the Ia, Ib, and Ic of the feeders to the panel ( which are approximately 30A, 29A, and 29A for the loads that I gave in the example). Instead you ran the calculation on the _subset_ of the panel loads that were connected to the neutral. IMHO a very reasonable approach, but a different one from simply saying 'use the equations on Ia, Ib, and Ic'.

If I had said 'I have a panel. Ia, Ib, and Ic are 30A, 29A, and 29A respectively. What is the load on the neutral?' Then you would have gotten a very different answer with the algebraic equation. Only by knowing the inner details of the load (the composite load presented to the feeder) does the algebraic equation give the correct answer.

-Jon
 

rattus

Senior Member
Not quite:

Not quite:

winnie said:
No, I agree with the results that you get.

However I note that you did not run the calculation on the Ia, Ib, and Ic of the feeders to the panel ( which are approximately 30A, 29A, and 29A for the loads that I gave in the example). Instead you ran the calculation on the _subset_ of the panel loads that were connected to the neutral. IMHO a very reasonable approach, but a different one from simply saying 'use the equations on Ia, Ib, and Ic'.

If I had said 'I have a panel. Ia, Ib, and Ic are 30A, 29A, and 29A respectively. What is the load on the neutral?' Then you would have gotten a very different answer with the algebraic equation. Only by knowing the inner details of the load (the composite load presented to the feeder) does the algebraic equation give the correct answer.

-Jon

Winnie, with the L-L load connected, the separation between line currents is not 120 degrees, therefore the algebraic equation does not apply.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
rattus said:
Winnie, with the L-L load connected, the separation between line currents is not 120 degrees, therefore the algebraic equation does not apply.

I agree completely. In my statement above, I should have said 'only by knowing the details of the load, and considering only the pure 120 degree L-N loads, does the algebraic equation give the correct answer.'

The point is that a panel with a mixture of line-neutral, line-line, and 3 phase loads is quite common. If the line-neutral and the line-line loads are not independently balanced, then you can very easily end up with a situation where the phase angles are so far off that the algebraic equation gives quite invalid results. The particular example that I gave could easily be built, and gives phase angles pretty close to the 0, +90,-90 example which you gave.

-Jon
 

rattus

Senior Member
A Graphical Solution:

A Graphical Solution:

Click on the attachment for a graphical solution to the OP's problem. Note that the phasors may be added in any order.

I used a scale of 10A/inch, but that may not be the case on your computer screen.

A drafting machine would be handy for the graphical solution.
 

Smart $

Esteemed Member
Location
Ohio
winnie said:
I agree completely. In my statement above, I should have said 'only by knowing the details of the load, and considering only the pure 120 degree L-N loads, does the algebraic equation give the correct answer.'

The point is that a panel with a mixture of line-neutral, line-line, and 3 phase loads is quite common. If the line-neutral and the line-line loads are not independently balanced, then you can very easily end up with a situation where the phase angles are so far off that the algebraic equation gives quite invalid results. The particular example that I gave could easily be built, and gives phase angles pretty close to the 0, +90,-90 example which you gave.

-Jon
Hmmm....

I don't believe line-line loads have any or, if so, have a quite small effect on neutral current... whether those line-line loads are balanced or not. I think the point you are making is more directly stated as...
Only line-neutral currents, or the portion thereof as it applies, are used in the formula to determine neutral current of standard 3?, 4-wire (wye) feeders or mwbcs.
The above has previously not been stated in this thread... explicitly, that is. I do not include line-line currents automatically, and forget that others may not know this.

However, what is more in question in my mind is: How do we determine neutral current when power factors are involved? If the power factors are identical on each phase I assume the formula is still valid for the phase angles of the line currents would all be shifted by the same amount. But what of differing power factors? For the sake of discussion let's use rattus' example of load currents at 0?, 90?, and 270? (or -90?). If the load current at 90? and -90? are equal, the current on the neutral would be the inverse of the current for the 0? load?i.e. same amperage at 180??regardless of the current amount, would it not?
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Smart $ said:
Hmmm....

Only line-neutral currents, or the portion thereof as it applies, are used in the formula to determine neutral current of standard 3?, 4-wire (wye) feeders or mwbcs.

Yes, I think that this is a fair statement.

The way that line-line loads come into play could be stated as:
A combination of unbalanced line-neutral and unbalanced line-line currents may result in a system where the _net_ current on each phase appears to be balanced, but the neutral carries significant unbalance current. If the goal is to _both_ balance phase current and minimize neutral current, then line-neutral and line-line loads need to be separately balanced.

How do we determine neutral current when power factors are involved? [...]

I believe that you are correct with respect to different power factors: if all phases have the same power factor, then the displacement is still 120 degrees, and the algebraic equation holds. I concur with your understanding of rattus' example. Finally, I beleive that the fallback answer for the general case is again: vector addtion :)

-Jon
 

rattus

Senior Member
I think he is right!

I think he is right!

Smart $ said:
Hmmm....

I don't believe line-line loads have any or, if so, have a quite small effect on neutral current... whether those line-line loads are balanced or not. I think the point you are making is more directly stated as...
Only line-neutral currents, or the portion thereof as it applies, are used in the formula to determine neutral current of standard 3?, 4-wire (wye) feeders or mwbcs.
The above has previously not been stated in this thread... explicitly, that is. I do not include line-line currents automatically, and forget that others may not know this.

However, what is more in question in my mind is: How do we determine neutral current when power factors are involved? If the power factors are identical on each phase I assume the formula is still valid for the phase angles of the line currents would all be shifted by the same amount. But what of differing power factors? For the sake of discussion let's use rattus' example of load currents at 0?, 90?, and 270? (or -90?). If the load current at 90? and -90? are equal, the current on the neutral would be the inverse of the current for the 0? load?i.e. same amperage at 180??regardless of the current amount, would it not?

Matter of fact, I zinged Bob on this point a few days ago. L-L loads do not affect In.

In the case of differing PFs, vector math may be used, or a graphical solution can be performed.
 

Smart $

Esteemed Member
Location
Ohio
rattus said:
Smart, I don't see your reasoning at all. Just evaluate the OP's problem with the algebraic formula and with phasor math. The result is,

In = 15.5A @ -63

It is just a matter of Kirchoff's current law.
First of all, the OP's problem was asked and answered. There is no need keep repeating the method (which I brought up first in this thread, so I do have an understanding of such) or the solution to that problem. Your responses as such are being taken as what amount to condescending remarks on this end.

Other than that, my reasoning is directed more at your response to charlie than it is the OP's problem. It does relate to the OP's problem in that I don't feel he/she fully comprehends the effect of power factors where current phase angles are out-of-phase with their respective circuit's voltage phase angles. Thus in stating the problem, he/she may have unwittingly used nominal voltage angles in describing current phase angles.
 

rattus

Senior Member
Smart $ said:
First of all, the OP's problem was asked and answered. There is no need keep repeating the method (which I brought up first in this thread, so I do have an understanding of such) or the solution to that problem. Your responses as such are being taken as what amount to condescending remarks on this end.

Other than that, my reasoning is directed more at your response to charlie than it is the OP's problem. It does relate to the OP's problem in that I don't feel he/she fully comprehends the effect of power factors where current phase angles are out-of-phase with their respective circuit's voltage phase angles. Thus in stating the problem, he/she may have unwittingly used nominal voltage angles in describing current phase angles.

Smart,

I was referring to the diagram and discussion which you recently posted. Makes no sense to me. You are, I think, trying to prove that In must be zero if the separation is 120 degrees. Straighten me out if I misinterpreted your post.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
rattus said:
Smart,

I was referring to the diagram and discussion which you recently posted. Makes no sense to me. You are, I think, trying to prove that In must be zero if the separation is 120 degrees. Straighten me out if I misinterpreted your post.

I've been thinking about Smart $'s point, and I agree with it.

Assume that you have only line-neutral loads, with unity power factor, and that the system supply has perfectly balanced voltages with 120 degree phase relation.

If the loads are not balanced, then there will be current flow on the neutral. Any current flow on the neutral will mean voltage drop on the neutral. This voltage drop will have a phase angle and magnitude that depends upon the loads and the resistance of the neutral conductor. The supply voltages, _relative to the neutral conductor at the loads_ will not be perfectly balanced, and will not have that perfectly balanced 120 degree phase relation.

Thus when you have a neutral conductor that has resistance, if the loads are not balanced, the load currents will not have the 120 degree phase relation necessary for the algebraic equation to be exact.

-Jon
 

K2500

Senior Member
Location
Texas
Smart $ said:
... I don't feel he fully comprehends the effect of power factors where current phase angles are out-of-phase with their respective circuit's voltage phase angles.

You are correct.
 

rattus

Senior Member
Sure, but:

Sure, but:

winnie said:
I've been thinking about Smart $'s point, and I agree with it.

Assume that you have only line-neutral loads, with unity power factor, and that the system supply has perfectly balanced voltages with 120 degree phase relation.

If the loads are not balanced, then there will be current flow on the neutral. Any current flow on the neutral will mean voltage drop on the neutral. This voltage drop will have a phase angle and magnitude that depends upon the loads and the resistance of the neutral conductor. The supply voltages, _relative to the neutral conductor at the loads_ will not be perfectly balanced, and will not have that perfectly balanced 120 degree phase relation.

Thus when you have a neutral conductor that has resistance, if the loads are not balanced, the load currents will not have the 120 degree phase relation necessary for the algebraic equation to be exact.

-Jon

Well yes, but as Larry Fine says, let's not confuse the OP any more than necessary. For this basic discussion, we should assume perfectly stiff sources. The OP is just starting to crawl, let's not make him run just yet.

OK with you k2500?
 
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