Unbalance neutral current in a 3 phase system

[Gary Shumaker]

I understand your position

I understand your position

Roger,
As an Engineer, I agree with your analysis of the proper way to derive the value present on the high leg. However, I am teaching Apprentice Electricians so I must use caution when presenting math problems. Since the √3 also works, that is what I present to these students.
An example of a side story, it has confused even Journeyman how is it that a single phase of 3 phase 480V can be 277V when 2x277=554. I have explained to those who were confused that the 3 phases were 120? out of phase. The sine of 120? is .866. If they multiply 554 by .866 the product is a close approximation of 480V.
Thanks again,
Gary Shumaker

 

[roger]

10a. That's easy to explain using just words.

If you start with the balanced 20/20/20a loading, and no neutral current (with which we all agree, no?), any current you subtract from one phase causes the neutral current to rise by the exact same amount.

Larry, however true, I don't think that this provides enough substantiation for this thread. :wink:

Roger

 

[Gary Shumaker]

10a. That's easy to explain using just words.

If you start with the balanced 20/20/20a loading, and no neutral current (with which we all agree, no?), any current you subtract from one phase causes the neutral current to rise by the exact same amount.

I believe that it is not that simple. The 10A load is still using the other two phases as part of the return path so the result must be less than 10A.
Gary Shumaker

 

[roger]

Gary, using the sqrt of 3 is fine for finding line to line or phase to phase voltages and currents, now you have been shown the correct ways for calculating imballanced neutral currents, and the sqrt of three does not play a direct part.

Roger

 

[charlie b]

May humbly point out that whenever 3 phase current calculations are expressed, √3 is always used (1.732).

May I humbly offer in return that that does not give you freedom to insert that number anywhere you please? It belongs where it belongs, and nowhere else.

May I also ask if you understood what I said earlier about your formula just giving an average value? May I point out that if you divide a number by 1.732, and then divide it again by 1.732, what you end up with is the original number divided by 3? May I also remind you of the ?Distributive Property of Multiplication Over Addition,? that we learn in the first year of Algebra? It tells us that you get the same answer if you, (1) Add a set of numbers, then divide the set by another number, or (2) Divide each of the numbers first, then add them. Your formula does nothing more than add three numbers, then divide the result by 3. That is an average value. It means absolutely nothing, no matter what your lab setup is telling you.

Please go back to post #2, and teach that formula, and no other. With all due respect, that formula is right, and yours is not.

Also, I am a big fan of Acums(sp) razor, ?The simplest solution is usually correct.?

Just FYI, it is spelled, ?Occam?s Razor,? and the text is as follows: "plurality should not be posited without necessity." That is not the same as saying ?the simplest solution is usually correct.? What it means is that if there is no good reason to offer more complex solutions to a problem, when you already have a simple one that seems to fit, then it would be best to go with the simple one. What was missed in the movie ?Contact,? if by chance that was where you first heard of Occam?s Razor (it was my first time hearing that phrase), is that there may well be a very good reason to offer more complex answers!

 

[charlie b]

The sine of 120? is .866. If they multiply 554 by .866 the product is a close approximation of 480V.

True, but not likely to be of much value to an apprentice. Why, for example, did you use sine, and not cosine or tangent or hyperbolic cosecant? What is it that makes ?sine? relevant? You need trigonometry to answer that question, and I suspect that apprentice electricians are not likely to bring a love of trig to the job site.

 

[Gary Shumaker]

May I humbly offer in return that that does not give you freedom to insert that number anywhere you please? It belongs where it belongs, and nowhere else.

May I also ask if you understood what I said earlier about your formula just giving an average value? May I point out that if you divide a number by 1.732, and then divide it again by 1.732, what you end up with is the original number divided by 3? May I also remind you of the ?Distributive Property of Multiplication Over Addition,? that we learn in the first year of Algebra? It tells us that you get the same answer if you, (1) Add a set of numbers, then divide the set by another number, or (2) Divide each of the numbers first, then add them. Your formula does nothing more than add three numbers, then divide the result by 3. That is an average value. It means absolutely nothing, no matter what your lab setup is telling you.

Please go back to post #2, and teach that formula, and no other. With all due respect, that formula is right, and yours is not.


Just FYI, it is spelled, ?Occam?s Razor,? and the text is as follows: "plurality should not be posited without necessity." That is not the same as saying ?the simplest solution is usually correct.? What it means is that if there is no good reason to offer more complex solutions to a problem, when you already have a simple one that seems to fit, then it would be best to go with the simple one. What was missed in the movie ?Contact,? if by chance that was where you first heard of Occam?s Razor (it was my first time hearing that phrase), is that there may well be a very good reason to offer more complex answers!

I agree completely. Perhaps I was trying to oversimplify the quote. I read the quote in one of Hawkings books and his expression was exactly as you stated it. I further agree that the simplest solution is not always the correct solution. However, I am attempting to find the simplest way to present an easy way to present the information to my students. Also, I will do as you suggest and revisit the #2 posting and re-run my testing in the lab. I will let you know how it turns out.
Regards,
Gary

 

[crossman]

May I also remind you of the ?Distributive Property of Multiplication Over Addition,? that we learn in the first year of Algebra? It tells us that you get the same answer if you, (1) Add a set of numbers, then divide the set by another number, or (2) Divide each of the numbers first, then add them. Your formula does nothing more than add three numbers, then divide the result by 3. That is an average value. It means absolutely nothing, no matter what your lab setup is telling you.

To say nothing of the actual algebra!

If we do ((A-B) + (B-C) + (C-A))/3 then that will always equal zero regardless of the phase amps.

If we do ((A-B) + (A-C) + (B-C))/3 we get (2A - 2C)/3 which has no meaning of value to the problem.


[quote = charlie b]Please go back to post #2, and teach that formula, and no other. With all due respect, that formula is right, and yours is not.[/quote]

I do not think it hurts to show what is happening with the phasors. Teaching a formula without at least showing something of where it comes from just doesn't feel legitimate to me.

 

[rattus]

Gary,

With all due respect, unless you can provide the derivation for your formula, it is baseless. Furthermore, there must be something wrong with your lab setup because the results disagree with no less than five members of this forum.

Now, there are at least four valid approaches to this problem:

1. The most difficult is the application of Kirchoff's current law and the use of phasor algebra to compute In, but this works for any set of currents and provides the phase angle as well. The other approaches assume a separation of 120 degrees.

2. The next is the algebraic formula posted by Roger which provides only the magnitude. This formula is derived from the first approach.

3. Then there is Crossman's method which can be easily explained from the phasor diagram of the currents.

4. Simplest of all is observation. We observe that if two phase currents are equal, the sum of the 3rd phase current and the neutral current is equal to the magnitude of either of the other two.

I suggest that you trace out your schematic and post it here. In the process you may find a surprise.

 

[Gary Shumaker]

Post #2 works

Post #2 works

Gentleman,
I am pleased to tell you that after re-calibrating my lab equipment and re-running the tests, the formula in post #2 is so far the most accurate and is an easier formula to teach my students.

Thank you for all of your help.
Gary

 

[crossman]

1. The most difficult is the application of Kirchoff's current law and the use of phasor algebra to compute In, but this works for any set of currents and provides the phase angle as well. The other approaches assume a separation of 120 degrees.

Hey Rattus, let me give this a shot just for fun!

Given: 208Y/120 system, Ia = 20amps, Ib = 20amps, Ic = 10amps
Find: Neutral current

solution:
I will choose the polar coordinates as follows, this is somewhat arbitrary as long as the 120 degree angle exists between phases.

Ia = 20amps @ 150 degrees
Ib = 20amps @ 30 degrees
Ic = 10amps @ -90 degrees

Convert to the complex plane

Ia = -17.32 + j10
Ib = 17.32 + j10
Ic = 0 + j(-10)

Find resultant by adding the real of each phasor and then add the imaginary of each phasor:

Resultant I = 0 + j10

This resultant is the net current leaving the neutral point due to the 3 phases. Kirchoff's law states that at any given point in a circuit, the current leaving must equal the current arriving. Therefore, the neutral current is the opposite of the resultant which will be:

In = 0 + j(-10)

Convert back to polar coordinates:

In = 10amps at -90 degrees

How about that? Any flaws in my computations?

 

[crossman]

Gentleman,
I am pleased to tell you that after re-calibrating my lab equipment and re-running the tests, the formula in post #2 is so far the most accurate and is an easier formula to teach my students. Thank you for all of your help.

Gary, I am glad you got that figured out. This has been a good thread!

 

[rattus]

Looks OK to me:

Looks OK to me:

Hey Rattus, let me give this a shot just for fun!

Given: 208Y/120 system, Ia = 20amps, Ib = 20amps, Ic = 10amps
Find: Neutral current

solution:
I will choose the polar coordinates as follows, this is somewhat arbitrary as long as the 120 degree angle exists between phases.

Ia = 20amps @ 150 degrees
Ib = 20amps @ 30 degrees
Ic = 10amps @ -90 degrees

Convert to the complex plane

Ia = -17.32 + j10
Ib = 17.32 + j10
Ic = 0 + j(-10)

Find resultant by adding the real of each phasor and then add the imaginary of each phasor:

Resultant I = 0 + j10

This resultant is the net current leaving the neutral point due to the 3 phases. Kirchoff's law states that at any given point in a circuit, the current leaving must equal the current arriving. Therefore, the neutral current is the opposite of the resultant which will be:

In = 0 + j(-10)

Convert back to polar coordinates:

In = 10amps at -90 degrees

How about that? Any flaws in my computations?

I think you've got it! Now it would be a bit easier to pull a balanced, 10A wye from the problem leaving you with 10A @ 30 and 10A @ 150, the sum of which is 10A @ 90. Therefore, In = 10A @ -90.

Now you can go and complete my derivation of the algebraic formula.

And, remember those arrowheads; they have meaning.

 

[rattus]

Yes, but:

Yes, but:

Gentleman,
I am pleased to tell you that after re-calibrating my lab equipment and re-running the tests, the formula in post #2 is so far the most accurate and is an easier formula to teach my students.

Thank you for all of your help.
Gary

Agreed, the algebraic formula should be taught because it is easily applied and can be utilized in a spreadsheet or programmable computer. However it is no more accurate than the other approaches. It should be adequate for most situations, but if the PFs are different among the phase currents, only the phasor/complex math approach will provide exact results. I would suggest that this point be emphasized.

 

[jim dungar]

Adequate and exact who cares?

In the very vast majority of electrical calculations we do, there is so much rounding and estimating that adequate is almost always sufficient.

For example, an exact 208Y/120V system does not exist. It is either a 207.85Y/120V or it is a 208Y/120.09V (and even these number have been rounded)

Now that I have further diminished all hope of world (or at least forum) peace.:wink:

Merry Christmas.

 

[Smart $]

To say nothing of the actual algebra!

If we do ((A-B) + (B-C) + (C-A))/3 then that will always equal zero regardless of the phase amps.

If we do ((A-B) + (A-C) + (B-C))/3 we get (2A - 2C)/3 which has no meaning of value to the problem.

I hate to say it, especially since the method is invalid... but neither combined formula is what Gary actually wrote, by my understanding...

(delta phase A - phase B) divided by 1.732 = I1
(delta phase A - phase C) divided by 1.732 = I2
(delta phase B - phase C) divided by 1.732 = I3

(I1+I2+I3) divided by 1.73 = neutral current

Assuming Gary meant delta as the difference in line currents, it would be:

((A?B)?√3 + (A?C)?√3 + (B?C)?√3) ? √3 = 2A?2C = In​

However, if by chance he meant delta as the absolute difference, what he wrote would reduce to:

|A?B|+|A?C|+|B?C| = In​

In neither case would the result be an average, or correct!

 

[rattus]

Adequate and exact who cares?

In the very vast majority of electrical calculations we do, there is so much rounding and estimating that adequate is almost always sufficient.

For example, an exact 208Y/120V system does not exist. It is either a 207.85Y/120V or it is a 208Y/120.09V (and even these number have been rounded)

Now that I have further diminished all hope of world (or at least forum) peace.:wink:

Merry Christmas.

Who cares? One should know the difference--especially P.E.s and instructors!

 

[Smart $]

...can be utilized in a spreadsheet...

Speaking of spreadsheets, there's an Excel neutral current calculator embedded in the attached .doc. To extract the spreadsheet, just edit or open it within Word, then use Excel's Save function to make it a native .xls(x) file.

The calculator accounts for power factor of the load (assuming a lagging pf).

View attachment 1081

EDIT: Attachment replaced with newer correct version. Old version reported a neutral angle result which was 180? out of phase (i.e. going in the opposite direction, phasor-wise :grin: )

 

[jim dungar]

Who cares? One should know the difference--especially P.E.s and instructors!

Knowing the difference was not my point. How many decimal points do you use when performing electrical power circuit calculations? How many significant digits? At which point do you perform your rounding to get the resulting voltages of 208Y/120.

I am not trying to pick any type of fight. My point was that the circuits that most electricians deal with can be handled with "adequate" equations and simplifications. When dealing with circuit ampacity to determine conductor size I am not often concerned with "1s" amps, I am usually satisfied with "5s" or even "10s" instead.

 

[crossman]

I hate to say it, especially since the method is invalid... but neither combined formula is what Gary actually wrote, by my understanding...


Assuming Gary meant delta as the difference in line currents, it would be:

((A?B)?√3 + (A?C)?√3 + (B?C)?√3) ? √3 = 2A?2C = I

I am thinking it would be (2A -2C)/3 because dividing the individual parts by √3 and then dividing by √3 again is actually dividing by 3. Check post #45 by charlie b.

However, if by chance he meant delta as the absolute difference, what he wrote would reduce to:

|A?B|+|A?C|+|B?C| = In

I'm thinking it would be (|A?B|+|A?C|+|B?C|)/3 for the same reasons.