Unbalance neutral current in a 3 phase system

[rattus]

We have some that are vehemetly opposed to symetrical comments, vectors, and rotating magnetic fields. I agree symetrical components will handle a general case better than most other methods. With the advent of calculators that will handle matrices(sp) and complex math, the method is much easier to apply.

carl

Now Carl,

It is not a matter of being opposed to symmetrical components. It is a matter of keeping the solution simple. My understanding is that this method is more or less required for problems involving rotating machinery, however it is too cumbersome and laborious for static problems. [Paraphrased from Kercnher & Corcoran, "Alternating Current Circuits", Wiley, 1951.]

Nor is it a matter of being opposed to phasors/vectors which will provide a solution just as accurate and is far easier to understand.

Neither is it a matter of being opposed to rotating magnetic fields because no one wants the motors and generators to stop working. Furthermore, I can see no reason to mention this subject in the calculation of neutral currents.

In practice, the algebraic method suffices for most situations, and it only requires a little bit of arithmetic and a trust in the formula.

In summary, we should fit the tool to the job, or "Keep It Simple!"

 

[Gary Shumaker]

Final solution

Final solution

Hello again guys,
Sorry it's been so long getting back to you. I took the formula from post #2 and had an entire class prove or disprove it in our lab. In every variation of values using the formula proved out and it is a simple formula to understand.

(AxA)+(BxB)+(CxC)-(AxB)-(AxC)-(BxC) = neutral current

 

[Smart $]

Hello again guys,
Sorry it's been so long getting back to you. I took the formula from post #2 and had an entire class prove or disprove it in our lab. In every variation of values using the formula proved out and it is a simple formula to understand.

(AxA)+(BxB)+(CxC)-(AxB)-(AxC)-(BxC) = neutral current

Umm... you forgot the square root/square relationship in your formula, i.e. square root left side or squared right side.

 

[jghrist]

Hello again guys,
Sorry it's been so long getting back to you. I took the formula from post #2 and had an entire class prove or disprove it in our lab. In every variation of values using the formula proved out and it is a simple formula to understand.

(AxA)+(BxB)+(CxC)-(AxB)-(AxC)-(BxC) = neutral current

Did you try any current values where the angle between phases was different than 120 degrees?

 

[mskaltz]

I have been told that to calculate the amount of current on the neutral of a three phase system, you simply take the square root of the highest phase current. I have an extensive engineering backgrond and I know that this is not possible. For example, if phase A = 49A and phase B = 47A and phase C = 46A, there is no way to convince me that there is 7A on the neutral. Likewise, if phase A= 49A and phase B = 3A and phase C = 5A, again I am not convinced that there will still be 7A on the neutral. I suspect it is more simple than that and involves the application of the square root of 3 (1.732) based on the difference between phases.

What is the formula to calculate the neutral current on unballanced 3 phase circuits?

Thank you for your time
Gary Shumaker
Instructor ABC So. Cal.

Resultant Vector is as following. Assume A to be 120, B to be 240, and C to be 0 degrees.

A = 49cos(120) + 49sin(120)
B = 47cos(240) + 47sin(240)
C = 46cos(0) + 46sin(0)

A = 49(-.5) + 49(.866)
B = 47(-.5) + 47(-.866)
C = 46(1) + 46(0)

A = -i24.5 + j42.435
B = -i23.5 - j40.7
C = i46 +j0
-----------------------------
= -i2 +j1.735

Take the square root of the sum of squares for magnitude of 2.65.

Take the arctan of j/i resultant magnitudes to find angle of -40.94 degrees.

However, note that resultant has negative i value and positive j value. This only occurs in the second quadrant, so add 90 to the absolute value of degrees to sum at 130.94 degrees.

Resultant Neutral Current is therefore 2.65 amps @ 130.94 degrees.

 

[rattus]

Nit picking:

Nit picking:

Aside from the error mentioned by Smart, there is the matter of symbology. I would use "I" as the symbol for current and use subscripts to identify the phase currents. Also, I would use parentheses only where needed. Then,

In = sqrt[Ia^2 + Ib^2 + Ic^2 -IaIb -IaIc -IbIc]

You should also emphasize that this formula is exact only for currents separated by 120 degrees.

 

[rattus]

Comments:

Comments:

Resultant Vector is as following. Assume A to be 120, B to be 240, and C to be 0 degrees.

A = 49cos(120) + 49sin(120)
B = 47cos(240) + 47sin(240)
C = 46cos(0) + 46sin(0)

A = 49(-.5) + 49(.866)
B = 47(-.5) + 47(-.866)
C = 46(1) + 46(0)

A = -i24.5 + j42.435
B = -i23.5 - j40.7
C = i46 +j0
-----------------------------
= -i2 +j1.735

Take the square root of the sum of squares for magnitude of 2.65.

Take the arctan of j/i resultant magnitudes to find angle of -40.94 degrees.

However, note that resultant has negative i value and positive j value. This only occurs in the second quadrant, so add 90 to the absolute value of degrees to sum at 130.94 degrees.

Resultant Neutral Current is therefore 2.65 amps @ 130.94 degrees.

Katz, Although you are on the right track, your first six equations are lacking the "j" operator, then both "i" and "j" appear out of nowhere. This is not mathematically rigorous. And, strictly speaking, "In" should be defined in the same "sense" as are the phase currents. That is, In is the NEGATIVE sum of the phase currents. Then,

In = 2 -j1.735

With the aid of a scientific calculator, or even a slide rule,

In = 2.65A @ -40.9 degrees

No need for Arctan or Pythagoras or quadrant correction.

One more thing,

Since the tangent of an angle is negative in quardants II and IV and is positive in quadrants I and III, one must use 180 degrees to select the correct answer--not 90.

 

[coulter]

...This is not mathematically rigorous. And, strictly speaking, "In" should be defined in the same "sense" as are the phase currents. That is, In is the NEGATIVE sum of the phase currents. ...

rattus -

You're being rather pedantic here. Interesting you insist on rigorous.

carl

To the rest of the crowd: Sorry, I just couldn't help my self:grin: :roll: :grin: :roll: (giggle, snork)

 

[rattus]

Worse than that:

Worse than that:

Carl,

It is more than lack of rigor. The solution is wrong! Now what is funny about that? Perhaps you can convince us that it is right? Or, perhaps you can add something of value to the discussion?

Tell us how you would solve this problem. I for one am waiting.

 

[coulter]

... Or, perhaps you can add something of value to the discussion?...

Not really. crossman pretty well nailed it with post 3 (for 120 deg angles). Smart $ gave a beautiful graphic rendition in post 13. I also agree with Jim D in post 75. The rest is just algebra and a little trig.

... The solution is wrong! ... Perhaps you can convince us that it is right?

I didn't see that it was all wrong. No, I wouldn't try to convince you it is entirely correct. I thought mskaltz did a pretty good job. One had to add the appropriate "j"s in the first six equations. And then shift to physics cartesians, i, j, k, for the last four. He messed up the angle in the last statement a bit, but I didn't have any trouble translating.

... Tell us how you would solve this problem. ...

Hummm, probably similar to Smart $, post 13.
1. Set calculator to complex numbers, polar coordinates.
2. Enter the three magnitudes with cooresponding angles. (assume 120 unless specified)
3. Press the "+" button twice
4. Multiply by -1

... Now what is funny about that? ...

Well, perhaps not "funny", but certainly "ammusing" to hear your switch to required rigor.

... I for one am waiting.

Really?? (I think not):roll:

carl

 

[Dennis Alwon]

After reading this entire thread I realized I must speak a different language then you all do. Egads was I lost. I am a simpleton so I took Rogers formula and stuck it in an excel spreadsheet.

=SQRT((A1^2)+(B1^2)+(C1^2)-(A1*B1)-(B1*C1)-(A1*C1))

In Cell A1, B1, & C1 I just enter the value of the phase current.

Put the formula in another cell and voila-- there is the neutral current. Life is so much easier now.

Thanks Roger

 

[rattus]

I don't get it:

I don't get it:

Carl,

I still don't understand your amusement. Tell me where I have been less than rigorous. Perhaps I could learn something new.

Now if this were a homework problem, the student would at best get partial credit for it because,

Six of the equations are wrong because they lack the operator "j".

"i" appears in other equations for no reason.

The sense of the currents is not defined.

The phase angle, which falls in quadrant II, is wrong.

If we do bother to use complex numbers, we should do it correctly and pay close attention to all the details--especially to the phase angle because we can get the magnitude from the algebraic formula with less effort. In my student days, I didn't understand why the profs were so particular. Now I do.

BTW, Smart's pretty diagrams are in error because he has drawn the phasors at arbitrary angles. Jim's comments are true, but have nothing do with this little exercise.

PS: Crossman's approach requires the use of complex numbers to verify its validity. Likewise, the algebraic formula posted by Roger requires complex algebra for its derivation.

 

[Gary Shumaker]

Oops. Your right. My most previous post didn't show the square root component. So just to revise:
The square root of (AxA)+(BxB)+(CxC)-(AxB)-(AxC)-(BxC)

Gary

 

[Gary Shumaker]

No. What would be the point in that? Utility 3 phase power is always 120 degrees out of phase.
Gary

 

[rattus]

Yes and no:

Yes and no:

No. What would be the point in that? Utility 3 phase power is always 120 degrees out of phase.
Gary

Can't tell which post you are referring to; please quote.

The voltages are separated by 120 degrees, but the separation between currents can deviate considerably.

You also need more parentheses in your expression to make it unambiguous. See Dennis's post. Roger's posted formula is also ambiguous. Humans may be able to interpret it, but computers cannot.

 

[Gary Shumaker]

Comment

Comment

Ooh. A math fight.
Gary

 

[rattus]

Not really!

Not really!

Ooh. A math fight.
Gary

An open neutral is wrong too. Wrong is wrong, whatever you may be doing.

 

[weressl]

I have been told that to calculate the amount of current on the neutral of a three phase system, you simply take the square root of the highest phase current. I have an extensive engineering backgrond and I know that this is not possible. For example, if phase A = 49A and phase B = 47A and phase C = 46A, there is no way to convince me that there is 7A on the neutral. Likewise, if phase A= 49A and phase B = 3A and phase C = 5A, again I am not convinced that there will still be 7A on the neutral. I suspect it is more simple than that and involves the application of the square root of 3 (1.732) based on the difference between phases.

What is the formula to calculate the neutral current on unballanced 3 phase circuits
Thank you for your time
Gary Shumaker
Instructor ABC So. Cal.

Could somebody untagle why the vector solution gives 1.79333A

 

[Smart $]

Could somebody untagle why the vector solution gives 1.79333A

Ummm... if you would be so kind as to give an expanation of the vectors in your diagram. As far as I can tell, they do not correlate to anything in the post you quoted.

 

[rattus]

It doesn't:

It doesn't:

Could somebody untagle why the vector solution gives 1.79333A

The easy way to do this is to determine a set of phasors which would balance the currents. By inspection, these currents are:

3A @ 0
2A @ 120

In is simply the sum of these two phasors,

In = 2.65A @ 40.9

This is essentially the same technique that Crossman posted earlier.