Unbalance neutral current in a 3 phase system

[Gary Shumaker]

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.

 

[roger]

SQRT I?A + I?B + I?C - (IA x IB) - (IB x IC) - (IC x IA)

Roger

 

[crossman]

Assuming you are speaking of a Wye system, there is a big ol' formula for doing it, but another way is to do this:

1. Draw a vector diagram with each of the three currents 120 degrees out of phase, like you would with a wye diagram. The tails should all be connected, the arrow going out from the center.

2. Take phase with the smallest current and subtract it numerically from each phase.

3. This will leave only 2 vectors that are 120 degrees out of phase.

4. Solve for the resultant vector using the parallelogram method and trig. This will be the unbalanced current.

Alternatively, you can accurately draw the wye diagram with a ruler and protractor, and then follow steps 2, 3, and 4 but instead of doing the math, just measure the resultant with the ruler.

 

[brian john]

Roger VS Crossman, Roger VS Crossman, Roger VS Crossman....Roger's looks similar on the face of it.....

 

[nakulak]

Assume phase angles = 0,120,240

Resultant vector = <x,y> where
x component = A + cos120 * B + cos240 * C= A - .5B - .5C
y component = sin120 * B + sin 240 * C= √3 /2 *B - √3 /2 *C

so for example, for A =49, B=47, C=46, the resultant vector is
x=49 - .5*47 - .5*46 = 2.5
y= 1.73 * 47/2 - 1.73 * 46 /2 = .87

the resultant vector = <2.5, .87>
the magnitude of the resultant current = √(x?+y?) = 2.65A

 

[crossman]

To show what I am talking about, let's do an example. Notice, I don't even need a calculator or any formulas. Now, if you want an eXaCt answer, well, you could always convert the vectors into horizontal and vertical components and do the math that way, or, you could use the old standby triangle formula which can calculate the side of a triangle without a right angle, I believe it is SQRT (A? + B? - 2ABcos theta)

The point is, there are generally multiple ways to do things, and this way is more instructive of WHY it works than just some big formula. Also, mathematically, I feel quite smug that the "big ol' formula" is actually derived from the vectors as I present them.

 

[rattus]

To show what I am talking about, let's do an example. Notice, I don't even need a calculator or any formulas. Now, if you want an eXaCt answer, well, you could always convert the vectors into horizontal and vertical components and do the math that way, or, you could use the old standby triangle formula which can calculate the side of a triangle without a right angle, I believe it is SQRT (A? + B? - 2ABcos theta)

The point is, there are generally multiple ways to do things, and this way is more instructive of WHY it works than just some big formula. Also, mathematically, I feel quite smug that the "big ol' formula" is actually derived from the vectors as I present them.

Crossman,

I would argue the opposite. That is, your method is developed from the phasor math, that is trig and complex numbers. It works because 2sin(30) = 1.0

And, since the lines represent phasors, you should put arrow heads on them.

One more thing, you should point out that your result yields the current INTO the neutral point whereas convention dictates that In should carry the same "sense" as do the phase currents.

That is, In + Ia + Ib + Ic = 0

You realize of course that you made me figure this thing out on a Sunday night after the Cowboys lost!

 

[crossman]

Woohoo Rattus!:grin: I've been wondering where you have been. I am a Cowboys fan too for many many years. Lived in Houston my whole life, but the Oilers were really bad when I was growing up, therefore the Cowboys were my team. Oh, the Landry years... Still, the Cowboys will be okay, but they really did look bad tonight.

Concerning the arrowheads, my diagram noted that I omitted them for clarity.

Concerning the direction of the neutral vector, I agree it should be pointing in to the middle. The sums of all 3 phases and the neutral should be 0 as you indicate.

What is your take on the origin of the formula posted by Roger?

And you agree that my method will give a fairly accurate answer?

 

[rattus]

Woohoo Rattus!:grin: I've been wondering where you have been. I am a Cowboys fan too for many many years. Lived in Houston my whole life, but the Oilers were really bad when I was growing up, therefore the Cowboys were my team. Oh, the Landry years... Still, the Cowboys will be okay, but they really did look bad tonight.

Concerning the arrowheads, my diagram noted that I omitted them for clarity.

Concerning the direction of the neutral vector, I agree it should be pointing in to the middle. The sums of all 3 phases and the neutral should be 0 as you indicate.

What is your take on the origin of the formula posted by Roger?

And you agree that my method will give a fairly accurate answer?

Crossman,

I have derived the formula in question in my past life and believe it to be correct. I need to do it again.

Your approach is an exact method, therefore it is as accurate as you want to make it.

I do believe though that both these methods depend on the phase separations being exactly 120 degrees which is not always the case.

 

[crossman]

I have derived the formula in question in my past life and believe it to be correct. I need to do it again.

I am sure that formula is correct, but I am asking where it came from. When you derived it, did you start with the triangles that are created from the phasors (vectors)? Is the formula more a geometric thing or based on the sine waves? And, I'm not sure if this question even makes sense.

 

[rattus]

You asked for it!

You asked for it!

I am sure that formula is correct, but I am asking where it came from. When you derived it, did you start with the triangles that are created from the phasors (vectors)? Is the formula more a geometric thing or based on the sine waves? And, I'm not sure if this question even makes sense.

Good thing I am retired! It is now 12:25AM, and I just doped it out. Do this

Let the currents be,

Ia @ 30
Ib @ -90
Ic @ -210

Draw the phasor diagram, arrows and all.

Now write expressions for the real and imaginary currents,

Ire = (Ia - Ic)cos(30)

Iim = (Ia + Ic)sin(30) - Ib

Now square both these expressions, add them together, and take the square root--Pythagoras.

I'm not going to bother my brain doing algebra this late.

 

[roger]

Crossman, the answer is 2.65.

I?A + I?B + I?C

2401 + 2209 + 2116 = 6726

(IA x IB) - (IB x IC) - (IC x IA)

2303 + 2162 + 2254 = 6719

6726 - 6719 = 7

sqrt 7 = 2.65

Roger

 

[Smart $]

I am sure that formula is correct, but I am asking where it came from.

I personally do not know.

When you derived it, did you start with the triangles that are created from the phasors (vectors)? Is the formula more a geometric thing or based on the sine waves? And, I'm not sure if this question even makes sense.

Again, I don't know (being that I'm not rattus :grin: ).

What I do know is that when I derived it, it was a "geometric thing" and I started with vectors. I'll use your problem of A=20A, B=60A, and C=50A as an example. Note in the diagram below that there are 4 vectors, one representing each of the three line currents and one for the neutral current. Also note the addition of a small section, colored cyan, which makes up the balance of an equilateral triangle. I then use the Law of Cosines to determine the magnitude of the neutral current using provided data as the side lengths of the triangular cyan section...

 

[rattus]

Crossman, the answer is 2.65.

I?A + I?B + I?C

2401 + 2209 + 2116 = 6726

(IA x IB) - (IB x IC) - (IC x IA)

2303 + 2162 + 2254 = 6719

6726 - 6719 = 7

sqrt 7 = 2.65

Roger

Try that again Roger. In is more like 36A.

 

[crossman]

What I do know is that when I derived it, it was a "geometric thing" and I started with vectors. ..... I then use the Law of Cosines to determine the magnitude of the neutral current using provided data as the side lengths of the triangular cyan section...

Smart, that is beautiful! Thank you for the analysis. When I was laying in bed last night, running it through my head, it dawned on me that my method is a sub-formula of the overall formula. What I mean is this: When I subtract out the smallest vector leaving only 2 vectors, the resultant of those gives a triangle with 2 known sides and an angle in between of 60 degrees. Using the triangle formula:

Side C = √(A? + B? - 2ABcos theta)

Side C = √(A? + B? - 2ABcos60)

Cosine of 60 = .5 which takes vare of the 2 in the 2ABcos term

so

Side C = √(A? + B? - AB)

which is exactly the original formula with one of the phase currents as zero

Woohoo!

 

[crossman]

Rattus:

Something else popped into my head last night while I was thinking in bed. I understand what the imaginary number i is but never knew why it would be used. What is its purpose? Where is it used? These things made me wonder.

With the vectors, I am thinking it is a "direction holder" thing. A vector itself is never really negative, I mean, the magnitude of a vector is positive. But the direction can be negative. So when we take the square root of a negative vector, we have to keep the direction intact and that is where i comes in?

 

[roger]

Rattus, I'm using the numbers the OP gave, not the numbers in the illustrations.

Roger

 

[rattus]

Rattus, I'm using the numbers the OP gave, not the numbers in the illustrations.

Roger

Sorry about that Roger. I knew you knew better. I had completely forgotten the post that started all this.

 

[rattus]

Rattus:

Something else popped into my head last night while I was thinking in bed. I understand what the imaginary number i is but never knew why it would be used. What is its purpose? Where is it used? These things made me wonder.

With the vectors, I am thinking it is a "direction holder" thing. A vector itself is never really negative, I mean, the magnitude of a vector is positive. But the direction can be negative. So when we take the square root of a negative vector, we have to keep the direction intact and that is where i comes in?

Crossman, imaginary numbers provide the vertical components (when plotted) of complex numbers. However, we use "j" instead of "i" to avoid confusion with the symbol for current. For example, in your example,

Ia = 60cos(30) + j60sin(30) = 60A @ 30 (complex and polar forms)

Ia is a constant. That is, this is a fixed or static phasor as opposed to a rotating phasor which is a function of time. The independent variable of a rotating phasor must include the frequency and time (wt) which the expression above does not.

Phasors in complex form are easily added and subtracted just by combining the real and imaginary components. Phasors in polar form are easily multiplied and divided.

Complex numbers are a means of expressing the magnitude and direction of vectors, but in phasors, they express the magnitude and phase angle.

See the difference?

 

[crossman]

You got my head spinning like them phasors.

Just to make sure: When you say "complex" numbers, we are talking about one component being the square root of negative one?