Unbalanced load on neutral 3 phase?

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ritelec

ritelec

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Hi
How do you determine the unbalanced load on a neutral on a 3 phase system?

A= 10 amps
B= 13 amps
C= 16 amps

do the 10 amps across the 3 phase cancel out?
then do the 3 amps between b and c?

would neutral be 3 amps ? 6 amps? neither?


(sorry, maybe this should be in education or calculations)
Wondering, thanks
 
ritelec said:
Hi
How do you determine the unbalanced load on a neutral on a 3 phase system?

A= 10 amps
B= 13 amps
C= 16 amps

do the 10 amps across the 3 phase cancel out?
then do the 3 amps between b and c?

would neutral be 3 amps ? 6 amps? neither?


(sorry, maybe this should be in education or calculations)
Wondering, thanks
Click to expand...
Neither. Someone here knows it by heart. I’ll have copy it from my spreadsheet.
 
The current among all three phases and neutral needs to add up to zero, assuming a circuit with no ground faults. Not a conventional addition of simple numbers, because you also need to track the phase. You construct the sine wave functions of time, with a phase offset term, and add them up, but using vector representations and the math of adding vectors makes it more convenient. You can even draw it out to scale, and measure the resultant.

Represent phase A on the x-axis, such that the following defines the vector for the current on A:
A = Ia*<1, 0>

Place phase B, 120 degrees clockwise from phase A. This means:
B = Ib*<-1/2, -sqrt(3)/2>

Place phase C, 120 degrees clockwise from phase B. This means:
C = Ic*<-1/2, +sqrt(3)/2>

Where did 1/2 and sqrt(3)/2 come from? These come from the special right triangle of the 30-60-90 triangle, that phases B and C make with the Y-axis. The 1/2 coefficient gives you the projection on the negative X-axis, and the sqrt(3)/2 is the projection on the Y-axis, where phase B is positive, and phase C is negative. You can also get this from the cosine and sine respectively, of 120 degrees for phase C, and 240 degrees for phase B.

Because all current vectors have to add up to zero:
A + B + C + N = 0

Thus:
N = -(A + B + C)

Add up corresponding components of the vector to carry out the sum. Add up the first (i.e. x) components, to find the x-component of the resultant. Then add up the second (i.e. y) components to find the y-component of the resultant. Then negate it, to find the neutral current.

Result:
N = <+4.5A, +2.6A>

Use the Pythagorean theorem to find the magnitude of neutral:
In = sqrt(Nx^2 + Ny^2)

And we get:
In = 5.2A
 
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I was following you up to here. You seem to be stating something you're not showing. Translation: Huh? :unsure:
Carultch said:
Add up corresponding components of the vector to carry out the sum. Add up the first (i.e. x) components, to find the x-component of the resultant. Then add up the second (i.e. y) components to find the y-component of the resultant. Then negate it, to find the neutral current.

Result:
N = <+4.5A, +2.6A>

Use the Pythagorean theorem to find the magnitude of neutral:
In = sqrt(Nx^2 + Ny^2)

And we get:
In = 5.2A
Click to expand...
 
LarryFine said:
I was following you up to here. You seem to be stating something you're not showing. Translation: Huh? :unsure:
Click to expand...

Remember that the capital letters A, B, C, and N all indicate vectors, rather than simple numbers (scalars). This is a 2-D problem, so that means they each have two components. One in the x-direction, and one in the y-direction.

The following vectors that define direction of the larger vectors, are called unit vectors. Because their length is one unit. I used these, as starting points to define the directions of A, B, and C.
a = <1, 0>
b = <-1/2, -sqrt(3)/2>
c = <-1/2, +sqrt(3)/2>

Distribute the magnitudes (Ia, Ib, and Ic) throughout the terms of the unit vector, for vectors A, B, and C:
A = <Ia, 0>
B = <-1/2*Ib, -sqrt(3)/2 * Ib>
C = <-1/2*Ic, -sqrt(3)/2 * Ic>

The first component of each vector is the part of the vector projected on the x-axis. That is, the terms between "<" and ",".
Ax = Ia
Bx = -1/2*Ib
Cx = -1/2*Ic

Likewise, the second component of each vector, is the part projected on the y-axis.
Ay = Ia
By = -sqrt(3)/2*Ib
Cy = +sqrt(3)/2*Ic

Add up the x-terms to get zero, leaving Nx as an unknown. And repeat with the y-terms, with Ny as an unknown.
Ax + Bx + Cx + Nx = 0
Ay + By + Cy + Ny = 0

Solve for Nx and Ny, which are the components of the neutral current vector:
Nx = -(Ax + Bx + Cx)
Ny = -(Ay + By + Cy)

When we get our results, this is the neutral current vector:
N = <Nx, Ny>

To get the magnitude of a vector, you combine the components in the Pythagorean theorem. Because placing the individual component vectors head-to-tail as we do when we add vectors, forms the two legs of a right triangle. The resultant of adding them, follows the diagonal.
 
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Here's the example problem drawn to scale.

The first sketch shows the given phase currents as vectors, each starting at the origin.
The second sketch shows the process of adding up vectors, by placing them head-to-tail, and drawing from the first head to the final tail. The violet vector shows the sum of A+B+C.
The third sketch shows how we reverse the direction of that sum, to get the vector of the neutral current (N). Placing vectors head-to-tail forms a closed path that returns to our starting point, indicating that they all add up to zero.

1622603859630.png
 
ritelec said:
Hi
How do you determine the unbalanced load on a neutral on a 3 phase system?

A= 10 amps
B= 13 amps
C= 16 amps

do the 10 amps across the 3 phase cancel out? ...
Click to expand...
If all the currents are 120 degrees from each other, you can start out by subtracting the lowest current (in this case 10 amps) from A, B, and C. So you're left with A' = 0 amps, B' = 3 amps, and C' = 6 amps. Then you can add just the B' and C' vectors since A' is zero.

Alternatively, you can use the formula in this thread when the currents are 120 degrees from each other:

Calulating Neutral Current

I am looking for a formula to calculate the neutral current on an unbalanced three phase system. I would need a reference from a published textbook so it does not look like I just came up with the data. I can calculate the X and Y components from each current based upon the phase angle.
forums.mikeholt.com

But if the angles are different due to power factor, etc., then the corresponding current vectors would need to be summed as Carultch did. That method will be valid for all cases.
 
All of the above is awesome stuff and helps visualize it, but if your confused and just want to plug and chug numbers here's the the simplified or direct version which was posted above.

Neutral current = The square root of this A^2 + B^2 + C^2 - (AB)-(BC)-(AC)

so your example

A= 10 amps
B= 13 amps
C= 16 amps

100+169 + 256 - 130 - 208 - 160 = 27

SQRT of 27 = ~5.19 A
 
mikeames said:
All of the above is awesome stuff and helps visualize it, but if your confused and just want to plug and chug numbers here's the the simplified or direct version which was posted above.

Neutral current = The square root of this A^2 + B^2 + C^2 - (AB)-(BC)-(AC)

so your example

A= 10 amps
B= 13 amps
C= 16 amps

100+169 + 256 - 130 - 208 - 160 = 27

SQRT of 27 = ~5.19 A
Click to expand...
Thanks, I couldn't find it quickly.
 
N=√ (A²+b²+c²)-(A*B - B*C - A*C). The square root sign goes over the entire right side of the equation.

Now you have a hundred ways to look at it...lol
 
mikeames said:
All of the above is awesome stuff and helps visualize it, but if your confused and just want to plug and chug numbers here's the the simplified or direct version which was posted above.

Neutral current = The square root of this A^2 + B^2 + C^2 - (AB)-(BC)-(AC)

so your example

A= 10 amps
B= 13 amps
C= 16 amps

100+169 + 256 - 130 - 208 - 160 = 27

SQRT of 27 = ~5.19 A
Click to expand...
If you plug in say 10 amps on A, 10 amps on B and zero for C and run calculations you will get 10 amps for the neutral.

Some ask why two phases and neutral doesn't cancel like it does for systems with 180 degree phase angles. This doesn't really explain why but does show you the result is the same value for the neutral if the two phase currents are the same.
 
kwired said:
If you plug in say 10 amps on A, 10 amps on B and zero for C and run calculations you will get 10 amps for the neutral.

Some ask why two phases and neutral doesn't cancel like it does for systems with 180 degree phase angles. This doesn't really explain why but does show you the result is the same value for the neutral if the two phase currents are the same.
Click to expand...
We know that fully-loaded 3ph MWBC has zero neutral current.

For each amp one given phase is reduced, the neutral current rises by the same amount.

Thus, if A=20, B=20, and C=15, then N=5. If A=20, B=20, and C=0, then N=20.
 
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