I have already asked this question, but I never received an answer to it. I have a 3 phase, 4 wire, multi-wire lighting circuit. If I have 10 amps in lighting load on each hotwire, how much does the neutral carry? If I turn off all of the lights on one phase, how much does the neutral carry? I would like this question answered in two forms: A.) If it was all incandescent lighting, and, B.) If it was all fluorescent lighting.
multiwire circuits
- Thread starter tjarus
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landelectric
Member
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- Colorado
Re: multiwire circuits
You are going to get all kinds of theoretical and very scientific answers to this question.
From an Electricians point of view, for designing and servicing real world installations, the answer is simply that the Neut current will fluctuate between zero and the highest of any one leg +/- depending on the loads at any given point in time.
Generally speaking, the Neutral of a multiwire will carry a current equal to the load imbalance .
Any goblygook beyond the above is just useless input.
A 4w with only two legs loaded will work something like a 3W +/- give or take a bit.
Eric Land, Electrical Contractor
[ February 25, 2003, 06:02 PM: Message edited by: landelectric ]
You are going to get all kinds of theoretical and very scientific answers to this question.
From an Electricians point of view, for designing and servicing real world installations, the answer is simply that the Neut current will fluctuate between zero and the highest of any one leg +/- depending on the loads at any given point in time.
Generally speaking, the Neutral of a multiwire will carry a current equal to the load imbalance .
Any goblygook beyond the above is just useless input.
A 4w with only two legs loaded will work something like a 3W +/- give or take a bit.
Eric Land, Electrical Contractor
[ February 25, 2003, 06:02 PM: Message edited by: landelectric ]
Re: multiwire circuits
Tjarus,
With your example of equal loads of ten amps on each phase A,B, and C the neutral would read zero on incandescent (linear) and take away the possible harmonics on the flourescents (non linear) it would still be zero.
If you turned off any one phase the neutral would still read zero .
If you turned off two phases, you would read 10 amps on what now is no longer a neutral but a grounded conductor.
Now if your example used different loads on each phase you would have a whole new formula to consider
This formula would be;
The square of---I^2A + I^2B +I^2C minus IA x IB minus IB x IC minus IC xIA
Of course you may tell anybody that ask you to do this, weather it be an exam or real life, that a contractor from Colorado said it was goblygook and not neccassary. :roll:
Roger
Tjarus,
With your example of equal loads of ten amps on each phase A,B, and C the neutral would read zero on incandescent (linear) and take away the possible harmonics on the flourescents (non linear) it would still be zero.
If you turned off any one phase the neutral would still read zero .
If you turned off two phases, you would read 10 amps on what now is no longer a neutral but a grounded conductor.
Now if your example used different loads on each phase you would have a whole new formula to consider
This formula would be;
The square of---I^2A + I^2B +I^2C minus IA x IB minus IB x IC minus IC xIA
Of course you may tell anybody that ask you to do this, weather it be an exam or real life, that a contractor from Colorado said it was goblygook and not neccassary. :roll:
Roger
- Location
- Illinois
- Occupation
- retired electrician
Re: multiwire circuits
Roger,
If you have two hots of a 3 phase 4 wire system loaded at 10 amps the current on the grounded conductor will also be 10 amps. You have to apply the forumula that you gave at the end of your post.
Don
Roger,
If you have two hots of a 3 phase 4 wire system loaded at 10 amps the current on the grounded conductor will also be 10 amps. You have to apply the forumula that you gave at the end of your post.
Don
landelectric
Member
- Location
- Colorado
Re: multiwire circuits
here we go ..................
here we go ..................
kevin
Member
- Location
- Post Falls, ID
Re: multiwire circuits
The neutral current in
a 3-phase, 4-wire wye
system is the square
root of:
Example 3,
below
A (amps) squared 15 x 15 = 225
plus
B (amps) squared 20 x 20 = +400
plus
C (amps) squared 25 x 25 = +625
minus
A (amps) x B (amps) 15 x 20 = -300
minus
B (amps) x C (amps) 20 x 25 = -500
minus
C (amps) x A (amps). 25 x 25 = -625
Total = 75
Square
Root = 8.66
1. If the current in all three legs are equal, neutral current is zero.
2. If the current in one leg is zero and the current in the other two legs are equal, neutral current is the same as the current in the current-carrying conductors. The final sentence in Section 220.22 implies as much where it prohibits a reduction in the capacity of a grounded conductor of a 3-wire circuit consisting of two phase wires and the neutral of a 4-wire, 3-phase, wye-connected system.
3.If the phase currents are unequal, the neutral current is solved by the above formula. For example, if A=15A, B=20A and C=25A the neutral current is the square root of 75, or 8.66A.
The neutral current in
a 3-phase, 4-wire wye
system is the square
root of:
Example 3,
below
A (amps) squared 15 x 15 = 225
plus
B (amps) squared 20 x 20 = +400
plus
C (amps) squared 25 x 25 = +625
minus
A (amps) x B (amps) 15 x 20 = -300
minus
B (amps) x C (amps) 20 x 25 = -500
minus
C (amps) x A (amps). 25 x 25 = -625
Total = 75
Square
Root = 8.66
1. If the current in all three legs are equal, neutral current is zero.
2. If the current in one leg is zero and the current in the other two legs are equal, neutral current is the same as the current in the current-carrying conductors. The final sentence in Section 220.22 implies as much where it prohibits a reduction in the capacity of a grounded conductor of a 3-wire circuit consisting of two phase wires and the neutral of a 4-wire, 3-phase, wye-connected system.
3.If the phase currents are unequal, the neutral current is solved by the above formula. For example, if A=15A, B=20A and C=25A the neutral current is the square root of 75, or 8.66A.
landelectric
Member
- Location
- Colorado
Re: multiwire circuits
The room's awash in gobblygook ! I told you this was going to happen
The room's awash in gobblygook ! I told you this was going to happen
landelectric
Member
- Location
- Colorado
Re: multiwire circuits
having said that........ I think Roger had it right the first time
having said that........ I think Roger had it right the first time
david
Senior Member
- Location
- Pennsylvania
Re: multiwire circuits
?If the phase currents are unequal, the neutral current is solved by the above formula. For example, if A=15A, B=20A and C=25A the neutral current is the square root of 75, or 8.66A.?
Or
You can take the total unbalance and apply the effect of 3 phase.
A= 15 B= 15+5 (20) C= 15+10(25) the total unbalance being 15 amps
5 from B phase + 10 from C phase
the effect of 3 phase = 1.732050808 or 1/X of that is .577350269
Take the total unbalance X .577350269
15 x .577350269 = 8.66
?If the phase currents are unequal, the neutral current is solved by the above formula. For example, if A=15A, B=20A and C=25A the neutral current is the square root of 75, or 8.66A.?
Or
You can take the total unbalance and apply the effect of 3 phase.
A= 15 B= 15+5 (20) C= 15+10(25) the total unbalance being 15 amps
5 from B phase + 10 from C phase
the effect of 3 phase = 1.732050808 or 1/X of that is .577350269
Take the total unbalance X .577350269
15 x .577350269 = 8.66
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: multiwire circuits
First, draw a circle around the secondary of the transformer, completely blacken in the circle, and treat it as a single node. Kirchhoff’s Current Law says that the current entering the node must equal the current leaving the node. One may conclude that the neutral current must be equal in magnitude, through opposite in direction, to the sum of the three phase currents.
Secondly, you must treat all four currents as vectors, meaning that they have not only a magnitude, but also a direction (or if you prefer, a phase angle). Therefore, when you find “the sum of the three phase currents,” you must use vector addition, and not just add the peak values. If two currents are out of phase from each other by 90 degrees (forming a right triangle, you can find the hypotenuse using the Pythagorean Theorem: C*2 = A*2 + B*2. But unbalanced currents are not typically out of phase by 90 degrees, but by an angle closer to 120 degrees. So you go to the Law of Cosines: C*2 = A*2 + B*2 – 2 x A x B x cos(<AB). But you would have to do this twice, since you have to add Ia to Ib, and then add the results to Ic, before you can get In.
Finally, as my former math teachers would often say, “The rest is left to you as a homework assignment.”
Not without bring back to mind the trigonometric identities, vector addition, and matrix algebra that I haven’t seen in a couple of decades. But I think I can easily give the basic physical and mathematical concepts from which it is derived.
First, draw a circle around the secondary of the transformer, completely blacken in the circle, and treat it as a single node. Kirchhoff’s Current Law says that the current entering the node must equal the current leaving the node. One may conclude that the neutral current must be equal in magnitude, through opposite in direction, to the sum of the three phase currents.
Secondly, you must treat all four currents as vectors, meaning that they have not only a magnitude, but also a direction (or if you prefer, a phase angle). Therefore, when you find “the sum of the three phase currents,” you must use vector addition, and not just add the peak values. If two currents are out of phase from each other by 90 degrees (forming a right triangle, you can find the hypotenuse using the Pythagorean Theorem: C*2 = A*2 + B*2. But unbalanced currents are not typically out of phase by 90 degrees, but by an angle closer to 120 degrees. So you go to the Law of Cosines: C*2 = A*2 + B*2 – 2 x A x B x cos(<AB). But you would have to do this twice, since you have to add Ia to Ib, and then add the results to Ic, before you can get In.
Finally, as my former math teachers would often say, “The rest is left to you as a homework assignment.”
landelectric
Member
- Location
- Colorado
Re: multiwire circuits
My only point is that as in David' example, the neutral current can be quickly "thumnailed" by subtracting the current of the lowest drawing phase from the highest drawing leg to get
10 Amps+/- Neut current.
Just as a pilot does not need to know the chemical formulation of jet fuel,
an Electrician has little use for this type of formula when designing, Installing and,servicing systems.
Multi-wire lighting circuits are a known science, ..........not unlike the wheel.
Eric Land
[ February 26, 2003, 03:13 PM: Message edited by: landelectric ]
My only point is that as in David' example, the neutral current can be quickly "thumnailed" by subtracting the current of the lowest drawing phase from the highest drawing leg to get
10 Amps+/- Neut current.
Just as a pilot does not need to know the chemical formulation of jet fuel,
an Electrician has little use for this type of formula when designing, Installing and,servicing systems.
Multi-wire lighting circuits are a known science, ..........not unlike the wheel.
Eric Land
[ February 26, 2003, 03:13 PM: Message edited by: landelectric ]
landelectric
Member
- Location
- Colorado
Re: multiwire circuits
One of the most often overlooked benifits of multi-wire circuits
serving lighting ballasts is;
Audible Noise Cancellation
Especially on split-phase (1ph.) systems with the opposed ballasts in close proximity to each other(often 2 within the same fixture). Although it's effective on 3ph. systems, it's harder to get true noise cancellation unless fixtures are side by side and number at least three or a multiple of three.
The most noticable results are in residential kitchens, but office environments can benifit too.
Try it, you'll like it. (use a 2 pole switch for control)
Eric Land
[ February 26, 2003, 03:47 PM: Message edited by: landelectric ]
One of the most often overlooked benifits of multi-wire circuits
serving lighting ballasts is;
Audible Noise Cancellation
Especially on split-phase (1ph.) systems with the opposed ballasts in close proximity to each other(often 2 within the same fixture). Although it's effective on 3ph. systems, it's harder to get true noise cancellation unless fixtures are side by side and number at least three or a multiple of three.
The most noticable results are in residential kitchens, but office environments can benifit too.
Try it, you'll like it. (use a 2 pole switch for control)
Eric Land
[ February 26, 2003, 03:47 PM: Message edited by: landelectric ]
hurk27
Senior Member
- Location
- Portage, Indiana NEC: 2008
Re: multiwire circuits
A good point Eric as there were some car manufactures that used this same princable to make there cars quieter on the inside while the car still sounded like a hot rod on the outside. useing the audio system and a few mic's
A good point Eric as there were some car manufactures that used this same princable to make there cars quieter on the inside while the car still sounded like a hot rod on the outside. useing the audio system and a few mic's
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