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Rick Christopherson

Rick Christopherson

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I was bored yesterday and reading through some old postings when I came across one that was similar to a situation I was examining a couple of weeks ago. My version is just a little more complicated, and since it took a dozen or so replies to answer the first version, I thought I would pose this version.

The original question was if you had a 120/208 wye system (with phase to neutral loads) 10 amps on phase A-N and 10 amps on phase B-N, what is the current in the neutral?

So here is the version that I am looking at, and this does apply to real-world application. The system is based on a 5-wire cable with each set of 120 volt outlets feeding from separate phases...blah, blah, blah.
  • On the A-phase outlet, you have a frosty machine which is 20 amps at 120 volts, and an inductive powerfactor of 0.8.
  • On the B-phase outlet you have incandescent tent lighting which is 20 amps at 120 volts and a unity powerfactor.
  • The C-phase outlet is not being used.

  1. What is the current in the neutral?
  2. For extra credit, what is the simplest way to reduce the current in the neutral, and what will that current end up being? (NO! You can't just unplug the loads.)
 
Rick Christopherson said:
I was bored yesterday and reading through some old postings when I came across one that was similar to a situation I was examining a couple of weeks ago. My version is just a little more complicated, and since it took a dozen or so replies to answer the first version, I thought I would pose this version.

The original question was if you had a 120/208 wye system (with phase to neutral loads) 10 amps on phase A-N and 10 amps on phase B-N, what is the current in the neutral?

So here is the version that I am looking at, and this does apply to real-world application. The system is based on a 5-wire cable with each set of 120 volt outlets feeding from separate phases...blah, blah, blah.
  • On the A-phase outlet, you have a frosty machine which is 20 amps at 120 volts, and an inductive powerfactor of 0.8.
  • On the B-phase outlet you have incandescent tent lighting which is 20 amps at 120 volts and a unity powerfactor.
  • The C-phase outlet is not being used.

  1. What is the current in the neutral?
  2. For extra credit, what is the simplest way to reduce the current in the neutral, and what will that current end up being? (NO! You can't just unplug the loads.)
Click to expand...

I come up with 7.92A. I did this quickly though so I may have made some errors.
 
I got 29.93A for the Neutral. An easy way to reduce would be to split the lighting load into 2 equal 10 amp loads each for phase B and phase C leaving a new Neutral load of 13.42 amps (or 10 amps when the frosty machine is not running).
 
090218-0849 EST

Without looking up the definition of phase rotation direction and by the nature of the question I get 29.93 A for (1), and interchange the phases for (2) and current equals 8.04 A. Assuming no calculation errors.

.
 
Yup Gar your way is much easier (wish I was thinking this morning). Just plug the lighting load into phase C instead of B and the neutral load drops to 8.02A (or 20 amps when the frosty is not running).
 
The two current phasors I used were:

Phase A current = 20A @ -323deg

Phase B current = 20A @ -120deg

I added these two phasors to arrive at my neutral current of 7.92A @ -41deg.

What did I do wrong that I'm coming up with a different answer.
 
090218-114 EST

mull982:

I got a lagging angle of 36.86 deg. This + 120 = 156.86. !80-156.86 = 23.14.

23.14/2 = 11.57 . sin 11.57 = 0.2006 . 20*0.2006 = 4.011 and 2 times this is 8.022 . I must have rounded at a different point to get 8.04 before.

.
 
gar said:
090218-114 EST

mull982:

I got a lagging angle of 36.86 deg. This + 120 = 156.86. !80-156.86 = 23.14.

23.14/2 = 11.57 . sin 11.57 = 0.2006 . 20*0.2006 = 4.011 and 2 times this is 8.022 . I must have rounded at a different point to get 8.04 before.

.
Click to expand...

For the current on phase A I also got a lagging angle of 36.86 so I figured this lagging current would be 36.86deg behind the A-N voltage angle which would be at 0deg. Therefore subtracting 360deg - 36.86deg gives a phase A current at -323.14deg.

Since the B-N phasor is at 120deg and the current is a unity pf then I assumed that the phase B current would be at -120deg. This is how I arrived at the current vectors in my previous post. I then added these two current vectors together.

A = 20A @ -323.14deg = 15.9 + 12.03j
B = 20A @ -120deg = -10-17.32j
C = 0A

A+B+C=N

N = (15.9 + 12.03j) + (-10 -17.32j)
= 7.92 @ -41deg

Am I calculating this incorrectly by doing it this way?
 
Mul, a lagging-anything means it is happening after some other event in time. It is therefore occurring farther into the positive time frame. I believe you are shifting yours into the past.
 
Rick Christopherson said:
Mul, a lagging-anything means it is happening after some other event in time. It is therefore occurring farther into the positive time frame. I believe you are shifting yours into the past.
Click to expand...

Yup you are right that was my mistake. I was shifting the lagging current the wrong way. Silly mistake.

I redid my calculations using the same approach as before and arrived at the same answers as others.
 
My math is so rusty, that when it comes to stuff like this I need to see it visually and turn to my CAD system to cheat a little. However, I cannot for the life of me follow the math that you are using here. Obviously it works, but would you mind explaining some of it?
gar said:
I got a lagging angle of 36.86 deg. This + 120 = 156.86. (Why did you add the 120 to A instead of leaving it at 0 degrees? Never mind. I realize you were working on the 2nd part. !80-156.86 = 23.14. I am at a complete loss for this. Are you just finding the compliment to the angle to make the numbers easier?

23.14/2 = 11.57 . sin 11.57 = 0.2006 . 20*0.2006 = 4.011 and 2 times this is 8.022 . Why do you take half the angle, get the sine function, and then multiply 2 back in?
.
Click to expand...
I know there are a lot of different ways of doing something, but this seems so overly complex that I cannot even follow it (no offense intended). For me, looking at this problem, I simply took 20<36.86 + 20<120. Even though I let the computer do the math the first time, I will duplicate this in rectangular coordinates:

20<36.86 = 16.002 + 11.997j
20<120 = -10 + 17.321j

Adding these I get:
16.002 - 10 +11.997j + 17.321j = 6.002 + 29.318j = 29.926<78.43

attachment.php
 
Last edited:
gar said:
23.14/2 = 11.57 . sin 11.57 = 0.2006 . 20*0.2006 = 4.011 and 2 times this is 8.022.
Click to expand...
It's too late to edit my previous posting again. I apologize. It wasn't until after I had written everything up that I noticed you were working on the part of reversing the two phases, and then I had to re-think everything, by the time I checked everything, I couldn't edit the post. However, I am still at a loss about dividing by 2, and why this actually works. But I have confirmed it graphically that it does in fact work. What is the function or basis behind this? I assume it must be some shortcut in the math, but I can't think of what that shortcut would be.
 
090218-2146 EST

If I have two equal length lines of length L intersecting at one end and an angle between them of X, then I can draw this as two right triangles sharing a side Y in length.

Thus, I divide angle X by 2 to get the angle between the common side Y and the hypotenuse L. (sin X/2)*L equals the length of the triangle's side opposite the angle X/2.

Last 2 times this opposite side of the triangle equals the distance between the non-intersecting ends of the two lines.

.
 
As a follow-up to this, I am wondering what everyone thinks about the underlying situation that caused me to examine this. According to 400.5 A and B, there is no need to derate this 5-wire cable because, as defined, I do not exceed 3 current carrying conductors, yet it is clear that the current through the neutral wire would exceed the ampacity of a 20 amp protected system.

Is this an oversight in the NEC, or am I missing another aspect that would otherwise require derating?
 
Rick,

I'm not sure you are truly violating a 20 amp circuit with the neutral.

You stated you had a 20 amp load on A-N and a 20 amp load on B-N (I believe both are actually protected by 20 amp breakers or don't even bother reading on). Since the load on A-N is a motor load, I would think to allow for starting the actual draw of this motor is quite a bit less than 20 amps. Also, the actual lighting load could be at max 80% of a 20 amp circuit breaker. If you were to use the actual loads vs the circuit breaker sizing, I think you would find that the neutral would fall under 25 amps and be just fine. Although #12 wire can not be protected by a circuit breaker larger than 20 amps, you can still use the 25 amp rating of the wire in regards to any derating factors. This was a nice little exercise, but in your real life situation I'm not sure you are using the exact right loads to start with. IMHO
 
Of course, there are times when the neutral IS overloaded and should be oversized. (See FPN 210.4(A), 210.19(A)(1) Exception No. 1 (since the neutral is connected directy to a bus) and 310.15(B)(4)(c) along with the explanation in the handbook. All references from 2008 NEC.
 
Beautiful. to pull or not to pull seperate neutrals to motor loads came up at work last week, and I remember the practice from my early apprenticeship in residential, but in single phase it never made sense. You have a really long way to go on a 120/240 system before you have a current of any concern on the neutral when at unity it is 0, 50% PF (60 degrees) in fact would be the limit with both hots carrying the maximum load.

I was aware that 2 phases plus a neutral in a 3 phase system resulted in full load current on the neutral conductor, because you have to count it as a current carrying conductor for derating in conduit and cable assemblies, but I don't think it's even been demonstrated to me what happens in this situation when you play with power factor. The instant that power factor goes towards the other phase on the circuit, you have more than the desired current on the neutral, be it a lagging PF on the leading phase, or a leading PF on the lagging phase. It makes perfect sense now, thank you for posting this!

Last week it was mentioned that it was in the Canadian Electrical Code to run separate neutrals for motor loads, likely for this very reason, but I am still looking for the specific rule. It must be addressed somehow, perhaps there is a similar stipulation in the NEC, or else I would agree, a significant oversight.
 
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