# Voltage Drop to Determine Wire Size on a 480/277 Line-to-Neutral Lighting Run?

#### winnie

##### Senior Member
Other than that, I finally get it for an unbalanced linear system... but now thinking about how to apply that to my 2nd diagram seems like a nightmare, and even more so when I actually try to apply it to the plans I am looking at.
Then don't. Figure out a simplified balanced system that you can demonstrate will be worse than the unbalanced system. Then do the calculation for that simplified system know will give larger wire sizes than the perfect calculation. Use the larger wire size and just figure that the extra cost of copper is cheaper than the design time to figure out the exact wire sizes to use, and knowing that you will also be saving a smidge of electricity over the life of the project.

-Jon

#### wwhitney

##### Senior Member
So the X from the neutral–4v–was subtracted from the X of Phase A–16v–and the Y of the neutral– -v√3–was added to the Y of Phase A–0?
Basically, although I would say that it's subtraction in both cases, using the proper sign conventions.

now thinking about how to apply that to my 2nd diagram seems like a nightmare,
As you have specified a single conductor size throughout, It's not that bad if you assume the voltage waveforms all stay 120 degrees apart and the load currents are constant and in phase with the voltages, as I did.

Just draw your network out as a one line graph, with a node at the source, at each fixture (indicate phase letter), and everywhere the conductors branch. Label each edge of the graph with the length of the run between nodes. If you have a fixture at a branching point, I think it will be simpler to separate those two conditions with an edge of length 0.

Now starting at the leafs of the tree, label edge with the lines currents (or rather number of unit currents for simplicity), using vector math on the neutral, where 1/a + 1/b + 1/c = 0. Each fixtures add a unit current pair, and the supply side of each branching node is the sum of the currents from the other edges incident to the node. [After you do this, if the segments with multiple downstream fixtures don't have neutral current near 0, adjust the phase choices of the downstream fixtures appropriately, and repeat.]

Finally, starting at the supply, add another set of labels to each edge, which is the cumulative total of the segment currents times the segment length (using vector math for the neutral). Then at each leaf of the tree, determine the phase X with label A for which A/X + the vector neutral label N has largest vector magnitude. [If the neutral currents have been kept near 0, I expect that will just be determined by the phase labels.] Take the biggest of these for all leafs, and plug that number as I * d into the unidirectional voltage drop formula Cmil = K * I * d / Vd.

Cheers, Wayne

#### nmonaco

##### Member
Then at each leaf of the tree, determine the phase X with label A for which A/X + the vector neutral label N has largest vector magnitude.

So I was able to follow up until this point:

where the highlighted portion is the cumulative total of the segment currents times the segment length without the vector magnitude from the neutral.

I was also unsure of how to direct the neutral when there weren't balances e.g. Phase A was 1, Phase B was 2, and Phase C was 3.

#### wwhitney

##### Senior Member
Very impressive chart. I think it's fair to say that you're in the process of computing the answer here with more precision that the project may individually merit, so I hope part of your motivation is in self-education, as mine is. (To which end once you have a precise answer, it would be interesting to see how the answers from various much simpler approximations compare, to know what sort of approximation is appropriate in the future.)

OK, I'm assuming the supply is the left end and the upper unhighlighted tuples are the currents (where maybe you chose to go with amps instead of multiples of the fixture unit draw, and the fixture unit draw is now 0.71A, and you have at least one typo, namely the second highest entry on the page, the neutral current. I find the bookkeeping easier if I just apply the 0.71A factor at the very end and use integers for the current tuples).

Then for the highlighted tuples I was proposing that (a) you include the neutral current times distance in your tuples and (b) I think you missed the word "cumulative" in my post. So the leftmost highlighted tuple is correct (assuming correct arithmetic), but the next one should be labeled with the sum of your current label and the earlier label, etc.

With those adjustments, the quoted sentence of mine in your last post should make sense.

Cheers, Wayne

#### wwhitney

##### Senior Member
I find the bookkeeping easier if I just apply the 0.71A factor at the very end and use integers for the current tuples.
Unless you have different sized fixtures, which I may have missed.

So the leftmost highlighted tuple is correct (assuming correct arithmetic), but the next one should be labeled with the sum of your current label and the earlier label, etc.
Bad word choice, I meant "the highlighted label you currently have in the diagram" not the upper labels of current tuples.

Cheers, Wayne

#### ptonsparky

##### Senior Member
This must be an exercise for education. No one in the field would apply such precision.

#### nmonaco

##### Member
Aside from truly learning how to do this, finding a good approximation for 3-phase is my goal.
This must be an exercise for education. No one in the field would apply such precision.
This a a real life application. I now realize an estimation would obviously make more sense, but at this point, I am too stubborn to stop.

#### nmonaco

##### Member
Very impressive chart. I think it's fair to say that you're in the process of computing the answer here with more precision that the project may individually merit, so I hope part of your motivation is in self-education, as mine is. (To which end once you have a precise answer, it would be interesting to see how the answers from various much simpler approximations compare, to know what sort of approximation is appropriate in the future.)

OK, I'm assuming the supply is the left end and the upper unhighlighted tuples are the currents (where maybe you chose to go with amps instead of multiples of the fixture unit draw, and the fixture unit draw is now 0.71A, and you have at least one typo, namely the second highest entry on the page, the neutral current. I find the bookkeeping easier if I just apply the 0.71A factor at the very end and use integers for the current tuples).

Then for the highlighted tuples I was proposing that (a) you include the neutral current times distance in your tuples and (b) I think you missed the word "cumulative" in my post. So the leftmost highlighted tuple is correct (assuming correct arithmetic), but the next one should be labeled with the sum of your current label and the earlier label, etc.

With those adjustments, the quoted sentence of mine in your last post should make sense.

Cheers, Wayne
You were right that there are different sized fixtures. I have run the cumulative sizes now. The cumulative neutral current is also included in brackets.

So based off of this. I can be conservative and not apply vector magnitude but instead use the full amp-ft on the neutral. Adding that to the total cumulative on Phase B, my final maximum l * d would be 1632+8947, which would equal 10579.

Cmil > 12.9 * 10579 / (0.03*277)
Cmil > 16422

for which a #8 AWG would work.

Does this sounds right?

#### winnie

##### Senior Member
Cmil > 12.9 * 10579 / (0.03*277)
Cmil > 16422

for which a #8 AWG would work.

Does this sounds right?
I would use a k of 10.6, which gets you down to a #9 AWG

Seriously though, I concur that a #8 would work.

Back in post #8 I assumed the full load all the way at the end of the 4200 foot run, and got that #8 would give a 3.8% voltage drop. A quick and dirty calc in the same ballpark as your detailed calc is a good double check.

This thread has been educational for me, especially since I learned that I was using a beautiful method that was fundamentally flawed Thanks for sticking with the calculation.

-Jon

#### nmonaco

##### Member
Back in post #8 I assumed the full load all the way at the end of the 4200 foot run, and got that #8 would give a 3.8% voltage drop. A quick and dirty calc in the same ballpark as your detailed calc is a good double check.
I've got a spreadsheet that does incremental, distance-based calcs similar to the exercise I just went through, but it does it for single phase. It produced a Vd of 4.37% for #8, your quick and dirty calc. was an even better estimator.

#### wwhitney

##### Senior Member
Does this sounds right?
I didn't check your arithmetic, but I believe the method is sound. Your final cumulative I*d vector (A,B,C,N) is (6160,8947,7408,1632b-1396a-325c)? The a and c components on the neutral look large enough to me not to ignore, although probably not so large as to change the final wire size. I'm also not 100% sure that the error due to the assumption that the load currents are always in phase with the line voltage isn't of comparable magnitude to those components.

Anyway, if you don't want to ignore the a and c neutral I*d components and if they are accurate, once you have your prospective wire size of #8, the voltage drop formula gives you a constant of proportionality between your cumulative I*d figures and the voltage delta. I'll call that P, I guess P = K/Cmil.

So if we choose different planar phase coordinates from earlier, where now b = (1,0); c = (-0.5, 0.5 √3), and a = (-0.5, -0.5 √3), then your voltage at the end on B is (277-8947P,0), and your voltage on N is P(1632+0.5*325+0.5*1396, (325-1396)*0.5*√3). Then you can take the difference of those two vectors, compute the length as usual to determine the magnitude of the B-N voltage, and confirm that it exceeds 97% * 277 = 268.7V.

Cheers, Wayne

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#### nmonaco

##### Member
Running through this, P=12.9/16510 = 0.000781

B = (270,0)
N = (1.946,-0.724)

The difference is (286.054,0.724), which comes out to a magnitude of 268.055V, which is < 268.7V. Therefore, #8 AWG would not be sufficient. Good catch and thanks so much for teaching me all of this.

I can't stress enough how helpful all of this was

#### wwhitney

##### Senior Member
I'm not sure that the method is accurate enough, or that the 3% rule is something that should be enforced strictly enough, to say that 268.1V vs 268.7V is reason enough to upsize the wire everywhere from #8 to #6. After all, #8 is good enough to 2 significant figures, and "3%" is only one significant figure. I'd stick with #8.

On the other hand, if you want head room to add loads in the future, go with the bigger size.

Cheers, Wayne

#### winnie

##### Senior Member
You also might try re-arranging your loads; you have 6 'B' lamps, and 5 'A' lamps and 5'C' lamps. B is your most heavily loaded phase and has a fixture at the longest distance.

I'd suggest changing the second lamp from the left from a B to a C, so that the shortest run has the 5.08A load on it.

-Jon

#### wwhitney

##### Senior Member
I'd suggest changing the second lamp from the left from a B to a C, so that the shortest run has the 5.08A load on it.
I'd suggest changing the last fixture before the split from B to A, so that the neutral current in the segment before that is 0. That will lead to lower neutral currents, as well as shift load from the phase with the highest I*d cumulative totals to the phase with the lowest I*d cumulative totals.

Cheers, Wayne

#### mbrooke

##### Batteries Not Included
You were right that there are different sized fixtures. I have run the cumulative sizes now. The cumulative neutral current is also included in brackets.

View attachment 2556039

So based off of this. I can be conservative and not apply vector magnitude but instead use the full amp-ft on the neutral. Adding that to the total cumulative on Phase B, my final maximum l * d would be 1632+8947, which would equal 10579.

Cmil > 12.9 * 10579 / (0.03*277)
Cmil > 16422

for which a #8 AWG would work.

Does this sounds right?

#8 on what size breaker?

#### mbrooke

##### Batteries Not Included
I would use a k of 10.6, which gets you down to a #9 AWG

Seriously though, I concur that a #8 would work.

Back in post #8 I assumed the full load all the way at the end of the 4200 foot run, and got that #8 would give a 3.8% voltage drop. A quick and dirty calc in the same ballpark as your detailed calc is a good double check.

This thread has been educational for me, especially since I learned that I was using a beautiful method that was fundamentally flawed Thanks for sticking with the calculation.

-Jon

#8 will take upwards of 60 seconds to open a 15 amp breaker.

#2 cu or 1/0AL would be a starting point.

#### ptonsparky

##### Senior Member
#8 will take upwards of 60 seconds to open a 15 amp breaker.

#2 cu or 1/0AL would be a starting point.
Yes. We’ve been down this road before. VD should not be the only concern.

#### mbrooke

##### Batteries Not Included
Yes. We’ve been down this road before. VD should not be the only concern.
VD is a secondary concern, if even worth considering.

#### romex jockey

##### Senior Member
on the tail end of 110.24A >>>

The calculation shall be documented and made available to those authorized to design, install, inspect, maintain, or operate the system.

one would think mag trip time would factor in here?

~S~