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

[nmonaco]

I'm not so familiar with the 3 phase voltage drop formula, but I would think the calculation should be pretty close to the correct answer.

However it seems to me that formula is for a single 3 phase load, rather than a series of single phase loads. So if you want to use it, I would suggest approximating the first 3 fixtures as a 3 phase load at 1200' out, and the second 3 fixtures as a 3 phase load at 3000' out. Maybe even wiring those first six as ABC CBA so that the total distances balance.

The seventh fixture is the odd one out (mod 3) and so I would think using the 2 wire circuit voltage drop formula would be appropriate for it.

So I guess I'm proposing:

C_mil = K / V_d * ( √3 * 1200 * 1.2 +√3 * 3000 * 1.2 + 2 * 4200 * 1.2)

Cheers, Wayne

ok. I think I am starting to get this. So would that final light still be added to Phase A on the Panel Schedule?

For the actual design, the distances and amperages are not nearly as uniform and there are separate branches off of the main run of conduit. I wouldn't be able to group them together so easily in terms of distance and amperage. Would I add each light at their actual distances and divide the distance and amperage each by 3?

What would I do for a branch where there are 2 extra lights (like if I had another B at a distance of 600' past the farthest A on this run)? Would I run 3 wires to A and then 2 wires to B and the calculation be: Cmil = K / Vd * ( √3 * 1200 * 1.2 +√3 * 3000 * 1.2 + 2 * 4200 * 0.6 + 2 * 4800 * 1.2)?

Thanks

 

[winnie]

So for the 3-hot, 1-neutral set up at 480/277 line-to-neutral, I am using the formula:

Vd > Sqrt(3) * K * La1 * (Ia1+Ia2+Ia3) / Cmil + Sqrt(3) * K * La2 * (Ia2+Ia3) / Cmil + Sqrt(3) * K * La3 * Ia3 / Cmil

I see Wayne figured out what you are doing in the formula, assuming a 3 phase balanced load that mimics your A phase arrangement, which should be the one with the worst voltage drop.

Several comments:

You used a voltage drop limit based on 3% of 277V, but have sqrt(3) in your equation. So you are counting the change from 480 to 277 twice.

Your K of 12.9 is the 75C value, which makes sense for conductors being used near their ampacity. But your conductors will be used at very low current and will be essentially at ambient temperature. The K at 20C is about 10.6 .

In your design the phase with the longest run is also the most heavily loaded. I'd suggest rearranging things so the most heavily loaded phase is the one going the least distance.

All of the issues I pointed out above lead to error on the conservative side. So your design would function just fine and at worst would have more copper than strictly necessary.

Jon

 

[nmonaco]

You used a voltage drop limit based on 3% of 277V, but have sqrt(3) in your equation. So you are counting the change from 480 to 277 twice.

I assumed if the voltage being used is 277V, it would be used for voltage drop percentage, and I thought 3 phase meant I had to use Sqrt(3) for the equation. So should I should be using 14.4V as the drop (0.03*480)?

 

[Strathead]

Are these lights staggered, so the shared neutral is common until the very end, or do they go to point and "star" out as individual circuit runs? I ask because there is the accurate way of getting VD, which is applying the formula to each run between each light fixture with either 277 or 480 as appropriate, and there is the one that gets you there quickly and covers you without being excessive. If they are a run, that is say along a street and every third pole is the same circuit, I would measure the distance to the first pole, use 480 volts full load of one circuit (all even anyway) then I would figure 1/2 of the remaining distance to the last pole use 1/2 of the full load of any one circuit. Using pone of many apps to do the calculation you can play with voltage drops and wire sizes until you get the best combination. The above allows you to use three different wire sizes, and arrive a figures in probably less time than reading all of our comments.

 

[Strathead]

ok. I think I am starting to get this. So would that final light still be added to Phase A on the Panel Schedule?

For the actual design, the distances and amperages are not nearly as uniform and there are separate branches off of the main run of conduit. I wouldn't be able to group them together so easily in terms of distance and amperage. Would I add each light at their actual distances and divide the distance and amperage each by 3?

What would I do for a branch where there are 2 extra lights (like if I had another B at a distance of 600' past the farthest A on this run)? Would I run 3 wires to A and then 2 wires to B and the calculation be: Cmil = K / Vd * ( √3 * 1200 * 1.2 +√3 * 3000 * 1.2 + 2 * 4200 * 0.6 + 2 * 4800 * 1.2)?

Thanks

another way is to work backwards. Figure out an acceptable voltage drop of the longest run assuming you start at the first fixture (no home runs) Like 1.5% Then figure out the wire size required between all three coming together and the source. There are so many ways you can arrive at a igure that you just have to pick one.

 

[winnie]

I assumed if the voltage being used is 277V, it would be used for voltage drop percentage, and I thought 3 phase meant I had to use Sqrt(3) for the equation. So should I should be using 14.4V as the drop (0.03*480)?

The issue is that you are mixing a couple of things up in order to get a good enough approximate answer..

The actual voltage that your loads care about is the 277V, so down at the basic level what you care about is the voltage drop relative to that 277V. You have single phase 277V line to neutral loads but the neutral current is almost balanced, so you don't want to use the normal single phase equation which assumes full current flowing on the neutral.

The equation that you actually used is for 3 phase loads, not single phase loads, but you ran the equation as if you had a balanced 3 phase load. In other words, the way you ran the equation was as if you had _3_ lamps at 600 feet, _3_ lamps 1800 feet further on.... with each set of 3 being distributed amongst the phases. This is a great approximation if you make the assumption of 0 voltage drop on the neutral. But if you use the 3 phae equation you have to use the 3 phase voltage.

So _if_ you use the 3 phase equation as you did to _approximate_ the single phase result, you use the _3 phase_ voltage even though what you actually care about is the single phase voltage, thus 14.4V as your allowed drop.

-Jon

 

[nmonaco]

Are these lights staggered, so the shared neutral is common until the very end, or do they go to point and "star" out as individual circuit runs?

These are not nearly as uniform as my original diagram showed. They branch off at various points and it is not at all balanced. This was designed showing all 4 wires running to the end of each branch as well. I am trying to correct it.

Here is a better example of how the design I am trying to get the wire size for. Additionally, they want the wire sizes to match, so I wouldn't be using different wire sizes between the phases even if that would work.

 

[wwhitney]

Trying to understand the √3 factor in the 3 phase voltage drop formula, would like to know if this is correct:

For the 2-wire (L1, L2) voltage drop formula, if we look at the drop on just L1, we get the formula with the one way length and no factor of 2. Similarly for L2, and so the total drop has the factor of 2. If we introduce a notional neutral wire N, which would be carrying no current, it would have no voltage drop/rise. So relative to N, the voltage drop formula has no factor of 2.

Similarly for a 3-wire (L1, L2, L3) voltage drop formula, without the factor of √3, we get the voltage drop of L1, L2 or L3 relative to a notional neutral N. But now if we want to express that as an L-L voltage drop, we need a factor of √3 as usual.

So for the OP's situation, if we model the loads as a 3-phase load at 1200', a 3 phase load at 3000', and a single phase load at 4200' [which maybe is non-conservative, as I'm not clear that the neutral rise will average out to zero after 3 fixtures distributed among the phases], and if we are working with the L-N voltage of 277V with a target drop of 3% of 8.3V, the formula we want is:

Cmil = K / Vd * (1200 * 1.2 + 3000 * 1.2 + 2 * 4200 * 1.2)

The factor of two in the last term reflects the voltage rise along the neutral from the unbalanced neutral current.

I think.

Cheers, Wayne

 

[nmonaco]

so if I wanted to find the voltage drop for Phase B for my most recent image, how would I apply this equation: Cmil = K / Vd * ( √3 * 1200 * 1.2 +√3 * 3000 * 1.2 + 2 * 4200 * 1.2)?

How are unbalanced branches added in?

 

[wwhitney]

model the loads as a 3-phase load at 1200', a 3 phase load at 3000', and a single phase load at 4200' . . .

Cmil = K / Vd * (1200 * 1.2 + 3000 * 1.2 + 2 * 4200 * 1.2)

A better model is as a single phase load at 600', a 3-phase load at 1800', and a 3-phase load at 3600', which would make the parenthetical sum (2 * 600 * 1.2 + 1800 * 1.2 + 3600 * 1.2), noticeably less, but now likely non-conservative due to the non-zero neutral currents.

Cheers, Wayne

 

[nmonaco]

So to find Phase B for my new image, a PF of 0.9, 300W will be 1.2A, 150W will be 0.6A, and 250W will be 1.0A. I would then group them where possible to create areas to average as a 3-phase load?



So the new formula for Phase B would be:

Cmil > 12.9 / 8.31 * (1033 * 1.0 + 2733 * 1.2 + 2 * 2450 * 1.0 + 2 * 4500 * 1.2)
Cmil > 31067, so a #4 AWG

Did I add in the branches correctly? Would there be any need for the extra wires to run to the last lights?

 

[winnie]

I started writing up a procedure for doing the calculation when you have a random map of lamps connected on a circuit (rather than all in a line). Unfortunately figuring out how to properly size the neutral makes things much more complex. So please treat the below as a very rough draft to support someone else figuring out the detailed method

-Jon

another way is to work backwards. Figure out an acceptable voltage drop of the longest run assuming you start at the first fixture (no home runs) Like 1.5% Then figure out the wire size required between all three coming together and the source. There are so many ways you can arrive at a igure that you just have to pick one.

I am going to take this as the jumping off point for describing a much more involved, but much more accurate approach to this.

For the actual design, the distances and amperages are not nearly as uniform and there are separate branches off of the main run of conduit. I wouldn't be able to group them together so easily in terms of distance and amperage. Would I add each light at their actual distances and divide the distance and amperage each by 3?

In this approach you are using the _one way_ voltage drop calculation. When you use the 'one way' equation you need to first figure out the total wire distance (both hot and neutral added up) that feeds a particular load. Since you have shared neutral portions, you also need to figure out which parts of the neutral 'don't count' in the distance.

Step 1: determine your allowed voltage drop. This is 277V * 0.03 = 8.3V.

Step 2: draw a map of your layout, with segment distances shown and labeled with the load current and phase. Make this map large because you will be putting a bunch of notations on it.

Step 3: for each segment, figure out the net neutral current, and label it with the 'phase' that dominates. So a 'leg' to a lamp on the B phase has neutral current that is 'B' dominant; a segment feeding 3 lamps distributed on phases A,B,C has 0 neutral current, etc.

Step 4: calculate the 1 way 'hot' distance to each load separately, by starting at the source and going through each segment in turn to the load and adding that segment length.

Step 5: calculate the 1 way 'neutral' distance to each load separately, by starting at the source and going through each segment in turn to the load, and adding that segment length if its neutral current is dominated by the phase of the load. (So if you are going to a phase B load, but the neutral current is net on phase A, phase C, or 0, then you don't count the length of that segment. This is an approximation for 2 reasons: if the neutral current is dominated by a different phase then you actually get a voltage increase, and here I am not accounting for reduced voltage drop when only part of the total B load shows up as net neutral current.)

Step 6: add up the total 1 way distance for each load. Apply your '1 way' voltage drop equation: CMil = K/Vd * 1 way distance. This is the required CMil to properly serve each load

Step 7: For each segment, total up the CM required for each load on each phase.

Step 8: For each segment, size the neutral to match the largest phase conductor.

Now you have a table of the minimum CMil requirements for voltage drop. Go back and adjust for _code_ requirements, eg minimum conductor size for OCPD, no smaller than required EGC, etc.

 

[wwhitney]

Going back to the diagram in the OP (trying to understand that simple model first before looking at the complexity of the latest layout), say we label the segments as 1 through 7 starting at the far end. And let's denote current by multiples of the unit fixture draw, 1.2A. Then the current tuples (A,B,C,N) would approximately be:

1) (1,0,0,1/a)
2) (1,0,1,1/-b)
3) (1,1,1,0)
4) (2,1,1,1/a)
5) (2,1,2,1/-b)
6) (2,2,2,0)
7) (3,2,2,1/a)

where I'm assuming the line currents on A, B, and C are in phase with the line voltages A-N, B-N, and C-N, and I've labeled the neutral current with the phase information (1/a is 1A in phase with A-N).

Then if v is the voltage drop from 1.2A through 600 ft of our chosen conductor, we can just multiply each tuple by +/-v and sum to get the voltage change from the starting (277/a, 277/b, 277/c, 0). At the far end, I get the sum v(-12,-7,-9,3/a-2/b). 3/a-2/b = (3,0) - (-1,√3) = (4,-√3), which I will approximate as 4.

So I get a voltage drop on phase A at the end of 16v, where v = K*1.2*600/C_mil. Which makes the length/quantity contribution a factor of 9600. Interestingly, that's the average of the length/quantity contribution of 12,600 from my initial overly pessimistic 3-phase approximation, and the length/quantity contribution of 6,600 from my non-conservative 3-phase approximation.

Cheers, Wayne

 

[winnie]

Did I add in the branches correctly? Would there be any need for the extra wires to run to the last lights?

If you don't have something connected to a phase you don't need to run those phase wires. So for your 'branches' you only need the phase and neutral for the individual lamp.

Also while it is fun to figure out the exact minimum wire to run for each segment, there is something to be said for running the same thing everywhere (including the extra wires for 'unused' phases): much less to get screwed up in the field.

-Jon

 

[wwhitney]

Step 7: For each segment, total up the CM required for each load on each phase.

I think this step doesn't work, as voltage drop is not linear in CM, since it appears in the denominator of the formula. The linearity assumption we have been using is in the context of a single wire size (which fortunately the OP recently specified as a requirement).

For example, suppose we have 2 fixtures and 2 wires, each drawing 1A, spaced at 1 ft and 2ft from the source, and let's call c the CM required by 1A over 1 ft for the allowable voltage drop. Then the farther fixture has a total length of 4 ft and is drawing 1A, so your procedure says we need 4c CM. The nearer fixture has a total length of 2 ft and is drawing 1A, so your procedure says we need 2c CM. That gives us a sum of 6c CM for the first segment and 4c CM for the second segment.

Then the voltage drop at the end is 2/6 + 1/4 + 1/4 + 2/6 = 7/6 of the allowable voltage drop.

Unless I've misinterpreted something in your procedure.

Cheers, Wayne

 

[nmonaco]

Thanks for the very thorough write-up. I cannot express how grateful I am for this.

So I followed everything up to this point:

At the far end, I get the sum v(-12,-7,-9,3/a-2/b). 3/a-2/b = (3,0) - (-1,√3) = (4,-√3), which I will approximate as 4.

How did you get 3/a to be (3,0) and 2/b to be (-1,√3)?

(4,-√3), which I will approximate as 4

Is the approximation of 4 actually √(4^2 + (-√3)^2 = 4.35?

So I get a voltage drop on phase A at the end of 16v, where v = K*1.2*600/Cmil. Which makes the length/quantity contribution a factor of 9600.

I think I understand the 16v voltage drop for A (-12 - the additional 4). I do not know how you got the length/quantity contribution of 9600, and I am not sure what that actually is.

 

[wwhitney]

How did you get 3/a to be (3,0) and 2/b to be (-1,√3)?

So a and b are unit vectors 120 degrees apart. I temporarily chose coordinates where a is (1,0) and then if b is CCW from a, b is (cos(120deg), sin(120deg)) = (-0.5, 0.5√3), so 2/b = (-1, √3)

Is the approximation of 4 actually √(4^2 + (-√3)^2 = 4.35?

No for proper vector math you carry the √3 orthogonal component through to the end. I.e. the source voltage on A is (277,0), the voltage drop on A makes that (277-12v,0) (under the assumption the current is in phase with A), and the voltage rise on N is (4v, -v√3). So the actual voltage for the last fixture is (277 - 16v, -v√3). Now I'm approximating that the orthogonal component is much less than 277 - 16v and ignoring it. But you could instead take the sqrt of the sum of squares of that vector. [And of course as that vector is not actually in phase with A, the load current for a purely resistive load won't actually be in phase with A, which is a complexity we are ignoring throughout.]

I think I understand the 16v voltage drop for A (-12 - the additional 4). I do not know how you got the length/quantity contribution of 9600, and I am not sure what that actually is.

"length/quantity contribution" is just the name I made up for the sums of products we are doing in the voltage drop formula, after factoring out everything else (including the 1.2A, which I haven't always factored out). So 9600 = 16 * 600 ft (part of how I defined v). And if you look back at the earlier 3-phase models, you get 6600 and 12600.

Cheers, Wayne

 

[wwhitney]

So the actual voltage for the last fixture is (277 - 16v, -v√3). Now I'm approximating that the orthogonal component is much less than 277 - 16v and ignoring it.

I.e. if 16v = 3% * 277 = 8.3 as desired, then v= 0.52, and v√3 = 0.9. So we are looking at (268.7,0.9), a vector whose length is negligibly different from 268.7.

Cheers, Wayne

 

[winnie]

Then the voltage drop at the end is 2/6 + 1/4 + 1/4 + 2/6 = 7/6 of the allowable voltage drop.

Unless I've misinterpreted something in your procedure.

Well pooh. You are correct. I made a core error.

The first steps are correct, determining the CM required for each load.

CM is another way of saying 'conductivity', and conductivity adds in parallel.

So I made the leap that if load A requires conductivity X and load B requires conductivity Y, then any shared segment can have conductivity X+Y and be fine.

The error is that load A and load B are at different distances, so the required voltage drop along each segment is different, and thus you can't simply add the conductivity, because once the segments are combined they must have the same voltage drop.

Back to the drawing board.

-Jon

 

[nmonaco]

the voltage drop on A makes that (277-12v,0) (under the assumption the current is in phase with A), and the voltage rise on N is (4v, -v√3). So the actual voltage for the last fixture is (277 - 16v, -v√3).

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?


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.