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

[nmonaco]

So I have been reading a few of these forums trying to get an understanding for how to run the voltage drop calcs for an unbalanced lighting run on 480/277v 4-wire line-to-neutral runs. I just haven't been able to grasp the concept of what to do. I am having trouble determining calculate the voltage drop to determine the wire size.

So I have 3 lighting runs.

• Phase A has 3-300W fixtures spaced every 600 ft starting 600 ft from the meter
• Phase B has 2-300W fixtures spaced every 600 ft starting 1200 ft from the meter
• Phase C has 2-300W fixtures spaced every 600 ft starting 1800 ft from the meter

I know I need to determine the voltage drop at each light along each phase and add them together to get my total voltage drop, but I am not sure what calculation to use to get the voltage drop at each location. Also, how do I determine the size of the neutral?

Here is the image of the run to better visualize it:

And a PF of 0.9 was applied as well. Here is the Panel:

I appreciate any guidance you can give and any time you take to explain this to my very dumb brain.

 

[ptonsparky]

Run individual branch circuits. Make neutral same as hot.

 

[nmonaco]

Run individual branch circuits. Make neutral same as hot.

So by this, use the Vd = 2 * K * L * I / Cmil for each light in each phase instead of Vd = Sqrt(3) * K * L * I / Cmil?

 

[ptonsparky]

'A' has 900 watts at each location? How many locations?
'B'. Same questions.
'C'. Same questions.

 

[nmonaco]

'A' has 900 watts at each location? How many locations?
'B'. Same questions.
'C'. Same questions.

A has 900 Watts total. 300W 600' from the panel, 300W 2400' from the panel, and 300W 4200' from the panel
B has 600 Watts total. 300W 1200' from the panel and 300W 3000' from the panel.
C has 600 Watts total. 300W 1800' from the panel and 300W 3600' from the panel.

 

[winnie]

There are various levels of approximation that you can apply to see where you get.

For example, if you assume that all the loads are at the end of your 4200 foot run, then you will maximum voltage drop. See what wire size this suggests, and if it is small enough then don't worry about the detailed calculation.

A better approximation is to assume that the current at each load is constant, and then figure out the current flow and voltage drop in each segment of wire, and then add up the voltage drops (considering phase angle in a simplistic fashion).

-Jon

 

[ptonsparky]

I figured 5 amps at the last of your longest circuit. #1 Al will have about a 3.8% VD.

You could run all separate neutrals or a MWBC. Pros and Cons for each.

 

[winnie]

I figured 3.6A at the end of 4200 ft, using a 3 phase 480V load, and got 3.8% drop using #8 Cu. So as a fast back of the envelope #8 Cu is more than sufficient. I think a detailed calculation would get the wire size down to #10 for heavily loaded portions and #12 for the rest. Even though the distance is huge, the current is small.

-Jon

 

[wwhitney]

The OP has an error in the description, I believe; on each phase, the fixture spacing is 1800 ft, rather than 600 ft, per the diagram. That gives an overall fixture spacing of 600 ft.

For example, if you assume that all the loads are at the end of your 4200 foot run, then you will maximum voltage drop. See what wire size this suggests, and if it is small enough then don't worry about the detailed calculation.

In the limit (large numbers of loads equally spaced, one wire size), that's going to be off by a factor of 2, right? Since the average load is only half way out. In the OP, on Phase A the fixtures are at 600 ft, 2400 ft, and 4200 ft, so the average is 2400 ft, or a bit over half. Phase B has fixtures at 1200 ft and 3000 ft, so the average is 2100 ft, exactly half. Phase C has fixtures at 1800 ft and 3600 ft, so the average is 2700 ft.

For a single wire size (constant ohms/ft), it should be enough to sum up the current * distance for each fixture individually, or when the fixtures are the same, just take the unit current * sum of the distances. E.g. on Phase B, if each fixture is 1 amp and the resistance is 1 ohm per 1000 ft (simplifications for discussion, those are not the correct numbers), then from V = I₁*R₁ + I₂*R₂ we have V = 2 * 1.2 + 1 * 1.8 = 4.2, which is the same as 1 * 1.2 + 1 * 3.0.

That certainly applies to the 2 wire circuit case; extending it to 3 phase requires a bit more thought.

Cheers, Wayne

 

[winnie]

I believe you are correct in the limit of large numbers of loads equally spaced and balanced on the 3 phase system.

I was being conservative rather than try to do the detailed calc of what is happening on the neutral in this situation where the system is not balanced and not equally spaced.

-Jon

 

[kwired]

The OP has an error in the description, I believe; on each phase, the fixture spacing is 1800 ft, rather than 600 ft, per the diagram. That gives an overall fixture spacing of 600 ft.


In the limit (large numbers of loads equally spaced, one wire size), that's going to be off by a factor of 2, right? Since the average load is only half way out. In the OP, on Phase A the fixtures are at 600 ft, 2400 ft, and 4200 ft, so the average is 2400 ft, or a bit over half. Phase B has fixtures at 1200 ft and 3000 ft, so the average is 2100 ft, exactly half. Phase C has fixtures at 1800 ft and 3600 ft, so the average is 2700 ft.

For a single wire size (constant ohms/ft), it should be enough to sum up the current * distance for each fixture individually, or when the fixtures are the same, just take the unit current * sum of the distances. E.g. on Phase B, if each fixture is 1 amp and the resistance is 1 ohm per 1000 ft (simplifications for discussion, those are not the correct numbers), then from V = I₁*R₁ + I₂*R₂ we have V = 2 * 1.2 + 1 * 1.8 = 4.2, which is the same as 1 * 1.2 + 1 * 3.0.

That certainly applies to the 2 wire circuit case; extending it to 3 phase requires a bit more thought.

Cheers, Wayne

I think it is 600 feet between fixtures, but by alternating phases connected to it will be 1800 feet between two fixtures on same phase.

 

[nmonaco]

I think it is 600 feet between fixtures, but by alternating phases connected to it will be 1800 feet between two fixtures on same phase.

This is correct

 

[oldsparky52]

I'm just an uneducated (well, self educated from reading, not school) wire puller, but it seems to me the most cost effective way to do this is to install 1 circuit and wire the fixtures to 480V. First 3 lights with #6 and the last four with #8. Should wind up just under 3% VD at the last fixture.

 

[nmonaco]

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

where

• Vd is voltage drop to be under 3%, so 0.03*277V = 8.31
• K = 12.9for Cu wiring
• La1 is the distance from the source to the first light pole of Circuit A = 600'
• La2 is the distance from the first light pole to the second light pole of Circuit A = 1800'
• La3 is the distance from the second light pole to the third light pole of Circuit A = 1800'
• Ia1 is the wattage/power factor to give me the VA, which is 300/0.9 = 333 VA, then the VA divided by the line-to-neutral voltage, (277), which is 1.2A
• Ia2 = 1.2A
• Ia3 = 1.2A
• Cmil is the area of the wire size I am solving for

Inputting all of this information:

8.31 > (Sqrt(3)*12.9*600*(1.2+1.2+1.2) + Sqrt(3)*12.9*1800*(1.2+1.2) + Sqrt(3)*12.9*1800*1.2)/Cmil
Cmil > (Sqrt(3)*12.9*600*3.6 + Sqrt(3)*12.9*1800*2.4 + Sqrt(3)*12.9*1800*1.2)/8.31
Cmil > 23230

This means I need a wire size greater than 23230, which would mean the smallest wire size I could use would be #6 AWG.

Does this check out? Is this fine?

 

[nmonaco]

I'm just an uneducated (well, self educated from reading, not school) wire puller, but it seems to me the most cost effective way to do this is to install 1 circuit and wire the fixtures to 480V. First 3 lights with #6 and the last four with #8. Should wind up just under 3% VD at the last fixture.

Unfortunately the design is already set. They want them on separate circuits so they don't go out all at once. I do agree that would be the most cost effective and based on my calcs, I agree it would work.

 

[oldsparky52]

nevermind.

 

[winnie]

Unfortunately the design is already set. They want them on separate circuits so they don't go out all at once. I do agree that would be the most cost effective and based on my calcs, I agree it would work.

But what you drew at the beginning is all lights on the same MWBC. Those require at least a handle tie between breakers, so if one sub-circuit goes out, they probably all do.

Jon

 

[ptonsparky]

Unfortunately the design is already set. They want them on separate circuits so they don't go out all at once. I do agree that would be the most cost effective and based on my calcs, I agree it would work.

If the design is set as you indicate, you have no choice but three hots, three neutrals, three CBs. Any other way will subject them to all going out at the same time..

 

[wwhitney]

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

 

[nmonaco]

If the design is set as you indicate, you have no choice but three hots, three neutrals, three CBs. Any other way will subject them to all going out at the same time..

oh right. ignore my comment about going out all at once. the design will have them go out all at once, but I know they have to use 277v.