Street Lighting Voltage Drop

[EE_rookie]

Having this office debate about how to properly calculate the voltage drop of a street lighting circuit for line to neutral circuits on 3 phase systems.

System info:

1. The power system is 480Y/277V, 3 Phase.
2. The branch circuits for the various lighting circuits are 20A, 1 pole breakers running line, dedicated neutral, and ground out to each fixture in the circuit, with an average of 10 fixtures per circuit.
3. The light fixtures themselves are 277V @ 100VA each and are spaced out 100’ apart

Formulas use to calculate voltage drop:
Option A:
V drop = 2 * I * R * L
I = current
R = resistance per 1000’
L = one way length from source to load, “2” doubles this for round trip length since it’s single phase.
Vdrop = 4.57 volts or 1.65% (assuming #10 AWG wires at .95 PF)

Option B:
V drop = I * R * L
Vdrop = 2.28 volts or .82% (assuming #10 AWG wires at .95 PF)

Looking at Table 9 in the NEC and the subsequent examples, it’s not any clearer (to me)

Note 2 under Table 9 indicates “Multiplying current by effective impedance gives a good approximation for line-to-neutral voltage drop”

If I simply do I * Zeff (effective impedance) as it states, on the above equation it’s without the “2” and V drop = I * R * L like in Option B. Unless there is something else not explicitly state to consider.

Now in a balanced 3 pole circuit, 3Phase, 4 Wire circuit, with 10 lights per phase (30 total), I can understand there being ~0 current on the neutral wire since the three phases are out of phase by 120° , offset each other and sum to 0. In that instance I would remove the “2” from the equation above since it’s only the source to the load distances that would have voltage drop.

Couldn’t really find definitive answers one way or another on the internet on how to truly calculate this voltage drop. Obviously there's a large difference in voltage drop as the quantity of fixtures get larger and farther apart - it effects wire size, conduit fill, and ultimately cost for the owner so it's important to get right!

 

[steve66]

Picture 3 lights on a run, each on a different phase , A, B, C.

For the 1st light (coming from the panel), the only voltage drop will be on the hot conductor. The 3 different currents coming back on the neutral will all add up to zero. So no current on that neutral, and no voltage drop across the neutral.

For the 2nd light, it will have neutral current for 2 lights. Those currents will be 120 degrees out of phase. So the neutral current will be less than the line current for one light, but not zero. So it will have some voltage drop on the neutral.

For the 3rd light, it has no other current offsetting its neutral current, so it can be treated as a single phase 277V line to neutral load.

If you have 10 lights, you are going to have this pattern 3 times over - A, B, C, A, B, C, A, B , C. Then you will have one additional light that will have to be treated as a single phase 277V load.

So you can basically try to calculate all that. Or you can try and approximate the voltage drop.

Note that it is best to alternate the phases, for each light, otherwise you will have sections with a much larger neutral current. You don't want to do the 1st 3 lights on phase A, and the next 3 on phase B. That would not work the same.

 

[EE_rookie]

That all makes sense to me, when you have a 3 phase circuit with lights alternating evenly down the run: A, B, C, A, B, C, etc. The neutral current is 0A since they all cancel out in a perfectly balanced system.

However, I'm referring to a situation where you have only a single pole breaker, running a lighting circuit that is just A, A, A, A.... with L-N-G wires.
How to best and most accurately calculate the voltage drop in that scenario?

 

[zbang]

(didn't somebody post a spreadsheet for this a while ago?)

Most accurate way is to make a spreadsheet that treats each connection between poles/lights as a separate calculation accounting for the increased current as you move towards the source. This also allow you to change the wire size from one end of the circuit to the other.

 

[steve66]

In engineering terms, it's basically the integral of the current with respect to distance & resistance of the wire.

So you just basically add up the voltage drop on each section using: Section #1 - current of all the lights, Section #2 - current of N-1 lights, Section #3 - current of N-2, ....., 3, 2, 1

And yes, its pretty easy to make a spreadsheet that calculates this.

 

[EE_rookie]

I have the spreadsheet all set up - see attached image.
In column X, I have highlighted the cell so you can see the formula - in which I have a 2X (the capture the outgoing wire and the return wire neutral).

Is this the correct formula since it's single phase L-N circuit, even though it's on a 3 phase system? or can I remove the "2" factor?

 

[steve66]

That's pretty close to what I get (I'm a little closer to a 3 % drop.)

I think most engineers usually include the reactance so they are calculating the AC voltage drop. And that probably explains why I get a slightly higher voltage drop.

But that equation usually needs more inputs, with the power factor being the main one. And sometimes differences in conduit (EMT, RGS, or PVC) and wire type (THW, THHN, etc...) are included.

 

[W@ttson]

I love voltage drop. It seems like an easy question but its anything but easy. There are different ways to wire it up and things get complicated. It ends up being what steve66 said in post #2. It got discussed in detail here:

Series Voltage Drop All Inclusive

Hello All, I would like to put the topic of series voltage drop in various types of three phase wiring arrangements for roadway lighting to rest. I have always had to look into this to get the right methodology, and I would like to get confirmation or explanation as to if my set up for the 4...

forums.mikeholt.com

 

[wwhitney]

In engineering terms, it's basically the integral of the current with respect to distance & resistance of the wire.

Yes for the 2 wire case in the OP. And if you use a single size of wire for the whole circuit, then the resistance per unit distance will be a constant and can be pulled out of the integral. That means for fixed current loads you can just add up (current * distance) for all the loads, and use that as your (current * distance) in the OP's "option A".

E.g. if you have a 5A load at 200' from the supply, a 10A load at 500' from the supply, and a 2A load at 1000' from the supply, just add up 5*200+10*500+2*1000 = 8000 amp-ft. The voltage drop at the end will be the same as if you had a single 8A load at 1000', or an 80A load at 100' or any such combination.

Because addition is commutative and multiplication distributes over it, this procedure will give the same answer as looking at the length of each segment and the current on that segment. I.e. in the example I made up, that would be 17A * 200' + 12A * 300' + 2A * 500' = 8000 amp-ft again.

Cheers, Wayne

 

[steve66]

Good input, and it reminds me that I planned on suggesting this:

Compare your results with using either 2 phases (each alternating light on a separate phase), or even a 3 phase circuit, again alternating the lights. See what size wire is required to get about the same voltage drop.

Then compare the cost of 1000' of #2 wire (hot and neutral) with the cost of the wire you need for 2 or 3 phases.

My guess is that it will be a substantial savings.