Lighting and Voltage Drop

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Jody Boehs

Jody Boehs

Member
Location
Fairview, Oklahoma, USA
Occupation
Controls and Electrical Manager
Hi,

My company is building a 380' x 320' warehouse. I have attached a photo of our lighting layout. My plan is to put each Bay of lighting on its own 20A circuit. 15 bays total. Each bay has 6-9 lights. The lights being installed are 240W 277V LED UFO fixtures. If my calculations are correct, each bay will pull around 8 amps assuming all lights are ON. Some of the lights will be on motion sensors. My question, however, is this.... I have always been told to increase the size of wire for each 100' of length to minimize voltage drop. The panels for this building will be on the left side and center of the attached image. If I am using 20A circuits (12 awg wire), that would mean I would have to use 6 awg wire to reach all the way across the building. There's got to be a better way to do this. Any ideas? Lights.png
 
There are several good voltage drop apps that give you that value without guessing. You just need the length, actual load and voltage. It will figure wire size automatically. I use the Southwire app, but there is many more out there too.
 
210303-1520 EST

Jody:

I would view it differently if it was my own building.

On that basis the first questions I would ask are:
1. What is the typical lowest no load voltage going to be at the main panel?
2. What is the lowest voltage that can be expected at the main panel under full load, in other words depends on source impedance?
3. Over what voltage range can the lights work with no abnormalities? Probability much less than 277.
4. What is the minimum voltage to the lights that can produce the required light intensity?

Suppose the answer 1 is 277.
Suppose the answer to 2 is 271.
Suppose 230 is the answer to 3 and 4. That is a full load voltage drop of 41 V.

Suppose your entire light load for one bay is concentrated at the end of the run, then for #12 copper resistance is 1.6 ohms, and at 8 A the voltage drop is 12.8 V for a 500 ft run (1000 ft loop). Thus, #12 could be used easily.

Pick your criteria and try this type of analysis.

.
 
Both replies are good information to have. Gar, your reply is very interesting. I can follow you most of the way through that analysis. I do not know the answer to number 2 for my case. I do not know much about impedance. The number 4 question is what caught my attention. The lights I will be using are advertised as 120-277 multivolt LED UFOs. So what you are saying in question number 4 is that voltage drop for these lights is not too critical?
 
Using VD=IR, I figure you can get by with #8 home runs on the farthest row (9 lights) and #10 from the 1st light to the others. Last light should be just under a 3% VD.

I used 400' from the 1st light to the panel and 30' light to light.
 
Jody Boehs said:
he lights I will be using are advertised as 120-277 multivolt LED UFOs. So what you are saying in question number 4 is that voltage drop for these lights is not too critical?
Click to expand...

I'll say that :D.

If the fixtures are truly rated for 120-277, then they won't care if they get 133, 197, or even 260. (There are some that one lead for 120-240 and another for 277; check that.) That said, they probably won't care if they get 277, 285, or even 265- LEDs are current sinks, so as long as the driver can put the design current through the LEDs, it'll be OK.

Another question is whether you mind the power loss from the drop; the answer is probably No.
 
2% VD is possible 500ft 1-way with #12, if wired for balanced loads.

240-Watt x 8 lights = 1,920 Watts for each circuit.
1,920 / 277 = 7A
Z x Amps = 5.3 VD
Z = 0.7533317, @ 0.85pf, Metallic conduit or MC Cable

Balanced loads keeps 3 ccc's, if the neutral remains close to Zero Amps
 
Jody Boehs said:
So what you are saying in question number 4 is that voltage drop for these lights is not too critical?
Click to expand...
Maybe.

Voltage drops because of wire resistance. Think of it as an in-line heating element.

How many circuits are going in your pipe?
Maybe they can double as heat strips for the warehouse
 
ramsy said:
2% VD is possible 500ft 1-way with #12, if wired for balanced loads.

Balanced loads keeps 3 ccc's, if the neutral remains close to Zero Amps
Click to expand...
Agreed. I would definitely wire these as MWBCs, starting with the longest runs with the same quantities of lights per circuit.

Lights with a voltage range will use a little more current on a lower voltage, similarly to motors, and maintain brightness.
 
Jody Boehs said:
The lights being installed are 240W 277V LED UFO fixtures.
Click to expand...
Per NEC 220.18(B) listed & labeled luminaries combine ballast factor to show Amps on nameplate, rather than Watts exclusively.

My clients are blowing Homeline AFCI's with LED's form Amazon that only show a CE Mark.
No FCC listing for noise interference, much less "IC" ratings, or UL label.

Unlike the UL Mark, or other NRTL marks, the CE Marking:
1) is not a safety certification mark,
2) is generally based on self-declaration rather than third party certification, and
3) does not demonstrate compliance to North American safety standards or installation codes.
 
210303-2016 EST

Jody:

First thing to do is take ONE fixture, 240 W rating, and run some experiments. These experiments are to determine if you see any objectional effects from making quick changes to the applied voltage.

At 1.6 ohms ( 1000 ft of #12 copper ) and 1 A current per fixture, then for 9 fixtures we get 14.4 V drop. And to this we need to add the voltage drop at the main panel of 6 V I previously assumed. So now assume a change in voltage at the light fixtures of 20 V. For 277 V at 240 W we get a current of 0.866 A Drop down 20 V from 277 V and voltage is 257 V. Assuming that the power load remains approximately constant, then current rises to about 0.94 A. To get a voltage drop of 20 V at 0.94 A we need a series resistance of 20/0.94 = 21.3 ohms.

A cold, room temperature 100 W, tungsten incandescent bulb has a resistance of about 10 ohms. But this rapidly goes up in resistance with current. A Silex 1000 W 120 V hot plate is about 17.5 ohms at room temperature. At about 20 V excitation I read about 1 A. This might be a useful series resistor for running a test on the LED. A toggle switch can be placed across the resistance to essentially instantaneously change voltage to the light. If you use tungsten bulbs as a series resistance you may find it hard, having to play with different bulbs in series and parallel, to get to the 20 V drop.

If you can show that rapid changes in voltage to the LED fixture does not cause unwanted problems from a light level perspective, t.hen you need to look at total inrush current to the breaker when switching a bank of lights on. If possible it would be interesting to see if you could drop down to a 15 A breaker, and #14 wire, without false trips on turn on.

Note: 9 A at 1.6 ohms is 130 W. Do you really care? Possibly.

Are there any other considerations relative to using #12 wire?

.




9 A,4444444 and ( 1 A and 277 V is 277 W, and thus more than 240 W, but convenient to use to make some guesses ) we it takes
 
One thought: If you'll be turning all the breakers on and off every day, I'd suggest a switched contactor on the feeder.
 
zbang said:
If the fixtures are truly rated for 120-277, then they won't care if they get 133, 197, or even 260. (There are some that one lead for 120-240 and another for 277; check that.) That said, they probably won't care if they get 277, 285, or even 265- LEDs are current sinks, so as long as the driver can put the design current through the LEDs, it'll be OK.
Click to expand...

These lights only have 1-3 wire pigtail coming out of them. Hot, Neutral, and Ground!
 
James L said:
Maybe.

Voltage drops because of wire resistance. Think of it as an in-line heating element.

How many circuits are going in your pipe?
Maybe they can double as heat strips for the warehouse
Click to expand...

Haha I think there are better ways to heat a warehouse. I have already decided to only put 2 circuits in each pipe to minimize the need for decreasing the amperage rating on the conductors.
 
Several thoughts:

1) The MWBC approach cuts voltage drop but costs redundancy. A single 3 pole circuit now feeds 3 bays of lights, rather than 3 separate single pole circuits each feeding 1 bay. As a design decision if I would use MWBCs I would 'distribute' each MWBC so that loss of one circuit does not kill light to adjacent bays. I might even arrange things so that adjacent lights were not on the same circuit.

2) On this site I learned a clever trick for dealing with voltage drop on long circuits with distributed loads.

Rather than selecting wire size and calculating voltage drop, the approach is to select a target voltage drop and then calculate the 'circular mils' needed to feed each load with that target voltage drop. Then for each stretch of circuit you add up the circular mils required by all the loads fed through that bit of circuit to get the required wire size for all the loads.

You use the voltage drop equation from this page: https://www.mikeholt.com/technnical-voltage-drop-calculations-part-one.php
But swap the CM (circular mil) term and the VD (voltage drop) term
CM = 2 * K * I * D / VD

K is the resistivity constant, here 12.9 circular mil * ohm / feet different values might be needed for lower temperature operation
The above formula is the single phase formula where D is the one way distance, thus the factor of 2

Here is an example:
9 lights in a bay, fed from the center.
Each light draws 1A at 265V
Target voltage drop of 12V
closest light is 300 feet from source, then 40 feet to each additional light
single pole circuit

light 1: 460feet CM= 2 * 12.9 * 1 * 460 / 12 = 989 (This means that you only need a copper cross section of 989 circular mils to feed this single lamp with the target voltage drop)
light 2: 420feet CM= 903
light 3: 380feet CM= 817
light 4: 340feet CM= 731
light 5: 300feet CM= 645
light 6: 340feet CM= 731
light 7: 380feet CM= 817
light 8: 420feet CM= 903
light 9: 460feet CM= 989

All of the lights are fed by the run from the panel to light 5. So that run needs wire size 645 + 2*731 + 2*817 +2*903 + 2*989 = 7525 circular mils
That wire size is bigger than #12 but smaller than #11. You can't buy #11 so you need #10.

The run from light 5 to light 4 carries the current for lights 1-4. So that wire needs to be 989 + 903 + 817 +731 = 3440 circular mils
That wire size is smaller than #14

Based on the above, I would consider using 15A circuits so that I could use 14ga wire. Increase the wire size only from the panel to the center light of a bay, then use 14ga for the rest.

-Jon
 
Jody Boehs said:
Haha I think there are better ways to heat a warehouse. I have already decided to only put 2 circuits in each pipe to minimize the need for decreasing the amperage rating on the conductors.
Click to expand...

When you increase the size of wire for voltage drop, that also compensates for required derating.

If you already need to have 10ga conductors for a 20A circuit because of voltage drop, then you can have 20 current carrying conductors in your pipe before derating is an issue. Far sooner you will want more pipes simply to have less mess coming out of each.

-Jon
 
winnie said:
When you increase the size of wire for voltage drop, that also compensates for required derating.

If you already need to have 10ga conductors for a 20A circuit because of voltage drop, then you can have 20 current carrying conductors in your pipe before derating is an issue. Far sooner you will want more pipes simply to have less mess coming out of each.

-Jon
Click to expand...
Someone suggested a #12 circuit, based on the idea that voltage could drop from 277 all the way to 136 volts and the lights won't care.

True, lights won't care. But ither factors have to be considered
 
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