LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Exactly what I said in post #38 using other words.
480V Light fixtures
- Thread starter Allegiantelectrical
- Start date
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
I'm unclear on whether the simple computation of 300W/480V/0.95 power factor = 658 mA per light is sufficiently conservative for all purposes, or whether the higher startup current of an LED driver will be an issue for some purposes. But 658 mA per light should be reasonable for voltage drop (not conservative, as I would think the load is constant power, so the current will go up with voltage drop).
With 32 lights per circuit, one option is to run a single 2 wire circuit. The total current is 21A. If this is a continuous load it would require a 30A breaker. For the longer run, if the first light is 473' away and the last light is 1353' away, and they are equally spaced, then the average distance is 913'. So if you want to run a single size wire for the whole run, then #2 AWG Al would give you a 2.45% voltage drop, or #4 AWG Al would give you a 3.82% voltage drop.
Or you could go with Larry's idea of reducing conductor size as you go. You could do that with 4 times as many computations, and more accurately by avoiding the constant current load assumption, as follows: with 480V at the panel, let's allow 4V of voltage drop per pole, so the final pole gets 464V (a little over 3%). That that pole draws 8*300W/464V/0.95 = 5.44A. Take the distance from the 3rd pole to the 4th pole, and figure the wire size for 4V drop and 5.44A. Then repeat for each earlier segment, updating the current draw per pole each time, and adding it to the downstream current draws.
Then another option is to run a 3 wire circuit. As 8 lights per pole is not divisible by 3, the simplest model is just to treat the pole as having 9 lights. So the current per pole is 9*300W/480V/0.95/sqrt(3) = 3.42A per line, and the total circuit current is 13.7A. If you go with a single size conductor, and 913' is the average distance on the longest run, then #4 Al would give you 2.16% voltage drop, or #6 Al would give you 3.38% voltage drop.
Not sure how the cost of 3x #2 or #4 Al for the 2-wire circuit option compares with 4x #4 or #6 Al for the 3-wire circuit option, respectively.
Cheers, Wayne
With 32 lights per circuit, one option is to run a single 2 wire circuit. The total current is 21A. If this is a continuous load it would require a 30A breaker. For the longer run, if the first light is 473' away and the last light is 1353' away, and they are equally spaced, then the average distance is 913'. So if you want to run a single size wire for the whole run, then #2 AWG Al would give you a 2.45% voltage drop, or #4 AWG Al would give you a 3.82% voltage drop.
Or you could go with Larry's idea of reducing conductor size as you go. You could do that with 4 times as many computations, and more accurately by avoiding the constant current load assumption, as follows: with 480V at the panel, let's allow 4V of voltage drop per pole, so the final pole gets 464V (a little over 3%). That that pole draws 8*300W/464V/0.95 = 5.44A. Take the distance from the 3rd pole to the 4th pole, and figure the wire size for 4V drop and 5.44A. Then repeat for each earlier segment, updating the current draw per pole each time, and adding it to the downstream current draws.
Then another option is to run a 3 wire circuit. As 8 lights per pole is not divisible by 3, the simplest model is just to treat the pole as having 9 lights. So the current per pole is 9*300W/480V/0.95/sqrt(3) = 3.42A per line, and the total circuit current is 13.7A. If you go with a single size conductor, and 913' is the average distance on the longest run, then #4 Al would give you 2.16% voltage drop, or #6 Al would give you 3.38% voltage drop.
Not sure how the cost of 3x #2 or #4 Al for the 2-wire circuit option compares with 4x #4 or #6 Al for the 3-wire circuit option, respectively.
Cheers, Wayne