voltage drop calculations and parking lot lights
Basic Information:
volunteer work for a local church parking lot lighting w/ 4 GE M400 400W fixtures (3.9A each w/ HPS bulbs) on 2 poles (2 fixtures per pole) fixture operating voltages: 108V  131V per GE tech. support 120V AC (1 PH) 3wire, copper conductor with PVC conduit One 20A circuit feeds the fixtures All 12 AWG wire (already is place when I arrived) Relevant Distances: segment 1: panel to center of parking lot (this point is geometrically between the poles): 175 feet segment 2: center point to base of pole 1 or 2: 60 feet segment 3: pole base to top of pole: 31 feet segment 4: pole out to fixture at arms end: 14 feet Wiring: 3 wires run from panel to center point of parking lot (175 feet) and feed: 3 wires running north to pole 1 (60 feet) and 3 wires running south to pole 2 These wires run to the base and up to the top of the pole (31 feet) where they feed: 3 wires running to fixture 1 (14 feet) and 3 wires running to fixture 2 My question: In calculating the voltage drop for each fixture via a spreadsheet, I did it in segments as follows: using Vd=2kIL/Cm (with 12.9 for k and 6530 for Cm) for each segment * segment 1 * load amperage: 4 fixtures*3.9A = 15.6A length: 175 feet * segments 2 and 3 * load amperage: 2 fixtures*3.9A = 7.8A length: 60 + 31 = 91 feet * segment 4 * load amperage: 1 fixtures*3.9A = 3.9A length:14 feet I estimated the total voltage drop (sum of voltage drop from each segment) at each fixture to be 13.8V or 11.5% of the supplied 120V. However, I have several "knowledgeable" people within this church arguing that #12 is adequate for the parking lot lighting. Based on the calculations (and I do realize they are mere estimates if they are indeed correct), the lights will burn dim if the even come on at all. Since I am probably rusty, I just wanted some advice on my calculations before proceeding. Thanks in advance 
It looks like you used the one way distance instead of the out and back distance of the circuit.

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Isn't the distance or length in the Vd equation (Vd=2kIL/Cm) the distance the load is located from the power supply and not the total length of the circuit conductors (i.e., out and back distance)? 
No way #12 is acceptable under those conditions. The voltage drop is too great even at the first pole. HPS lights won't really burn dim like incandescents will, since the ballast will try to operate at rated current no matter the input voltage. However, running low input voltage will make the fixture harder to start (reducing effective lamp life) and make the ballast run hotter. It's not a good thing. #12 wire would probably be acceptable if you use 240V and ballasts that are rated 208240V.

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For the life of me WHY would anyone run FOUR 400W MH parking lot fixtures at 120V??? Being a parking lot automatically screams voltage drop. 240V has 1/4 as severe VD as 120V. 
It really should not be much of an issue to change this over to a 240V circuit. (If you know what you are doing)
Either that or pull larger wire. 
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Whoops, missed that. Sorry, my mistake. Eyes glossed over at all the calculations:) 
:laughing:Wire Size Calculator You can try this...and pick up an "Ugly's"

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I will not run with 4.0mm2 ( 12 AWG ) on 120 volts like that distance you will cook the conductors and will have issue with luminaries. For myself I will just run much larger conductors they can handle the voltage drop far much better than what ya got there. But have a electrician assit you on this one. This is one of few spots I will recomened that the church volunteer work off limit on that part. I am sorry if I come out little harsh but I have deal quite few Church projects before so I know what the outcome is. ( there is few set of codes we have to follow btw it is not the same as resdentail is ) Merci, Marc 
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1200W load at 120V is 10A (as seen at the source) 1200W load at 240V is 5A (as seen at the source) Say the wire resistance is 1 ohm. Since Vd=I*R, the Vd for the 120V circuit is 10V and the Vd for the 240V circuit is 5V. Of course this is only valid for relatively small Vd, since the actual load voltages would then be 110V and 235V respectively and the power would be 1100W and 1175W. I think you are confusing voltage and power. The power lost by voltage drop is I^2*R, so dropping the raising the voltage to 2 times, drops the current to 1/2 and the power lost in the wire to 1/4 (as is seen by the 1100W versus the 1175W). However, the voltage drop is only 1/2 (10V versus 5V). Mark 
It has half the voltage drop, but 1/4 of the voltage drop *problem*. The important parameter with voltage drop is not the absolute drop in voltage (5V or 10V) but the percentage reduction in supply voltage (2% vs. 8%). Doubling the voltage reduces the percentage voltage drop by a factor of four.

If I had been forced to use 120 volts I would have compensated in the first run to the center of the parking lot between poles with larger wire, then used 12 awg to each pole. I think a most 20 amp breakers would takeup to a #8 awg. Anyway I would play with wire sizes to see if that would get me reasonable Vd using 120 volts.
All this 12 awg and poles are in place but I take it that it is not in operation yet. 
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