Install lighting 1000 feet away


  #1  
Old 07-27-04, 06:08 AM
ramm
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Install lighting 1000 feet away

I want to install lighting at my entrance gate, but it is 1000 ft. from my electrical source. I want pole lights, with 3 6o watt bulbs in each. Any ideas? Solar lighting doens't seem to be sufficient.
 
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Old 07-27-04, 07:32 AM
J
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1000 feet is very, very far to run 120-volt power, and very far to dig a trench. So every amp is costly. Exactly how many post lights are we talking about? For one post, use 12-gauge wire. For two, use 10-gauge. For three, use 8-gauge. For four, use 6-gauge. You can fudge on these numbers if you don't mind that the bulbs are dim and half your power is consumed by the wire.

You might decide that solar is better after all.
 
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Old 07-27-04, 08:13 AM
M
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Maybe one or more of these:

http://www.siliconsolar.com/portable_power_system.htm

combined with high-efficiency lighting (CFL or LED) might do the trick.

I cringe to think of the time/cost to bury 1000' of wire just for a few lights...
 
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Old 07-29-04, 06:29 AM
ramm
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Would it make a difference if I were to run 240V line from a main panel to a sub-panel? I want two pole lights with 3 60 watt bulbs each.
 
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Old 07-29-04, 07:00 AM
M
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You still need to dig 1000' feet of trench 18" deep. This is difficult and expensive. If you are OK with that then read on.

Running 240V with a shared neutral will cut your voltage drop by a factor of four if you put one pole light on each phase and the wattage of each pole light is the same. With 2 180W fixtures (1.5A each) the voltage drop with 12 gauge (1.6 ohms @ 1000') is 2.4V; with 14 gauge (2.5 ohms @ 1000') the voltage drop is 3.75V. Either value is OK for lighting, but don't plan on adding much more load to the circuit.
 
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Old 07-29-04, 07:43 AM
J
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Running 240V with a shared neutral will cut your voltage drop by a factor of four
I can understand a factor of two. I don't see a factor of four.
 
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Old 07-29-04, 09:14 AM
M
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John:

Think of it this way:

(1) Put both fixtures on a single circuit, producing a voltage drop of X volts split equally between the hot and the neutral.

(2) Now put each fixture on its own circuit with separate neutrals. Because the current is halved, the voltage drop is cut in half to X/2 volts, again split equally between hot and neutral (X/4 volts drop on hot, X/4 volts on neutral).

(3) Now share the neutrals. Since the loads are perfectly balanced, the currents in the shared neutral cancel - no current, no voltage drop. The only voltage drop is in the hot (X/4 volts).
 
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Old 07-29-04, 11:01 AM
J
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Okay, I see your point. Thanks.
 
 

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