Voltage Drop For Light Poles

[TheGingerElectrician]

I also don't know what the values in that formula represent.

Actually I found the formula in 310.14(B)

The formula that you used calculates the minimum wire size needed to serve a particular load with the required voltage drop.

In this case, because you are using a very lightly loaded circuit (6A on perhaps 8awg wire) you can use the resistance value for _cool_ wire. I would use 10.4 circular mils per ohm foot for copper at 20C, rather than the 12.9 that you used.

Running the calculation for the full load at 1000 feet we get:
CM = 2 * 10.4 * 1000 * 5.74 / 7.2 = 16582 circular mils which is a scotch larger than 8awg. I would run 8awg and call it good since the customer will probably want to add more load in the future....

However there is a slightly different way you can use this same equation. Run the calculation for each load _separately_, and then add up the copper cross section required wherever the loads actually share a wire.

So for the pole right at 1000 feet you would need 2370 circular mils to supply just that one load with a 3% voltage drop.
Now for the next pole in, at say 900 feet you would need CM = 2 * 10.4 * 900 * 0.82/7.2 = 2130 circular mils to supply just that one load with a 3% drop. However since you are running both loads on the _same_ wire for part of the run, that shared part of the run would need to be 4500 circular mils.
Do the same calculation for all 7 loads. The run from the panel to the first pole would need a cross section calculated for all 7 poles added together. Then the run from the 1st pole to the 2nd would need a cross section for the remaining 6 poles added up, and so on.

Of course you can't go smaller than 14ga wire on a 15A circuit, and I'd probably not bother trying to perfectly optimize jumping from 8 to 10 to 12 to 14ga in the circuit.

-Jon

So In my particular application looks like #8 is a good and safe bet... I just came by it on accident... They will be adding more later but it will be a separate run around the other side of the house so I think we will be good. Does the ambient temperature at the connections at the pole base make any difference to this equation or is the effect minimal? I might look into an extra class or something that explains all this. Mike holts books have been great for most applications but this is definitely getting on the engineering side of things. I may be running into this more in the future. Most of the time I've had to calculate VD it's been for a sub panel 300' from the house and so forth. Nothing like this where there are multiple stops and varying loads.

 

[winnie]

I also don't know what the values in that formula represent.

Good question.

The 12.9 (or 10.4) is the resistance constant for the wire. It tells you how many circular mils you need to get a resistance of 1 ohm over 1 foot of length at a particular temperature. Since the resistance is lower at lower temperature, when the wire is cool you need fewer circular mils to keep the resistance down.
Distance is just distance, and you multiply by 2 because in a single phase circuit you have to go there and come back again. With a pole 1000 feet away you have 2000 feet of wire in the circuit.
If you just calculated 10.4 * 2 * distance you would calculate the cross section needed to give your circuit 1 ohm of resistance.

The next values are amps, which is your load and '3% max VD' which is the voltage you are permitted to lose in the resistance of the wire.

3% of 240 is 7.2, so you are permitted to lose 7.2V in the wire. '3% max VD' / amps gives you the maximum allowed resistance to get your maximum allowed voltage drop.

Take your 1 ohm circular mil area and divide my your maximum allowed resistance, and you get your required cross section.

In your equation you have amps/ '3% max VD', the inverse of resistance. But this is simply the mathematical trick of multiplying by the inverse instead of dividing (you know, multiplying by 1/2 instead of dividing by 2; gives the same result).

Hope this helps
Jon

 

[winnie]

Heh. I am all about the engineering side of things. Not enough practical experience.

 

[zbang]

Depending on the lights, can you feed them with 480 via a transformer? Might not be worth it but it's probably worth a little time to check it out.

 

[powerpete69]

Lick your thumb, put it up in the air and here ya go...
Use #8 wire the whole way to account for future lighting.

If you really want to get crazy, run two sets of #8's. Use the second set as a spare. You customer may love the thought process. At least give them the option and the relatively small increase in price for future flexibility.

 

[petersonra]

why not change to 480 volt lights?

 

[TheGingerElectrician]

Lick your thumb, put it up in the air and here ya go...
Use #8 wire the whole way to account for future lighting.

If you really want to get crazy, run two sets of #8's. Use the second set as a spare. You customer may love the thought process. At least give them the option and the relatively small increase in price for future flexibility.

That'd be great but they want three zones Now and that changes everything! Blah!

 

[petersonra]

That'd be great but they want three zones Now and that changes everything! Blah!

not really. previously you would have run 2 wires. now you need 4.

 

[oldsparky52]

I've always used this method for the simple things I've worked on.

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www.mikeholt.com

 

[TheGingerElectrician]

not really. previously you would have run 2 wires. now you need 4.

True but I think we are going to route the one in the front differently since they want a 3rd zone but we are still good because the other distances are still there and I've got the wire sized appropriately now.

 

[oldsparky52]

and I've got the wire sized appropriately now.

What size did you decide on using?

 

[oldsparky52]

Here is a shot of a spreadsheet I put together using VD=IR. Distances are guessed. I used #10 the whole circuit.


ETA: Yea, ... I'm bored.