Voltage Drop For Light Poles

[TheGingerElectrician]

I've got a lighting post project I'm quoting for a customer in the back of their property that includes (7) 20' tall posts. The furthest one is right at about 1000' away.

I have a 5.74 amp calculated load and am going to run them on 240v single phase. Just want to make sure I did my calculations correct for quoting out wire size and conduit.

If I am correct I will have to run a #3 cu. in order to avoid going above the 3% max for a branch circuit. Can someone tell me I've done this correctly? I don't do this very much.

 

[tom baker]

the voltage drop decrease past each pole, so you may start with a 3 AWG but end up with 12. Make a spread sheet for each run
Home run would be 7 luminaires, then 6, then 5. Lots of VD calculators on line. or from mike holt

 

[TheGingerElectrician]

We are planning on daisy chaining them. Can you provide a simple explanation of how to do that? Your saying to calculate the home run with it's load, then run 6 with it's lower load, the 5 the same way and so on? How do you do that without making the VD% additive?

 

[TheGingerElectrician]

Do you do this only based on the total load at the end of the entire run? In that case the total load at the end of the run is right at .82 amps.

 

[PaulMmn]

Can you feed to the center post then send to the 3 on each side? Less overall VD...

 

[oldsparky52]

Can you feed to the center post then send to the 3 on each side? Less overall VD...

That might be possible, but it really depends on the layout.

 

[oldsparky52]

My question was answered already, sorry.

 

[oldsparky52]

The way I have always done something like this is in a spreadsheet so I can play with the numbers as I go down the line.

 

[kwired]

We are planning on daisy chaining them. Can you provide a simple explanation of how to do that? Your saying to calculate the home run with it's load, then run 6 with it's lower load, the 5 the same way and so on? How do you do that without making the VD% additive?

it is additive. If you drop 1 volt between the start and the first load then you have one more volt drop between first and second load that is two volts drop at the second load.

So figure drop on each segment then subtract that from starting voltage. If each segment is same length there will be somewhat a pattern to it. If you have one long segment somewhere in the middle you might actually have a larger drop there than over any other segment, just depends on conditions. Nothing wrong with running a smaller conductor to first pole if it is close and then increasing conductor size after that point either, the VD just is not there.

 

[oldsparky52]

We are planning on daisy chaining them. Can you provide a simple explanation of how to do that? Your saying to calculate the home run with it's load, then run 6 with it's lower load, the 5 the same way and so on? How do you do that without making the VD% additive?

The VD are additive (if I understand what you are saying).

 

[kwired]

Do you do this only based on the total load at the end of the entire run? In that case the total load at the end of the run is right at .82 amps.

You calculate each segment separately based on that segment load.

 

[TheGingerElectrician]

So calculate load at each distance and then use the largest wire for every location or size the wire based on each individual run from pole to pole. For example I may start out with 12 cu. but then end up with #3 by the end and so on?

 

[oldsparky52]

We are planning on daisy chaining them. Can you provide a simple explanation of how to do that?

You calculate the VD to the 1st light using a 5.74 amp load. Then you subtract the VD from the starting voltage and calculate the voltage drop to the 2nd light using a 4.92 amp load, then repeat.

 

[oldsparky52]

So calculate load at each distance and then use the largest wire for every location or size the wire based on each individual run from pole to pole. For example I may start out with 12 cu. but then end up with #3 by the end and so on?

Typically you would use the larger conductor with the highest loads.

 

[TheGingerElectrician]

That might be possible, but it really depends on the layout.

I totally would but we have to go around their septic system so that makes it one run...

 

[TheGingerElectrician]

the voltage drop decrease past each pole, so you may start with a 3 AWG but end up with 12. Make a spread sheet for each run
Home run would be 7 luminaires, then 6, then 5. Lots of VD calculators on line. or from mike holt

Here is what I did. I used Mike Holts Formula for determining wire size for each location and calculated that based on each run. Does this look correct? If so I think we will just run it all in #8.

 

[winnie]

Here is what I did. I used Mike Holts Formula for determining wire size for each location and calculated that based on each run. Does this look correct? If so I think we will just run it all in #8.

I don't think your calculations are quite correct, since if you daisy chain things, then the voltage drop to pole 2 is the drop from pole 1 to pole 2 _plus_ the drop from the panel to pole 1; similarly at pole 3 you add the drop from pole 2 to pole 3 to the drop at pole 2 and so on.

So if you figure the drop at pole 1 using the full load current (5.74A) and then use the full 3% drop for the distance to pole 1, then you will get a small wire size result. But now the drop at pole 2 will be greater than 3% because you already used up the allowed 3% just getting to pole 1.

I'd suggest 2 things:
1) Just assume that the full 5.74A is located out at 1000 feet, but do the calculation using resistance values for wire at 20C (the NEC tabulates the wire resistance at 75C, but for your application the wire will run cool and you can use the lower 20C resistance). If this gives you a comfortable wire size just run it.

2) If you need to optimize the results because the wire size is huge, see my next post.

-Jon

 

[TheGingerElectrician]

I've

I don't think your calculations are quite correct, since if you daisy chain things, then the voltage drop to pole 2 is the drop from pole 1 to pole 2 _plus_ the drop from the panel to pole 1; similarly at pole 3 you add the drop from pole 2 to pole 3 to the drop at pole 2 and so on.

So if you figure the drop at pole 1 using the full load current (5.74A) and then use the full 3% drop for the distance to pole 1, then you will get a small wire size result. But now the drop at pole 2 will be greater than 3% because you already used up the allowed 3% just getting to pole 1.

I'd suggest 2 things:
1) Just assume that the full 5.74A is located out at 1000 feet, but do the calculation using resistance values for wire at 20C (the NEC tabulates the wire resistance at 75C, but for your application the wire will run cool and you can use the lower 20C resistance). If this gives you a comfortable wire size just run it.

2) If you need to optimize the results because the wire size is huge, see my next post.

-Jon

I've never done this paticular calculation for a change in ambient temperature although I understand what you are talking about. I see the formula in table 8. Do I figure the wire size first based on the CM and then do the temperature change formula?

 

[TheGingerElectrician]

I've

I've never done this paticular calculation for a change in ambient temperature although I understand what you are talking about. I see the formula in table 8. Do I figure the wire size first based on the CM and then do the temperature change formula?

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

 

[winnie]

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