Voltage drop for light pole circuit

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Dsg319

Dsg319

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Wv Master “lectrician”
I cannot for the life of me find out how to attach an image to a thread could someone please help? I have some calculations for practice for voltage drop on a pole lighting circuit that goes a total of 725ft to the last light. 5 pole lights total each light Has 240watt bulb (2amps). In the photo I want to post have it drawn all and calculations written down with more info. Just want to make sure I have this right and if not I would appreciate the correction. Thanks I’m advance.
 
On the reply screen toolbar at top , see the rectangle with the mountain, tap that, it will say drop image here. If you have an image on your phone you want to post, again tap rectangle with mountain, and tap the drop image here rectangle, take photo or photo library should pop up , hope this helps
 
Okay so disregarding the photo I will just jot it all down here to see if my numbers are correct. Say I have a 120volt power supply, supplying a lighting pole circuit with the first pole being 325ft away(length of the pole included) and the remaining 4 pole are 100ft apart. Using this formula 2xKxIxD/VD. (Kept voltage drop at 3percent which is 3.6volts. Came up with using 6awg wire from panel all the way to the 4th light pole and 8awg from 4th to the 5th pole. Is this correct?
 
We;d need to know each load to check your math.

The last run carries only the last light's current, the fourth carries two, etc.
 
Yes sorry I forgot to mention each light pole will draw 2 amps. Back at the panel would be 20amp overcurrent protection
 
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Dsg319 said:
Okay so disregarding the photo I will just jot it all down here to see if my numbers are correct. Say I have a 120volt power supply, supplying a lighting pole circuit with the first pole being 325ft away(length of the pole included) and the remaining 4 pole are 100ft apart. Using this formula 2xKxIxD/VD. (Kept voltage drop at 3percent which is 3.6volts. Came up with using 6awg wire from panel all the way to the 4th light pole and 8awg from 4th to the 5th pole. Is this correct?
Click to expand...
That looks like 3.7V drop just to the first pole.
 
Without doing the math, you could use #12 from pole 4 to 5, #10 from 3 to 4, #8 from pole 2 to 3, #6 from pole 1 to 2, and # 4 for the home run.

Now that I've said that, I wonder how the numbers work. Too bad I'm too lazy to figure it out. But, it sure looks elegant. :cool:
 
LarryFine said:
Without doing the math, you could use #12 from pole 4 to 5, #10 from 3 to 4, #8 from pole 2 to 3, #6 from pole 1 to 2, and # 4 for the home run.

Now that I've said that, I wonder how the numbers work. Too bad I'm too lazy to figure it out. But, it sure looks elegant. :cool:
Click to expand...
Voltage drop be damned, ain’t nobody wiring it like that 😎
 
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If you stick to a single wire size, the math becomes much simpler. Then the voltage drop is linear in (I * D). So for a series of loads at different distances, you can just add up (I * D) for each load (I is the current for just one load at a time, ignoring all other loads). Putting that into the voltage drop formula gives the voltage drop at the end of the line.

Where all the load currents are the same, as in the OP, it's even simpler (with a single wire size). You can just use the total current and the average of the distances. Or the current for a single load and the sum of the distances, either one gives you the sum of (I * D) when all of the I values are the same.

Cheers, Wayne
 
Optimizing for multiple sizes is more complicated. A given layout can be analyzed piece-wise as above for each segment of wire of a given size, bearing in mind that the load current at the end of the segment should include all downstream loads.

A heuristic for sizing might be to calculate the single wire size that would work for the whole circuit. Then since wires come in discrete sizes, the actual voltage drop at the end will be less than the target. If you'd like to take advantage of the unused allowable voltage drop to downsize for the last so many segments, you can rephrase that perturbation problem as a voltage drop problem just like the original.

In the perturbation problem, the far end of end of the run is now the starting point for distance measurements, and each load is associated with the length of the conductor segment preceding it. And for a resistance per unit length you take the difference between that of the larger size and the smaller wire size. Then you can take the unused allowable voltage drop, divide it by that differential resistance per unit length, and come up with a maximum allowable sum of perturbation I*D values. Then add up the perturbation I*D values starting at the far end until you just exceed the maximum allowable; that load is the point at which you can downsize the wiring downstream of it, but not upstream of it.

For further optimization, the perturbation problem can be repeated for another smaller wire size, using whatever remaining allowable voltage drop there is.

Cheers, Wayne
 
Now that's the way an applied mathematics expert would size the wires ;).
Now you're stirring up old memories of doing matrix perturbation problems in school.
 
There is a simpler approach to solving for the minimum wire size that meets the voltage drop requirements.

Treat each load completely separately and calculate the wire cross section (circular mils or mm^2) meet the voltage drop requirements for that one load.

Then for the portions of the circuit shared by multiple loads just add up the individual load cross section. Bingo: enough cross section to supply all the loads with acceptable voltage drop.

Jon
 
That approach is equivalent to what I stated in my first post, as the required cross sectional area for a single load is proportional to (I * D). So you can sum up the (I * D) values either before or after applying the proportionality.

I first misread your suggestion as using multiple wire sizes, where each segment is sized as the sum of the required single load areas for all the loads that segment carries. For any arrangement of non-zero loads with more than one segment, that approach is guaranteed to give more voltage drop than the target (for the mythical wires of the exact calculated cross sectional areas). I expect that last statement can be shown to follow from the arithmetic/geometric mean inequality or something like that.

Cheers, Wayne
 
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