Voltage drop on a string of lights

[peter]

I got involved in an interesting question question on another website [Taunton/Breaktime]. This question involves a string of five outdoor lights, each being 1 amp apiece. The arrangement is as follows:
"Fuse" 60' 1st light 175' 2nd light 175' 3rd light 190' 4th light 220' 5th light. He is using the voltage drop calculator on Electrician's Toolbox.com [3% vd] and needs to know the wire size.
I pointed out that the wire size could diminish as the amp loads got further out. But using the calculator, I ended up with some wierd results: Namely, #14 for the first 60' to feed 5 amps, #10 Al between light 1 and 2, #12 to carry 3 amps between light 2 and 3, #12 3 to 4; and #14 for the last run.
But just inputing 1 amp for the last fixture at 820', I got #10 guage Aluminum.
Oh yeah!: 120 volts.
Would appreciate some wisdom.
~Peter

 

[chris kennedy]

Would appreciate some wisdom.
~Peter

Your not going to get that here.

What difference does it make to an electron if the 14 wire is at the source or the end?

 

[beanland]

Too Many Solutions

Too Many Solutions

I have done these. You need to decide on some rules. Do you want all the wire the same size to keep the job simple? Do you want the voltage drop to be uniform (so each light is 1V lower than the previous)? You can build this system any number of ways and still meet the 3% at the last light. But, other considerations are needed in the design.

Personally, I like the "all wire the same size" because that makes for a cleaner design. Now, calculate backwards to find out what is the allowable resistance per foot for the system and select the wire accordingly.

 

[beanland]

Calc

Calc

If we assume a uniform wire for the entire distance.

220 ft x 2 x 1A +
190 ft x 2 x 2A +
175 ft x 2 x 3A +
175 ft x 2 x 4A +
60 ft x 2 x 5A =
4250 ampere-ft

120V x 3% = 3.6V

3.6V / 4.25A-kft = 0.8471 ohms per kft

Per Uglys, you need #8 copper

 

[don_resqcapt19]

Peter,
Doing it the way you did it, you have an increasing voltage drop as you run down the circuit. The drops are additive if you calculate them one at a time. The #14 has about a 1.3% or 1.56 volts, the #10 Al has about 2% or 2.4 volt drop, making the drop at light 2 about 3.96 volts. The same as you move on to the next lights

 

[480sparky]

As I recall, the American Electricians Handbook has a formula you can use to calculate the 'center' of the load. You can then calculate the VD from there.

 

[peter]

I was thinking more in terms of doing it the most elegant way. Obviously he wouldn't need #8 wire to carry 1 amp for 220'. So I will toss in the stipulation that this would have to be the lowest material cost.
But the responses so far has got me thinking that he would have to deliver a little bit more amps to light 4 to split between 4 and 5.
~Peter
The Rolls Royce engine on the V-22 weighs about a ton and produces 6,000 horsepower.

 

[Bob NH]

I was thinking more in terms of doing it the most elegant way. Obviously he wouldn't need #8 wire to carry 1 amp for 220'. So I will toss in the stipulation that this would have to be the lowest material cost. . . .~Peter

I did a bit of calculus to determine that the optimum to minimize copper is if the area of the conductor (circular mils) is proportional to the square root of the current (amps). I based the calculation on two equal lengths of wire with one having 2 times the current of the other.

Then I set up a table using the given lengths and calculated the voltage drop with the theoretical conductors.

Since it is necessary to select conductors that are actually available, that can only give you an idea of the ratios of the sizes. It might be useful if you had a string of 20 lights or loads.

From a practical point of view one can select the smallest single size conductor that meets the voltage drop requirement (8 AWG copper in the cited case), and then start from the far end and replace the last sections with the next samller conductor until the voltage drop requirement is just met.

In the example case the last 220 ft. section can be replaced with 10 AWG copper to result in a voltage drop of 3.45 Volts and only 7% more copper than if the optimum (and unobtainable) conductors are used.