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Joined: Mar 2001
Posts: 2,056
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Redsy Offline OP
Member
Assume the following:

(10) 120-volt, incandescent lamp posts @100 watts each', spaced 50' apart.
1,000 ft. conductor length (50 x 10 x 2 conductors)
Total line current 8.33 amps.

The conductors from the panel to the first post will see all 8.33 amps, and each successive segment of conductors will see .83 amps less, correct?
However, each successive segment will see an additional increase in line resistance due to conductor length. ie, the first pole will see more voltage drop due to line current, but less due to the conductor resistance.

There will be more voltage drop at the last post due to conductor length, but it will be somewhat offset by reduced line current, correct?
I may run a multi-wire ckt. to help offset voltage drop, but changing to 240 volt fixtures are out of the question.
Bottom line--- #6, or #4?

Thanks.



[This message has been edited by Redsy (edited 10-29-2005).]

Joined: Jun 2004
Posts: 1,273
T
Member
Split the load across two hots and one neutral. The effect, due to balance is to 'drop the neutral out' of the voltage drop calculation.

The effect is to cut the distance in half. Just make sure to alternate the load at each light.


Tesla
Joined: Sep 2003
Posts: 650
W
Member
I like to approach such questions by starting with simple approximations, and then get more complex only as needed.

To really asses the voltage drop accurately, you would proceed as you suggested; calculating the voltage drop from the source to the first light with the full current, then calculating the drop to the second light with the slightly lower current, then the third light, etc. But you would then need to include the fact that the current drawn by the lights would change with the voltage; the lamp might draw .833A at 120V, but will draw less than this at 115V. If you really want to get picky, you would need to include the fact that the resistance of the copper wire will change with temperature [Linked Image]

On the other hand, if you simply assume a lumped load of all of the lights at the end of the line, and assume that the wire is at its 75C resistance, then you will over-estimate voltage drop. If the voltage drop is low enough in this case, then you will be certain that it is low enough with the more complex distributed load.

I'd use #12 conductors and a multi-wire circuit with the lamps evenly distributed on each side. You have a total load of 4.16A at 240V. Resistance of #12 stranded conductors is 2.05 ohms/kft (Chapter 9, table 8, stranded coated conductors). Voltage drop of 3.6% in this worst case; better in reality. 3.6% is a touch high for a branch circuit, but this is a very stable load, so no flickering.

Although you might want to factor in that 100W every 50feet is not that much, and consider the chance that the owner will want to increase the wattage later on.

-Jon

Joined: May 2002
Posts: 1,716
R
Member
Redsy, you could also use loop or "Ring" (not in the over seas sense) circuit. [Linked Image]

This might start something. [Linked Image]

Roger

Joined: Sep 2005
Posts: 33
N
Member
redsy.. whats up
you are asking if you should use #4 or #6 copper i assume. well # 6 would get you within code #210-19 fpn no. 4. This came from 1993 code book by the way

States you cant have 3%drop from the farthest outlet of power. and no more than a total of 5%total VDrop, from the fedders to the farthest outlet.

calculating yours
(calculation for #4 )2kil/cm 2x11=22x8.3=182.6x1000'=182600/41740(cm of #4) =4.37volts. .03x120v=3.6 volts so # 4 wouldnt work

(calculation for #3 ) 2kil/cm
2x11=22x8.3=182.6x1000'=182600/52620=3.4v
#3 is imo the single size wire you would need to stay under that three percent voltage drop. would this be to big of wire you think. do you agree with me.

Joined: Mar 2001
Posts: 2,056
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Redsy Offline OP
Member
Thanks for the input guys!

I am probably going to install 23-watt compact fluorescents anyway, so the incandescents are a "worst case" scenario.
I do like the idea of a multi-wire ckt., but with the timer/photocell control scheme I plan to use, and the added labor & materials, I'd prefer to use a straight 120-volt circuit.
I'll do some more thinking.

Joined: Sep 2005
Posts: 33
N
Member
in this case you can run a #8 copper for this circuit. you agree?

Joined: Mar 2005
Posts: 38
D
Member
Redsy,

Once again I'm not trying to sound like a commerical but I've always felt that it is better to know than to guess. Please download Volts and use it's Series Voltage Drop module. It takes all factors into consideration and automatically sizes your conductor sizes and resulting voltage drop in less time than it took me to write this post. http://www.dolphins-software.com


Dolphins Software
Joined: Mar 2001
Posts: 2,056
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Redsy Offline OP
Member
Dan,

Doing it this way is more fun and keeps us sharp. I don't want to "dumb it down" by relying on software.



[This message has been edited by Redsy (edited 10-29-2005).]

Joined: Sep 2005
Posts: 33
N
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me either

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