calculating voltage drop multiple loads derating

[howardrichman]

I'm wiring a group of horse waterers near a horse barn I'm near done wiring. two fields, each w/4-5 waterers 250 watt load each@ 120 volts located 400+ft from source. If I figure Five in 1st field is 10.42 amps @100% load ; Ide have to run at least a #8(5+%drop) or a #6(3.4%drop) copp. conductors when voltage drop is calculated. What would be an acceptable E-drop to these waterers? Do I calculate the waterers @100% or derate these loads? waterer company told me these waterers operate only 10% an hour; but couldn't most or all the waterers kik-in simutaniously? I'm using chart and sample in DR Watts pocket guide for E-drop calc.

If can Help;
please Reply;
HR...

 

[dj94]

clarify, did they tell you %10 an hour running OR %10 acceptable voltage loss to the motors?????

 

[Smart $]

Here's a recent thread regarding similar situation:

http://forums.mikeholt.com/showthread.php?t=141444

I recall reading initial posts, but didn't participate or read in its entirety... and thus cannot offer any opinion on its merits.

You mentioned two fields. Is there one field to each side of the barn? Or, the better question here is: what is the distance of each heater to the source and relative each other?

Quite often, consideration for powering many remote loads is a series or "chained", meaning running power to the closest, then continuing on from there to the next closest, and the next closest thereafter... similar to street lighting. However, if the bearing also changes, it could be more efficient to run to a central remote point, then branch off to each individually, or a pair.

 

[gar]

111228-2030 EST

You really don't care what is the voltage drop in this application.

The real criteria is that you do not overload any wires when all heaters are on simultaneously, and that when all heaters are on simultaneously that sufficient heat is produced at the most remote heater. I think the second part of this criteria can be ignored because I think it will automatically happen because of the duty cycle you imply.

Thus, I would add up the total current from all heaters with 120 V applied to them. Your current calculation is correct at 10.4 A. #12 wire would work, and would not be overloaded. The resistance of 5 heaters in parallel is 120/10.4 = 11.5 ohms. 800 ft of #12 at 20 deg C is 1.27 ohms. Thus, the voltage at the end of the 400 ft run is about 120 * 11.5 / 12.77 = 108 V . Thus, power at the load is 81% and the time to heat water will be about 1.24 times what it would be at 120 V. No big problem with a 10 % duty cycle. And all heaters will not be on at the same time. Also note that the actual maximum current would be 108 / 11.5 = 9.4 A.

Really you could use #14 wire.

.

 

[kwired]

gar explained it fairly well.

With resistance heat all you lose with voltage drop is some heat output at the element. This will only make it take a little longer to reach setpoint than if there was no voltage drop. Your 250 watt heater maybe becomes 225 or 230 watts. If you are lucky enough to have a supply voltage of 123 -125 volts you make up for some of the loss just with an increase in supply voltage.

Small decrease in power to these elements will extend the life of them anyway. That is better than a small increase in power to them which could make them operate at a temperature higher than designed for and shorten life of them.