Help me understand voltage drop

[Bluedog225]

I’ve got 500 feet of 1 inch conduit buried above my waterline. I stubbed it up every 20 feet. My plan is to run some copper wire through there for some sort of low-voltage lighting system. I can split this distance in half at the cabin, resulting in approximately two, 250 foot runs. I’ll use whatever works with my final lighting design and abandon the rest.

I understand the blue seat capacity charts. And I see the distinction between 3% voltage drop for critical loads and 10% voltage drop for non-critical loads.

Some of the led lights I’m looking at have a pretty wide range of acceptable voltages. E.g.: 10-60 volts.

What happens when there is greater voltage drop? Does the wire get unacceptably hot?

I’m trying to grok what happens if I have a long run of wire, starting at 48 volts, but delivering say 10 volts at the end of the run.

As I type this, I’m thinking there is some relationship between ampacity an voltage drop I don’t get.

Can anyone point me in the right direction to clear up my muddy thinking?

Thanks

 

[timselectric]

Amperage has everything to do with voltage drop.
More Amperage means more voltage drop.
Without Amperage, there is no voltage drop. (VOC)
There are a few of ways to deal with voltage drop.
1. Increase conductor size.
2. Lower Amperage at end of circuit.
3. Increase voltage at source of circuit.

#3 can be dangerous if the load can't handle the VOC.

 

[Mattb4]

...

What happens when there is greater voltage drop? Does the wire get unacceptably hot?

I’m trying to grok what happens if I have a long run of wire, starting at 48 volts, but delivering say 10 volts at the end of the run.

As I type this, I’m thinking there is some relationship between ampacity an voltage drop I don’t get.

Can anyone point me in the right direction to clear up my muddy thinking?

Thanks

Keep in mind VA=W.

The lower your voltage goes the higher the amperage goes to maintain the same load. Motors compound this during startup since if the voltage becomes insufficient to start rotation they draw max current and yes the supply wires and motor gets hot fast.

Ampacity is the ability of a conductor to safely carry a current level at a particular voltage and distance. So if a wires resistance is such over a distance it acts to create a load that varies with amount of wattage being needed. You only see the voltage drop happen when a load is present. A bit like pressure , flow rate and volume in a water pipe. Without flow the pressure in a system is equal.

Not sure if this helps your conception of the process. Especially since it has been a lot of years since I was educated on the principle.

 

[hwy17]

What happens when there is greater voltage drop? Does the wire get unacceptably hot?

Not necessarily because it's spread out. If the wire is rated for the amps, then that's how you know it is acceptably spread out.

But it was key for me understand that all voltage drop is due to heating, that's where the voltage "goes".

If you have 48v 1a at the supply end and 10v 1A at the load then the load is receiving 10 watts and 38 watts of heat is dissipating along the wire. The load and the heating always sum up to the supply.

 

[Bluedog225]

I was under the impression that the ampacity of a wire doesn’t change. Therefore, a wire rated for 15 amps would be able to carry 15 amps at 10 feet or 100 feet. Or I guess 10,000 feet. But at some point the voltage approaches zero.

Since the wire resistance is greater at 100 feet, the voltage would be lower.

If I design a long run of direct current lights, starting at 50 volts and dropping to 10 volts, and keep the loads below the ampacity of the wire, would that be ok?

The blue sea ampacity chart limits the length of the wire to maintain at least 90% voltage. Is it ok to work beyond this limit?

Here is an example of a light with the wide voltage input I’m seeing.

 

[timselectric]

If I design a long run of direct current lights, starting at 50 volts and dropping to 10 volts, and keep the loads below the ampacity of the wire, would that be ok?

Yes
As long as the device voltage range covers the maximum input voltage.
But keep in mind that if the voltage is lower, the device will draw more current.
You have to factor this into the circuit.

 

[hwy17]

I was under the impression that the ampacity off a wire doesn’t change. Therefore, a wire rated for 15 amps would be able to carry 15 amps at 10 feet or 100 feet. Or I guess 10,000 feet. But at some point the voltage approaches zero

This is correct. The blue sea chart is not a wire amps rating chart it's an acceptable voltage drop chart. As far as the wire is concerned, it can carry its full amp rating unlimited distance. Complicated by the fact that inductive loads like motors will increase their amp draw proportionate to voltage. So a 10amp 100v motor would become a 20 amp 50v motor if it's only getting 50v at the end of the wire. The way it changes its draw like this could either exceed the wire rating if it's 10A wire, or it could overheat the motor. LED lights are also an inductive load I think.

If I design a long run of direct current lights, starting at 50 volts and dropping to 10 volts, and keep the loads below the ampacity of the wire, would that be ok?

The problem is when the lights are off, the 10v light at the end will get 50v. It only drops to 10 under load. So the light at the end of this high voltage drop string has to be rated to accept 10-50v. Edit: I see you already understand this part.

The blue sea ampacity chart limits the length of the wire to maintain at least 90% voltage. Is it ok to work beyond this limit?

Yes, if you respect all of the above.

 

[Bluedog225]

That makes sense. With the above light, it would theoretically draw about 1 amp at a 60 volt source but about 5 amps at the point along the line where voltage was 10 volts. All to maintain a 60 watt load per light.

And I would size the wire to handle however many amps (lights) I put on the string.

Thanks!

 

[Hedges]

Wire is a resistor, dissipates power according to I^2R.
The thermal path from wire through insulation, air, conduit, earth is a thermal resistance. Temperature rises linearly with power dissipated.
Voltage drop occurs as IR

Enough current to drop 10V out of 240V would still let things work. 10V out of 12V would not.
Incandescent lamps would glow more dimly at reduced voltage. A long pair of wires with lamps between them, further lamps would be more dim. Voltage graph resembles cable of suspension bridge.
If you fed the wires from opposite ends, distributed lamps would see closer to same voltage but not identical.

LED lamps could have a resistor to convert voltage to current, then would sag. Or could have DC/DC converter, try to maintain brightness. Some try to the point of burning themselves out with excess current during "brownout", low voltage. I got a pair of LED driving lights from Harbor Freight, which work at 12V or 24V but say not to run at low voltage. I need to make a voltage qualification circuit before using them on an old motorcycle.

Higher voltage is preferred. Twice the voltage, half the current, half the drop in volts but 1/4 the drop in percentage.

Some of the led lights I’m looking at have a pretty wide range of acceptable voltages. E.g.: 10-60 volts.

What happens when there is greater voltage drop? Does the wire get unacceptably hot?

I’m trying to grok what happens if I have a long run of wire, starting at 48 volts, but delivering say 10 volts at the end of the run.

That sounds like the way to go.
Just obey ampacity for the total current, keep unloaded voltage within spec of the lights, minimum within spec.

If the layout can return to the cabin, drive both ends. (Because you don't care about voltage difference between lamps, don't need to drive +/- from opposite ends, rather both wires from both ends.)

If the lights accept AC, you could distribute high voltage and periodically step down with transformer. Use GFCI. That would mean running primary and secondary wires, but secondary could be thinner for just a few lights, reducing total copper. Optionally rectify AC to DC.

If these were dumb incandescent, you could string in series. Like some Christmas lights, also some PA systems speakers. Don't think it is a good idea for SMPS of LED!

 

[Vigo]

If you can afford the voltage drop, the heat created in the wire run is usually not very serious. The insulation of a given wire will be rated up to a certain temperature, and as long as you arent melting the wire insulation or anything that insulation touches, how hot you allow it to get is really up to you. But in the case of a 500ft wire run, even if you are losing a couple hundred watts as heat in the wiring, that heat will be so dispersed over that wide area that, as long as there are no 'bad connections' in there getting particularly hotter than anything else, it will not matter because no single point will approach the melting temp of the insulation or of the conduit.

All of this should be verified with math but i think the big picture is that yes, you can choose to give away watts in the form of voltage drop in the wiring converting to heat, as long as said heat is dispersed over a wide enough area that no single point gets particularly hot.