First post here. I've searched and seen some similar issues on the boards, but couldn't find the answer to my question. I am considering the installation of an actuated valve that is located about 2000' away from my power source. I currently have a 2 pair 18AWG cable buried the entire distance. The first pair is a 4-20mA loop and I'd like to use the second pair to control the actuated valve motor. The motor draws only .7amps at 24vDC. I've looked at voltage drop calculators and it appears that I could increase the voltage to 42volts to get the proper 24v to the motor. I understand the best solution would be a larger cable, but that isn't practical. I also can't install a wireless system due to visual constraints. Do I risk damage to the cable by having such large voltage losses?

 

If you have a spare pair, you could use a self regulating power supply, these are made by a few manuf.
What they have is a small gauge sense pair connected at the P.S. at one end and the device end at the other and sense the resultant volt drop, which then is used at the supply end to increase the supply automatically to conform to a stable load-end voltage.
The beauty is they auto regulate with changing load.
Max.

 

To some extent, it depends how thermally insulated the cable is but I doubt there will be a problem. The power dissipation would be 6.3mW per foot.

 

I've looked at voltage drop calculators and it appears that I could increase the voltage to 42volts to get the proper 24v to the motor.

The disadvantage to this is the voltage may swing due to a changing load, with a straight DC solenoid this most likely would be no problem, but a motor current changes with changing load.
On the other hand, a motor generally has a relatively wide voltage range.
Max.

 

I assume the valve motor will only be actuated periodically?
Why not put 24V worth of battery at the end by the motor, to supply the sudden current needs, and whenever the valve motor isn't on, the battery is trickle charging

 

I appreciate the different solutions and input! The self regulating power supply would be a good choice, but I can't give up the second pair that is using the 4-20mA loop. I also considered the battery idea, but couldn't find a DC trickle charger that required less than the .7amps used by the motor itself. The second concern with a charger is the possibility of noise introduced by the 24v alongside the 4-20mA loop. The wire is a communication cable from Paige P7171D and doesn't have individually shielded pairs. Just noticed in the link that it says 30V in the surface printing section. Not sure if that is a concern?

 

Resistance of 1000 ft 18AWG wire is 6.385 Ohm, Maximum current is 2.3 A. allohttp://diyaudioprojects.com/Technical/American-Wire-Gauge/
Resistance of pair of wires is 6.385 x 2 = 12.77 Ohm.
Voltage drop on pair is 12.77 Ohm x 0.7 A = 8.939 V.
Voltage you need send to motor is 24 V + 9 V = 33 V.
Power dissipated on 1 ft = (9 V / 2000) x 0.7 A = 3,15 mW.
Pairs of cable are twisted, than coupling between pairs will minimal possible and power current will not affect to 4...20 mA pair.
EDIT: Calculation above is for 1000 feet of cable
For 2000 feet it is: 18 V voltage drop; voltage sent to motor 42 V; power dissipated on 1 ft of cable is 6.3 mW.

 

Power dissipated on 1 ft = (9 V / 2000) x 0.7 A = 3,15 mW

But for each foot there are two wires so the total power per foot is 6.3mW.

 

But for each foot there are two wires so the total power per foot is 6.3mW.

One wire have resistance 6.385 Ohm.
Power dissipation for one wire is 6.385 x 0.7^2 = 3.12865 W
Power dissipation for one wire, one ft = 3.12865 / 2000 = 0.001564325
For two wires, one ft of cable 0.001564325 x 2 = 0.00312865 W
EDIT: Calculation above is for 1000 feet of cable. For 2000 feet power dissipation is 6,3 mW per ft.

 

You would likely be just fine with using any common DC power supply that is between 36 and 48 volts given most general purpose DC motors are not all that fussy about having the exact voltage they are rated for.

 

You can use 3 small cheap wall adapters, 12 V + 12V + 9 V, with rated current 1 A or more. Connect their outputs in series for 33 V.

 

Hello,

You can use 3 small cheap wall adapters, 12 V + 12V + 9 V, with rated current 1 A or more. Connect their outputs in series for 33 V.

This will only work if the wall adapters are fully isolated.

Bertus

 

A linear trickle charging circuit will not induce any noise into the comm pair, and would need much less than 0.7 A; that's the thing about trickle charging, it needs only a trickle.

Separate from that, a way to guarantee that the motor voltage is not too far above or below 24 V under all conditions is to raise the source voltage from 42 V to to 48 Vdc and put a small linear regulator circuit (LM317) at the motor. Kits are on ebay for something like 2-3$ and that includes a heatsink. The 317 has an in-out difference spec of 43 V, so a 48 V input would be no problem when the output is 24 V, and after the long run voltage drop the input will be much less.

ak

 

Resistance of 1000 ft 18AWG wire is 6.385 Ohm, Maximum current is 2.3 A. allohttp://diyaudioprojects.com/Technical/American-Wire-Gauge/
Resistance of pair of wires is 6.385 x 2 = 12.77 Ohm.
Voltage drop on pair is 12.77 Ohm x 0.7 A = 8.939 V.
Voltage you need send to motor is 24 V + 9 V = 33 V.
Power dissipated on 1 ft = (9 V / 2000) x 0.7 A = 3,15 mW.
Pairs of cable are twisted, than coupling between pairs will minimal possible and power current will not affect to 4...20 mA pair.

Note that the installation is 2000' from the power source. Your calculations assume it is 1000' away.

 

This will only work if the wall adapters are fully isolated

Usually they are. We everywhere use them and do not afraid of electrical shock.

 

If you want to trickle charge the battery then that will need to use your pair of leads that you also need to use to control the motor. I'm assuming the 4-20 mA current loop isn't what is controlling the motor, it sounded like that was separate.

You can double team them by putting in a circuit that draw power from the control signal at, say, 12 V (at the control end) and uses a small DC-DC converter to pump it up to trickle charge the battery. When the voltage rises above 20 V (again, at the control end) the circuitry at the motor end senses this then it powers the motor from the battery.

 

Note that the installation is 2000' from the power source. Your calculations assume it is 1000' away.

I am sorry. It was my big mistake.
Seems I need delete my calculations.

 

I am sorry. It was my big mistake.
Seems I need delete my calculations.

No need to delete them -- in fact please don't delete them because subsequent posts refer to them. Deleting them introduces chaos in the thread.

Either post a follow up with the corrections or edit the post and add the corrections. But don't delete the post or the original calculations

 

No need to delete them. Either post a follow up with the corrections or edit the post with the corrections. But don't delete the post or the original calculations because subsequent posts refer to them. Deleting them introduces chaos in the thread.

I see. Thank you.

 

I also considered the battery idea, but couldn't find a DC trickle charger that required less than the .7amps used by the motor itself.

I shouldn't have used the term "trickle charge" because it implies a device, called a trickle charger. What I meant was just a 24V power supply at you base station, and a 24v battery at your remote station. The 24V battery would require very little current to keep topped up.