1
\$\begingroup\$

I have purchased two 24V AC solenoid valves to control the irrigation in my greenhouse. The plan is to install the two valves inside the greenhouse and power them using relays in the garage which is 50 meters away.

Having access to 50 meters of ethernet cable, can I use two of the four pairs of wire inside the cable to power the solenoids valves? The solenoids use 370mA of inrush current.

Edits:

  • Cable says 23AWG which apparently is even thicker than 24AWG.
  • It works! Water immediately started to flow even when using a single pair of wires per valve

Bottom line:

If you consider using 50 meters of network cable to power solenoid valves, just go ahead, it will most likely work 👍

\$\endgroup\$
3
  • 1
    \$\begingroup\$ It would depend on what kind of cable you have and what are it's ratings, such as single core or stranded, wire copper diameter (AWG), and if it is even rated for outdoor use. Wire has resistance and when current flows there will be a drop in load side voltage, so you need to know how much voltage the solenoid needs to operate properly, as it won't be 24 VAC at the load. \$\endgroup\$
    – Justme
    Commented May 6, 2022 at 19:42
  • 3
    \$\begingroup\$ You could use four of the four pairs to halve the resistance. If you already have the cable and the solenoid valves, I suggest that you try it - you don't have to unroll the 50 m of cable for a test. \$\endgroup\$
    – Andrew Morton
    Commented May 6, 2022 at 19:58
  • 2
    \$\begingroup\$ at 7ohm/100m, between supply and return you have 100m. So you'll have 2V drop during inrush and less during normal operations. That sounds likely to work when you start with 24V. \$\endgroup\$
    – Ben Voigt
    Commented May 6, 2022 at 19:58

1 Answer 1

Reset to default
1
\$\begingroup\$

Think of wire as a low-value resistor. The thicker the wire, the less its resistance. The question then becomes "Is adding a resistor to the load going to cause a problem?" The voltage drop across the resistor will increase proportionally with its resistance. If you think of it this way, the goal is to reduce resistance to a usable (and safe) point.

You can look up wire resistance values on various wire gauge charts and calculators like this one. Network cable is ~24 AWG (you didn't specify what type of network cable). 24 AWG has a resistance of about 84 mΩ per meter. If your cable length is 50 meters, then you're adding 100 meters (both supply and return), or 8.4Ω to the circuit.

Treating the solenoid as a fixed resistance of 65Ω: at 24 V the total current will be limited to 327 mA instead of 370 mA, and the voltage drop by the wire will be about 2.75 V. In that case, the solenoid will "see" 21.3 volts instead of 24.

If you use two conductors instead of one, you'll halve the resistance. Total current would be 347 mA, voltage drop on the wire 1.46 V, and the solenoid "sees" 22.5 V.

You'll need to test to see if either of these configurations allows the solenoid to operate properly. As a valve or any mechanical thing ages/oxidizes/collects debris, what works now may not work later.

Andy Aka wrote a great answer discussing inrush current on AC-powered solenoids. The figures I used above are based on the initial state as you mentioned in the question. As the solenoid settles, current should decrease a bit, which means the values I calculated should be "worst case."

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.