Greetings,
I have a set of solenoid valves that need 24V +/- 10% to operate properly. Unfortunately, my PLC is about 1000' away (with existing 16 gauge wires). My electric circuits is a little rusty. Will the following diagram work with the voltage requirements at the solenoids? Seems like it would if I put the power supply close to the solenoid. But I'm not sure how that 1000' to the PLC on the "-" side will affect the voltage.

Option B would be to use an interposing relay. I'm hoping to avoid if possible (this is a SIF and would prefer to not add components to affect the achieved SIL). Any input would be much appreciated. Thank you!

 

It doesn't matter where you have the power supply, you would still have a voltage drop, and of course it's " about 1000' " twice.

Using a relay would have a minimal effect on SIL rating. If you have sufficient spare cores of existing wiring you could wire the solenoid back to a spare PLC i/p to confirm energisation [and de-energisation].

The third alternative is PLC remote I/O. This would be more advantageous if it's a Safety PLC; but would probably need additional cabling for the comms, link.

 

2000 ft of 16g wire has 8.03 Ohms of resistance, according to the first on-line calculator that comes up in my search engine.

If the valve coil needs 24Vdc ± 10%, the minimum voltage needed is 24 - 2.4 = 21.6V; a maximum voltage drop of 2.4V from a nominal 24Vdc power supply.

Here's the dreaded Ohms law: I = E/R = 2.4V/8.03Ohms = 0.3A.
In English, 0.3A will drop 2.4V over 2,000 feet of 16AWG wire.

If your solenoid valve draws more than 0.3A, then the wiring voltage drop will exceed the maximum voltage drop the specs say that the valve will operate at.

 

Thanks for that David_2.

I took a glance at voltage drop calculations - and based current at 0.5Amp on the basis that:
- We don't know the actual solenoid current
- We don't know the distance and " about 1000' " a little vague, so trust voltage drop is an issue.

 

Gentlemen...
How can you calculate voltage-drop without knowing the rated Amperes, or rated Ohms of the Solenoid valve?
1) David2's guess was 0.3 A, thus establishing the Solenoid's rated-power as 8 Watts !
2) Oneye's guess was 0.5 A, thus establishing the Solenoid's rated-power as 12 Watts !
Stop guessing and determine the Solenoid's Namelate data !
Regards, Phil Corso

 

Thank you, all @oneye14 @David_2 @PhilCorso
I might've been not clear in my depiction and definitely did not include the information about the solenoid. It is rated for 21W. The distance between the "+" side of the power supply and the solenoid is about 50 feet (using 12 AWG wire). Then the wire loops to the PLC contact which is 1000 feet away before looping back to the power supply (so yes, 2000 feet there). Question was: Will the solenoid still see about 24V (minus the drop in that 50 feet) since it's only 50 feet from the "+" side of the power supply?

 

It's a loop, so the whole loop "sees" the same current. Replace each wire with a resistor and calculate the voltage drop of each "resistor" based on the solenoid current. Subtract the sum of all such voltage drops from the PS voltage and you'll have the voltage "seen" by the solenoid coil. It makes zero difference whether the extra resistance is in the "+" side or the "-" side. Either way, it will subtract from the available voltage for the solenoid.

 

Jurutera...
Now the rest is easy Q.E.D. !
Phil Corso

 

Jurutera...
Now the rest is easy Q.E.D. !
Phil Corso

Correct me if I'm wrong but isn't Jurutera looking for a solution to overcoming voltage drop over a long distance, not fancy calculations....as indicated in my posting last Friday.

 

Oneye14...
Of course you are correct regarding Jurutera's Goal. And your loop-approach is correct.
However, I must apologize for my earlier mistaken calcs. Following are the right values:
a) Now that the rated Power and rated Voltage are known, rated Amperes, A = P/V or 0.8575.
b) Additionally, since rated A is now known, rated Ohms, R = P/(A^2) or 28.55.
Regards to all,
Phil Corso

 

Use the existing PLC output to close an ice-cube relay in the local panel.