what wire is best for 12vdc solenoid valve at 800 feet? 2

[trainpilot]

We have built an air actuated gate valve that needs to be controlled by a radio receiver located 800 feet away. The receiver puts out 12vdc. 460vac and 120vac are also available where receiver is. I designed an air operated control system which works OK to close the valve on application of pressure, but is too slow to open the valve on pressure bleed-off. I had an air line available to valve location, and have an open pvc conduit available for control wiring. Primary wires for 3-phase 460vac runs in conduit next to open conduit. Two vfds are supplied off this circuit. What is best wire type, voltage to run controls at this distance for our Vectorr atmospheric train? Can I run a solenoid air valve directly off of the radio output, or should I, as SCOTTYUK suggested in an earlier thread use a SSR?

 

[cranky108]

Some of this depends on the current draw of the solenoid.

If you have 120 Vac and a reciever that can deal with 120 Vac, then use the reciever to operate a 120 Vac solenoid.

 

[trainpilot]

I could use a control relay to send 120volts to the valve, though I would have to change coils on the valve. I am concerned about voltage drop and about interference from the VFDs. I'm trying to decide what type of wire to pull given the length of the run and the fact that we may eventually want to automate the entire system with some kind of PLC.

 

[MikeHalloran]

Could you use a quick-exhaust valve?


Mike Halloran
Pembroke Pines, FL, USA

 

[ScottyUK]

The SSR would work, but you would likely want to use a screened twisted pair for the interconnection and add some burden onto the load end of the cable to prevent false triggering. Are the VFD cables shielded? At least tell me they aren't conduit singles!


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If we learn from our mistakes I'm getting a great education!

 

[trainpilot]

Thanks Cranky. Mike, I tried a quick-release valve (two actually). This helps, but the valve doesn't dump the air from the control (supply) quickly enough. I'll try to post a schematic of the air circuit later, but a quick written description follows: air valve opens at receiver, sends pressure down 800' of 3/8 nylon hose. At other end of hose pressure builds and extends air cylinder. At full extension of cylinder a valve sends air to a resevoir through a check valve. When the valve by the receiver dumps, line pressure drops and when it gets to near zero, a spring return air control valve uses the pressure in the resevoir to blow the air cylinder back down. It takes about 12 seconds for the pressure to drain down out of the 800' of hose and the quick-release valve does not seem to help this. It does however, help vent the air cylinder as it is blown back down. I've pretty much given up on trying to actuate this cylinder just on air over this distance, so I'm ready to pull wire for electric control. I'm just trying to avoid mistakes in selecting cable and voltage. I wouldn't want the VFD's to interfere with the operation of the air cylinder.

"...engineering is the art of doing for 10 shillings what any fool can do for a pound". Duke of Wellington Arthur Wellesley

 

[trainpilot]

Thanks Scotty, Its the lines that supply the VFDs that run with the air hose through a bigger pvc conduit in the same trench. I added a separate 1" conduit thinking I might have to run some control wires eventually. The VFDs run on remote radio control and when I run them I can hear some electrical noise from the panel where those lines originate. The VFDs are located close to the motors and supply the motors through metallic flex. Would a small indicator lamp be enough of an addtional load to keep the SSR from misbehaving?

"...engineering is the art of doing for 10 shillings what any fool can do for a pound". Duke of Wellington Arthur Wellesley

 

[Compositepro]

Did you put the quick exhaust valve directly on the valve actuator? If you can supply adequate air through 800 feet of line the it does not make sense that a quick exhaust valve at the valve actuator won't work. I think you may have put it on the wrong end of the 800 ft. line.

 

[waross]

The quick release valves that I am familiar with may be installed as close to the cylinder as is physically possible. A quick release valve mid point on the 800 foot line may help if that is possible.
When the line is exhausted the pressure will drop on a curve similar to a capacitor discharge curve. You may want to try a pressure switch that will operate an electric dump valve when the pressure in the air line starts to drop. You may be able to get the response time down to about a second. You may have to install a local reservoir and use a two way valve and check valve.
A pressure rise may be used with the same pressure switch and solenoid valve to extend the cylinder.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[trainpilot]

Thanks to all of you for your suggestions. Waross, I had not thought of using a pressure switch. I'm sure that would work. I'm going to try repositioning quick-release valves. I wish I could put one in the middle of the run, but it's all buried. The boss wants to add a compressor near the valve, so my one air line will instead be used to increase capacity down the track. I'm going to have to pull some control wiring anyway. When I laid our system out, I put a small compressor at the far end of the track, thinking it would be less noticeable there. As we've added track that has become the midpoint and we've added an open control tower nearby, it would be nice not to hear that compressor cycling. I'm not going to complain about growing pains as I'm lucky, in this economy, to have a good job.

"...engineering is the art of doing for 10 shillings what any fool can do for a pound". Duke of Wellington Arthur Wellesley

 

[waross]

Some simple calcs:
It is assumed that a system will reach equilibrium in about 5 time constants.
So at 12 seconds to empty the system, one time cycle will be about 2.4 seconds. At the end of one time constant the pressure will have dropped 63%. If we take 100%-63% = 37% as the equilibrium then our time constant becomes 0.48 seconds. 100% -(63% x 37%) = 77%
In about 1/2 second the pressure should drop to 77%.
If we go one step further, in about 0.1 second the pressure will have dropped to about 86%.
I hope I haven't made any mistakes and I hope this helps.

Bill
--------------------
"Why not the best?"
Jimmy Carter