Solenoid questions

Technical Pneumatics
1

I am new, couldn’t find much information on Chief Delphi or Google.

Here is my situation: I have 1 double acting, 10 inch stroke, 2 inch bore pneumatic cylinder and needs to generate a 60 psi. I have the cylinder, now I need to choose a solenoid valve to control the cylinder (and this valve will be controlled by PLC). To confirm with you guys, does it sound right that I should have at least 1 inlet, and at least 2 outlets? With this conclusion, I am planning to get a 4/5 port solenoid valve.

I understand 2 port valves, where there is just one inlet and one outlet, but how exactly does a 4/5 port valve work? Does it have to work with a manifold (because I am not planning to)? Any valves you guys recommend by say SMC (SYJ 3000 series?)?

2

I believe the valves provided by SMC and Festo in the FRC Kit of Parts can be classified as 4/5 port valves. They have 1 input, 2 outputs (A and B), and then a relief for each output (EA and EB). The block had two positions, one which would feed the input to A and put B to relief EB, and vice versa. SYJ5000 and SYJ3000 look very similar to the valves we’ve had in the past. Searching SY3140-6H-S (the part number of an '08 valve) on their website came up with the SY3000 series.

Hope that helps.

3

I understand a little better. So on SMC’s website, when it says 5 ports, you mean there are two that are just relief ports (1 input, 1 output A, 1 output B, 2 relief ports)? In other words, is this a 1 port in, 2 port out?

So say I needed a solenoid valve with 1 in, 4 out, then this SYJ series is incompatible with what I am looking for?

I am just trying to understand the lingo. Also, is there a good tutorial I can read up on so I can understand those diagrams of the valves?

—Edit—
Sorry… I thought this forum was open to everyone.

4

Don’t worry, it is. Anybody that tells you otherwise doesn’t quite understand that it is.

Many answers, however, will be from the perspective of a competitor/robot builder in the FIRST Robotics Competition.

Question for you: You say that you need to generate 60 PSI. Is that generated by or used by the cylinder in question?

5

Yes to the first question. As for the second, you may be able to do something like if you strangely connect things to work off the relief off the valve, however you won’t independent control of anything this way. The defining factor is the number of positions the solenoid can be in. The ones I was talking are all examples of 2 position solenoids, where there are really only two useful outputs. The relief ports are where you might put something like a flow control if you want to vent slowly, or stuff like that.

Here is a document on reading the schematics.
Here is a site talking about the principles behind the operation of a solenoid valve.

And yes, this forum is absolutely open to everyone and anyone who wants to talk about anything. We’re more than happy to help out. However, as Eric said, it will be dominated by FIRST robotics participants talking about robots. In this case, and in many others, your solenoid question fits in perfectly.

6

I need an axial force of 180 lbs. Now, I have a 2-inch bore piston in the cylinder. Therefore, it will generate roughly 60 PSI correct? - Pressure = force x area of bore.

I just need the cylinder to have 1 stroke extension, and 1 stroke back to it’s initial position. With that down, I think I have a basic idea of what solenoid I need, but am still unsure of one thing. My cylinder is a double-acting cylinder, so it will require pressure to extend, and to retract, correct? In conclusion, I here is the SMC part number I came up with: SYJ5120-5G-C4-F. I am still trying to do my research on 2-position single/double and the 3-position closed/exhaust/center. I think I will need a 2 position single valve though, but just wanted to verify so I don’t screw up.

Thanks for helping me, a rookie rookie, out. greatly appreciated!!! :slight_smile:

7

You keep talking about “generating” pressure with the piston. That’s got me confused. Can it be done? Yep. I can think of ways to do that, involving hooking it up to the solenoid in reverse. But generally a piston does not generate air pressure, it uses it. The pressure generated at the non-pressurized end due to cylinder motion leaves via a vent valve, or the piston stops halfway, and we can’t have that now can we?:wink: And when I say leaves, it’s virtually impossible to keep it compressed and use it again. You can slow down/stop the cylinder with it–see below–but even that use is tricky.

Your PSI estimate is correct, and a bit generous (never a bad thing). A=Pi*r^2, and with a 2-inch bore, r=1 inch, so A=Pi, which is about 3.14159 in^2. To generate 180 lbs of force on said piston, you need 57.296 PSI at a minimum, but 60 is a much easier number to play with, and a good overestimate, so we’ll go with that. The tough part is maintaining that PSI–I’d try to do the math for flow rates to maintain pressure in a given volume and all that, but it’s late and I don’t think I’m quite up to trying calculus at this point in time.

I’d look at using some of the FRC-type solenoids, such as the FESTO VUVG-L10-B52-T-M5-1P3-566458 or VUVG-L10-B52-T-M7-1P3, as those work pretty reliably. The SMC 3000 series was also used in FRC with a manifold, IIRC–some of the older FRC veterans might have a few laying around. SMC 5000 series should work also.

If I remember correctly, a single-acting solenoid causes the cylinder to default to one end or the other; tripping the solenoid causes it to move and untripping it causes the cylinder to return to the previous position. A double-acting solenoid has to be tripped both ways. So if you’re simply doing a quickish out-and-back motion, a single solenoid might be the better way to go; if you need to hold position for a while (and especially if you think there might be a power failure), a double-acting is probably better.

Bonus: some FRC teams have rigged a multi-position cylinder by attaching a single solenoid to the exhaust port(s) of a double solenoid, closing the single to generate pressure, and opening it to allow for full travel.

Just out of curiosity, what are you working on?

8

Your calculation is correct for the extend direction.

For the retract direction, you have to subtract the cross-sectional area of the shaft. So the required pressure (to achieve the same force) is slightly higher.

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