Would this Valve Move? 1

[jm-darcy]

Hi there,

I was hoping someone may be able to help with a query that has been dividing opinions of the people I ask to comment on it.

If I were to place a dual actuator valve filled with a compressible fluid into a compression chamber (or drop it to the bottom of the sea), with the open and close ports connected (as per sketch in attachment), would the valve move?

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https://res.cloudinary.com/engineering-com/image/upload/v1484926474/tips/Untitled_ckn62m.png

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If we consider the valve in the picture is (almost) open, it is noted that the close actuator has a smaller volume due to the valve shaft passing through it.

Therefore, given that the volume in the open side (RHS) is greater than that of the close side (LHS), and given the difference in volume and surface area of the two chambers; would the volumetric change of the compressible fluid from hydrostatic head, drive the piston to move at all to equalise?

Thank you ever so much for any advice you could provide.

 

[bimr]

The extreme pressure at the bottom of the sea would tend to compress the valve actuator housing and force the piston to move. The housing will collapse more on the right side. So yes.

 

[LittleInch]

Absolutely nothing would happen.

You've connected the two ports together so any change in pressure would be the same on both sides of the piston so there is no differential force and no movement.

Volume is totally irrelevant - its differential pressure that moves things.

also there is no path for external pressure to influence anything in the piston.

Now if the port on the right was open to the sea and the port on the left sealed, then there would be a very small movement to the left, but it would be very small if this whole thing was filled with liquid.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[NBrink]

LI, do you not have to consider the area of the stem exposed to higher external pressure? Would think it would open...

Nathan Brink

 

[Compositepro]

If you have dual actuator cylinders opposing each other they should not move. A single cylinder will move because the cylinder rod area will apply more pressure to the gas in the cylinder and compress it. But with a dual cylinders ambient pressure will have no effect if the cylinders are identical.

 

[NBrink]

I'm not sure I follow... Why would the piston not move all the way to the right? The piston may as well not exist (since the chambers can communicate with each other). Differential pressure across the stem should move it? No?



Nathan Brink

 

[Compositepro]

For a single cylinder it will move. How far it moves would depend on how much the ambient pressure rises and what is the pressure in the cylinder to start with. As the piston rod enters the cylinder the gas volume in the cylinder must decrease (the gas gets compressed).

 

[NBrink]

Right, but since they're communicated, it'll behave just like a single cylinder.

Nathan Brink

 

[Drexl]

It will move until the forces equalize.

 

[Compositepro]

"Right, but since they're communicated, it'll behave just like a single cylinder."

I don't understand your point. Please elaborate. But what ever way you mean this, I think you are wrong. If you elaborate you may see why.

 

[NBrink]

Alright, here goes. With the equalization line between both chambers, the piston may as well not be there (excepting of course a small amount of additional friction).

Go ahead and picture removing the piston (since it's not doing anything). All you have is a rod going through some packing into a chamber of compressible fluid, effectively creating a single acting piston. Now, if there's a pressure difference across that rod, it'll move, no?

Not my bailiwick, but I can't see it any other way. Why isn't this moving?

 

[LittleInch]

The diagram isn't great so it's not easy, However once you connect both sides of a piston like that together its not going anywhere.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Compositepro]

Nbrink, my previous post addressed your concerns.

"For a single cylinder it will move. How far it moves would depend on how much the ambient pressure rises and what is the pressure in the cylinder to start with. As the piston rod enters the cylinder the gas volume in the cylinder must decrease (the gas gets compressed)."

Yes, the piston in the cylinder acts as though it is not there except for friction. So the rod will move until the pressure in the cylinder equals the pressure out side. The rod moving into the cylinder will increase the air pressure in the cylinder.

LittleInch, I have not seen the diagram but we are talking about an air cylinder not a hydraulic one, so it will move.

 

[BigInch]

Sea water pressure acts on the piston rod area.
Internal pressure acts on the annular area on the left of the piston.
Internal pressure acts on the total area of the piston on the right of the piston.

Charge the cylinder with some mass amount of compressed air. The pressure will be P = nRT/V
V doesn't change.
R gas constant doesn't chagne.
T might change, in which case P might change, however if T remains constant and

the charge's pressure is,

1) less than sea water pressure, the piston moves all the way right.
2) is equal to sea water pressure, the piston moves not.
3) greater than sea water pressure, the piston moves all the way left.

 

[JohnGP]

Type of valve is not indicated, but I'm not seeing that sea water pressure is acting on the piston rod area. Unless it can, then the piston won't move with the ports connected like that. I suppose if the external pressure was sufficient to collapse the cylinder, the piston may move a little to the left, but what's the point of that?

 

[hydtools]

The compressible fluid (gas) acts as a spring on the piston rod. The force of the gas spring is the product of the gas pressure times the area of the rod. With no force acting on the left end of the rod, the rod will extend. If some force is added to the end of the rod, it will push the rod back into the cylinder if the applied force is greater than the internal force pushing the rod to the left. Since the internal and external pressures are acting on the same rod area, the rod will not move if both internal and external pressures are equal. If the external pressure resulting from submergence is higher than the internal pressure, the rod will be pushed into the cylinder until the resulting volume reduction increases the internal pressure to equal the external pressure.

With both cylinder volumes connected together, the net operating area of the piston is the area of the rod.

Ted

 

[BigInch]

JohnGP the rod is exposed to the ambient pressure outside of the cylinder which will force the piston into the cylinder until (if ever) internal pressure x (piston_area-annulus_area) equalizes with the seawater_pressure x piston_rod_area. That might happen before the piston reaches the end of the cylinder.

hydtools, I think we agree.

 

[LittleInch]

Ok, it was abit of friday afternoon response and I missed a couple of things.

SO

The connection of the cylinder ports in essence makes this a rod being inserted into a cylinder of fixed volume Vc. - As pressure acts on the remaining area in equal measure.

The force on the rod is therefore Pc x Ar = Pressure in the cylinder times area of the rod

If the fluid is compressible (something I missed) then as the rod enters the cylinder, the pressure inside the cylinder will change due to the volume of the rod reducing the volume available for the fluid. We don't know the relative volumes Vc and Vr, but unless Vr was close to Vc then the pressure rise wouldn't be big. If however this was virtually incompressible fluid then the pressure rise would be much much greater.

Hence if the pressure acting on the end of the rod Pr, is less than Pc, or force is leass than Pc x Ar then the piston/rod will move left in the diagram. If pressure Pr > Pc( a free end) or force on end of rod > Pc x Ar, then it will move right until it either hits the end of the cylinder or PC becomes equal to Pr or the force is equal to Pc x Ar.

SO if it is just the actuator cylinder dropped into the compression tank then yes, it will move right.

BUT, if the end of the rod is either fixed to something or is not exposed to the external pressure then it won't move. You also need to factor in friction and the forces on the end of the rod (an unknown unknown).

The question was "will this valve move", not will a rod move. I actually don't know because there are too many unknowns.

I note the OP calls this actuator cylinder thing a "dual actuator valve". I don't see any valve anywhere so it's all a bit confusing....

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[hydtools]

To answer the OP question, it will move depending on the balance or imbalance of forces on the rod.

Ted

 

[JohnGP]

My take on it was that the piston rod end may not be exposed to external pressure. If it is then I agree the piston could move.