Hydraulic vs pneumatic piston/cylinder design

[LaurenceSachs]

Hi there,

I need to understand something that I hope you can help with!

I work with water powered machinery as well as pneumatic machinery and I have found that there are specific designs required for each instance. Obviously this is due to the incompressibility of water compared to air's compressibility.

Note: I have attached a quick drawing for reference.
The left hand drawing is of a pneumatic cylinder and the right one is of a hydraulic cylinder.

Now as you can see the pnuematic cylinder is common sense. The stroke is related to the length of the cylinder up to the exhaust where pressure is released. The air decompresses very quickly and so only needs a small exhaust at the end of the stroke for this.

However the hydraulic cylinder, the piston move even though there is no pressure difference in the second portion of movement where the cylinder enlarges allowing the water to flow over the piston.

Due to the fact that the water is "incompressible" the whole pressurised volume needs to be removed so that the piston can return to the start of the stroke.

How on earth do I calculate the force exerted by the water piston at the end of its stroke?

Is it related to momentum of the piston?

How do I calculate the correct length of the initial pressurised area?

Please help me understand this!!!

Regards
laurence

 

[hydtools]

The force at the end of the stroke is the water pressure times the area of the piston rod.

'How do I calculate the correct length of the initial pressurised area?' To do what? What do you want to have happen at what stroke?

Ted

 

[LaurenceSachs]

Hi there,

How can the force be related to pressure at the end of the stroke when the water is running over the piston and being exhausted?

This design is used to turn ratched a rotor, however it is also being used for the piston in a percussion drill.

Laurence

 

[israelkk]

The pressure of the water can be calculated if you know/measure the mass flow rate of the water around the piston (can be calculated from the piston velocity). Use the annular area around the piston as the "orifice" area to calculate the pressure.

 

[hydtools]

The hydraulic cylinder could look like the pneumatic cylinder, where the piston uncovers the exhaust port.

How do you return the piston? In both cases.

Is there valving involved in the circuit? Some means to cause the piston to move back and forth?

Is flow stopped in both pneumatic and hydraulic cases? Does the pneumatic piston move because you allow a quantity of air to just expand after flow is stopped?

Some pressure exists in the hydraulic case as flow exits the exhaust port.

Are you driving a racheted rotation to turn a drill bit as in a rock drill? Does the piston rotate as well as reciprocate and hammer?

We need more information about what you are trying to do.

Ted

 

[LaurenceSachs]

Thanks for the replies!

To clarify:

In the hydraulic case, the piston physically closes a valve which then allows water to flow into the front of the piston making it return. I am still unsure how this works because the valve is only open for a very short amount of time.

For the pneumatic case, the exhaust is in the middle of the cylinder. The back chamber is pressurised until the piston passes the exhaust at which point a valve at the back of the back chamber closes and air is then directed to the front chamber, returning the piston. ie Secan 250 percussion rock drill. The piston rotates on a rifle bar.

In both cases the flow is not stopped, just redirected.

Any ideas how to explain this clearly?

Please explain how mass flow is used to move a piston?

Laurence

 

[hydtools]

Hydraulic drills generally have very short strokes, ~.75inches, compared to pneumatic drill stroke, ~ 1.5 to 3 or 4 inches. Valve open time will be much different between the two. These would be handheld drills.

Since pneumatic drills exhaust to atmosphere, they do not need the auto-valve to handle the exhaust, only the pressured air. The exhaust can be in the middle of the cylinder and the piston performs the exhaust valving function.

Hydraulic drills do not exhaust to atmosphere and need an auto-valve and internal passages to handle return flow. Constant pressure applied to the rod end returns the piston when the head end is exhausted to return. Hydraulic drills generally do not have rifle bar rotation. They use independant rotary motors to drive the drill steel rotation.

Ted

 

[LaurenceSachs]

Thanks Ted,

The hydraulic drill in question has a piston stroke of just under two inches however the pressurised section as per drawing is only about half an inch.

Do you know of any reference material I can have a look at that will fully explain the hydraulic workings?

Who don't hydraulic drill use rifle bar rotation? As you correctly mentioned, the drill uses an independent system that rotates the chuck.

I have many questions about this section too.


Laurence

 

[LaurenceSachs]

Israelkk,

Tell me, how do you calculate the pressure once you know the mass flow rate passing the piston? And what do you do then?

Laurence

 

[hydtools]

The rifle bar mechanism uses two sliding splines, one with a twist that is part of a ratchet and one that is straight that is part of the piston. On the backstroke, the ratchet locks and the twisted spline causes the piston to rotate which drives the steel chuck rotation. Sealing for hydraulic application would be problematic as would moving fluid in and out of the spline cavity in the piston. Pnuematic drill has no piston seals.

Ted

 

[israelkk]

Standard flow rate formula of liquids through orifice.

Q=Constant*SQRT(Pressure difference over orifice)

 

[LaurenceSachs]

I think I have the mass flow rate around the piston but I don't have the pressure difference, its not easy to measure.

Is there a law that states that the piston will move the same speed as the water in that cavity? I can't see that being true??

I do not know how the piston is continuing to thrust.

L

 

[LaurenceSachs]

Ah yes, One can use the continuity equation in bernoulli's to get the pressure difference across the piston. If you have the flow rate, which I have!

F=P.A so we can work out the force being exerted on the piston.

Am I on the right track?

Laurence

 

[LaurenceSachs]

Are we talking the same concept as Rotameters (variable area meter)?

http://en.wikipedia.org/wiki/Rotameter

 

[hydtools]

Not quite the same as a rotameter. The flow around your piston exits to one side, flow will not be uniform around the piston.

How do you expect reversing flow to retract the piston in the hydraulic case? Pressure on the rod side of the piston will be the same as pressure on the head side of the piston and net force will be to hold the piston extended.

Ted

 

[hydtools]

Another observation. In the case of the hydraulic cylinder, the piston speed may very well increase since the effective area of the piston after it passes the constant area portion of the cylinder is the area of the rod. Assuming constant flow into the cylinder, the reduced effective area will have to speed up so that the displaced rod volume rate equals the incoming fluid volume rate. Depends on how much flow exits the cylinder as the piston continues to extend. Exiting flow rate and rod dispacement rate must equal incoming flow rate, the fluid being incompressible.

Ted

 

[LaurenceSachs]

Hi Ted,

There is another chamber further down the piston rod that is only pressurized when the piston is fully forward. This is because of the valve that is opened by the piston in the full forward position.

What about the friction and counteractive forces that act against the piston? how do I take those into account?

Laurence

 

[hydtools]

To take all forces into account, you need to determine what they are and begin working with free body diagrams to resolve the forces. Then work on the dynamic response.

Does the main piston have to move fluid out of the other chamber as the main piston moves forward? You then need to account for flow losses and resulting pressure/force acting against the movement of the main piston.

Is the hydraulic case an existing device or an idea for a device?

Ted

 

[LaurenceSachs]

Hi,

Yes the piston has to push the water in the opposing chambers away, its complex.

This is an existing device.