[Olivia]

Does anyone know how to calculate the pressure that a air cylinder can generate. I want to use a cylinder to generate pressure by forcing the shaft down to force air out the port, will this vary with temperature.

[diyengineer]

Area X Pressure = overall force. I believe. google it.

[Olivia]

I am using a air cylinder to make pressure ie. compressing the rod down forces air out the 1/4 port what will that pressure be if it is 20mm tube with 400mm stroke.

[RICHARD ZASTROW]

Area of circle = 20mm (dia.)= r (radius) 10mm
pi = (3.1416) X r squared (100) = area (314.2 sq.mm)
Force being applied is missing from the equation. How much force will you apply to the rod? This will be the PRESSURE at the outlet.

Dick Z

[pointcloud]

In this senario you will need to divide the input pressure by the surface of the piston in the cylinder.

Example your surface is 314.2 if you use 100psmm your pressure @ the 1/4" outlet will be a whopping 0.31826862psmm

[vidio1]

Radius squared x pi x air pressure. IE a 2 1/2in cylinder on 90psi air pressure would 1.25 x 1.25 x 3.14 x 90 = 441.56 lbs of force.

Use the same formula to subtract the piston rod dia. for the retraction force.

[pointcloud]

vidio1 you would be correct if they were useing the cylinder that way.. lol.. they say they want to make pressure with the cylinder... so your math is way off for this application.. Although if you put 441.56 pounds of pressure on the rod then the cylinder would infact produce 90 psi...

No not reverse inside the cylinder... but on the rod, as if it were an air compressor...

[vidio1]

I see......That's what I get for skimming over a post an assuming I know the answer.

[pointcloud]

lol... That's nothing I have ever done! Yea, sure!! I still do that...

[diyengineer]

How much force are you looking to develop?

[Olivia]

So volume of air in the stroke of the piston plays no part in the equation but would play a role in the volume of air available at that pressure is this correct.

[pointcloud]

Yes.
Whatcha trying todo?

[pointcloud]

I ment no it doesn't matter on psi/mm...

[diyengineer]

So volume of air in the stroke of the piston plays no part in the equation but would play a role in the volume of air available at that pressure is this correct.

The volume of air is just there for the overall piston stroke length. You could have a 100" stroke with a 1" diameter piston @ 100psi and have the same power as a 10" stroke with 1" diameter @ 100PSI.

Look @ the link. select round body stainless cylinders. then look at the force say one has 12 pounds of force at 1/2" bore @ 2". Well the 4" 10" and 12" all have the same power but, have different volumes.

Mc MasterCar:
http://www.mcmaster.com/#standard-air-cylinders/=8x72zh

EDIT: "BUT" If they all have the same fill port size, then the cylinder with the greater volume is going to take LONGER to extend and retract. Do you need speed to or just power?

[vegipete]

I am using a air cylinder to make pressure ie. compressing the rod down forces air out the 1/4 port what will that pressure be if it is 20mm tube with 400mm stroke.

I don't think we have enough information yet. The rest of the system matters. For example, if the cylinder is open to the atmosphere, pushing the 20 mm piston through a stroke of 400 mm will generate no pressure. If the cylinder is corked right at the end, theoretically infinite pressures 'could' be generated. The ideal gas law says that pressure times volume divided by temperature is constant. In your case, approximately, the ratio that the pressure increases will be the same as the ratio that the system volume decreases. Note 'system volume', not merely bore times stroke. Note also that compressors have check valves. A bicycle pump is very similar to what you are trying to do.

[Andre' B]

Note that if this compression happens fast enough and to a high enough pressure you could easily generate temperatures that will quickly destroy the seals.

Or if your luck tends to the more exiting you have an oil mist inside the cylinder which ignites and everything blows up.

[mechtech10]

Olivia,

I think Vegipete is on the right track. Are you simply capping off the exit port on the cylinder? Or are you directing flow into a container or some other closed system? Ambient air pressure is 14.7psi. So use the ratio of initial volume of the fully extended cylinder (before any load applied) and your 'containment system' to the final volume of your system after you've applied your load or force. The ratio of initial voume to final volume gives your factor to multiply the ambient pressure by... If the ratio is 4, you should have developed 58.8 psi in your containment system - whatever that might be. Does your containment expand with the applied air pressure? - then this won't be accurate.

Also, if you are using air pressure to push from the rod end, then you must allow for the rod diameter - subtract rod area from cylinder area to calculate the force being developed by your cylinder.

It would help to know the application.

Brad

[Olivia]

The rod in the cylinder is being forced down on the rod end into the cylinder @ 390fps the force pushing the rod is 275lbs. I am burning seals out every time check valves are also burning out. It smells like burning oil after a single cycle. Does any one know of high temp seals for these applications, check valve and cylinder.

[vegipete]

390fps !!!!!!!!!!!!!

And you are surprised by burning? For rough comparison, a Chevy small block engine (3.75" stroke) develops maximum piston speeds of around 85 feet per second at 5000 rpm. No rubber piston seals there!

[handlewanker]

WTF......that's a canon he's designing, or perhaps a Diesel engine...LOL....on second thoughts with a 20mm bore and 400mm stroke and a lead ball traveling at 390FPS, that's a short barrel muzzle loader, LOL.

BTW, how do you get a 273lb weight to travel at 390FPS from standing start over a distance of 400mm and hit a 20mm shaft without missing it?

Why the 1/4" diam outlet hole...is that a touch hole for the fuse?
Ian.