pneumatic question

[JimboJones21]

Hello All, I'm making a little pneumatic press, I want to limit the speed of the ram so I put an adjustable restrictor in the line to the cylinder. I have set it to the speed I want, but my question is have I also decreased the amount of power the ram can do because I've decreased the speed? Thanks.

 

[bluesman0007]

JimboJones21,

A little ditty that might make sense in your application. "When in doubt meter out" If you are trying to control the speed you want to put the restriction (flow control, metering device, etc.) on the exhaust side. The line pressure acting upon the cylinder piston will be whatever it is when it stops no matter if it's restricted or not. Are you trying to use speed + pressure to do the job? You may need to increase the size of the cylinder. Maybe tell us a little more about your application.

Dennis

SolidWorks 2006 SP3.4
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[MintJulep]

"Power" involves units of time, so yes, if you have changed the speed you have changed the power.

However, if you really meant to ask about force, then no, you haven't changed the force because force is a function of pressure and you haven't changed the final pressure.

 

[Barry1961]

Blueman0007 is right. The common flow control for an air cylinder will be free flow in and meter out. They do this by putting a check valve in parallel with a needle valve. These are made by several companies and not expensive.

As far as the force available it will be the same once the piston has stopped and the pressure has built up to line pressure. While the piston is moving in either case there will be a pressure drop. How much pressure drop will depend on speed of the piston, flow controls and line loss.

Barry1961

 

[JimboJones21]

Thanks everyone, this is my first mechanical project. I am making an electrical connector that has a cam lock holding system. So the connector must be pressed into its sleeve. The connector is rated at a couple 100 amps, so it's big.

My question about power was because my colleague kept saying that power will not change if we change the speed of the ram, but I told him he was confusing that with force. Argument basically, just need to confirm. (P=Force*Velocity)

When in doubt meter out! Cool. Thanks

 

[Ashereng]

SMC makes a free flow in and meter out devise. It is called a speed controller (needle valve in parallel with a check valve).

http://www.smcetech.com/CC_host/pag...cfm?&cc_nvl=((searchPart,),(CC,SMC,CNN,3158))

Take a look.

"Do not worry about your problems with mathematics, I assure you mine are far greater."
Albert Einstein
Have you read FAQ731-376 to make the best use of Eng-Tips Forums?

 

[Fabrico]

Sounds like you already had the problem fixed. Just remember it will take a little longer to reach full pressure.

 

[zekeman]

I might add that the velocity will be stable depending on the orifice size you set manually. For this project, you should not be concerned with power, only that the final ram force at stoppage will be the area of the cylinder *pressure, independent of the orifice size.(you might note to tell your colleague that the power is zero at stoppage in any event) Before stoppage, the wonderful phenomenon of choked flow assures that the velocity remains virtually constant for a particular setting.
Don't bother with flow controllers, just change the orifice ( needle valve)by trial and error, and once set, it will be very consistant.

 

[zekeman]

On rethinking your friend's comments, they might have some merit if you want to use the impact forces of the ram mass to increase the instantaneous pressure on impact. Then you might want more speed and thus more power during the stroke to increase the kinetic energy of the ram mass which is equal to power times time. Then the equations(coupled accleration and fluid) that develop the velocity become a a little more complicated since the velocity doesn't reach the final equlibrium value immediately, nor does it have to in your case.