Force of a pneumatic piston

[JaysAuto]

How much force would a 2.5 inch diameter 4 lb. piston traveling 6 inches pressurized by 150 psi. be hitting a stop? I know the force of a 2.5 inch piston with 150 psi is 736 lbs.

Thanks
Jay

 

[hydtools]

Use work = energy to calculate velocity at impact.
Use impulse = momentum to calculate force acting over time to stop. The time to stop is the tricky variable to guess or want by design.

Ted

 

[JaysAuto]

Ted, that may help. the intention is for that piston to hit then end of it's cylinder as hard & fast as possible. Something like an encased jack hammer.

Thanks
Jay

 

[FeX32]

To push in the right direction mentioned above.

W=integral(F*ds)
W=736.2lb*(6/12)ft=368.1 lb-ft
Ke=0.5*m*v^2
368.1lb-ft=(0.5)*(4/32.2)*v^2 => v=76.98 ft/s

736.2/(4/32.2)=5926.41 ft/s^2

s=0.5*a*t^2 => t=1.687*(10^-4) sec.

momentum. assuming linear force not varying with time

m*v=integral(F*dt)

(4/32.2)*(76.98)/(time to stop)=F

pick a time to stop as Ted said. Ex. pick half the time it takes to go from rest to 6in in distance give about 113.4 kips.

Note: this assumes a completely constant pressure behind the piston with absolutely no losses.
Also, the assumption about the constant linear force not varying with time is not really correct but the analysis cannot be concluded without making something of the force.

This is not the only way to do this, there are better more extensive methods that are not worth the time in a lot of cases, its just the only way that takes 3min. and can give you a feel for the problem you have.
ttyl,





Fe

 

[FeX32]

I forgot to mention that the force also depends on the stiffness/damping of the 2 materials that come into contact.
Keep this and how you might use this in the above simple analysis in mind. (to give you a hint, you can play with the dt by playing with the k)



Fe

 

[hydtools]

When we designed impact hammers we were only concerned about the impact energy and not the force at impact. We knew that impact energy and impact rate caused some level of rock or concrete destruction which was our goal.
Not all impact energy is absorbed, some is returned in rebound.

Ted

 

[JaysAuto]

hydtools & Fex32 thanks for the time, knowledge and examples. I'm sure I can take it from here.

Thanks Again
Jay

 

[FeX32]

np

Fe