Force, Area, Pressure Question: Positive Displacement Pump 1

[gorilla1969]

I have an exercise I'm working on and had a question wiht regards to P = F/A. If I have a pump used to compress oxygen in order to increase pressure to an external cylinder, which surfaces would incur force as the piston in the pump compresses the gas?

Within the chamber holding the gas (there would be a tube/cylinder wall) and a piston within being pulled back to compress the gas. Would the force to be overcome (which needs to overcome the pressure in the cavity as the piston moves back) be only on the surfaces (area) in which the piston is moving (in that specific direction) or does the pressure on any opposite walls help cancel the forces seen in the direction of movement and this should be taken into account.

Any insights would be appreciated. Thanks.

 

[hydtools]

Pressure is the same on all surfaces therefore, forces are the same on all surfaces. Only the surface which is moving is doing work of compressing the oxygen. Compression takes place when the piston moves toward the opposite fixed surface. There is no balancing, cancelling of forces to reduce force required to push the piston.

Ted

 

[gorilla1969]

Thanks a bunch.

So, just to clarify, the force needed to be overcome the pressure would be related only to the area of the walls inside the cylinder in the direction of the movement.

The attached graphic will give a better idea of the exercise. So the surfaces outlined in green will be the only areas (A) of concern needing to overcome the pressure using P = F/A and pressures on other surfaces do not play a part in determining the force (F) to move the piston?

Thanks again. Just want to be clear on this.

 

[JJPellin]

That is not correct. The configuration of your piston is more complex and makes it appear as if the problem is complex. The area to be considered is the area of the bore of the cylinder minus the area of the rod that passes out of the cylinder. The other small surface is irrelevant as it does not seal to the bore of the cylinder. And even the large surface on the piston is only partially relevant. The part of it that is within the circle of the rod diameter is cancelled out by an equal and opposite force. I modified your drawing. Red cancels out red. Blue cancels out blue. All that is left is green.

If you want to prove this out do a free body diagram of the piston assembly. For each axial surface, determine the pressure acting on that surface and the area it acts upon. Sum it all up being careful of your sign convention.


Johnny Pellin

 

[hydtools]

Pulling the piston outward will not compress the gas. More material volume is leaving the space, therefore the cylinder space is increasing not decreasing. See the modified illustration by JJPellin. The red volume, the net effective volume is increasing, expanding the cylinder volume. The cylinder volume must be reduced with piston movement in order to compress the oxygen.

Ted

 

[btrueblood]

Hmm, maybe hydtools, but I interpreted the entire blue piston to be moving as one piece, i.e. no axial seperation of the two red surfaces occurs...

 

[hydtools]

The larger rod is exiting and the smaller rod is entering the cylinder volume. Net change in cylinder volume is increasing.

Ted

 

[hydtools]

Disregard my previous post. I missed seeing that you have a piston at the left end of the multiple diamter piston rod. See JJPEllin's description for the correct analysis. Unless the flange feature between the two rod features is required for some mechanical reason, say as a stop, it serves no purpose relative to the compression function. The large diameter rod could be extended to the piston and you would see JJPellin's analysis.

Ted