Think of a 4" cyl say 3 ft long. The piston will have a hollow shaft pushing it in and out with a 3/4" ID. The air will escape through the hole as it is pushed down and will reenter the chamber when it is pulled back out through the same hole. My question is will the effort needed to operate the cyl be about the same on both strokes even though one is pressure and one is suction? So if 50# pressure applied with depress it in 3 seconds how much force and time to retract it?
My first wild guess is it might be a little harder to pull the piston back outward. Just because the area around the shaft towards the open end of the piston always has ambient air pressure upon its surface area. The inner surface has more or less than ambient pressure when moving back and forth. Just a guess though.
Depends how fast you move it. The effort required will be proportional to the difference in pressures on the two faces of the piston. If you push the piston in extremely fast, you could generate very high pressures in the cylinder, potentially many times the atmospheric pressure on the outer surface of the piston. On the other hand, if you’re pulling the piston out very fast, the strongest vacuum you can produce inside the cylinder would be only 14.7 PSI below atmospheric pressure.
The slower the speeds you are using, the more equal the efforts for compression and expansion will be.
There’s a principle in steam engine design that vacuum gets you more work than positive pressure. Something on the order of -10psi vacuum is equivalent to +100psi steam, such that early steam engines worked on vacuum alone. I could be completely wrong about the physics, especially since it’s comparing atmospheric pressure/vacuum versus steam pressure, whereas the OP is talking only about air. Still, could there be a similar factor at work here?
Not seeing how this is possible. Consider engine braking, in which a vehicle’s engine is used to provide deceleration. This works on a gasoline engine because it’s difficult for the engine to suck air past a closed throttle plate, and they end up generating high vacuum in the combustion chamber during each intake stroke. A diesel engine doesn’t have a throttle plate on the intake, so without any modifications, it doesn’t provide good engine braking. To get good engine braking, big diesel trucks are not fitted with an intake throttle valve; instead, they use compression brakes, which let the engine build high pressure during the intake stroke, and then they suddenly release the pressure at top dead center so that they don’t get all that absorbed energy bring transferred back into the driveline:
Compression brakes allow an engine to absorb far more energy from the driveline then a throttle plate.
I am building a little cylinder for exercise, Going to use several size holes depending on what I am doing. I am hoping I can use the same hole both ways so I don’t have to fool with a check valve.
How big is the cylinder bore? If it’s going to be hand-actuated, then you’ll want it small enough so that you can actually pull a hard vacuum. A bore area of just four square inches means you’ll need to pull with almost 60 pounds to experience the limit of what’s possible under vacuum, and it will be difficult to feel the difference between that and pushing as hard as you can when the cylinder is under pressure. If you want to make it easy to tell the difference, shoot for a bore area of maybe one square inch; you’ll be able to pull a hard vacuum with just 15 pounds of force.
I was thinking of starting with around 4" diameter. I would think the smaller bores would compress too much too early in the stroke and I would have less effective stroke.I will probably have to put a one flapper in it for the return stoke so there is no resistance.
You can calculate the pressure due to flow rate your hollow shaft from a simple equation , given a flow rate…
see
Flow Rate And Pressure: Features, Relationship & Applications - Supmea Automation Co.,Ltd.
You can just set P1, V1, h1 , h2 as 0, and realise that your P2 is then the difference in pressure.
(Or just write it as P2-P1=PD pressure difference to keep it clear.)
Its for fluid ?? oh, well gas is just a thin fluid.
There is a difference between pressure and vacuum, if the differential pressures are more than a small fraction of the absolute pressures. The reason is that resistance to flow depends strongly on air velocity, which comes from the volume per time flowing through the system, and the same mass of air has a bigger volume when at low absolute pressure than when at high absolute pressure. If you look at compressed air systems you see big mass flows through small plumbing, whereas vacuum systems have tiny flows through huge plumbing. Very high vacuum systems might have, for example, a 6" bore going less than 6" to get enough flow between a vacuum chamber and a turbomolecular pump.
At those low pressures achieved by a diffusion or turbo vacuum pump (e.g. starting around 10^-2 torr or below) gas flow moves from viscous to molecular flow and “normal” gas behavior is gone.
You could be comparing the difference between negative pressure (vacuum) and ambient air pressure to the difference between positive pressure (steam) and ambient air pressure. If some equivalent level between the negative and positive pressure can be established then the difference with the ambient air would clearly be greater with vacuum.
True. It gets into the Knudsen regime where the gas molecules bounce off pipe walls more often than they bounce off each other. But that’s not the only thing going on. Vacuums nowhere near the Knudsen regime still need big pipes to transport much flow. And of course compressed gasses are the other way around.
This is where my math is lacking. I don’t know the terminology I need to research the formulas I need to be playing with. I will give an example. Suppose I have a 30 sq inch cylinder say 24" long. So I have 720 cubic inches of air I want to push through a 1 sq inch hole. I would like to be able to figure out how long it would take to do a full stroke on that cylinder with specified amounts of pressure against a specified variable hole size.
“Compressible gas flow rate through orifice” should get you started.
I’m thinking the OP probably needs the simple “spherical cow” Physics 202 algebra version of the formulas, not the real differential equation ones applicable to actual engineering calcs involving appreciable pressures, non-linear effects, compression heating, less-than-ideal gasses, turbulence, and all the rest.
At least if my guesses about what this magic secret device is for are correct. Something human-powered is not going to have enough oomph (technical term) to get into the regime where any of that stuff matters.
I will tell you exactly what I am oing, I won’t go into the configurations or geometry of how it will be used just the pistion and check valve assemblies. I want to build a piston with at least 30 sq inches, possibly as much as 120 sq inches. This will be extremely low pressure, hopefully low enough that instead of a piston with seals it will simply have an easy glide surface for less friction and a low clearance enough fit that some escaping air won’t affect the working resistance that I will need. The air will escape through the handle and in the T of the handle will be suction and pressure check valves with an adjustable orifice following. The idea is that the user will be able to adjust the flow on both strokes so he can maintain the tempo and force desired. Low tec and inexpensive. I am also working on an auto adjust where if the user is pumping faster than his setting it will close the orifice or if he is going slower it will open the orifice. The proto type models willl simply use a hose connected to a valve body where I can figure out all my hole sizes. I think PVC will work fine for the cyl but I may have to boree them for a cleaner surface.
Do you mean a cylinder with 30-120 cubic inch volume?
Sq inches, the cubic inches will vary with the length. With a 12" cyl I would have less than 5#
with with 500# applied to it.
The larger cyl makes the system much less sensitive to compression losses in the stroke.