Knitel's InfinityPump

[Thaelin]

   In reality on the 8" tube, all you need to do is use a short piece of PVC and then cut out
a clear window just to see the top entry point of the side pipe. Two 8" caps and a few fittings
will do it. Just have to figure out a way to trigger the valves inside the sealed unit.
   I am almost wondering if light oil would be useful here. The seal will still be a concern for me
to come up with. Felt for a quick try but will not last unless in a light oil. "O" rings would be good
but where to get one that big? 

thaelin

[Onevoice]

Thaelin, you and I are thinking in parallel. I was figuring a light mineral oil instead of water. More viscous, better sealing and less evaporation. That would affect the buoyancy equations. I also had an idea for a pressure compensator. I we put a hollow rubber ball inside the piston, it could compensate for the difference in pressure.

When I was a kid, we had an old tirepump that had a leather gasket. This and the backflow valves would work better in oil.

[hansvonlieven]

@ Onevoice,

You are missing the point. In a hydraulic system the pressure, apart from the gravity component, is uniform in all directions throughout the system That means that the pressure beneath the piston is exactly the same as the pressure in the top reservoir.

That is not the point though.

In order to get the system to cycle you must get rid of the air in the top reservoir to make room for the air beneath the piston.

That you cannot do as you cannot vent from a low pressure area into a high pressure area without some sort of a compressor. A compressor capable of doing this would consume many times the energy you are likely to generate.

Hans von Lieven

[Onevoice]

Simple experiment. Take a closed cylinder filled with water with a flexible membrane across the middle. Heat up one side, the membrane will deform due to the differences in pressure.

By the same token, a rigid membrane being shifted by gravity should be able alter the pressure on either side. It won't really be higher on the bottom because there is an outlet flow. Now, the interesting thing is that there is an intake flow at the top too but since that one is essentially acting like a siphon, the air pressure would be a little lower at the top. Not uniform on both sides.

My idea was that the evacuation is being done by a compressor. The piston has a positive buoyancy, so it rises. how high it rises is the question. It needs to rise high enough above the top level in the tank to displace a volume of air equal to the amount of air under the piston, give or take the relative pressure differences in the two areas. The math for this is way of my knowledge level.

If its not enough, there are two ways I can think of to mitigate this:

The first would be some kind of compressible object under the piston, like a small balloon that can absorb some of the pressure differential.

Second, since new fluid is pulled into the system during the downstroke. If an insufficient amount of air is pushed out, then that air will displace fluid each cycle and the pump will slowly go dry and stop. If there was a secondary pump that was actuated by the rising of the piston, it could make up for the difference by pumping a small amount of additional water into the top during the up stroke. It seems like almost all of the potential energy of the rising piston is wasted. If we could capture a little to do a little work, it would be easier to close the loop.

[hansvonlieven]

You are already expending far more energy than the system is capable of generating. A loss situation!

Hans von Lieven