Big try at gravity wheel

[MarkE]

Thank you, "minnie."  Is there any chance that you can assist by checking my work as I have requested?  Because unless I've made some mistake in the math, or the analysis process, I think I have uncovered something very interesting.  At least it is interesting to me.

M.

Mondrasek, I am looking at it now.

[MarkE]

What a load of rubbish MarkE.

The issue I have is with you "enhancing" what I put forward.

The transfer problem is yours,YOU created it when YOU took 2 discrete volumes of air and made them into 1 shared volume.

Webby I have faithfully followed what you have described.  I am afraid that you simply do not understand the physics.  There are but the two cylinders where you can store energy in the problem as you set it up.  The pressure lock in the pump does not change the fact that once you have transferred enough "air" to drive the meniscus in the "A" cylinder down by 67.5mm, the meniscus in the "B" cylinder has risen by 67.5mm and the two meniscus are even.   At that point there is no pressure differential across the pressure lock in your pump.  At that moment, the total stored potential energy in "A" cylinder is 790uJ, and in the "B" only 3.110mJ of the original 7.408mJ remains. You have lost nearly half the energy originally stored in the "B" cylinder.  Go ahead and try to diagram a different result.  Find a way to make your transfer pump preserve the pressure in the "B" side as you withdraw the "air" and the meniscus rises.The pressure lock in the pump is of zero consequence until you start pumping beyond the equilibrium pressure point. 

The pressure lock only serves a useful function past the equilibrium point and we perform even more work to increase the pressure on the "A" above the falling pressure in the "B" side.  Up to this point a wide open pipe between the two cylinders is equally effective as the pump. 

[MarkE]

This is the 3rd choice you have made that changes the design I am putting forward, each of those choices change the operation of the system into a non-functional arrangement.

You are choosing to ignore that with 2 discrete volumes of air that the ratio of volume change is set by the transfer pump,, unless you are going to tell me that the volumes can only be a 1:1, in which case I will need to see the law that states that, but that does not exist.

Since this appears to be your trend, I conclude that the choices are deliberate.

No webby you are ignoring the physics.  The "air" is incompressible.  At all times the volume of "air" in the system is constant at 5.0363cc.  The transfer pump cannot create or destroy the amount of "air" in the system.  It simply displaces it from one port to the other.  Draw 0.01cc out of "B" into the transfer pump and the pump then drives that same 0.01cc into "A".  Since both "A" and "B" are held in place up to this point, the things that move are the "air" and the water in the annular rings.  For every um we drive the meniscus down the "A" annular ring, the meniscus in the "B" annular ring rises identically by 1um.  This remains true until we get to the point that the meniscus reaches the top of the piston in "B".  After 67.5mm movement in each, the meniscus is at the same height in each.  The volumes and pressures are as shown in the diagrams, as are the stored energies.

It is on you to show the configuration you claim will perform as you claim.  By all means be my guest and show that you can get to the point where the meniscus' are even without losing nearly half the stored potential energy.

[minnie]

     Webby,
              all we need from you is a diagram of a single ZED that gives back more than
   the pre-charge. If you can't get more out than is put in that's the end of the story.
   period.
            John.

[mondrasek]

Mondrasek, I am looking at it now.

That you are looking at it is much appreciated.  Please let me know if I have been attempting an incorrect analysis method or if we have uncovered an apparent anomaly.  When you can.

Again, thank you so much for your participation and efforts.

M.