Hydro Differential pressure exchange over unity system.

[wildew]

Neither are sitting perfectly flat.
That is / was a quick setup, mainly to get a feel for how things need to be arranged.
First item is a label for the "Tank" as per Wayne's request.
Then a real stand instead of stacks of scrap lumber
And relocating the screw-jack holding the little input cylinder lower so I can increase the volume of water in it.
M. and Webby were right, even with just 1 riser and 3 water levels it's tough to really "see" what's going on.
After adding a couple more of the risers ( all the same length ) it will be nearly impossible....

Thanks for the compliment
Dale

[TinselKoala]

OK, so the problem will be to see where the water levels are within all the layers. I would suggest the following, if possible. Try to find or make some transparent line gauges. You could lay them out with a graphics program, print them onto transparency film and cut them out in strips, glue them right to the individual tubes, staggered around the circumference so they don't block each other's view. Then use some kind of different colored "floaters" like tiny little plastic beads from the crafts department at WalMart. Just a few or even one in each gap space will show where the surface is, and the color coding will help tell them apart, and the line gauges will allow a reference for both the water level and the height of the floating parts.

Oh, ETA: and I'd suggest not discarding or neglecting data just because you don't think it's useful now. I think you should measure everything you can, even the ambient temperature of the air and the water temperature. Heights of all moving parts before and after lift, certainly.

ETA 2: a laser pointer, aimed around in there at various places, may make things stand out where you couldn't see them before.

[wildew]

Some good suggestions in those last few posts TK.
All noted.
Thanks
Dale

[mondrasek]

I'm a few pages back in the thread still trying to catch up, but I wanted to post the reply below before I forgot where this was...

OK... so start with M in the full up position (call this H_M) and some small amt. of water in the gauge. Mark the CofM of the water in the gauge, height h=0. It's OK, I think, to neglect the water in the tube at this point.
Now open your drain valve and let M sink and settle, collecting the water expressed in your cup or whatever. Close your valves. Weigh this water and record the weight. Lift up the cup of water using a hydraulic forklift, chain hoist, a jetpack, or your hand and arm. Pour the water into the rain gauge. This will raise the level of the rain gauge's water CofM by some amount to height h=h_W. Measure this amount and record it.
Now open your valve to allow this water to run back into the Zed pod chamber, raising the M mass from height H=0 to height H=H_M. Record.
Lather, rinse, repeat 20 times to get good data for stats. Make a data table; crunch to means and SDs, calculate PEs and deltas. Report.

TK, what you describe is not necessary, IINM.  The fact that I am removing water by allowing it to vent from the bottom of the Pod chamber is only due to the construction of the test system.  In this case, yes, the vented water must be raise to be reintroduced into the fill tube.  And that water is required to be vented in order for the ZED to sink and return to the initial pre-lift starting condition.  However, the venting of the water could also have taken place directly from the top of the water in the fill tube and at that same level if the system was designed for that more difficult venting option.  And if vented from the top of the fill tube it is clear that the vented water does not need to be raised or lowered (change of PE) to do so.

I believe the total of all input energy required to cause the previously described output is defined by the water volume and change in fill tube water level head that was reported.  Please let me know if you disagree.

I think that what can be confusing in this case is that I vent "high pressure" water from the bottom of the Pod chamber.  That pressure is lost from the system (and not accounted for) when this is done.  So that energy must be reintroduced by raising the vented water before reintroduction.  When we consider venting at the same level the water is introduced as input there is no difference in the pressures and no energy is lost.

Thanks and good morning from Corolla, North Carolina!

M.

[neptune]

This is a problem that needs to be given very careful thought to arrive at the right answer.
Here is a thought experiment that is an equivalent. Imagine a tall cylindrical tank with a valve in the side wall near the bottom.The tank is 12 inches tall and is filled . Problem. How do we remove the top 2 inches of water from the tank, and then replace it, using least possible energy. We could open the valve and run the water into a vessel at floor level, then raise that vessel to the top of the tank and pour the liquid back into the tank. Or we could attach a hose to the valve, with its open end 2 inches below the top of the tank, and place the vessel just under the hose outlet. We then open the valve and fill the vessel. This time we only have to lift the vessel a small distance to pour its contents back into the tank. And we did not have to add extra plumbing to the tank to achieve this.
        Another half formed idea is this . Fill and empty the Zed simply by lifting and lowering a water reservoir connected by a flexible hose to the Zeds input. If we lift it with a pulley and weight, measuring input is easy. Just some random thoughts...