Quote from: see3d on September 08, 2012, 12:08:34 PM
An interesting experiment would be to measure the difference in operation between a 1, 2, 3, or 4 riser system, keeping everything else the same -- assuming you don't blow the skirt of the last one.

It would be indeed interesting, it is also not clear to me how adding layers helps here. I think it helps increase COP of device  so for same amount of water injected you get more work out but I still need to prove it in numbers. I'm bit stuck in calculating work done for injection of water from bottom.
respect,
Marcel

Quote from: webby1 on September 08, 2012, 12:21:25 PM
I will do what I can to help.  I need to know what the setup needs to be...

Thanks webby,

I can input any dimensions into the simulation for a single riser ZED at this point. 
I would have to know the exact diameters of your parts (or average if not quite round):
    Pod
    First Tank ring ID or OD, and wall
    First Riser ring ID or OD, and wall
    Exit Tank ring ID
    Internal Heights of parts
    The gap at the top and bottom of the walls when at the bottom most position
    The riser/pod combined dry weight

My sim usually starts with:
    The Pod at the bottom
    Water just touching the bottom of the Pod, or as deep a needed to make it start to float
    Zero air PSI
    Pocket water at about 75% of the way up the total riser height

However, I can adjust things in the sim as needed.

The measurements taken would be for a load weight that gives a meaningful stroke without blowing the skirt:
    3 water heads and riser stroke for each of the following -- the amount of water added would be calculated from the Pod head:
        When the Pod first moves off the bottom
        1 or 2 partial stroke points.  These could not be averaged, so must be just individual single point observations
        At the top of the stroke

Averaging several measurements at the bottom (just lifting), and top (just touching top) of stroke would help with accuracy.

I would have to back calculate the actual input force and air PSI from the geometry and internal states.  I would in essence play with inputs to the sim to match the observations.

Thanks,
Dennis

Quote from: TinselKoala on September 06, 2012, 08:04:13 PM
So the Zeds aren't "transitive" then? Are the two Zeds different, so you couldn't take the one on the right and put it where the one on the left is?

Because if A=B in the first Zed,  and D is cycled around between the Zeds..... then why doesn't C =B as well...... that is, C must equal A. So if C+D = A.... then therefore.... D = 0.
If the two Zeds are the same, and a cycle is being performed, it seems to me that this must be true.

@TK, let us take another look at the Diagram By MT. This is my analysis. Figures quoted are just examples.Losses are ignored.
Let A = 12 units of energy.
B= 12 units.
C=6units-  .These 6 units come from the B above Zed one, leaving 6 units as output.
D= 6units which are added to C to form the total input to Zed Two.
That makes sense to me . Does it make sense to you?

@MT: Thanks for your drawing, I know how hard "simple" computer graphics can be.  I think you are right about the PE change being equal to the input work, this is as it should be.

However.... you are showing water flowing from one "simple zed" into the other. We've been told several times that this isn't the case in the actual device. The water flows out into an external bag for one zed and each zed has its own bag, and these bags are pressed on, or they press back against, some mechanical/hydraulic leverage system, I think, right?
There must be a reason for this, otherwise the simplification you have shown would have been used in the real device. I don't know what difference it makes but there must be one, and if we keep ignoring the bags in our sims and sketches we might never understand what the bags are actually for.

@Webby: You've described the perfect system for getting actual work in/work out measurements, using your elevated reservoir, going up and down to get the water in and out of your zed and making the load mass rise and sink.

When you repeat the cycle, do you find that you are lifting and lowering the elevated reservoir the same amount each time? (This is an important question, not just from the standpoint of repeatability of a cycle. If there is excess energy being created, or evolved or whatever.... where does it go to, in your system?)

So, if you've got a repeatable setup, and the practice in making it work, you could take some measurements, like I suggested for mondrasek. Load and "prime" your plumbing so that you have a little water left in the elevated reservoir at the end of the input, so you can get a good measurement of the overall quantity being transferred. Then do the cycles, raising and lowering the reservoir to the required height smoothly and quickly, and holding it there until the water is transferred. If we know the amount of water and the heights of the centers of mass of the "slug" of water in the reservoir at the bottom and the top, we can calculate the input work, I think, if I am understanding your arrangement correctly. I think we can neglect the water that remains in the plumbing tube from the elevated reservoir to the zed input port.
So it's as if the reservoir you are raising and lowering is the "bag" and you are providing pressure by manipulating the head height, instead of pressing on the bag with levers, I think, right?

Quote from: neptune on September 08, 2012, 01:47:49 PM

@TK, let us take another look at the Diagram By MT. This is my analysis. Figures quoted are just examples.Losses are ignored.
Let A = 12 units of energy.
B= 12 units.
C=6units-  .These 6 units come from the B above Zed one, leaving 6 units as output.
D= 6units which are added to C to form the total input to Zed Two.
That makes sense to me . Does it make sense to you?

No, not quite.
Let A = 12 units, fine.
B₁ = 12 units, also fine--- until the next step.

C = 6 units.... from the "B above Zed one, leaving 6 as output." so 6 are in C, input to Zed two, and six are in B₁, the output, right? So we've input twice the amount of energy units than we get in this Zed's B output, and half of this input energy is actually going over to the second Zed. Right? Half of our input 12 units winds up in B₁ and the other half winds up in C, going into Zed two.

D = six units which are added to C..... six units coming from where? Out the side of Zed One? Or from B₁, leaving nothing in B₁?  Where does this D come from, if not from the "D" that is coming in from the left side, the recycled D from Zed two?

I agree that C+D should equal A. But you are either using the same D to add to C in the second zed that you have input into the first Zed before the cycle, or you are making it from nothing. Or.... it is equal to zero _in this schematic_, which makes more sense to me. In other words, I see no mechanism for any gain in what has been described in this sketch. Not only that, you seem to be left with only six units in B₁ unless you add D to them, no matter where D comes from ....