Hydro Differential pressure exchange over unity system.

[see3d]

AND darn it any way,, here my numbers are getting converted into metric so I give metric and you want them in SAE.

Gads!  If your measurement tools and you are used to working in SAE, then by all means report them in SAE.  Let others who want them in metric do the conversions.  The reason I say that is because you are more likely to make a clerical conversion error to different units than you measure and it might not be caught by others.  I add a second error possibility when I convert them back to your original units, and I am pretty bad about making clerical errors, transposing digits, etc.  I will still look into having a metric output button though, because that is just a one time cost for me.

[TinselKoala]

@TK,

I've made no attempt to manufacture a complete miniature self running (dual) ZED system.  I have only constructed a single three layer ZED for testing of the input energy vs. output energy ratio as so many times has been requested and stated as necessary.  And I posted one admittedly hurried and sloppy data point prematurely just for those that wanted to try and make that calculation.  Why have you or anyone else not done so?

I have posted measured input values.  Are they not enough to calculate input energy?
I have posted measured output values.  Are they not enough to calculate output energy?

The input causes the extra ~2.5 lbs of "load" mass to lift.  This "load" mass is to simulate the hydraulic rams that produce the pressurized hydraulic oil that is accumulated in Wayne's system.  This "load" is then removed and not used while the system is reset to initial start conditions.  I think it is a pretty good first test of the model I have built.  Don't try to pretend that the model or the test is anything more or less than it is, please.

Could you please just run the numbers or do you just want to pick at it/me?  I'm on vacation, remember?

M.

Getting a little "tetchy" there, are you? I don't think I'm "picking" at you at all. I am responding to what you posted, and pointing out that you can't drain water out without a differential in pressure, and you can't drain water out and put it back in at the same point at the same pressure and still get work out of it,  which is what it appears to me that you have suggested in your post.

If I am not understanding you... don't PICK AT ME, just explain what you mean.

When you first described your apparatus I thought that the water went INTO the chamber itself at the bottom center, just like the Zeds and just like I have been told is correct. Pouring it in your fill tube at the top, letting it run down and then come into the chamber at the bottom, is not the same as pouring the water into the chamber itself at the top.

You are on vacation, I remember. What am I on? I'm strictly on my own time and my own dime.

Why haven't I made a measurement model of my own? I think you know the answer to that. Why haven't I run a calculation on your numbers like I did with Webby's? Perhaps it's because I am still not clear on what your numbers are and just how they have been gathered. Perhaps I'm not happy working with "sloppy and premature" data points. Are you going to make me look back through the thread, to find and decode the numbers you would like me to multiply together and compare? Or would you just list them again, or perhaps even do the calculations yourself.  I'm not averse to calculations.... it seems that perhaps I'm the only one that actually ran the numbers on wildew's nice little problem... but I really cannot recall seeing numbers from you that I could easily interpret and work with. Maybe I missed them, and maybe instead of flaming me, you could just point them out again, with explanations in one of the languages I actually understand.

And while you are at it, maybe you or somebody else could give me comparable numbers on MrWayne's simple three layer system that is clearly overunity by itself.

Or maybe not.

[TinselKoala]

On units: if you use Imperial units, it's easiest to use ounces and inches, or pounds and feet, and not mix them like "ounce-feet" or "pound-inches". 

If you use "metric" please use either SI, which is kilograms meters and seconds, or cgs, which is centimeters grams and seconds. The very best is SI because then you can use the named units of energy and force like Joules, Newtons, and so on, but for small weak systems cgs is fine.

Conversions are relatively easy to do, for example:

1 pound (force)  foot = 1.35581795 newtons meter

and here's a calculator that will do any unit system conversion you need:

http://www.unitconversion.org//unit_converter/energy-ex.html

[TinselKoala]

A cycle isn't complete until all parts and pressures and so on are back to the start state. If you have moved up and then back down and all your physical parts are back where they started from AND you still have excess pressure somewhere in the system, more than you started with.... please let me know right away.

And in the MrWayne Zeds, some energy IS expended to reset the system, by the hydraulic assist applied to the bags. It's not resetting just from the weight alone. In my way of thinking the assist is making up for losses; in the official Zed way of thinking, this is what you do with the extra energy or work produced by the first zed: you use it to help reset the second zed.

[fletcher]

@mond: why can't you just take your outlet "drain" from the top two inches, and plumb it directly into the bottom inlet, and have it flow around perpetually, extracting work from a little turbine wheel on the way? That is, I believe, essentially what you are proposing, with the complication of doing the work extraction a little more remotely. And there is also no work that can be performed by the water in this way.  You are proposing letting water run out.... which MUST occur from a region of higher pressure to a region of lower pressure, otherwise no flow. And then you are proposing letting water run back in at the same pressure and volume..... not physically possible, I don't think.


Besides, I thought the input water had to be introduced at the center bottom. You are actually removing it under pressure from the bottom, lifting it up -- and pouring it back _into the bottom_ by pouring it into the top of the input tube which is plumbed to the bottom of the chamber, right?


Nobody is working through the numerical problem that wildew presented? It's an interesting exercise, took me an hour to do it, several pages of sketches and figuring, and I learned a few things on the way to an answer. Did I do it right? "yes, I think, correct so far" is pretty much all the feedback I've gotten on this interesting problem.

TK .. saw your post so thought I'd whip up a quick pic regarding your last paragraph.

I used metrics because I'm used to them & they are SI - instead of using different volumes I used the same volume containers & starting volumes of water in each [same principle] - blue water in left vessel becomes green water in right vessel after transfer - you can see the sum of the PE's above each twin set - the pic flows from left to right as you would read across a page in incremental steps [for expediency] until all water is transferred.

The objective is to show the transfer of blue water from one vessel to another [connect tube not shown] by lifting the left vessel contents - the container mass itself can be ignored because it could be counter balanced so all we are interested in is the volume & weight transfer.

It occurs as a series of incremental steps - the left contents are lifted an incremental step height - the water flows from left to right container [water always finds its own level & flows downhill] - both water columns readjust heights i.e. the blue sinks a little, the right rises a little, until water levels are equal - then the process repeats until all blue water column is transferred across to become green water column.

At the conclusion of the process the PE is doubled, so WORK must have been done on lifting the blue water so that it could run downhill to become green water in one container - the amount of Net WORK/ENERGY is the same as the gain in PE.

In reality, if the lifting is done very slowly there will appear to be no incremental steps & the water just flows across but in fact work is being done so that it can run downhill even if very slowly.

To view it another way the blue water packet could have been raised quickly so that its bottom was level with the top of the green container - that's a lot of work done as it has raised the PE considerably - then the water could flow across until the right vessel was full & the left empty - the difference in work done/energy in this scenario & final resting PE's at equilibrium is energy lost to viscous & drag forces of fluid dynamics - IOW's the KE of the steep flowing water, after accounting for losses, does not equal the extra PE put into raising the left container to create that gradient of fall & therefore velocity for KE of flowing water.

Bottom line - Work is done just as you predicted to transfer water across by lifting one container.