Hydro Differential pressure exchange over unity system.

[minnie]

Hi TK,
        thank you for answering my question. The idea behind it was to show that it wouldn't be easy to make even a modest amount
of power from water and gravity.
    Talk about getting blood out of stone, that would be easy compared to getting some meaningful info. out of mrwayne!
I give up- I'll wait and see if the laws have to be reworked when the machine is revealed.
    Tinsel, you do answer the question asked, explain things really well and seem to have limitless patience, I really admire you
for that.
                         John.

[TinselKoala]

Now... to continue with Minnie's question.
We have calculated that it is going to take about 21473.483 Watts to pump 22 gallons per minute at 2250 psi.

But we are asked to use an input head of 10 meters, and we need to determine the flow rate of water at that head, to make the power for our elephant bathtub Jacuzzi pump.

Running the pump power equation backwards, we see that

Water horsepower = (Flow rate in gpm x total head in feet)/3960
so converting units and inserting the known values and solving for flow rate in GPM, we have
21473 Watts / (743 Watts/horsepower) = 28.9 horsepower, and 10 meters = 32.8 feet
SO
Flow Rate in gpm = (28.9 x 3960) / 32.8
Flow rate in gpm =  3489 gallons per minute

So in answer to your original question,

How much water (gravity) would it take to run a 2250 p.s.i. 22 g.p.m. pump? You can have a head
of say, 10 mtrs.

I arrive at an answer, neglecting losses and assuming 100 percent conversion efficiencies, of a flow of just under 3500 gallons per minute, at a head of 10 meters, to make the 21.473 kW of power necessary to pump 22 gallons per minute at a pressure of 2250 psi.

I am prone to misplacing decimals.... I've checked these calculations several times and I'm still getting this rather large number for an answer. Is anyone interested in checking my work? Or will insults be more satisfactory?

(Astoundingly, and as a check.... 22 gallons per minute x (1584.2 meters/10 meters) = 3485.2 gallons per minute. QED.)

Now... let's see if we can make our "input" stage 960 percent efficient. Then we get
3489 gpm / 9.6 = a bit over 363 gallons per minute, or about 6 gallons per second, at ten meters head pressure.
That's firehose territory, I think.

[TinselKoala]

Hi TK,
        thank you for answering my question. The idea behind it was to show that it wouldn't be easy to make even a modest amount
of power from water and gravity.
    Talk about getting blood out of stone, that would be easy compared to getting some meaningful info. out of mrwayne!
I give up- I'll wait and see if the laws have to be reworked when the machine is revealed.
    Tinsel, you do answer the question asked, explain things really well and seem to have limitless patience, I really admire you
for that.
                         John.

Thanks, we crossed posts there.

I guess you've seen my final answer of around 3500 gpm at 10  meters head, and that it can be obtained in two different ways, the full derivation and the easy check by head ratios and the pump power equation.

My patience isn't limitless, though, and I'm running out of it.

ETA: Note to the novice calculators out there: I have specified the units (feet inches meters seconds gallons whatever) at every step of the calculations, and the units can be seen to agree algebraically throughout ("pounds/square inch" x "square inches" equals an answer in "pounds", for example). This is also another important check of validity and accuracy that should be applied to _every_ calculation of this type: the units, as well as the numbers used, must agree from start to finish.)

[TinselKoala]

No sarcasm TK.

I am very aware of the method to determine efficiency.  The value is compared to an outside input, what if there is not one, then how can you find relationships and limits and all that?

My system ran in a single setup condition for weeks without me needing to recharge it, I put the system in a safe run condition so I could do a lot pf playing, put it away and just take it back out and play some more, I lost no air and no water,, that started later after beating on the side of the risers to help with the sticky sink, so now I have to set it up with the water level lower than the extension on the outside retainer.

M's system does not loose anything either and nor does Dales,, so where do you keep coming up with this fantasy loss, you took a leak and made it a normal part, why?

Your system does not operate on NO INPUT, does it? I don't think it does, and neither does Dale's. You are compressing and uncompressing a spring, adding and recovering work, but you aren't "creating" any extra usable work. And I'll bet your arms get tired after a while nevertheless.
I have referred to a possible "leak" being the cause of Mister Wayne's observation of short periods of "self-running" of his systems, giving him the benefit of the doubt.... which I am less and less prone to do these days, but there it is. I'm sure you could imagine a pressure leak that could result in a dual-cylinder system running for a while on stored "setup" pressure as the pressure leaks past seals and so on, into the right (or wrong) chambers during the cycling. Have you ever built and operated a simple Stirling engine? You should, they are easy and enlightening.
Your system doesn't leak. Fine, that's good. Assemble two of them and make your system self run. I'll bet you cannot. But I'll also bet that if you put the right kind of leak in there... it would run for a few cycles, until it went flat.

Mister Wayne has to give some kind of efficiency number to satisfy "bottomline" people like the accountants of investors. How can you do this if you claim to have no input? You cannot, so he has to come up with a new and different definition of efficiency that allows him to come up with a number other than zero or infinity. But since he actually does NOT have a self running system, and won't cite the original data from which his numbers come, nor even describe the system ..... the whole thing is still a big red herring, and is beginning to smell.

[Red_Sunset]

I just got back from a evening on the town
Am I seeing this right, is Tinsel writing now posts to himself ? Pretending to be a Zed expert !
Handling in & out posts single handed
An amazing twist of events.