Hydro Differential pressure exchange over unity system.

[TinselKoala]

Switch and bait by you or my mistake, I was refering to the quote from the "Our system explained" page ..."What is the input and output and how are they measured?
We have our method:

How do we measure input - the simple answer is - We don't have an input - and the output is measured like any other device - work - or joules - power etcetera - we have an output system - not an input system.

Now - our engineers use the internal operating cost - the internal resistance to the systems own output - to determine the optimizations of our ZED.

So when we say - we have a 10-1 system - that means we have 1 unit of resistance - while producing 10 equivalent units - leaves us with 9 to provide to the consumer.

Yet we are completely allowing the Validation teams to measure anyway they choose - no restrictions and no secrets - we have nothing to hide - and any true validation needs to be as thorough."


TK, You have made countless GIANT LEAPS regarding our work and efforts - so I am not sure if you are playing games here or this was an honest mistake. I do not know how you could disect every sentence on my web site and miss the explanation you railed on about - as being missing - but OK.
I will accept the blame for this one,  I apologize for any insult or misunderstanding I might have caused regarding my assesment of your intentions in that case.
Wayne

So... it was a mistake, not a deliberate lie, then, for you to say this:

p.s. the quote on efficiency that TK posted from my web site - he conveniently left off the next line which explained how we calculated efficiency - get the picture - he needed a story, he wrote a story - pun intended. Wayne Travis

Right? That's what you are saying now, and you are apologising to me for this mistake of yours. Right?
OK, apology noted and accepted, and I'll be watching for the true sign of repentance: a change in your behaviour.

Now.... how is efficiency REALLY calculated? It is the ratio of output work or energy, to input work or energy. You can look this up in many different places. You claim repeatedly, including above, to have NO INPUT. NO INPUT means ZERO energy or work going in. So your efficiency calculated CORRECTLY is, again, just what I said it was: INFINITE.
Or, using your "960 percent figure" one may conclude, if you have NO INPUT, that your output is 960 percent of that.... in other words 9.6 times nothing. Your "no input" statement and your "960 percent efficiency" claim cannot possibly both be true, unless you are redefining "input" or "efficiency"..... and of course that is just what you are doing.
Can you justify, by reference to standard scientific and engineering work, your method of obtaining an "efficiency" number like the 160 percent or the current 960 percent? Your description of having one unit of resistance and providing 9 to the consumer.... makes no sense as any measure of efficiency. It is equivalent to saying that you have a wheel inside that turns one time, and you provide an output that turns ten times. Big deal, any clockwork mechanism can do the same thing. You are explaining nothing with that "explanation", it is not how efficiency is calculated by ANYONE ANYWHERE but you,  and you've not even provided any evidence that the numbers are even real!

Feel free to demonstrate, by ACTUAL MEASUREMENTS ON A REAL SYSTEM, your claims of zero input with usable work output, and your claim of "960 percent" of something. Once again, I assert that you cannot do these things, which is why you always reply with word salads.... salads that contain little nuggets that reveal that you still don't really have what you claim to have:

[TinselKoala]

Hi Tinsel,
              it's no use asking mrwayne or Red_Sunset a technical question, they're more into politics.
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.
                       Thank you John.

I'm not sure exactly what you are asking. How much _water_? Why, it will take 22 gallons,  per minute of operation. Somewhat more, to make the power from a head of only 10 meters, to pump that 22 gpm at that pressure.

How much _gravity_? In my world, gravity isn't a consumable item, it doesn't run anything. Elevated heads of water are a way of storing energy in a lifted mass, which can be recovered by allowing the mass to fall back down. But something has to lift the mass in the first place.

So, with the restriction of 10 meters of input head, your question then apparently becomes "how much water flowing _in_ to a system at a head of 10 meters, is required to produce an output flow of 22 gallons per minute at 2250 psi?"
Right?

Mixed unit systems there... but OK. 2250 psi is equivalent to a head of (2250 x 2.31) = 5197.5 feet ! And that's 1584.2 meters, roughly. So you want an output flow of 22 gallons per minute at a head of 1584 meters, with an input of  "X" amount of water at 10 meters head.
http://www.engineeringtoolbox.com/pump-head-pressure-d_663.html

Am I right so far? This 1584.2 meters number seemed really large to me so I recomputed it several times. But taking it as correct and proceeding:

Now we need the power required to pump 22 gallons per minute to a height of a kilometer and a half. The specific work, in Joules per kilogram ( or m²/s²),  required to raise water to a head of 1584 meters, neglecting losses in the pump itself like shaft losses, etc, is given by ( Head x gravitational acceleration), or (1584 meters x 9.8 meters/sec²) = about 15523 Joules per kilogram. 22 gallons is (22 gallons x 3.78 liters per gallon) = about 83 kilograms of water and you want to pump this much per minute, so that's going to cost you 83 x 15523 Joules per minute,  or about 1.3 megaJoules per minute, or about 21473.483 Joules per second, or about 21 and a half kiloWatts. This is true no matter where the power or energy or work comes from.... it is simply represents the gain in energy of lifting that much water that high, and the power required to do it that fast.
http://www.engineeringtoolbox.com/specific-work-turbo-machines-d_629.html

Am I right so far?

So now the question becomes, how much water at 10 meters head do we need to run thru a 100 percent efficient hydroelectric turbine to make 21.47 kiloWatts of electrical power? (Your superJacuzzi-for-elephants pump is electrically powered, right?)
And the next question will be to do the same problem using a 960 percent efficient Hydro Differential Pressure Exchange system. Right?

Well, I'll pause here for another cup of coffee and let you contemplate these numbers and point out any errors so far.

[TinselKoala]

TK,

After that response this one should be a no brainer for you

Since you do not like Waynes method of calculating performance, setting limits on what the system can do so that people will not jump to the conclusion that you could just build a 2 inch widget and run the world, may I suggest that you provide such a method.

I suppose then you would also need to define what your version of input is, since when started this system has no input from the operator other than to control the timing of events.

Output is usable work out,, right??

Wayne measures input volume and pressure, that would be fluid into the pod chamber, and then measures work out from the risers lifting, then subtracts the value of input fluid and pressure from that to come up with a relationship,, so you can not do it that way because you have clearly stated that does not work.

Please cite a reference OTHER than Mister Wayne for his method of computing "efficiency". I can cite dozens that do it by comparing the ratio of output work to input work. I am not making this stuff up!

If there's no input but usable output.... well, you are dividing by zero. On the other hand, if the output is 960 percent of the input... and there's no input, then you are multiplying by zero.

I'd like to see some real refutation of this, instead of pointing to Mister Wayne's imaginary undocumented "measurements" on a fantasy system that doesn't actually exist.

Next, please give me some experimental data that shows any system, Mister Wayne's, yours, monds, ANYONE's.... producing usable output power with no input power..... and not "running down" as its stored energy is depleted.

When you've done that, I'll see about providing the method you ask for so sarcastically.

Would you care to address the question minnie asked, or my beginnings of a real answer to it? Now that it's becoming evident just what a power output of, say, 20 kW actually MEANS in a hydraulic system, thanks to minnie's excellent questions ....... it's got to make you think, doesn't it?

[TinselKoala]

I have here a TKbrand Automagic Bollard. The bollard itself is a sturdy steel cylinder, painted by the local Boy Scout troop to resemble a giant pencil shaft. It weighs three hundred pounds, and is set into the ground so that it is flush when retracted. Normally it would take a lifting force of 300 pounds, applied over the full 36 inch height of the bollard, to lift it up into place to work as a security barrier, or thorn in the side.
However, my patented (er....sorry, patent soon to be applied for) system of TKsprings takes one unit of resistance from the consumer and returns 99 times that as NET to the consumer to be used as work.
Thus, the TK Automagic bollard only needs a lifting force of THREE POUNDS, applied over the full 36 inches of lift, to be raised up and locked into position. When you are done with the bollard, that same THREE POUNDS is used to gently lower the heavy 300 pound shaft back into the ground. Thus, the work you put in at the beginning is given back at the end of the cycle of using the bollard.... ZERO INPUT in total, and an advantage to the consumer of (300-3)/3 or 99, or 9900 percent efficiency.

[TinselKoala]

LarryC said,

It only takes a few crazies especially those with MMD (multiple member disorder) to completely disrupt, what could be a helpful learning process.

I believe he is here saying that he thinks that some people posting here on this thread have "multiple member disorder" or multiple accounts and usernames.

I challenge him to provide ANY evidence for this baseless and paranoid accusation. If he cannot.... I suggest that he withdraw it and apologise for making it.
Will he do either? What do you think, gentle reader?

I also ask Stefan Hartmann, the owner of this site, to provide any evidence he might have, that I or anyone else on this thread has more than a single account/username.