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Overunity Machines Forum



Rosch taking orders on OU Bouyancy device.

Started by ramset, April 26, 2015, 09:52:03 AM

Previous topic - Next topic

0 Members and 6 Guests are viewing this topic.

thngr

Quote from: MarkE on May 14, 2015, 11:38:22 PM
1) A Striling or any other kind of heat engine needs both a hot and a cold reservoir.
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yes you are right.


2) Rosch / Gaia claim that the system is self sustaining.  The compressor and each process that follows it is lossy.  The claim is false.

that system is lossy mine also but if you take out the heat of compressed air to stirling pump than there will be efficiency increase four times, this is what makes it going but they may lie about it.

conradelektro

Why the calculation in http://revolution-green.com/rosch-smoking-guns-and-the-scientific-explanation/ or http://pesn.com/2015/05/13/9602617_scientific-explanation-for-Rosch-KPP-buoyancy-system/ is wrong:

The first power calculation in the article is:

== Formula 1 ====   Thus, we get: N = 9.81 x 2 x Q x 0.5 x H x 3 = 9.81 x Q x H x n ====
Further down one calculates:
== Formula 2 === N = 9.81 x 0.167 m3/sec x 2 m x 5 x 0,9 = 14.7 kW  =====
There is the error of putting "2 m" in again, because 2 was cancelled by 2 * 0.5 in Formula 1.
Therefore the corrected Formula 2 is:
== Formula 2 corrected === N = 9.81 x 0.167 m3/sec  x 5 x 0,9 = 7.35 kW  =====
But there is a further error in "Formula 2 corrected", which is the factor 5, because of  "5 working wheels above each other".
If there are 5 wheels above each other the water rises from one wheel to the other only 0.4 meters and not each time 2 meters. Therefore we can calculate with one wheel a height H of 2 meters, or for 5 wheels a height of H 0.4 meters each. We can calculate with one wheel with H = 2, or we can calculate with 5 wheels with a H = 0.4. So, either H = 2 or H = 0.4 x 5, which is the same.
Finally we end up with a correct "final Formula 2":
== final Formula 2 === N = 9.81 x 0.167 m3/sec  x 0,9 = 1.47 kW  =====
In addition we have to factor in a bigger loss than 0.9 because of the very optimistic height H = 2. If the water rises 2 meter it encounters friction at the wheel and some water is lost because it runs through gaps.
Further there will be losses in the gear connecting the wheel with a dynamo and the losses in the dynamo itself.
Therefore we will get much less than 1.47 KW.
Please note: I have not criticised the calculation itself (although one should do that), I only corrected obvious errors.


Criticism of "Formula 1":
The formula stems from water turbines http://en.wikipedia.org/wiki/Water_turbine#Power and assumes that water is falling down through the turbine. But one is not allowed to assume the same efficiency if "bubbly water (water air mixture with 50% air) rises through a turbine" rather than "smooth water is falling through a turbine". Much higher losses by friction of "bubbly water rising" in contrast to "smooth water falling" have to be assumed. And if things happen slowly (as is necessary in the "bubbly water rising case"), the formula is overly optimistic, because the formula only works with rather fast moving water.

Greetings, Conrad

d3x0r


(for the TL;DR 'So ya, still can't happen.') [/size]


Updated spreadsheet.https://docs.google.com/spreadsheets/d/1c4rZq5sB6CeXymY2mMU1JjbZCuY-WLawfC1VsMNkDbY
Light blue/cyan cells are primary input parameters - everything else is derived from them and constants.


white on Red cells are weight (given in pounds)
white on Orange cells are length (given in inches)
(first sheet is freely editable, a copy is protected readonly for original reference)


At the current state, the floats travel a greater distance than the pump is required to travel and the lift-weight of displacement is greater than the weight required on the pump head to pressurize the air...
so it's a 22% excess of power from the floats than required to pump 1 bottom float full of air.


- The excess gets greater by just adding separation between floats... since less air is required at greater depth, the stroke of the pump is less... but I guess even with a double action pump the excess distance is consumed to repressurize the second action from slightly less than 1atm(vacuum against valves to pull in new air) to the required pressure and then the additional distance to fill the float....


So I guess that's where the 'excess' would be going if accounted...


---
Note: There are 2 'required stroke to fill' the top one is to fill the float to capacity to full volume, the bottom one is required to fill a bottom float and have that be 100% full at the top; without losing air to expansion as pressure decreases on rise.


Hmmm... how to determine total stroke - oh just use the 1atm volume... so yes
any possible gain number turns into a loss


Still not satisfied that the final ratios don't turn out to be closer to 1; I guess it would if I assume that the double-action pump chamber will also self-balance a little... that is the compressed side will help the head move to pressurize the unpressurized side slightly...


So; again... really 1:1 best... unless you have a storage tank that gets depleted with time.
------
was listening to pesn video so decided to clean up the sheet and take another look to see if I missed something.  At least the input/output is in the same units now :)


I guess even on a continuous pump process the input at 1atm to X atm is a loss... cause really it's calculated from already having pressurized air... some sort of differential is required otherwise (integral over time?).


So ya, still can't happen.

MarkE

Quote from: thngr on May 14, 2015, 11:32:17 PM
buoyancy is not a source of energy but with one addition to compressor it going to be; if one know how use that heat with stirling pump.(stirling engine but without a work piston insted check valves) but please for give me; I do not have to teach you to design a perfect stirling pump.[/size]
Your statement self contradicts itself.  All the energy that if any that reaches the output shaft of te buoyancy machine comes from energy supplied by the compressor.  The compressor that Rosch/Gaia used had a capacity of ~75W puping air to 5 meters depth.

MarkE

Quote from: thngr on May 14, 2015, 11:48:32 PM
[/font][/size]
yes you are right.


2) Rosch / Gaia claim that the system is self sustaining.  The compressor and each process that follows it is lossy.  The claim is false.

that system is lossy mine also but if you take out the heat of compressed air to stirling pump than there will be efficiency increase four times, this is what makes it going but they may lie about it.
The efficiency is less than unity, far less.