Recirculating fluid turbine invention

[quantumtangles]

Thanks Frii143.

I know that tank B, ceteris paribus, will maintain its pressure of 500,000 Pascals once it has been pressurised by the air compressor. Equally I am aware that it could indeed be "pre-pressurised" to this level.

But I must consider how the system is intended to work. If we were to change one part of the system, we would change something else as well.

There are really two (connected) parts of the system.

First, water flows from tank B to tank A in the lower connecting pipe. This means that tank B pressure must be higher than tank A base pressure, or no such flow would take place.

Secondly, the upper siphon must flow from tank A to tank B. In contrast, this requires tank A pressure to be greater than tank B pressure.

At first sight, this seems impossible. How can tank B pressure be higher and lower than tank A pressure at one and the same time?

Ordinarily it would be impossible, even before considering the consequences of P1V1= P2V2.

However, by applying the outlet of the air compressor into the nozzle of the upper siphon, the nozzle pressure can exceed 500,000 Pa.

This means the siphon is at 600,000 Pa and the rest of tank B can be maintained at 500,000 Pa by the pressure relief valve triggered at 501,000 Pa.

And as well as this, tank B is still more highly pressurised than the base of tank A (400,000 Pa).

So my design works in terms of recirculation of fluid (before looking at energy output and input).

Sealed pre-pressurised connected tanks would not work because pressure and volume always equalise in connected systems. So tank A must be open to atmospheric pressure to prevent pressure building up in it as a result of tail gate water being forced into it from tank B.

Your view seems to be that one need not keep applying pressure to tank B.

Once it has been pressurised to 500,000 Pa, it will maintain that pressure if sealed. I agree completely.

But unless further pressure is applied to tank B (by the air compressor outlet located inside the nozzle of the siphon outlet) then the siphon will be unable to flow into tank B in the first place.

This is because the 100,000 Pa pressure on the surface of tank A will be the same pressure as the siphon water, and this must be amplified to 600,000 Pa by the air compressor if it is to overcome the 500,000 Pa pressure in tank B (else it will be unable to enter tank B).

Accordingly, despite the fact tank B could maintain constant pressure of 500,000 Pa, it is nevertheless essential to pressurise the exit nozzle of the siphon to 600,000 Pa or more.

This pressurisation has three effects.

First, it will enable the siphon to flow into tank B in the first place (the whole point of the system).

Secondly it will result in a gradual increase of pressure in tank B, which must be dissipated by a pressure relief valve triggered at 501,000 Pa.

Thirdly, the tailgate water at the base of tank B will be forced back into tank A.

This is the only way the system can work mathematically.

It is far from clear that connected hermetically sealed pre-pressurised tanks would prevent equalisation of pressure and volume. Well established principles of physics (P1V1 = P2V2) indicate this to be impossible.

If Tank A were under negative (vacuum) pressure applied to an air gap above the water surface, the system would be interesting, but water would be unable to circulate because of P1V1 = P2V2, and as the tanks are connected (as tank A would no longer be open to atmospheric pressure) all the extra pressure in tank B needed to force tailgate water back into tank A, would equalise pressure and volume.

Accordingly, the only possibility I can think of at the moment is for tank A to be open to atmospheric pressure, and for tank B to be pressurised to 500,000 Pa.

Although no further pressure need be applied to tank B itself (once it has been pressurised to 500,000 Pascals) nevertheless, extra pressure must be applied to the siphon exit nozzle. So extra pressure in tank B is not needed. But extra siphon nozzle pressure is needed, and the inevitable result of this is that tank B pressure will rise unless prevented by the tank B pressure relief valve.

This in turn involves work being performed constantly by the air compressor.

If the siphon exit nozzle is not pressurised to over 500,000 Pa, water will not be able to exit the siphon nozzle into tank B (because tank B pressure will be 500,000 Pa and siphon nozzle pressure will be 100,000 Pa).

Kind regards,

[andrea]

Thanks Frii143.

I know that tank B, ceteris paribus, will maintain its pressure of 500,000 Pascals once it has been pressurised by the air compressor.

Hi quantumangles, I have another question. If you maintain the pressure in the tank B by pumping air on it, where will the plus of air that you pump go ? If the system is closed, the air that you pump inside should remains inside, I think. Isn't this a problem? Thank you if you could clearify this point. Andrea

[quantumtangles]

Hi Andrea,

The air pumped into tank B by the air compressor is released by a pressure relief valve when the pressure in tank B exceeds 501,000 Pascals.

This pressure relief valve is situated on top of tank B.

The air compressor must work continually to keep tank B at 500,000 Pascals.

At the same time, the pressure relief valve in tank B must work continually to release all pressure in excess of 501,000 Pascals.

This is the only way tank B pressure can be maintained. Remember that the pressure in tank B has two jobs to do.

First it must expel tailgate water at the bottom of tank B into tank A. This means tank B pressure must exceed 400,000 pascals.

Secondly, the pressure in tank B cannot be so high that it prevents the upper siphon working (hence the pressure relief valve has to prevent pressure exceeding 501,000 Pascals).

For this reason, tank B has a pressure relief valve which releases all excess pressure above 501,000 Pascals into the environment.

This allows tank B pressure to exceed 400,000 Pascals, but to remain below 501,000 Pascals.

[pese]

here an collection (part of ma homepage)
http://alt-nrg.de/pppp/HHO_AIR_ELECTRO_%28Cars%29_EM12.html

specialle you can give attention ti this here:

motor with compressed refrigant !!
http://www.innovativetech.us/FutureProd-np.htm

---------------------  if you have interest, look ober the next lines
end of list more in englisch laguages....

(and my URL).  I am sure , you will finde some "extras" thas was
collected over years... 
only simple html nitices to remember speyialle myself,
so it was not collected as (for) professionals.

Gustav Pese

-----------------------




AIR  LUFT

http://pesetrier.150m.com/luft.html Verweis zu AirAccess


http://www.aladin24.de/Mazenauer/mazenauer.htm
MAZENAUER (System Schauberger Clem vergleichbar.)

http://peswiki.com/index.php/Directory:Generator_from_Ambient_Air_by_Kim_Zorzi
Zorzi   GENERATOR (Luftkompression- Wärmepumpenkonzept!!)

http://www.rexresearch.com/feg/feg1.htm LUFT/ WÃ,,RME(KÃ,,LTE) SYSTEM   Housten 1925
---------------------------------------------------  Febr.2010



Thermoakustische Wärmepumpe

http://pesetrier.150m.com/luft.html Begleittext zu AirAccess

http://aircaraccess.com  !!

Hitzeenergie erzeugen durch Luft-Komprimierung.

Durch Komprimierung von Luft entstehen Temperaturen bis 350- 500° F  170-260°C   (Rechner hierzu:    http://www.celsius-fahrenheit.de/ )

(Bekanntes Beispiel zur Klärung der Wirkung (Fahrradluftpumpe wird heiss.)

60-90 % des erzeugten Luftdruckes gehen nach Abkühlung wieder verloren

Der Luftdruck ist also hier nur eine kleinere Energiequelle , die jedoch
einfach gespeichert werden kann (und abrufbar bleibt)
(Werkzeugantriebe. Strom-Generatoren (Licht oder Wärme))
--------------------------------------------
46-00 Thermoacoustic Heat Pumps

 
http://www.aircaraccess.com/pdf/thp%20cover.pdf  thp cover.pdf  26kb
http://www.aircaraccess.com/pdf/thp.pdf                thp.pdf       4.84 mb
-----------------------------------------------------

Hello Gustav,


you write me, for your air-wheel that produce air pressure
(no realy efficient you sayd.
BUT i find 1000 page now over air compression
and this here  is an almanach , specially for you:

http://www.aircaraccess.com/download.htm   so i think , i MUST informe you


specially http://www.aircaraccess.com/pdf/racd%20i-ii,%201-50.pdf

let kow you that 60-90% of the pressure go lost
if tge 350-500°F hot aire will coling in the boilers.

So yüu can frofuce an air powered (wheel) heating system, also
collect as Hot wather in an boiler system.

Not necesary to say, that the stored pressed air, can als  uses
(-also after collect the pressed air
for mechanical applications , Air driven electric converter can
produced also in germany (ar even othe first countries
Use your system to hear rge house, ar even an swimmimgpool
http://72.14.205.104/shttp://www.innovativetech.us/FutureProd-np.htmearch?hl=en&q=cache%3Awww.ultralightamerica.com%2Fair_power.htm&btnG=Google+Search
http://www.aircaraccess.com/download.htm  1000+ pages air compressed power

http://72.14.205.104/search?hl=en&q=cache%3Awww.ultralightamerica.com%2Fair_power.htm&btnG=Google+Searchp://peswiki.com/index.php/Directory:Generahttp://alt-nrg.de/pppp/HHO_AIR_ELECTRO_%28Cars%29_EM12.htmltor_from_Ambient_Air_by_Kim_Zorzi

https://www.abbeon.com/store/item.cfm?code=2082  vortex-tube  hot cold  without moving parts


Jim




http://www.innovativetech.us/FutureProd-np.htm
Hier von 2 "Vorstellungen DEN 2. ansehen.

Hydraulic Motor run under pressured refrigant !!
folglich Fluid/Gas Wandlung da thermisch Wandlung

http://www.innovativetech.us/-np.htm

http://www.google.de/patents/about?id=uLZPAAAAEBAJ&dq=1781062  1925  Houston




http://www.google.de/patents?id=uUxFAAAAEBAJ&printsec=drawing&zoom=4#v=onepage&q=&f=false  1940

http://www.google.de/patents?id=uNdbAAAAEBAJ&printsec=drawing&zoom=4#v=onepage&q=&f=false  ++++
mocullum

http://www.google.de/patents?id=VAseAAAAEBAJ&printsec=drawing&zoom=4#v=onepage&q=&f=false
---- LUFT
full page   and HOME  you find here:


http://alt-nrg.de/pppp/HHO_AIR_ELECTRO_%28Cars%29_EM12.html

[quantumtangles]

This is very interesting Pese.

I would not have thought of using refrigerant in the compressor.

If for example we were to use 1,1,1,2-Tetrafluoroethane, otherwise known as R-134a, (Genetron 134a, Suva 134a or HFC-134a), this has a density of about 4.25kg/m3.

The refrigerant would be used to generate very high pressure fluid. This would be an adiabatic process causing the temperature of the compressed R134a to increase significantly under pressure.

This I understand.

But I do not know what sort of flow rate would be involved (normally, refrigerators have very narrow nozzles and thus very high pressure and low fluid flow rates cause cooling in the 'cool box' pipes).

Do you envisage only replacing compressed air with compressed R134a?

Or were you thinking of completely replacing water as a working fluid with R134a (R134a would have much lower density even if highly pressurised).

Lower density working fluid would mean less force in Newtons being applied to the turbine per F = m*a

The mathematics of the system would have to take into account significant 'delta h' changes (thermal changes).

I will have a look at equations relating to thermodynamic changes (in open systems with non-reversible processes).

What are your thoughts as to energy output versus energy input using R134a instead of compressed air?

Can you let me have any preliminary calculations?

What changes would occur in the system if R134a were used instead of compressed air?

This is really interesting work. It may be (fingers crossed) that a combination of our ideas would allow the system to heat water in one part of the system, and use heat energy to help a water pumping process generate electricity from the kinetic energy of water striking the turbine in another.

Brilliant stuff if it can be made to work. I do not yet have any understanding of the mathematics (flow rates, thermal changes etc) and hope you will explain mathematically how our ideas can be integrated.

Kind regards and thanks,