Recirculating fluid turbine invention

[pese]

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 low high pressure flow rates are used to 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,

I am not sure, i tink in frigidaires ist the refrigant gas mixed with water. The water flow thriu compressor ex compress als the refrigant (that ist only an smal part of the total filling)
you will find out the best way


-also think on butane
was uses in east ggermany sinc 40 years ore mor in frigidaires.
(low cost and any thome avaiable.

Now it ist used in "green- fridigaires" that contains no more dangerosly fillings


I can not calculate this out, but i hope that the "users" can
find some ideas to becomes better results in there experiences

Gustav Pese

[quantumtangles]

Thanks again Pese. Very interesting 

[quantumtangles]

If you cannot speak English properly, please do not post replies to this thread.

If you have been mercifully untouched by the ravages of mathematics, please do not post replies to this thread.

If you feel an overwhelming urge to share esoteric non-mathematical ideas about perpetual motion, please do not post replies to this thread.

There are lots of other places to post threads (eg for people who think circles of wood covered in magnets do not suffer from back emf). This particular thread is not one of them.

This thread is concerned only with precise mathematical calculations concerning a recirculating fluid turbine.

The electrical output in watts has already been calculated using trivial mathematics.

Only two questions remain:

1. Can fluid recirculate within this system if unlimited power is available to enable recirculation?

2. If so, how much electrical power will be consumed by a siphon pump and air compressor to enable this to happen?

From this point onwards, only mathematical contributions using well established formulae and SI units will be helpful.

Our species faces an energy catastrophe. If this system is beyond your comprehension or you lack the mathematical apparatus to make an articulate contribution to the thread, please do not post any replies. For all our sakes.

Many thanks.

[quantumtangles]

Lets get back to the mathematics of the system.

Some people will always try to sabotage mathematical posts. It is not their fault they do not understand the maths, but it is nonetheless irritating and diverting.

This thread is all about mathematics and engineering. Please do not read any further unless English is your first language and you have a primary degree from a top 100 (world ranking) university in maths or sciences.

We have two 30m high steel cylinders 1m in diameter.

Tank A is full of seawater. Tank B contains only 10% by volume of seawater. An impulse turbine is placed at the base of tank B above the water line. Two pipes connect the cylinders. An upper siphon leading into tank B from the open surface of tank A, and a lower connecting pipe leading from the base of tank B to the base of tank A. Water flows from tank A into the siphon by means of a water pump. The water enters the siphon and is accelerated by compressed air from the air compressor before it leaves the siphon nozzle. This water strikes the impulse turbine 25m beneath with a flow rate of 1m3/s.

The tailgate water is ejected from tank B into tank A by means of higher pressure in tank B. A pressure relief valve in tank B prevents pressure exceeding 501kPa. Tank B pressure always exceeds tank A base pressure (401kPa). Tank A surface pressure is always Patmos (101325Pa). The fluid recirculates and the turbine produces electricity.

Summary:

Steel Cylinder A: Height 30m Diameter 1m (full of seawater)
Steel Cylinder B: Height 30m Diameter 1m (3m depth of seawater)

Pelton Impulse Turbine: Diameter 0.9m (PCD 0.87m) (Efficiency 0.9)
Abac Genesis 1108 air compressor (11kW 800kPa @ 1.67 m3/m = 0.027833 m3/s)
ESP-10 150 siphon water pump (37.77kW 1m3/s flow rate)
Drawn Copper Siphon Pipe: Diameter 0.38m
Drawn Copper Lower Connecting Pipe: Diameter 0.38m
Relative pipe roughness: 0.0000125

Back flow prevention valve
http://www.3ptechnik.co.uk/en/backflowpreventionvalve.html

Pressure relief valve
http://webwormcpt.blogspot.com/2008/01/useful-documents-related-to-pressure.html


Working fluid: Seawater of density 1020kg/m3
Fluid viscosity 0.00108 Pa-s

Pressure in Tank A
Surface Pressure = Patmos = 101350 Pa
Base Pressure
= 30m x 1020kg/m3 x 9.81 m/s/s
= 300,186 Pa Gauge Pressure + Patmos (101350 Pa)
= 401,536 Pa Absolute

Operating (Absolute) Pressure of tank B 500kPa

Water Jet Velocity Vjet

Bernoulli's equation gives us the velocity of the water jet applied to the turbine.

P = ½ r . V2

P = Pressure (401,536 Pa)
rho = density (1020kg/m3)
V = velocity (m/s) mystery value

401536 = ½ 1020 . V2
401536 = 510 . V2
V2 = 401,536 / 510
V = 28 m/s

Force of water Jet (Fjet)

1 m3/s of seawater (1020kg/m3) represents a flow rate of 1020kg/s.

F = 1020kg/s x 28 m/s/s
= 28560 Newtons

Fjet Momentum Change Calculation

Turbine speed may not exceed 50% of water jet speed

Vjet = 28 m/s
Vrunner = 14 m/s

Delta Mom = mass flow rate x Delta V
Delta Mom = mass flow rate x (Vjet â€" Vrunner)
Delta Mom = 1020kg/s x (28 m/s â€" 14 m/s)
Delta Mom = 14280 N

Fjet = 14280N.

Turbine RPM

We can now calculate the RPM figure for the turbine based on runner velocity
of 14 m/s.

First we need to know the circumference of the turbine.

Diameter = 0.9m
radius = 0.45m
2.pi.r = 2.827433388m circumference

Vrunner = 14 m/s
RPS = 4.951487 revolutions per second x 60
= 297 RPM

Power Output

Applying 297 RPM and Fjet = 14280 Newtons to the Pmech equation:

Pmech = Fjet x Njet x pi x flowrate x RPM x 0.9 x 0.87m / 60
= 14280N x 1(jet) x pi x 1m3/s x 297RPM x 0.9(eff) x 0.87m / 60
= 173Kw

Cross referencing this output figure with the conventional equation for electrical power output in watts:

Pwatts = h(25m) x g(9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW

It would be highly conservative and I think reasonable to take the lower of these two figures to represent maximum electrical output of the turbine (173kW).

Calculations of electrical output from hydroelectric facilities always rely on the second (higher output) equation based on head and flow rate (because it always correctly predicts output).

But being conservative to help you debunk the system, I am happy to give you a 52kW head start.

Maximum total system power output = 173,000 watts

Tank Pressure Dynamics

When the system is operating, tank A base pressure is always 401,350 Pa. Tank A surface pressure is always Patmos (101,350 Pa).

But tank B is the place of interest in terms of pressure dynamics.

The air compressor (Abac Genesis 1108) has output of 800,000 Pa at a rate of 59 cubic feet per minute = 1.67 m3 per minute = 0.027833 m3/s.

The volume of tank B (h=30m d=1m)
V= pi.r2.h
V= pi x (r x r) x h
V = 3.141592654 x (0.5 x 0.5) x 30m
= 23.56m3.

The volume of air in tank B (after deducting the 3m tailgate water taking up 10% by volume of the cylinder = 2.356m3)
= 21.2m3.

The air compressor takes 12.697 minutes to pressurise the 21.2m3 of air inside tank B to 800,000 Pa. We can reduce the volume of tank B by 86% by using just a pipe with a shorter cylinder beneath it, but I have stayed with a full size tank B with 3m of water in the tailgate.

The air compressor has to do the work of exceeding the pressure at the base of tank A (401,536 Pascals) to force fluid out of tank B.

There is a back flow prevention valve in the lower connecting pipe. The working fluid must move water from a high pressure to the lower pressure area.

Once air pressure has built up in tank B, it can only leave tank B through the upper siphon (which would stop the system working) or the lower connecting pipe (which would keep the system working).

The problem seems to be that the siphon may not work because of the high pressure in tank B. Surprisingly this is not the case.

First, the tailgate water (3m in depth) supplies pressure of 30,018.6 Pa.

P = 3m x 1020kg/m3 x 9.81 m/s/s
=  30,018.6 Pa

Secondly and much more importantly the output pipe of the air-compressor is connected to the siphon exit nozzle (this is the only place the air compressor output exits into tank B).

The force applied by the compressed air is not going to increase just because we apply it inside a thin siphon pipe.

Nonetheless pressure increases dramatically. Not because the force has increased, but because the area over which the force is applied in the nozzle has decreased.

The siphon pipe is 0.38m in diameter. Delta pipe (a cross sectional area of the siphon pipe) = pi*r2
Delta pipe = 0.1134 m2

This is the surface area to which pressure from the air compressor is applied to the water exiting the siphon (whilst it is still inside the siphon).

The 800,000 Pascal output of the air compressor (at a air flow rate 0.027833 m3/s) is delivered to the siphon nozzle over an area of 0.1134m2.

We know the area of delta pipe is 0.1134 m2 (the siphon nozzle).

In comparison the surface area of the inside of tank B (height 30m) is:

A = 2*pi*r2 + 2*pi*r*h
A = 1.57 + 94.24778
A = 95.8177 m2

So if the maximum pressure applied by the air compressor to the walls of tank B is 800,000 Pa, the pressure applied over the much smaller area in the siphon nozzle will be 845 times greater.

P = 422500000 Pa = 422500000 N/m2 = 42250 N/cm2.

Not because the force has increased. But because the area has decreased.

In practical terms, the pressure applied to the walls of tank B and the fluid column at the bottom of it will be no more than 500kPa (because of the pressure relief valve built into tank B).

But the pressure relief valve will not prevent very high pressure being generated inside the siphon nozzle because this pressure of 4222500 kPa will decrease as soon as it enters the wider chamber of tank B.

Not because the force has decreased, but because the area has increased.

In summary, the pressure applied in the siphon nozzle will significantly exceed 600Kpa.

So pressure in tank B is 500kPa. Pressure in the siphon nozzle is over 600KPa (actually it would be 4222500kPa) and pressure at the base of tank A is 401350 Pa.

The water must recirculate from tank B into tank A, and the siphon nozzle must continue to operate despite 500kPa pressure in tank B.

Conclusion

Power output is 173kW. The air compressor consumes 11kW. The water pump consumes 37.66kW. Net power output is 124kW.

You may be thinking the maths don't make sense because right at the beginning I assumed water pressure of 401350 Pa for the purpose of calculating fluid velocity.

But if you consider that the the water pump and in particular the air compressor increase siphon nozzle pressure to over 600kPa, velocity per Bernoulli and power output would be even higher:

P = ½ r . V2

P = Pressure (600,000 Pa)
rho = density (1020kg/m3)
V = velocity (m/s) mystery value

600,000 = ½ 1020 . V2
600,000 = 510 . V2
V2 = 600,000 / 510
V = 34.3 m/s

F = m*a
= 1020kg/s x 34.3m/s/s
= 34986 N

Vjet = 34.3 m/s
Vrunner = 17.15 m/s

Delta Mom = mass flow rate x Delta V
Delta Mom = mass flow rate x (Vjet â€" Vrunner)
Delta Mom = 1020kg/s x (34.3 m/s â€" 17.15 m/s)
Delta Mom = 17493 N

Fjet = 17493N

Vrunner = 17.15 m/s
Diameter = 0.9m
radius = 0.45m
2.pi.r = 2.827433388m circumference
RPS = 6 revolutions per second x 60
= 360 RPM

Pmech = Fjet x Njet x pi x flowrate x RPM x 0.9 x 0.87m / 60
= 17493N x 1(jet) x pi x 1m3/s x 360RPM x 0.9(eff) x 0.87m / 60
= 258Kw

Cross referencing this output figure with the conventional equation for electrical power output in watts:

Pwatts = h(25m) x g(9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW

Whichever way you look at it, output would be over 200kW. Power consumption is less than 25% of output.

It does not breach the laws of thermodynamics because it is an open system in which both mass and energy may pass through the system boundaries. Note also that the process is irreversible, and that entrophy increases due to frictional heat losses.

It is tantamount to a giant electric pressure jet but has a high mass flow rate.

The siphon nozzle has become the nozzle of the water jet. 500kPA pressure in tank B cannot stop the 1020kg/s juggernaut entering tank B because the siphon water is pressurised to over 600kPa by the air compressor (in the narrow confines of the siphon exit nozzle).

Equally, the tailgate fluid must leave tank B through the one way flow valve in the lower connecting pipe because tank A base pressure (401.35kPa) is lower than tank B overall pressure (500kPa).

Interestingly, tank B pressure, once raised by the air compressor, will remain at that pressure despite fluid being evacuated back into tank A. This is because every cubic metre of water that leaves tank B is replaced by another cubic metre of water entering tank B through the siphon nozzle.

Tank B will stay at 500kPa once pressurised (because it has a pressure relief valve triggered at 501kPa).

The pressure will only fall if there is a reason for it to fall. Evacuation of one cubic metre of water per second through the lower connecting pipe is balanced by one cubic metre of water per second flowing from the upper siphon. So the mass flow in itself does not cause air pressure to fall in tank B (in the sense lower water volume reduces pressure).

The gas fluid pressure in tank B will not find its way into tank A because only aqueous fluid is forced into tank A. Not air.

A pressure relief valve on tank A releases any compressed air that manages to get into tank A and also releases any gases released by the agitated water.

Higher fluid pressure in tank A would only help the siphon, not hinder it. A pressure relief valve prevents excessively high water pressure in tank A. But the turbine also removes kinetic energy from the water before it re-enters tank A.

The one way flow valve in the lower connecting pipe prevents the water at the base of tank A flowing into tank B.

Lets see some mathematics in the replies. Not semi-literate ramblings from uneducated primitives.

By way of indication, a primary degree in physics or mathematics is a prerequisite for a coherent response. A doctorate in applied physics or fluid dynamics would be preferred.

The energy crisis is real and will affect all of us. Please do not contaminate this thread with intellectual pollution. Provide a mathematical response or leave the thread to better educated respondents.

[quantumtangles]

The density of mercury is 13546 kg/m3. If tank A were full of mercury, seawater, having a lower density (1020kg/m3), would float to the top of tank A.

No energy would be required to 'lift' working fluid (seawater) to the top of tank A.

However, pressure at the base of tank A would be higher, due to a 30m column of mercury.

P = 30m x 13546kg/m3 x 9.81 m/s/s
=  3986.58 kPa = 3986587 Pa

High pressure and high toxicity are not a good combination.

A more practical option would be Perchloroethylene (the solvent used in dry cleaning). The density of Perchloroethylene is 1622kg/m3.

Tetrachloroethylene or Perchloroethylene is still toxic, but much less so than mercury. It dissolves only slightly when mixed with water (0.015 g/100 mL @ 20 °C) however it evaporates when exposed to air.

Provided no air entered tank A, this could be used as a dense substrate to lift working fluid.

In that event, pressure at the base of tank A would be as follows:

P = 30m x 1622kg/m3 x 9.81 m/s/s
=  477 kPa = 477354 Pa

The Abac Genesis air compressor would be sufficiently powerful to force tailgate working fluid into tank B, where it would rise to the surface of tank B and fall once again onto the impulse turbine.

Michael Faraday discovered this compound. It would be interesting if his discovery can enable a new method of electricity generation.