Recirculating fluid turbine invention

[quantumtangles]

System specifications:

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.

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 impulse turbine thereby helps prevent a P1V1=P2V2 pressure/volume equalisation.

If you represent an engineering company and would like to build the system, please message me.

[quantumtangles]

Pressure in tank A cannot be allowed to exceed pressure in tank B or the system will stop working.

A pressure relief valve on top of a sealed tank A is one solution, but working fluid would be lost.

For these reasons, a pressure relief valve on top of a sealed tank A is not optimal.

However, by ensuring tank A is left open to the elements, it will be subject to atmospheric pressure.

Just as water at the bottom of the sea loses pressure as it nears the surface, so too the pressure at the surface of tank A may never exceed atmospheric pressure (101350 Pa).

If the pressure at the surface of tank A cannot exceed 101350 Pa, the pressure at the bottom of tank A (which is 30m high) can never exceed 401350 Pa absolute.

Therefore the pressure at the base of tank A (401350 Pa) can never exceed the pressure in tank B (500,000 Pa).

Accordingly tank A pressure can never exceed tank B pressure and the system fluid will recirculate.

[quantumtangles]

It is not enough to assert that a machine will produce electricity without considering all of the forces acting upon the system.

Ordinarily air resistance does not greatly effect turbine RPM, and therefore has negligible effect on turbine output in kW.

However the turbine under review will be operating at over 500,000 Pascal = 5 bar = 5 atmospheres of pressure.

At this pressure, air density might materially affect system performance because air density increases with pressure.

The turbine will have to rotate in a sort of air soup. Much thicker than air at Patmos.

The density of air at sea level (101325 Pascals) at 15 degrees Celsius = 1.22521 kg/m3.

The density of dry air can be calculated using the ideal gas law expressed as a function of temperature and pressure:

D = P  / Rspecific x T
D = Density (kg/m3)
P = Absolute pressure (Pa)
Rspecific = the specific gas constant for dry air = 287.058 J/(kg.k)

We need to calculate the density of air in tank B when it is subject to 501325 Pascals of pressure at a temperature of 15 degrees celsius (288 degrees Kelvin).

D(kg/m3 = 501325 (Pa) / 287.058 J/(kg.k) x 288
D = 6.06 kg/m3

Immediately we can see that the density of air at 501325 Pa is 6.06 (kg/m3) whereas the density of air at 101325 PA is 1.2251 (kg/m3).

Humid air is indeed less dense than dry air. This counter-intuitive fact seems to be at variance with our perception. But it is indeed true because the molecular mass of water (18 g/mol) is less than the molecular mass of dry air (29 g/mol). For any gas at a given temperature and pressure the number of molecules present is constant for a particular volume (Avogadro's Law).

So when water molecules (vapour) are added to a given volume of air, the dry air molecules must decrease by the same number to keep the pressure or temperature from increasing.

Hence the mass per unit volume of the gas (the density of the gas) decreases.

Inside the turbine housing, water vapour will mix with dry air emitted by the air compressor. The water vapour content of air just above the surface of an ocean is approximately 4% by volume.

Ordinarily we might reasonably assume that the water vapour content of the air in the turbine housing would be 4% by volume, but I am concerned that the industrial air compressor will reduce the water vapour content of the air in the system.

For that reason, it would be safest, contrary to interest, to assume the density of the air in the turbine housing will not be reduced by water vapour content.

So the turbine in the system under review must move through water spray (as would any other impulse turbine) and also high density air (air that is approximately 5 times as dense under 500kPA pressure in tank B).

This will affect system output because the denser air will create a counter-force to the force of the falling working fluid, and by so impeding movement of the turbine will result in lower RPM and lower system output in Kilowatts.

Of greater concern is the fact that lower absolute siphon nozzle pressure (with reference to the high pressure in tank B) will reduce the water jet velocity as calculated using Bernoulli's equation, and thus the acceleration figure upon which the entire force figure in Newtons of the water jet striking the turbine was based (per F = m*a).

But to what extent will high air pressure in tank B create a counter-force to the water jet, reduce the Fjet figure in Newtons, and reduce RPM leading to lower electrical output?

The negative force exerted by this high density air (6.06kg/m3) can be calculated.

To begin with I will deduct the density of air at atmospheric pressure from the density of air under 500,000 Pascals of pressure (because turbine output equations broadly take into account atmospheric air pressure in the unit-less fraction representing system efficiency).

6.06 kg/m3 - 1.22521 kg/m3 = 4.83479 kg/m3

Taking an agricultural approach, I will first deduct this mass of air from the water mass striking the turbine.

The mass of water striking the turbine is 1020kg per second (1 m3/s of seawater per second).

Assuming an entire cubic metre per second of this high density air impedes the rotation of the turbine buckets (creating a counter-force to the force exerted by the mass of falling seawater) the counter-force may be calculated as follows:

F = m*a
F = 4.83479 kg/s x 14 m/s/s
F = 67.68 Newtons

If we now deduct this negative Fjet figure in Newtons from the Fjet figure resulting from 1 m3/s of seawater striking the turbine, this results in the turbine Fjet calculation of 14280 Newtons being reduced by 67.68 Newtons. A negligible figure.

The lower force applied to the turbine would be 14280 â€" 67.68 Newtons = 14212.32 Newtons.

Applying this Fjet figure to the Pmech equation:
Pmech = Fjet x Njet x pi x flowrate x RPM x 0.9 x 0.87m / 60
= 14212.32N x 1(jet) x pi x 1m3/s x 297RPM x 0.9(eff) x 0.87m / 60
= 173Kw

But I am not satisfied this accurately reflects the true losses in power that high air pressure will cause in tank B.

This is because RPM was originally calculated with reference to the pressure in the upper siphon. Specifically, the pressure of the working fluid which in turn enabled us to calculate the velocity of the working fluid.

Bernoulli's equation:

P = ½ r . V2

So can high air pressure in tank B significantly reduce the velocity of the water jet entering tank B?

Yes it most certainly can.

At pressure of 401535 Pa as orginally projected (in the absence of counter-pressure), the water flow velocity was 28 m/s.

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

But in the presence of counter-pressure the velocity will decrease.

The pressure in tank B will be 500,000 Pa.
The pressure of fluid as it exits the upper siphon was originally projected to be 600,000 Pa

The absolute pressure of fluid exiting the siphon would only be 100,000 Pa.

In that event, applying Bernoulli's equation:
P = ½ r . V2
100,000 = ½ 1020 . V2
100,000 = 510 . V2
V = 14 m/s

However because this is an impulse turbine where turbine speed may not exceed 50% of water jet speed if maximum efficiency is to be maintained, we have to do some subtraction:

Vrunner may not exceed 50% of Vjet
Vjet = 14 m/s
Vrunner = 7 m/s

Delta Mom = mass flow rate x Delta V
Delta Mom = mass flow rate x (Vjet â€" Vrunner)
Delta Mom = 1020kg/s x (14 m/s â€" 7 m/s)
Delta Mom = 7140 N

Accordingly the velocity of the turbine runner = 7 m/s
If the turbine runner travels at 7 m/s, this translates to RPM as follows:

Turbine diameter = 0.9m
Radius = 0.45m
Circumference = 2*pi*r = 2.827433388 m
= 2.475743559 revolutions per second x 60 = 148.5446 RPM

Applying this lower RPM value and the reduced Fjet value of 7140 Newtons in the Pmech equation gives us the following power output:

Pmech = Fjet x Njet x pi x flowrate x RPM x 0.9 x 0.87m / 60
= 7140 N x 1(jet) x pi x 1m3/s x 148RPM x 0.9(eff) x 0.87m / 60
= 43.32Kw

A significant reduction in output power.

I iterate that this really is the worst possible case scenario, but nonetheless power output of 43.32kW would not exceed power consumed by the air compressor and water pump (48.66kW) leading to a net power loss of 5.34 kW.

If on the other hand the pressure of working fluid leaving the small (0.1134 m2) area of the siphon nozzle is 800,000 Pa (the maximum pressure output of the air compressor), gauge pressure at the siphon nozzle would be 800,000 Pa less tank B pressure of 500,000 Pa = 300,000 Pa absolute.

P = ½ r . V2
300,000 = ½ 1020 . V2
300,000 = 510 . V2
V = 24 m/s

Vrunner may not exceed 50% of Vjet
Vjet = 24 m/s
Vrunner = 12 m/s

Delta Mom = mass flow rate x Delta V
Delta Mom = mass flow rate x (Vjet â€" Vrunner)
Delta Mom = 1020kg/s x (24 m/s â€" 12 m/s)
Delta Mom = 12,240 N

Accordingly the velocity of the turbine runner = 12 m/s

This translates to RPM as follows:

Turbine diameter = 0.9m
Radius = 0.45m
Circumference = 2*pi*r = 2.827433388 m
= 4.244131816 revolutions per second x 60 = 254.649 RPM

Applying this RPM value and the reduced Fjet value of 12,240 Newtons in the Pmech equation gives us power output as follows:

Pmech = Fjet x Njet x pi x flowrate x RPM x 0.9 x 0.87m / 60
= 12,240 N x 1(jet) x pi x 1m3/s x 254.649 RPM x 0.9(eff) x 0.87m / 60
= 127.786 Kw

So everything depends on the pressure at the output nozzle of the siphon.

If the nozzle pressure is 100,000 Pa absolute, the system will not be worthwhile.

If on the other hand the nozzle pressure is 300,000 Pa, the machine will be a profitable electrical generator.

If the nozzle pressure is 600,000 Pa absolute, the machine will be remarkable.

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

Vrunner may not exceed 50% of Vjet
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

This translates to RPM as follows:

Turbine diameter = 0.9m
Radius = 0.45m
Circumference = 2*pi*r = 2.827433388 m
= 6.06557 revolutions per second x 60 = 363.93 RPM

Pmech = Fjet x Njet x pi x flowrate x RPM x 0.9 x 0.87m / 60
= 17493 N x 1(jet) x pi x 1m3/s x 363.93 RPM x 0.9(eff) x 0.87m / 60
= 261 Kw

I reasonably believe the absolute pressure in the siphon nozzle can exceed 800,000 Pa (not because the force delivered by the air compressor can increase, but because the area over which that force can be applied in a narrow nozzle at the end of the siphon can be decreased. Pressure = Force / area.

[pese]

Good thread !

Another
fluid (liquid) working system is (was) its here..
(but surpressed beginnig from asked patent  (Richard Clem Motor)
http://www.keelynet.com/energy/clemreborn.html

fluid system (compressed air) with look-like
wheel system was used (working) from Mazenauer
(sure on  Schauberger´s Knowledges).

Air -Overunity- System  with strokes. find out "Zorzi" (google)
you will also "impressed".

Pese

http://alt-nrg.de/pppp
my german/englisch
(unprof.) link collection
(started 2002)


google (german) zorzi:  (see the oic´s)
http://www.google.de/#sclient=psy&hl=de&site=&source=hp&q=pesetrier+zorzi&aq=f&aqi=&aql=&oq=&pbx=1&bav=on.2,or.r_gc.r_pw.&fp=5a4a001e44ffdadc&biw=641&bih=273

[quantumtangles]

Many thanks Pese for the extremely helpful references.

I will check them out and see what I can learn from them.

Kind regards,