Thanks again Pese.

The whole idea of an 'air bubble' in which combustion takes place (rather than inside a cylinder) combined with the idea of using the fuel itself as a working fluid (as well as fuel for combustion) is very interesting.

In the first mathematical model of this machine, I considered using light crude oil to power an impulse turbine and using the same oil as both a working fluid for the turbine and also to fuel a combustion engine to recirculate the fluid striking the turbine.

But I abandoned the idea for a variety of reasons. I did not like the idea of using light crude oil (I would be pleased to invent anything than another way to harm our planet) and I knew that working fluid used to power the combustion engine would somehow have to be replaced.

I much preferred the idea of using seawater, despite its corrosive tendencies, because it is plentiful and relatively dense compared to fresh water (giving more punch per cubic metre).

Even castor oil as a buoyant working fluid floating up through tank A (due to its lower density) did not appeal to me because the energy gained from its positive buoyancy (floating to the top of tank A) was lost when its lower density as a working fluid was taken into account. Counter-intuitively there was no net energy gain.

Oil also mingles with water to a certain extent, decreasing flow rate. Flow rate problems also arise from high viscosity at low temperatures. So oil is a horror story as a working fluid or fuel (if you want to generate clean energy).

But yes I like the idea of a combined combustion engine turbine, with combustion chambers in vortices actually inside the working fluid, possibly generated by shock waves or thermal anomalies. Very interesting.

I am not sure if the working fluid in the second device you refer to is dichlorodifluoromethane (freon 12) or a noble gas (such as Neon, argon, krypton or xenon). They all have low boiling and melting points (useful in refrigeration).

Kind regards and thanks,

Interesting idea Frii32.

The only thing that matters in this particular system is not so much 'pressure' as it is 'relative pressure'.

The key question is this: What is the pressure at the base of Tank A relative to the pressure in Tank B.

This value (the difference in pressure) is important because it determines whether fluid can move back into tank A from tank B. Pressure differences also determine the velocity of the working fluid travelling through the upper siphon. If the siphon fluid moves slowly or not at all, due to higher pressure in tank B than siphon nozzle pressure, the machine will produce little or no electricity.

By way of example, if the pressure at the base of tank A is 400,000 Pascals absolute, and the pressure in tank B (due to the air compressor) is 500,000 Pascals absolute, then clearly water should flow through the lower connecting pipe (from the high pressure area) back into tank A (the relatively lower pressure area).

But what about the upper siphon?

At the top of the tanks, the surface of Tank A will be at atmospheric pressure (100,000 Pa). But the inside of tank B will be 500,000 Pa absolute. The water in the siphon will be unable to flow from the lower pressure area at the surface of tank A to the higher pressure region in tank B.

That is why the air compressor nozzle is released 'inside' the siphon nozzle to create 600,000 Pa pressure inside the siphon. This causes the siphon pressure (600,000 Pa) to exceed the pressure inside tank B (500,000).

The velocity of the siphon fluid is relative to the overall pressure (per Bernoulli's equation) and velocity determines force (per Newton's equation F = m.a).

So if you line the usual suspects up against a wall, pressure in Pascals (force divided by area) determines velocity (m/s), which determines acceleration (m/s/s), which determines force (newtons) which determines power output (watts).

Your idea of having submerged tanks is very interesting. The only observation I make is that pressure at the base of tank A (however much it increases at depth by placing it under the sea) still has to be relative to tank B pressure, and siphon pressure must also be considered in terms of the relative pressure in tank B.

In other words, when we increase the value of pressure in one part of the system, we must compensate for this in other parts of the system.

The energy used to crush tanks and expel water would have to be matched by energy expenditure to un-crush the tanks at some later point.

So I am not sure depth would help for this particular system. Although by placing the tanks in the sea, waves of seawater entering the system could supply both external energy and mass. So your idea is an interesting one.

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,

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.

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,

Thanks again Pese. Very interesting 

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 – Vrunner)
Delta Mom = 1020kg/s x (28 m/s – 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 – Vrunner)
Delta Mom = 1020kg/s x (34.3 m/s – 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.