Recirculating fluid turbine invention

[quantumtangles]

This publication by the author and inventor is not subject to a patent application or copyright. It belongs to you now. I hope my work has been articulated clearly enough to enable you to build the system.

The invention relates to an impulse turbine contained in one of two large cylinders. The short end of a siphon  at the top of the first cylinder delivers working fluid into the longer end of the siphon in the second cylinder.

A pipe also connects the bases of the cylinders, allowing fluid to travel back into the first cylinder after striking the turbine.

Recirculation of fluid is achieved with compressed air and also a conventional water pump. The compressed air prevents equalisation of pressure and volume in the two vessels and also increases the velocity of the working fluid before it strikes the turbine.

In other words the compressed air (from a pulse powered float activated compressor) forces spent working fluid back into the cylinder from which it originated (as fluid always moves from high pressure to low pressure areas and tries to equalise in connected systems unless prevented from doing so at the expense of energy).

The system described below generates 160kW of electricity using a Pelton impulse turbine connected to an alternator motor.

The water pump and air compressor consume 41kW of electricity. However the turbine generates 160kW of electricity.

The net electrical output is 119kW.

If decommissioned (coal powered) power station cooling towers were used as the system cylinders (200m high) and the flow rate were increased to 20 cubic metres of working fluid per second, the system would generate 36 megawatts of electricity minus power consumed by the water pumps and air compressors:

Pwatts = 200m x 20m3/s x 9.81m/s/s x 1020m3/kg x 0.9 (efficiency fraction).

This is possible because it is an open system (where both mass and energy flow into and out of the system boundaries).

It would not be possible if the system were isolated (where neither energy nor mass may pass the system boundary) or if it were a closed system (where energy may pass the system boundary but not mass).

The drawbacks are that high fluid flow rates of at least 1 cubic metre per second are required for meaningful power output,and also, because it is gravity based, cylinders of 25m or more are required to enable working fluid to fall at least 20 metres onto an impulse turbine to obtain meaningful output.

So the scale and expense of the system are significant, though it is intended for industrial electricity generation (rather than domestic use) or to be shared by a community.

Specifications:

Two Cylinders A and B, each 25m high and 1m in diameter, stand side by side.

Cylinder A is 90% full of seawater of density 1020kg/m3. It also contains approximately 10% by volume of castor oil (density 961kg/m3) which floats on top of the seawater. A small air gap at the top of cylinder A and a pressure relief valve are required.

A siphon of diameter 0.12m leads from the top of cylinder A down into cylinder B. An electric pump primes the siphon and begins fluid flow into cylinder B.

Cylinder B contains only air to begin with, but there is an impulse turbine connected to an alternator motor at the base of cylinder B.

A 3m space at the bottom of tank B (underneath the turbine) is needed to allow tailgate oil to accumulate without interfering with the movement of the turbine.

The siphon (the short end of which ascends from the oil on the surface of tank A) has its longer end in Cylinder B, so that the longer end of the siphon allows oil to flow into tank B and strike the turbine.

The flow rate of the oil is 1 cubic metre per second and it falls 20m (after exiting the nozzle at the long end of the siphon) before striking the turbine.

An electric pump is used to prime the siphon flowing at a flow rate of 1 cubic metre per second. The pump consumes 30kW but requires only pulsed power because once the siphon has started working, it will continue working without help from the electric pump until the level of working fluid in tank A drops below the input nozzle of the siphon.

For reasons that will become clear, the level of the working fluid in tank A cannot fall below the level of the input nozzle of the siphon.

Now we come to the critical factor. An air compressor at the top of tank B (as well as being directed to increase the velocity of the working fluid as it travels down with the help of gravity to strike the turbine) is also used to pressurise the oil that accumulates at the bottom of tank B (in the tailgate area after having struck the turbine).

This higher pressure in tank B is needed to prevent the 350kPA (absolute) pressure at the base of tank A flooding tank B through the lower connecting pipe (also of diameter 0.12m) and it is also needed to force tailgate oil back into tank A (which is full of seawater and oil).

The pressure at the base of tank A is approximately 350,000 Pascals absolute, whereas the pressure at the base of tank B (which only contains a small height of oil) would only be 130,000 Pascals absolute before the air compressor operates.

The air compressor consumes 11kW and can pressurise the volume of air inside tank B (which is hermetically sealed) to 800,000 pascals within 7 minutes.

A constant pressure of at least 350Kpa must be maintained in tank B to prevent water from tank A forcing its way into the turbine tank (into tank B) and flooding the turbine housing.

This pressure is also needed to evacuate tailgate oil from tank B. So we are using cheaply generated pressure to control the flow of fluid, and we know the fluid must flow towards the lower pressure area (in other words it must flow where we want it to flow).

There is a pressure release valve at the top of tank A to prevent P1V1 = P2V2 equilibrium in the air gap at the top of tank A.

Preventing pressure equilibrium is critical as the system will want to equalise fluid and pressure levels immediately (and would most certainly do so were it not for the pulsed power air compressor, the air gap in tank A, and the pressure relief valve at the top of tank A).

The air compressor is float activated, so that when the level of oil at the base of tank B gets too close to the rotating turbine, the air compressor is then triggered, pressurising tank B from 350Kpa to in excess of 350Kpa and forcing tailgate oil through the lower connecting pipe back into tank A where it floats to the surface of the tank.

The flow of oil onto the turbine generates 160kW.

Power (watts) = 20(m) x 961(kg/m3) x 9.81 m/s/s x 0.85 (efficiency fraction)

However the pumps used to recirculate the working fluid (the siphon pump and air compressor) together consume only 41kW if operated continuously.

Note that neither the pump nor the air compressor must work continuously. Both use pulsed power when required. A computer controlled pressure/valve regulator can prevent pressure equalisation and maximise efficiency. I would be grateful if someone would be kind enough to send me a schematic for a board to regulate pressure automatically (can we build it...yes we can).

The siphon at the top of the cylinders minimises (to zero) the work that has to be done to move oil from tank A to tank B.

The principles underlying siphons are well established and do not need expansion here.

However the positive buoyancy of the oil leading it to float back to the top of tank A does not in fact provide energy benefit.

I used the example of a less dense working fluid (castor oil) simply to illustrate that work does not have to be done to 'lift' working fluid to the top of tank A (in the sense of work performed during the lifting process up through the height of tank A).

Of course work has to be carried out to force tailgate fluid back into tank A, but that is actually the only work the system must perform in order to operate. So although gravity is a conservative force and is not path dependent, the recirculation of fluid in this system is path dependent in terms of efficiency,and pressure regulation is an efficient way of carrying it out.

Strong ionic bonds between water molecules, if expressed (allbeit inelegantly) in terms of pressure, attract one another with relative pressure of 3000kpa. In other words, the water molecules are virtually chained together. This enables processes in nature to work as well, including the phenomenon of Giant Redwood trees being able to lift water 115m into the air at a rate of 160 gallons per day.

In these trees, water evaporating from the leaves creates a partial vacuum pulling water up from the roots.

In this system, water leaving tank A in the siphon pulls more water towards the input nozzle of the siphon.

Seawater throughout tank A (and also used as a working fluid) actually performs more efficiently because it has a higher density. We know from Newton's formula (F=m.a) that a fluid of higher density will cause higher energy output from the turbine because the mass in kg will be higher and we will not have to worry about the higher viscosity of oil at low temperatures decreasing the velocity of the working fluid.

My point is that the energy ostensibly gained by having a less dense working fluid (that uses buoyancy to float up a more dense substrate) is matched and neutralised by the lower energy generated by less dense working fluid striking the turbine.

These large cylinders can form the pillars of energy pyramids. Platforms at the top of the cylinders would make excellent locations for wind turbines. Solar panels can cover the pyramids enclosing the cylinders. Geothermal energy may be generated from beneath the cylinders.

I hope I have been of service.

May the peace without opposite be with you.

[guruji]

Hi quantum can you please post a diagram?
Thanks

[quantumtangles]

Sure thing. I will scan a diagram and post it early next week. I am sorry I cannot provide one immediately as I do not have a scanner in my home.

Kind regards and thanks

[eisnad karm]

Many thanks for you post
There is lot of original thinking in this concept.
I look forward to the drawings.
I am interested in how you derived your calculations for pressures needed and power needed for pumps and compressors. Have you been able to test any of these in a practical manner?
Kind Regards
Mark

[quantumtangles]

Many thanks for the response. Detailed schematics, detailed specifications (for the pump, air compressor and turbine) including cup/bucket calculations will be posted from early next week. The initial post is a concept summary.

I used the advertised specifications of a commercially available water pump and air compressor, in terms of their capacity, power consumption etc.

Although I calculated turbine output using the conventional equation in the posting, I have also done some cross checking calculations using angular velocity and torque when under load with different types of alternator motor as well as turbine pitch circle diameter and turbine cup dimension calculations.

The upshot is that a vertically mounted (horizontal) turbine using neodymium magnet bearings and multiple jets (though no more than 4 jets) would be optimal.

I have already checked the x=0.5 speed limit (the ratio of the speed of the turbine to the speed of the water jet, such that the speed of the turbine may not exceed 50% of the speed of the water jet if maximum efficiency is to be obtained).

We are dealing with a high torque situation, and superficially low RPM or angular velocity (in radians per second) that does not cause me concern.

The upshot is that the Pmech equation figures (see below) using Fjet or force in Newtons to calculate mechanical power output in watts comply with F= m.a and dovetail with the conventional equation output figures as well.

So the only real area of interest is P1V1 = P2V2. In other words, the pressure calculations.

The Pmech equation is very useful indeed (provided one avoids unit errors when calculating the value of Fjet in Newtons from the pressure values.

Pmech (watts) = Fjet x Njet x pi x h x w x d / 60


Fjet = Force in Newtons

Njet = number of jets eg 1 jet nozzle

pi = 3.141592654

h = efficiency coefficient (unit-less fraction between 0 and 1) = eg 0.85 (it is always going to be a figure between 0.69 and 0.94 for Pelton turbines. The larger they are, the more efficient they become)

d = diameter of turbine in metres (circle representing the pitch circle diameter of the turbine, or in other words a circle whose diameter represents the point where the jet strikes the circular turbine which will not necessarily be the outermost edge of the turbine)

w = rpm (here not rad/s) = eg for example 3000 rpm

Note that the derivation of this equation in the fabulous book by Jeremy Thake entitled "The Micro-hydro Pelton Turbine Manual" published by Practical Action Publishing in 2000 (2009 edition) contains an error. He got the final equation correct but the derivation is not (probably a typo).

Kind regards,