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Author Topic: Recirculating fluid turbine invention  (Read 41211 times)

quantumtangles

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Recirculating fluid turbine invention
« on: May 07, 2011, 03:38:20 AM »
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.
« Last Edit: May 07, 2011, 01:34:03 PM by quantumtangles »

guruji

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Re: Recirculating fluid turbine invention
« Reply #1 on: May 07, 2011, 01:01:42 PM »
Hi quantum can you please post a diagram?
Thanks

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #2 on: May 07, 2011, 01:15:43 PM »
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

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Re: Recirculating fluid turbine invention
« Reply #3 on: May 07, 2011, 02:09:55 PM »
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

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Re: Recirculating fluid turbine invention
« Reply #4 on: May 07, 2011, 03:09:24 PM »
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,

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #5 on: May 07, 2011, 04:08:42 PM »
Hi again Mark,

I did not answer your question fully. I have not built or tested this particular system. I no longer have funding.

I got £10,000 from a major UK university last year for research equipment.

Once they figured out I was going for gold (over-unity) they dropped the project like a hot potato  :D

I have spent pretty all my time and money buying different types of electric water pressure jets and custom built turbines. I just love it.

I quickly learnt that mucking about with turbines is the most fun you can have with a calculator.

I learnt that mass flow rates (of water) from electric pressure jets are way too low to obtain meaningful power output from impulse turbines. This despite impressive RPM when not under load (eg 3500 rpm).

This is an inevitable consequence of F=m.a

The mass flow rates of even the most powerful household electric water jets (3000w models) is about 0.00016 m3/s, which translates to about 28.6 Newtons of force (despite the fact that a 160 bar pressure washer is delivering 16,000,000 pascals = 16,000,000 N/m2 of pressure at the nozzle.

So I designed a high flow rate system. And I'm gonna use it  :)

If the pressure in the two connected tanks can be prevented from equalising by low expenditure of energy via an air compressor, it will work, and if it works people will build it.

And I will be even happier than I already am.

eisnad karm

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Re: Recirculating fluid turbine invention
« Reply #6 on: May 07, 2011, 10:17:53 PM »
@Quantumtangles
Having worked at a university for many years they can get a little funny about what they spend their money on, which is ironic since they are suppose to be free thinking and imaginative institutions.
There is another simple way of getting a fluid up those towers. Osmosis. It is possible with the right film to build up head pressures of over 40 meters. But alas the catch....you consume the fresh water.
I wonder if there is a membrane or a catalyst that would allow the oil and water to mix the separate? That's probably a bit abstract for now.
looking forward to the drawings and eventually the evidence you have and calculations to arrive at the numbers you have..email them if you don't want them in public.
Thanks for activating my grey cells again
Kind Regards
Mark

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #7 on: May 08, 2011, 12:53:49 AM »
Osmosis is an interesting idea. It could probably cope with head in meters (in terms of raising fluid) but my concern would be the flow rate in cubic meters per second.

I do not know of osmotic materials capable of delivering more than 0.00001 cubic metres per second of water flow. It would be interesting to know if they exist.

I am digging out the pump specifications (buried somewhere in my bunker). I had this awful habit of grabbing any available scrap of paper to make manuscript notes so they are scattered everywhere.

I hope to get some critical analysis of the pressure calculations in particular.

The turbine output calculations are trivial, but the pressure calculations (whether the air compressor, air gap and pressure relief valve prevent P1V1=P2V2 equalisation) are not so trivial.

A comprehensive parametric equation would be good.

But comprehensive mathematical models of thermodynamic systems can be tricky. Especially if you are the author or inventor of a system as this can lead to bias, oversight and error.

That is why I would really appreciate critical analysis of the maths next week.

Even though inevitable heat dissipation from the pumps can largely be ignored (other than to the extent positive values for entrophy are a good thing in that they at least indicate the system to be possible and also because the unit-less fraction representing overall system efficiency can be adjusted to take into account thermal energy dissipation), it is preventing pressure equalisation within the two vessels that is the key.

Kind regards,


quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #8 on: May 08, 2011, 11:05:29 AM »
An impulse turbine made of copper alloy surrounded by adjustable neodymium magnets would be helpful for variable head variable flow situations.

In other words, the turbine and turbine housing itself becomes the alternator motor. It is a fully adjustable turbine/alternator motor in one unit.

The alloy composition has to be considered carefully. Pure copper would conduct electricity well but it is too soft and will bend or fail at high Fjet (force) values. Copper buckets would also become pitted, increasing friction and reducing efficiency.

I considered using teflon coated turbine cups in conjunction with a highly conductive copper alloy (adding tin etc).

By adjusting the distance between the turbine cups and the magnets, one can vary required torque (rotational force required) and therefore RPM to suit the Fjet value of the water jet.

A combined turbine/alternator assembly also provides a solution to the x =0.5 ratio speed limit problem (of turbine speed in m/s to water jet speed in m/s) so in very high velocity flow situations, you could reduce the distance between the copper alloy turbine and the magnets in the turbine housing to decrease the angular velocity of the turbine, thus increasing torque and keeping electrical output at maximum efficiency.

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #9 on: May 09, 2011, 03:12:55 PM »
I have tried to download diagrams of the system but the maximum download allowed by the site is half the size of one of the 5 diagrams, so I will figure out how to compress and post the diagrams.

In the meantime if anyone would like the diagrams sent by to them by email please let me know.

eisnad karm

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Re: Recirculating fluid turbine invention
« Reply #10 on: May 09, 2011, 03:19:17 PM »
Hi
many thanks for the posts
there are osmotic membranes that now have a high flow rate. I will send some links when I find them.
Will send a longer post latter.
mark

quantumtangles

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Re: Mathematics of recirculating fluid turbine Part I
« Reply #11 on: May 09, 2011, 06:16:46 PM »
Mathematics

The specifications of the final system are as follows:

Cylinder A = height 25m and diameter 1m
Cylinder B = identical to cylinder A
Working fluid = seawater of density 1020kg/m3
Flow rate of working fluid: 1 cubic meter per second
Height water falls before striking turbine = 20m
Diameter of turbine = 0.9m
Pitch Circle Diameter of turbine = 0.87m

First of all, we must calculate the maximum total electrical power in watts the system is capable of generating.

To do this we multiply the density of the working fluid (seawater = 1020kg/m3) by the height the fluid falls before hitting the turbine (20m), then by acceleration due to gravity (9.81 m/s/s) by the flow rate in cubic metres per second of the seawater (here 1 m3/s) and finally by a unit-less fraction representing turbine efficiency (85% = 0.85).

Pw ([power in watts) = 1020kg/m3 x 20m x 9.81 m/s/s x 1m/3/s (flow rate) x 0.85 (efficiency of turbine)

Pw = 170,105.4 watts = 170.1054 kW

So this is the maximum output we can get out of the turbine if 'the turbine' is 85% efficient. Other inefficiencies in the system must later be taken into account, but this is a good starting point based on head and flow rate.

We can calculate the force applied to the turbine by using Newton's equation F = m.a

If we assume for the moment that the acceleration of the working fluid will be the same as acceleration due to gravity (9.81 m/s/s), then we can calculate the force in Newtons that will be applied to the turbine by the flow rate.

F (Newtons) = Mass (kg/s) x Acceleration (m/s/s)
F = 1020kg/s x 9.81 m/s/s
F = 10006 Newtons

This is a useful piece of information.

We now know that over 10,000 Newtons of force will be applied to the buckets of the turbine by the flow of fluid striking it from above. This is an enormous amount of force.

Once we have decided the diameter of the impulse turbine (which must be less than 1m in diameter to fit inside the cylinder and must be greater than 0.2m in diameter if we are to avoid serious inefficiencies), we can also use this force figure (Fjet) in Newtons to calculate the angular velocity of the turbine in radians per second (which I converted to RPM below).

So we should be able to calculate how fast the turbine will rotate just from knowing the value of Fjet (which we calculated to be 10,006 Newtons) as well as other variables we established earlier.

I have chosen to use a Pelton impulse turbine of diameter 0.9 metres. I could have chosen a smaller diameter. I decided not to. However I could not have chosen a much larger diameter because it would not have fitted inside the system cylinder.

In fact a 0.9m diameter turbine is a giant by Pelton turbine standards. Lets look at the maths.

The equation for determining the mechanical power output in watts of a turbine is as follows:

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

Explanation:

Pmech = 170,000 watts (from the first calculation of maximum power output)

Fjet = Force in Newtons of the water striking the turbine = 10,006 Newtons (calculated above)

Njet = number of water jets = 1 jet nozzle

pi = 3.141592654

h = efficiency coefficient (unit-less fraction between 0 and 1). This 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, so 0.85 efficiency for a large turbine is an acceptable estimate = 0.85)

d = pitch circle diameter of the turbine in meters (this will be a slightly smaller diameter than the outer diameter of the turbine = 0.87m)

w = rpm (here not rad/s) which is the mystery value

The mechanical power output in watts is going to be approximately the same as the electrical power output in watts (if we make allowance for heat dissipation and the inefficiency of components other than the turbine itself which we can do at any later point).

Accordingly, applying the Pmech equation:

170,000 (watts) = Fjet (10,006 Newtons) x Njet (1) x pi (3.141592654) x h (0.85) x w (mystery value in rpm) x d (0.87m) / 60

= 10,006 X 1 X Pi x 0.85 x RPM x 0.87 / 60
= 23246 x w / 60
170,000 = 23246w / 60
170,000 = 387.4333w
w = 438.78 RPM

So the RPM figure look reasonable. It we had obtained a very high RPM figure this would been worrying because high RPM values would cause a turbine under this sort of force to fail after a few months of use.

Please stick around for part II of the maths relating to the system which I will post shortly.



quantumtangles

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Re: Recirculating fluid turbine mathematics Part II
« Reply #12 on: May 09, 2011, 07:28:17 PM »
In part I of the mathematics posting relating to this system, we established electrical output of the turbine would be 170kW using a 0.9 metre diameter turbine rotating at 439 RPM.

This is all very well provided the system can recirculate the working fluid.

If it cannot do so, the turbine will only work for a few seconds before getting flooded by tailgate water that builds up in cylinder B.

When considering the pressure in the two cylinders, the first thing we need to calculate is the pressure at the base of cylinder A (which is 25m high and full of seawater).

We can ignore the small air gap at the top of cylinder A for the moment because the air gap would reduce base pressure rather than increase it.

The formula for calculating the pressure at the bottom of a cylinder is as follows:

P = height(m) x density(kg/m3) x gravity (9.81m/s/s)
P = 25m x 1020kg/m3 x 9.81m/s/s
P = 250155 Pascals

However this is gauge pressure. We need to add atmospheric pressure to obtain the absolute pressure value of the fluid in the base of cylinder A.

Adding 101,325 Pascals of atmospheric pressure gives us an absolute pressure value at the base of Cylinder A of 351,480 Pascals.

This means that the tailgate water in Cylinder B must somehow force its way back into Cylinder A despite there being a pressure of 351.5 Kpa in cylinder A.

The pressure in Cylinder B, which is hermetically sealed and is not subject to atmospheric pressure, is due only to the  height of the tailgate water contained in it. We have not switched on the air compressor yet.

The tailgate water is only 2.5m high. So the pressure at the base of cylinder B is only 25,000 Pascals. Even if it were also subject to atmospheric pressure (which it is not) it would only have a pressure of 125,325 Pascals.

However, the air compressor at the top of cylinder B comes to the rescue. It can pressurise the volume of air in tank B to 800,000 Pascals in 10.58 minutes.

The compressor in question is the Abac Genesis 1108 air compressor which can provide a maximum pressure of 800,000 Pa at a rate of 59 cubic feet per minute = 1.67 m3 per minute.
 
The volume of tank B (h=25m d=1m)
I calculated using the formula V= pi.r2.h
V= pi x (r x r) x h (where r = radius in metres and h= height in metres)
V = 3.141592654 x (0.5 x 0.5) x 20m = 15.7m3.

The volume of air in tank B (after deducting the tailgate water taking up 10% by volume of the cylinder) is 14.13m3.

The air compressor takes just over 8 minutes to pressurise the 14.13m3 of air inside tank B to 800,000 Pa.

In fact, the air compressor does not need to create 800,000 Pascals of pressure inside tank B.

It only needs to exceed the pressure at the base of Cylinder A (351,480 Pascals).

Once 351Kpa pressure has been exceeded, the system will try to equalise pressure in both of the connected cylinders as per the formula:

P1V1 = P2V2

This formula means that the pressure multiplied by the volume in one cylinder (P1 x V1) will always equal P2 x V2 in a connected vessel (unless some force prevents equalisation).

Here the force preventing equalisation is provided by the air compressor. The pressure relief valve in Tank A breaks the equalising pressure circuit from continuing its journey into tank B.

Note that the output of the pressure relief could be used to perform work if I need to make amendments to the schematic.

In any event, once the air compressor kicks in, tailgate water must move from the area of higher pressure (at the base of tank B) into the base of tank A (which has now become the lower pressure area).

So the tailgate water is forced through the lower connecting pipe back into tank A, whereupon the siphon recirculates it back into tank B.

The pressure relief valve at the top of tank A prevents the pressure in the air gap exceeding 350Kpa. So any compressed air or excess fluid forced into tank A (which will try and cause the pressure in tank A to become the same as in tank B) will be released by the pressure relief valve, thus ensuring no equalisation of pressure in the two tanks (or dangerous pressure build up in tank A).

This air compressor consumes 11kW of electricity when operating at maximum capacity.

Maximum capacity involves generating 800,000 Pascals of pressure.

The air compressor should be able to expel tailgate water from tank B at less than 50% of its operating capacity. In other words, it will consume approximately 5.5kW of power to pressurise tank B to just over 351,480 Pascals.

Only when the pressure in tank B falls below 351,480 Pascals will the float trigger the air compressor (only when the water level in tank B nears the turbine will the air compressor be activated).

However, even if the air compressor continuously consumed 11kW, it would still consume only a small fraction of the 170kW output of the turbine.

In part III of the system mathematics I want to look at the pressure calculations in more detail.

« Last Edit: May 10, 2011, 03:05:28 AM by quantumtangles »

quantumtangles

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Re: Schematic I
« Reply #13 on: May 09, 2011, 09:36:27 PM »
Here are images of the schematic for the recirculating fluid turbine. The first image shows the entire system. Clearer pictures of part of the system follow as the labels on the main image are hard to read.

quantumtangles

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Re: Recirculating fluid turbine invention Mathematics Part III
« Reply #14 on: May 15, 2011, 09:29:46 PM »
We know from sections I and II of the earlier calculations that:
Tank A has base pressure of 351480 Pascals
Tank A water volume = 15.7m3
Tank B has base pressure of 25,000 Pascals.
Tank B water volume = 1.57m3

The system will try to equalise pressure and volume. Ceteris paribus:
P1V1 = P2V2

If P2 is the mystery value
351480 x 15.7 = P2 x 1.57m3
5518236 = P2 X 1.57m3
P2 = 3,514,800 Pascals
This is an enormous figure (3514 Kpa)

Who can explain why tank B does not need to be pressurised to 3514kPa?