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Gravity powered devices => Gravity powered devices => Topic started by: quantumtangles on May 07, 2011, 03:38:20 AM
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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.
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Hi quantum can you please post a diagram?
Thanks
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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
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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
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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,
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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.
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@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
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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,
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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.
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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.
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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
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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.
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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.
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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.
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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?
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Obviously there must be a one way pressure activated flow valve connecting the bases of tanks A and B.
Hence water flows into tank A from tank B through the lower connecting pipe when pressure in tank B exceeds 350kPa.
However water cannot flow into tank B other than through the upper siphon.
The type of valve I have in mind to prevent back-flow is as follows:
http://www.3ptechnik.co.uk/en/backflowpreventionvalve.html
Apologies for omitting reference to this lower pipe valve in the diagram and item description.
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Very interesting. Have you tried the project on a small scale, for example using two great tanks? Just for looking if it the process works. Thank you for sharing your idea with the world, this is big.
Andrea
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Thanks for your post Andrea. I have not built a scale model.
I am an inventor and theoretician. Unfortunately I am not an engineer.
Of course it would be possible to build a scale model of the machine, but it will not work.
Friction in thin siphon tubing will slow the flow of water to a trickle. The velocity of the water exiting the siphon and the mass flow rate will be very low indeed.
If for example soda bottles are used as cylinders, and a balloon is used to provide external pressure, the points where the siphon tubing enters the soda bottles will leak under pressure and the bottles themselves will crumble.
I considered commissioning a glass instrument maker to build a scale model of the system, but decided not to because of the risk of vessel failure (exploding glass).
1. A full scale prototype would be preferable to a scale model in terms of functionality and economics.
2. A large diameter siphon of 0.12m to 0.16m will reduce water/siphon friction and allow a flow rate capable of generating meaningful electricity.
3. Miniature turbines and small alternator motors are extremely inefficient. Even if a scale model was precision built, it would generate much less electricity than that consumed by the miniature air compressor and miniature water pump. It would be prohibitively expensive to build a miniature turbine, miniature pressure relief valve and miniature back-flow prevention valve, as opposed to using full size 'off the shelf' components.
4. Stainless steel or tough perspex tanks must be used for safety reasons to reduce the danger of pressure failure. Machining such tanks for a scale model would cost almost as much as building a full scale version (using 'off the shelf' cylinders)
5. A full scale system is essential because at low head parameters (less than 25m) electrical output will be nominal. There can be no substitute for real height in a gravity based system. Scale models of this particular machine will not work but they will cost as much if not more than a full scale working prototype.
6. Numerous modifications and experiments may be carried out on a full scale prototype that would be impossible on a scale model.
7. From an academic funding perspective, the turbine, alternator motor, inverter and air compressor can be used for a variety of different alternative energy experiments. Accordingly the main expenses involved in building the system (aside from the two cylinders) can be used by other inventors and scientists for different experiments.
In summary, a fully operational scale model will not work and will cost a fortune. Materials used to make a scale model will further deplete rapidly diminishing natural resources.
Only a full scale prototype, after various modifications and adjustments, will be capable of demonstrating viability or experimental failure.
A prototype should not be built until my electrical power output and pressure calculations have been rigorously examined.
If mathematical objections cannot properly be raised, the rewards for building the system would greatly exceed the risk of experimental failure.
Kind regards,
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Who can explain why tank B does not need to be pressurised to 3514kPa?
...no one reply to this, what is the response Quantumangles? Thank you :)
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In answer to your question, tank B does not need to be pressurised to 3514kPa because the pressure need only exceed the pressure in tank A, and the pressure in tank A is 351kPa.
We know for sure that Tank A has base pressure of 351kPa.
This pressure has been calculated solely with reference to the height of the column of seawater, its density and gravity.
(height(25m) x density (1020kg/m3) x gravity (9.81 m/s/s)
= 250,155 Pa gauge pressure
Adding atmospheric pressure of 101,350 Pa gives absolute pressure of 351505 Pa.
So base pressure in tank A = 352kPa (+/- 1kPa)
Pressure at other points in tank A will be lower as one moves up the tank (as the height of the column of fluid decreases).
If for example we wanted to connect a pipe from tank B to a point half way up tank A, the pressure at that point (in tank A) would be half the base pressure (175.5kPa).
We know fluid must travel from areas of high pressure to areas of lower pressure.
Accordingly, if the base pressure in tank A is 351kPa, and the pressure in tank B is greater than 351kPa (because of the air compressor) fluid must then travel through the lower connecting pipe from B to A.
Upper Siphon Pressure
The problem with the siphon topside is that water must somehow flow into tank B after tank B has been pressurised to more than 351kPa.
How can this happen?
First, a powerful 30kW water pump pulls water along the siphon tube. Gravity also works through the longer 'end' of the siphon tube.
As well as this, water exiting the output nozzle of the siphon creates a partial vacuum, causing water to flow up the input nozzle of the siphon and into tank B (similar to what happens in the trunks of Great Redwood trees when water evaporates from leaves).
Another factor is the weight of the working fluid, which falls towards the base of tank B at a rate of 1020kg per second. That sort of mass is hard to stop. Only very high air pressure in tank B would prevent this.
Fortunately, the air pressure from the compressor is also fed into the exit nozzle of the siphon (as in the nozzle of an electric pressure jet washer) thus increasing fluid velocity and fluid output pressure.
We know from Pascal's law that pressure applied to a confined fluid is transmitted undiminished with equal force on equal areas at 90 degrees to the container wall.
In other words, the pressure applied by the air compressor in tank B will be transmitted to the fluid in tank A. That pressure however will zero when the fluid strikes the turbine in tank B (which will remove all kinetic energy from the fluid then in tank B but not the fluid still pushing to leave tank A and enter tank B).
Thus the turbine itself breaks any circuit of ever increasing re-transmitted pressure (more so than the air gap at the top of tank A, the pressure relief valve and the back-flow prevention valve).
The turbine itself is the perfect antidote to P1V1=P2V2.
The main factor preventing fluid exiting the siphon nozzle in tank B (the air pressure in tank B) is also a source of nozzle fluid propellant.
I will come back to the question of hydraulics at some future point, because changing the diameter of the respective cylinders and therefore changing the distance over which force is applied by the air compressor is an interesting subject in itself.
In any event the upper siphon allows water to flow into tank B at the same time as the lower connecting pipe forces water to flow out of tank B.
The 'connected pressure vessel circuit' is completely broken when the turbine itself removes all pressure (indeed all kinetic energy) from the working fluid.
For the above reasons, working fluid must flow into tank A from tank B through the lower connecting pipe due to air compression in tank B, and from tank A into tank B when re-pressurised by the air compressor in tank B and fed through the pump assisted siphon.
Erratum - Lower connecting pipe diameter
I now realise the connecting pipes must have greater diameter. I stumbled on this when calculating pipe friction.
Relative pipe roughness is a factor. This is calculated by dividing the absolute roughness (e) of the material used to make the pipe by the pipe diameter D (m).
Relative roughness = e / D
Drawn copper pipes have absolute roughness of 1.5 microns = 0.0015mm.
If the pipes are 0.12m in diameter, relative roughness can be calculated by dividing absolute roughness (e) (0.0015mm) by D in mm (0.12m = 120mm)
0.0015 / 120 = 1.25 x 10(-5) = 0.0000125.
Relative roughness = 0.0000125
Dealing with the lower connecting pipe first,
at an elevation of -1m (flowing from tank B down to tank A through a 1m section of 0.12m diameter pipe), and knowing relative roughness of the pipe to be 0.0000125 m/m, it is possible to calculate the change in pressure allowing for friction in the pipe.
Assuming average fluid velocity in the pipe is 9 m/s/s (mass flow in must equal mass flow out), assuming that the connecting pipe is 1m in length, that fluid density is 1020kg/m3 and fluid viscosity is 0.00108 Pa-s, then the pressure difference (allowing for friction) in a 1m length of copper pipe of diameter 0.12m at an elevation of 1m above the entry point into tank A would be approximately 8900 Pa.
However, in that event the flow rate would only be approximately 0.1m3/s.
I now realise the lower connecting pipe must have a diameter of 0.38m (not 0.12m) to maintain the required flow rate of 1 m3/s.
The upper siphon pipe will need to be 0.38m in diameter as well to maintain the required flow rate.
They are heavy duty pipes containing significant flow rates. The lower connecting pipe must work because of the pressure differential.
The upper connecting siphon cannot be stopped by higher air pressure in tank B because the force applied by the mass flow rate (even if the fluid velocity were nominal) is too great. At a velocity of 9.81m/s/s, 1m3/s of seawater delivers a force of 10,006 Newtons. It simply cannot be the case that tank B air pressure of just over 352kPa can possibly prevent the upper siphon working.
But it would be interesting to calculate how much air pressure would be needed to prevent the upper siphon working.
Delta pipe = cross sectional area of 0.38m diameter pipe comprising the siphon. Radius = 0.19m
Delta pipe = pi x r x r
Delta pipe (m2) = 3.141592654 x 0.0361m
= 0.113411494 m2
Delta pipe = 0.113411494 m2.
F = M.a
Force = 10006N (1020kg/m3 x 9.81 m/s/s)
Pressure (Pa) = Force (N) / area (m2)
Pressure = 10006N / 0.113411494 m2
= 88227.389 Pascals = 88227.389 N/m2 gauge pressure.
However, the velocity of the fluid will not be 9.81 m/s/s when it exits the siphon (leaving aside the water pump and air compressor acceleration for the moment). More likely it will be 1.5 m/s.
F = 1020kg/m3 x 1.5
F = 1530 Newtons
Pressure (Pa) = Force (N) / area (m2)
Pressure (Pa) = 1530N / 0.113411494 m2
= 13490.69 Pascals = 13490.69 N/m2 gauge pressure
This is the gauge pressure at the exit nozzle of the siphon in tank B before the water pump and the air compressor nozzle itself assists it.
To calculate absolute pressure we must add atmospheric pressure (1 bar = 101350 Pa) to the fluid exiting the siphon nozzle = 114840 Pa.
So a good approximation is that the pressure at the siphon nozzle output in tank B will be 115,000 Pa before the water pump and air compressor nozzle apply further pressure it.
The air compressor will pressurise this fluid to above 352kPa throughout all parts of the flow circuit except at the moment when all its kinetic energy is removed by the impulse turbine.
At that point, the water falls under its own weight to the base of tank B, before immediately being pressurised once again and expelled to tank A.
The air compressor outlet feeds into the siphon nozzle to increase the velocity (and pressure) of the working fluid exiting the nozzle and striking the turbine.
So whatever pressure might be thought to prevent fluid exiting the siphon nozzle is matched by the pressure of fluid in tank A.
The same pressure that forces fluid to exit the siphon nozzle in the first place (not yet taking into account the siphon water pump).
The key is the turbine acting as a pressure circuit breaker by removing all kinetic energy and therefore all pressure from the water.
Focusing on Potential Problems
What we are creating here could be thought of as an enormous electric pressure jet washer.
Absent debunking of my maths so far, I will try to act as devil's advocate myself (always tricky for inventors because objectivity tends to hurt).
Domestic electric pressure jets (such as you might use to wash your car) output massive pressure through very small nozzles (eg 160 bar = 16,000,000 Pa = 16,000,000 N/m2).
They still have high water jet velocity (eg 180 m/s) but all have very low mass flow rates (typically less than 0.00016 m3/s).
This means the force produced in Newtons by a 16,000,000 Pascal pressure jet (160 bar = 16 million pascals!) consuming 3000 watts with a fluid flow rate 0.00016m3/s and jet velocity of 179 m/s will only be about 28 Newtons (top end).
Bernoulli's equation gives us jet velocity.
P = ½ r . V2
P = Pressure (Pa)
rho = density (kg/m3)
V = velocity (m/s)
P = 160 bar = 16,000,000 Pa = 16,000,000 N/m2
rho fresh water = 1000 kg/m3
The mystery value is velocity (m/s)
16,000,000 = ½ 1000 . V2
V = 178.8854382 m/s
Newton gives us Force (per mass x acceleration)
F = m.a
F = 0.16kg/s x 179 m/s
F = 28.64 Newtons
Note that the Turbine will be at its most efficient when the runner is travelling at half the jet speed ie 89.5 m/s (see note below marked **)
The Change in momentum of the jet (assuming the water jet leaves the cups with zero absolute tangential velocity) will balance the force applied to the cup.
Accordingly:
Delta Mom = mass flow rate x Delta V
Delta Mom = mass flow rate x (Vjet – Vrunner)
Delta Mom = 0.16 x (179 – 89.5)
Delta Mom = 14.3 N
The x= 0.5 speed limit (the ratio of water jet speed to turbine speed such that turbine speed may not exceed 50% of water jet speed) is a subject unto itself which I would like to return to later*.
But what a raw deal. 16 million pascals of pressure output at the nozzle and only 14.3 Newtons of force to show for it. Pathetic.
Turning now to the system under review:
Bernoulli's equation gives us jet velocity.
P = ½ r . V2
Assuming pressure of 352kPa throughout the connected vessels.
P = Pressure (352,000 Pa)
rho = density (1020kg/m3)
V = velocity (m/s) mystery value
352,000 = ½ 1020 . V2
352,000 = 510 . V2
V2 = 352,000 / 510
V = 26.2 m/s
Higher system fluid pressure eg 450,000 Pa would give fluid velocity of 29.7 m/s.
The mass flow rate of 1 m3/s (1020kg/s) provides the following Force figures in Newtons depending on velocity:
F = 1020kg/s x 26.2 m/s/s
= 26,724 Newtons
F = 1020kg/s x 29.7 m/s/s
= 30,294 Newtons
Applying these figures to the Pmech equation for mechanical power output in watts based on estimated minimum RPM of 100 RPM at Fjet values of 26724N and 30294N respectively
Pmech = Fjet x Njet x pi x flowrate x RPM x 0.85 / 60
= 26724N x 1 x pi x 1m3/s x 100 x 0.85 / 60
= 118Kw
= 30294N x 1 x pi x 1m3/s x 100 x 0.85 / 60
= 134.8Kw
These figures seem reasonable based on the assumed RPM value under load, though they are both lower than the original calculation of 170kW output.
Note that the Turbine will be at its most efficient when the runner is travelling at half the jet speed (see note below marked **)
The Change in momentum of the jet (assuming the water jet leaves the cups with zero absolute tangential velocity) will balance the force applied to the cup.
Accordingly, revisiting the pressure washer jet example:
Delta Mom = mass flow rate x Delta V
Delta Mom = mass flow rate x (Vjet – Vrunner)
Delta Mom = 0.16 x (179 – 89.5)
Delta Mom = 14.3 N
However, by combining a high mass flow rate of 1 m3/s with median velocity (20-30 m/s) and high Fjet (force = 26724N), we have created an electric pressure jet capable of generating meaningful electrical output from a large impulse turbine, given that the pressure calculations seem to permit recirculation of the working fluid in an energy efficient path dependent manner.
In the system under review (1m3/s flow rate: Fjet = 26724N: Turbine 0.87m effective diameter) pressure would be approximately 352kPa, but that pressure would zero as the water strikes the turbine anyway. It will not help to drive tailgate water from tank B.
The kinetic energy of the water is transferred to the turbine. The water falls under its own weight to the base of tank B before being re-pressurised by the air compressor.
I am optimistic the turbine itself breaks any circuit of never ending pressure build up in the fluid (by removing all kinetic energy and therefore all water pressure).
But there are other considerations concerning power output that must be examined.
* If the x=0.5 speed limit applies to this turbine, as I reasonably believe it does, Fjet would effectively be 13362N instead of 26724N.
My idea of using a copper alloy turbine contained inside an array of adjustable neodymium magnets should prevent the x = 0.5 speed limit problem.
This would combine a turbine and alternator motor in a single unit and more importantly would allow us to reduce RPM (or runner speed in m/s or rad/s) simply by increasing torque (N.m).
Using this set up, torque can be increased by 'decreasing the distance' between a copper alloy turbine and the neodymium magnets surrounding it.
Thus we would be back to an Fjet value of 26724N because runner angular velocity (rad/s) relative to water jet speed could never exceed 50% of water jet speed.
It would ensure maximum efficiency regardless of the angular velocity ratio (regardless of how slowly the water jet was moving relative to turbine speed).
It would mean we could slow down the speed of the turbine to 50% of the speed of the water jet to ensure maximum efficiency.
We know from Bernoulli that higher pressure (counter-intuitively) means lower velocity and vice versa.
If we adjusted or moved the copper alloy turbine so close to the neodymium magnets surrounding it that 9486 N.m of torque were required to rotate the turbine at 4.9 m/s to produce 170kW of electricity, we could in that way ensure we never exceed 50% of the water jet speed of say 9.81 m/s.
RPM and torque always go hand in hand. When you increase the RPM you reduce the torque and vice versa. But without a high fluid flow rate and high Fjet in Newtons, power output will never be significant.
I suggest the flow rate and Fjet values of the proposed system are sufficiently high (1 m3/s and 26724N respectively) to generate significant electrical output
I suggest the turbine itself breaks any equalising P1V1=P2V2 pressure circuit in the connected vessels by removing all kinetic energy and therefore all pressure from the working fluid.
**
The relative runner speed (m/s), relative to the speed of the water jet Vjet (m/s) (whether under load or at runaway speed) should be a number between zero and 0.5.
In a jet pressure washer situation, it can probably never practically exceed 0.5 (because the runner or turbine speed of a small turbine will never really exceed 30 m/s at runaway rpm levels, in comparison with a constant Vjet of about 178 m/s).
But in high flow low velocity situations (such as the large system under review) the x=0.5 speed limit ratio is potentially a problem.
It can effectively half an Fjet value of 10,000N and leave you in practice with only 5,000N of force being applied to the buckets of the turbine (you may argue Fjet is a fact, and cannot be other than what it has been calculated to be, but that may not be the entire picture).
Rotational force or torque is at a maximum when the turbine runner is stationary and the jet first strikes it (when x = 0). It might equally well be said that inefficiency is at a maximum when the runner is stationary.
However, at any value of x above where x = 0, the system should generate some electricity. It is simply the efficiency of the system that declines as the value of x exceeds 0.5.
In conventional systems runner speed cannot exceed half jet speed (optimally x =0.5).
Conventional systems rely primarily on head and flow. The mass of water entering the turbine penstock has to be matched by the mass leaving it.
If the turbine moves too slowly relative to the flow of water, mass flowing into the penstock cannot escape, and the penstock and turbine become clogged with stationary water. Cross flows may act as a brake on the turbine (because of the high flow). In other words, the water may hit the back of the turbine cups and act as a brake should the turbine move too slowly relative to Vjet. The same sort of thing happens if the turbine speed exceeds half the water velocity.
I invented the combined turbine alternator unit in the hope turbine velocity can never exceed half water jet speed, regardless of how low the value for water jet speed is.
By increasing the torque, RPM may be precisely controlled to match 50% of water jet speed.
This in the hope of avoiding unpleasant consequences when subtracting vb from vj in the equation: Mb = Ajet . (vj - vb) . pwater
Derivation:
x = vb / vj
x = ratio
vb = Cup velocity at pitch circle diameter of turbine
vj = Jet velocity
F = mb. vj . (1-x) (1+ z.cos g)
h = mb . (vj . vj) . x . (1-x) . (1+z.cosg) / ½ . mb (vj . vj)
P = F . vb = mb . vj . (1-x) . (1+z.cos g) . vb = mb . vj . x . (1-x) . (1+z.cos g)
dh / dx = 2(1-2x). (1+z.cos g) = 0
x = 0.5
h = system efficiency as a unit-less fraction between zero and one
F = force of water on cups (N)
mb = mass flow rate into cup (kg/s)
vj = Jet velocity (m/s)
vb = runner tangential velocity at pitch circle diameter (m/s)
z = efficiency factor for flow in buckets (unit-less fraction between zero and 1)
g = angle of sides of cups
x = speed ratio of vj to vb
I hope these pros and cons have been of service.
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The most frustrating thing about energy generation is not the three laws of thermodynamics. It is when inventors omit to tell you the power output of their system in watts, or the power consumption of their system in watts. This is the only information of any real consequence.
Or when they show you a shaky video of a voltmeter vibrating on a table beside a wheel covered in magnets, and you can never be quite sure whether one of the many wires connected to the device is also connected to a hidden power source.
So I am going to set out here the final system specifications including power output and power consumption in watts, together with detailed mathematical calculations.
Your role here? Debunk it or build it. If you can't debunk it mathematically, build it.
I have chosen to use 30m high cylinders of 1m in diameter and a vertically mounted (horizontal) turbine with low friction neodymium magnet bearings to which a water jet with a flow rate of 1m3/s is directed.
The efficiency of such a turbine could theoretically be 0.94. I have assumed turbine efficiency of 0.9 because of the low friction bearing and size of the turbine.
I have assumed for the moment all system inefficiencies are reflected in this unit-less fraction (0.9).
Working fluid: Seawater of density 1020kg/m3
Pressure at the base of tank A:
= 30m x 1020kg/m3 x 9.81 m/s/s
= 300,186 Pa Gauge Pressure
Adding Patmos of 101350 Pa
= 401,536 Pa absolute
Bernoulli's equation gives us the velocity of the water jet applied to the turbine.
P = ½ r . V2
At pressure of 401.536kPa (in the absence of counter-pressure) the water flow velocity is as follows:
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
The mass flow rate through of 1 m3/s (1020kg/s) provides the following Force figures in Newtons:
F = 1020kg/s x 28 m/s/s
= 28560 Newtons
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 = 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
This limits the Fjet force figure in Newtons to 14280N, but we can now calculate the RPM figure for the turbine based on runner velocity of 14 m/s.
First we need to calculate 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
Applying this figure to the Pmech equation used to calculate mechanical power output in watts based on RPM of 297 RPM and the Fjet value of 14280 Newtons:
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 very 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 lets be exceptionally conservative to help the debunkers. Lets give the debunkers a 52kW head start. They're going to need it.
The water siphon primer pump consumes 36.77kW. The air compressor consumes 11kW.
I selected the water pump from flo-pump online (they also have a linked pump catalogue at:
http://www.eng-software.com/pml/
The 50 hp electric ESP-10 150 water pump can move 12000 US gallons of fresh water per minute up 0.0475 feet of head at 1150 rpm through a 10 inch diameter pipe. Apologies for the non-SI units.
Converting to SI units, this is 60,000 litres per minute = 1000 litres per second = 1m3/s.
The specifications of this and many other low head high flow rate pumps can be seen on the flo-pump website. They provide free software for pump selection based on flow rate etc.
It cannot lift fluid up very well, but it can move fluid sideways pretty well at a flow rate of 1m3/s. Which is all we need it to do as a siphon pump.
It consumes 36.77 kW of electricity.
I realise the pump specification relates to fresh water and that the working fluid in this system is salt water, but we can make allowance for that later on if need be.
The flow in the siphon is helped by the siphon pump but also by the fact there is no negative pressure in tank A (base pressure is 401kPa), as well as the air compressor outlet being fed into the siphon exit nozzle to increase the velocity and pressure of the water jet and to pressurise tank B.
Flow rate in water pumps decreases depending upon how many metres water is pumped upwards, but that is not a factor here.
So we don't have to ask the debunkers for their 52kW head-start back just yet (so far as the water siphon pump is concerned).
The 11kW air compressor is the Abac Genesis 1108 air compressor, which provides a maximum pressure output of 800,000 Pa at a rate of 59 cubic feet per minute = 1.67 m3 per minute.
The volume of tank B (h=30m 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 30m = 23.56m3.
The volume of air in tank B (after deducting the tailgate water taking up 10% by volume of the cylinder = 2.356m3) is 21.2m3.
The air compressor takes 12.697 minutes to pressurise the 21.2m3 of air inside tank B to 800,000 Pa. Note we can easily reduce the volume of tank B by 70% if we need to (by having only a pipe descending from the siphon into a cut off (shortened) tank B cylinder).
For sure the air compressor does not need to create 800,000 Pascals of pressure inside tank B.
It only needs slightly to exceed the pressure at the base of Cylinder A (401,536 Pascals). The working fluid will move from a high pressure to a lower pressure area. In other words it will move in the direction we want it to.
Accordingly the air compressor may use pulsed power (it need not run continuously).
At most it must run 50% of the time to force tailgate water back into tank A, consuming a maximum of 5.5kW in the process.
Remember that tank B is not directly open to atmospheric pressure. It is sealed save for the input of the upper siphon pipe (and the lower connecting pipe protected by the back-flow prevention valve).
So once air pressure has built up in tank B, it can only leave tank B either through the upper siphon (which would stop the system working) or the lower connecting pipe (where we would like air pressure to expel tailgate water from underneath the turbine).
But what is sauce for the Goose is sauce for the gander. The pressure we need to expel the tailgate water from tank B may also prevent the siphon from working.
The pressure difference between the top of tank A (forget the base of it) and the top of tank B is almost 300,000 Pa = 300,000 N/m2.
But this is not as bad as it seems. Household water pressure varies from area to area from 207,000 Pa to 550,000 Pa.
The critical fact is that the output 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 combined pressure of the siphon water pump and the air compressor are both focused on the exit nozzle of the siphon.
The water entering the siphon is not a tank A base pressure of 401kPa. It is at Patmos which is 101350 Pa.
The air compressor acts like a household electric jet washer when applied to the fluid in the siphon. Household electric pressure jets have typical output pressures of 160 bar = 16,000,000 Pa = 16,000,000 N/m2.
The air compressor temporarily raises the pressure in the siphon exit nozzle (in the direction of fluid travel) to 800,000 Pa (within the narrow confines of the space of the siphon nozzle) before the fluid is ejected. In the same way that an electric pressure jet washer works.
This raises the pressure of the fluid exiting the siphon nozzle first to a higher pressure and then, when released into the tank, to the same pressure as the air in tank B.
The highest pressure in tank B is at the end of the siphon nozzle (the end of the siphon nozzle is the source of the high pressure in tank B).
This increases the kinetic energy of fluid exiting the nozzle, and combined with the weight of seawater exiting the nozzle (1020kg/s) allows the siphon to keep working. There is also a vacuum effect because water that has left tank A creates a vacuum that keeps pulling water up the siphon (as do the leaves of a Giant Redwood tree).
The point of least resistance (for the higher pressure in tank B) is the lower connecting pipe because there is 10% by volume of seawater in tank B, creating extra pressure because of the height of fluid in that column.
This extra pressure (of the column of seawater in tank B) is as follows:
P(Pa) = 3m x 1020kg/m3 x 9.81 m/s/s
= 30,018.6 Pa
This pressure additional pressure in itself will be enough to drive tailgate fluid back into tank A (If we add the "Ptank(atmos)" of 451kPa to it to give us a relative absolute pressure...even if we use a shortened lower volume tank B as it will still contain this tailgate fluid column.
We can also manipulate the lower connecting pipe shape and diameter to increase pressure flow back into tank A (venturi tube).
So a combination of high mass flow rate (1020kg/s), siphon gravity dynamics stemming from there being a longer pipe descending into tank B, the siphon water pump, pressure from the fluid in tank A, the air compressor and the fact nature abhors a vacuum; all these factors will propel working fluid in the direction of the intended circuit, down the siphon into tank B and out again into tank A.
Fluid will leave tank B through the lower connecting pipe protected by the one way flow valve.
Equally, the water pump need not operate continuously.
It should only need to prime the siphon.
The siphon may keep functioning under the weight and pressure of the working fluid until the fluid level in tank A falls below the input nozzle of the siphon (which can never happen because the fluid recirculates).
A float switch below the turbine activates the air compressor when the water level rises sufficiently close to the turbine. Thus keeping power consumption to a minimum. Tank A retains much of the pressure generated by the air compressor earlier on.
So far, power expenditure is a maximum of 47.77kW though in reality it may be as little as 5.5kW.
The turbine is generating 173kW, thus providing net power output of between 125kW and 167.5kW.
The bad news is that I want the 52kW head start I gave the debunkers back as well (why not).
This means net power output is between 177kW and 219.5kW. This is 'net' power output.
The net output figures get bigger and bigger depending on the height of the cylinders used. If higher flow rates and multiple water jets are used (as well as 200m cylinders converted from old coal power station cooling towers) we enter national power production territory.
Before and after 451Kpa pressure has been exceeded in the connected cylinders, the 'connecting pressure medium' (the aqueous working fluid) will try to equalise volume and pressure in the connected cylinders as per the formula:
P1V1 = P2V2
Only a slight pressure differential (tank B having slightly more internal pressure than the base pressure in tank A) is needed to cause the P1V1 = P2V2 algorithm to do the otherwise difficult work of fluid recirculation.
The water back flow prevention valve in the lower pipe (connecting cylinder B to cylinder A) prevents the higher column of seawater in tank A ever flowing into tank B.
More importantly, all kinetic energy (and thus all "gauge" pressure) in the aqueous working fluid is removed by the impulse turbine itself (and if necessary by the pressure relief valve at the top of tank A) therefore preventing completion of the P1V1=P2V2 cycle that would otherwise require ever increasing pressure to allow the system to recirculate working fluid.
I hope I have been of service.
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Originally I thought the air compressor could operate sporadically (using pulsed power). I was wrong.
It must operate continuously.
The air compressor is essential. It has two important jobs to do.
First it must increase pressure throughout the entire volume of tank B to force seawater back into the base of tank A.
Secondly, it must also pressurise (to an even greater extent) a small volume inside the exit nozzle of the siphon, temporarily supercharging the pressure of the exiting water.
So there are really two pressure regions in tank B. First the siphon exit nozzle and secondly the rest of tank B. Similarly there would be two pressure regions when a water tap fills a bucket. First the pressure of the water at the nozzle of the tap and secondly the pressure of the water in the bucket.
The main pressure region extends over the entire area of tank B (like the bucket). But there is also a second pressure region in tank B. It is the tiny volume inside the siphon nozzle where the compressed air actually strikes the exiting water (like the nozzle of the tap in the bucket example).
This siphon exit nozzle pressure always exceeds the pressure in the rest of tank B (because the force per unit area is higher in the confined space of the siphon nozzle).
This means that a pressure relief valve will have to be placed on tank B, to prevent the rest of the air pressure in tank B equalling pressure in the confined space at the end of the siphon nozzle. I have since realised such a pressure equalisation (of tank B pressure to nozzle pressure) to be impossible as you will see if you read later posts because the air compressor will create surprisingly high pressure in the nozzle due to the much smaller area over which the same force is applied in the nozzle.
For example, let us say tank B has overall pressure of 500kPA. This pressure will exceed tank A base pressure of 400kPA. So the lower connecting pipe will flow from tank B to tank A.
At the top of the system, the air compressor will have pressure of 600kPA or more inside the confined space at the end of the siphon nozzle.
A pressure relief valve built into tank B must ensure pressure in tank B does not exceed 510kPA, so as to ensure there are always pressure differences in the three key areas in the system.
1. Tank A base pressure (400kPA)
2. Siphon nozzle exit pressure (600kPA)
and 3. Tank B pressure (500 kPa).
Think of the siphon nozzle as the small space at the end into which the entire outlet of the air compressor pumps all the compressed air.
Because the compressed air is released into a small space inside the siphon nozzle, it creates a very high pressure zone that allows water to enter the relatively high pressure environment of tank B (high relative to tank A base pressure).
Once the compressed air has done the job of supercharging the siphon nozzle, it must join the rest of the air in tank B and contribute to providing higher overall pressure in tank B than exists at the base of tank A.
In practical terms, the air compressor outlet will have to be carefully positioned inside the exit nozzle of the siphon.
As in all useful systems, it is an open system. Mass enters and leaves the system boundary (to a greater extent than I first thought).
Energy also enters and leaves the system boundary.
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It is now clear that a pressure relief valve must be fitted to the top of tank B.
This is because there are effectively two different pressure regions in tank B.
There is a 500kPa region in almost the entirety of tank B, but there is also a much smaller 600kPa+ region in the confined space of the siphon nozzle aperture (where the compressed air feed pressurises the exiting water jet and force per unit area is higher in that confined space than in the rest of tank B).
To prevent all of tank B becoming pressurised to 600kPa (which would slow nozzle exit velocity) there must be a 510kPa pressure relief valve built into the side of tank B.
This may not be as wasteful as it seems.
This 'excess' air pressure (expelled by the pressure relief valve in tank B) does not have to be wasted if cylinders are built in arrays or collections of cylinder pairs.
Which is to say the excess pressure from the largest pair of cylinders can be used to pressurise the B tanks of the next smallest pair of cylinders.
This idea makes the construction of consecutive sets or arrays of cylinder pairs (of incrementally decreasing height) viable.
It would mean an array of cylinder pairs would be more efficient than any isolated pair of cylinders.
Larger cylinder pairs would in this way 'help' smaller pairs (as one sometimes finds in naturally occurring open boundary systems including mammals).
The excess pressure that must be released by the largest cylinder pair can be used by the B tank of the second pair of cylinders and so on and so forth.
Only the final (smallest) pair of cylinders in an array will have to release excess pressure from tank B directly to the environment. Only that would be wasted.
I foresee forests of cylinder pairs, all seeming to be the same height, unless viewed from a distance.
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The 11kW air compressor (Abac Genesis 1108) provides maximum pressure output of 800,000 Pa at a rate of 59 cubic feet per minute = 1.67 m3 per minute = 0.027833 m3/s.
But can we increase the pressure of the water flowing from the siphon using low flow rate compressed air?
For sure the force applied by the compressed air is not going to increase just because we apply it inside a thin siphon pipe.
Pressure is force per unit area. Pressure will increase not because the force has increased, but because the area over which force is applied 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
So this is the maximum surface area to which pressure from the air compressor can be applied to the water exiting the siphon (whilst it is still inside the siphon).
Ideally we would like to apply 800,000 Pa to an area of 0.1134m2 (siphon cross section area).
The 800,000 Pascal output of the air compressor (0.027833 m3/s) can be delivered to the siphon nozzle over an area of 0.1134m2.
The extra pressure and velocity should be enough to allow the siphon to work, despite 500kPA of pressure in tank B.
It will be applying the same force (as the air compressor always applies) over a smaller area in the narrow siphon tube. The same force it later applies over a greater area in the wider cylindrical expanse of tank B (forcing tailgate water back into tank A).
It is rather like an hydraulic cylinder in reverse.
Usually, a force F is applied a long distance over a small area in a narrow cylinder, enabling that same force F (which has not increased) to lift a heavy weight a short distance in a connected wider cylinder.
When you press the brake pedal in your car, you move the pedal perhaps 0.1m. The brake pads pressing against the wheels only move perhaps 0.01m.
The same force applied over greater distance in a narrow cylinder leading to the same force being applied over a shorter distance throughout a wider cylinder.
The force did not increase. The pressure increased because the area over which it was applied decreased.
Here, a force F is applied a short distance to a narrow (0.19m diameter) cylinder (the siphon nozzle) and later that same force is applied over a much greater distance to a much wider cylinder (tank B of diameter 1m).
Quite confusing. But essentially it means the pressure (force per unit area) applied by the air compressor to the siphon nozzle will be at least 5 times greater than the pressure applied by the air compressor to the base of tank B.
That would mean 2,500,000 Pa being applied inside the siphon nozzle as compared to 500,000 Pa to the walls of tank B.
But we have only compared the ratio of the diameter of the siphon nozzle to the diameter of tank B (about 0.2m to 1m). 5 to 1.
What we should really compare is the relative area, because pressure is force per unit area.
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) will be approximately:
A = 2*pi*r2 + 2*pi*r*h
A = 1.57 + 94.24778
A = 95.8177 m2
So if the maximum pressure that can be applied by the air compressor to the walls of tank B is 800,000 Pa, the pressure that can be 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).
But the pressure relief valve will not prevent very high pressure being generated inside the siphon nozzle, because this pressure 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 exceed 600Kpa. (greatly exceed it)
The same force (which can never be more than it was to begin with) applied over a smaller area will increase pressure.
So I think this is the answer. The air compressor can apply 500,000 Pa to the walls of tank B, having already applied much greater pressure (in terms of force per unit area) to the water exiting the siphon nozzle.
This means siphon water will flow into tank B, despite 500kPa pressure in tank B.
The air compressor outlet and siphon nozzle must be connected together in an efficient way, but it should work.
**Pressure conversions
The pressure applied by the air compressor in the siphon nozzle can be more than 800,000 Pa. Once again, not because the force has increased, but because the area over which the force is applied has decreased.
Counter-intuitively, pressure of 800,000 Pa = 800,000 N/m2, when applied to a square centimetre, would be 80 N/cm2. One divides by 1000, rather than multiplies.
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Operating Instructions
Ensure all electrical fittings comply with local electrical safety requirements and have been installed by fully qualified industrial electricians.
Ensure all electrical fixtures are turned off at the mains supply.
Take extra precautions against the possibility of electrocution. Water and electricity are extremely dangerous in combination. Wear rubber boots and rubber gloves at all times when preparing the machine for operation.
Check that the two cylinders are firmly affixed to the ground with strong securing bolts throughout the circumference of each cylinder base. If any of the securing bolts are missing or damaged, do not operate the machine.
If there is any possibility of one of the cylinders becoming unstable during operation, if for example high winds could destabilise the cylinders, do not operate the machine.
Remember that if one of the cylinders falls over, it could cause serious injury. In addition, it could fall onto the other cylinder, introducing a serious risk of traumatic injury or electrocution.
Ensure the turbine has been disconnected from the alternator motor.
Before turning on the electrical mains supply (for the air compressor and water pump) ensure tank A is full to the brim with water.
Check the one way water valve. This is situated at the base of tank A where the lower connecting pipe connects the two cylinders.
Disconnect the lower connecting pipe, and press the one way water valve with your hand from the outside of tank A. It should open and allow a little water to flow from tank A onto the floor.
When you release the one way pressure valve, it should shut firmly and the rubber seal should not allow any further water to flow from tank A onto the floor.
Then reconnect the lower connecting pipe and ensure it is well secured both to tank A and tank B.
Replace any water that has flowed from tank A during the valve test. Tank A should be full to the brim. Mop up any water that has fallen on the floor during the valve test.
Then check the Pelton turbine. The buckets of the Pelton turbine should not be pitted or marked. This will reduce efficiency. Pitted or marked buckets should be replaced. This will involve removing the turbine and fitting new buckets.
There should be no need to replace the entire turbine unless it has been cast. Ensure all buckets are very tightly secured to the turbine hub (unless it is a cast turbine in which event it will be a single unit) and that the turbine itself is well secured to the shaft.
If the turbine fails during operation, it could cause serious injury or electrocution. A rapidly spinning turbine can rupture the walls of the cylinders causing serious injury by trauma or electrocution.
Check the turbine shaft for stress fractures. It should be perfectly straight and not to any extent damaged. The turbine should turn easily in your hand unless it has already been connected to the alternator motor (it should not yet have been connected to the alternator).
There should be no perceptible wobble when you rotate the turbine, either on the circumference of the turbine itself or the shaft of the turbine. Even the slightest wobble will become greatly amplified at high RPM and may cause catastrophic system failure.
Once you are satisfied the turbine is operating correctly, it should be connected to the shaft (or belt drive) of the alternator motor.
Ensure the alternator motor has been correctly connected to a recommended inverter for that type of alternator by a qualified industrial electrician.
Check the seals where the shaft enters tank B for deterioration. If the seals could cause the shaft to become unstable at high RPM, do not operate the machine.
If you are using a belt drive to connect the shaft of the turbine to the alternator motor, check the wheels, bearings and tension of the belt drive in accordance with the manufacturers instructions.
If the shaft becomes disconnected or the belt drive fails during operation, this could cause serious damage to the turbine. Turbines are not designed to operate at high RPM without being under load.
Fill the upper siphon tube with water so there are no air gaps in it. This is the most difficult part of preparing the machine for use.
Consideration should be given, when building the machine, to how the upper siphon can most easily be filled with water before operation.
Two small holes could have been made on either side of the crest of the siphon to fill it before use.
In that event, water should be poured into these holes after sealing the 'tank B end' of the siphon (the tank A end will already be immersed in water and will not need to be sealed).
Alternatively the siphon tube should have been made in three sections to make it easier to fill with water.
First, there should be a section of tube leading up from tank A. This can easily be filled with water when the tube is disassembled simply by pouring water into it from above.
The fact that tank A is already full of water will keep section 1 of the siphon pipe full to the brim.
The second section, a straight section connecting the shorter to the longer end of the siphon, must be completely filled with water and then each end sealed with plastic and waterproof tape until final assembly.
Section 3 leading down into tank B (the longer end of the siphon) must also be filled with water in the same way. The sections should be assembled whilst keeping the exit nozzle (the longer end of the siphon) sealed until final assembly. Otherwise water will flow out of it and the siphon will not work when the machine is switched on.
Depending on how the siphon pipe has been made, it may be possible to detach it from the machine completely, turn it upside down, fill it with water, seal the ends, and then assemble it, taking care not to let water spill from the tank B end of the siphon after placing the 'tank A end' in tank A first and then rotating the pipe until it has been fitted onto tank B.
Then fill Tank B with water to a level of 3m.
Once the siphon pipe and tank A are full of water, and tank B has been filled with water to a level of 3m, turn on the mains electrical supply to the machine.
Then switch on the air compressor and allow pressure in tank B to reach 500kPa (5 bar). This will take about 8 minutes.
The correct tank pressure should be visible from the pressure gauge on top of tank B.
Then, turn on the siphon pump at the precise moment the exit nozzle of the siphon is unsealed, allowing water to flow from tank A to tank B and onto the turbine.
The machine is now generating electricity.
The mains supply connected to the water pump and air compressor can now be disconnected as the inverter will now be supplying power to the system.
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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 – 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.
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.
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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.
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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 – 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 – Vrunner)
Delta Mom = 1020kg/s x (14 m/s – 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 – Vrunner)
Delta Mom = 1020kg/s x (24 m/s – 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 – Vrunner)
Delta Mom = 1020kg/s x (34.3 m/s – 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.
-
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
-
Many thanks Pese for the extremely helpful references.
I will check them out and see what I can learn from them.
Kind regards,
-
Tks,
In overview both systems
(air and liquis fkid()
it is possible that the forces comes from
shock-changing its termal souronding
between time of compression and expanding
(like heat-pump?)
so i remember that also anthirs usefully (?) system
was found , that worked both wit gas and liquid fluid.
-an motor that worked with compressed "FRIGEN",
it is know ?. It was not "PAPP" ! This ist enother
stroke Motor wit "nobel-gas" ....
I look to find out and add..
G. Pese
-
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,
-
I'm wondering what if you used a body of water deep enough that had a crush depth equal to the pressure in the tank and redesigning the tanks into a system to produce energy? (may be submarine design out of floats that take on water. Then crushes and shoots the water it takes on to sink, through a tube to a generator, then back out into the ocean again. then has springs that push it back open above the crush depth drawing in water like a pendulum movement.)
I'm wondering also what would a pipe that zig zag's have reduce the pressure inside the pipe so water would flow through it to a higher less pressure water depth?
-
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.
-
This is a link to water pressure for depth: http://www.calctool.org/CALC/other/games/depth_press
Do you know what a pressure tank is it use in plumbing? Its a tank with pressure the water is force in the tank and it keeps pressure allowing water to flow without having the motor running all the time.
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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,
-
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.
Hi quantumangles, I have another question. If you maintain the pressure in the tank B by pumping air on it, where will the plus of air that you pump go ? If the system is closed, the air that you pump inside should remains inside, I think. Isn't this a problem? Thank you if you could clearify this point. Andrea
-
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.
-
here an collection (part of ma homepage)
http://alt-nrg.de/pppp/HHO_AIR_ELECTRO_%28Cars%29_EM12.html
specialle you can give attention ti this here:
motor with compressed refrigant !!
http://www.innovativetech.us/FutureProd-np.htm
--------------------- if you have interest, look ober the next lines
end of list more in englisch laguages....
(and my URL). I am sure , you will finde some "extras" thas was
collected over years...
only simple html nitices to remember speyialle myself,
so it was not collected as (for) professionals.
Gustav Pese
-----------------------
AIR LUFT
http://pesetrier.150m.com/luft.html Verweis zu AirAccess
http://www.aladin24.de/Mazenauer/mazenauer.htm
MAZENAUER (System Schauberger Clem vergleichbar.)
http://peswiki.com/index.php/Directory:Generator_from_Ambient_Air_by_Kim_Zorzi
Zorzi GENERATOR (Luftkompression- Wärmepumpenkonzept!!)
http://www.rexresearch.com/feg/feg1.htm LUFT/ WÄRME(KÄLTE) SYSTEM Housten 1925
--------------------------------------------------- Febr.2010
Thermoakustische Wärmepumpe
http://pesetrier.150m.com/luft.html Begleittext zu AirAccess
http://aircaraccess.com !!
Hitzeenergie erzeugen durch Luft-Komprimierung.
Durch Komprimierung von Luft entstehen Temperaturen bis 350- 500° F 170-260°C (Rechner hierzu: http://www.celsius-fahrenheit.de/ )
(Bekanntes Beispiel zur Klärung der Wirkung (Fahrradluftpumpe wird heiss.)
60-90 % des erzeugten Luftdruckes gehen nach Abkühlung wieder verloren
Der Luftdruck ist also hier nur eine kleinere Energiequelle , die jedoch
einfach gespeichert werden kann (und abrufbar bleibt)
(Werkzeugantriebe. Strom-Generatoren (Licht oder Wärme))
--------------------------------------------
46-00 Thermoacoustic Heat Pumps
http://www.aircaraccess.com/pdf/thp%20cover.pdf thp cover.pdf 26kb
http://www.aircaraccess.com/pdf/thp.pdf thp.pdf 4.84 mb
-----------------------------------------------------
Hello Gustav,
you write me, for your air-wheel that produce air pressure
(no realy efficient you sayd.
BUT i find 1000 page now over air compression
and this here is an almanach , specially for you:
http://www.aircaraccess.com/download.htm so i think , i MUST informe you
specially http://www.aircaraccess.com/pdf/racd%20i-ii,%201-50.pdf
let kow you that 60-90% of the pressure go lost
if tge 350-500°F hot aire will coling in the boilers.
So yüu can frofuce an air powered (wheel) heating system, also
collect as Hot wather in an boiler system.
Not necesary to say, that the stored pressed air, can als uses
(-also after collect the pressed air
for mechanical applications , Air driven electric converter can
produced also in germany (ar even othe first countries
Use your system to hear rge house, ar even an swimmimgpool
http://72.14.205.104/shttp://www.innovativetech.us/FutureProd-np.htmearch?hl=en&q=cache%3Awww.ultralightamerica.com%2Fair_power.htm&btnG=Google+Search
http://www.aircaraccess.com/download.htm 1000+ pages air compressed power
http://72.14.205.104/search?hl=en&q=cache%3Awww.ultralightamerica.com%2Fair_power.htm&btnG=Google+Searchp://peswiki.com/index.php/Directory:Generahttp://alt-nrg.de/pppp/HHO_AIR_ELECTRO_%28Cars%29_EM12.htmltor_from_Ambient_Air_by_Kim_Zorzi
https://www.abbeon.com/store/item.cfm?code=2082 vortex-tube hot cold without moving parts
Jim
http://www.innovativetech.us/FutureProd-np.htm
Hier von 2 "Vorstellungen DEN 2. ansehen.
Hydraulic Motor run under pressured refrigant !!
folglich Fluid/Gas Wandlung da thermisch Wandlung
http://www.innovativetech.us/-np.htm
http://www.google.de/patents/about?id=uLZPAAAAEBAJ&dq=1781062 1925 Houston
http://www.google.de/patents?id=uUxFAAAAEBAJ&printsec=drawing&zoom=4#v=onepage&q=&f=false 1940
http://www.google.de/patents?id=uNdbAAAAEBAJ&printsec=drawing&zoom=4#v=onepage&q=&f=false ++++
mocullum
http://www.google.de/patents?id=VAseAAAAEBAJ&printsec=drawing&zoom=4#v=onepage&q=&f=false
---- LUFT
full page and HOME you find here:
http://alt-nrg.de/pppp/HHO_AIR_ELECTRO_%28Cars%29_EM12.html
-
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,
-
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
-
Thanks again Pese. Very interesting ;D
-
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.
-
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.
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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.
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If Glycerol were used to fill tank A (density 1261kg/m3), in that event, water (of lower density than the Glycerol) would be able to rise to the top of tank A through the denser fluid.
Gravity would propel working fluid (water) onto the impulse turbine in tank B, but Glycerol would cause working fluid to rise back to the top of tank A.
In this way, gravity can be used to perform work on the working fluid, whilst at the same time, the positive buoyancy of water in Glycerol would render otiose the fact gravity is a conservative force.
We have a down (gravity) and an up (buoyancy).
Obviously energy must be spent to force working fluid back into the base of tank A (a 30m high cylinder 90% full of Glycerol). The amount of energy we have to spend is directly proportionate to the pressure that must be overcome.
The pressure at the base of tank A is the critical issue.
P(Pa) = 1261kg/m3 x (30m x 0.9) x 9.81 m/s/s
= 334,001 Pascals Gauge
= 334,001 + 101325 Patmos
= 435,326 Pascals Absolute.
So the air compressor must perform sufficient work to force working fluid through the lower back flow prevention valve at the base of the tanks (overcoming pressure of 435.3kPa) so it ends up back in tank A.
If tank A contains 90% Glycerol the air compressor will have to overcome 435326 Pa of pressure to force tailgate fluid from tank B back into tank A, plus pressure exerted by 3m of working fluid at the top of tank A (approximately a further 30,000 Pa).
I am not sure how the flow rate of working fluid (water) would be affected by having to rise up through the dense substrate of Glycerol in tank A.
However, by widening the diameter of cylinder A (increasing its volume) one might thereby ensure that sufficient working fluid remained available to the input end of the siphon (so that water floating on top of the Glycerol cannot fall below the input level of the siphon).
This would enable the pump assisted siphon to continue delivering working fluid to tank B, which in turn would enable the impulse turbine in tank B to continue generating electricity.
The Glycerol would end up doing most of the hard work (raising water to the top of tank A).
Please read my earlier posts and throw some maths at me.
This is beginning to feel like a solo attempt (to solve the energy crisis).
There is another reason I would value input.
It is because in quiet rooms, greedy people read the threads on this site, and calculate how they may try to profit from them personally.
NO. These ideas are open source. They cannot be patented by individuals or corporations.
Lets keep it that way. Lets share information, debunk our errors, and spread innovative ideas to protect the poor and vulnerable from the greed of others.
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Ah yes .. this is where I joined this thread a few months ago & saw your post 2 back.
I'm out of time today - tried to take a look at your picture of the proposed system on page 1 but couldn't wait for the download - dang, I'm on dial-up speed for the rest of the month as my son used up my allocation last night downloading movies.
I've gisted the rest of your posts in this thread which was no easy task ;7)
Let's stick with the logic of your system first & stay away from math supporting it until later down the track, if required.
In your latest iteration above [& correct me if I'm wrong] glycerol is in tank A & the working fluid is sea water - picking the water at top of A as a starting point for the cycle, water falls via a pipe to enter tank B [tank B maybe already filled with water so that volume entering must also leave via the down pipe to the turbine (pelton wheel) so that tank B always has a constant volume.
N.B. I'm going to leave aside the compressed air used to accelerate the falling sea water to a greater velocity than gravity alone can achieve for the moment - also I'm going to leave it out of the system where it is also used to push a volume of water thru the one-way valve into the bottom of tank A where it rises due to less density than glycerol.
Falling water [gravity acceleration alone] hits a turbine near the bottom which is connected to a generator to produce electricity.
N.B. After a certain height loss the water 'packet' has Kinetic Energy [Ke] - it also has momentum - the traditional way of explaining how a turbine works is that it is a partial momentum transfer process - turbines efficiency of transferring momentum are bound by Betz Law which says that a maximum of 59.3% of KINETIC ENERGY can be converted into mechanical energy to do Work - this sounds correct because we also see by observation that it is only a partial momentum transfer from the water molecules to the impellers because if it were a 100% transfer of momentum then the water would be stopped - it isn't & after turning the turbine still has velocity downwards.
So already there is going to be an energy shortfall, before we take into account other losses like internal pipe friction losses, generator to motor losses thru heat etc - just establishing benchmarks here.
Ok, the turbine, coupled to the generator makes electrical Pe [Joules] - this energy output we know is insufficient to lift an equivalent mass per sec [1 m^3] the required height to recirculate the system - it is also insufficient energy to pass that same volume thru the check valve in the same time period into the bottom of tank A against the outward pressure of the internal liquid in tank A - as I said earlier, if it could push past the check valve it could also lift the water the full restoration height & we know it can't do that.
So, the achilles heel of the system is that the outward pressure from the bottom [side wall] of tank A is too great to push the required replenishment volume thru the check valve in the same time period, as the volume delivered to the turbine in the same time period - we just don't have enough mechanical energy which is in physics terms Work Done [force x distance in Joules].
So we have to find ways to mitigate the system or augment it - you have proposed both - in one you propose to use compressed air to super accelerate the the delivery volume to the turbine so that more Ke is transferred into mechanical energy - in the second you propose that the same source of compressed air is used to act as a piston to push the replenishment volume thru the check value to close the system.
I won't discuss the relative merits of either yet as they muddy the water at this stage - both are downstream effects [excuse the liquid metaphors] of assuming you can charge a tank full of compressed air & get it to do mechanical work [we know how many Joules are required] for less energy in Joules than the Input required to charge the tank - IOW's, the Input Work Done mechanical Joules required to compress air is assumed to be less than the Output Work Done mechanical energy Joules to mitigate & augment the system into a state of recirculation equilibrium at least.
Putting aside math & theory for a moment - ask yourself does this basic premiss look & sound logical ?
So, IMO, practicalities, engineering & math issues aside, you are hoping to find a way to draw extra energy from an external source to top up the system energy budget so that it at least balances.
A readily available source of heat energy is from the atmosphere - ambient air can be used as a heat sink for both cooling & heating purposes - according to thermodynamics there are only two ways to increase a systems energy quotient - do Work on it e.g. mechanical work compressing a gas, or to have ambient heat enter the system - in one you supply the extra energy from an unrelated source & in the other ambient energy enters the system because you create a heat differential & heat flows from hot to cold re entropy.
This is where there is perhaps possibility to augment your energy deficit with heat energy drawn into the system from ambient air at the compressor stage - compressed gas has mechanical work done on it - the volume decreases & the temp increases - this is adiabatic heating [quick heating] where PV&T are proportional etc [Boyle's & Charles's Laws]- heat flows from the compressed gas tank to the atmosphere [perhaps aided by a water jacket] & the compressed gas cools to ambient & looses a little pressure because it now has a lesser volume - but .. when that gas is released to do Work it expands & gets even colder - this has potential to draw heat energy into a system by Isothermal [slow] warming, where the ambient heat enters re_expanding the air volume creating additional pressure to do Work - otherwise I'm at a loss to know how you can do mechanical work to compress gas & expect it to do mechanical work on water [i.e. volume per time period] & have surplus energy left over ?
Gotta go.
P.S. in case you're wondering - can it take less mechanical energy Joules to push water past the check valve than it takes to lift the same mass of water the full height of water ?
NO, the two are equivalent - when you open the tank side the pressure pushes against your piston or whatever - since water's pressure is derived by it's density [mass] then a certain frontal area will have exerted on it a certain pressure - since pressure is normal to faces presented to it then it is the equivalent of raising the same volume the full 30 m's or the whole tank mass by 10 cm for instance - that is when you push something into the tank sideways, or from any direction actually - this is height dependent - think of it like a L tube from bottom to the top - when you push sideways the top level raises - so you are still having to find enough force & energy to lift 1 m^3 30 meters.
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I've gisted the rest of your posts in this thread which was no easy task ;7)
Let's stick with the logic of your system first & stay away from math supporting it until later down the track, if required.
In your latest iteration above [& correct me if I'm wrong] glycerol is in tank A & the working fluid is sea water - picking the water at top of A as a starting point for the cycle, water falls via a pipe to enter tank B [tank B maybe already filled with water so that volume entering must also leave via the down pipe to the turbine (pelton wheel) so that tank B always has a constant volume.
N.B. I'm going to leave aside the compressed air used to accelerate the falling sea water to a greater velocity than gravity alone can achieve for the moment - also I'm going to leave it out of the system where it is also used to push a volume of water thru the one-way valve into the bottom of tank A where it rises due to less density than glycerol.
Falling water [gravity acceleration alone] hits a turbine near the bottom which is connected to a generator to produce electricity.
N.B. After a certain height loss the water 'packet' has Kinetic Energy [Ke] - it also has momentum - the traditional way of explaining how a turbine works is that it is a partial momentum transfer process - turbines efficiency of transferring momentum are bound by Betz Law which says that a maximum of 59.3% of KINETIC ENERGY can be converted into mechanical energy to do Work - this sounds correct because we also see by observation that it is only a partial momentum transfer from the water molecules to the impellers because if it were a 100% transfer of momentum then the water would be stopped - it isn't & after turning the turbine still has velocity downwards.
So already there is going to be an energy shortfall, before we take into account other losses like internal pipe friction losses, generator to motor losses thru heat etc - just establishing benchmarks here.
Ok, the turbine, coupled to the generator makes electrical Pe [Joules] - this energy output we know is insufficient to lift an equivalent mass per sec [1 m^3] the required height to recirculate the system - it is also insufficient energy to pass that same volume thru the check valve in the same time period into the bottom of tank A against the outward pressure of the internal liquid in tank A - as I said earlier, if it could push past the check valve it could also lift the water the full restoration height & we know it can't do that.
So, the achilles heel of the system is that the outward pressure from the bottom [side wall] of tank A is too great to push the required replenishment volume thru the check valve in the same time period, as the volume delivered to the turbine in the same time period - we just don't have enough mechanical energy which is in physics terms Work Done [force x distance in Joules].
So we have to find ways to mitigate the system or augment it - you have proposed both - in one you propose to use compressed air to super accelerate the the delivery volume to the turbine so that more Ke is transferred into mechanical energy - in the second you propose that the same source of compressed air is used to act as a piston to push the replenishment volume thru the check value to close the system.
I won't discuss the relative merits of either yet as they muddy the water at this stage - both are downstream effects [excuse the liquid metaphors] of assuming you can charge a tank full of compressed air & get it to do mechanical work [we know how many Joules are required] for less energy in Joules than the Input required to charge the tank - IOW's, the Input Work Done mechanical Joules required to compress air is assumed to be less than the Output Work Done mechanical energy Joules to mitigate & augment the system into a state of recirculation equilibrium at least.
Putting aside math & theory for a moment - ask yourself does this basic premiss look & sound logical ?
So, IMO, practicalities, engineering & math issues aside, you are hoping to find a way to draw extra energy from an external source to top up the system energy budget so that it at least balances.
A readily available source of heat energy is from the atmosphere - ambient air can be used as a heat sink for both cooling & heating purposes - according to thermodynamics there are only two ways to increase a systems energy quotient - do Work on it e.g. mechanical work compressing a gas, or to have ambient heat enter the system - in one you supply the extra energy from an unrelated source & in the other ambient energy enters the system because you create a heat differential & heat flows from hot to cold re entropy.
This is where there is perhaps possibility to augment your energy deficit with heat energy drawn into the system from ambient air at the compressor stage - compressed gas has mechanical work done on it - the volume decreases & the temp increases - this is adiabatic heating [quick heating] where PV&T are proportional etc [Boyle's & Charles's Laws]- heat flows from the compressed gas tank to the atmosphere [perhaps aided by a water jacket] & the compressed gas cools to ambient & looses a little pressure because it now has a lesser volume - but .. when that gas is released to do Work it expands & gets even colder - this has potential to draw heat energy into a system by Isothermal [slow] warming, where the ambient heat enters re_expanding the air volume creating additional pressure to do Work - otherwise I'm at a loss to know how you can do mechanical work to compress gas & expect it to do mechanical work on water [i.e. volume per time period] & have surplus energy left over ?
Gotta go.
P.S. in case you're wondering - can it take less mechanical energy Joules to push water past the check valve than it takes to lift the same mass of water the full height of water ?
NO, the two are equivalent - when you open the tank side the pressure pushes against your piston or whatever - since water's pressure is derived by it's density [mass] then a certain frontal area will have exerted on it a certain pressure - since pressure is normal to faces presented to it then it is the equivalent of raising the same volume the full 30 m's or the whole tank mass by 10 cm for instance - that is when you push something into the tank sideways, or from any direction actually - this is height dependent - think of it like a L tube from bottom to the top - when you push sideways the top level raises - so you are still having to find enough force & energy to lift 1 m^3 30 meters.
1. Apologies for the rolling thought process. I was trying to refine the idea as I went along, with the result it lacked focus.
2. The most recent posts mention fluids other than seawater (such as Glycerol) but this is not important. The working fluid is seawater and the fluid in both tanks is seawater.
3. You understand the flow system correctly (eg tanks A and B always having a constant volume despite water flow).
4. I agree about partial momentum change. In the calculations I always took care when calculating actual force in Newtons to do a Delta Mom calculation (essentially dividing the force figure by 2). Whatever figure in Newtons I began with ended up being divided in two to account for Delta Mom (which you refer to as Betz's law). Entirely agree with you about this. I do not view it as an energy shortfall. I view it as a fact.
5. We also agree there will be nowhere near enough output to lift water 30m. but my first point of disagreement is here when you go on to say that the same amount of energy would be needed to move water sideways through the check valve as would be needed to move the water up 30m. This is true in one sense but not in another. It is true over time Ti (initial), but not true over time Tc (continuous operation).
The entire principle of operation of this machine is that LESS energy is needed (over Tc) to move fluid through the lower valve than would be needed to move it up 30m. By less energy I mean less energy over the course of a defined time frame. We must come back to this idea of time in a moment.
Of course you are right to point out that pressure in tank A is just as high at the sides as at the bottom (at a depth of 30m). So pressure will be circa 400kPa everywhere at the bottom of tank A including at the sides.
What I am saying is that to overcome this 400kPa pressure (which is the only thing that must be overcome at energy expense), higher pressure needs to be exerted inside tank B.
The mere fact of higher pressure will do the same job as lifting the water 30m (at lower energy cost once time is factored in).
If I am wrong about this (wrong about there being different (path dependent and time dependent) ways of achieving the objective, some with lower energy costs than others) then the machine is of no value. It is this central point that requires analysis.
By way of example (to explain the time factor) suppose it takes 12 minutes and 300kW to pressurise tank B to 500kPa.
In the short term, the machine has spent more electrical energy than it can generate. Quite so.
But once tank B has been pressurised to 500kPa, recirculation of fluid will take place, and much less energy is required to maintain that level of pressure in tank B than was required to create it.
So in the short term, there can never be useful energy generation (when the system is being primed). But in the medium to long term (after 12 minutes has elapsed) there should be an overall energy profit due to work being performed on the system by the environment (by gravity).
You have rather brilliantly focused on the core issue. This is a key point for discussion and clarification.
Another key point is how high air pressure in tank B will retard rotation of the turbine. I have done some maths on this and this is also something that needs further analysis.
If I had to summarise the machine, I would give the example of pointing a household water jet cleaner at a Pelton turbine. The effect is remarkable. Very high RPM is acheived, but the mass flow rate (the flow of water) is very low and this leads (per F=m*a) to low force in Newtons (regardless of how fabulously high the pressure at the nozzle is). This low force in Newtons means that despite high RPM, such devices (I built and tested two of them) lack the torque to drive alternator motors effectively. RPM under load plummets because of low rotational force (low torque).
In contrast, this system has a high mass flow rate, leading to high force levels in Newtons. So the system works, but the question is one of time. How long will it work for, and does the initial energy needed to pressurise tank B (which will exceed output) have to be maintained continously (leading to net energy loss) or on a pulsed basis via a float activated switch connected to the air compressor (leading to net energy gain).
I iterate that open systems (where both mass and energy transcend the system boundary) may generate useful energy provided the environment (gravity/heat from the sun etc) performs work on the system.
This would not be the case were it an isolated system (neither mass nor energy would be able to pass through the system boundary) or a closed system (mass would be unable to pass through the system boundary).
Finally, it has just occured to me that banks of batteries could store solar charge sufficient to provide a tailgate purge boost (if needed).
In this way solar power (low on kW save in the largest installations) could augment the system in hot countries, ensuring longer operating times.
This would be particularly useful for desalination of seawater. In fact the machine could be combined with desalination equipment to reduce the energy costs of desalination (the main long term barrier to producing fresh water from the sea).
The turbine could provide a high kW bang, and solar energy could keep it working for longer with pulsed tailgate purges coming via solar batteries to the float activated air compressor switch in tank B.
The larger the tank B tailgate water space, the longer the period between purges of tailgate water (perhaps giving time for solar powered batteries to store the required charge should there be a shortfall in my calculations).
Looking at it even more simplistically, if the tanks were built 'over' seawater (by the coast of a desert country), tailgate water could be expelled directly into the ocean to increase operating times (depending upon how the tanks are sited and how large the tanks are). Tidal flows could help refill the tanks.
Tidal flow into a 1 million cubic metre tank would contain sufficient water flow continuously to operate the machine for 11.5 days, or for the time it takes for the next tide at a flow rate of 20 cubic metres per second (3,460kW of output at that flow rate).
If the tanks were 'buried' in a beach cavity (installed on a tidal beach), regular tidal flows might also refill the tanks in that way (perhaps tailgate water could be purged into a deeper cavity beneath tank B). (sorry, just thinking out loud).
Kind regards and thanks,
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We could trade thoughts back & forwards but I don't think we could reach agreement - time is critical, as you've said, but I don't believe there is any way to pressurize tank B [top up the operating pressure] for less mechanical energy input simply because you have created an energy sink [battery analogue] of compressed gas that can do mechanical work - the battery needs periodic topping up & every time you top it up it takes the same amount of mechanical work aka mechanical energy.
If you are able to supply known examples to support your contention then that would be worth considering ? - bearing in mind that the alternator/generator in this set_up can at max utilize about half the Kinetic Energy of the falling water that arrives at the pelton wheel.
I was going to go thru the points listed below but I think for expediency I'll just sum up at the end & let you think about it some more - I have no doubt about your enthusiasm & commitment to your idea but unfortunately 'wishing it were so does not make it so' - I hope that doesn't sound dismissive but the way I see it your whole proposals premiss is that it takes less energy [considerably less btw] to recharge the compressed air battery once the recirculation machine is in operation - IOW's the dynamic situation requires less input energy to sustain the system than the static situation - this is marginally true for mechanical devices with bearings etc because once in motion the dynamic friction is less than the static friction, but that is quite different to what you are proposing.
5. We also agree there will be nowhere near enough output to lift water 30m. But my first point of disagreement is here when you go on to say that the same amount of energy would be needed to move water sideways through the check valve as would be needed to move the water up 30m.
This is true in one sense but not in another.
It is true over time Ti (initial), but not true over time Tc (continuous operation).
The entire principle of operation of this machine is that LESS energy is needed (over Tc) to move fluid through the lower valve than would be needed to move it up 30m. By less energy I mean less energy over the course of a defined time frame. We must come back to this idea of time in a moment.
Why should dynamic require less input energy than static ?
What I am saying is that to overcome this 400kPa pressure (which is the only thing that must be overcome at energy expense), higher pressure needs to be exerted inside tank B.
The mere fact of higher pressure will do the same job as lifting the water 30m (at lower energy cost once time is factored in).
If I am wrong about this (wrong about there being different (path dependent and time dependent) ways of achieving the objective, some with lower energy costs than others) then the machine is of no value. It is this central point that requires analysis.
Yes, you need to provide real examples of where this is obviously happening ?
By way of example (to explain the time factor) suppose it takes 12 minutes and 300kW to pressurise tank B to 500kPa.
In the short term, the machine has spent more electrical energy than it can generate. Quite so.
But once tank B has been pressurised to 500kPa, recirculation of fluid will take place, and much less energy is required to maintain that level of pressure in tank B than was required to create it.
So in the short term, there can never be useful energy generation (when the system is being primed). But in the medium to long term (after 12 minutes has elapsed) there should be an overall energy profit due to work being performed on the system by the environment (by gravity).
See above - gravity is a force when acceleration & mass combine - the mass acquires Ke.
If I had to summarise the machine, I would give the example of pointing a household water jet cleaner at a Pelton turbine. The effect is remarkable. Very high RPM is acheived, but the mass flow rate (the flow of water) is very low and this leads (per F=m*a) to low force in Newtons (regardless of how fabulously high the pressure at the nozzle is). This low force in Newtons means that despite high RPM, such devices (I built and tested two of them) lack the torque to drive alternator motors effectively. RPM under load plummets because of low rotational force (low torque).
In contrast, this system has a high mass flow rate, leading to high force levels in Newtons. So the system works, but the question is one of time. How long will it work for, and does the initial energy needed to pressurise tank B (which will exceed output) have to be maintained continously (leading to net energy loss) or on a pulsed basis via a float activated switch connected to the air compressor (leading to net energy gain).
I iterate that open systems (where both mass and energy transcend the system boundary) may generate useful energy provided the environment (gravity/heat from the sun etc) performs work on the system.
General Observations :
Force is not Energy - it is something that pushes or pulls something, usually associated with a mass.
Mass in movement has Kinetic Energy.
Knowing the force [torque is rotational force] is not indicative of the energy status.
When we want to know a systems likely performance we must do an energy budget.
Force has a part to play in that because it is integral to Work Done where Work Done = Force x Distance.
WD is also in Joules as for energy types, therefore they are interchangeable.
Power [the rate of doing work] is Work Done / time or Joules /sec [Watts, which also equal amps x volts].
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Your whole proposal appears to me to suggest that that Power [WD/t] required to top_up tank B is less than you would normally need if a compressed air tank were not in the circuit at all ?
I would need to see real examples where there is a significant difference in energy consumption to do mechanical work to top up a 'buffer/battery' which then releases energy via a pulse, to cause a self sustaining system, that can cover internal frictional losses, do some modicum of external work, & replenish its own energy usage, all at the same time ? - IOW's Perpetual Motion - N.B. this assumes that time has no bearing because it is accounted for in the Power terminology.
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Great post. This is the first time there has been any real debunking/analysis, for which I am most grateful.
1. You ask me for examples to support the assertion.
Lifting water requires much more energy than moving it sideways. I hope we are in agreement about this at least.
As less energy is needed to move water sideways than upwards, then the machine may work depending upon how the water is moved sideways.
Admittedly water is being moved sideways into a higher pressure environment (the base of tank A).
But base pressure in tank A can never exceed about 400,000 Pascals. So my hypothesis is simple.
I contend the energy needed to force water sideways into the base of tank A is less than the energy generated by the turbine.
The turbine generates about 173kW. I contend an air compressor (and siphon pump) can recirculate water for less than 173kW.
How?
It is all about pressure. Water always moves from high to low pressure areas. Once tank B has been pressurised to 500,000 Pascals, water MUST move into the base of tank A. It has no choice.
So I cannot point to a specific example as a proof of concept other than a high mass flow electric water jet cleaner. I point therefore to trivial physics and common sense.
Many household taps have higher water pressure than 400,000 Pascals.
If tank A were full of water, you could easily push your hand through the flap (the one way valve) at the base of tank A.
Of course water would flood the floor, but my point is that you would be able to do it. And you do not have 173kW at your disposal.
2. "Your...premise is that it takes less energy...to recharge the compressed air battery once the recirculation machine is in operation.
Why should dynamic require less input energy than static?
Yes it does. When tank B has been pressurised to 500,000 Pascals, the machine (when operating) will not require it to be "re-pressurised" to 500,000 Pascals.
in other words, when it has been pressurised, it will stay pressurised unless something causes it to lose pressure.
What in your view would cause tank B to lose pressure when the machine is operating?
You may point to the pressure relief valve in tank B that prevents pressure exceeding 500kPa. Quite so, but maintaining 500kPa is rather different to generating 500kPa from scratch. Note also that air compressor nozzle pressure is used to hyper-accelerate siphon water.
Imagine a hypothetical situation in which tank B at Ti is pressurised to 1000kPa (even though it need not be).
The time 'T' it would take the pressure in tank B to fall to 500kPa would be time during which the machine could operate without high energy expenditure (without the air compressor being on full blast).
3. "Your...proposal...suggest(s)...Power [WD/t] required to top-up tank B is less than you would normally need if a compressed air tank were not in the circuit at all?
I am sorry to say I do not understand this sentence. It is not willful misunderstanding (eg where I do not like what you say, feigning ignorance). Rather, a genuine inability to understand what you mean.
Incidentally, 'knowing the force' IS indicative of the energy status (so far as Pelton turbines are concerned).
The Pmech equation accurately forecasts mechanical power output in watts for impulse turbines.
This equation largely relies on the value for Fjet (force in Newtons applied to the turbine buckets after allowing for the 50% Delta Mom reduction).
If you get the Fjet value correct, the other variables (RPM, turbine diameter, efficiency co-efficient, mass flow rate) are trivial. In other words, knowing the force value means you can accurately predict the output in watts.
Finally, with reference to 'only half' the kinetic energy from falling water being available, I agree.
But I took this into account when performing the Delta Mom calculations.
The Fjet figures used in these equations (force in Newtons) were always halved in consequence of the Delta Mom calculation.
So if I had water jet velocity of 34 m/s and a mass flow rate of 1 m3/s (1020kg/s), I always applied 17.5 m/s as my acceleration figure (not 34 m/s), thus ending up with 17850 Newtons of force rather than 35,700 N).
Notwithstanding Delta Mom (50%) deductions (Benz's law), 17850 Newtons is a heck of a lot of force.
Thank you again for your excellent post. Please respond if you have time as it would be nice to get to the truth of the matter.
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@ Fletcher
Can we agree the power output of the machine (please double check my figures if you have time).
If we can agree notional output, going on to establish input will resolve the matter.
How much power will the air compressor and siphon water pump consume?
I have already calculated power output (+/- margin of error). It is somewhere between 173kW and 225kW.
Fine. But how much power will be consumed by the air compressor?
This question can be answered mathematically. So far I have failed properly to address it.
I would much appreciate your views or calculations concerning the consumption of the air compressor.
Put simply, if air compressor consumption exceeds 173kW, the machine cannot work.
If air compressor consumption is less than 173kW, the machine may work.
Accordingly, this is the critical question.
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Here is a website that may help you find what you are looking for.
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
This is the Hyper_Physics website - it loads on Mechanics > go to far right side to Fluids > Pressure [read/scroll down the page].
Read about Fluid Potential Energy & how to calculate it, Fluid Kinetic Energy, & Bernoulli Equations.
You can enter inputs into a calculator & find resultants etc.
There is a Work/Energy Equivalence principle so a fluid [which air is] will have an energy density - the equivalence principle means that no less energy can be spent than got from the fluid [or gas] using these equations.
If I have time I will look thru a few old text books I have to see if I can find an example you can relate to & use for your scenario.
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http://www.physicsforums.com/showthread.php?t=68977
Physics Forum discussion that relates to your question.
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@ Fletcher
You are a star in the fluid dynamics firmament. Many thanks once again for your kind help ;D
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This invention of mine does not work and cannot possibly ever work, but I would like to explain why I was wrong, after acknowledging that Fletcher's observations were, after all, perfectly correct.
First of all, using the simpler example of magnets, it has been tempting for millennia to imagine that, because magnets exert a force, that this necessarily means magnets 'contain' energy. Not so.
There is a critically important distinction between a 'force' and 'energy'. Only energy is capable of performing work.
A force is capable of performing work if energy is added to it. Magnetic PM inventions cannot work for this reason. Some sort of mechanical external energy is always needed to start them off and keep them moving.
Although (mercifully) I already knew magnetic force cannot perform work (unless energy is added...normally from someone's arm as they marvel about 'how it almost works...so I better keep trying' etc), I fell into error here in a very similar way (when it came to water and gravity).
The central fact, and it seems so obvious to me now, is that gravity, like magnetism, is a force. But like magnetism, it does not contain energy. Energy must always be added (from some external source) in order to enable either magnetism or gravity (both of which are energy free zones unless energy is externally added) to perform work.
I am truly sorry not simply for being wrong concerning the potential usefulness of a useless invention. I am mostly sorry because I perpetuated a pervasive misunderstanding commonly applied both to gravity and magnetism; a misunderstanding that may have led other people to waste time and resources trying to build a machine that in reality always required external energy to operate.
I published the idea on an open-source basis without hope or expectation of profit. I did not seek 'investors' or offer to sell 'plans'. I wanted to share what I hoped was a good idea. But I now realise I was labouring under a misapprehension. I thought my machine could help the poor and vulnerable have considerably cheaper electricity. Not so. I confused a force (gravity) with something completely different. I confused a force with 'energy', and I should have known better.
Sincere and contrite apologies.