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Author Topic: Recirculating fluid turbine invention  (Read 41204 times)

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #15 on: May 17, 2011, 04:46:39 PM »
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.
« Last Edit: May 18, 2011, 02:51:58 AM by quantumtangles »

andrea

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Re: Recirculating fluid turbine invention
« Reply #16 on: May 18, 2011, 10:29:54 AM »
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

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #17 on: May 18, 2011, 02:12:03 PM »
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,

andrea

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Re: Recirculating fluid turbine invention Mathematics Part III
« Reply #18 on: May 18, 2011, 04:37:58 PM »

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 :)

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #19 on: May 18, 2011, 06:20:18 PM »
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.
« Last Edit: May 19, 2011, 05:19:35 AM by quantumtangles »

quantumtangles

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Final System Specifications
« Reply #20 on: May 19, 2011, 01:54:18 PM »
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.
« Last Edit: May 19, 2011, 10:55:49 PM by quantumtangles »

quantumtangles

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Re: Supercharging the siphon exit nozzle
« Reply #21 on: May 20, 2011, 04:21:18 PM »
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.
« Last Edit: May 21, 2011, 01:06:08 AM by quantumtangles »

quantumtangles

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Re: Forests of Cylinders
« Reply #22 on: May 20, 2011, 05:57:52 PM »
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.


quantumtangles

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Re: Air compressor analysis
« Reply #23 on: May 20, 2011, 07:23:19 PM »
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.
« Last Edit: May 21, 2011, 06:57:29 AM by quantumtangles »

quantumtangles

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Operating Instructions
« Reply #24 on: May 21, 2011, 06:40:23 AM »
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.


« Last Edit: May 21, 2011, 04:07:16 PM by quantumtangles »

quantumtangles

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Re: System summary, specifications and calculations
« Reply #25 on: May 22, 2011, 12:44:18 PM »
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.
« Last Edit: May 22, 2011, 01:44:55 PM by quantumtangles »

quantumtangles

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Addendum
« Reply #26 on: May 23, 2011, 02:26:31 PM »
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.

« Last Edit: May 23, 2011, 03:41:16 PM by quantumtangles »

quantumtangles

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Consideration of all forces acting on the turbine
« Reply #27 on: May 24, 2011, 04:10:16 AM »
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.
« Last Edit: May 24, 2011, 08:47:40 AM by quantumtangles »

pese

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Re: Recirculating fluid turbine invention
« Reply #28 on: May 24, 2011, 06:25:39 AM »
Good thread !

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

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

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

Pese

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


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

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #29 on: May 24, 2011, 08:41:13 AM »
Many thanks Pese for the extremely helpful references.

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

Kind regards,