Recirculating fluid turbine invention

[quantumtangles]

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 â€" Vrunner)
Delta Mom = 1020kg/s x (28 m/s â€" 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.

[quantumtangles]

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.

[quantumtangles]

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]

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.

[quantumtangles]

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.