Recirculating fluid turbine invention

[quantumtangles]

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.

[andrea]

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]

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]

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]

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 â€" Vrunner)
Delta Mom = 0.16 x (179 â€" 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 â€" Vrunner)
Delta Mom = 0.16 x (179 â€" 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.