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Author Topic: Recirculating fluid turbine invention  (Read 41203 times)

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #45 on: June 11, 2011, 09:15:40 AM »
If Glycerol were used to fill tank A (density 1261kg/m3), in that event, water (of lower density than the Glycerol) would be able to rise to the top of tank A through the denser fluid.

Gravity would propel working fluid (water) onto the impulse turbine in tank B, but Glycerol would cause working fluid to rise back to the top of tank A.

In this way, gravity can be used to perform work on the working fluid, whilst at the same time, the positive buoyancy of water in Glycerol would render otiose the fact gravity is a conservative force.

We have a down (gravity) and an up (buoyancy).

Obviously energy must be spent to force working fluid back into the base of tank A (a 30m high cylinder 90% full of Glycerol). The amount of energy we have to spend is directly proportionate to the pressure that must be overcome.

The pressure at the base of tank A is the critical issue.

P(Pa) = 1261kg/m3 x (30m x 0.9) x 9.81 m/s/s

= 334,001 Pascals Gauge
= 334,001 + 101325 Patmos
= 435,326 Pascals Absolute.

So the air compressor must perform sufficient work to force working fluid through the lower back flow prevention valve at the base of the tanks (overcoming pressure of 435.3kPa) so it ends up back in tank A.

If tank A contains 90% Glycerol the air compressor will have to overcome 435326 Pa of pressure to force tailgate fluid from tank B back into tank A, plus pressure exerted by 3m of working fluid at the top of tank A (approximately a further 30,000 Pa).

I am not sure how the flow rate of working fluid (water) would be affected by having to rise up through the dense substrate of Glycerol in tank A.

However, by widening the diameter of cylinder A (increasing its volume) one might thereby ensure that sufficient working fluid remained available to the input end of the siphon (so that water floating on top of the Glycerol cannot fall below the input level of the siphon).

This would enable the pump assisted siphon to continue delivering working fluid to tank B, which in turn would enable the impulse turbine in tank B to continue generating electricity.

The Glycerol would end up doing most of the hard work (raising water to the top of tank A).

Please read my earlier posts and throw some maths at me.

This is beginning to feel like a solo attempt (to solve the energy crisis).

There is another reason I would value input.

It is because in quiet rooms, greedy people read the threads on this site, and calculate how they may try to profit from them personally.

NO. These ideas are open source. They cannot be patented by individuals or corporations.

Lets keep it that way. Lets share information, debunk our errors, and spread innovative ideas to protect the poor and vulnerable from the greed of others.


« Last Edit: June 11, 2011, 09:50:10 AM by quantumtangles »

fletcher

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Re: Recirculating fluid turbine invention
« Reply #46 on: August 05, 2011, 02:46:14 AM »
Ah yes .. this is where I joined this thread a few months ago & saw your post 2 back.

I'm out of time today - tried to take a look at your picture of the proposed system on page 1 but couldn't wait for the download - dang, I'm on dial-up speed for the rest of the month as my son used up my allocation last night downloading movies.

I've gisted the rest of your posts in this thread which was no easy task ;7)

Let's stick with the logic of your system first & stay away from math supporting it until later down the track, if required.

In your latest iteration above [& correct me if I'm wrong] glycerol is in tank A & the working fluid is sea water - picking the water at top of A as a starting point for the cycle, water falls via a pipe to enter tank B [tank B maybe already filled with water so that volume entering must also leave via the down pipe to the turbine (pelton wheel) so that tank B always has a constant volume.

N.B. I'm going to leave aside the compressed air used to accelerate the falling sea water to a greater velocity than gravity alone can achieve for the moment - also I'm going to leave it out of the system where it is also used to push a volume of water thru the one-way valve into the bottom of tank A where it rises due to less density than glycerol.

Falling water [gravity acceleration alone] hits a turbine near the bottom which is connected to a generator to produce electricity.

N.B. After a certain height loss the water 'packet' has Kinetic Energy [Ke] - it also has momentum - the traditional way of explaining how a turbine works is that it is a partial momentum transfer process - turbines efficiency of transferring momentum are bound by Betz Law which says that a maximum of 59.3% of KINETIC ENERGY can be converted into mechanical energy to do Work - this sounds correct because we also see by observation that it is only a partial momentum transfer from the water molecules to the impellers because if it were a 100% transfer of momentum then the water would be stopped - it isn't & after turning the turbine still has velocity downwards.

So already there is going to be an energy shortfall, before we take into account other losses like internal pipe friction losses, generator to motor losses thru heat etc - just establishing benchmarks here.

Ok, the turbine, coupled to the generator makes electrical Pe [Joules] - this energy output we know is insufficient to lift an equivalent mass per sec [1 m^3] the required height to recirculate the system - it is also insufficient energy to pass that same volume thru the check valve in the same time period into the bottom of tank A against the outward pressure of the internal liquid in tank A - as I said earlier, if it could push past the check valve it could also lift the water the full restoration height & we know it can't do that.

So, the achilles heel of the system is that the outward pressure from the bottom [side wall] of tank A is too great to push the required replenishment volume thru the check valve in the same time period, as the volume delivered to the turbine in the same time period - we just don't have enough mechanical energy which is in physics terms Work Done [force x distance in Joules].

So we have to find ways to mitigate the system or augment it - you have proposed both - in one you propose to use compressed air to super accelerate the the delivery volume to the turbine so that more Ke is transferred into mechanical energy - in the second you propose that the same source of compressed air is used to act as a piston to push the replenishment volume thru the check value to close the system.

I won't discuss the relative merits of either yet as they muddy the water at this stage - both are downstream effects [excuse the liquid metaphors] of assuming you can charge a tank full of compressed air & get it to do mechanical work [we know how many Joules are required] for less energy in Joules than the Input required to charge the tank - IOW's, the Input Work Done mechanical Joules required to compress air is assumed to be less than the Output Work Done mechanical energy Joules to mitigate & augment the system into a state of recirculation equilibrium at least.

Putting aside math & theory for a moment - ask yourself does this basic premiss look & sound logical ?

So, IMO, practicalities, engineering & math issues aside, you are hoping to find a way to draw extra energy from an external source to top up the system energy budget so that it at least balances.

A readily available source of heat energy is from the atmosphere - ambient air can be used as a heat sink for both cooling & heating purposes - according to thermodynamics there are only two ways to increase a systems energy quotient - do Work on it e.g. mechanical work compressing a gas, or to have ambient heat enter the system - in one you supply the extra energy from an unrelated source & in the other ambient energy enters the system because you create a heat differential & heat flows from hot to cold re entropy.

This is where there is perhaps possibility to augment your energy deficit with heat energy drawn into the system from ambient air at the compressor stage - compressed gas has mechanical work done on it - the volume decreases & the temp increases - this is adiabatic heating [quick heating] where PV&T are proportional etc [Boyle's & Charles's Laws]- heat flows from the compressed gas tank to the atmosphere [perhaps aided by a water jacket] & the compressed gas cools to ambient & looses a little pressure because it now has a lesser volume - but .. when that gas is released to do Work it expands & gets even colder - this has potential to draw heat energy into a system by Isothermal [slow] warming, where the ambient heat enters re_expanding the air volume creating additional pressure to do Work - otherwise I'm at a loss to know how you can do mechanical work to compress gas & expect it to do mechanical work on water [i.e. volume per time period] & have surplus energy left over ?

Gotta go.

P.S. in case you're wondering - can it take less mechanical energy Joules to push water past the check valve than it takes to lift the same mass of water the full height of water ?

NO, the two are equivalent - when you open the tank side the pressure pushes against your piston or whatever - since water's pressure is derived by it's density [mass] then a certain frontal area will have exerted on it a certain pressure - since pressure is normal to faces presented to it then it is the equivalent of raising the same volume the full 30 m's or the whole tank mass by 10 cm for instance - that is when you push something into the tank sideways, or from any direction actually - this is height dependent - think of it like a L tube from bottom to the top - when you push sideways the top level raises - so you are still having to find enough force & energy to lift 1 m^3  30 meters.
« Last Edit: August 05, 2011, 05:21:19 AM by fletcher »

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #47 on: August 05, 2011, 10:57:33 AM »

I've gisted the rest of your posts in this thread which was no easy task ;7)

Let's stick with the logic of your system first & stay away from math supporting it until later down the track, if required.

In your latest iteration above [& correct me if I'm wrong] glycerol is in tank A & the working fluid is sea water - picking the water at top of A as a starting point for the cycle, water falls via a pipe to enter tank B [tank B maybe already filled with water so that volume entering must also leave via the down pipe to the turbine (pelton wheel) so that tank B always has a constant volume.

N.B. I'm going to leave aside the compressed air used to accelerate the falling sea water to a greater velocity than gravity alone can achieve for the moment - also I'm going to leave it out of the system where it is also used to push a volume of water thru the one-way valve into the bottom of tank A where it rises due to less density than glycerol.

Falling water [gravity acceleration alone] hits a turbine near the bottom which is connected to a generator to produce electricity.

N.B. After a certain height loss the water 'packet' has Kinetic Energy [Ke] - it also has momentum - the traditional way of explaining how a turbine works is that it is a partial momentum transfer process - turbines efficiency of transferring momentum are bound by Betz Law which says that a maximum of 59.3% of KINETIC ENERGY can be converted into mechanical energy to do Work - this sounds correct because we also see by observation that it is only a partial momentum transfer from the water molecules to the impellers because if it were a 100% transfer of momentum then the water would be stopped - it isn't & after turning the turbine still has velocity downwards.

So already there is going to be an energy shortfall, before we take into account other losses like internal pipe friction losses, generator to motor losses thru heat etc - just establishing benchmarks here.

Ok, the turbine, coupled to the generator makes electrical Pe [Joules] - this energy output we know is insufficient to lift an equivalent mass per sec [1 m^3] the required height to recirculate the system - it is also insufficient energy to pass that same volume thru the check valve in the same time period into the bottom of tank A against the outward pressure of the internal liquid in tank A - as I said earlier, if it could push past the check valve it could also lift the water the full restoration height & we know it can't do that.

So, the achilles heel of the system is that the outward pressure from the bottom [side wall] of tank A is too great to push the required replenishment volume thru the check valve in the same time period, as the volume delivered to the turbine in the same time period - we just don't have enough mechanical energy which is in physics terms Work Done [force x distance in Joules].

So we have to find ways to mitigate the system or augment it - you have proposed both - in one you propose to use compressed air to super accelerate the the delivery volume to the turbine so that more Ke is transferred into mechanical energy - in the second you propose that the same source of compressed air is used to act as a piston to push the replenishment volume thru the check value to close the system.

I won't discuss the relative merits of either yet as they muddy the water at this stage - both are downstream effects [excuse the liquid metaphors] of assuming you can charge a tank full of compressed air & get it to do mechanical work [we know how many Joules are required] for less energy in Joules than the Input required to charge the tank - IOW's, the Input Work Done mechanical Joules required to compress air is assumed to be less than the Output Work Done mechanical energy Joules to mitigate & augment the system into a state of recirculation equilibrium at least.

Putting aside math & theory for a moment - ask yourself does this basic premiss look & sound logical ?

So, IMO, practicalities, engineering & math issues aside, you are hoping to find a way to draw extra energy from an external source to top up the system energy budget so that it at least balances.

A readily available source of heat energy is from the atmosphere - ambient air can be used as a heat sink for both cooling & heating purposes - according to thermodynamics there are only two ways to increase a systems energy quotient - do Work on it e.g. mechanical work compressing a gas, or to have ambient heat enter the system - in one you supply the extra energy from an unrelated source & in the other ambient energy enters the system because you create a heat differential & heat flows from hot to cold re entropy.

This is where there is perhaps possibility to augment your energy deficit with heat energy drawn into the system from ambient air at the compressor stage - compressed gas has mechanical work done on it - the volume decreases & the temp increases - this is adiabatic heating [quick heating] where PV&T are proportional etc [Boyle's & Charles's Laws]- heat flows from the compressed gas tank to the atmosphere [perhaps aided by a water jacket] & the compressed gas cools to ambient & looses a little pressure because it now has a lesser volume - but .. when that gas is released to do Work it expands & gets even colder - this has potential to draw heat energy into a system by Isothermal [slow] warming, where the ambient heat enters re_expanding the air volume creating additional pressure to do Work - otherwise I'm at a loss to know how you can do mechanical work to compress gas & expect it to do mechanical work on water [i.e. volume per time period] & have surplus energy left over ?

Gotta go.

P.S. in case you're wondering - can it take less mechanical energy Joules to push water past the check valve than it takes to lift the same mass of water the full height of water ?

NO, the two are equivalent - when you open the tank side the pressure pushes against your piston or whatever - since water's pressure is derived by it's density [mass] then a certain frontal area will have exerted on it a certain pressure - since pressure is normal to faces presented to it then it is the equivalent of raising the same volume the full 30 m's or the whole tank mass by 10 cm for instance - that is when you push something into the tank sideways, or from any direction actually - this is height dependent - think of it like a L tube from bottom to the top - when you push sideways the top level raises - so you are still having to find enough force & energy to lift 1 m^3  30 meters.

1. Apologies for the rolling thought process. I was trying to refine the idea as I went along, with the result it lacked focus.

2. The most recent posts mention fluids other than seawater (such as Glycerol) but this is not important. The working fluid is seawater and the fluid in both tanks is seawater.

3. You understand the flow system correctly (eg tanks A and B always having a constant volume despite water flow).

4. I agree about partial momentum change. In the calculations I always took care when calculating actual force in Newtons to do a Delta Mom calculation (essentially dividing the force figure by 2). Whatever figure in Newtons I began with ended up being divided in two to account for Delta Mom (which you refer to as Betz's law). Entirely agree with you about this. I do not view it as an energy shortfall. I view it as a fact.

5. We also agree there will be nowhere near enough output to lift water 30m. but my first point of disagreement is here when you go on to say that the same amount of energy would be needed to move water sideways through the check valve as would be needed to move the water up 30m. This is true in one sense but not in another. It is true over time Ti (initial), but not true over time Tc (continuous operation).

The entire principle of operation of this machine is that LESS energy is needed (over Tc) to move fluid through the lower valve than would be needed to move it up 30m. By less energy I mean less energy over the course of a defined time frame. We must come back to this idea of time in a moment.

Of course you are right to point out that pressure in tank A is just as high at the sides as at the bottom (at a depth of 30m). So pressure will be circa 400kPa everywhere at the bottom of tank A including at the sides.

What I am saying is that to overcome this 400kPa pressure (which is the only thing that must be overcome at energy expense), higher pressure needs to be exerted inside tank B.

The mere fact of higher pressure will do the same job as lifting the water 30m (at lower energy cost once time is factored in).

If I am wrong about this (wrong about there being different (path dependent and time dependent) ways of achieving the objective, some with lower energy costs than others) then the machine is of no value. It is this central point that requires analysis.

By way of example (to explain the time factor) suppose it takes 12 minutes and 300kW to pressurise tank B to 500kPa.

In the short term, the machine has spent more electrical energy than it can generate. Quite so.

But once tank B has been pressurised to 500kPa, recirculation of fluid will take place, and much less energy is required to maintain that level of pressure in tank B than was required to create it.

So in the short term, there can never be useful energy generation (when the system is being primed). But in the medium to long term (after 12 minutes has elapsed) there should be an overall energy profit due to work being performed on the system by the environment (by gravity).

You have rather brilliantly focused on the core issue. This is a key point for discussion and clarification.

Another key point is how high air pressure in tank B will retard rotation of the turbine. I have done some maths on this and this is also something that needs further analysis.

If I had to summarise the machine, I would give the example of pointing a household water jet cleaner at a Pelton turbine. The effect is remarkable. Very high RPM is acheived, but the mass flow rate (the flow of water) is very low and this leads (per F=m*a) to low force in Newtons (regardless of how fabulously high the pressure at the nozzle is). This low force in Newtons means that despite high RPM, such devices (I built and tested two of them) lack the torque to drive alternator motors effectively. RPM under load plummets because of low rotational force (low torque).

In contrast, this system has a high mass flow rate, leading to high force levels in Newtons. So the system works, but the question is one of time. How long will it work for, and does the initial energy needed to pressurise tank B (which will exceed output) have to be maintained continously (leading to net energy loss) or on a pulsed basis via a float activated switch connected to the air compressor (leading to net energy gain).

I iterate that open systems (where both mass and energy transcend the system boundary) may generate useful energy provided the environment (gravity/heat from the sun etc) performs work on the system.

This would not be the case were it an isolated system (neither mass nor energy would be able to pass through the system boundary) or a closed system (mass would be unable to pass through the system boundary).

Finally, it has just occured to me that banks of batteries could store solar charge sufficient to provide a tailgate purge boost (if needed).

In this way solar power (low on kW save in the largest installations) could augment the system in hot countries, ensuring longer operating times.

This would be particularly useful for desalination of seawater. In fact the machine could be combined with desalination equipment to reduce the energy costs of desalination (the main long term barrier to producing fresh water from the sea).

The turbine could provide a high kW bang, and solar energy could keep it working for longer with pulsed tailgate purges coming via solar batteries to the float activated air compressor switch in tank B.

The larger the tank B tailgate water space, the longer the period between purges of tailgate water (perhaps giving time for solar powered batteries to store the required charge should there be a shortfall in my calculations).

Looking at it even more simplistically, if the tanks were built 'over' seawater (by the coast of a desert country), tailgate water could be expelled directly into the ocean to increase operating times (depending upon how the tanks are sited and how large the tanks are). Tidal flows could help refill the tanks.

Tidal flow into a 1 million cubic metre tank would contain sufficient water flow continuously to operate the machine for 11.5 days, or for the time it takes for the next tide at a flow rate of 20 cubic metres per second (3,460kW of output at that flow rate).

If the tanks were 'buried' in a beach cavity (installed on a tidal beach), regular tidal flows might also refill the tanks in that way (perhaps tailgate water could be purged into a deeper cavity beneath tank B). (sorry, just thinking out loud).

Kind regards and thanks,
« Last Edit: August 05, 2011, 04:24:19 PM by quantumtangles »

fletcher

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Re: Recirculating fluid turbine invention
« Reply #48 on: August 06, 2011, 10:21:39 PM »
We could trade thoughts back & forwards but I don't think we could reach agreement - time is critical, as you've said, but I don't believe there is any way to pressurize tank B [top up the operating pressure] for less mechanical energy input simply because you have created an energy sink [battery analogue] of compressed gas that can do mechanical work - the battery needs periodic topping up & every time you top it up it takes the same amount of mechanical work aka mechanical energy.

If you are able to supply known examples to support your contention then that would be worth considering ? - bearing in mind that the alternator/generator in this set_up can at max utilize about half the Kinetic Energy of the falling water that arrives at the pelton wheel.

I was going to go thru the points listed below but I think for expediency I'll just sum up at the end & let you think about it some more - I have no doubt about your enthusiasm & commitment to your idea but unfortunately 'wishing it were so does not make it so' - I hope that doesn't sound dismissive but the way I see it your whole proposals premiss is that it takes less energy [considerably less btw] to recharge the compressed air battery once the recirculation machine is in operation - IOW's the dynamic situation requires less input energy to sustain the system than the static situation - this is marginally true for mechanical devices with bearings etc because once in motion the dynamic friction is less than the static friction, but that is quite different to what you are proposing.

Quote from: quantumt

5. We also agree there will be nowhere near enough output to lift water 30m. But my first point of disagreement is here when you go on to say that the same amount of energy would be needed to move water sideways through the check valve as would be needed to move the water up 30m.

This is true in one sense but not in another.

It is true over time Ti (initial), but not true over time Tc (continuous operation).

The entire principle of operation of this machine is that LESS energy is needed (over Tc) to move fluid through the lower valve than would be needed to move it up 30m. By less energy I mean less energy over the course of a defined time frame. We must come back to this idea of time in a moment.



Why should dynamic require less input energy than static ?


Quote from: quantumt

What I am saying is that to overcome this 400kPa pressure (which is the only thing that must be overcome at energy expense), higher pressure needs to be exerted inside tank B.

The mere fact of higher pressure will do the same job as lifting the water 30m (at lower energy cost once time is factored in).

If I am wrong about this (wrong about there being different (path dependent and time dependent) ways of achieving the objective, some with lower energy costs than others) then the machine is of no value. It is this central point that requires analysis.



Yes, you need to provide real examples of where this is obviously happening ?


Quote from: quantumt

By way of example (to explain the time factor) suppose it takes 12 minutes and 300kW to pressurise tank B to 500kPa.

In the short term, the machine has spent more electrical energy than it can generate. Quite so.

But once tank B has been pressurised to 500kPa, recirculation of fluid will take place, and much less energy is required to maintain that level of pressure in tank B than was required to create it.

So in the short term, there can never be useful energy generation (when the system is being primed). But in the medium to long term (after 12 minutes has elapsed) there should be an overall energy profit due to work being performed on the system by the environment (by gravity).



See above - gravity is a force when acceleration & mass combine - the mass acquires Ke.


Quote from: quantumt

If I had to summarise the machine, I would give the example of pointing a household water jet cleaner at a Pelton turbine. The effect is remarkable. Very high RPM is acheived, but the mass flow rate (the flow of water) is very low and this leads (per F=m*a) to low force in Newtons (regardless of how fabulously high the pressure at the nozzle is). This low force in Newtons means that despite high RPM, such devices (I built and tested two of them) lack the torque to drive alternator motors effectively. RPM under load plummets because of low rotational force (low torque).

In contrast, this system has a high mass flow rate, leading to high force levels in Newtons. So the system works, but the question is one of time. How long will it work for, and does the initial energy needed to pressurise tank B (which will exceed output) have to be maintained continously (leading to net energy loss) or on a pulsed basis via a float activated switch connected to the air compressor (leading to net energy gain).

I iterate that open systems (where both mass and energy transcend the system boundary) may generate useful energy provided the environment (gravity/heat from the sun etc) performs work on the system.


General Observations :

Force is not Energy - it is something that pushes or pulls something, usually associated with a mass.

Mass in movement has Kinetic Energy.

Knowing the force [torque is rotational force] is not indicative of the energy status.

When we want to know a systems likely performance we must do an energy budget.

Force has a part to play in that because it is integral to Work Done where Work Done = Force x Distance.

WD is also in Joules as for energy types, therefore they are interchangeable.

Power [the rate of doing work] is Work Done / time or Joules /sec [Watts, which also equal amps x volts].

-----------------------------------------

Your whole proposal appears to me to suggest that that Power [WD/t] required to top_up tank B is less than you would normally need if a compressed air tank were not in the circuit at all ?

I would need to see real examples where there is a significant difference in energy consumption to do mechanical work to top up a 'buffer/battery' which then releases energy via a pulse, to cause a self sustaining system, that can cover internal frictional losses, do some modicum of external work, & replenish its own energy usage, all at the same time ? - IOW's Perpetual Motion - N.B. this assumes that time has no bearing because it is accounted for in the Power terminology.

« Last Edit: August 06, 2011, 11:44:57 PM by fletcher »

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #49 on: August 10, 2011, 08:44:00 PM »
Great post. This is the first time there has been any real debunking/analysis, for which I am most grateful.

1. You ask me for examples to support the assertion.

Lifting water requires much more energy than moving it sideways. I hope we are in agreement about this at least.

As less energy is needed to move water sideways than upwards, then the machine may work depending upon how the water is moved sideways.

Admittedly water is being moved sideways into a higher pressure environment (the base of tank A).

But base pressure in tank A can never exceed about 400,000 Pascals. So my hypothesis is simple.

I contend the energy needed to force water sideways into the base of tank A is less than the energy generated by the turbine.

The turbine generates about 173kW. I contend an air compressor (and siphon pump) can recirculate water for less than 173kW.

How?

It is all about pressure. Water always moves from high to low pressure areas. Once tank B has been pressurised to 500,000 Pascals, water MUST move into the base of tank A. It has no choice.

So I cannot point to a specific example as a proof of concept other than a high mass flow electric water jet cleaner. I point therefore to trivial physics and common sense.

Many household taps have higher water pressure than 400,000 Pascals.

If tank A were full of water, you could easily push your hand through the flap (the one way valve) at the base of tank A.

Of course water would flood the floor, but my point is that you would be able to do it. And you do not have 173kW at your disposal.

2. "Your...premise is that it takes less energy...to recharge the compressed air battery once the recirculation machine is in operation.
Why should dynamic require less input energy than static?

Yes it does. When tank B has been pressurised to 500,000 Pascals, the machine (when operating) will not require it to be "re-pressurised" to 500,000 Pascals.

in other words, when it has been pressurised, it will stay pressurised unless something causes it to lose pressure.

What in your view would cause tank B to lose pressure when the machine is operating?

You may point to the pressure relief valve in tank B that prevents pressure exceeding 500kPa. Quite so, but maintaining 500kPa is rather different to generating 500kPa from scratch. Note also that air compressor nozzle pressure is used to hyper-accelerate siphon water.

Imagine a hypothetical situation in which tank B at Ti is pressurised to 1000kPa (even though it need not be).

The time 'T' it would take the pressure in tank B to fall to 500kPa would be time during which the machine could operate without high energy expenditure (without the air compressor being on full blast).

3. "Your...proposal...suggest(s)...Power [WD/t] required to top-up tank B is less than you would normally need if a compressed air tank were not in the circuit at all?

I am sorry to say I do not understand this sentence. It is not willful misunderstanding (eg where I do not like what you say, feigning ignorance). Rather, a genuine inability to understand what you mean.

Incidentally, 'knowing the force' IS indicative of the energy status (so far as Pelton turbines are concerned).

The Pmech equation accurately forecasts mechanical power output in watts for impulse turbines.

This equation largely relies on the value for Fjet (force in Newtons applied to the turbine buckets after allowing for the 50% Delta Mom reduction).

If you get the Fjet value correct, the other variables (RPM, turbine diameter, efficiency co-efficient, mass flow rate) are trivial. In other words, knowing the force value means you can accurately predict the output in watts.

Finally, with reference to 'only half' the kinetic energy from falling water being available, I agree.

But I took this into account when performing the Delta Mom calculations.

The Fjet figures used in these equations (force in Newtons) were always halved in consequence of the Delta Mom calculation.

So if I had water jet velocity of 34 m/s and a mass flow rate of 1 m3/s (1020kg/s), I always applied 17.5 m/s as my acceleration figure (not 34 m/s), thus ending up with 17850 Newtons of force rather than 35,700 N).

Notwithstanding Delta Mom (50%) deductions (Benz's law), 17850 Newtons is a heck of a lot of force.

Thank you again for your excellent post. Please respond if you have time as it would be nice to get to the truth of the matter.
« Last Edit: August 10, 2011, 11:40:53 PM by quantumtangles »

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #50 on: August 11, 2011, 11:10:01 AM »
@ Fletcher

Can we agree the power output of the machine (please double check my figures if you have time).

If we can agree notional output, going on to establish input will resolve the matter.

How much power will the air compressor and siphon water pump consume?

I have already calculated power output (+/- margin of error). It is somewhere between 173kW and 225kW.

Fine. But how much power will be consumed by the air compressor?

This question can be answered mathematically. So far I have failed properly to address it.

I would much appreciate your views or calculations concerning the consumption of the air compressor.

Put simply, if air compressor consumption exceeds 173kW, the machine cannot work.

If air compressor consumption is less than 173kW, the machine may work.

Accordingly, this is the critical question.

fletcher

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Re: Recirculating fluid turbine invention
« Reply #51 on: August 17, 2011, 02:06:02 AM »
Here is a website that may help you find what you are looking for.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

This is the Hyper_Physics website - it loads on Mechanics > go to far right side to Fluids > Pressure [read/scroll down the page].

Read about Fluid Potential Energy & how to calculate it, Fluid Kinetic Energy, & Bernoulli Equations.

You can enter inputs into a calculator & find resultants etc.

There is a Work/Energy Equivalence principle so a fluid [which air is] will have an energy density - the equivalence principle means that no less energy can be spent than got from the fluid [or gas] using these equations.

If I have time I will look thru a few old text books I have to see if I can find an example you can relate to & use for your scenario.

-------------------------------

http://www.physicsforums.com/showthread.php?t=68977

Physics Forum discussion that relates to your question.
« Last Edit: August 17, 2011, 06:17:35 AM by fletcher »

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #52 on: August 17, 2011, 11:01:43 AM »
@ Fletcher

You are a star in the fluid dynamics firmament. Many thanks once again for your kind help ;D

quantumtangles

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Re: Recirculating fluid turbine invention
« Reply #53 on: March 09, 2012, 03:01:26 AM »
This invention of mine does not work and cannot possibly ever work, but I would like to explain why I was wrong, after acknowledging that Fletcher's observations were, after all, perfectly correct.



First of all, using the simpler example of magnets, it has been tempting for millennia to imagine that, because magnets exert a force, that this necessarily means magnets 'contain' energy. Not so.


There is a critically important distinction between a 'force' and 'energy'. Only energy is capable of performing work.


A force is capable of performing work if energy is added to it. Magnetic PM inventions cannot work for this reason. Some sort of mechanical external energy is always needed to start them off and keep them moving.


Although (mercifully) I already knew magnetic force cannot perform work (unless energy is added...normally from someone's arm as they marvel about 'how it almost works...so I better keep trying' etc), I fell into error here in a very similar way (when it came to water and gravity).


The central fact, and it seems so obvious to me now, is that gravity, like magnetism, is a force. But like magnetism, it does not contain energy. Energy must always be added (from some external source) in order to enable either magnetism or gravity (both of which are energy free zones unless energy is externally added) to perform work.


I am truly sorry not simply for being wrong concerning the potential usefulness of a useless invention. I am mostly sorry because I perpetuated a pervasive misunderstanding commonly applied both to gravity and magnetism; a misunderstanding that may have led other people to waste time and resources trying to build a machine that in reality always required external energy to operate.


I published the idea on an open-source basis without hope or expectation of profit. I did not seek 'investors' or offer to sell 'plans'. I wanted to share what I hoped was a good idea. But I now realise I was labouring under a misapprehension. I thought my machine could help the poor and vulnerable have considerably cheaper electricity. Not so. I confused a force (gravity) with something completely different. I confused a force with 'energy', and I should have known better.


Sincere and contrite apologies.