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Water wave energy usage => Wave (from the beach) energy => Topic started by: TommeyLReed on January 12, 2011, 11:01:18 PM

This is another project I'm working on, is using displacement of water to convert it into direct rotational energy.
I should have a prototype done soon.
What do you think?
http://www.youtube.com/watch?v=AnraqeoGaNU
http://overunitynow.com/page26
Tommey Reed

This is another project I'm working on, is using displacement of water to convert it into direct rotational energy.
I should have a prototype done soon.
What do you think?
http://www.youtube.com/watch?v=AnraqeoGaNU
http://overunitynow.com/page26
Tommey Reed
When you provide air to a cone at the bottom, you replace a certain water volume with air. As water is incompressible, this air volume can be made only by raising the water above, what is strictly equivalent to raise to the surface the same volume of water as the volume of air in the cone. Work is v*d*g*h where v is the air volume in a cone at the bottom, d the water volumetric mass, and h the height.
It follows that the work needed to compress air in order to fill a cone is exactly what the system will provide back when the cone will rise up, restoring towards the bottom the volume of water that had been displaced upwards.
No extra energy. Buoyancy and Archimedes principle is a well known phenomenon.

As the air rises, it expands and displaces more water assuming you did not fill the cup all the way.
Does a cubic foot of displacement have the same amount of buoyancy 100 feet below the surface as it does just below the surface? Yes because the water is incompressible.
So the energy you gain is due to expansion of the water as the cup rises. The taller the bucket line the more efficient it gets.
Sound about right?

Tommey Reed,
i think you have overlooked something. 436.96 lb of upward force sounds like a lot, but you should look closer.
let's say the compressor is a simple box, 1 foot x 1 foot x 1 foot (1 cubic foot). the top is moveable and when weight is added to the top, it would go inside compressing the air. to get your 4.4 psi (pounds per square inch), you would need 144 (1 sq ft) x 4.4 = 633.6 lbs. of course the volume inside will be reduced also. this sounds like a bit more than your 436.96 you predict.
maybe you should have another look at this.
tom

My data say that 4.4(salt water), psi is needed to force air under water at a depth of 10ft. fill the cube (12^3) 1728 cu/in having a displacement of 1cu/ft of are. This will produce at a 10ft lift:(62.3*10)=623lb
pressure ((4.4 psi x 1728)/12)=633.6lb is what I get.
Now as you may know to compress air at a rate of 3cfm@90 psi=about 1hp, this could also much more volume at (90psi/18)=5 psi@(3*18)=54cfm.
Would not a 1hp make a total of 54cfm@5psi?
If 6.28 cu/ft in needed to turn a 2' wheel to turn 1 revolution, this will have a constant load of 62.3lb*10ft=623ft/lb.
At 60 rpm's or 376.8 cu/ft of air is need to make ((623*60)/5252)=7.12hp
If 7.11hp is used to pump the air pressure is it takes 1hp for each 54cfm@5 psi. Would this not also say that 7.11*54cfm=384.3?
384.3/376.8=1.02,102% efficient?
I do know that the less psi to compress, the more efficient the pump can be...

As the air rises, it expands and displaces more water assuming you did not fill the cup all the way.
...
I agree that we would have to assume the change of the air volume in the calculation. But it doesn't really a matter. The volume expands because the water pressure is decreasing, but an expanding gas loses its energy. This wasted energy is the energy provided for raising up to the surface a increasing volume of water, that one corresponding to the increasing volume of air.

tommey,
it looks like you are changing the numbers from your website.
This will produce at a 10ft lift:(62.3*10)=623lb
pressure ((4.4 psi x 1728)/12)=633.6lb is what I get.
other numbers just seem to appear out of nowhere...
If 6.28 cu/ft in needed to turn a 2' wheel to turn 1 revolution, this will have a constant load of 62.3lb*10ft=623ft/lb.
i don't think you meant 10ft. maybe 10 containers?
maybe a good idea to make a list of the givens you want us to use.
tom

exnihiloest
I agree that we would have to assume the change of the air volume in the calculation. But it doesn't really a matter. The volume expands because the water pressure is decreasing, but an expanding gas loses its energy. This wasted energy is the energy provided for raising up to the surface a increasing volume of water, that one corresponding to the increasing volume of air.
i think you are a bit confussed. the air expanding inside the cone does make a difference, if the cone only has 1 cubic foot of air to start. at 10 feet it will look like only 83% of the full size. thus the lift will be less. you would have to use the average (91.5%) over the distance to figure how much work it can do.
if the cone is in a tube and sealed so water can't pass by, then your statement....
As water is incompressible, this air volume can be made only by raising the water above, what is strictly equivalent to raise to the surface the same volume of water as the volume of air in the cone.
would have some truth to it. you forget about the void left behind by the cone as it goes up. the water simply fills in behind. i will grant you that if the air is allowed to expand, that small amount of water will have to go some place. i don't think it will make a measurable difference in the water level.
btw, the last quote, is a very poor way of explaining your point. i'm sure not many people get it. maybe you want to try again.
tom

This is a prototype of the belt drive system.
http://www.youtube.com/watch?v=oX6QLaYbN2E
This system will be running soon.
The theory to this system is, how much energy is need to produec 2.5 psi to fill the cans vs the output of work?
Can this recompress enough air to fill each can while it moves upwards?
Could this turn into overunity?
Tom

This is a prototype of the belt drive system.
http://www.youtube.com/watch?v=oX6QLaYbN2E
This system will be running soon.
The theory to this system is, how much energy is need to produec 2.5 psi to fill the cans vs the output of work?
Can this recompress enough air to fill each can while it moves upwards?
Could this turn into overunity?
Tom
Hey, nice engineering! I'd like to be there with you, sacrificing myself for emptying all those cans.... Beer, I presume? No? Damn..
I believe it will work, but, sadly, it won't be 'OU'.
Far from UNITY, actually.
I'd say, some 15% efficiency? Not bad, it is quite better than, e.g., Steorn's electrical Orbo from the year ago....

Hi Tommy
I am always in full admiration of your projects so well done.
However they are correct that what you have here will not work and has been tried many times.
If you email me markdansiebigpond,com or Skype me mark dansie (nsw austraia) I will show you two ways that should work. One is a maybe but it uses gravity to get the air in under presure the other will work and when I explain it you will see why (uses existing technology that will create a vacuum and suck the air in.
There is a third way.
I never got around to building them but your welcome to te information as I think your a decent person
Mark

This is what I have come up with, and how much energy is needed to make over unity.
If you only pump down at a depth of 10ft, .5(864cu/in)cu/ft
At 1 cu/ft of displacement of water is about 62.3lb at 1ft at depth, does this force not increase at a depth of 10 ft?
Pumping .5 cu/ft at a depth of 10 ft will increase the displacement each foot as it rise to the surfice, this will this also increase on lift too?
Its all theory, this is why it really needs to be look at.
Remember this 4.4 psi compare to a higher pressure in more efficient too make then compressing 90psi
Thanks you all the comment, now its almost ready to do real testing. This is where theory's become facts.
If i'm wrong so be it, but a prototype is better then any theory's!
Tommey Reed

exnihiloest
i think you are a bit confussed. the air expanding inside the cone does make a difference, if the cone only has 1 cubic foot of air to start. at 10 feet it will look like only 83% of the full size. thus the lift will be less. you would have to use the average (91.5%) over the distance to figure how much work it can do.
if the cone is in a tube and sealed so water can't pass by, then your statement....
Do the maths.
Let V(h) the air volume of a cone, depending on the pressure. The variation of pressure is inversely proportional to the variation of height, thus we can write:
V(h)=V0+(V1V0)*h/H
where h is any height where the cone is, V(h=0) = V0 is the inital volume at the bottom, V(h=H) = V1 is the volume near the surface at height H.
At height h, the mass of the volume of displaced water is:
M(h) = V(h)*d where d is the water volumetric mass (d=1 kg/dm^{3}).
Its weight is P = M(h)*g = (V0+(V1V0)*h/H)*d*g
The work of the Achimedes force is thus:
W = int[P*dh] to be integrated from h=0 to h=H
W = int[(V0+(V1V0)*h/H)*d*g*dh] = g*d* int[(V0+(V1V0)*h/H)*dh]
int[h*dh] being 1/2*h^{2}, the calculus is straightforward:
W = g*d*(V0*H + 1/2*H*(V1V0)) = g*d*(1/2*V0 + 1/2*V1)*H
This result is obvious because the volume is proportional to the pressure thus to the height, it follows that we can replace the variable pressure and the variable volume by a constant pressure and a constant volume equal to the mean values of the variable case. We can operate at constant pressure with a constant volume V= 1/2*V0 + 1/2*V1 and get strictly the same result.
This is explained by the Archimedes force and the weight of displaced water which are always balanced. Therefore the problem reduces to a pure problem of potential energy. The energy to compress air in the cone at the bottom, being the only energy to be considered as potential energy that can be recovered when the cone rises up, without extra gain.

This is two basic system.
One using gravity and the other using displacement.
http://www.youtube.com/watch?v=rn0yt0mylW8
http://www.youtube.com/watch?v=jC2kmKsfEQ
Tom.

Excellent videos ;D ;D ;D

Hi Tommey,
Great work.
Just a idea. I think a torpedo shaped collector would be better than the cans bowed bottoms. Maybe plastic or glass test tubes?

Yes you're right, this is just a prototype at a very low cost to me.
Less then 40$ to prove a theory sounds great to me, what do you think?
This is another test I just did with a smaller hose to feed air into the cans.
http://www.youtube.com/watch?v=TnSIZJ61wI
Tom

If a hydrogen cell was used to make the bubbles, then when all the lifting was done, you could still burn the hydrogen, though how well the hydrogen cell works under pressure I don't know.
just a thought

...
how well the hydrogen cell works under pressure I don't know.
just a thought
It is a known phenomenon. The hydrogen pops out in bubles along the electrode only when it can overcome the ambiant pressure. The more the external pressure, the more the electrical energy needed for the electrolysis.

It is a known phenomenon. The hydrogen pops out in bubles along the electrode only when it can overcome the ambiant pressure. The more the external pressure, the more the electrical energy needed for the electrolysis.
Thanks exnihiloest
oh well another thought bites the dust.

It is a known phenomenon. The hydrogen pops out in bubles along the electrode only when it can overcome the ambiant pressure. The more the external pressure, the more the electrical energy needed for the electrolysis.
Does that mean that hydrogen cells in a partial vacuum would have a much higher output?
Furthermore, since hydrogen would have an obvious buoyancyadvantage, why not use a vacuum pump to suck the air & hydrogen out, feeding it either directly to another compression tank to increase pressure, or maybe straight to the submersible pump itself would do. And as powercat said, you could then also use the hydrogen to produce energy  maybe help run the vacuum pump!

Does that mean that hydrogen cells in a partial vacuum would have a much higher output?
...
Of course, it is the corollary.
Hydrogen is lighter than air but both have a quite similar density when compared to water (air density is about 1/900th that of water at atmospheric pressure). So in water the buoyancy advantage of hydrogen over air is very very weak.
In any case, I don't see the interest of more and more complicated setup when the functioning is conventional and can be described by a hamiltonian which implies energy conservation.
Either a system is supposed to work according to the physics laws, and here it seems to me that it is the case, thus extraenergy can't be produced without a hidden source that would have to be specified, or there are new laws of physics (but why and which?) or there is a flaw in the laws or equations of mechanics, that however have been proved relevant and internally consistent for more than two centuries.

Lets start over again.
12^3=1729 cu/in
7.48 gal = 1728 cu/in
7.48*8.33=62.3 lb.
I forgot that the deeper you go, it will also increase upward force?
So really each cube at a deeper depth would have different force?
I'm trying to find the basic formulas to test each cube at 1ft,2ft,3ft,,,,,,10ft depth.
This would also mean less volume at a depth of 10ft, because as the cube rise the pressure make more displacement as it rise to the top.
This sound simple, but many factor to think about.
Archimedes' principle
A system consists of a wellsealed object of mass m and volume V which is fully submerged in a uniform fluid body of density Ïf and in an environment of a uniform gravitational field g. Under the forces of buoyancy and gravity alone, the "dynamic buoyant force" B acting on the object and its upward acceleration a are given by:
Buoyant force
B=2gmpfV/m+pfV
Upward acceleration
a=g(pfVm)/m+pfV
Tom

This submersible engine goes beyond Archimedes' Principle:
Archimedes' principle is a fluid statics concept. In its simple form, it applies when the object is not accelerating relative to the fluid. To examine the case when the object is accelerated by buoyancy and gravity, the fact that the displaced fluid itself has inertia as well must be considered.[5]
This means that both the buoyant object and a parcel of fluid (equal in volume to the object) will experience the same magnitude of buoyant force because of Newton's third law, and will experience the same acceleration, but in opposite directions, since the total volume of the system is unchanged. In each case, the difference between magnitudes of the buoyant force and the force of gravity is the net force, and when divided by the relevant mass, it will yield the respective acceleration through Newton's second law. All acceleration measures are relative to the reference frame of the undisturbed background fluid.
Atwood's machine analogy
Atwood's Machine Analogy for dynamics of buoyant objects in vertical motion. The displaced parcel of fluid is indicated as the dark blue rectangle, and the buoyant solid object is indicated as the gray object. The acceleration vectors (a) in this visual depict a positively buoyant object which naturally accelerates upward, and upward acceleration of the object is our sign convention.The system can be understood by analogy with a suitable modification of Atwood's machine, to represent the mechanical coupling of the displaced fluid and the buoyant object, as shown in the diagram right.
The solid object is represented by the gray object
The fluid being displaced is represented by dark blue object
Undisturbed background fluid is analogous to the inextensible massless cord
The force of buoyancy is analogous to the tension in the cord
The solid floor of the body of fluid is analogous to the pulley, and reverses the direction of the buoyancy force, such that both the solid object and the displaced fluid experience their buoyancy force upward.
[edit] ResultsIt is important to note that this simplification of the situation completely ignores drag and viscosity, both of which come in to play to a greater extent as speed increases, when considering the dynamics of buoyant objects. The following simple formulation makes the assumption of slow speeds such that drag and viscosity are not significant. It is difficult to carry out such an experiment in practice with speeds close to zero, but if measurements of acceleration are made as quickly as possible after release from rest, the equations below give a good approximation to the acceleration and the buoyancy force.
A system consists of a wellsealed object of mass m and volume V which is fully submerged in a uniform fluid body of density Ïf and in an environment of a uniform gravitational field g. Under the forces of buoyancy and gravity alone, the "dynamic buoyant force" B acting on the object and its upward acceleration a are given by:
Buoyant force
Upward acceleration
Derivations of both of these equations originates from constructing a system of equations by means of Newton's second law for both the solid object and the displaced parcel of fluid. An equation for upward acceleration of the object is constructed by dividing the net force on the object (B âˆ’ mg) by its mass m. Due to the mechanical coupling, the object's upward acceleration is equal in magnitude to the downward acceleration of the displaced fluid, an equation constructed by dividing the net force on the displaced fluid (B âˆ’ ÏfVg) by its mass ÏfV.
Should other forces come in to play in a different situation (such as spring forces, human forces, thrust, drag, or lift), it is necessary for the solver of problem to reconsider the construction of Newton's second law and the mechanical coupling conditions for both bodies, now involving these other forces. In many situations turbulence will introduce other forces that are much more complex to calculate.
In the case of neutral buoyancy, m is equal to ÏfV. Thus B reduces to mg and the acceleration is zero. If the object is much denser than the fluid, then B approaches zero and the object's upward acceleration is approximately âˆ’g, i.e. it is accelerated downward due to gravity as if the fluid were not present. Similarly, if the fluid is much denser than the object, then B approaches 2mg and the upward acceleration is approximately g.
Tommey Reed