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'Free Energy' DIY Anyone?

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posted on May, 31 2009 @ 08:48 PM
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reply to post by TheRedneck
 


You just need to get 12 volt DC appliances. Why are you married to AC?




posted on May, 31 2009 @ 10:41 PM
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reply to post by dooper
 


Dooper you say that if someone announced a real free energy device, fossil fuel resources and assets based on them would become worthless overnight. I disagree. The reason why is that the new technology can't be deployed everywhere overnight. It will take time for everyone to obtain the new technology and in the meantime, they will have to keep on using gasoline, natural gas, oil or coal or power generated from those fuels. I've wondered if someone should set up a hedge fund specifically aimed at going short on the five largest oil company stocks. Then as demand for conventional fuel declines, so will the price of oil and oil stocks will plunge. So that destruction of old money could generate the creation of new wealth by selling short their stocks. As an added bonus, some of the initial capital used to set up the hedge fund, could be deployed as seed money to bring promising technology to a practical point for example Steven Greer's Orion Project stuff.

Anyway, I'm impressed by your knowledge and hope you keep sharing it with us.



posted on Jun, 1 2009 @ 09:01 AM
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RodgerT-

Looks like your Bendini guy is going the way Tesla did as far as diversifing into other areas-

www.abovetopsecret.com...

Are the principles the same with what you're working on?



posted on Jun, 1 2009 @ 04:13 PM
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reply to post by geo1066
 


No nothing like that, or at least not that I'm aware of.

Bedini seems to be a real live 'mad scientist genius' so I wouldn't be surprised to see him take his ideas in all sorts of directions.

I also think he seems to be the real deal, who has gained credibility and exposure, so others are willing to open up to him and give him access to the old stuff and the new ideas.

But I know almost nothing about the man, just what I glean from watching a few videos.



posted on Jun, 1 2009 @ 07:49 PM
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I haven't read the whole thread so I may be wrong, but it seems like you need to include watts into your measurements.

I have been experimenting off and on with free energy for a few years. My idea uses input to get the device spinning, once it reaches the designed rpm, the input watts are decreased say 50%. The design of the mechanics allows the matter in motion to take advantage of several forces of nature and increase the torque. So when the input wattage is reduced the system maintains its RPM and output allowing you to loop the output back to the input, with an excess of watts to run whatever. This will work with any voltage.

Thomas Young coined the term "energy" and defined it as "work stored within". Air/gas vapor, water/fluids, solid objects/mass all have potential energy within them, some in more forms than one. There are several forms of energy chemical, nuclear, electrical, mechanical etc. all of which must expend some of its potential energy to become kinetic energy so it can do work. But mechanical energy is unique to all other forms of energy, science does not define its soure/fuel. Today we can only convert any form of energy into ME. Convert, in the context of energy, means to go from one form to another and in doing so the original form is lost, its chemical make-up, volume etc. You burn a log for heat, the log turns to ash and can only burn once, you put gas in your car and it must be replaced after you go so far. This may seem elementary, but if my theory is correct the law of energy conservation and its use of the word convert is flawed.

My theory. All objects have work stored within. You put them in motion, use complex mechanics to invoke the forces of nature to increase the force of their motion. Force= mass x velocity2. That means if you double either one , you quadruple the force, double both and you increase the force 8x, fact, not my theory. Back to my theory. The fuel/source for mechanical energy is any matter, air, water, solid objects. The difference is when you put matter in motion, harness some of its KE for useful work, the original form and volume of the matter is not affected. All other forms of energy lose their original form. In otherwords, any matter, the source/fuel for ME, can go from PE to KE and back to PE infinately without ever losing its PE. No other form of energy can do that.

Lets take it one step further. The fact that we are using the forces of nature to increase the force, it should be called free developed mechanical energy. Develop is bringing forth something from something else, but the original form is maintain and able to bring forth the same thing again or even bring forth something else with out ever losing its original form.

Here is a real life example of this and physists will make themselves look like literary idots trying to invalidate it. Imagine 2 systems, system A has a reservoir that holds x amount of water, its not raining and there is no other supply to add to or refill it. The reservoir has a 100' head, distance from surfice to the headgate. When the head gate opens the water free falls 1000' to a lower reservoir where it collects and settles. the elevation of the lower reservoir is the point where water will reach terminal velocity/max KE in a 100' free fall. The total elevation of both systems is 1100'.

System B Has identical upper and lower reservoirs and the same x amount of water in the upper reservoir. The difference is we take 10 turbines and ten more reservoirs every 100' of the system. These additional reservoirs have the same head so when the water exits the first turbine it enters the second reservoir and begins to fill it. This repeats 9 more times untill all the water reaches the bottom. All 10 turbines are hooked to batteries to convert the mechanical energy into electrical energy and store it as potential eletric energy.

Then we release the water. System A, the water accelerates and at 100' reaches terminal velocity/max KE and falls the remaining 1000' at a constant velocity/max KE, until it impacts the lower reservoir. When the water settles it has x amout of PE. System B the water is released, spins the turbine and enters the 2nd reservoir. Once the 1st reservoir is empty and te 2nd one is full, the process is repeated. When the water exits the last turbine, there is no reservoir just 100' free fall, where it reaches terminal velocity/max KE just before it impacts the lower reservoir. When the water settles we have the same PE in the water as system A, but we have also stored electrical energy in the batteries. The PE in system B is greater.

How can this be? Scientist will try to say KE was converted but they cannot tell you what form it was in. Some might try to say gravitational potential energy, but for 1, how can you convert a constant? The loss in GPE is directly proportionate to its height the amount of work performed is irrelevant to GPE. Regardless of how much work gravity does in a gravity field, the force is always constant. That sounds like free energy to me. #2, the KE, at the bottom just prior to impacting the lower reservoirs in both systems is at its max and = in both systems, when the water settles it PE is also =. Proof #2 nothing was converted. If they insist its GPE then it must be reclassified as free energy, because the forces of nature are free and ever present. What was converted was the path the water travelled, its shape due to compression and the speed was increased and decreased several times in one cycle. It went from PE to KE 10 times instead of just 1.

We didn't create or convert any energy. We developed it in the form of mechanical energy and since we used the forces of nature to do this it is free developed mechanical energy.

What are you more interested in use air/wind, water or solid objects? Let me know and I'll give you some ideas that can be experimented with fairly cheap.



posted on Jun, 1 2009 @ 08:41 PM
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Originally posted by mhc_70
My theory. All objects have work stored within. You put them in motion, use complex mechanics to invoke the forces of nature to increase the force of their motion. Force= mass x velocity2. That means if you double either one , you quadruple the force, double both and you increase the force 8x, fact, not my theory.


Actually F=0.5 x m x v^2
IE velocity is the only factor squared

For the 'waterfall' example you gave, the appropriate formula is P = QGH
where P is the energy available for conversion by a hydro turbine in kW, Q is the flow rate in m^3/sec, G is gravity 9.8m/sec^2 and H is the head in metres. Those 2 examples would produce the same amount of energy in principle but efficiency problems would make the multi-stage system somewhat less in practise. I will agree that the energy is essentially 'free' only because it uses solar energy to place the fuel in the top reservoir (rain) but the infrastructure cost to collect that energy is astronomical (dams, pipes, tunnels, power stations, transmission system, switchyards etc etc)



posted on Jun, 1 2009 @ 10:25 PM
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reply to post by Pilgrum
 



You are correct, I was confusing it with Einsteins E=mc^2, which is irrelevant to this discussion. My mistake.

You say higher efficiency, I say freely developed. Can you tell me what form of energy you are refferring to when you say we are converting it more efficiently?

Edit to add: principles and logic often do not apply to physics.


[edit on 1-6-2009 by mhc_70]



posted on Jun, 1 2009 @ 10:40 PM
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reply to post by Studenofhistory
 



You'd be amazed at how easy and fast it would be to convert over.

Folks think in terms of power grids, since that's what we've become accustomed to. Grids would rapidly fall. Very rapidly.

Folks could easily and quickly take their homes off the grid, and do so in one afternoon.

Cars? No fuel whatsoever, no recharging. In fact, when your car was at home, it would also input additional power into the home.

Form follows function, and cars would be redesigned rapidly to take advantage of the new form of energy.

So, even the big boys know this, and their stock would plummet, even though there would be a multi-year transition time.



posted on Jun, 2 2009 @ 12:14 AM
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reply to post by mhc_70
 


In E=m.C^2, C (light speed) is the only factor squared

Q.g.h is an adaptation of M.g.h which is the expression for gravitational potential energy (mass x gravity x height) but efficiency must be considered and for a typical hydro application an overall figure of about 80% or even a little higher is the norm when conduit friction (head loss), machine losses (mostly heat), runner cavitation, transformer and conductor losses are all factored in. Cascaded schemes are dictated by topography where using the total head at a single point isn't practical but they suffer in overall efficiency compared to using the total head on a single turbine. As the turbines peak in efficiency at approx 70-90% of rated output, better efficiency over a wide range of outputs is achieved by using multiple turbines in parallel.



posted on Jun, 2 2009 @ 12:55 PM
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Originally posted by Pilgrum
reply to post by mhc_70
 


In E=m.C^2, C (light speed) is the only factor squared

Q.g.h is an adaptation of M.g.h which is the expression for gravitational potential energy (mass x gravity x height) but efficiency must be considered and for a typical hydro application an overall figure of about 80% or even a little higher is the norm when conduit friction (head loss), machine losses (mostly heat), runner cavitation, transformer and conductor losses are all factored in. Cascaded schemes are dictated by topography where using the total head at a single point isn't practical but they suffer in overall efficiency compared to using the total head on a single turbine. As the turbines peak in efficiency at approx 70-90% of rated output, better efficiency over a wide range of outputs is achieved by using multiple turbines in parallel.


Your getting ahead of me. I guess my thought process works different, but I cannot understand what you are saying until I know what form of energy you are referring to when you talk about the efficiency of its conversion.



posted on Jun, 3 2009 @ 03:10 AM
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reply to post by mhc_70
 


Sorry if I was drifting off a little there


I'm just working on your 'hydro' example given above to put some actual quantities to it like energy available compared to energy actually obtained. The energy is initially gravitational potential energy which gets converted to kinetic by falling, then mechanical (spins a turbine) and finally electrical energy at the terminals of an alternator.

This is a very good example of 'free' energy as the whole system is powered by the sun's heat evaporating water to form clouds which rain into the headwater catchment. It's also an excellent example of how expensive it is to collect 'free' energy in usable quantities.

You'd be correct about your proposal of a 1000' freefall scheme being inferior to a 10 stage system of the same overall height using 10 freefalls but the reason for that is not mysterious at all - it's caused by the freefall interaction with air (huge losses) in the single fall example which causes the stream to spread out a lot (eg spray) making it impractical to apply it to a turbine. Put that water in a high strength tapered steel pipeline and exclude all air from the pipeline so the full force of the column can be applied to a turbine and the single 1000' fall becomes much more efficient than the 10 x 100' system. The water in that tapered pipe (called a 'penstock') has virtually no terminal velocity and the only losses are in friction and turbulence caused by irregularities in the pipe wall so the smoother the pipe, the better the result.



[edit on 3/6/2009 by Pilgrum]



posted on Jun, 3 2009 @ 05:44 AM
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Originally posted by Pilgrum
reply to post by mhc_70
 


Sorry if I was drifting off a little there


I'm just working on your 'hydro' example given above to put some actual quantities to it like energy available compared to energy actually obtained. The energy is initially gravitational potential energy which gets converted to kinetic by falling, then mechanical (spins a turbine) and finally electrical energy at the terminals of an alternator.

[edit on 3/6/2009 by Pilgrum]


KE is describing the state the form of energy is in. Then you went on to say mechanical energy is converted to electrical energy, which is correct, those are forms of energy, KE is a descriptor of those forms or a type of energy, there is a difference.

GPE is a constant force of nature that cannot be converted. Regardless of how much work GPE does, its force is directly proportianate to its height, not how much work it does, therefore, it is never converted. The second problem with GPE it is an invisible force that requires a medium to show itself, so by definition GPE cannot be converted, although it is a constant force of nature we can use to develop energy.

What form of energy are you reffering to when you discuss the efficiency of its conversion into mechanical energy?



posted on Jun, 3 2009 @ 06:10 AM
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reply to post by mhc_70
 


The key to the 'potential energy' is the term 'potential'. It can only be used if the mass or part thereof in m.g.h is allowed to fall where it becomes kinetic energy. Basically potential energy is a form of energy storage.

In your example of a pond with 1000' head and no inflows, how much potential energy does it have once all the water is drained out (falling 1000')? Where does the potential energy go?

When I speak of efficiency it's in terms of the overall energy conversion and in practise a portion of that initial potential energy is always lost in the conversion process(es) most commonly as wasted heat.



posted on Jun, 3 2009 @ 12:16 PM
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Originally posted by Pilgrum
reply to post by mhc_70
 


The key to the 'potential energy' is the term 'potential'. It can only be used if the mass or part thereof in m.g.h is allowed to fall where it becomes kinetic energy. Basically potential energy is a form of energy storage.


Potential energy is not a form of energy storage. PE , in this case, defines the water as motionless, KE defines it as in motion. There are only 2 TYPES of energy, PE and KE. There are several forms of energy, nuclear, electrical, chemical, mechanical etc. all of which can store PE. PE and KE only define the state they are in.

So what FORM of energy are you reffering to being converted?



In your example of a pond with 1000' head and no inflows, how much potential energy does it have once all the water is drained out (falling 1000')? Where does the potential energy go?


Niether system has a 1000' head, both have a 100' head. System A the water exits the headgate and free falls 1000'. System B the the water exits the headgate and impacts the turbine and then begins to fill the next reservoir. The volume of water is the same, the decrease in PE, in both systems, is due to lower elevation, not the work it performed on the turbines.



When I speak of efficiency it's in terms of the overall energy conversion and in practise a portion of that initial potential energy is always lost in the conversion process(es) most commonly as wasted heat.


The fact is you cannot tell me what form of energy is being converted, because there is no conversion taking place.

Disagree? Then end this and tell me what form of energy you are talking about.



posted on Jun, 4 2009 @ 04:58 AM
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Originally posted by mhc_70

Niether system has a 1000' head, both have a 100' head. System A the water exits the headgate and free falls 1000'. System B the the water exits the headgate and impacts the turbine and then begins to fill the next reservoir. The volume of water is the same, the decrease in PE, in both systems, is due to lower elevation, not the work it performed on the turbines.



I see you're acknowledging the reduction in PE caused by the reduction in height and if you consider the principle of conservation of energy you should realise that the lost potential energy didn't just disappear, it had to be converted to another form of energy and that is kinetic energy. The potential energy can be restored by simply raising the water back to its original height but that requires input of energy in some other form doesn't it? Consequently storing energy - potential energy which can be re-used.

The amount of energy recovered from your hypothetical system is paltry in comparison to the losses because of the highly inefficient setup.

I'll put some actual numbers to it (using metric because it's more convenient for conversion to actual kW available)

The 'magic' formula for hydro applications is P=Q.g.h (kW)
1000' head = 304.8m
100' = 30.48m
g=9.8m/sec^2
and let's suppose you can release water from the top pond at 1 m^3/sec until the pond is empty.

First - the most efficient arrangement would be as I suggested earlier
IE a turbine fed by a penstock at the bottom of the 1000' drop.
That output figure amounts to 1 x 9.8 x 304.8 = 2987kW or nearly 3 megawatts.
Not mysteriously by any means, that is the minimum amount of energy you'd need to supply to lift that water back up to the top pond (by pumping) if the pump system was 100% efficient (they never are). The act of pumping is a good example as it's the whole system in reverse IE using electrical energy, converted to mechanical energy, then kinetic energy and finally potential energy once the water is back in the top pond.

Next - let's see how much energy is recovered when the water falls, say, just 1m to the turbine, then free-falling the rest of the drop with the same constant flow of 1m^3/sec:
P = Q.g.h = 1 x 9.8 x 1 = 9.8kW
10 such drops with turbines at each outlet with the same 1m effective head on each turbine would, at best, generate only 98kW.

So what happens to the lost 2987kW-98kW = almost 2.9MW gone?
It's because your system is only making use of 10m of the available 304.8m of head and the missing 2.9MW is simply being wasted with your comparative efficiency of merely 3.3% which is woeful.

Once the water reaches the lowest level its potential energy has been exhausted no matter how it gets there but the energy has not been destroyed, it was simply converted on the way down. The variance in the examples reflects only the efficiency of the systems.

Now there is a way to extract the full potential of the fall with the turbine located somewhere other than the bottom and that involves utilising what we call 'suction head' and I could explain it for you if you're interested. It is subject to some stringent mechanical limitations though. I work with these systems every day.

[edit on 4/6/2009 by Pilgrum]



posted on Jun, 4 2009 @ 09:19 PM
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Originally posted by Pilgrum

I see you're acknowledging the reduction in PE caused by the reduction in height and if you consider the principle of conservation of energy you should realise that the lost potential energy didn't just disappear, it had to be converted to another form of energy and that is kinetic energy. The potential energy can be restored by simply raising the water back to its original height but that requires input of energy in some other form doesn't it? Consequently storing energy - potential energy which can be re-used.


I have never said the PE was not less in the bottom reservoirs, I agree that it is, what I did say is that it was a result of lower elevation and not the amount of work it performed. I also said that the PE in the lower reservoirs is equal, because the water volume is equal.



The amount of energy recovered from your hypothetical system is paltry in comparison to the losses because of the highly inefficient setup.

I'll put some actual numbers to it (using metric because it's more convenient for conversion to actual kW available)

The 'magic' formula for hydro applications is P=Q.g.h (kW)
1000' head = 304.8m
100' = 30.48m
g=9.8m/sec^2
and let's suppose you can release water from the top pond at 1 m^3/sec until the pond is empty.


The reason for the design was not because I think it is efficient, it was to demonstrate that regardless how many times or how much work the water performs once it reaches the lower reservoirs the PE is the same(in both lower reservoirs, it has decreased because of lower elevation) due to having the same volume of water.



First - the most efficient arrangement would be as I suggested earlier
IE a turbine fed by a penstock at the bottom of the 1000' drop.
That output figure amounts to 1 x 9.8 x 304.8 = 2987kW or nearly 3 megawatts.
Not mysteriously by any means, that is the minimum amount of energy you'd need to supply to lift that water back up to the top pond (by pumping) if the pump system was 100% efficient (they never are). The act of pumping is a good example as it's the whole system in reverse IE using electrical energy, converted to mechanical energy, then kinetic energy and finally potential energy once the water is back in the top pond.

Next - let's see how much energy is recovered when the water falls, say, just 1m to the turbine, then free-falling the rest of the drop with the same constant flow of 1m^3/sec:
P = Q.g.h = 1 x 9.8 x 1 = 9.8kW
10 such drops with turbines at each outlet with the same 1m effective head on each turbine would, at best, generate only 98kW.

So what happens to the lost 2987kW-98kW = almost 2.9MW gone?
It's because your system is only making use of 10m of the available 304.8m of head and the missing 2.9MW is simply being wasted with your comparative efficiency of merely 3.3% which is woeful.

Once the water reaches the lowest level its potential energy has been exhausted no matter how it gets there but the energy has not been destroyed, it was simply converted on the way down. The variance in the examples reflects only the efficiency of the systems.


I understand the increased efficiency in the system with the 1000' head vs. 100' head ie more pressure. But in the two systems I described one is doing nothing, just letting the water fall, the other is using the kinetic energy in the falling water to spin the turbine. When you compare efficiences in this context, you compare how well the systems use the falling water ie which one makes more electricity. One system is doing nothing, would'nt it have to be accoplishing something before you start talking about how efficient it is? Its like pouring gas on the ground and burning it or pouring it in a car and burning it using it to drive. Then saying pouring gas on the ground is a less efficient way to power your car.

If you were comparing my 10 turbine system to your one turbine system, with a 900' higher head, then yes one system is more efficient than the other.

If I burn a log, its PE is converted into heat and it becomes ash, 0 PE.

If I throw that log, the KE in my arm converted the logs PE to KE. Once the log hits the ground, it still has the same PE when it settles as it did before I threw it. All forms of energy lose their original form when they go from PE to KE. Surely, you understand this fundemental difference?

All forms of energy are either PE of KE, these describe the state they are in. When you say KE you are describing a form of energy, it is being stored or it is expending itself, KE must be in some form to do work.

Do you disagree with that?




Now there is a way to extract the full potential of the fall with the turbine located somewhere other than the bottom and that involves utilising what we call 'suction head' and I could explain it for you if you're interested. It is subject to some stringent mechanical limitations though. I work with these systems every day.

[edit on 4/6/2009 by Pilgrum]


Sure, sounds interesting.



posted on Jun, 5 2009 @ 06:13 AM
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I'd say the answer the the problem you're proposing is that the potential energy is released regardless of whether you collect it or not. It could be concentrated into a very small cross-section (in a low friction penstock) and applied to a turbine in order to do useful work such as generating electricity or running a mill, or it can fall naturally allowing that energy be dissipated over a much wider area in a chaotic fashion such as generating wind, noise, eroding rocks etc.

The potential energy released either way is the same whether it's collected or not and you'd need to input the same amount of energy to restore the water to its original height in both cases.

Your petrol example is another good one - I could put 40 litres of petrol in the tank of my car and drive it 500km in 5 hours which is a measurable amount of chemical potential energy being converted to kinetic energy (mainly) and other forms of energy like heat and overcoming friction which is wasted in the conversion. Or I could simply spill the 40 litres on the ground and light it in which case the energy of that fuel is released in minutes and in a spectacular fashion. Both ways release the same amount of potential energy whether I choose to make use of it or not.

The suction head I mentioned earlier is utilised with 'reaction' type turbines (eg Francis or Kaplan turbine runners) mounted higher than their tailrace level. The water falling away from the turbine creates a negative pressure which augments the positive pressure above the turbine allowing the full height to be utilised but the design and construction needs to be very precise for this to work. The suction height below the turbine is limited to less than about 30' because, above that height, a vacuum will start to be drawn which absolutely kills the efficiency. The suction section needs to be designed to perform with a negative pressure unlike the section above the turbine which operates at positive pressure.



posted on Jun, 5 2009 @ 06:53 PM
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reply to post by Pilgrum
 


What I am saying is regardless how much work(spinning the turbines) the PE in the water does, once it starts falling and becomes KE, It does not affect the PE. The only thing that decreases the PE is the lower elevation, not the work it performed spinning the turbines.

This is unique among all forms of energy, no?

If I throw that log, the KE in my arm converted the logs PE to KE. Once the log hits the ground, it still has the same PE when it settles as it did before I threw it. All forms of energy lose their original form when they go from PE to KE. Surely, you understand this fundemental difference?

All forms of energy are either PE of KE, these describe the state they are in. When you say KE you are describing a form of energy, it is being stored or it is expending itself, KE must be in some form to do work.

Do you disagree with that?



posted on Jun, 5 2009 @ 09:58 PM
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Sorry but I don't agree

PE exists in many forms EG a mass raised to a relative height, a wound spring, chemical etc but whatever type it is a static form of energy in storage IE the potential to do work. Using it to do work depletes it through conversion to other forms of energy. Some types of PE can be easily restored EG raising the relative height of the mass or winding up the spring by putting energy back into the system (kinetic energy
converted back into potential energy).

If h=0 then Mgh = 0 (no potential energy remains) but it wasn't destroyed, it was converted to other forms.

There's not much more I could say about it and we're probably wrecking this thread about DIY 'free' energy by continuing this.

Hydro is an excellent example of 'free' energy that works, is well understood, and has been utilised worldwide for centuries. It's also 'doable' on a hobby type scale for enthusiasts if you have access to the right physical situation (a stream of water falling).



posted on Jun, 5 2009 @ 11:48 PM
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reply to post by Pilgrum
 


I agree, at this point its semantics...Back to free energy or something close to it...



The modern hydrodynamic turbulent water circulation will be eliminated as good as possible. On the other hand we developed the world wide first so called gravitation water vortex power plant which uses the rotation energy of a single gigantic water vortex. It is a milestone in the hydrodyamic development because in the past we needed energy to aerate water - now we have a water aeration process which produce electric energy. Already in the first operation year the invention of the Austrian engineer Franz Zotlöterer produced 50.000kWh - electric energy for 14 average european households. The prototype plant supply the public electricity network with current. You can find the gravitation water vortex power plant on the millstream in Obergrafendorf (10km southwest from St.Pölten in Austria in Central Europa). It is public and can be reached on the cycle track along the pielach river:

Presently used falling height 1,3m
Presently used flow rate 1m³/s
Diameter of the rotation tank 5,5m
Hydraulical power 13kW
Elektrical power 8kW
Effectiveness of the turbine 80% at 3/3, 83% at 2/3 and 76% at 1/3 of the maximum flow rate
Turbine speed 25rpm
Investment minus financial support around 40.000€
Joyful working capacity of 50.000kWh in the first year of operating - since February 2006 the actually total production of electricity is over 120.000kWh


Pictures and the rest...www.zotloeterer.com... vortex_engineering/water_vortex_power_plant.php



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