Simple Physics Question

Only if the air bubbles are originating from the ocean floor....

A1. Integral calculus can be used to approximate an infinite number of circles whose circumferences describe the locus of a sphere. The energy needed is the sum of that of displacing the circle's volume of water at that depth multiplied by an infinite number of circles. The calculus is a little involved and it will respond to an hour or so with a pencil and paper for the interested Mathematician. However, if the balloon is filled with a gas and not a liquid and the pressure inside the balloon is constant, and that outside the balloon varies with depth below the surface, the balloon will never be spherical.

A2. That depends upon whether the reduction of external pressure causes the balloon to burst or not as it rises.

A3. There will always be a net energy loss in any real-world process. The Laws of Thermodynamics say so.

Just off the top of my head, I'm thinking: Assuming the pressure is constant, isn't the integral of an expanding sphere just equal to the volume of the final sphere? So the work is P*V.

(Or so it seems to me. I'll think about it some more.)

The outside pressure varies with depth. If the tensile strength of the balloon is uniformly distributed in whatever material comprises its skin then even if the intent was for it to be spherical, it cannot be. It follows that any analysis of a spherical bubble containing a gas, however straightforward mathematically, can only be an approximation.

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Who cares if it's homework? Many of us like the challenges, around things we may have forgotten long ago. Good mental exercise when we're a little rusty.

Only a little rusty ?? Bless you...

Only A LITTLE rusty ...

Yeah. For me it's probably more oxides than original metal mettle.

That's fine. Do the work to your heart's content. Just don't give the student the answer unless you are sure they've learned the concepts behind how to solve the problem.

Many students, and many gainfully employed adults, come here hoping to have the forum do the work for them, give them the answer so they don't have to put out the effort themselves.

That's all I'm saying. That's all Admin is saying, too. They made the rules, I didn't.

We've all helped students with homework.

Shouldn't be too difficult. How many atmospheres does 10 m equate too? Would you get the same result if you inflated a balloon in the atmosphere to the same equivalent?

The pressure change in water is one bar for each 10 meters. In air it's very small almost negligible for this purpose.

The balloon expands roughly 2x when moving from 10m depth to "0" meter depth giving it double force up after 10 meter travel upward.

This gives you twice the buoyancy force since than at 10 meters depth. The expansion creates added force up. This expansion part seems to give it the surplus exceeding the energy which was used when pumping up the balloon 10m under water by about 50% for this scenario if I'm not wrong.

I thought we weren't supposed to give the OP that info, he was supposed to look it up. My question to the OP, was if you apply the same bar to the balloon at sea level, what will be the resulting volume?

Well, the Laws of Thermodynamics say that there is no way that an equal or greater energy can come out than went in, so the analysis document requires up-revving.

There is a related method of off shore energy storage that has been proposed by some researchers at MIT....It's a different concept, and looks totally outrageous to me....The costs of deployment and maintenance seem to be overwhelming...

"The key to this concept is the placement of huge concrete spheres on the seafloor under the wind turbines. These structures, weighing thousands of tons apiece, could serve both as anchors to moor the floating turbines and as a means of storing the energy they produce.

Whenever the wind turbines produce more power than is needed, that power would be diverted to drive a pump attached to the underwater structure, pumping seawater from a 30-meter-diameter hollow sphere. (For comparison, the tank's diameter is about that of MIT's Great Dome, or of the dome atop the U.S. Capitol.) Later, when power is needed, water would be allowed to flow back into the sphere through a turbine attached to a generator, and the resulting electricity sent back to shore.

One such 25-meter sphere in 400-meter-deep water could store up to 6 megawatt-hours of power, the MIT researchers have calculated; that means that 1,000 such spheres could supply as much power as a nuclear plant for several hours - enough to make them a reliable source of power."

http://cleantechnica.com/2013/04/30/mits-innovative-floating-wind-energy-storage-technology/

What am I missing here?

Are we not just compressing and releasing air?

Yes. It's akin to filling a reservoir atop a mountain using excess power when the wind is blowing and using it to generate when the wind slows.

The ingenious part of this scenario is that air is easier to store because it's compressible.

Interesting.

That's where I'm lost. It takes energy to compress air, and to pump water uphill.

Water will do a lot more work than air, so I'm having problems putting this into perspective.

I assume it (the sphere) will be pressurized when deployed? What turns the turbine, air or water?

The dome starts out full of water, the excess energy produced by the turbine is used to run a pump that pumps water out drawing air from the surface....then when needed the air is released and the water then flows back into the dome driving the turbine....the thing is the domes are like 12 million per, and the're talking about 1000's....

The idea is that the sphere is connected to a drive, pulley, and pulls it as it goes up producing energy going up. Going down happens by gravity using a system with one balloon going up and down like a piston so to speak. The charge of air happens at the lowest point after which it goes up producing more energy than was used to fill the balloon. The rest is done by the law of buoyancy and the expansion of the balloon as it goes up.

I don't think when all is said and done that there will be any net energy gain....

I figured the air was compressed and pumped into the concrete sphere displacing the water in the sphere. This is all done with the excess energy from wind turbines.

When it's time to use the 'stored' energy in the concrete spheres, the pressure is released through a turbine connected to another generator.

Concrete is used so that the spheres don't float away. Doesn't need to be concrete, but what do you find that's as dense and as strong?

No, there's no net energy gain that would be over-unity. There's some serious losses as compressors are not efficient, but this is simply a storage medium. As I said before, pretty nifty because it's a similar principal to using storage water reservoirs on mountain tops, but better because compressed air takes up less space than water in a reservoir.

Wikipedia - Pumped-storage hydroelectricity

Wikipedia - Compressed air energy storage

I think you're thinking right, Solar.

With the system proposed, and in fact with any closed system, there is NEVER a net energy gain.

Energy Out = Energy In - 'Waste Heat'

There are ways to store energy, pumping water up a mountain stores some of the energy used to pump the water in the potential energy of the water being uphill, with the rest of the energy used lost to heat and friction in the pumping equipment. Letting the water flow downhill through a turbine recovers some of the potential energy stored in the water, with the rest lost to heat and friction in the turbine equipment, and the kinetic energy of the water as it continues to flow to an even lower level.

This balloon system is, in essence, nothing more than the 'pumping water up a hill' system, only standing on its head.

When looking for a 'net gain' in an energy storage system/perpetual motion machine/over-unity amplifier, the original poster needs to remember the Three laws of Thermodymanics.

1) Energy and matter cannot be created or destroyed, only converted from one to the other.

2) In every conversion process, some energy will be lost to waste heat.

3) In a closed system, over time all the energy will be converted to waste heat, ending the processes in the system.

Or to state it in ten words or less: You can't win. You can't break even. You can't quit.

I'm feeling a wet sensation in my boot, and it ain't raining.

The last time I checked, concrete balloons don't expand, or contract.

You can't get more out than goes in, except for hot air in your case.

You say pulleys, they say turbines.

Do not pass Go, do not collect $200.00.

Obviously, Lyn, for any hydraulic energy storage scheme to work, regardless of the working fluid (oil, water), there must be a volume of air or other gas capable of expanding to fill the withdrawn volume during the energy storage half of the cycle. Water admission must therefore, recompress this gas. There is no free lunch, but donuts and coffee are possible.

I think this energy storage in submerged gas bags might have some merit. Some of us could volunteer as gas bags.

I'm fine with all that. It's the part about an obviously very heavy concrete structure floating up and down in the ocean (like a balloon) that I'm having trouble with.

I have no issues with a fluid compressing a gas inside a rigid structure.

I think the real idea is that the very heavy structure sits permanently on the bottom, whether filled with air or water.

The real problem I see with this scheme has nothing to do with 'free energy' but with the enormous (compressive) explosive forces stored cyclically on the spheres structure. Any form of energy storage has the potential to become a ..

Actually no, the actual pressure loading will only be very high at the top of the pumping pipe. The main force on the sphere will be the upward buoyancy created by the air trapped inside it (30m dia = 14,137 m³ ) or displacing 14,000 metric tons of water. - If nothing else that's a lot of mechanical stress.

Not, exactly.

The concrete ball, "weighing many tons" was a part of SE's #7 response to the original post.

Not to speak for the OP, but I believe that they meant some undefined expanding bubble, not made of concrete.

I've given up trying to understand or make sense of the whole mess.

This is horribly reminiscent of that thread a while ago that was full of twaddle about extracting energy from two coupled vacuum-driven elevators, or some such drivel.

Would it be possible to use AC Clarke's idea in "The Ghost from the Grand Banks"? He used electrolysis to breakdown sea water, negating the need for pumps and piping. Granted the result is H₂ and CL, but each has industrial applications.

Question?

Are we supposed to include the energy required to make the balloon, too?

Good point. And the energy needed to transport it to where it is to be installed, and the energy needed to anchor it in place, and the energy needed to create the filling system and the equipment that charges it, and....

Damn, reality gets in the way, again.

Indeed nifty idea, but it's gotta be pretty expensive to implement. But when you think about it, it's gotta be cheaper than building a lake on a mountain.

Lo Volt, you say "it's gotta be cheaper than building a lake on a mountain"

http://news.bbc.co.uk/2/hi/asia-pacific/1530182.stm

There's a very inexpensive lake on a mountain for you

First ask where you are pumping air from. Presumably it is the surface no higher up. The 10m depth has a pressure difference of rho_water*g*h from the surface where rho=10^3 kg/m^3 and h=10. If the size of the balloon is small compared to h then this is a good approximation. This all assumes you are in an infinite body of water where the height of the water is negligibly raise by the balloon expansion AND that the elastic energy of the balloon is zero by comparison. The energy is then E=P*Vol+E_compression where E_compression is the compression energy of the air from the surface to that depth. If one adiabatically compresses, it heats up and this rapidly exchanges energy with the water to cool it. If one isothermally compresses, use PV=nRT where T is the temperature of the water at depth to integrate pdV. Near the surface the elastic energy will become important so probably is not ignorable any more.

The balloon will expand and, if the elastic energy is relatively zero, then the air expands back to the 1 atm pressure. Adiabatic expansion might be a good approximation since the balloon is pretty big and rises fast. Inviscid won't be a good approximation so most E will be lost as heat if you just release it. If you let it rise slowly then you might have isothermal expansion which helps you since the balloon would otherwise cool and give as much expansion energy.

The last point is how do you expect to collect the energy? By using just buoyant force? Expansion force? I think the last question might overspecify things. The expansion to the surface over 10 m will not give a 2cbm expansion unless you have a very particular temperature profile and balloon elasticity.

The question is to calculate with real numbers using standard physics.

How many Joules does it take to pump up the balloon at 10m depth to 1 cbm volume.

Then how many Joules will be produced by the force (which increases when the balloon rises) as it goes up. (For simplicity think with it connected to a dynamo)

So there should be 2 numbers at the end of the calculation: 1: Joules for pumping air into the balloon. 2: Joules as the balloon rises up 10 meters.

No assumptions allowed:

That's the whole exercise.

This is sounding like a homework problem. Everyone here can put numbers into a calculator. I would suggest looking at the book chapter. It will give you cues and to what assumptions are reasonable or assumed by the author. A lot of such problems are poorly considered and the poor student has to infer what the writer would have considered reasonable. For example, if you have not been talking about thermo you can probably assume constant temperature.

Not it's not home work. Try doing the math with whatever reasonable assumptions you care to enter into it. How much can the rising cycle produce and how much does filling the balloon require? Temperature should remain pretty constant as the water temp is the biggest controlling factor.

Sort of allows use gravity both ways, up and down.

Thought it's a neat concept.

No, it doesn't use gravity both ways.

It uses magic if you think you can generate more power with this dream that goes into it.

Look here: The Gallery of Ingenious, but Impractical Devices

Waiting for the "but you can't PROVE it doesn't work" retort.

"Is it possible that this generates more energy in going up than it takes to pump it up initially?"

No. In a theoretical perfect system you could expend an amount of energy, x, filling the balloon that would be reclaimed by harnessing the lift energy, y. But y will never be greater than x. In the real world everything from friction upwards will interfere and the energy that you harness from the rise will be less than that expended in filling the balloon.

… and the theoretical perfect system assumes that there has been no energy expended in making any of the parts. So if you want to be really pedantic it doesn't even work in a perfect world.

Don't stop thinking though. All new things are brought about by people who start with "I wonder if this would work".

Somehow I can't account for the increase of up-force because of the expansion of the balloon as it drifts up? It's just more and quite a bit, more than friction etc. would eat up. The force increases with the power 2 as it goes up and if you go further down you'd get more if you can mechanically realize even more expansion of the balloon. Do you have an explanation other than it's not possible? Did you do a calculation?
Just trying to find my error if it is - of course you can't get out more than you put in.

Patent US20100107627 - Buoyancy energy storage and ... - Google

Do Giant Compressed Air Underwater Balloons Store Energy Well ...

This is a good link. Claims on price are BS which makes me doubt some of the rest. Nice to see people doing calculations.

Interesting patents.

But I don't mean energy storage. Just trying to get someone to do an independent math of what the rising balloon would produce and what energy it takes to pump up the balloon below. There were many comments but nobody has yet answered the question really. Just teasing.

Well,

We've had many here like you. You have an idea, no way to validate/verify/prove that it is feasible, and yet it "can't miss".

Your response to evanmjones says it all. In so many words, "prove me wrong". Do the math for me.

So, even though you have no idea if your system will work, and no way to prove it will, until someone runs through some calculations that you can understand, you seem hell bent to say it is so. That's fine by me.

I say forge ahead with total abandon, but don't be disappointed if it doesn't pan out.

There may be other patents of sources, I'm out of time to play.

Look mate, we're not doing all your homework for you. Particularly when you post anonymously.

Many years ago I remember a maths lecturer telling me, "a good rough answer is a really useful start", and he was right. So break out the elements and see how it works for you.

You've got three elements to harness some energy: fill the riser; rise the riser; harness the rise.

Forget the balloon for a moment, start with a hard cube, 1M3, at depth 10M and calculate the force upwards. So you've now got a defined force. You've also got a defined distance (10m) so what's the maximum amount of energy available if the cube rises 10m?

Now remember the balloon. You've already got the formula for filling a balloon to a given volume at a given pressure. How much energy did that take?

Harnessing.... actually I'm not going to go there. I started with giving you a scenario involving a 2M circumference wheel revolving against a 5cm circumference wheel attached to a shaft powering a generator.... but lost interest while I was defining the generator. Plus my wife has put the kettle on.

So. I'm just going to say, you've got your in and out energy from the central bit of your contention. Think some more about the compressed air you made to fill the balloon - how much energy did it take to run the compressor; and how much energy would the generator actually produce in watts?

Finally, loosen up a bit, stop posting anonymously, go out and mod it up for the heck of it. A model will get you through many of the challenges but I can think of a lot worse things to do with a Summer day than mess about with a boat, a lake, a balloon, a weight, some string, a pulley force measuring thing I've forgotten the name of ..... and my son so I can pretend that's it's educational and not me having a laugh. Beer is optional. Actually, scratch that, beer is compulsory.

W=∫Vdp = VxP, where P is the differential pressure (1 atm) at 10 m. The buoyancy force at any depth is the volume of the balloon at that depth x the difference in weight of water and air. If you integrate the buoyancy force over depth, you should get the same answer.

OK. So you want a large balloon that is raised very slowly (so isothermal) to store energy from the surface to 10m (where the pressure doubles to 2 atm).At the surface the balloon is V_i. The water is temperature T. Isothermal assumption means that as the pressure doubles the volume decreases by half. Let atmospheric pressure (in whatever units you use) be P_0. We can assume no viscosity for this. h=10m. rho_w=10^3 kg/m^3.

The particle number in moles is n=P_0*V_i/RT.
The work to compress the balloon to depth is W_c=nRT Log(Vf/V_i)=nRT Log(2)
The work to push the water out of the way is W_g=Vf*rho_w*g*h=0.5*V_i*rho_w*g*h=5x10^4*V_i

W=W_c+W_g gives the work to depress the balloon each cycle and what you will get back. The one thing I neglected here is the expansion energy of the balloon itself and its effect on the pressure. This is usually not insignificant. One can get ~atm changes in pressure in a balloon or tire. This is probably a heavy balloon so this changes the final volume and also gives an elastic energy that must be accounted for.

When the balloon goes down again all air is drained and there is just the dead weight going down, once it's down again it's pumped up to 1cbm and then travels up.
The going up cycle has the addition of the expansion of the balloon plus the original size.
So theoretically to pump up the balloon to 1 cbm would take 100,000 Nm energy not considering any losses.
This is returned in going up by the "reverse" gravity force pulling up now plus 50,000Nm as the balloon expands from approximately 2 cbm in it's travel up of 10 meters. Not including all the small inefficiencies.
So there seems to be the 50,000Nm additional from the expansion of the gas and larger buoyancy. Does that make sense?

Wait. You are draining the air out as it descends???

I don't understand what you are doing. I thought it was a fixed particle number balloon. I guarantee there is no free energy here. You can store it and retrieve it. That is all. You can get an order of magnitude estimate just from the water displaced. One liter per ten meters gives 100 Joules. One cubic meter gives 10^5 J. I liter of fuel has about 30MJ. This sets the scale of about 300 cubic meters or a ball of radius about 4m for you to get energy of about a quarter of a gallon of gas.

Once the balloon is up all air gets released(there would be no pressure anyway) and down it goes by gravity for another cycle of pumping it up removing the energy on the way up. If the up cycle produces more energy as the pumping cycle per basic physics I think I learned in school then this seems at least promising.

When the balloon goes down, what work is it doing? 0

When the balloon is at the bottom, what pumps it up again? Good will?

You show no work product from the balloon going up and what is this "original size" and what buoyancy could it add? 0

Where are you getting this "theoretical" 100,000 Nm figure?

How is the energy "returned in going up" when no work is being done?

The "small" inefficiencies you mention are not small at all.

You get NO work from the, "additional [energy] from the expansion of the gas" because that is the result of the decreasing pressure exerted by the water as the balloon rises.

Yes, we've been there and done this before. No amount of wishful thinking will make this work.

Going down is sort of the return stroke by gravity.

At the bottom it's pumped up with a normal air compressor.

When you compress 2cbm air to 1cbm gives you that. Isothermal. Pumped with a normal compressor.

It was 1cbm displacement at 10m depth = 1 bar which is the pressure at 10m depth. Correct?

Then 1cbm displacement equals 10,000 Nm Energy = Force x distance.
1000kg*9.81 x 10m. = roughly 100,000Nm(joules) correct?

Now add to that the balloon expanding as it goes up which averages to 50,000N x 10m = 50,000Nm(Joules). You can use calculus to derive it.

So the total work product going up is 150,000Joules. Minus the 100,000Joules for pumping the air leaves 50,000Joules in one such cycle. Do you think there would be that much losses to eat that up?

Down it goes empty and uses no energy as gravity will pull it down. Once down it's pumped up again with some mechanism doing a new cycle of going up.

That's the cycle it goes through. Imagine the whole into a rotating device with many balloons pulling up where the air gets released and then back down Once they get to the bottom again refilled.

Not considering, frictions or losses for this, do you see a flaw in this logic? Please point it out.

I do prefer constructive criticism as the other type of criticism makes this world
not a nice place to be. Engineers like to built and create things, right? If it doesn't work drop it no need to point out how wrong it was - that's self evident.

Hope you get my train of thought - I like to think positive, critics do not create things, but quite the opposite it in many cases!

Oh, I think positive, also.

I'm positive this is folly.

You provide a narrative with numbers at the end of sentences. Nothing more than wishful thinking.

As I said earlier, we've been here before. Joe-Fordham comes to mind. Don't confuse him with facts, he KNOWS it works.

Now, you'll say to me, "prove you are wrong".

I say to you, build it, make it work and prove all the naysayers wrong.

"Engineers like to built and create things, right?"

Yep, we do. Love to. But hate to build things that can't work, and hate it even worse when a doofus who can't understand simple physics insists WE are the ones who should waste OUR energy and our time (limited as those are) to build HIS ideas and make them work.

You are like the Liberal Politician who can find the energy and money to do anything he wants to do, as long as it's the other guy's energy and money.

Put YOUR energy and money where YOUR mouth is, and show us why we're wrong. After all, we've studied, and many of us have taught, simple (and not so simple) physics pretty much all our lives. Obviously we need to be taught where we went wrong by you!

Please, I stand (way off) in awe of your superior intellect. Enlighten me! But do it by demonstration, please. I eagerly await.

"Not considering, frictions or losses for this, do you see a flaw in this logic? Please point it out."

You ask, I answer. "Not considering, frictions or losses for this." There is your flaw.

But go ahead. Build it and show us how it's done. We'd like to see the frictionless, lossless machine you plan to build.

Oh, wait. I keep forgetting. You want US to build it.

Bwah, hah, hah. That was good. Tell another one. Pull the other leg. What? You are serious?

Nope. Done. Not interested.

You do it. Don't forget to budget for "frictions and losses" in your engineering and design work. We won't.

Please read the referenced thread.

This describes you perfectly.

#14

Your process is ambiguous. I think this is why this is not making sense. Try sorting out exactly what happens at each stage and between them. I think you are cobbling together inconsistent assumptions. That is where your missing energy is from. With what you have written it is so unclear I can't point to the first mistake but is certainly occurs by the line "Now add to that the balloon expanding as it goes up which..."

There are two components of energy to inflating the balloon underwater. First, you have to compress 2 cu m of air to 1 cu m and then you have to inflate the balloon against 1 atm of overpressure. The compression requires 50000 J of energy and the inflation requires 100000 J. This is why you get 150000 J if the balloon rises and expands.
If the balloon were rigid and rose to the surface without expanding, you would recover 100000 J in your tether. You could then recover the 50000 Joules by releasing the compressed air.

Tell me if this is the sequence of events you mean

1. Start with empty balloon at 10m down

2. pump up with air from surface

3. allow to rise slowly and collect the energy as work

4. evacuate balloon at surface

5. push empty balloon back down

That's how I see it. (May've screwed up a bit in my last analysis). <smilie>.

Ok. That makes sense to me. I haven't double checked the numbers but it seems right. Not sure why this would be better than just pushing a fixed mass bladder up and down in the waters. I guess it does use some less space but the efficiency of pumping air down a tube seems poor. Either way, you get out what you put in, at best.

Empty balloons don't sink very well. So, add weight to make it sink, heavy objects don't float very well. Add more air to make it float.

I don't see any net energy gain here, as is the case with so many of these dreams.

"Either way, you get out what you put in, at best" Only in a perfect world.

This contraption wouldn't generate enough energy to run the compressor.

Yes, there are two energy inputs: compress the air and inflate the balloon. Inflating the balloon pushes a cubic meter of water out of the way. If your surface area is 1 square meter, the water level will be 1 meter higher. If it's 100 square meters, the water level will be 1 cm higher, and so on. But that's the work you are doing inflating the balloon. If you float the balloon to the surface and it expands, you get both of these energy inputs back (except for losses).

Thanks for supplying this interesting problem. It has been fun thinking about this.

I left the part that you have to displace the water 1 bar 1cmb which uses another 100,000Nm from my assumption, so there is no gain from that with this approach. That answers the question I had.

But what inspired me to check it out is, there is a plant operating in Belgrad and in Germany which operates in some way on that idea. Done by a German/Swiss company. Plants doable in Megawatt size. It's quite intriguing. Can't quite figure how they do it. I have little chance to go there as I'm in the Caribbean. Check it out.

http://rosch-energy.ag/downloads/Rosch_Pr%C3%A4sentation.pdf

"So theoretically to pump up the balloon to 1 cbm would take 100,000 Nm energy not considering any losses."

There's the first big flaw. Your 100,000 Nm would compress 1m³ of air at 1atm to 2atm. But the volume at 2atm would only be 0.5m³. So to inflate the balloon 10m below the surface to 1m³ you would need twice as much energy!.

That would explain it. Let me think about that. I had calculated to compress 2 cbm atmospheric air to 1cbm being 100,000Nm.

This goes in the category of hydropump storage. Unfortunately it takes a lot of water to get a fair amount of PE. Go deep maybe. Some suggest undersea caverns. I like the hollowed out island ideas.

The factor is not 2. The pressure changes as you compress the air. At constant temperature PV=nRT is constant. P is inversely proportional to V, the volume of air you put in the tank. For a final pressure of 2 atm (.2 MPa) the correct formula for the amount of work = ∫Pdv is:

W=.2 MPa x 1 m3 x ln(.1 MPa / .2 MPa) = -.1386 MJ

The assumption that the balloon varies in volume linearly with depth is incorrect so you can't calculate the energy derived from allowing the balloon to rise by taking the average volume to calculate the buoyancy force. The volume varies inversely with the depth just as the pressure in the tank varies inversely with the volume pumped in. Integrating under the curve ∫Vdp gives the same amount of energy derived from the expansion as was expended in compressing the air.

Minus any friction losses and compressor/motor/generator inefficiencies.

If you want free energy, harness the tidal change in the ocean....that's something that does work....at high tide you fill a reservoir, and when the tide goes out you release the water through a turbine generator....they can be made to work both ways, coming and going....moon power baby

http://en.wikipedia.org/wiki/Tidal_barrage

Ya mean like Long Island Sound, or Great South Bay?