Pressure Harvesting - from ocean depths

[billvon]

I know that .. How much energy can 1 Kpa deliver?
I get the impression 1 Kpa m^3 delivers 1 Joule.

Pressure is just pressure. That's not enough to answer the question.

Is there a cubic meter of air at 1000 KPa, released into the atmosphere (100 KPa?) Then the available energy with an ideal motor (which doesn't exist of course) is about 291 KJ, which is about .08 kilowatt-hours. (Enough to run a typical home for about 5 minutes.)

Suppose you fill Hang Son Doong cave with 1000KPa air? (38.5 million cubic meters) Then you will get 3 megawatt-hours, enough to run about 100 homes for a day.

Is there a cubic meter of water at 1000KPa, released at atmospheric pressure? Since water is incompressible, the available energy is close to zero.

 

[billvon]

There is no pumping involved. Nothing is stored .

You can store water high, or you can store a vacuum low. Then let the water flow. Exactly the same. Same flow, same pressures.

. . .the guy who wrote the review . . .

Two of those quotes were not from the guy who wrote the review, but from the guy who invented it. Again, read the article.

[he] did not fully understand the system .

So you went from claiming that I had not read the article, to saying that I DID read the article, but what I read was wrong?

You're one of those people whose ego is so big that you can never, ever admit you are wrong. Good luck with that.

 

[Quantum Quack]

No different in overall principle - only in particulars. In every 'gravity fueled' scenario possible, there is simply exchange of one form of energy for another. With no cyclic gain possible. Let's run through the specifics quoted in p9 #174:
Initially a fully expanded cylinder containing air is poised just above the ocean surface. There experiencing just gravitational force of magnitude mg. Lowering it till just fully submerged has steadily reduced the effective weight to zero when fully submerged. There has been a loss of cylinder PE and a concomitant gain in ocean PE, since it's level must rise against gravity during this process. Further lowering the cylinder - now having neutral buoyancy - entails no appreciable further energy exchanges (finite compressibility of the cylinder is entirely material dependent thus not fundamental to the basic physics, and at any rate will be very small).
At the bottom, the cylinder is allowed to compress piston-like. The ocean surface level now drops back to almost the level initially before cylinder submersion - reducing it's PE accordingly. But that merely accounts for almost the initial PE gain when the cylinder went from dry to just fully submerged. It does not account for the huge driving pressure at the ocean depth creating much more pneumatic PE than that involved in the initial submersion energy exchanges. Where then does the extra energy come from?

The answer is there in plain sight. Compressing the cylinder at depth has simultaneously lowered the PE of the cylinder - it's no longer at neutral buoyancy and weighs almost as much as when suspended in air. Hauling it back up accounts for all the apparent magical gain in pneumatic energy of compressed air.

To repeat from p7 #128, for a gravitational field
∇ x -∇φ = 0. Zero curl. The latter is a fundamental mathematical fact and excludes any possibility of net gain or loss of energy over a closed path. One-way trips are pointless, as stated many times this thread.
Are we finally done here?

Unfortunately when it comes to math I am quite dyslexic. (Seriously) But I do try.
To work out the energy cycle in this scenario including all energy potentials would be great.
Further,
Retrieving the VVSS:
Like your hole in the ground scenario it can provide energy via cables connected to a generator as the VVSS descends.
So electricity is generated as the VVSS compresses the air while descending. In a perfect situation that energy gained should be significant. When used in addition to added energy, whether solar or other in retrieving the VVSS from the depth that it rests (probably less than 50%) using that combined electricity source to power a high torque slow lift winch. (Time taken to raise the VVSS is not a factor)

So,
Energy derived from descent =?
Plus
Energy derived from compressed air, bottled at the surface.= ?
Less
Any non renewable Energy needed to supliment retrieval of the VVSS= ?
Equals
Net energy converted to a useful form =?

Any other factors to consider other than infrastructure costs?

Iam not even sure i am asking the right questions...

(Mobile)

 

[Q-reeus]

Unfortunately when it comes to math I am quite dyslexic. (Seriously) But I do try.
To work out the energy cycle in this scenario including all energy potentials would be great.
Further,
Retrieving the VVSS:
Like your hole in the ground scenario it can provide energy via cables connected to a generator as the VVSS descends.
So electricity is generated as the VVSS compresses the air while descending. In a perfect situation that energy gained should be significant. When used in addition to added energy, whether solar or other in retrieving the VVSS from the depth that it rests (probably less than 50%) using that combined electricity source to power a high torque slow lift winch. (Time taken to raise the VVSS is not a factor)

So,
Energy derived from descent =?
Plus
Energy derived from compressed air, bottled at the surface.= ?
Less
Any non renewable Energy needed to supliment retrieval of the VVSS= ?
Equals
Net energy converted to a useful form =?

Any other factors to consider other than infrastructure costs?


(Mobile)

At this stage I suggest contacting one or more university physics lecturers and see who straightens who out. Cheers.

 

[Quantum Quack]

At this stage I suggest contacting one or more university physics lecturers and see who straightens who out. Cheers.

Ok.. thanks...

 

[Quantum Quack]

Pressure is just pressure. That's not enough to answer the question.

Is there a cubic meter of air at 1000 KPa, released into the atmosphere (100 KPa?) Then the available energy with an ideal motor (which doesn't exist of course) is about 291 KJ, which is about .08 kilowatt-hours. (Enough to run a typical home for about 5 minutes.)

Suppose you fill Hang Son Doong cave with 1000KPa air? (38.5 million cubic meters) Then you will get 3 megawatt-hours, enough to run about 100 homes for a day.

Is there a cubic meter of water at 1000KPa, released at atmospheric pressure? Since water is incompressible, the available energy is close to zero.

So 35000kPa could power a lot of homes for a day.....(3500)

Edit: Sorry i screwed that up. Shall post again later when I get back to my desktop...pc

 

[Write4U]

You can store water high, or you can store a vacuum low. Then let the water flow. Exactly the same. Same flow, same pressures.

OK from the illustration, what part of the system is used for storing electricity? Nothing! It identifies the ocean as the only energy storage medium.
But that does not make the ocean, nor the generating system a battery. The energy potential of the ocean has not even been converted, until it activates the turbine generator/pump, at which time electricity is produced, but not shown to be stored in any batteries. In fact the illustration also fails to shown the direct connection to the surface grid, but it mentions that in the article so we can assume that was an aesthetic deletion by the illustrator.

Two of those quotes were not from the guy who wrote the review, but from the guy who invented it. Again, read the article.

Right and he is German and probably spoke German and the translator was using the term "storage" incorrectly, just as the illustrator neglected to shown the electrical connection to the surface grid.
If you substitute the term storage with "processing", and the term battery with "electricity transmission", it clears up the entire picture, and makes perfect sense of the system's function and utility.

So you went from claiming that I had not read the article, to saying that I DID read the article, but what I read was wrong?

No , I have read it three times now and the word storage is not used correctly except as a vague generality to indicate that the system can produce electricity over and over again. But that is hardly storing electricity as in a battery, which I am willing to bet is another mis-translation of a German term.

If the system does use batteries to store the electricity produced by the turbine (which I stipulated that could be done) it is not shown in the illustration. Moreover it does state that the system is directly connected to the surface power grid, which it feeds and occasionally uses to pump the water out when the tanks are saturated and cannot accommodate additional incoming water.

If the system does use batteries the article does not mention that at all. It merely states that the system is a battery and a power storage devise, which it is not as illustrated. As illustrated any batteries are located on the surface and only the ocean itself is the only pressurized energy storage device. Is "storage device" the correct term here? I'd prefer to see the term "energy potential", just as a mountain lake has "energy potential" until the kinetic force of the water turns a generator and the energy potential is converted in to electricity which is usable and/or may be stored in batteries.

The article does not make any mention of electricity being stored in an on-board battery.

You're one of those people whose ego is so big that you can never, ever admit you are wrong. Good luck with that.

Why on earth should I admit to being wrong when you cannot point to anything I said that is in fact wrong. But I can identify several flaws in the article's translation and illustration of the system's function. Which is not surprising being that we are dealing with a translation of a German article.
The author;

BRIAN WESTENHAUS
Brian is the editor of the popular energy technology site New Energy and Fuel. The site’s mission is to inform, stimulate, amuse and abuse the news and views across the emerging field of energy and fuels in our future. You will find the most exciting and useful news, guides and tips for making and saving money in energy and fuel, just how things work or not, where you might want to invest or get involved in a brainstorming session with other readers.

I rest my case.

p.s. Of all the nonsense about rocks and artificial pressure creation, this is the only discussion which directly addresses the OP; "Pressure Harvesting from ocean depths".

 

[Xelasnave.1947]

I have it.
You could put flowers in books lower them an use the pressure to press your flowers.
Alex

 

[Quantum Quack]

I have it.
You could put flowers in books lower them an use the pressure to press your flowers.
Alex

But wouldn't that be in conflict with the laws of thermodynamics. The energy cost of retrieving your flowers and book will stop it from being a perpetual energy device. lol...

 

[exchemist]

You don't seem to have had the seeming conundrum presented in p9 #174 resolved.
This is simply a variant on say lowering a mass down a hole in the ground, via a winch at the surface attached to say an electrical generator. A one-way conversion of potential energy to electrical. In your conundrum it's not immediately obvious where the trade between potential energy and pneumatic energy arises. It occurs at the bottom when the air is being compressed, not on the way down. Hauling the compressed cylinder back up to the surface now takes energy since the vessel is now far less buoyant, despite a small overall reduced mass given air has escaped to the surface via the attached air-line. I won't bother doing explicit sums to prove it - what I wrote in #128/#134 covers all possible variations within the basic theme. If the source is a conservative field no net changes in overall energy are possible.

Also note that a bird's-eye view must take into account a very tiny lowering of ocean depth when the submerged cylinder is compressing (it was initially raised when the cylinder was completely submerged near the top - Archimedes principle). Overall then merely trading one source of PE for another - compressed gas. With the attendant losses thrown in.

Yes your explanation is exactly the same as the one I have been trying to get across. The problem is that I can't make the maths add up in my simplified example, so either I have made an arithmetical error or I have neglected some element in my analysis. At the moment the work done by the ocean in compressing the air is only a quarter of the work needed to pull the less buoyant chamber back to the surface. So there is a rabbit away somewhere. I was hoping someone here might be able to point out the mistake.

 

[Xelasnave.1947]

But wouldn't that be in conflict with the laws of thermodynamics. The energy cost of retrieving your flowers and book will stop it from being a perpetual energy device. lol...

No conflict. Pressing flowers is not about saving money it's about producing the best pressed flowers ever created...anyways I have solved the potential energy loss concern...we use volunteers to haul them up...it's so crazy it must work.
Alex

 

[Write4U]

I have it.
You could put flowers in books lower them an use the pressure to press your flowers.
Alex

An octopus would steal it and use the book and flower as a decorated home..

....

 

[Xelasnave.1947]

An octopus would steal it and use the book and flower as a decorated home..

....

Why must you be so negative?
But what you raise can only help marketing...these flowers had to be produced under difficult conditions, and hence the high price, ...
Alex

 

[Write4U]

Why must you be so negative?
But what you raise can only help marketing...these flowers had to be produced under difficult conditions, and hence the high price, ...
Alex

Octopuses have a very keen sense of value and they are master thieves, able to navigate the most intricate mazes and open triple locked vaults .

and a pretty female

 

[Xelasnave.1947]

Octopuses have a very keen sense of value and they are master thieves, able to navigate the most intricate mazes and open triple locked vaults .

I know that's why I never allow them in my house.
Clearly they are the aliens in the UFO s we don't notice diving into the mid Atlantic...an interesting creature..brains in their tenticals I think...I propose we negotiate with them ..the pressed flower market is big enough to pay many players...but at the annual general meeting of the ocean pressed flower corporation we take them out ok.
Alex

 

[geordief]

I know that's why I never allow them in my house.
Clearly they are the aliens in the UFO s we don't notice diving into the mid Atlantic...an interesting creature..brains in their tenticals I think...I propose we negotiate with them ..the pressed flower market is big enough to pay many players...but at the annual general meeting of the ocean pressed flower corporation we take them out ok.
Alex

I read "testicles" at first. The mind boggles (or boogies)

 

[Q-reeus]

Yes your explanation is exactly the same as the one I have been trying to get across. The problem is that I can't make the maths add up in my simplified example, so either I have made an arithmetical error or I have neglected some element in my analysis. At the moment the work done by the ocean in compressing the air is only a quarter of the work needed to pull the less buoyant chamber back to the surface. So there is a rabbit away somewhere. I was hoping someone here might be able to point out the mistake.

You have presented explicit calculations this thread? If so point to the post. It's a bit involved since only an adiabatic process can even in principle achieve 100% efficiency over a cycle. The relation between piston stroke position and pressure is non-linear. And energy change in turn requires integrating of that. Maybe try working through this article to find what you need:
https://en.wikipedia.org/wiki/Adiabatic_process
The thread has degenerated into a shitty mess.....

 

[exchemist]

You have presented explicit calculations this thread? If so point to the post. It's a bit involved since only an adiabatic process can even in principle achieve 100% efficiency over a cycle. The relation between piston stroke position and pressure is non-linear. And energy change in turn requires integrating of that. Maybe try working through this article to find what you need:
https://en.wikipedia.org/wiki/Adiabatic_process
The thread has degenerated into a shitty mess.....

Post 96. This contains one mistake in that I used the gas equation to calculate the work done by the sea (i.e. isothermal) whereas I should have used an adiabatic one to avoid the issue of heat loss to the seawater. However the difference is not that great - I gave the adiabatic figure in the footnote (and I see billvon has used a similar figure in a later post). But even using that figure I can only account for about a third of the work done in recovering the chamber, once the buoyancy has been reduced. I think I must have overlooked something.

Any ideas?

 

[Q-reeus]

Post 96. This contains one mistake in that I used the gas equation to calculate the work done by the sea (i.e. isothermal) whereas I should have used an adiabatic one to avoid the issue of heat loss to the seawater. However the difference is not that great - I gave the adiabatic figure in the footnote (and I see billvon has used a similar figure in a later post). But even using that figure I can only account for about a third of the work done in recovering the chamber, once the buoyancy has been reduced. I think I must have overlooked something.

Any ideas?

Interesting. I'll have to sleep on it but will just point out the pneumatic energy storage must be less for an adiabatic process as the pressure rises more steeply with compression viz a viz isothermal. Hence less swept volume occurs. Apparently worsening the 'crisis'. Well maybe not as in turn the reduced buoyancy is itself reduced also, tending to negate the discrepancy. Not enough I fear. Dear dear!

 

[Q-reeus]

Stop Press:
OK I think I now know where the discrepancy lies. When the piston is suddenly released there is initially little resistance from the air inside. Consequently there is a rapid buildup of inrush KE that has not been factored in. All along the assumption was of a quasi-static compression where only an exchange of PE's in involved. Which is not realistic.
So you will have to work out the added KE resulting in a final higher compression than expected till now. Because it is dynamical there would normally be a bounce back somewhat, which can in principle be obviated by a one-way camming action say. Regardless, the final correct calculations must be consistent with zero net change over a cycle.