Pressure Harvesting - from ocean depths

Quantum Quack said:
No I haven't made a mistake. Pressure is in fact considered as energy. KE Kinetic energy and PE Potential energy and a couple of other more obscurely labeled energies.
All we are doing is transferring the potential energy of the ocean weight and VVSS's weight by way of compression of air into the VVSS.
The compressed air in the VVSS then has the PE of the ambient PE.
That PE is then ported to the atmosphere still as Potential Energy where it can be deployed later at leisure as energy.

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

Again there is no free lunch.
If we were not able to do the above the laws of thermodynamics would fail.
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This is wrong and you even posted a formula that says it is wrong. The formula was P=W/V, or, in the form most familiar to us, PdV=dW. Notice it does not say P=W. Also notice the unit of pressure are not units of energy.

"PV work" is a standard concept in physical science. What this means is that Pressure combined with a change in volume involves work being done, either by the pressure against something, or by something against it. But if the volume does not change no work is done.
 
Write4U said:
I took Alex's compliment as an honest expression.
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You are correct.
Let me ask you what you think about this statement.
The op asks if energy can be harvested from the pressure under water.
I say that any energy taken out in any exercise will be less than the energy put into the system that enables the harvesting system to work.
Alex
 
billvon said:
OK then do the math. I have 1000psi. How much energy do I have?
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depends on how much volume you have.... let's see....an ocean of volume amounts to a hell of a lot of potential energy.....

so I suggest you word your challenge so it can be answered...
 
Neddy Bate said:
A better version would be to connect the free end of the hose to a tank (mounted above the surface of the water), and put a check valve there so that no pressure from the tank can ever come back down the hose. Now put your thumb over the other end of the hose, swim to the bottom, and take your thumb off the hose. The water pressure will push water up into the hose, forcing air into the tank. Swim back up to the surface, disconnect the hose from the check valve/tank and then pull the hose up so that the water empties out of it back into the pool. Reconnect the hose and repeat. You get a tank of compressed air, but you also get tired expending all of that energy yourself.
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yes now you are thinking...
the displacement of the air filled hose raises the sea level so that when you open the end of the hose at depth the water pressure will force water into the hose and expel the air under pressure to the surface....
The VVSS system I have described is more or less the same principle but involves a much larger amount of high density potential energy potted at the surface.
The cost of exploiting potential energy from the ocean depths is the only thing yet to be properly determined ( in a set of ratios and formula) and could be considerably better than solar etc.

The key benefit is that the PE acheived is high density energy. PE at 3000 meters is about 36000kPa which is high value potential energy.

Once this hysteria about free energy is sorted maybe just maybe we may find out about the economics of making use of this PE displacement principle...
 
exchemist said:
Also notice the unit of pressure are not units of energy.
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of course not.
Pressure as a material phenomena has potential energy. Units of measurement have no potential energy.
I'll try to explain how I see it...
Quite simply put, as billvon is attempting to stress for pressure to deliver an energy it must flow.
Potting high density pressure from ocean depths is simply transferring potential, not flow.

What happens at depth is just transferring by way of compression PE from the ocean and weight of the Vessel to the vessel. Then it gets a transferred again to the surface as PE.
Once the pressure is made use of by flowing, then and only then is it able to be converted to energy that does work.
 
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billvon said:
Good! We are making progress.

OK so you have 1000psi and a volume of 1 cubic meter. How much energy do you have?
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you only have potential to do work.
how do you calculate PE?
DaveC426913 said:
You still haven't explained how this is any better than an air pump.
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Until a proper assessment of the energy costs I can't address that. can you? Or are you happy to just guess?
You seem quite happy to mix cost with principle and get no where doing either but I am not..
 
Quantum Quack said:
Until a proper assessment of the energy costs I can't address that. can you? Or are you happy to just guess?
You seem quite happy to mix cost with principle and get no where doing either but I am not..
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Yours is orders or magnitude more costly, not to mention incredibly inconveniently located, requiring a crew and ships.
 
DaveC426913 said:
You still haven't explained how this is any better than a pump.

Here's a pump that delivers 41,000 kpa for as long as you need it to. It weighs 325 lb and can be transported in the back of a pickup to anywhere you want.

https://ph.parker.com/ca/en/premier...e-industrial-piston-pump-premier-piston-pumps
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to pump 35000kPa using a mechanical pump would be something to work out and could prove interesting when compared to exploiting 35000kPa from ocean depths using and efficient displacement system.
 
DaveC426913 said:
Yours is orders or magnitude more costly, not to mention incredibly inconveniently located, requiring a crew and ships.
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easily fully automated., self contained and to be honest probably low cost once established.

You need to be more creative than obstructionist.
 
Quantum Quack said:
to pump 35000kPa using a mechanical pump would be something to work out and could prove interesting when compared to exploiting 35000kPa from ocean depths using and efficient displacement system.
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Electric pumps are already very efficient, quiet, clean and very cheap to operate, and maintain. Not to mention they can be easily place wherever you want them.
 
DaveC426913 said:
Electric pumps are already very efficient, quiet, clean and very cheap to operate, and maintain. Not to mention they can be easily place wherever you want them.
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well then, shut down nuclear reactors and coal fired power stations if that is true.... by all means. I look forward to it..
 
DaveC426913 said:
Oh, so we're fantasizing are we?

A cargo shuttle to Jupiter and back can "easily" be fully automated too.
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When the time comes, if ever, it makes sense to run automated runs of cargo to Jupiter, don't you think...
I hear Titan is great this time of the year... lol

We are already running probes and landers more or less automated to Mars...Running automated cargo to Mars is a foregone.
 
Quantum Quack said:
well then, shutdown nuclear reactors and coal fired power stations if that is true.... by all means. I look forward to it..
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Wow. You continue to confuse energy converters with energy generators.


Look, I don't have a problem with you imagining novel solutions to energy problems, but Jeez, you gotta know the first thing about the principles involved. You don't understand the very basic rule zero of thermodynamics and hydraulics - you're just talking out of your hat.

You would do better to stay in your lane. What are you knowledgeable about?
 
Quantum Quack said:
yes now you are thinking...
the displacement of the air filled hose raises the sea level so that when you open the end of the hose at depth the water pressure will force water into the hose and expel the air under pressure to the surface....
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Okay, so I found some information online that says that for every 33 feet (10.06 meters) you go down into the ocean, the pressure increases by 14.5 psi (1.00 bar). So at 6*33=about 200 feet (60 meters) the pressure would be 87 psi (6 bar) which is a decent amount of pressure for storing air pressure in metal tank, so let's use that as an example.

A hollow metal tube with an inner diameter of 2 inches would have about 2piR=3 square inches of area. So the force required to push that tube down to where we need to harvest our pressure would be 87 (pounds per square inch) multiplied by 3 (square inches) equals 261 pounds of force. That means you would have to use 261 pounds of force to push the tube down 200 feet into the ocean.

So, if you use a 2 inch diameter pneumatic rod cylinder to push it down, you would require 87 psi which is exactly the amount you hope to harvest. As for the volume of compressed air involved, since we used the same diameter tube for both harvesting and pushing the tube down, you would obtain the same amount of air volume that you would expend. You end up right back where you started, except there is energy lost to friction and other effects. You don't gain anything.

You could also just use a 261 pound weight to push the tube down. But picking up that weight, for example using a 2 inch bore pneumatic activator, would also require 87 psi, and you would have to pick it up about 200 feet to get it over the 200 foot tube. You just don't gain anything.
 
Neddy Bate said:
Okay, so I found some information online that says that for every 33 feet (10.06 meters) you go down into the ocean, the pressure increases by 14.5 psi (1.00 bar). So at 6*33=about 200 feet (60 meters) the pressure would be 87 psi (6 bar) which is a decent amount of pressure for storing air pressure in metal tank, so let's use that as an example.

A hollow metal tube with an inner diameter of 2 inches would have about 2piR=3 square inches of area. So the force required to push that tube down to where we need to harvest our pressure would be 87 (pounds per square inch) multiplied by 3 (square inches) equals 261 pounds of force. That means you would have to use 261 pounds of force to push the tube down 200 feet into the ocean.

So, if you use a 2 inch diameter pneumatic rod cylinder to push it down, you would require 87 psi which is exactly the amount you hope to harvest. As for the volume of compressed air involved, since we used the same diameter tube for both harvesting and pushing the tube down, you would obtain the same amount of air volume that you would expend. You end up right back where you started, except there is energy lost to friction and other effects. You don't gain anything.

You could also just use a 261 pound weight to push the tube down. But picking up that weight, for example using a 2 inch bore pneumatic activator, would also require 87 psi, and you would have to pick it up about 200 feet to get it over the 200 foot tube. You just don't gain anything.
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yep... an exercise in futility for sure...
so what's the difference between a variable volume vessel and your fixed volume vessel?
Answer: you don't need to push it down...
 
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