Pressure Harvesting - from ocean depths

Quantum Quack said:
This puzzles me...
By the time it gets to 1000 meters it's crew are living in approx. 1472psi of compressed air. If it were not the sub would implode.
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Sorry. What do you think the pressure is inside a sub at-depth?

OK, Ex beat me to it. The pressure inside a sub is 1 atmosphere. That's why they have to be rigid, and why they implode if you look at them funny.
 
Quantum Quack said:
This puzzles me...
A submarine with uncompressed air for the crew only has to fill it's ballast with water and it sinks no problemo...
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Yes. You can build an arbitrarily strong container; then the internal pressure does not change (much) no matter what the external pressure does. And you can make it arbitrarily heavy so it sinks.

Now - how do you think it rises again?
By the time it gets to 1000 meters it's crew are living in approx. 1472psi of compressed air. If it were not the sub would implode.
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At first when you posted stuff like this I assumed you were joking. I have come to realize you simply have no clue.
 
billvon said:
Yes. You can build an arbitrarily strong container; then the internal pressure does not change (much) no matter what the external pressure does. And you can make it arbitrarily heavy so it sinks.

Now - how do you think it rises again?

At first when you posted stuff like this I assumed you were joking. I have come to realize you simply have no clue.
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How do you think a submarine rises?
 
Quantum Quack said:
How do you think a submarine rises?
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The problem here is you're off on a wild goose chase. A submarine is not an analogy to a pressure harvesting setup. A submarine is not a source of energy - it's a sink.
Let's not waste dozens of posts convincing you you can't make a sub into an energy source.
 
DaveC426913 said:
The problem here is you're off on a wild goose chase. A submarine is not an analogy to a pressure harvesting setup. A submarine is not a source of energy - it's a sink.
Let's not waste dozens of posts convincing you you can't make a sub into an energy source.
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Agreed. Let's wait for him to get back on topic.

Ball is in your court, QQ.
 
exchemist said:
The sub sinks due to its weight minus the buoyancy of its hull. When the tanks are flooded the buoyancy is reduced to close to neutral (ie almost matching the weight), allowing the sub to go up or down in response to the angle of the hydroplanes. When it surfaces, air is pumped into the tanks, displacing water out and increasing buoyancy again, so that it rises.
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ok... so... how does knowing that effect the ability to sink a Variable volume storage vessel (VVSS) and deliver air compressed on descent to a bulk storage facility and then return to the surface to do it again if wanted.
 
Quantum Quack said:
ok... so... how does knowing that effect the ability to sink a Variable volume storage vessel (VVSS) and deliver air compressed on descent to a bulk storage facility and then return to the surface to do it again if wanted.
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Whut?

Try again?
 
Now you clever chaps are on line I have a puzzle for you.

I imagine a simple scenario with a 1 litre container weighing 1kg, so its buoyancy exactly matches its weight. This can descend, without any work being done at all, say to 100m depth, at which point the ambient pressure is about 10 atm. QQ then opens the valve. The air is compressed to 10atm, isothermally, because of the water all around. The work done by the water to compress it is given by W=nRTln(V0/V) which I work out to be about 230J*. This is then the stored energy QQ can "harvest" when it gets back to the surface.

However the volume of water displaced by this device is now only 0.1litres, so its apparent "weight" is now 0.9kg, i.e. 9N. So to pull it up to the surface one has to expend 9x100 = 900J. Nearly 4 times the stored energy. But I can't think where in the energy balance for this operation, the rest appears. So I'm wondering if I have made an error somewhere. Any ideas?


(*I found an engineering source on line that quotes the equivalent of about 270J, but that was adiabatic, I think)
 
exchemist said:
I imagine a simple scenario with a 1 litre container weighing 1kg, so its buoyancy exactly matches its weight. This can descend, without any work being done at all, say to 100m depth, at which point the ambient pressure is about 10 atm. QQ then opens the valve. The air is compressed to 10atm, isothermally, because of the water all around. The work done by the water to compress it is given by W=nRTln(V0/V) which I work out to be about 230J*. This is then the stored energy QQ can "harvest" when it gets back to the surface.

However the volume of water displaced by this device is now only 0.1litres, so its apparent "weight" is now 0.9kg, i.e. 9N. So to pull it up to the surface one has to expend 9x100 = 900J. Nearly 4 times the stored energy. But I can't think where in the energy balance for this operation, the rest appears.
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I strongly suspect it is present in the additional heat the now-compressed air contains.
 
billvon said:
With a hose.
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well done!
So stored pressure is delivered to the surface with out having to add energy to the system.

All we have to do now is get the air down to the storage facility also with out adding energy.
So we use a specially designed submarine that sinks down to the storage unit compressing on board variable volume storage while it does so.
It docks with the storage unit and transfers the pressurized air to supplement the bulk storage unit.
The sub then blows it's ballast and surfaces to do it all over again...
 
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