# Pressure Harvesting - from ocean depths

Yes I did. Explicitly.
Sorry you did to....

You still have to re-inflate your collapsed storage vessel and drag it back to the surface. That will cost more energy than you get from the pressurized air.
why would you reinflate the vessel? If you did that you may as well just pump the 35000kpa on the surface...
At the moment this thread has yet to properly deal with the issue of retrieval and it probably wont... because the energy gained from the vessel's descent is not allowed for some inexplicable reason.

I'd attempt to explain again but you are not reading my posts fully as it is... so whats the point...?

If you don't wish to explore the methods possible for retrieval then fine... don't.

There is no need. I was just hinting how to get your air down there with as little energy as possible.
Alex

Sorry you did to....
If you are not getting it free why are you proposing to do it? It just becomes a pointlessly elaborate way of getting compressed air, instead of running a compressor.

If you are not getting it free why are you proposing to do it? It just becomes a pointlessly elaborate way of getting compressed air, instead of running a compressor.
Because the weight of the VVSS and the weight of the ocean are doing the compressing not a pump. The ability to compress air to 35000kpa using a mechanical pump is not easy and very energy intensive I would presume. ( Coal, Nuclear or Hydro)
The rest is just how you retrieve your vessel to make it economical.

Because the weight of the VVSS and the weight of the ocean are doing the compressing not a pump. The ability to compress air to 35000kpa using a mechanical pump is not easy and very energy intensive I would presume. ( Coal, Nuclear or Hydro)
The rest is just how you retrieve your vessel to make it economical.
The process you propose costs just as much energy as simple mechanical compression.

If you are not getting it free why are you proposing to do it? It just becomes a pointlessly elaborate way of getting compressed air, instead of running a compressor.
Besides I was learning how to do an energy cycle assessment but now I am not sure I could be bothered...

The process you propose costs just as much energy as simple mechanical compression.
where is all the energy cost?

Unfortunately because you refuse to accept the down cycle is able to generate electricity...which can be used to help bring the vessel back to the surface...you will never know how much it actually costs to retrieve the vessel.

where is all the energy cost?

Unfortunately because you refuse to accept the down cycle is able to generate electricity...which can be used to help bring the vessel back to the surface...you will never know how much it actually costs to retrieve the vessel.
Nope, I am not that stupid, oddly enough.

What you need to do is calculate the difference in buoyancy between the lowering and the raising. The buoyancy is reduced by the reduction in air volume at the bottom. So you expend more in raising it than you get from lowering. And this difference is more than the energy you can get from the compressed air at the surface.

Nope, I am not that stupid, oddly enough.

What you need to do is calculate the difference in buoyancy between the lowering and the raising. The buoyancy is reduced by the reduction in air volume at the bottom. So you expend more in raising it than you get from lowering. And this difference is more than the energy you can get from the compressed air at the surface.
ok.... we are getting somewhere finally..
Now I got to go away and learn how to do the math so it includes deep water density changes in weight (displacement) and exactly how much energy it would take to bring the VVSS to the surface less the energy it generated going down.

ok.... we are getting somewhere finally..
Now I got to go away and learn how to do the math so it includes deep water density changes in weight (displacement) and exactly how much energy it would take to bring the VVSS to the surface less the energy it generated going down.
There is almost no change in density of water with depth, because water is virtually incompressible.

The only factors increasing water density with depth are changes in temperature and salinity. These are negligible as far as this exercise is concerned, less than 1% or so.

There is almost no change in density of water with depth, because water is virtually incompressible.

so I need to work out the rest of the problem before we can get an energy cost assessment...

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There is almost no change in density of water with depth, because water is virtually incompressible.
I found this earlier which is why I included it...
Remember we are talking about very deep 3000 meters or more
Where d is the density of water, g is the acceleration due to gravity, v is the volume of the immersed object.
Now in practical situation, as we go to more and more depths in an ocean or something, the density of water increases with increase in pressure.
Also the acceleration due to gravity changes with the depth. So bouyant force does change with the depth but in our day to day life one can assume bouyant force to be constant for small depths.

https://www.quora.com/Does-buoyant-force-depend-on-depth
Density = mass/volume

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Why wouldn't this system be as efficient, say, as pumped hydro ( 60%- 70%)?

or Write4u's deep water turbine at 80%

I found this earlier which is why I included it...
Remember we are talking about very deep 3000 meters or more
Where d is the density of water, g is the acceleration due to gravity, v is the volume of the immersed object.
Now in practical situation, as we go to more and more depths in an ocean or something, the density of water increases with increase in pressure.
Also the acceleration due to gravity changes with the depth. So bouyant force does change with the depth but in our day to day life one can assume bouyant force to be constant for small depths.

https://www.quora.com/Does-buoyant-force-depend-on-depth
But this small effect is not quantified. And if you read on, there is also this reply:
"
The way I think you mean the question, the answer is no. If you have a rigid object with a fixed volume, it will have the same buoyancy force just below the surface of the water as it would at the bottom of a swimming pool.

If you submerge it to the bottom of the ocean, then there is a small increase in the density of the water because it is very slightly compressed by the weight of all the water above it pressing down. But this is a small effect. It would very slightly increase the buoyancy force on the object. "

But this small effect is not quantified. And if you read on, there is also this reply:
"
The way I think you mean the question, the answer is no. If you have a rigid object with a fixed volume, it will have the same buoyancy force just below the surface of the water as it would at the bottom of a swimming pool.

If you submerge it to the bottom of the ocean, then there is a small increase in the density of the water because it is very slightly compressed by the weight of all the water above it pressing down. But this is a small effect. It would very slightly increase the buoyancy force on the object. "
sorry but I am reading conflicting info, that just confuses me.
Density = mass/volume
Deep water has greater mass (?)
so wouldn't that mean that deep water has a higher density.

Say we take a rigid 1m^3 container open it at 3000 meters, seal it and bring it to the surface full of deep water. Certainly the volume will stay the same...
Would it weigh the same as a similar container that has captured water from the surface?
Intuitively I would say no... but hey...

sorry for being a pain

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why would you reinflate the vessel?
You'd do that when it was on the surface, to get it ready for the next drop to the ocean floor. Otherwise, what are you going to do? Construct a new vessel for each drop?

If you did that you may as well just pump the 35000kpa on the surface...
Bingo! That's what I've been saying all along. Pumping the pressure at the surface will be a much more cost effective way to get pressurised air.

At the moment this thread has yet to properly deal with the issue of retrieval and it probably wont... because the energy gained from the vessel's descent is not allowed for some inexplicable reason.
What? I discussed the energy gained from the vessel's descent in a previous post, along with the energy lost during that descent.

I'd attempt to explain again but you are not reading my posts fully as it is... so whats the point...?
What makes you think I'm not reading your posts fully? Did I miss something in one of your replies to me?

If you don't wish to explore the methods possible for retrieval then fine... don't.
I didn't say that.

Do you think you've got some retrieval process that will get you your 70% efficiency?

Where does that number even come from, anyway? Did you just make it up?

I believe I'm the only one who addressed the OP question of direct harvesting of deep water pressure.

Using Water Pressure at the Bottom of the Ocean to Store Energy

To use the water pressure at the seabed in practice, the mechanical energy is converted by a reversible pump turbine, as in a normal pumped storage hydroelectric plant.

https://oilprice.com/Energy/Energy-...-the-Bottom-of-the-Ocean-to-Store-Energy.html

In short:
Ocean high pressure Headwater ~~~> inlet/outlet valve ~~~> reversible turbine/generator ~~~> low pressure tailwater tanks/reservoir ----> passive airvent equalizer.
This is free electric generation. ~~~~~~~> 100%

Ocean <~~~ inlet/outlet valve <~~~ reversible turbine/pump <~~~ filled tailwater tanks/reservoir <---- passive airvent for equalization.
This is electric pumping use <~~~~~~~~ 80% ?

The only pumping that needs to be done is to evacuate the accumulated water from the tailwater tanks.

My question is if the high pressure inflow can generate more electricity than will be used to pump the water back to the ocean. Even if the net gain is small, this would still be beneficial over the long run. Net gain 20%? would be awesome.

At first impression, this seems a simple but functional system. Question is Net yield.

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sorry but I am reading conflicting info, that just confuses me.
Density = mass/volume
Deep water has greater mass (?)
so wouldn't that mean that deep water has a higher density.

Say we take a rigid 1m^3 container open it at 3000 meters, seal it and bring it to the surface full of deep water. Certainly the volume will stay the same...
Would it weigh the same as a similar container that has captured water from the surface?
Intuitively I would say no... but hey...

sorry for being a pain
Yes I'm afraid it is rather a thankless task trying to explain this to you. I may come back to it later, if I have enough energy and nothing better to do.

The only pumping that needs to be done is to evacuate the accumulated water from the tailwater tanks.
That's true. And the only pumping that needs to be done in pumped hydro is evacuating the accumulated water in the lower reservoir.
My question is if the high pressure inflow can generate more electricity than will be used to pump the water back to the ocean.
Answer is no. Just like pumped hydro.