Pressure Harvesting - from ocean depths

Quantum Quack said:
but your assessment of the pump is wrong and if you can not see that then we have little more to discuss..
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Okay, you're right. If one side of the piston is the sealed off 'air' side, and the other side of the piston is the sealed off 'water' side, then it should work. But it's not because of the handle at the top, because that has equal pressure on both sides of the handle. It's because of the open port that lets the water pressure in to one side of the piston. I realise now that you never said the handle was the reason why, but I mistakenly thought that was your idea. Sorry.
 
Neddy Bate said:
Okay, you're right. If one side of the piston is the sealed off 'air' side, and the other side of the piston is the sealed off 'water' side, then it should work. But it's not because of the handle at the top, because that has equal pressure on both sides of the handle. It's because of the open port that lets the water pressure in to one side of the piston. I realise now that you never said the handle was the reason why, but I mistakenly thought that was your idea. Sorry.
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No problemo.. forum posting can be sooo confusing...
At least we are on the same page now...
 
Quantum Quack said:
but you do...
1m^3 of 1000psi at external atmospheric pressure has a significant potential to do work.
Doesn't it?
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If released into a vacuum - yes. (You can do the calculation yourself; I gave it to you in the last post.) If released into atmospheric pressure - slightly less.
 
To find a way to estimate the cost of retrieval of the VVSS we can compare it with Pumped Hydro which has an energy efficiency of about 70-80%

Assume capitalization on the VVSS descent using a surface electrical generator, cabled to the VVSS.
If energy efficiency was at 70% in this part of the system then only 30 % supplemental energy would have to be added to retrieve the VVSS.
billvon said:
If released into a vacuum - yes. (You can do the calculation yourself; I gave it to you in the last post.) If released into atmospheric pressure - slightly less.
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so what would you call the energy that is stored in a high pressure vessel with out release? ( assume ambient atmospheric pressure at sea level)
 
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Write4U said:
That's true, there will always be loss
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Thank you for that.
That is the point that seemed so difficult to nail down.

Write4U said:
We are already using several natural pressure energy sources with good result. Tidal pressure , wind pressure , gravity fed lake water pressure, river pressure.
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I know however all your examples exploit change and although there is indeed pressure under the ocean there is no change that can be harvested such that a claim that one can is wrong.
Write4U said:
I see no reason why underwater pressure cannot be used if a low pressure area can be made underwater so that a kinetic flow force can be facilitated.
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Yes but getting more out than you put in is not possible.
Write4U said:
As I understand it, in the mountains there are several valleys which provide an constant increased uni-directional windflow , which drive high yield windmills.
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That is harvesting a flow.
Write4U said:
Can we build underwater "funnels" with generators in places of fast gulf stream flow. Kind of underwater funnelling dams? That would certainly create subsurface ocean high pressure headwaters and low pressure tail waters. Place several generators in the middle of that increased flow and presto cheap electricity!
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Certainly but this is not harvesting pressure it is harvesting flow.

Nevertheless I thank you again for confirmation of the statement that I presented.
Alex
 
billvon said:
Well, I gave you the equation above. Is there something specific you wanted help with?
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yes you did.. thanks..
And I have no idea how to work it out...(math dyslexic or dyscalculia to be specific doesn't help)


is it similar to the following:
formulacompressedenergy.png
 
Quantum Quack said:
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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Wrong. Pressure does not have potential energy. Potential energy is....a form of energy! So the simple fact that pressure has the wrong units shows you it cannot be potential energy.

If you pressurise air by compressing it, i.e. changing its volume, then, sure, the compressed air has potential energy, as given by PdV=dW. But if you pressurise water, which is virtually incompressible, its volume does not materially change and there is practically no stored energy in it.

This is why this thread is all about compressing air. You need something to change in volume, before pressure results in any energy loss or gain. That is the meaning of the equation you found.
 
exchemist said:
Wrong. Pressure does not have potential energy. Potential energy is....a form of energy! So the simple fact that pressure has the wrong units shows you it cannot be potential energy.

If you pressurise air by compressing it, i.e. changing its volume, then, sure, the compressed air has potential energy, as given by PdV=dW. But if you pressurise water, which is virtually incompressible, its volume does not materially change and there is practically no stored energy in it.

This is why this thread is all about compressing air. You need something to change in volume, before pressure results in any energy loss or gain. That is the meaning of the equation you found.
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ok...
so there fore the ocean has no potential energy but the air I have compressed using that oceans pressure does? Is that what you are saying?
 
exchemist said:
Spot-on.
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ok... that's great. Learning the language is always tough for me...
thanks...
ok so we place a vessel into the ocean and displaced it with the weight and volume of the vessel. Have we induced potential energy into the ocean equivalent to the displacement?

and no, I am not trying to be smart arsed. It is about learning the limitations of the term potential energy.
 
exchemist said:
Wrong. Pressure does not have potential energy. Potential energy is....a form of energy! So the simple fact that pressure has the wrong units shows you it cannot be potential energy.

If you pressurise air by compressing it, i.e. changing its volume, then, sure, the compressed air has potential energy, as given by PdV=dW. But if you pressurise water, which is virtually incompressible, its volume does not materially change and there is practically no stored energy in it.

This is why this thread is all about compressing air. You need something to change in volume, before pressure results in any energy loss or gain. That is the meaning of the equation you found.
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True, and reminded me of a clash over that basic fact I had way back in 2013 with principally a then resident and venerated GR expert Markus Hanke. You can start here:
http://www.sciforums.com/threads/gravitational-collapse.135084/page-2#post-3080539
It gets quite acrimonious and continues in fits and starts over a number of pages. An example imo why one should never simply trust experts as purveyors of Gospel Truth.
 
OK I understand now that we have an issue with the more precise meaning of terms, like comparing speed with velocity or quantity with quality etc...
I also wish to thank all those who have been trying to explain this factor and now can understand why the notion of harvesting pressure is technically nonsense. This was not intended.

However it appears as this idea has evolved that displacing the ocean with a weighted volume does allow the ocean depths to gain potential energy to the value of that displacement. A potential energy that can be exploited especially at depth.
Am I correct in claiming the above?
Have I used the right terms or have I confused the issue even more so..?
There is also the issue that at depth due to increased water density, displacement increases making the vessel less heavy the further down it goes, but simultaneously as it compresses the air it's displacement reduces mitigating the increase due to water density.
So we have both an increase in displacement (PE) due to density increase and a reduction in displacement (PE) due to the reduction in volume ( air compression )
The question then is what relationship the gain in PE has with the simultaneous reduction in PE and whether they are equal or not. ( is there a net gain or loss in PE as the VVSS sinks?)
Does that sound right?
 
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Question:
If one were to place a device on the seabed in 100 feet of water, and then a 6 ft tide came in, would the pressure on the device increase by 2.6011 psi(the pressure of a 6 ft. column of water) or by a fraction of that 6 ft increase due to the gravitational pull of the moon?
 
sculptor said:
Question:
If one were to place a device on the seabed in 100 feet of water, and then a 6 ft tide came in, would the pressure on the device increase by 2.6011 psi(the pressure of a 6 ft. column of water) or by a fraction of that 6 ft increase due to the gravitational pull of the moon?
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Tidal Power
Although not yet widely used, tidal energy has potential for future electricity generation. Tides are more predictable than the wind and the sun. Among sources of renewable energy, tidal energy has traditionally suffered from relatively high cost and limited availability of sites with sufficiently high tidal ranges or flow velocities, thus constricting its total availability. However, many recent technological developments and improvements, both in design (e.g. dynamic tidal power, tidal lagoons) and turbine technology (e.g. new axial turbines, cross flow turbines), indicate that the total availability of tidal power may be much higher than previously assumed, and that economic and environmental costs may be brought down to competitive levels
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The original power plants were large and costly.
220px-Sihwa_Lake_Tidal_Power_Station_01.png
https://en.wikipedia.org/wiki/Tidal_power

Today we are looking at smaller individual units , based on the same principle as wind mills. They are just located under water and the water drives the propellers.
upload_2020-3-13_9-13-12.jpeg
images

The first picture shows the windmill principle with several foundation options.

I like the second one. Note how the incoming water is forced through the turbine tunnel below the dam. The pressure generated this way has the entire ocean behind it. This is brute power. Millions of pounds of pressure which can generate enormous power even at low turbine speeds.
 
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Alternative bi-directional methods
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and the toilet idea of a rising float bulb which leverages a pressure generator.
images
 
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sculptor said:
Question:
If one were to place a device on the seabed in 100 feet of water, and then a 6 ft tide came in, would the pressure on the device increase by 2.6011 psi(the pressure of a 6 ft. column of water) or by a fraction of that 6 ft increase due to the gravitational pull of the moon?
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Nope. The pull of the Moon is so small that it's effectively unmeasurable.

Consider it this way. By adding 6 feet to a 100 foot water column, its pressure increases by 6%. The tidal forces (both moon and sun) change that pressure by .000 016% by changing the level of gravity in the area.
 
Write4U said:
I like the second one. Note how the incoming water is forced through the turbine tunnel below the dam. The pressure generated this way has the entire ocean behind it. This is brute power. Millions of pounds of pressure which can generate enormous power even at low turbine speeds.
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Well, yes. But there are costs.

The more water you have to touch, and the slower the flow you are dealing with, the more expensive the turbine is. (And the heavier it is, and the more prone to damage, and the more likely to foul etc.)
This has to be built on a coastline, so you lose coastline.
The sea is about the most corrosive environment on Earth that we see with any regularity, so it's hard to see long turbine life.
It's somewhat destructive to sea life.
 
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