Pressure Harvesting - from ocean depths

Q-reeus said:
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.
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Thanks for the idea. However I don't think I buy it. I think one should be able in principle to do the compression quasi-statically and avoid kinetic effects (for example by allowing the seawater in through an arbitrarily small nozzle, or something). It seems to me the total thermodynamic energy balance of this should not depend on the "kinetics", i.e. variable losses like this, friction etc.

No, there's got to be something fundamental that I've missed, or else it is in the arithmetic. But where?
 
Write4U said:
OK from the illustration, what part of the system is used for storing electricity?
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The empty space in the tank. The pumps move water from the tank into the ocean, thus increasing the level of the ocean (at the surface) by some infinitesimal fraction of an inch.
Why on earth should I admit to being wrong . . .
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Reasons you might want to admit it:

1) Your claims run counter to the laws of thermodynamics.
2) Your claims run counter to the laws of hydraulics.
3) You are trying to do the same sort of mechanical tricks that thousands of people before you have tried, and they've all been wrong, too
4) Both the author of the article and the inventor of the storage device stated the opposite of what you said

Reasons you might not want to admit it:
1) Your ego is so big that you cannot ever admit that you've been wrong, and cannot learn from your mistakes.
 
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billvon said:
The empty space in the tank. The pumps move water from the tank into the ocean, thus increasing the level of the ocean (at the surface) by some infinitesimal fraction of an inch.
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And what is creating the electricity being stored?
And how is this electricity stored in the empty tank that makes it a battery?
 
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exchemist said:
Thanks for the idea. However I don't think I buy it. I think one should be able in principle to do the compression quasi-statically and avoid kinetic effects (for example by allowing the seawater in through an arbitrarily small nozzle, or something). It seems to me the total thermodynamic energy balance of this should not depend on the "kinetics", i.e. variable losses like this, friction etc.

No, there's got to be something fundamental that I've missed, or else it is in the arithmetic. But where?
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Think about it some more. If it is done quasi-statically then something other than gas pressure is resisting inward motion. And that something has to be friction of some kind.
Use a very small nozzle and turbulence is set up, which means conversion to heat loss above that owing to say isothermal compression. Recall we agreed only purely adiabatic compression can avoid inherent loss via compression/expansion of the air. Unless the piston + column of in rushing water is allowed to initially accelerate under assumed constant seawater pressure, frictional forces, rather than inertia, must do most of the resisting. I stick with that as the fundamental reason for the large discrepancy.
 
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billvon said:
The head. (Google the term; it's a term used in hydropower.)
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No, you have this completely backward. In this system the tanks are not the head! In this system the head is the ocean (pressure) and the tanks are the downhill space for the water from the head to flow into.
Read this quote again and you'll see what it says.
In this pumped storage power plant a turbine will be connected to a tank on the seabed at a depth of 400-800 meters. The turbine is fitted with a valve, and when this is opened, water flows in and starts turning the turbine. The turbine drives a generator to produce electricity. One can connect as many tanks as one might wish. In other words, it is the number of water tanks that decides how long the plant can generate electricity, before the energy storage capacity is exhausted.
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The red is a false statement, the same error you are making. This should read "before the (downhill) water storage capacity is exhausted". I have never heard of waste water being an energy storage. It's potential energy has been expended and it is now downhill from the head.

The 'head", the energy storage (potential) is in the exterior water pressure of the ocean, which flows (downhill) into the empty tanks (replacing the air via the vent pipe). The kinetic force of the inflowing water from the Ocean (head) activates a generator, which produces the electricity, which is not stored but transferred to the surface electric grid.

Once the tanks (the downhill reservoirs) are full with water, this water needs to be pumped out to provide new downhill space for the inflowing water from the ocean (head). This out-pumping of the water, via the reversible turbine, uses electricity, the very opposite of storing it.

Just to make my logic clear. If we have turbine generator in a river, would you call the river downstream from the generator the "head" (energy storage), or is the "head" upstream from the generator?

Now apply this logic to the ocean being upstream from the generator and the tanks are downstream from the turbine. Get it?
 
billvon said:
Then you don't understand head.
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Yes I do. The "head is upstream (uphill) from the generator.
Hydraulic head
The hydraulic head is a value that measures the amount of mechanical energy available in water in a river, stream or even lake. The hydraulic head is equivalent to the water level in a static (non-flowing) water body.[1] In its simplest form hydraulic head is a measurement of the height of a static water column above an arbitrary point, usually expressed in meters (or feet in the US). The higher the water level or hydraulic head, the more energy that the water at a specific location has.
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In this case the Ocean (pressure) is the hydraulic head In this system the tanks contain the "tailwater".
Tailwater refers to waters located immediately downstream from a hydraulic structure, such as a dam (excluding minimum release such as for fish water), spillway, bridge or culvert. Generally measured and reported as the average water depth downstream of an hydraulic structure, tailwater can vary based on the outlet from the structure as well as downstream influences that may restrict or advance the usual flow of water from
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When the tanks are full, it restricts the capacity to hold tailwater. The llustration clearly indicates the ocean as the head and the tanks as the tailwater.
In a hydroelectric facility, the amount of energy harnessed depends on the difference between the headwater level in the reservoir upstream of the dam and the tailwater level below the hydroelectric dam.
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This is known as the hydraulic head difference and it represents the amount of energy that can be transformed into electricity by the turbines and generators. More detailed calculations show that not only the difference in height of the water determines how much energy can be harnessed, but there are a variety of losses known as head losses. These head losses occur as a result of friction in pipes. When head losses are accounted for, the actual amount of energy that can be harnessed is reduced. This reduced value for hydraulic head with the losses accounted for is known as the effective head.
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https://energyeducation.ca/encyclopedia/Hydraulic_head
 
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billvon said:
You keep telling yourself that. With your ego, there is no room for you to learn. Good luck with that.
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Look on the bright side: at least he hasn't managed to divert the thread onto functions, potentials or microtubules. Though I thought we were getting close when he introduced octopuses.....:biggrin:
 
Q-reeus,
Is there any reason you are not using a variable volume vessel that compresses the air AS it sinks and not waiting till you get to the maximum depth?

If it compresses the air as it sinks the displacement should steadily reduce, and the vessel maintain a slow acceleration as the vessel descends ( little to no heat loss?)
eg.
--[--------------]
---[-------------]
----[------------]
-----[-----------]
------[----------]

There should be no resistance to ambient pressure as the chamber descends other than the air being compressed.
 
Quantum Quack said:
Q-reeus,
Is there any reason you are not using a variable volume vessel that compresses the air AS it sinks and not waiting till you get to the maximum depth?

If it compresses the air as it sinks the displacement should steadily reduce and maintain a slow acceleration as the vessel descends ( little to no heat loss?)
eg.
--[--------------]
---[-------------]
----[------------]
-----[-----------]
------[----------]
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No reason you can't do that.

A rock being lowered slowly off a cliff will compress a can just as much as if lowered quickly.

Once.
 
DaveC426913 said:
No reason you can't do that.

A rock being lowered slowly off a cliff will compress a can just as much as if lowered quickly.

Once.
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Unfortunately the rock's descent isn't governed by displacement.
Also , do you know what the efficiency is for capitalizing on the PE of the falling weight ( rock) using a generator and cable attached as it falls.
I would be confident there is a energy gain percentage using a the rock as an energy source as it falls (once)
If I am not mistaken it should be about 60-80 % in a vertical arrangement.
A pendulum scenario has a higher efficiency probably around 90%
The drag caused by the generator does not effect the compression of the air in the chamber.
But this is only a guess on my part, not math.
Question:
If we capitalize on our sinking vessels descent using an electrical generator on the surface, how far can we lift the vessel after it has reached it's maximum depth using that stored energy ( in percentages)
 
I see it as a energy system of at least 3 phases.

Phase 1
Sinking the VVSS and capitalizing on it's descent using a cable and gen set.
Phase 2
Delivering the compressed air to the surface or to a large bulk variable volume storage unit, once it has reached it's maximum depth.
Phase 3
Retrieving the VVSS using both capitalized energy and added green energy ( eg. 60%,40% )


then repeat as often as you wish - fully automated and totally green.
At a depth pressure of 35000 kpa we have a high density energy system.
 
Quantum Quack said:
I see it as a energy system of at least 3 phases.

Phase 1
Sinking the VVSS and capitalizing on it's descent using a cable and gen set.
Phase 2
Delivering the compressed air to the surface or to a large bulk variable volume storage unit, once it has reached it's maximum depth.
Phase 3
Retrieving the VVSS using both capitalized energy and added green energy ( eg. 60%,40% )


then repeat as often as you wish fully automated and totally green.
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Phase 3 will consume all the energy generated in phases 1 and 2.
 
exchemist said:
Phase 3 will consume all the energy generated in phases 1 and 2.
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are you sure?
Or are you saying so because you haven't seen how this system can gain energy from ambient pressure ( potential energy) with out conflicting the laws of thermodynamics?
 
exchemist said:
Phase 3 will consume all the energy generated in phases 1 and 2.
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Put it this way...
Compare the rock of a cliff scenario with the VVSS scenario.
How are they different?
Rock: the compression of the can is delivered by the weight of the rock. ( rock + can = Kpa)
VVSS: the compression is delivered by the weight of an ocean and the VVSS (rock + can (VVSS) + ocean = Kpa)

or
Claim:
We are using the rock (VVSS) weight to capitalize on the weight of the ocean.

Like all evolving ideas it takes a while for the fog to clear...
 
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say the VVSS had two variable volume storage chambers installed, one for assisting the retrieval when it reaches a lesser depth while ascending ( compressed air) and one for the delivery to the bulk VV storage or surface storage.

Edit : second thoughts: gain would be neutral sorry...
 
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