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Author Topic: buoyancy cycle: mg where the h is free  (Read 5078 times)

Offline sm0ky2

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buoyancy cycle: mg where the h is free
« on: September 13, 2009, 05:55:54 AM »
electrolysis + fuel cell technology is comming very close to a reversable process. (very few losses when care is taken)

i propose the following buoyancy cycle:

1) Water - a container sitting "low" in a gravitational field.
2) electrolysis to break the liquid into two buoyant gasses
3) tanks to capture these gases placed "high" in said gravitational field.
4) a fuel cell to convert the energy from these gasses back to continue the electrolysis process.
5) a generating mechanism to convert the (free) gravitational potential energy gained during the gasseous stage of the cycle.

sort of a perpetual waterfall.

class dismissed, have a great weekend.

Free Energy | searching for free energy and discussing free energy


Offline sm0ky2

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Re: buoyancy cycle: mg where the h is free
« Reply #1 on: September 13, 2009, 06:17:49 AM »
graphical

Offline spoondini

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Re: buoyancy cycle: mg where the h is free
« Reply #2 on: September 13, 2009, 06:49:51 AM »
Very clever mind.

Implying COP=1 (or really close)for the electrolysis and reverse process. Since the gasses will naturally rise, you harness the trip back down to replenish any losses and the remainder is pure OU.

Free Energy | searching for free energy and discussing free energy

Re: buoyancy cycle: mg where the h is free
« Reply #2 on: September 13, 2009, 06:49:51 AM »
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Offline spoondini

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Re: buoyancy cycle: mg where the h is free
« Reply #3 on: September 13, 2009, 07:03:35 AM »
http://en.m.wikipedia.org/wiki/Electrolysis?wasRedirected=true

ok, if it's in wiki it must be true (lol)

electrolysis realistic efficiency is around 50%
theoretical max between 80-94%

Correct my logic but this should work as long as COP > 0.

You simply need to solve for the height of the chamber to dtermine how much gravitational energy needed to plug the gap(including all associated losses).

I can already picture a future with water tower like power plants, but fuel cells on top.  All increases of efficiency in the conversion processes allows for reductions in required height and/or increases in existing towers when retrofitted.  Very cool.

Offline onthecuttingedge2005

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Re: buoyancy cycle: mg where the h is free
« Reply #4 on: September 13, 2009, 08:10:12 AM »

5) a generating mechanism to convert the (free) gravitational potential energy gained during the gasseous stage of the cycle.

sort of a perpetual waterfall.

class dismissed, have a great weekend.

Hi Smoky.

gravity is only a carrier of kinetic energy, the only energy an object has when falling is the energy that was put in to it to lift it up. it stores kinetic energy, no OU.

it requires more force to lift up then there is force when it's falling down, this is why there are no working gravity devices. for steady state dense objects, falling is the path of 'least' resistance.

as far as recycling the gases in the electrolysis reaction have you tried a big condenser coil to re-condense the gases into a pressure canister, this process takes place all the time in a refrigerator condenser system.

have fun anyways.
Jerry :)

Free Energy | searching for free energy and discussing free energy

Re: buoyancy cycle: mg where the h is free
« Reply #4 on: September 13, 2009, 08:10:12 AM »
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Offline mscoffman

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Re: buoyancy cycle: mg where the h is free
« Reply #5 on: September 13, 2009, 09:46:39 PM »
electrolysis + fuel cell technology is comming very close to a reversable process. (very few losses when care is taken)

i propose the following buoyancy cycle:

1) Water - a container sitting "low" in a gravitational field.
2) electrolysis to break the liquid into two buoyant gasses
3) tanks to capture these gases placed "high" in said gravitational field.
4) a fuel cell to convert the energy from these gasses back to continue the electrolysis process.
5) a generating mechanism to convert the (free) gravitational potential energy gained during the gasseous stage of the cycle.

sort of a perpetual waterfall.

class dismissed, have a great weekend.

Couple of problems;

a)Hydrogen gas will compress at the bottom of a tall water column -
this will lessen it's bouncy. So the fluid columns can only be so tall.

b)Oxygen will be absorbed into the water, unbalancing the stochiometric
reaction balance, at least at first.


:S:MarkSCoffman

Offline gsmsslsb

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Re: buoyancy cycle: mg where the h is free
« Reply #6 on: September 13, 2009, 11:37:04 PM »
electrolysis + fuel cell technology is comming very close to a reversable process. (very few losses when care is taken)

i propose the following buoyancy cycle:

1) Water - a container sitting "low" in a gravitational field.
2) electrolysis to break the liquid into two buoyant gasses
3) tanks to capture these gases placed "high" in said gravitational field.
4) a fuel cell to convert the energy from these gasses back to continue the electrolysis process.
5) a generating mechanism to convert the (free) gravitational potential energy gained during the gasseous stage of the cycle.

sort of a perpetual waterfall.

class dismissed, have a great weekend.

Excelent
  I had the same idea but using the bingo type arc for the gas production.
Does anyone have the efficiency data for underwater arc HHO production.

Free Energy | searching for free energy and discussing free energy

Re: buoyancy cycle: mg where the h is free
« Reply #6 on: September 13, 2009, 11:37:04 PM »
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Offline spoondini

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Re: buoyancy cycle: mg where the h is free
« Reply #7 on: September 14, 2009, 08:00:43 PM »
Did a few minutes of research and back of napkin calculations:

Theoretical maximum efficiency for H2 production via electrolysis:
32.9 kwh/kg of H2
or 118,440,000 joules

When converted back to water, the 1 kg of H2 will be about 5 kg's of water

Using earth's gravity to acellerate the 5 kg of water at 9.8 m/s^2, I believe we would need to drop it from 2,417,143 meters to recover the energy used to produce the hydrogen gas.

Another observation - 02 is not buoyant.  It will require energy to elevate.
My thought was to let the H2 rise, combust it at altitude to reconstitute the water (and recapture some energy) with atmospheric 02.  Even with energy recapture, we're looking at some really tall water return systems to make enough hydro power to even meet energy unity.

These numbers are also using theoretical maximum efficiencies, in reality we would be lucky to net 50% via all the conversion processes.

Great idea, not sure how practical.  Please let me know if I made any serious math flaws.

Offline gsmsslsb

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Re: buoyancy cycle: mg where the h is free
« Reply #8 on: September 15, 2009, 03:46:21 AM »
Did a few minutes of research and back of napkin calculations:



Using earth's gravity to acellerate the 5 kg of water at 9.8 m/s^2, I believe we would need to drop it from 2,417,143 meters to recover the energy used to produce the hydrogen gas.

Another observation - 02 is not buoyant.  It will require energy to elevate.
My thought was to let the H2 rise, combust it at altitude to reconstitute the water (and recapture some energy) with atmospheric 02.  Even with energy recapture, we're looking at some really tall water return systems to make enough hydro power to even meet energy unity.

Great idea, not sure how practical.  Please let me know if I made any serious math flaws.

Why not do it underwater in a column, or lake, or in the sea.
The greater specific gravity difference should allow the energy to be regained at a fraction of the height.
I am not sure but maybe even enhance the efect by using the pressure to displace a greater volume of water. see the attachments

Free Energy | searching for free energy and discussing free energy

Re: buoyancy cycle: mg where the h is free
« Reply #8 on: September 15, 2009, 03:46:21 AM »
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Offline utilitarian

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Re: buoyancy cycle: mg where the h is free
« Reply #9 on: September 15, 2009, 03:53:11 AM »
Why not do it underwater in a column, or lake, or in the sea.
The greater specific gravity difference should allow the energy to be regained at a fraction of the height.
I am not sure but maybe even enhance the efect by using the pressure to displace a greater volume of water. see the attachments

I believe that producing HHO under pressure decreases the efficiency of producing said gas.

Offline exnihiloest

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Re: buoyancy cycle: mg where the h is free
« Reply #10 on: September 15, 2009, 09:21:45 AM »
electrolysis + fuel cell technology is comming very close to a reversable process. (very few losses when care is taken)

i propose the following buoyancy cycle:

1) Water - a container sitting "low" in a gravitational field.
2) electrolysis to break the liquid into two buoyant gasses
3) tanks to capture these gases placed "high" in said gravitational field.
4) a fuel cell to convert the energy from these gasses back to continue the electrolysis process.
5) a generating mechanism to convert the (free) gravitational potential energy gained during the gasseous stage of the cycle.

sort of a perpetual waterfall.

class dismissed, have a great weekend.

Not perpetual. The energy for the electrolysis is pressure dependant. The higher the glass tubes, the heavier the pressure at the bottom and the more energy you need to break H2O against the pressure. You cannot expect a gain.





Free Energy | searching for free energy and discussing free energy

Re: buoyancy cycle: mg where the h is free
« Reply #10 on: September 15, 2009, 09:21:45 AM »
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Offline mscoffman

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Re: buoyancy cycle: mg where the h is free
« Reply #11 on: September 15, 2009, 05:39:45 PM »
I believe that producing HHO under pressure decreases the efficiency of producing said gas.

Actually, it doesn't work like one would think. To some extent HHO
likes to be electrolysed from water under pressure, it takes less energy
then compressing it after the fact...Within limits.

:S:MarkSCoffman

Offline synchro1

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Re: buoyancy cycle: mg where the h is free
« Reply #12 on: June 25, 2020, 09:37:01 AM »
The electrolysis unit can be kept at any pressure under water. Suppose we put one in a submarine?
Imagine water proof copper coil stacks rising miles from the sea floor. We can drive oil drums filled at depth by HHO through them. The oil drums could have huge magnet collars that would generate electrical power in the coils as they rise to the surface. The HHO would generate power through a cell at the surface; The drum would fill with water and generate more power as it sank back down.


This would work in a stationary displacement stack on any scale!.

Offline Floor

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Re: buoyancy cycle: mg where the h is free
« Reply #13 on: June 27, 2020, 07:31:45 AM »

It would require the approximately same amount of energy, to force a buoyant drum out the door of a
submerged submarine as it requires to sink that same drum from the surface, down to the submarine's depth (unless we let an equal volume of water into the sub as the barrel exits).


 

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