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Author Topic: Overunity electrolysis - 31 times more effective gas production than with DC  (Read 143634 times)

Offline Paul-R

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Simple attempt to explain IMPLOSION working principle.  Practically no exhaust gas comes out!
I remember disclosing this idea in the hydroxy yahoo group years ago, (I was pleased that Smack of Smack's Booster fame was impressed) and although I am sure it would work, the noise would be so dreadful that it would never get off the ground. Also, I proposed that the firing stroke should generate work as well as the suction.

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Offline ltseung888

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I remember disclosing this idea in the hydroxy yahoo group years ago, (I was pleased that Smack of Smack's Booster fame was impressed) and although I am sure it would work, the noise would be so dreadful that it would never get off the ground. Also, I proposed that the firing stroke should generate work as well as the suction.
 
@Paul-R,
 
One problem with HHO implosion is that it requires relatively pure mixture of the gases in the right proportions and with little impurity.
 
If we introduce water vapor as in the continuous Implosion example, the result will be explosion.  See the Youtube video mentioned in reply 101:
https://www.youtube.com/watch?v=bqlbb_ojAdI

At present, I focus only on the theoretical aspects of “Implosion giving rise to Energy being brought-in from the OUTSIDE environment”.   Some engineering ideas include
https://www.youtube.com/watch?v=ndpk3qAJS-8&list=PL7AF455C0A317C691&index=3
 
I have not studied them fully yet.



There is likely to be many more hurdles to overcome before the final successful implementation.  Please do share your ideas and experience.

Offline Paul-R

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It is the stoichiometric mix of H2 and O2 that does the business, and of course, the end product, a very small amount of water, is what results in the high vacuum which does the suction.

Introducing water vapour will make everything more manageable, but may well ruin the performance of the suction.



(People should remember that the stoichiometric HHO mix of gases is VERY DANGEROUS INDEED and should not be stored, only generated as needed).

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Offline MarkE

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We do not use springs.  We use the kinetic energy of air molecules.  We cool the environment.
 
God provides us with the Environmental Energy.  We must employ our Physics Knowledge to use it...
 
God also provides us with sunlight.  At one time, we did not know solar panels...
Whether you realize it or not, yes what you propose is a gas spring.  Work out a complete cycle in some ultimately finite volume:

Some amount of energy stored in a gas spring at the operating temperature:  PV = nRT, where n = sum of H2, O2, and other gasses, we'll just call that N2.

Ignite the mixture and the energy increases by QCOMBUSTION, increasing T and as a consequence P.

The gas quantity n decreases by the the number of moles H2 and O2combusted.  P now falls from its elevated value.

To complete the cycle, the water has to be converted back into H2 and O2 gas either by draining the water and introducing fresh gas, or electrolyzing the water.

(n1 N2 + 2n2*H2 + n2 O2)*R*T => n1 N2 + n2 H2O + QCOMBUSTION - (2n2*H2 + n2 O2)*R*T => (n1 N2 + 2n2*H2 + n2 O2)*R*T

The starting and ending conditions are the same with the same number of moles gas in the same volume at the same temperature with the same energy.

Offline ltseung888

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Whether you realize it or not, yes what you propose is a gas spring. Work out a complete cycle in some ultimately finite volume:
 
 Some amount of energy stored in a gas spring at the operating temperature: PV = nRT, where n = sum of H2, O2, and other gasses, we'll just call that N2.
 
 Ignite the mixture and the energy increases by QCOMBUSTION, increasing T and as a consequence P.
 
 The gas quantity n decreases by the the number of moles H2 and O2combusted. P now falls from its elevated value.
 
 To complete the cycle, the water has to be converted back into H2 and O2 gas either by draining the water and introducing fresh gas, or electrolyzing the water.
 
 (n1 N2 + 2n2*H2 + n2 O2)*R*T => n1 N2 + n2 H2O + QCOMBUSTION - (2n2*H2 + n2 O2)*R*T => (n1 N2 + 2n2*H2 + n2 O2)*R*T
 
 The starting and ending conditions are the same with the same number of moles gas in the same volume at the same temperature with the same energy.
 

If the result at some stage is a colder temperature of the initial air and apparatus, there will be an exchange of energy from the surrounding air to the apparatus. This is regarded as Lead-out or Bring-in energy.
 
 
The simple water bottle rocket gets colder after firing. The Physics says - the rapid expansion of the compressed air used some of the kinetic energy of the air molecules. The molecular velocity will be lower. Hence the temperature becomes cooler. This is referred to as adiabatic expansion. Energy will be Brought-in from the surrounding to restore the temperature.
 
 
The initial pumping of the compressed air uses X unit of energy. In the firing, the total energy used is X + Y units.  Y is initially borrowed from the sir molecules – making the air and apparatus cooler. Y units of surrounding energy come to the "Lead-out Energy" system to restore the temperature.
 
Please study the attached lead-out energy system diagram more.  It is correct Physics.
 

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Offline MarkE

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If the result at some stage is a colder temperature of the initial air and apparatus, there will be an exchange of energy from the surrounding air to the apparatus. This is regarded as Lead-out or Bring-in energy.
It is silly to speak of energy balance in only a portion of a cyclical process.
Quote

 
The simple water bottle rocket gets colder after firing. The Physics says - the rapid expansion of the compressed air used some of the kinetic energy of the air molecules. The molecular velocity will be lower. Hence the temperature becomes cooler. This is referred to as adiabatic expansion. Energy will be Brought-in from the surrounding to restore the temperature.
That is all fine and well but just like the simple water bottle rocket, if you are looking for a power generator, you need to evaluate the entire cycle.
Quote

 
The initial pumping of the compressed air uses X unit of energy. In the firing, the total energy used is X + Y units.  Y is initially borrowed from the sir molecules – making the air and apparatus cooler. Y units of surrounding energy come to the "Lead-out Energy" system to restore the temperature.
 
Please study the attached lead-out energy system diagram more.  It is correct Physics.
You keep neglecting to evaluate the entire cycle.  Your proposal does not lead to a workable free energy machine.

Offline ltseung888

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One experiment that needs to be confirmed is:
 
If Browns Gas were to be left alone for some hours in an enclosed bottle, it will no longer show the Implosion Effect.
 
One explanation is that the Browns Gas absorbs Electromagnetic Wave of some sort to convert itself to more like a mixture of hydrogen and oxygen gases.  Browns Gas can be thought of as a "half-way" product of electrolysis.  It will require additional energy to become a "full electrolysis" product.
 
Can anyone confirm that?

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Offline ltseung888

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It is silly to speak of energy balance in only a portion of a cyclical process.That is all fine and well but just like the simple water bottle rocket, if you are looking for a power generator, you need to evaluate the entire cycle.You keep neglecting to evaluate the entire cycle.  Your proposal does not lead to a workable free energy machine.
The complete cycle for the water rocket is:
1. Fill bottle half full with water at room temperature.
2. Pump air in at room temperature (isothermal process) until pressure is P.  Energy supplied X units is PV.  V is the remaining volume of bottle.
3. Fire water rocket.
4. Get water bottle back and allow temperature to warm up to room temperature.  Energy (Y unit) is brought-in from the surrounding to warm up the bottle and the escaped cold air.
5. Back to step 1 to repeat the experiment.
 
The most important step to note is number 4.  External energy comes in to complete the cycle.  The Hidro (James Kwok) of Australia works on this principle.
http://pesn.com/2011/04/17/9501811_James_Kwoks_Hidro_Tech_Floating_to_the_Top/
A chamber is pumped with air isothermally.  It is fired up inside a water tube.  Energy is extracted.  The chamber is allowed to fall back, warmed by surrounding and repeat the cycle.  The isothermal pumping supplied X units of energy.  The Energy extracted theoretically can be X + Y where Y is the brought-in energy.

Offline MarkE

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The complete cycle for the water rocket is:
1. Fill bottle half full with water at room temperature.
2. Pump air in at room temperature (isothermal process) until pressure is P.  Energy supplied X units is PV.  V is the remaining volume of bottle.
3. Fire water rocket.
4. Get water bottle back and allow temperature to warm up to room temperature.  Energy (Y unit) is brought-in from the surrounding to warm up the bottle and the escaped cold air.
5. Back to step 1 to repeat the experiment.
 
The most important step to note is number 4.  External energy comes in to complete the cycle.  The Hidro (James Kwok) of Australia works on this principle.
http://pesn.com/2011/04/17/9501811_James_Kwoks_Hidro_Tech_Floating_to_the_Top/
A chamber is pumped with air isothermally.  It is fired up inside a water tube.  Energy is extracted.  The chamber is allowed to fall back, warmed by surrounding and repeat the cycle.  The isothermal pumping supplied X units of energy.  The Energy extracted theoretically can be X + Y where Y is the brought-in energy.
James Kwok's device does not work.  Where did you get the idea that it does?

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Offline ltseung888

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James Kwok's device does not work. Where did you get the idea that it does?
 

James Kwok had a tough engineering job. I, as a Physicist, can just write equations on paper and talk about the adiabatic expansion.  He had to capture the energy on the way up and make sure that the losses were as low as possible.
 
 
 
He built a massive machine to capture a small amount of energy.  I congratulate him for his efforts. 
 

Offline MarkE

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James Kwok had a tough engineering job. I, as a Physicist, can just write equations on paper and talk about the adiabatic expansion.  He had to capture the energy on the way up and make sure that the losses were as low as possible.
 
 
 
He built a massive machine to capture a small amount of energy.  I congratulate him for his efforts.
He built a machine that did not and does not operate as he claims.  He set-up a very creative corporate structure that was not shall we say: investor friendly.

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Offline Paul-R

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He built a machine that did not and does not operate as he claims.
What is the evidence for this claim of yours?

Offline ltseung888

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 Let us talk about the different scenarios that may be possible to use water as fuel.  We shall use entire cycles in every scenario.

A.    The complete cycle for a normal electrolysis can be:
1.    Start with water at room temperature.
2.    Use the water as electrolyte in a normal electrolysis setup.
3.    Use X units of Energy to turn water into hydrogen and oxygen gases in the normal electrolysis way.
4.    Feed the hydrogen and oxygen gases into a fuel cell (reverse electrolysis) and generate X units of electricity.  Or feed the hydrogen and oxygen gas as fuel for combustion.  X units of energy will be available to do work.  (Ideal case with no loss assumed.)
5.    The result of 4 will be water again in the case of combustion. The result of 4 will be water as electrolyte in the case of fuel cell.
6.    Repeat from 1 or 2 again.
The above is a well known process.  Water can be changed into hydrogen and oxygen gases as fuel.  The problem is the large amount of X energy needed for electrolysis and the danger of transporting hydrogen and oxygen gases.
 
B.    DC Pulsing can break up the water molecules into some form of HHO with much less energy.
1.    Start with water at room temperature.
2.    Use the water as electrolyte in a special Stan Meyer, HHO or some special type device.
3.    DC pulse the Electrolyte with X1 units of Energy.  Some kind of resonance occurs. Produces some “HHO gas mixture” that is different from normal hydrogen and oxygen gas mixture resulting from normal electrolysis. X1 is much less than X as in scenario A.  (May be 1/8 or 1/31 as discussed in the Indian Paper in this thread.)
4.    The “HHO gas mixture” is used as fuel, releasing X1 units of energy.  This “HHO gas mixture” is then turned into water.  (There is some justification that the “HHO gas mixture” may be different from Hydrogen and oxygen mixture in that – Browns Gas, if left for hours, will lose its implosion property.  One explanation is that electromagnetic waves from outside will alter the structure of the Browns Gas.)
5.    Resulting water is fed back to start the process at 1.
In this case, the amount of energy X1 to turn water into some “HHO gas mixture” can be much less than X.  There may be some electrical energy source needed to supply X1.  This is still acceptable as it makes it practical to use water as fuel.
 
C.    Scenario B PLUS lead-out or bring-in energy Y1 via Implosion
1.    Start with water at room temperature.
2.    Use the water as electrolyte in a special Stan Meyer, HHO or some special type device.
3.    DC pulse the Electrolyte with X1 units of Energy.  Some kind of resonance occurs. Produces some “HHO gas mixture” that is different from normal hydrogen and oxygen gas mixture resulting from normal electrolysis. X1 is much less than X in scenario A.  (May be 1/8 or 1/31 as discussed in the Indian Paper in this thread.)
4.    The “HHO gas mixture” is used as fuel in an Implosion fashion.  The Implosion will bring in Y1 Units of Energy.
5.    The total energy available to use is X1+Y1.  This Bring-in System may be able to self-sustain.  X1 units of energy is fed back to generate the “HHO gas mixture” and Y1 units used to do work. The “HHO gas mixture” is then turned into water.
6.    Resulting water is fed back to start the process at 1.
This becomes a very attractive option.  The additional Y1 energy from the Implosion process may be able to supply the electrical energy in step 3.  The System becomes a Bring-in Energy System continuously bringing-in the kinetic energy of air molecules to do work and cool the environment at the same time.
(More to follow)

Offline TheCell

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<The most important step to note is number 4.  External energy comes in to complete the cycle. 
 The Hidro (James Kwok) of Australia works on this principle.>

http://www.rosch.ag/index.php/de/entwicklungen/aufriebskraftwerk
buoyancy power plant (Auftriebskraftwerk)

While the shovels, that are mounted on a large chain rise up to the surface,
the air in each of them expands and therefor drops in temperature.
If the air , that is pressed into the lowest shovels has the same temperature
as the surrounding water a steady heat flow from the water to the air occurs
while rising to the surface thereby resulting in additional buoyancy.
(Special kind of heat pump)

Offline MarkE

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What is the evidence for this claim of yours?
In 2011 he changed the company from FTD Corporation Ltd to Centium Electric.  In late 2012 he changed his company Centium Electric to Hidro Co.  In late 2013 Hidro Co withdrew their investment prospectus. Hidro Co has been moribund for months.  They have removed all mention of Kwok's free energy machine claims:  http://www.hidroco.com

His claim is for a perpetual motion machine.  PMM's have never worked for reasons that should be obvious:  Energy/matter is concserved.  You might also be interested to know that Kwok was convicted of securities fraud a few years ago:  http://www.asic.gov.au/asic/asic.nsf/byheadline/07-32+Former+director+of+Sydney+power+company+jailed?openDocument

 

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