Overunity electrolysis - 31 times more effective gas production than with DC

[ltseung888]

 Let me quickly review our discussion here.

1.    The Lead-out or Bring-in Energy Machine diagram is in reply 109.  It is the most important slide in our discussion.  The Lead-out Energy Machine is NOT the impossible Perpetual Motion Machine.  It requires energy to run.  It brings-in energy from the environment to run and contributes to Global Cooling.
The simplest example is the water bottle rocket where the temperature of the bottle is cooler after firing.  James Kwok tried to build a machine Hidro (reply 112) to take advantage of this principle.

2.    Browns Gas has the property of Implosion.  On ignition, the gas mixture implodes instead of explodes.  However, adding water vapor or other gases will turn the gas mixture to exhibit explosion.
A simple Implosion working diagram was presented on reply 130.
Some engineering ideas on how to modify the standard 4 stroke engine to use implosion is in reply 106:

https://www.youtube.com/watch?v=ndpk3qAJS-8&list=PL7AF455C0A317C691&index=3

3.    Three scenarios related to electrolysis and Browns Gas are presented in reply 117.  Scenario A is the classical electrolysis where the cathode and anode are well separated to collect the resulting gases.  The reaction can be reversed by feeding the resulting hydrogen and oxygen gases into a fuel cell.  No Lead-out or Bring-in Energy from the environment is involved.  Energy for combustion must be supplied to do the electrolysis.  Scenario B is to use DC Pulsing to break the water molecule up into some form of HHO gas with much less energy.  Again, no lead-out or Bring-in Energy from the environment is involved.  Scenario C is to add Implosion to Scenario B.  Implosion brings-in the kinetic energy of the air molecules, cooling the environment.  Scenario C can become the basis of a Bring-in Energy machine running forever after starting as in reply 109.  However, there is much engineering difficulty in ensuring that Implosion does not change into Explosion.

In the next post, I shall describe the more interesting Scenario D of "resonance breaking" of the dipole water molecules.

[MarkE]

@MarkE,
Thank you very much for spending much time in doing the diagram.  I obviously do not understand some of your terms or accounting.  Thus I simplified it to my understanding.  Please explain your diagram in detail so that all can benefit.

Lawrence, _ATM are quantities of the surrounding atmosphere environment, _CYL are quantities of the cylinder, _ST1 are quantities at State 1, _ST2 are quantities at State 2, and _ST3 are quantities at State 3.  P is pressure, V is volume, E is energy.  You incorrectly misaccount the amount of work performed displacing the air in the cylinder in State 1 with HHO in State 2.  The actual work that must be performed is not just the change in energy state within the cylinder, but the change in energy state of the SYSTEM, which includes the environment. 

It is total nonsense that you count on the environment for energy as part of your system to provide what you call LEAD IN energy, but ignore the energy that you have to drive into the environment when filling the cylinder with HHO.  When you drive the HHO into the cylinder, the energy in the cylinder increases by only the pressure difference times the cylinder volume, but the energy in the surrounding atmosphere into which you expel an identical volume of air increases by the starting pressure of the air times the cylinder volume.  By filling the cylinder with HHO you compress a gas spring in the surrounding atmosphere.  When the HHO burns and the gas volume changes to a many times smaller liquid volume, the atmospheric gas spring pushes the piston back to the top of the cylinder and in doing so relaxes.  Ignoring the water volume, the system: volume, pressure, and energy at State 3 is the same as at State 1.  I have explained your error a number of times now.  The diagram explains the actual energy relationships in each state in sufficient detail that any high school physics student can easily understand.

[MarkE]

      Actually want to increase the water molecule polarization.  The two "shared" electrons move within the oxygen valence orbital.  So you could say the hydrogen and oxygen atoms become ionized.  The hydrogen atoms take on a plus one charge while the oxygen atom takes on -2 charges.  [size=78%]The oxygen atom itself becomes polarized with excess electrons moved towards the positive plate. This leaves the two hydrogen protons and the positive pole of the oxygen atom all working to fracture the molecule.[/size]

Well I think you have a big problem.  You can apply so much field that you partially or fully ionize the molecule.  What I don't see any means to do is to apply an external field that increases the polarization within the molecules.  How do you think that you can increase the polarization within individual molecules by applying an external field?

[MarkE]

Let me quickly review our discussion here.

1.    The Lead-out or Bring-in Energy Machine diagram is in reply 109.  It is the most important slide in our discussion.  The Lead-out Energy Machine is NOT the impossible Perpetual Motion Machine.  It requires energy to run.  It brings-in energy from the environment to run and contributes to Global Cooling.
The simplest example is the water bottle rocket where the temperature of the bottle is cooler after firing.  James Kwok tried to build a machine Hidro (reply 112) to take advantage of this principle.

Post 132 clearly illustrates the fallacy of post 109. James Kwok was convicted of securiteis crimes.  His HidroCo is moribund and no longer promotes his claims.

2.    Browns Gas has the property of Implosion.  On ignition, the gas mixture implodes instead of explodes.  However, adding water vapor or other gases will turn the gas mixture to exhibit explosion.
A simple Implosion working diagram was presented on reply 130.
Some engineering ideas on how to modify the standard 4 stroke engine to use implosion is in reply 106:

https://www.youtube.com/watch?v=ndpk3qAJS-8&list=PL7AF455C0A317C691&index=3

3.    Three scenarios related to electrolysis and Browns Gas are presented in reply 117.  Scenario A is the classical electrolysis where the cathode and anode are well separated to collect the resulting gases.  The reaction can be reversed by feeding the resulting hydrogen and oxygen gases into a fuel cell.  No Lead-out or Bring-in Energy from the environment is involved.  Energy for combustion must be supplied to do the electrolysis.  Scenario B is to use DC Pulsing to break the water molecule up into some form of HHO gas with much less energy.  Again, no lead-out or Bring-in Energy from the environment is involved.  Scenario C is to add Implosion to Scenario B.  Implosion brings-in the kinetic energy of the air molecules, cooling the environment.  Scenario C can become the basis of a Bring-in Energy machine running forever after starting as in reply 109.  However, there is much engineering difficulty in ensuring that Implosion does not change into Explosion.

In the next post, I shall describe the more interesting Scenario D of "resonance breaking" of the dipole water molecules.

The energy associated with the difference in energy between the gas volume at any pressure and disassociated water is a source of energy loss in any electrolysis system.  In the most ideal case of a machine that were to completely recover that energy difference, an electrolysis / combustion machine would be limited to the Carnot efficiency of the heat engineutilysing the hydrogen combustion as its input heat source.

[ltseung888]

 

Lawrence, _ATM are quantities of the surrounding atmosphere environment, _CYL are quantities of the cylinder. P is pressure, V is volume, E is energy. You incorrectly misaccount the amount of work performed displacing the air in the cylinder in State 1 with HHO in State 2. The actual work that must be performed is not just the change in energy state within the cylinder, but the change in energy state of the SYSTEM, which includes the environment.

It is total nonsense that you count on the environment for energy as part of your system to provide what you call LEAD IN energy, but ignore the energy that you have to drive into the environment when filling the cylinder with HHO. When you drive the HHO into the cylinder, the energy in the cylinder increases by only the pressure difference times the cylinder volume, but the energy in the surrounding atmosphere into which you expel an identical volume of air increases by the starting pressure of the air times the cylinder volume. By filling the cylinder with HHO you compress a gas spring in the surrounding atmosphere. When the HHO burns and the gas volume changes to a many times smaller liquid volume, the atmospheric gas spring pushes the piston back to the top of the cylinder and in doing so relaxes. Ignoring the water volume, the system: volume, pressure, and energy at State 3 is the same as at State 1. I have explained your error a number of times now. The diagram explains the actual energy relationships in each state in sufficient detail that any high school physics student can easily understand.

@MarkE,

Thank you for your explanation. 



From a pure Physics point of view of work performed by the piston, it is clear:

1.   Work done by the piston to fill the cylinder with HHO is the (HHO-Atm)pressure x Volume of the Cylinder.  HHO pressure can be 101% of Atm pressure.  The work done can then be 1% of Atm x Volume of Cylinder.

2.   Work done by the piston to fill the vacuum in the implosion process is Force x Displacement.  The force is Atm, Pressure x area of the Cylinder.  The Displacement is the height of the cylinder.  The work done is then equal to Atm pressure x area x height which is equal to Atm Pressure x Volume of Cylinder.  This is 100% times of stage 1.

From the pure Mechanical Consideration, the work done by the HHO gas to fill the cylinder is 1% of the work done by the Atm Pressure.  We assume the HHO gas has been generated and stored in a tank with HHO pressure 1% more than Atm.  The Energy used to generate the gas is NOT considered in this example.



May be we should let others comment on such calculations.  I consider the subject closed.  The work done or energy supplied by the Environment in one cycle = Atm Pressure x Volume of Cylinder.  This is the Bring-in Energy.