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

[Paul-R]

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.

[ltseung888]

 

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.

[Paul-R]

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).

[MarkE]

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 Q_COMBUSTION, 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 + Q_COMBUSTION - (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.

[ltseung888]

 

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 Q_COMBUSTION, 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 + Q_COMBUSTION - (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.