Atomic combustion of hydrogen H and oxygen O

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Atomic combustion of hydrogen H and oxygen O




by Christophe » 19/11/09, 10:57

By making some storage in my archives I found a piece of paper with some notes on the combustion of hydrogen and atomic oxygen (H ^ + and O ^ 2-) that I had made a few years ago.

I put you raw copy so check before debate.
I simplify the writing by turning the electronic charges:

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2H --> H2 + 437,6 kJ/mol
O --> 1/2 O2 + 248,4 kJ/mol


These 2 first enthalpies would be to check I do not know where I found them!

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H2 + 1/2 O2 --> H2O + 242,7 kJ/mol (celle là, tout le monde connait)


So if we start from atoms to get water here are the enthalpies released:

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2H + O --> H2 + 1/2 O2 (+ 437,6 + 248,4 kJ/mol) --> H2O (+ 242,7 kJ/mol)


Total of the energy released by this reaction of "combustion" of the atomic elements: 437,6 +248,4 + 242,7 = 928,7 kJ / mol

You know that obviously thinks about doping with water.

So 1 L of fractional water at 100% in a motor (under the effect of heat of combustion) in H ^ + and O ^ 2- would give:

928,7 * 1000 / 18 = 51 600 kJ / L (because molar mass of water = 18 g / L and 1 L of water = 1000 g) and it is more than oil! It is about 1L of water = 1,3L / oil : Cheesy:

If we make the same reasoning but starting from the H2 dihydrogen we have: 120 000 kJ / kg H2 (121 300 according to the enthalpy)

1 L of water = 2 / 18 * 1000 = 111,1 g of H2 is 120 000 * 0.111 = 13 330 kJ / L of water.

We see that it is much less than if we started with atomic elements!

Conclusion: 1 L of fractional water with 100% in di-hydrogen would thus contain the equivalent of 0.37 L of oil it is 3.5 times less than passing through the atomic states.

In other words: atomic hydrogen is 3.5 times more energy than diatomic hydrogen.


Question: do we arrive at atomic states in an engine? This is likely at some high T ° so load and since the reduction in consumption is related to the (high) load engine ... there is surely a rational explanation track (one more) to doping at the water.

In short: reasoning to check and track to follow ...
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by Capt_Maloche » 19/11/09, 11:51

I'll have to check, I had taken a similar calculation note, must I put my hand over

but I do not follow you in your atomic and diatomic variations ...
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by Christophe » 19/11/09, 12:02

Ben it is simple one no longer reasons in diatomic chemical elements H2 or O2 = already a molecule (2 atoms) but (mono) atomic = atom of matter ...

The essential question is to know if we reach the T ° sufficient to significantly break the molecules of H2O into atomic elements, but in my opinion yes ...

Diatomic atom breaking is the basis of hydrocarbon combustion.

All the carbon to pass from CyHx to CO2 must necessarily go through its atomic phase, which is not necessarily the case of the Hx (which can remain in form H2 before combining in H2O).


But then it is possible to consider that part of the H2O injected in surplus into the engine would pump "chemically" thus joules in the engine cycle normally lost in heat .... to pass to the atomic state, favors combustion and increases cycle efficiency.

And especially the higher the T °, ​​the more atomic elements there are.

This explains, for example, that there is a lot more NOx at high temperatures because we have to break the N2, a neutral gas in a combustion, in atomic nitrogen to get there ...

Pkoi doping works better at high chamber temperature? It would not be, in others, for the same thing?
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by Christophe » 19/11/09, 12:27

Capt_Maloche wrote:Yep,

Thermolysis of water from 1000 ° C (slow) and total to 2300 ° C (unlikely in a cylinder) because the T ° of combustion in a cylinder is of the order of 1600 to 1800 ° C


Oh yes! The T ° of combustion far exceeds the 1800 ° C ... when one taps in recess!

Capt_Maloche wrote:I bet more on a reaction with unburned, incandescent carbon particles what


Yes toutafé for the reaction of carbon (hot) + H2O -> H2 but the 2 explanations are not incompatible if?

That of the passage in atomic elements would explain why one has better results with strong load thus T ° engine ...
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by Remundo » 19/11/09, 14:50

You should not forget that to have the "atomic element", you need the same energy that it can give by putting itself back into a molecule.

For example H2 -> 2H * is not free and takes energy from the reaction medium to break the chemical bond, as much as it gives back by making 2H * -> H2 (to re-establish the same bond)

Where there may be a small gain is if the H * "excite / accelerate" the reaction chains converting the hydrocarbon into its ultimate combustion products, namely CO2 and H2O.

The presence of these radicals H *, O *, HO * can perfect the combustion efficiency ... : Idea:

That said, if we do 6L / 100km, we will not make 2 ... Maybe 5,5L? It should be reproducible experimental data on the engine test bench. :?
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by Capt_Maloche » 24/11/09, 22:11

EH remu, what does the enthalpy balance of the "combustion" of H2 + O2 give

we are at -242 KJ / mol for electrolysis dissociation

how much for thermolysis?

How much for "combustion"?

not want to poke around the numbers tonight :D
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by Christophe » 25/11/09, 10:04

Capt_Maloche these are the same figures whatever the method although it surprises me a little that it is exactly the same in electrolysis ...

Remundo wrote:You should not forget that to have the "atomic element", you need the same energy that it can give by putting itself back into a molecule.


Yes (although there is a "rumor", see langmuir, like what there would be an asymmetry on hydrogen ...) but do not forget that there is 60 to 70% of thermal losses in an engine. .so if we take part of these losses to "create fuel" it's all good, isn't it?

Same if it promotes combustion.
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by Capt_Maloche » 25/11/09, 10:25

It is a loop whose energy source is fuel (100% primary energy)

to improve the combustion of 10% for example (optimistic value) on a machine (engine) whose yield is 30% maxi change you from 30% to 30x1.10 = 33%

- it does not improve the efficiency of the alternator 90%
- it does not improve the efficiency of electrolysis 70% in the best case

the result remains at 0.33x0.90x0.70 = 20% on the string
it's still 2% energy won will you tell me : Cheesy: and 10% in less fuel

provided it can be produced in sufficient quantity
moreover the load of a motor varies from 10 to 100%, the flow should follow, which is not the case of all these montages

I strongly doubt that the energy returned compensates the energy expended, on the other hand I believe very strongly in the improvement of the combustion by the reaction of the water on the carbon, produced by the combution of the H2 at the beginning of the explosion
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"Consumption is similar to a search consolation, a way to fill a growing existential void. With, the key, a lot of frustration and a little guilt, increasing the environmental awareness." (Gérard Mermet)
OUCH, OUILLE, OUCH, AAHH! ^ _ ^

 


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