Hydrogen generator

[Remundo]

what matters in fine, these are the destroyed and formed bonds,

thus if C + H2O -> CO + H2 is involved in the reaction chain, but the balance is C + O2-> CO2, then you only recover 393 kJ / mol of "pure carbon soot": this results from the difference in energy between C and O2 alone and CO2 with 2 stabilizing chemical bonds.

And this is the case:
(1) C + H2O -> CO + H2
(2) CO + 1/2 O2 -> CO2
(3) H2 + 1/2 O2 -> H2O
(4) C + O2 -> CO2
And (4) = (1) + (2) + (3)

[Capt_Maloche]

Let's see

if we are well into the subject Hydrogen generator and we want to inject H2 + O2 into the bouzin

it's first:

H2 + 1/2 O2 -> H2O + C + O2 -> CO2 + (CO small quantity)
C + H2O -> CO + H2 (> 950 ° C)
CO + H2O -> CO2 + H2 (> 950 °) + H2 + 1/2 O2 -> H2O and so on until O2 is exhausted
CO + 1/2 O2 -> CO2

Fun to note that if the reaction with H2 occurs before the combustion of the diesel fuel (even safe) it amounts to sending water vapor

The following reactions show that H2O reacts with C and CO under the action of T ° to produce H2 again which will react as long as there is O2

I would be curious to know if this cycle of H2O occurs several times in an explosion
in fact it will work the better the lower the engine speed

Interesting huh?

[Christophe]

I know where we are and I did the HS!

So sorry I thought it was induced in my explanations: these equations are valid only for water doping: there is no H2 in the "basic" reagents at the start of combustion!

H2 acts as an intermediate reagent!
That's all.

thus if C + H2O -> CO + H2 is involved in the reaction chain, but the balance is C + O2-> CO2, then you only recover 393 kJ / mol of "pure carbon soot": this results from the difference in energy between C and O2 alone and CO2 with 2 stabilizing chemical bonds.

And this is the case:
(1) C + H2O -> CO + H2
(2) CO + 1/2 O2 -> CO2
(3) H2 + 1/2 O2 -> H2O
(4) C + O2 -> CO2
And (4) = (1) + (2) + (3)

Hey he ises not Prof! Mwarf! Hihihiih

(4) = what missed in the combustion cycle!
We never come back to it: the reaction will no longer reproduce.

We are therefore talking about converting this soot C back into energy via H2. So via another route !!

So we replace it with (1): endothermic reaction, energy = motor losses

Then (2) and (3), both of them being exothermic and supplying the energy that (2) could not supply ... C pure being a waste of combustion ...

Good as it is like that, I will finish the calculation of the enthalpies myself ... any!

[Remundo]

well you see, I told you that your reasoning was mysterious to me ...

Unroll your reasoning and the following calculations will

[Remundo]

Let's see

if we are well into the subject Hydrogen generator and we want to inject H2 + O2 into the bouzin

it's first:

H2 + 1/2 O2 -> H2O + C + O2 -> CO2 + (CO small quantity)
C + H2O -> CO + H2 (> 950 ° C)
CO + H2O -> CO2 + H2 (> 950 °) + H2 + 1/2 O2 -> H2O and so on until O2 is exhausted
CO + 1/2 O2 -> CO2

Fun to note that if the reaction with H2 occurs before the combustion of the diesel fuel (even safe) it amounts to sending water vapor

The following reactions show that H2O reacts with C and CO under the action of T ° to produce H2 again which will react as long as there is O2

I would be curious to know if this cycle of H2O occurs several times in an explosion
in fact it will work the better the lower the engine speed

Interesting huh?

Hello Maloch,

you're right in the chemical kinetics of combustion.

Don't worry, nobody knows how it works.

On the other hand, we know that radicals are formed, and that the presence of "good" initiator radicals considerably accelerates the decomposition of carbon chains. When we know that an engine at 4000 rpm leaves only 7 or 8 ms of combustion to burn the charge in each room ...

Inject H2 or H2O, only empirical methods will say which is better.

Ante combustion of the hydrocarbon, when H2 gives H2O during compression, the H * radicals are probably more numerous with H2 alone than with H2O alone (which does not react or very little).

With H2O, HO * H * is probably produced in a similar amount at the start of hydrocarbon combustion.

In fact, one can also wonder about the interest of a mixed H2O / H2 injection.

[the middle]

Good evening,
I have a silly question, but it has been bothering me for too long:
When you boil water, the bubbles rising from the bottom of the pan, what is gas?

[Remundo]

Hello The Just,

This is perlin pinon, rare gas of the 8th row and 19th column of the periodic table.

PS: It's steam water (gaseous H2O)

[the middle]

Ha bin yes, it makes sense,
Thanks, I won't die silly

[Remundo]

to be 100% rigorous, there are traces of air (4N2 / O2) dissolved in cold water, which when heated, degassed (like when heating coca, CO2 degassed).

But this is negligible compared to the gaseous H2O.

[Christophe]

PS: It's steam water (gaseous H2O)

Oh not that !! It’s true at the end when it boils with a big ball but before from 70-80 ° C it’s before all the gases dissolved in the water which are released: CO2 and O2 being the most abundant I think .but there are surely many others ...