Hydrogen generator

[Christophe]

Forget Jonule is "out"

On the same subject but on the theoretical side, the "advantages" of atomic combustion:

https://www.econologie.com/forums/combustion ... t8780.html

[Remundo]

Hi Maloche and Christophe,

The balance sheet is not so simple because there is an energy loop in the system, namely the mechanical power taken to electrolyze, which is likely to improve combustion, and therefore the available mechanical energy;

Of course it will not diverge

Where H2 can help is that when we say that there is X MJ / kg of fuel, with combustion imperfections, there can be, for example, only 0.92 X MJ / kg.

if 100 Wmeca for H2 saves 4% on 100 Wmeca ...

After that, when we reach 100% combustion efficiency, the H2 produced on board is no longer useful and even becomes a loss because electrolysis also has its losses.

[Capt_Maloche]

I bet on 5 to 10%, no more

[Remundo]

I was scribbling some equations on the subject ...

The energy advantage occurs if:

Pcarb * etamot * (1-etacomb)> Pelec

where
Pcarb is the thermal power of the fuel supposed to burn ideally: for example 400 kW

etamot is the thermodynamic efficiency of the motor e.g. 25%

Pelec mechanical power drawn for the electrolyser

etacomb is the combustion efficiency (actually obtained heat / ideally obtained heat).

There or it gets gratin is that etacomb depends on Pelec and we don't know how ... At most we can offer the following look

where I took 90% of comb output without electrolysis and 98% with massive injection of "H2" (we could also take 100% to be optimistic) ...

Say Maloche, how much is the combustion efficiency without electrolysis? I put 90% ... no precise idea and fluctuating with the engine, the speed, the load, the state of the engine ...

Oh it's soup time

[Remundo]

I think it's possible, but it's long

do you have time?

[Capt_Maloche]

No not too much

Come on ! when burning fuel oil we have at the input:
Combustion air composed of 21% O2 + 78% N2
fuel oil: approximate composition 2 C16H34 (without sulfur and other additives ...)


output
2C16H34 + 49O2 + 182N2 = 32CO2 + 34 H2O + 181N2 + 1NO2 + 1CO + xC?

If with 1.25L of fuel I get around 25 / 30g of Carbon (TCB test of white cloths), what does unburnt matter represent?

[Remundo]

Hello Maloche,

Ah, I thought I was done with the equations ...

The fuel oil weighs in at 860 kg / m3, if you have 1,25L, that's 860/1000 * 1.25 = 1.075 kg of fuel oil.

Based on the formula C16H34, carbon is 12g / mol and hydrogen is 1g / mol.

The "fictitious alkane" C16H34 is at 16 * 12 + 34 = 226 g / mol

in% by mass, C therefore represents 16 * 12/226 = 85% and H, 15%

In your 1.25L of fuel, there is therefore 1.075 * 0.85 = 913.3 g of carbon.

If you get 30 g of carbon, it means that the unburnt pure carbon represent 30 / 913.3 = 3.2%

But be careful because other unburnt materials are not detected, especially the gaseous gases: CO and organic residues HC

With probe measurements, we can have access to% CO and HC in the exhaust gases ...

Let's say we can maybe count on 5% unburnt, and 10% by being pessimistic.

that said, 5% is not so negligible: on a tank of 50L, it is 2.5L which can typically cross 50 km more for a Diesel at 5L / 100 km.

[Christophe]

Good estimates +1.

Except that 10% is by being optimistic (for the benefit of water injection / electrolysis) right?

So let's go on 5%! How much energy gain would that be compared to the initial fuel if say 100% (ideal case) of these 5% of carbon were rebrulated according to the following reactions: H2O + C -> H2 + CO then on the one hand H2 + O2 - > H2O + 242,7 kJ / mol and on the other CO + 1/2 O2 -> CO2 + 283 kJ / mol?

Remundo how hot you are, I let you conclude

It would be nice to reach something between 10 and 20%!

But nothing says that it is these reactions that are made directly. See the possible reactions between H2 + CO for example: http://fr.wikipedia.org/wiki/Monoxyde_d ... C3.A8ne_H2

ps: flat anyway:

that said, 5% is not so negligible: on a tank of 50L, it is 2.5L which can typically cross 50 km more for a Diesel at 5L / 100 km.

Ah no, we are talking about 5% by mass of unburned carbon, not fuel. It would be true if the fuel was pure carbon. But H2 provides more energy than carbon on an L of fuel ... you can easily find the value.

Did you know there was a carbon-based nuclear reaction? http://fr.wikipedia.org/wiki/Combustion_du_carbone

[Christophe]

Good could not prevent me from trying to find the contribution of H2 in% and I advanced too quickly ...

I take your figures by taking the "theoretical" GO: C16H34 = 226 g / mol

Now we have:
- H2+ 1/2 O2 -> H2O + 242,7 kJ / mol
- C + O2 -> CO2 + 393 kJ / mol

We have 34/2 moles of H2 and 16 moles of C for 226 g of fuel.
Or an energy intake of:

242.7 * 34/2 = 4126 kJ for H2 and 393 * 16 = 6288 kJ for carbon. Total = 10414 kJ / mol (we find 39 570 kJ / L)

% energy from H2 = 4126/10414 = 39.6%
% energy from C = 60.4%

However, diesel is already a "heavy" fuel since:

% by mass of H2 = 15%
% by mass of C = 85%

In lighter fuels the energy% of H2 rises higher.

The highest being methane: 242.7 * 2 = 485 kJ against 393 * 1 = 393 kJ for carbon. Total = 878 kJ.

Or an energy contribution of H2 of 55% (for a mass% of 25%)

So H2 provides more energy than carbon only for the lightest alkane fuels!

[Remundo]

Hi Christophe,

for water, it's H2 + 1/2 O2
but since air costs nothing, it changes nothing in the following reasoning.

I rather knew the value 286 kJ / mol, at 298K, liquid water. You must therefore probably assume it in the gaseous state (and rightly so).

We were wondering if the "artificial" injection of hydrogen into the engine reduced unburnt.

And what is roughly the possible gain since it would seem that currently, the unburnt mass fraction of the fuel is approximately 5%.

According to the handkerchiefs of Captain 'Maloche who has a big cold