electrolysis improved

[nlc]

Well waiting to receive all the stuff for my pantone, I look at other solutions to save money. Direct doping with hydrogen in admission would be good, but how to generate hydrogen?

Since a few days I studied about improved electrolysis (frequencial) or high voltage, but I do not believe it anymore. On the other hand I found a way to reduce the yield of a normal electrolysis close to 100%.

So I'm copying here what I've already put on the forum engine so that a maximum of people can enjoy the idea:

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I saw on the net that the threshold to exceed for electrolysis of water is 1.47V, and then the amount of gas produced depends only on the current ...

Well, in our tests this afternoon, we had 4A in our electrodes, under 30V. This makes us a resistance of electrodes of R = U / I is 7.5ohms. It does not bubble masses, for a total power of 120W (30V x 4A).
But the problem is that on these 120W, there is only 5.88W useful (1.47V x 4A). The rest, 114W, escapes pure calorific loss. So we have a deplorable performance ....

Now, let's say that I pass the resistance of my electrodes to 0.5ohms (I do not take the value at random you will see ), by considerably increasing their surface. To pass the same number of ampere, so 4, I need how many volts? Simple: U = RxI = 0.5ohm x 4A = 2V.

I spend so much current, so although I have more electrode surface than before, I make exactly the same amount of gas.
BUT, if I calculate the power, I now have: P = U x I = 2V x 4A = 8W.

You see where I am coming from ..... there is almost no more heat loss .....
We can even reduce them again if we make the electrodes so that they pass 4A under 1.5V (they would thus 0.375ohms). It gives us a power of 6W.

What exactly did we do? Well thinking a little we just passed a poor performance, 114W for 5.88W useful, has a performance close to perfection, 6W for 5.88W useful! : Very Happy:

What's left now? And simply increase the power to be able to charge a lot more gas!
Let's say we want 120W useful power, to do as before (except that we lost 114W in heat and we had that 5.88W useful).

And very simple, I make a supply of 1.5V and 80A (I = P / U = 120 / 1.5) And I have to dig up then for my electrodes to make 0.018ohms. So I make electrodes with a huge surface. And now I have more heat loss, I consume 120W as at the beginning, but I leave 20x more gas than before (I let you check the calculation).

But the worst part of all this is that with a switching power supply, which has good efficiency, often> 70%, we will only draw on the 12V: 1.5 / 12x80 + 30% or 13A ....

It leaves dreamer no?

In fact, we have just eliminated heat losses by using a voltage barely above the useful threshold of 1.47V. By cons, it is necessary to increase enorment the surface of the electrodes, but it does not care! : Wink:

There's just a power supply 1.5 or 2V able to get a lot of amperes ...
But know that it is not the sea to drink ...

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I found DC-DC converters quite 2V / 80A, 87% yield. So it would be 160W max, so a consumption on 13.6V 2 / 13.6x80 + 13% = 13.3A...

We could have fun making a management box with 8 10A outputs, and as it can connect more or fewer reactors consuming each 10A.

I still think that 80A electrolysis ca must start to make some bubbles no? : Wink:
While on the battery we only shoot 13.3A, it's beautiful no?

For those who want to see the original post, I do not know if I have the right to put the link to the other forumbut you'll find it good.

A+

Cyril

[rpsantina]

Indeed,

to make a power supply that develops several amperes is often described in the application notes of mosfets and power transistors.
See the post of anto-genyus on the model reduced electric on the old forum.

Where I am dreamer is the surface needed to get 0.015 ohms.

If someone finds the relationship between the trading surface and the resistance we will have a good idea about the feasibility (if the whole thing can come under a hood)
Personally, I found this:

Resistivity and electrical conductivity

It is known that the conductivity in siemens per centimeter is the inverse of the resistivity in
ohm-centimeter. In the case of drinking water, it is expressed in micro-siemens per centimeter;
it is proportional to the mineralization, whereas the resistivity varies as the inverse of this one.

Conductivity μS / cm
Resistivity Ohms / cm
Pure water:> 23 µS / cm and> 30 Ohms / cm

Mineralized fresh water: 100 to 200 μS / cm and 5 000 to 10 000 Ohms / cm

Medium mineralization water: 250 to 500 μS / cm and 2 000 to 4 000 Ohms / cm

Very mineralized water: 1 000 to 2 500 μS / cm and 400 to 1 000 Ohms / cm

Knowing that the more water will be mineralized the more there will be impacts on the longevity of the system.

keep us informed of your progress ...

[nlc]

My idea is rather to make several reactors capable of pumping 10A under 2V. So 0.2ohms resistance.

There are several solutions to decrease the resistance between the electrodes. Increase the area, or reduce the spacing, or even the 2 at the same time. I think we need a compromise of the 2, because if the electrodes are too close, when it will bubble there will be less water between the 2, and so it takes a good circulation for the water to come from below . (Sucked by the rising bubbles).

For the DC-DC power supply, although being electronic, I would not venture into its manufacture! It exists quite, with super efficiency, an extremely small footprint, with all the protections, etc ...

Of course, what I mentioned above with the 80A was just one example. We can also ensisager the following thing:

Manufacture a reactor that pulls 10A under 2V (20W), and its associated control card, much smaller since it only has 10A to go out.
But the principle is the same, one would shoot the alternator only 2 / 13.6x10 + 15% = 1.7A.

1.7A on the battery, and 10A electrolysis, with a very good performance because very little thermal losses.

And if you want more gas, just make other reactors and their associated control charts.

In fact we can really do as we want, as soon as the most important is done: excessively increase the efficiency of electrolysis.

In addition, a DC-DC converter 2V / 10A is immediately less difficult to do (or is cheaper) than 80A ....
So on a car with several control / reactor cards, if one of the sets breaks down, it remains the others for anyway to save fuel;)

My opinion is that with a total of 80A electrolysis, we must still be able to get enough gas for it to feel on the vehicle.
What do you think ?

Knowing that yesterday with a friend we did experiments, in our reactor there were 3 or 4 1cm diameter bubbles that had formed in the dirt of the water, I approached a torch on fire, I can to tell you that I was surprised so much the explosion was powerful and the sound deafening!
My ears whistled a little while ...

In addition to the bubbles, there was necessarily the perfect ratio of oxygen / hydrogen !! From where the explosion I think. Because of the hydrogen alone it burns it seems to me, combining with the oxygen of the air?

Besides, we have to try an experiment I thought of tonight. Fill a syringe with this perfect mixture, and set the needle on fire. I'm pretty sure the syringe will fit completely. Indeed, the gas takes a lot of volume, but once transformed back into water, it takes almost no more! So vacuum will be created in the syringe and it will close again?

What do you think ?

A+
Cyril

[nlc]

Au (x) moderator (s)

In fact I think I was wrong place for this post, could there be a way to move in "Tips for consuming less and special engines"?

Thank you so much!

Cyril

[MobyleX]

In addition to the bubbles, there was necessarily the perfect ratio of oxygen / hydrogen !! From where the explosion I think. Because of the hydrogen alone it burns it seems to me, combining with the oxygen of the air?

Besides, we have to try an experiment I thought of tonight. Fill a syringe with this perfect mixture, and set the needle on fire. I'm pretty sure the syringe will fit completely. Indeed, the gas takes a lot of volume, but once transformed back into water, it takes almost no more! So vacuum will be created in the syringe and it will close again?

What do you think ?

A+
Cyril

Me what I think is that when you burn your mixture of O2 and H2, your syringe will fart. What is combustion? It is a chemical transformation that gives off heat. And a gas that warms what does it do? It expands pardis !! And your H2O that you will get after combustion, it will be in the form of steam at 2500 ° (do not worry the temperature will drop very quickly due to the expansion of the gas, no burns) and not liquid. So, in a nutshell: You fill your syringe, you mix the mixture, at the beginning of the combustion (of the order of the millisecond) your gas is with 2500 °, then it expands (and cools considerably in the passage), and your syringe farting under the pressure. The expansion of the gases by combustion of these is what makes the internal combustion engines run for 100 years. And any fuel containing hydrogen atoms (all hydrocarbons for example) will form water during their combustion. But it is not liquid when it comes out of the muffler (except in some cases where it has condensed in the pot).

[Rabbit]

On a car, you can put several electrolitic cells in series, right?

With 6 or 7 cells will have your 1.5 2 v.

For electrodes, a stainless steel mesh should do the job.
from Tienen a hardware store sells it to the meter. stitches of
1mmX1mm basically, I have not measure. Anyway do not have that
it is too fine otherwise the bubbles will remain glue and will not allow the electrolise.

[nlc]

Putting them in series actually is not interesting.
Because what makes generate more gas is the current. By putting them in series, we will actually have our 1.5V if we want, or 2V, but we will have a lower current. So it will be necessary to make electrodes with even more surface.

And let's say I buy a DC-DC 2V / 80A power supply. At first I make a reactor that pulls me 10A. Not enough gas, I make a second that I plug in // etc, etc ....

It is the current that plays on quantity.
The big advantage is that a DC DC power supply 2V / 80A, I found one with a yield of 87%, so we only draw 13.3A on the 13.6V battery.

In addition, I think we can even lower the voltage, to 1.6V.
Because I saw that electrolysis works between 1.23 and 1.47V. And with this tension, the electrolysis does not produce heat, on the contrary it absorbs it. So even if we pass 80A, there will be no warm-up.

And to pump a lot of current without needing too much surface, or without bringing too close the electrodes, an idea given on another forum would make the electrolyte more conductive.

[nlc]

Well, I made 2 or 3 experiences:
60V / 1A electrolysis and 12V / 1A electrolysis (bringing the electrodes closer).

We generate as much gas in the 2eme case, but the power delivered by the power supply is 12W instead of 60 ....

I tried to feed 2V but impossible to consume much, my 2 electrodes are too small!
With adequate electrodes, I would have released as much gas by putting 1A, but I will have consumed only 2W!

[nlc]

I found at Farnell, a DC DC converter module 10A, adjustable from 0.75 to 5V.
Set to 1.8V, its yield is 90%.

So an electrolysis of 10A will consume 1.45A on my 13.6V (20W) power supply.

Well I order it to see what debite an electrolysis of 10A!

[pluesy]

in electricity and electronics we find more easily devices that gives a lot of tension than a lot of current because of the losses by joule effect to my knowledge the only "current" devices that can deliver amperage are instant soldering irons that come out on their resitance in U a low voltage under a high intensity and post at the arc

the advantage of putting in serie is to decrease the current either you put a cell under 1.5v and 100 amperes!

either you put 10 cells in series under 15v and 10amperes

you will produce the same amount of gas but a power supply of 15v 10 A is still easier to find an 1.5v power supply 100 A

to have a yield of 100% you have to apply the 1v per cell
2v is 50%
1.5v 75%
in clear the efficiency and equal 100 / (the voltage at the terminal of a cell)
if you manage to go down to 0.5v you get a return of 200% (utopian!)
what makes the voltage rise is mainly the fuel cell effect and the greater the gas release and important and the more it is present
leave a voltmeter connected to a cell of electricity when you cut the juice and you will see that it persists a voltage of about 2v over several minutes