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