good cells and bad cells

[Darrell]

In the simplest terms using 6"x6" plates and one cell stack like this -NNNNNN+ you can only apply 11.25 amps before you waste power and run into over heating problems.

If you want more gas you have to add more plates or another cell stack next to the first one like this -NNNNNN+NNNNNN- Now you can run up to 22.5amps with out over heating.

Hope this helps. "D"

[danser75]

well at the current time we don't have a calculation. I will be watching youtube videos and scouring this forum for the data i need to find the equation. once i find the equation i will post it on a new topic and Ill send you a PM letting you know where it is at. I will put it on a new topic just so that the discussion that it will surely start up doesn't get mixed into this thread.

So at this time all I can tell you is there is more to come.


Also I would like to say thank you to those who have been trying to help me and maybe after all this we will all learn something.

[danser75]

In the simplest terms using 6"x6" plates and one cell stack like this -NNNNNN+ you can only apply 11.25 amps before you waste power and run into over heating problems.

If you want more gas you have to add more plates or another cell stack next to the first one like this -NNNNNN+NNNNNN- Now you can run up to 22.5amps with out over heating.

Hope this helps. "D"

Yes but keep in mind that that is only according to one source and no one seems to understand why or how that rule came into play.

[Darrell]

Several of us have tested and saw this first hand. They were just discussing the differnt messurements to get you to those numbers. "D"

[sixpack127]

you would think (I am not being crass - sorry if it sounds that way - but I am not a mathamatician or a scientist) that after all the years people have been working on this there would be a formula to use to figure out the mass of the cell and the power input (amps/watts) - temperature - etc - to come up with the size of cell that is needed to produce the best mmw

Again - sorry for sounding glibb

[myoldyourgold]

you would think (I am not being crass - sorry if it sounds that way - but I am not a mathamatician or a scientist) that after all the years people have been working on this there would be a formula to use to figure out the mass of the cell and the power input (amps/watts) - temperature - etc - to come up with the size of cell that is needed to produce the best mmw

It is not the size of the reactor that makes for better MMW's. You have got it all wrong and obviously do not know what you are talking about. At least you admit it that I complement you on! It makes no difference what size the reactor is. Why spend the money to build a large reactor when you only need a few LPM of HHO. The better MMW's are made by getting the design of the plates right, plate preparation, plates magnetically aligned, electrolyte concentration right, amount of heat right, atmospheric pressure matched to your reactor, back pressure right, flow right, and at least 20 plus more things that have to be exactly right and kept right. There is so many variables that maybe some math genius who understands all of the variables could come up with a formula but no one like that exists. If there was someone like that he would God. He would have to be a chemist, electrical engineer, mechanical engineer, physicist, and be an expert in thermodynamics besides a good math genius with quite a few years of experimentation in building reactors to even at best would be just close.

Because of the volume of variables there are many different methods or methods of calculating things not to mention types of reactors. Some times things are not as clear as they should be but it always gets sorted out in the end. There are many things that some of us disagree with each other over and based on our own experience think we are right. I think in most cases both schools of thought are right but are applied in a different manner. This field is very complex and most of what is discussed on this forum is about just one narrow part and that part is a bipolar series reactor that is composed of a number of cells that are run with brute force. No fancy frequencies or other methods like steam/pressure or cavitation to separate the water, just simple brute force. AMPS

Now let us say you get everything right which is almost impossible, then comes the hard part and that is using it so you really do get some gains in a vehicle if that is your goal. That is not an easy task as many here will attest to and is even harder than making a reactor. So if you think this is all about a formula and size/mass, of the reactor and heat, you need to do a lot more studying before you make crazy statements like that. Most of all the information is on the forum and it is a shame that after spending 100s of hours by hundreds of people the new guys are to lazy to read what has been recorded to help them but want someone to hand feed them like they were physically challenged or just born. I think I have said enough and I sure you will agree!! LOL

[danser75]

This is exactly why I started the tread data collection. But even without the data of people here I have found that some of the numbers on here may not be too far off because 1 sq. inch seems to be good for up to .375 amps and 25 amps seems to produce between 1.5 and 2 lpm. at least from the youtube videos I have watched and been able to collect my required data from. So theres a starting point but I am by NO means calling this a set equation as I just don't have enough data to base it off if yet. the .375 comes from the people here and theres a few articles I have read and everyone seems to agree on that. At least with it not heating up.

If you want to boil water you can do as many amps as you want to trow at it.

[myoldyourgold]

Calculating Faraday Efficiency for HHO Production
With an emphasis on temperature correction

By: Nick Stone

Let's get a couple of things straight right out of the chute. First, an electrochemical reactor is NOT "a cell"!

Your reactor might consist of "a cell", but, more than likely, it consists of multiple cells. A "cell" consists of two electronically conducting phases (at the plates) connected by an ionically conducting phase (through the aqueous electrolyte between them). In a single "cell", as an electrical current passes, it must change from electronic current at the surface of the positive face of one plate to ionic current through the electrolyte and back to electronic current at the negative face of the other plate. These changes of conduction mode, in each and every single "cell", are always accompanied by oxidation/reduction reactions. A single "cell" can consist of either two bipolar plates (improperly refered to as "neutral plates") or two monopolar plates (plates connected, externally, to positive and/or negative terminals) or a combination of a monopolar plate and a bipolar plate. It makes no difference which is which. Furthermore, this holds true for nested tubes or any other types of electrode configurations you can dream up.

Next,

Faraday Efficiency is a term that crops up in the HHO community over and over. There have been claims made that some of the best and the brightest in the HHO community have calculated it every way from Sunday and they swear that, as it turns out, 100% Faraday Efficiency is equal to somewhere in the neighborhood of 7.5 MMW (Milliliters per Minute per Watt). That is PURE UNADULTERATED HOGWASH! It is simply NOT true and the reason why should be blatantly obvious to you once you have read and understand this article, but if it still isn't clear to you, in part two of this article, I will use the data from an actual YouTube video, that you can watch, to explain exactly why it's not even anywhere close to being true, but, first, you need to understand how Faraday Efficiency (Refered to as Current Efficiency in Electrochemistry reference manuals and materials) is calculated.

For the purposes of this article, I am going to round off some numbers, but I would like to encourage you to do the math yourself to double check any and all of my numbers.

.627 Liters per hour per amp is representative of 100% Faraday Efficiency at 32 oF and 1 atm pressure. The ratio of .627 liters per hour per amp per cell is the same thing as .627 divided by 60 minutes to get .01045 liters per minute per amp per cell and then, because there are 1000 milliliters in a liter, multiplied by 1000 to get 10.45 milliliters per minute per amp per cell.

So, let's take a look at another way to get there and try to construct a proof along the way.

We'll use Faraday's Laws and perform the calculation for the half reactions to find the theoretical volumes of Hydrogen and Oxygen produced per minute per amp per cell during the electrolysis of water, but there are a few preliminary considerations to get out of the way first.

The volume of Hydrogen, Oxygen, Air, HHO or any gas for that matter, per mole is a given value. At standard pressure ( 1 atmosphere) and temperature (273.15 oK or 298 oK depending on who you ask), the volume of Hydrogen per mole is 22.414 Liters or 22,414 Milliliters at 273.15 oK which, by the way, is the Ideal Gas Constant (0.0820574587) multiplied by that particular "Standard Temperature"(represented in degrees Kelvin). Also, this is the point in the calculations where, by compensating for the volume per mole, temperature corrections are made.

For example, some people will use what is commonly referred to as "room temperature" (25 oCentigrade (synonymous with Celsius) = 77 oFahrenheit = 298oKelvin) as the Standard Temperature to make these calculations which makes the volume of gas per mole = Ideal Gas Constant (0.0820574587) multiplied by room temperature in Kelvins (298oK) = 24.4531226926 Liters or 24,453 Milliliters per mole. In order to make this more clear, I will carry out the example for both Standard Temperatures throughout these calculations, but we could insert any temperature at all, as long as it is represented in degrees Kelvin, and it would work exactly the same.

Anyway, here are a couple more things to understand first. Electrical Charge in Coulombs (C) = t (time) multiplied by I (current)...(60 seconds x 1 Amp) = 60 Coulombs Also, 1 mole of Hydrogen yields 2 moles of electrons. The electrical charge of one mole of electrons (Faraday's Constant) is given as 96,485 Coulombs (1 Faraday). Since there are two moles of electrons in one mole of Hydrogen, the electrical charge delivered by one mole of Hydrogen = 2 x 96,485 Coulombs or 192,970 Coulombs.

Hydrogen volume per minute per amp per cell = Electrical charge in Coulombs (60 Coulombs) / (divided by) Electrical charge delivered by one mole of Hydrogen (192,970 Coulombs) multiplied by the Ideal Gas Constant (0.0820574587) multiplied by the temperature (represented in degrees Kelvin) which, fo example, is equal to 22,414 milliliters (at 273.15oK or 0oC or 32oF) or 24,453 milliliters (at 298oK or 25oC or 77oF).

Now, when you perform those half reaction calculations for Hydrogen at a temperature of 273.15oK (0 oC or 32 oF), it turns out like this:

60 Coulombs / 192,970 Coulombs = .000311 Coulombs

And,
.000311 Coulombs x 22,414 milliliters = 6.97 milliliters (Hydrogen per minute per amp per cell)

Or, at room temperature (298oK):
.000311 Coulombs x 24,453 milliliters = 7.60 milliliters (Hydrogen per minute per amp per cell)


Hydroxy contains 100% more or twice as much Hydrogen than Oxygen, so 50% or half of 6.97 milliliters (3.485 Milliliters) should equal the volume of Oxygen produced per minute per amp per cell and when we add the two together, that brings us up to 10.45 milliliters per minute per amp per cell.

Okay, now, let's go ahead and carry out our proof and verify that again by performing the calculations for the other half reaction for Oxygen and adding it to our results for Hydrogen.

Instead of 2 moles of electrons like we had for Hydrogen, we have 4 moles of electrons for Oxygen, so 4 x 96,485 C = 385,940 C/mole.

Now, again, when you perform those half reaction calculations for Oxygen at a temperature of 273.15 oK (0 oC or 32 oF), it turns out like this:

60 Coulombs / 385,940 Coulombs = 0.000155 Coulombs

And,
0.000155 Coulombs x 22,414 milliliters = 3.48 milliliters (Oxygen per minute per amp per cell)

Or, at room temperature (298 oK):
.000155 Coulombs x 24,453 milliliters = 3.80 milliliters (Oxygen per minute per amp per cell)


Now,
6.97 milliliters Hydrogen + 3.48 milliliters Oxygen = 10.45 milliliters of Hydroxy per minute per amp per cell.
Or, at room temperature (298 oK):
7.60 milliliters Hydrogen + 3.80 milliliters Oxygen = 11.4 milliliters of Hydroxy per minute per amp per cell.



Presto! There's your proof!

To correct for pressure, you just divide that final number by the atmospheric pressure represented in units of atm (atmospheres). Most local weather stations report atmospheric pressure in millibars or hectopascals (both the same thing). You can convert barometric pressure to atmospheres by multiplying the value given in millibars or hectopascals by .0009869233

That's the nub of it, but keep in mind that although I rounded off many of the figures for the purpose of slightly improving readablility of this article, for the sake of precision and accuracy, my calculator does not round off anything during the calculations. It's only the final output that gets rounded of!

This by far is the best done article you will find. Full credit and thanks to Nick for this!!

http://nicksrealm.com/HHO/Library/faraday.html

[Darrell]

I haven't seen that in a while!

Thanks for the post Myold. "D"

[madman]

danser

I am not an expert but I am going to give my 2 cents worth or maybe 3. First off the guys that are trying to help you out here are some of the most knowedgeable in this field of building reactors. They have been at it for years, spent a lot of cash and time. They do not mind being asked questions BUT it is a good idea to try and read up on it first, like use the search button or look at other sources.

It seems like you have done some research. I am all for verification or at least colaberation like 2 or 3 guys saying the same thing. Sometimes you will get 3 or 4 diff. opinions about a certain aspect. This is ussually just like diff techniques. Basics is basics and if you can get a basic understanding of the diff. components one at a time it will be less confusing. 99% of the info. is here for free. Also if you know some smart guys like engineer or even a mechanic or an electrician they will be a very big help.

Now the question. AMPS vs. SQ. INCH Reg. bipolar one side (plate) per stack measuring only the active area or wet surface Over .5 will kill your plates, under .5 will give you less output and .5 is the sweet spot. Rule of thumb for gasoline engines is .5 littres of gas for every littre of engine size. Also you might want to add maybe 20% more cause your engine might need it to find the sweet spot.

Be conservative and say 14 amps per littre of gas. Do the math. Note: better to have plates that are wide and less tall like 2/3 say 4x6.

Good luck