good cells and bad cells

[danser75]

Well guys if this information is on the forum i can't find it and i am sorry for bringing it up again. But if its on the internet and no one is trying to sell it to me i can't find it.

I want to know how to find a good dry cell and tell it apart form a junk cell. Almost all of them seem to be built with 316L stainless steel using a configuration of +nnnnn-nnnnn+ so how do I tell if the one am about to purchase is good.

Also while we are at it since I am new to this and everyone claimes that their system will produce 50,000 lpm hho at 5 amps running 12 volts. How do I do the math to figure out what the output of a cell should be?

I am going to be using this for a car install and so i am working between 11-14 volts. I have also wondered why do they only use 5 neutral plates not 6? yes the car battery is 12 volts but the running car has an alternator putting out 13.8 volts thus it only seems resonable to me to include 13.8 volts in the calculation for the cell. If you take 12 and divide it by the now 7 cavaties you will get 1.72 volts which is low but should produce hho correct?

[lhazleton]

I am going to be using this for a car install and so i am working between 11-14 volts. I have also wondered why do they only use 5 neutral plates not 6? yes

Neutral plates are up to the builder. I prefer using 6 (13.8 / 7 cells = 1.97vdc per cell), while others use 5 neutrals for 2.3vdc. Either is acceptable, though I personally don't like going over 2v per cell gap.

[BioFarmer93]

Danser75,
The math is approximate. Rule of thumb is 13-14A per liter per minute, 4 square inches per amp (wet working area)... So, 52 to 56 sq.in. x 13-14A = 1LPM. Neutral (bipolar) plates- most folks at this forum agree that six is best. Alternators typically provide between 13.5 & 14.5 volts, and a six bipolar plate seven cell electrolyzer will operate at 1.92 - 2.07 VpC (volts per cell). This is pretty much considered the sweet spot. If a seller is claiming 4LPM @ 30A with a five neutral plate reactor, you can bet the farm that one of those L'sPM is steam.

[danser75]

Ok thanks for the replies. This ability to figure out the output will be really helpful but i am wondering with the cost being so high, I have found a locally owned sheet metal company that has a laser cutter and i called them to get a rough idea of cost and looks like for 20 plates ready to use (cut to my specs with three having a tab to connect the power and the rest neutrals, and the holes all being laser cut) made of 316L 18 gauge stainless steel, it will cost me 200. I was wondering how would it be for me to make my own. If i made my own plates in this manner could I come out with something as good as the mass produced ones?

I know I don't need all 20 I actually would need 15. Also I found 1/4 inch acrylic and for those 2 pieces (cut, holes drilled for me) would be 45. the holes were the expensive part but they said with no drill press, it was unlikely that i would be able to drill the pieces myself without cracking one. Also is this true? Has anyone else drilled the acrylic plates themselves?

To get these estimates I told them it would be 6 inches by 8 inches for the plates and then 8 inches by 10 inches for the acrylic. Which now that I have the calculations for how to determine the size I need I can go back and find out the actual size. Which since I am paying by the sq. inch for the materials if I change the size it will either cost a little more or less. I just want to get this going as soon as possible so I wanted to see what cost would look like.

But one other question I had was the 4 inches per amp is that 4 inches for both anode and cathode, or is that 2 inches of anode and 2 inches of cathode?

[lhazleton]

I know I don't need all 20 I actually would need 15. Also I found 1/4 inch acrylic and for those 2 pieces (cut, holes drilled for me) would be 45. the holes were the expensive part but they said with no drill press, it was unlikely that i would be able to drill the pieces myself without cracking one. Also is this true? Has anyone else drilled the acrylic plates themselves?

Don't use acrylic! Also, 1/4" is way too thin. You need to use 3/4" HDPE.


But one other question I had was the 4 inches per amp is that 4 inches for both anode and cathode, or is that 2 inches of anode and 2 inches of cathode?

The inches are measured one side of one plate, buy only where the electrolyte is actually in contact.

[danser75]

Ok thanks again guys.

[BioFarmer93]

OK, we're kinda going about this backwards.. I had to go back and read your first posts to see that you are boosting a 3.3L Buick, so you need about 1.7LPM of HHO. Using the formula from earlier today where we established the averages to make 1LPM, we just multiply 54 sq.in. by 1.7 to see that you need 91.8 (92) sq.in.. If you make a two stack reactor [-nnnnnn+nnnnnn-], that's 92/15 = 6.133~ sq.in. of working area on each of those plates. The square root of 6.1333 is 2.4765~ (so call it 2.5" x 2.5") plus a half inch of gasket all around, so 3.5" x 3.5" plates. It has been reported lately that efficiency can be increased by leaving the center power plate intact (no holes) Anyway, add one more inch for end plate overlap and that brings you up to a whopping 4.5" x 4.5". At this size you can get away with a 1/2" thick HDPE Walmart cutting board for end plates and they don't break when YOU drill them. Also you might save a few shekels by going with 20ga. instead of 18ga.- the plates really don't need to be that heavy. This little jewel will be at it's most efficient right around 23A.:cool:

[myoldyourgold]

Gus, I am very tired tonight but I am lost here. In my calculation for this Buick shows it will need at least a 6" X 6" [-nnnnnn+nnnnnn-] 2 stacks, .5 amps per active square inch on one side of one plate. This comes to 25 amps in the above example. That means that each plate would get 12.5 amps which is .5 amps per active 25 square inches with out deduction for ports and dead area at the top. This should make a little less than the 1.7 LPM but would work. I will revisit this in the AM when I a firing on all cylinders and hope it isn't foggy. LOL

[danser75]

Biofarmer I dont mean to correct or offend you but my looking at it says I have 14 cells not 15. therefore its 6.9 sq in. I need right? am i missing a cell in +nnnnnn-nnnnnn+ =
+,1,n,2,n,3,n,4,n,5,n,6,n,7,-,8,n,9,n,10,n,11,n,12,n,13,n,14,+ Right?

sorry if thats complicating but it shows how i am counting the cells.

But if that's true to be on the safe side I would like to go with 3x 3.5 inch active area cells making the plates 4x 4.5 inch total with three tabbed and drilled for power and the rest neutral with only the 1/4 inch hole at bottom and 3/8 inch hole at top. I was reading on another website if you give a larger hole in the top it allows for better flow and i have seen you guys use 2-3 holes in the top so i figure its go to be worth something. and the reason i am using one bigger hole not 2-3 smaller holes is because I have to pay by the hole ($1.00 per hole) because i having them laser cut when they make my plates.

But also what would happen if i built and ran 6x6 inch cells at 23 amps. would i have too much hho or would the unit just run cooler? If it never gets hot it seems it would never have a chance for thermal runaway. thus eliminating the need for a pwm. I am however planning to use a thermal circuit breaker to control it should it get excited and decide to act up.

I have used them on other projects and they are great. If you overload it it will "blow" then once it cools down then it reconnects. In case you guys haven't used them before.

[BioFarmer93]

Gus, I am very tired tonight but I am lost here. In my calculation for this Buick shows it will need at least a 6" X 6" [-nnnnnn+nnnnnn-] 2 stacks, .5 amps per active square inch on one side of one plate. This comes to 25 amps in the above example. That means that each plate would get 12.5 amps which is .5 amps per active 25 square inches with out deduction for ports and dead area at the top. This should make a little less than the 1.7 LPM but would work. I will revisit this in the AM when I a firing on all cylinders and hope it isn't foggy. LOL

Carter, I was PM'ing with someone else about a unipolar build just before I wrote that and based it on .25A per sq.in., (sorry!) not .5A. BUT- at 5x5 working area (not considering holes at the moment) that's 15 plates x 25sq.in. per plate = 375sq.in.. At .5A per sq.in. that's 187.5A! In your explanation above it appears that you have based the amperage need on just two plates- when did this method develop? I've always calculated amperage (for bipolar) based on the working area of one side of every plate, not just a single set of power plates. No disrespect intended my friend! This is just a HUGE difference in method that I must clear up and understand completely the math behind.

[BioFarmer93]

Biofarmer I dont mean to correct or offend you but my looking at it says I have 14 cells not 15. therefore its 6.9 sq in. I need right? am i missing a cell in +nnnnnn-nnnnnn+ =
+,1,n,2,n,3,n,4,n,5,n,6,n,7,-,8,n,9,n,10,n,11,n,12,n,13,n,14,+ Right?

sorry if thats complicating but it shows how i am counting the cells.

But if that's true to be on the safe side I would like to go with 3x 3.5 inch active area cells making the plates 4x 4.5 inch total with three tabbed and drilled for power and the rest neutral with only the 1/4 inch hole at bottom and 3/8 inch hole at top. I was reading on another website if you give a larger hole in the top it allows for better flow and i have seen you guys use 2-3 holes in the top so i figure its go to be worth something. and the reason i am using one bigger hole not 2-3 smaller holes is because I have to pay by the hole ($1.00 per hole) because i having them laser cut when they make my plates.

But also what would happen if i built and ran 6x6 inch cells at 23 amps. would i have too much hho or would the unit just run cooler? If it never gets hot it seems it would never have a chance for thermal runaway. thus eliminating the need for a pwm. I am however planning to use a thermal circuit breaker to control it should it get excited and decide to act up.

I have used them on other projects and they are great. If you overload it it will "blow" then once it cools down then it reconnects. In case you guys haven't used them before.

Danser, I don't mind being corrected when I'm wrong, otherwise I would never learn anything- and I'm not offended at all. The "15" I used as a divisor was plates, not cells. 15 plates makes 14 cells.. As for the larger square inch version, I think I'm going to wait and see what myoldyourgold says about my previous post before I comment, because I have just discovered that my calculation method may be in error.

[myoldyourgold]

Carter, I was PM'ing with someone else about a unipolar build just before I wrote that and based it on .25A per sq.in., (sorry!) not .5A. BUT- at 5x5 working area (not considering holes at the moment) that's 15 plates x 25sq.in. per plate = 375sq.in.. At .5A per sq.in. that's 187.5A! In your explanation above it appears that you have based the amperage need on just two plates- when did this method develop? I've always calculated amperage (for bipolar) based on the working area of one side of every plate, not just a single set of power plates. No disrespect intended my friend! This is just a HUGE difference in method that I must clear up and understand completely the math behind.

Gus, amps are the same on every plate and only split by the stacks. If you measured the amps going into each plate you would find it is identical which I know you know. You only need to calculate the active area of one side of one plate of a bipolar plate reactor but both sides for a unipolar plate reactor but still on just one plate to get the right number. I measured it and found this to be true. Here is a picture of my test reactor that I measured this with and a number of other things for both uni and bi reactors. I was able to move the shunts around and measure every single possible combination. Some of the wires were not on when I took this picture for a particular reason.

http://i57.photobucket.com/albums/g215/yenom_1945/Testreactor3.jpg

I need to come up with a better way to explain this and will give it some thought and see if I can write it up so it makes sense.

In a 2 stack reactor the calculation is simple just take the active area of one side of one plate and that is your maximum amps for .5 amps or divide that by 2 for .25amps.

I could have miss interpreted or measured wrong and would be extremely happy to be wrong because I could then make much smaller reactors or get a lot more gas out of the current ones.

[myoldyourgold]

But also what would happen if i built and ran 6x6 inch cells at 23 amps. would i have too much hho or would the unit just run cooler? If it never gets hot it seems it would never have a chance for thermal runaway. thus eliminating the need for a pwm. I am however planning to use a thermal circuit breaker to control it should it get excited and decide to act up.

I have used them on other projects and they are great. If you overload it it will "blow" then once it cools down then it reconnects. In case you guys haven't used them before.

Based on my calculation you will need to have the 6x6 and could run it at 23 amps and get about 1.65 LPM more or less. Remember it is the amps that make the gas. Surface area is necessary so the amps do not eat up your plates by running to many amps per active area and helps control heat preventing thermal runaway. Here is what happens when you push small plates to hard. In this case the water was being separated faster than it could fill the cells and the level of electrolyte in the reactor was lower than it should have been making the active area much less and destroying the plates in about 1500 miles. This was a very poor design but is a good picture to demonstrate what can happen.

http://i57.photobucket.com/albums/g215/yenom_1945/eatenplates.jpg

[BioFarmer93]

Now I am beyond confused... This guy only figures the working area of 1 plate.. http://www.hho4free.com/amperage_understanding.htm

[danser75]

Ok so if I am understanding correctly I am able to run the 6x6 cells at 25 amps and be fine making up to 1.7 lpm.

If so I am misunderstanding the calculations we used yesterday. Because that was:

4 sq inches per amp.

14 amps per lpm

15 plates in my series

So a 6x6 cell if I figure it at about 5x4.5 (to calculate in the holes for sake of argument) that is 22.5 sq active inches

22.5sq in times the 15 plates = 337.5 total active sq in

337.5 total inches divided by 4 inches per amp= 84.375 amps

84.375 amps divided by 14 amps per lpm= 6 lpm of hho.

Which would be nice to have and might even come closer to running the car but I am not ready for that big of an overhaul. Nor am I ready to install a new beefy alternator.

But the main concern i have is what if I got too big of a cell? Could I use it to produce less than its "rating?"

Also What size should the holes in the cell be to allow flow? Is 1/4in. in the bottom enough or should it be 1/2in.?

[danser75]

So I read some of the article you just put up biofarmer but it was way more detailed than I wish to deal with at this second.

So Biofarmer have you run the calculations you gave me yesterday and found them to produce a gentle running good production cell, that had the desired output and amps?

Also myoldyourgold Could you give the calculations to the rest of the class. I don't understand the formula you are using to get your numbers.

[BioFarmer93]

Danser,
Please disregard anything I may have written or any calculations I have made. I am going to go commit suicide now by pulling my head out of my ass. I'm sure the sudden vacuum created will cause me to implode, and die.

It's been real,
BioFarmer

[myoldyourgold]

In my calculation you would be limited to 22.5 amps with 2 stacks. I am at a loss to explain this and even reading what the guy wrote on the link that Bio gave really does not clear this up in my mind. Bio you might be looking at the bipolar plate as a wire just like the one that connects your groups. One end is positive and the other end is negative but the amps going across are the same even though the polarity changes. Using this logic then the total area is both sides and when I say .5 of one side it is actually .25 of the total plate. That is not what is happening in my understanding. In my mind I consider the bipolar plate as two plates each having only one side active. Full amps go across each side separately. It is hard to get a handle because a bipolar plate by definition gets its current horizontally through the electrolyte not from a hard wire so each side is different and could be two separate plates connected by a connector. This is where even I am confused. I still think that you use only one side of one plate in a bipolar reactor to calculate the maximum amount of amps per square inch but can not explain it to my satisfaction. Lets have some more feed back.

Here is my calculation. Active area of one side of one plate is 22.5. You divide that by 2 (or x .5) which is 11.25 but because you will have 2 stacks you then have to multiply it by 2 which is 22.5 amps. Remember stacks divide the incoming current so you can now use twice the amount of current and still stay with in the bounds. 22.5 x .5 = 11.25 x 2 = 22.5 amps. Now if you had 3 stacks (+nnnnnn-nnnnnn+nnnnnn-) the calculation would be 22.5 x .5 = 11.25 x 3 = 33.75

Here is the portion of Bio's link that is related to this discussion.


It has been determined that 1 square inch of plate surface can produce 6.27 milliliters of HHO gas per minute - using 0.5 amps (Faraday). That statement is not exactly true. You see, the hydrogen needs one square inch and the oxygen needs one square inch. Faraday is describing a cell (two plates separated by water). The plates are generally spaced 1/16 or 1/8 inch apart. So actually we need one square inch of sandwiched surface area (one square inch from each side of the water). Look at it as if you have a one inch square sandwich; hydrogen is made on one side and oxygen on the other. It will be much easier to understand. Here is why. Our HHO generator is going to need a lot of plates; in pairs. Each pair needs to have the same amount of surface area (sandwiched). The total of all of that surface area is going to be a big factor in how much hydrogen and oxygen gas we make. If one square inch makes 6.27 milliliters per minute, then 100 square inches should make 627 milliliters, using 1/2 of an amp per square inch (sandwitched). That is the theory.

Bio if you did that you would just join me. LOL

[danser75]

Well j hope I didn't offend anyone with thks thread it just really convoluted. And har
d to follow so I was mearly trying to break down the math.

[sixpack127]

Maybe it's my lack of brain power at this juncture in my life - but - what exactly is the calculation needed to figure out how many amps I will need to run my generator? (I have not built it yet - as I am trying to get materials fro free from different sources - ie - appliance dumps and the like) - anyway - after reading most of the posts - I got completely lost as to which one was correct - help me pleeeeaaaasssseee :eek:

[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 :eek:

[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

[sixpack127]

I already said that I am not smart enough to figure all the differences out - I have been researching HHO generators for only a few weeks - thought I would build one first. Anyway - as to the title - Engineers et al - My Brother-in-Law is an engineer with Ford Motor Company - he has worked for GM also - as well as Dale Jarret's racing team - he is an Engine designer - so I told him about this - told him to do some research also - he basically says that it will not work - so I was looking for a math equation that I could shove in his face - I guess I will have to do the brute force method and prove him wrong by building my generator - tweaking it and installing in my truck - and get better gas mileage from it - I was not trying to disrespect anyone here - I was just trying to see if there was a formula - cest le vie - got to do it the hard way - just proof for the doubting Thomas

Thanks for all your help - I plan on doing a great deal more research as I build my generator (when I get the plates I will be just about ready - trying to do it on the cheap through scrap from a couple of places my sister-in-Law works)

[myoldyourgold]

I already said that I am not smart enough to figure all the differences out - I have been researching HHO generators for only a few weeks -

Do not under estimate how smart you are. I think you have plenty of smarts. The fact that you joined the forum shows you have plenty. Keep studying and you will be as good at this any anyone. Please do not think a lack of experience is because you have no smarts and do not take some of my remarks to negatively. I some time get a little testy in my old age. You are doing just fine!! There are plenty of documents that support the positive side of HHO. PM me if you are interested in some to help you or better yet here is a link.

http://reduceyourfuelbill.com.au/forum/index.php?topic=444.msg2810#new

[hhofox]

Hi all,

As was said, you gotta just keep at it and compare the info that you have amassed. I have a home built 5.5"x7" dry cell which puts out 2LPM easily. Turns out that it's overkill for my 1.3L engine, so I just lower the amps and run it cool for the results I want.
Just be glad that you didn't start out with wet cells like many have! I will post a link below so you can have even more reference material. Enjoy!

http://2hho.com/hho_plans.htm

[danser75]

Well guys thanks for the help on this. I hadnt been on alot in the last couple days but I did get caught up here. I really hope that i didn't upset anyone here. I was trying to figure something out that seems to be just a little over my head.

It started as a thought to myself, "there must be a formula for what size cell I need and the amp draw I will have." but then i got way carried away with it and started obsessing alot about it when I saw there wasn't one. But Now that my head is back out of my ***, I am ready to regain a normal role as a newbie and say that I think for my first cell I am going to get a 6x8 316L ss plate setup with 15 plates for 2 stacks. I am planning to leave the middle plate with no holes and give an in and an out on each side.

I was also talking with a friend of mine who hasn't done this before but has done alot with electronics and when he saw the sketches of my plates I am planning on, and the plates alot of other ppl are using, he asked, "why are you only using one power input in the corner?" This got me thinking should we use 2 to better distribute the power across the plate? So for the cost I am going to go ahead and add a second input on my power plates and going to try moving the power to the middle instead of the corner unless someone can give me an explaination of why not to. I figure even if it only helps a little bit its worth a shot. It can't hurt anything. Finally its only going to be like a 5 dollar differance on the plate cost.

But again if anyone was offeneded or thoguht I was argueing or overstepping my bounds I appoligize. I didnt mean to in anyway.

[ultra_efficient]

It has been reported lately that efficiency can be increased by leaving the center power plate intact (no holes)

wow you can notice an increase in efficiency with only one plate that has no holes, imagine a cell where none of the plates have any holes and 6 neutrals?

http://www.youtube.com/watch?v=szLJFl0x2bc :rolleyes:

[hhofox]

Regarding that "no holes" design on Youtube. Wouldn't the same or even better results be achieved by putting the same insulating material he has around the holes on his setup, on the holes in regular dry cell setup? I mean, people have used Weld-on 15(?) for this right? :confused:
Let me know if I am missing something here, cause I thought up a similar design a while back, but scrapped it because of this reasoning.

[danser75]

wow you can notice an increase in efficiency with only one plate that has no holes, imagine a cell where none of the plates have any holes and 6 neutrals?

http://www.youtube.com/watch?v=szLJFl0x2bc :rolleyes:

There is a wet cell setup that uses 6 wet cells wired in series each carrying 2 volts. it is suppose to be the mother of all cause it is suppose to have a very low amp draw and very high output. The only reason I am not using that setup is because of the cost. I would need 14 cells, (because of my choice to run 6 neutrals not just 5) for a total of 28 plates. Almost doubling the cost of my system. Plus then I would have to assemble and maintain 14 cells not just one. Maybe if I convert my wife's car I will see about this system But I think trying to tackle it as my first build would be a little too much for me.

[myoldyourgold]

There is a wet cell setup that uses 6 wet cells wired in series each carrying 2 volts. it is suppose to be the mother of all cause it is suppose to have a very low amp draw and very high output. The only reason I am not using that setup is because of the cost. I would need 14 cells, (because of my choice to run 6 neutrals not just 5) for a total of 28 plates. Almost doubling the cost of my system. Plus then I would have to assemble and maintain 14 cells not just one. Maybe if I convert my wife's car I will see about this system But I think trying to tackle it as my first build would be a little too much for me.

Dancer, and Richard there is no reactor that will exceed 11.4 LPM per amp per cell if measured correctly. Just not possible. Gas has a direct relation to amps. Lower amps and you lower the volume of gas. Anyone claiming different doesn't know what they are talking about and are just measuring moisture, are just lying, or using some other method other than brute force which is yet to proven scientifically.

[Weapon_R]

There is a wet cell setup that uses 6 wet cells wired in series each carrying 2 volts. it is suppose to be the mother of all cause it is suppose to have a very low amp draw and very high output. The only reason I am not using that setup is because of the cost. I would need 14 cells, (because of my choice to run 6 neutrals not just 5) for a total of 28 plates. Almost doubling the cost of my system. Plus then I would have to assemble and maintain 14 cells not just one. Maybe if I convert my wife's car I will see about this system But I think trying to tackle it as my first build would be a little too much for me.

danser75

could you provide more info on this setup (wet cell).

[danser75]

Just skip the explaination at the top if you wish. Its a chapter from a book and the chapter is 122 pages if printed. Its got alot of interesting ideas in there and some that are really out there that being said the web address is http://www.free-energy-info.com/Chapt10.html

Just so everyone is clear I have not built this unit nor do I know if it will work it was just a site a friend of mine sent me and I think it looks like if someone could/ would build it it would work. Maybe not at 6 amps but I think its an interesting concept.

I did place my order at the sheet metal shop today for my plates of 316L and I am getting them with the neutrals to have power tabs on them so that i can try multiple dry cells and eliminate current leakage. I am not trying to exceed 11 amps per cell or anything I just wanted to try this. If it doesn't work then I will be out the cost of the bolts and plastic cutting boards used for the end caps and the hose fittings but once I have my cell done I have 2 friends wanting their cars adapted to this as well so I will just use them on those conversions. What I am planning is to have six dry cells wired in series making it 2 volts do each of the cells with each cell only having 2 plates in it. Then each cell will have its own hoses back to the tank. I have not seen anything like this done but I am hoping to see what happens. Another plus if it works is that I have seen some (not all but some) cells on youtube getting filled with gas faster than it can be released into the tank. In my design since the only product flowing through the hoses would be the gas/unprocessed water of just the 2 plates in that cell there shouldn't be the build up of gases waiting to escape. I know the way I am doing this would be fine to go with a wet cell but I chose to try a dry cell because if it works I will have 7 cells to place in the car under the hood and that would be alot of space but the dry cells will take up less room. I am giving up surface area of plates for the gaskets but gaining room to be able to install this in my car. In a couple weeks when I have it running I will post a youtube video and let everyone see it.

I don't know that it will be better than any of the others but I am doing this project as something to use my time on so I am willing to do whatever assembly and tear down is necessary to come out with a reliable and well working system.

[myoldyourgold]

Dancer75, the open bath reactor that you reference in the chapter 10 pdf is not what most people are talking about when they reference a open bath (wet cell) because electrolyte is not shared between the cells. The problem with this type of reactor is the method of filling is complex or you just dump water in the top and flood the reactor to make sure each compartment is filled. As the level goes down the active surface area decreases, heat goes up and efficiency goes down. It is efficient when levels are equal compared to the normal open bath where all cells share the same electrolyte. It also is space consuming compared to a series sealed flow through reactor. Keeping all cells equal is also a problem compared to a sealed series flow through reactor. If you take the average efficiency between fill times, the sealed series flow through reactor wins hands down, unless you want to refill this type of open bath very often which is not practical. It is also more complicated to build. For ease of build and efficiency the sealed series flow through reactor is the winner. Lots of good information in the chapter 10 pdf and the DB9 pdf that everyone interested in HHO should read.

[danser75]

myoldyourgold

This is why I am not building that reactor at this time. I only mentioned it as another idea that was on the table in response to ultra_efficient's remark that imagine the gains if you had an entire reactor with no holes in it.

What I am going to test it to make a total of 6 small dry reactors. Each reactor will have 2 volts and 2 plates in it. Then they will be wired in series to take the 12 volt supply on the workbench. Each reactor will have 2 hoses running straight to the tank. This is merely a test I want to do for myself and wanted to share with the forum. If it works great I have figured out a more efficient way to make hho, if not I will be out about 12 bucks for hosing and whatever time it took to build the setup. The plates I ordered will be able to be used in either the cells I am trying or I can remove the power tabs on the neutral plates and use them in a normal dry cell. I feel and hope that my approach to this build will give the best of both worlds. I am hoping this will have the efficiency of the wet cell shown in that chapter. But still have the constant flow of electrolyte and smaller size of a dry cell. Like I said I have no idea if it will work but the only way we ever find this stuff out is by tinkering and trying new thoughts and new ideas.

[myoldyourgold]

What I am going to test it to make a total of 6 small dry reactors. Each reactor will have 2 volts and 2 plates in it. Then they will be wired in series to take the 12 volt supply on the workbench.

This will work just fine but takes to much room in a vehicle. That is why all the plates are put in a stack. For bench testing and experimenting many of us have setups similar to this or have had them. You learn a lot with less. I use a 3 cell test reactor now and very the voltage between 1.5 and 3 volts per cell to collect data on heat, wear, flow, production, and a lot of other information at various voltages and amp draw. It is just a lot easier to assemble and disassemble to examine thing under a microscope etc. Saves a lot of time and money too!!

[madman]

danser,

Let me add to the confusion. I am not an expert so I would like to see verification of my thery here. Here goes. First if you are going to run this on a car the voltage will probably be closer to 13.8 volts. Divided by 7 is 1.97 which is good for the e. process and does not cause overheating by excessive voltage. 13.8 divided by 6 is 2.3 volts which will create more amps/ heat. Mike (a member here) has power supplies resonable, see hho connection.com.

If you go with 22 sq. inch per plate you can push 11 amps at that 1 reactor. Basing this on you are going to use only 2 plates per reactor. So to get voltage divided you are going to have 7 of these reactors tied together in series thus 1.97 per reactor. Now since they are in series the VOLTAGE also divides( or multiplies) . 7 reactors x 11amps is 77 amps. Or 77 amps divided by 7 reactors is 11 amps per reactor and 11 amps per plate.

Now if everyone is not confused since you have all your plates in each reactor with an electrical charge it is UNIPOLAR meaning there are no BIPOLAR plates ( no direct el. charge) Bipolar plates get their charge from contact with the electrolite. SO you can DOUBLE the amps 77x2is 154 amps.With 15 amps per littre of gas being conservative you are looking at 10 LPM.

Where is Gus when you need him? Lee? Chief? Carter? Steve? HELP.

I am not at all sure that this is right. I built a mini beast (unipolar) with 7 groups of 7 plates each(49 plates total) and I am still not 100 percent understanding the theory. I have been researching and racking my brain for a reactor design that will be simple,small,efficient, that will produce about 3 LPM. Got to get over this part.

Lets see what everyone has to say. I can not wait.

[myoldyourgold]

Let me add to the confusion. I am not an expert so I would like to see verification of my thery here. Here goes. First if you are going to run this on a car the voltage will probably be closer to 13.8 volts. Divided by 7 is 1.97 which is good for the e. process and does not cause overheating by excessive voltage. 13.8 divided by 6 is 2.3 volts which will create more amps/ heat. Mike (a member here) has power supplies resonable, see hho connection.com.

If you go with 22 sq. inch per plate you can push 11 amps at that 1 reactor. Basing this on you are going to use only 2 plates per reactor. So to get voltage divided you are going to have 7 of these reactors tied together in series thus 1.97 per reactor. Now since they are in series the VOLTAGE also divides( or multiplies) . 7 reactors x 11amps is 77 amps. Or 77 amps divided by 7 reactors is 11 amps per reactor and 11 amps per plate.

Now if everyone is not confused since you have all your plates in each reactor with an electrical charge it is UNIPOLAR meaning there are no BIPOLAR plates ( no direct el. charge) Bipolar plates get their charge from contact with the electrolite. SO you can DOUBLE the amps 77x2is 154 amps.With 15 amps per littre of gas being conservative you are looking at 10 LPM.

Where is Gus when you need him? Lee? Chief? Carter? Steve? HELP.

I am not at all sure that this is right. I built a mini beast (unipolar) with 7 groups of 7 plates each(49 plates total) and I am still not 100 percent understanding the theory. I have been researching and racking my brain for a reactor design that will be simple,small,efficient, that will produce about 3 LPM. Got to get over this part.

Lets see what everyone has to say. I can not wait.

Let me see if I can straighten this out. Here is the data.
11 square inches on one side of one plate or 22 on both sides. Only 2 plates in each of seven groups totaling 14 plates. 7 unipolar cells/pairs/groups in series (+-_+-_+-_+-_+-_+-_+-)does not split amps but will split volts. Unipolar cells act like stacks but only if both sides of the plate are active which is not the case with only one unipolar cell in each of the seven groups. So if you only have one in a group it will not split the amps. Only one side is active. It is the same as +nnnnnn- 7 cell bipolar reactor because it only splits voltage using only one side of each plate. I also see no advantage but the opposite. You have to have more than one Unipolar cell (2 unipolar plates) in a group in order to split amps. Next you are only using one side of each of the plates so you can only count one side of the plate. In my calculation you will be limited to only 5.5 amps to stay within .5 amps per active square inch. You would do much better if you added one plate and made a bipolar reactor with two stacks -NNNNNN+NNNNNN-. In this case you could use 11 amps and make more gas. The center plate is a unipolar plate and splits the amps, half to one side and the other half to the other side. To keep within the .5 amps per square inch of active area you will have to take 22 sq in (both sides of the unipolar plate which are active) 22 x .5 = 11 which is split by 2 making each side of the plate have 5.5 amps per active area and each bipolar plate will also have 5.5 amps per square inch through the electrolyte.

In a single stack -nnnnnn+ you would take only one side of one plate active area because the unipolar plates (1 Positive plate and one Negative) are only used on one side and then pass the amps through the electrolyte to the bipolar plates. In the same example you would be limited to 5.5 amps because you only have 11 active inches on one side.

If you are not totally confused I will be surprised!! LOL

[madman]

Let me see if I can straighten this out. Here is the data.
11 square inches on one side of one plate or 22 on both sides. Only 2 plates in each of seven groups totaling 14 plates. 7 unipolar cells/pairs/groups in series (+-_+-_+-_+-_+-_+-_+-)does not split amps but will split volts. Unipolar cells act like stacks but only if both sides of the plate are active which is not the case with only one unipolar cell in each of the seven groups. So if you only have one in a group it will not split the amps. Only one side is active. It is the same as +nnnnnn- 7 cell bipolar reactor because it only splits voltage using only one side of each plate. I also see no advantage but the opposite. You have to have more than one Unipolar cell (2 unipolar plates) in a group in order to split amps. Next you are only using one side of each of the plates so you can only count one side of the plate. In my calculation you will be limited to only 5.5 amps to stay within .5 amps per active square inch. You would do much better if you added one plate and made a bipolar reactor with two stacks -NNNNNN+NNNNNN-. In this case you could use 11 amps and make more gas. The center plate is a unipolar plate and splits the amps, half to one side and the other half to the other side. To keep within the .5 amps per square inch of active area you will have to take 22 sq in (both sides of the unipolar plate which are active) 22 x .5 = 11 which is split by 2 making each side of the plate have 5.5 amps per active area and each bipolar plate will also have 5.5 amps per square inch through the electrolyte.

In a single stack -nnnnnn+ you would take only one side of one plate active area because the unipolar plates (1 Positive plate and one Negative) are only used on one side and then pass the amps through the electrolyte to the bipolar plates. In the same example you would be limited to 5.5 amps because you only have 11 active inches on one side.

If you are not totally confused I will be surprised!! LOL

Carter BRILLIANT
I thought I was wrong about the voltage splitting. Here is how I see it . One single stack(8plates,6 bipolar) is 5.5 amps. No voltage divide. A double stack(15 plates,12 bipolar with charged plate in middle) Divide voltage(11) is5.5 x2 is 11 amps total. Triple stack 18 bipolar ,+BBBBBB-BBBBBB+BBBBBB- 22 plates is 5.5 x3 is16.5 amps. In each case voltage is sent to each stack and then divided between the 7 cells(area between a pair of plates)

If this is not correct you will be charged with murder tomorrow because I will jump off the bridge into traffic on the interstate if I read that I GOT IT WRONG AGAIN.

Now 2 questions about unipolar.
1) At .5 amps per ACTIVE sq. inch how many amps can I run my MINI Beast? I have 7 seperated stacks with 7 plates each total 49 plates. Each plate has 10 sq. inch active area. My calculation is 10x.5 is 5. Now multiply that times 2 because unipolar where both sides of plates are charged is 10amps per plate. so each stack is 10 x7 stacks because they are in series so I come up with 70 amps max for the MB. What do you come up with?

2) If it takes more than 1 cell to divide voltage in a unipolar system then what is the least amount of plates I can use per stack?

As always THANK YOU

[myoldyourgold]

There is not much discussion on the pros and cons of a bipolar reactor compared to a Unipolar reactor. Now that, in this thread, the two are being discussed and compared I thought I would bring this up. This discussion is limited to sealed flow through series reactors and not an open bath reactor and does not take into consideration plate design, preparation, conditioning, etc, but considers all of this to be equal when comparing the two types.

Bipolar sealed series flow through reactors (one stack -NNNNNN+) is the most efficient METHOD in relation to size and the space, that the reactor consumes physically, that you can build. Now as you start adding stacks in series the advantage becomes less and less. The other advantage is that the amps are basically reused through each cell in the bipolar single stack and if done right has only small losses but still has some losses in the process.

In a unipolar sealed series flow through reactor each cell has its own voltage and current supply. Because of this there is less loss in each group than in a bipolar stack. The down side is that it requires lots of space but produces some amount more HHO per amp than a bipolar reactor when everything is equal but requires more plates. The only way you could run a unipolar reactor with only one group is to feed it with 2 volts or what ever voltage close to 2 volts that you wanted to run each cell with. This will require a buck converter which even though they can be in the 90% efficient range still have some loss. Next the unipolar reactor will require 1 additional plate than the bipolar single stack for the same number of cells. To avoid a buck converter in the unipolar reactor you need to have 6 or 7 groups with a minimum of 4 plates 3 cells in each group. Space is the problem. Where it begins to get ahead of the bipolar reactor is when you need large amounts of HHO efficiently. Because each group can have its own input and exit ports, as a result of the groups being separated they run cooler and more efficiently than say a 2 stack bipolar reactor and get even better if compared with a 3 stack or 4 stack bipolar reactor. One more advantage with the unipolar reactor that is very important, is the ability to split amps by adding 1 additional plate per group that is active on both sides compared to the bipolar reactor where you have to add a complete stack thus allowing you to decrease the size/active area of the plates in the unipolar reactor. This is I think a big advantage in how the electrolyte flows.

If space is not an issue then there is evidence that the Unipolar is more efficient everything else being equal. If large amounts of HHO is required then a Unipolar reactor would also have an advantage. The bipolar reactor will be much more suited to small vehicles where space is non existent and the volume of HHO needed is small.

One could argue that one could build a number of efficient bipolar reactors and run them in parallel and compete with a large unipolar reactor when large volumes of HHO are required. This might be possible and a debatable point.

[madman]

This got posted after your last post but was written befor I saw it. Basis uni polar reactor with 10 sq. inch active on one side.

7 plates is 5 active plates both sides. So 10 amps x 5 = 50 amps.

9 plates is 7 active plates both sides. So 10 amps x 7 = 70 amps. This is an increase of 40% in amperage.

Going from 7 plates to 9 plates is a 29% increase in steel, why is the percentage in amperage and the steel not the same?

Last question: Does total sq. inch of steel (combining all the groups together)have a bearing on total output of gas? If amperage determines amount of gas why can't I have 1 group, 7 plates (controlling voltage to .2) and produce 50 amps at 14 LPM for3.85 litres total?

Seems to me that the other 6 groups are there only to divide the voltage. Why not use 6 pair of steel bolts and wire them in series or use a PWM.

[danser75]

Could someone list the steps I need to go through when I pick my plates up next Friday? I will have new 316L plates. I know I need to media blast them but what kind of sand do I use? My uncle has a sand blaster he bought at an auction and has never used.

After looking into it I have decided I would like to use NAoH. Since its winter time I would will be using it in 20% concentration.

One more thought. Has anyone tried putting the system in the trunk and piping the HHO to the engine? Is it too far or would it work? If that would work then we all have space for the multi-cell reactor. Then all I would have is a bubbler under the hood. I have looked under the hood seriously to find a place to put this but I can't find a place big enough for this setup. If the trunk is too far what is the longest Hose I can use for the HHO? I know Gus is putting his reactor in the bed of his truck but not sure if he is pushing the limits. I haven't seen anywhere how far we can pipe HHO before it recombines to make water.

[myoldyourgold]

Basis uni polar reactor with 10 sq. inch active on one side.

7 plates is 5 active plates both sides. So 10 amps x 5 = 50 amps.

9 plates is 7 active plates both sides. So 10 amps x 7 = 70 amps. This is an increase of 40% in amperage.

Going from 7 plates to 9 plates is a 29% increase in steel, why is the percentage in amperage and the steel not the same?

I am not a math guy but I think you need to compare all the active area to make this comparison but that will not be the same either. Even though the two plates that have one side inactive and do not split amps they still have one side active. I think you might be trying to compare apples to oranges but really have no idea.

Last question: Does total sq. inch of steel (combining all the groups together)have a bearing on total output of gas? If amperage determines amount of gas why can't I have 1 group, 7 plates (controlling voltage to .2) and produce 50 amps at 14 LPM for3.85 litres total?

The number of cells and the amps per cell does have a bearing on how much gas is made so is related to the total sq inches of steel because it limits the amount of amps. Remember Faraday's max is 11.4 ML per amp per cell of moisture free HHO. You will be lucky to come anywhere close to that in real life with brute force. Sorry but the following makes no sense to me. .2 volts? 14 LPM for 3.85 liters total?

Seems to me that the other 6 groups are there only to divide the voltage.

See above comment.

Why not use 6 pair of steel bolts and wire them in series or use a PWM.

It will make gas but not efficiently many have tried it and end up making a good method of boiling water

[madman]

there are a few threads on media blasting. The most important aspect seems to be not to warp the plates. Whether you do it yourself or have it done pratice on a spare one if you have it. After blasting and cleaning you will need to PASSIVATE the plates using citrric acid . There are some threads on this also. After that keep clean(wear gloves) and assemble.

What configuration and size did you decide to go with?

AS FAR AS GOING FROM YOUR TRUNK allI can tell you is you will need larger gauge wire depending on length and amps .

[madman]

Basis uni polar reactor with 10 sq. inch active on one side.

7 plates is 5 active plates both sides. So 10 amps x 5 = 50 amps.

9 plates is 7 active plates both sides. So 10 amps x 7 = 70 amps. This is an increase of 40% in amperage.

Going from 7 plates to 9 plates is a 29% increase in steel, why is the percentage in amperage and the steel not the same?

I am not a math guy but I think you need to compare all the active area to make this comparison but that will not be the same either. Even though the two plates that have one side inactive and do not split amps they still have one side active. I think you might be trying to compare apples to oranges but really have no idea.

Last question: Does total sq. inch of steel (combining all the groups together)have a bearing on total output of gas? If amperage determines amount of gas why can't I have 1 group, 7 plates (controlling voltage to .2) and produce 50 amps at 14 LPM for3.85 litres total?

The number of cells and the amps per cell does have a bearing on how much gas is made so is related to the total sq inches of steel because it limits the amount of amps. Remember Faraday's max is 11.4 ML per amp per cell of moisture free HHO. You will be lucky to come anywhere close to that in real life with brute force. Sorry but the following makes no sense to me. .2 volts? 14 LPM for 3.85 liters total?

Seems to me that the other 6 groups are there only to divide the voltage.

See above comment.

Why not use 6 pair of steel bolts and wire them in series or use a PWM.

It will make gas but not efficiently many have tried it and end up making a good method of boiling water

Sorry to have been so confussing. I got a couple of things mixed up. Untill you showed us the Faraday equation I only computed volume by amount of amps. This makes much more sense now.

As far as the end plates in a unipolar group I understand they are still charged but active only on one side.I assume they are charged on both sides but the side to the outside(facing the endplate) does nothing to produce because it does not create a cell.

Here is the kicker. You stated before that a 7 plate group has 5 plates (charged both sides) so they divide amps so 5 x 10 amps is 50. My brain says there are 6 cells that get charged from both side so 10 x 6 is60amps.

Thought about the bridge jumping and decided that is way to messy . Thinking HARD about an alternative.

[myoldyourgold]

Here is the kicker. You stated before that a 7 plate group has 5 plates (charged both sides) so they divide amps so 5 x 10 amps is 50. My brain says there are 6 cells that get charged from both side so 10 x 6 is60amps.

Thought about the bridge jumping and decided that is way to messy . Thinking HARD about an alternative.

Ya I though about jumping too after I realized how many mistakes I had made but agree it is to messy. You are right about the 60 amps.

Here is a revised post that I deleted.

I must apologize for the wrong information and have gone to join Gus. LOL I told him he would find me there. LOL I deleted my post like this and am reposting it here.


I thought I was wrong about the voltage splitting. Here is how I see it . One single stack(8plates,6 bipolar) is 5.5 amps. No voltage divide.

I apologize if I misled you. Bipolar plates split voltage. So in the 8 plate reactor the voltage is split by 7(7cells) Stacks or unipolar plates when both sides are active split amps and will explain further along.

A doublestack(15 plates,12 bipolar with charged plate in middle) Divide voltage (amps not voltage)(11) is5.5 x2 is 11 amps total. Triple stack 18 bipolar ,+BBBBBB-BBBBBB+BBBBBB- 22 plates is 5.5 x3 is16.5 amps. In each case voltage is sent to each stack and then divided between the 7 cells(area between a pair of plates)

Voltage is only split by the number of bipolar cells in ONE stack when in series. Makes no difference to voltage if it is more than one stack.

If this is not correct you will be charged with murder tomorrow because I will jump off the bridge into traffic on the interstate if I read that I GOT IT WRONG AGAIN.

Remember the basic rules. Bipolar cells split voltage by the number of cells in one stack no mater how many stacks. Amps are split by stacks and unipolar reactors split like this. 1 active plate splits by 2. 2 active plates splits by 3. 3 active plates splits by 4 and so on. Now you do not have to jump!!

Now 2 questions about unipolar.
1) At .5 amps per ACTIVE sq. inch how many amps can I run my MINI Beast? I have 7 seperated stacks with 7 plates each total 49 plates. Each plate has 10 sq. inch active area. My calculation is 10x.5 is 5. Now multiply that times 2 because unipolar where both sides of plates are charged is 10amps per plate. so each stack is 10 x7 stacks because they are in series so I come up with 70 amps max for the MB. What do you come up with?

Unipolar plates that are active on both sides split amps by the above listed amount. In a seven plate group amps will be split by 6 (5 plates that are active on both sides. Now because there are 7 groups voltage will be split by 7). Max amperage is 20 x .5 = 10 x 6 = 60 amps max. A Group of unipolar cells when in series with another group of unipolar cells act the same as a bipolar cell in series and split voltage

2) If it takes more than 1 cell to divide voltage (divides amperage not voltage) in a unipolar system then what is the least amount of plates I can use per stack?
The lowest number of plates in a unipolar group (a group is defined as a number of Unipolar cells, it is different than a stack) will be 2 plates, but if you want to split amps it will be 3 plates 2 cells 1 plate active on both sides. This will split amp by 2. Remember the number of groups in series splits volts. Here is an example: +-+-/+-+-/+-+-/+-+-/+-+-/+-+-/ This will split voltage by 6(6 groups) and amps will be split by 3.

NOW NO JUMPING OFF BRIDGES JUST ASK, IT IS EASIER TO ASK QUESTIONS. YOU MIGHT NOT GET THE RIGHT ANSWER THOUGH BUT STILL NO EXCUSE TO JUMP LOL

[danser75]

Well I have decided to go with a couple tests before I decide what is going in the car. I want to try the multi-cell reactor and see how it does compared to the same plates in a normal dry cell.

I have the plates ordered and they are 6x6 giving me 5x5 active area. but they wont just be squares I left 2 1x1 tabs on opposite corners to allow me to connect power to both sides of the plates. Then the holes are 3/8 at top and 1/4 at the bottom. they will be in the opposite corners as the power tabs.

I will post pics of them as I get them and as I am doing the steps to make them the reactor.

[madman]

Carter,

We are going to be rich. I am going to advertise for a HHO class. It is going to be called Mark's college of HHO knowledge. Sat. noon till 4. Cost will be $50 Thousands will enroll and I will split the cash with you.

Since I only know enough to take up about 5 min. I guess the rest of the time will be tellling jokes. The problem is I can only remember enough jokes to take up another 5 min.

Do you know any good jokes?

Thanks for all the help. It has been VERY helpfull, more than you realize and there are many others that must feel the same way. I feel much more confident now that I have a much better understanding.