good cells and bad cells

[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.



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.

[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