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.

[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.