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hhonewbie
02-25-2011, 11:24 AM
What is the maximum optimum Amps per square inch @ 12V please?

myoldyourgold
02-25-2011, 01:21 PM
The total amps is based on .5 amps per active surface area of one side of one plate. You do not add them up!!. Lee do you agree? LOL This might not hold true for media blasted plates which have increased surface area. There is still a lot of testing in this area to be completed. Some recommend .25 amps and not more than .5 amps. Now that should confuse you. To be safe do not exceed .5 amps per active surface area of one side of one plate. Again you do not add all the plates up either!!

lhazleton
02-25-2011, 02:17 PM
Yeah Carter, I agree with you 100%. How often does that happen?:D

Darrell
04-07-2011, 05:48 AM
I think alllot of people have talked about .25 amps per sqaure inch being the sweet spot and .5 being the max using unblasted plates. Bad built reactor can cause heat misleading that number though.

I will tell you that the sweet spot for blasted plates is much higher than .25 from what I have tested.

But is always good to have a base line to start from.

"D"

myoldyourgold
04-07-2011, 06:12 AM
Darrell I am trying to establish what the max amps / voltage on media blasted plates is by examining the plates, measuring heat, and HHO production. This is going to take some time and many tear downs thus the smaller test reactor. Just keep a record of what the max amps you ran through your reactor and for how long and then take some pictures of a plate and I will add it to my data. I have Larry's great videos and Uno's pictures to compare with. It is interesting what I am finding. I will post the information when completed.

There will be some differences possibly, because of the method and media used in blasting. There will also be difference in plate design that will play into the picture. This is also part of the test to see if that makes any measurable difference in gas production/heat or any visible difference in the plates. I guess time will tell all. If we pool all the information it will happen faster and cheaper for us all.

This will just be a ball park figure because of so many other variables but maybe we can see some kind of trend.

Darrell
04-07-2011, 06:41 AM
Doing some testing this weekend and will keep you updated to the progress.

"D"

alphatonic
04-09-2011, 07:28 PM
The total amps is based on .5 amps per active surface area of one side of one plate. You do not add them up!!.


I'm missing something. Surely a 4x4 25 plate gen and a 4x4 5 plate gen wouldn't be the same....?

myoldyourgold
04-09-2011, 08:09 PM
Yes you are missing something. The 5 plate reactor is setup like +nnn- I assume. (Not a good setup) How are you setting up the 25 plate reactor? Answer that and I will explain what you are missing. Basically I need more information in order to explain it to you. It will depend on how the 25 plate reactor is setup.

keiththevp
04-09-2011, 11:40 PM
I think what he was trying to get at was the .5 amps per sq figure is good PER CELL not per device. By cell I mean a pos a series of bipolar plates and a neg. The next cell of theses grouped plates could receive another .5 amps per sq in.

So if your dry cell has two cells and you use the sq in. from one plate out of the two cells then you can now run 1 amp per sq. in. because the amperage will be divided by the number of cells in a dry cell. Just as the voltage will be divided by the number of plate gaps within a cell.

myoldyourgold
04-10-2011, 05:44 AM
Lets get some definitions down. The gap between 2 plates is a cell. A series of these like +nnnnnn- is a reactor. Next +nnnnnn-nnnnnn+ This is a reactor with 2 stacks. A stack is between the positive and the negative connections from the power source regardless of the number of plates. OK now one stack of 8 plates (7cells) +nnnnnn- where the plates are 4 X 4 and have a 1/2 inch gasket will give you 3.5 X 3.5 active surface area minus the ports. Lets call it 12 square inches of active surface area. Now the 8 plate one stack reactor will be limited to a max of 6 amps (3.5x3.5=12.25-.25x.5= 6). Lets us make a reactor with 3 stacks +nnnnnn-nnnnnn+nnnnnn- 22 plates. The rule is that stacks divide the current (amps) and the cells (gaps) divide the voltage. If you just ran 6 amps in the 3 stack reactor you would be running 6 divided by 3 equals 2 amps through each stack but you want to run 6 to be .5 amps per square inch of active surface area on one plate. This means that you now have to run 18 amps to get to .5 amp/sq/in of active area of one side of one plate. This applies to series reactors with biopolar (neutral) plates. You can add more stacks and you have to add more amps to keep it at 6 amps per stack. The voltage is divided by 7 because there is 7 cells (gaps) in each stack. Hope that confuses everyone. LOL

lhazleton
04-10-2011, 05:53 AM
Good definition, Carter. How many times have we posted this formula over and over again?:rolleyes: It should really be a sticky since it's so important.

Mhz250
05-17-2012, 03:46 PM
Hi, I have a doubt...

in this math: (3.5x3.5=12.25-.25x.5= 6)

why is this? .25x.5????

.5 is the amps per sq inch?
and what is .25 for?

Stevo
05-18-2012, 12:02 AM
From the formula, I would say it is a buffer to ensure you run just under .5 Amps per square inch during most all conditions. Shoot for 6 and occasionally end up with ~6.2 and it's still acceptable. At least according to the example above.

Of course if you really wanted to get crazy with numbers then plug in The Golden Mean 1.618:

( (3.5x3.5) / 1.618 ) / 1.618 = 4.68 Amps

which is 0.38 A/in sq.

Which is right smack in between 0.25 and 0.5 A/in. sq.