New Unipolar Setup

[keiththevp]

I am going to setup a new unipolar system as follows:

7 stacks

each stack has 10 plates that have an active surface are of 12 sq. in each

Now how do you calculate the max amperage draw at full concentration?

[myoldyourgold]

I am going to setup a new unipolar system as follows:

7 stacks

each stack has 10 plates that have an active surface are of 12 sq. in each

Now how do you calculate the max amperage draw at full concentration?

Keith I should let Bio or Larry answer this but maybe I can learn something here. Bio/Larry please correct me if I am wrong!! My understanding which could be wrong is you count both sides of one plate in unipolar setups. In your example 12 + 12 = 24/2 = 12 being the 50% active surface area. Then you have 10 plates that is 9 stacks (they divide the amps) so 12 X 9 = 108 amps would be your max. That is a lot!! Now you are going to have 7 sets of these which will divide the voltage by 7.

[BioFarmer93]

I am going to setup a new unipolar system as follows:

7 stacks

each stack has 10 plates that have an active surface are of 12 sq. in each

Now how do you calculate the max amperage draw at full concentration?

Hey Kieth,
The way you phrased that made me do a double take... Take a look at the two .jpg's in this post and see if your description still comes out sounding the same. http://www.hhoforums.com/showthread.php?t=6737&highlight=biofarmer93
As for figuring the amps, it's wet area (yes, both sides of the plates) times 1/4 of an amp. Now this is by no means the maximum amperage that it will handle, but going over this starts to degrade your plates and causes Cr6 to leach from the 316L.

[keiththevp]

Bio,

Ok, so in the "terms" listed in your pic I would have the following:

7 groups

1 group = 10 4" x 4" plates or 9 cells

The reactive surface area of the plates after gaskets would be 3.5" x 3.5". Making the reactive surface are of one side of one plate 12.25 sq. in.

So in myold's formula I would have (12.25 X 2)/2 = 12.25
Then 12.25 x 9 cells = 110.25 amps max per group

In Bio's formula I would have (12.25 x 2) x 9 = 220.5
Then 220.5 x 0.25 = 55.125 amps max per group

I understand that this unit can take a lot of amps but I really would like to keep it at the safe and running cool levels so I guess if I want this thing to be able to do 100 amps then I will have to up the plate size a bit or add more cells to each group, right?

[myoldyourgold]

Bio,

Ok, so in the "terms" listed in your pic I would have the following:

7 groups

1 group = 10 4" x 4" plates or 9 cells

The reactive surface area of the plates after gaskets would be 3.5" x 3.5". Making the reactive surface are of one side of one plate 12.25 sq. in.

So in myold's formula I would have (12.25 X 2)/2 = 12.25
Then 12.25 x 9 cells = 110.25 amps max per group

In Bio's formula I would have (12.25 x 2) x 9 = 220.5
Then 220.5 x 0.25 = 55.125 amps max per group

I understand that this unit can take a lot of amps but I really would like to keep it at the safe and running cool levels so I guess if I want this thing to be able to do 100 amps then I will have to up the plate size a bit or add more cells to each group, right?

Keith the difference between my calculation and Bio is that mine is stated as the max. The range is .25 to .50. I use heat to determine where to really run it at. In my reactors it is always under .5 amps per active square inch. In fact you should look for the spot where there is the least heat but the most gas per amp. The right amount of heat is a good thing. To much is very BAD.

The 55.125 amps would be the total amps your system would draw using .25 amps per active square inch of area not per group but all seven groups/sets, each with 10 plates.

Now due to media blasted plates all this might change and is requiring a lot of testing which is going on everyday.

[BioFarmer93]

Bio,

Ok, so in the "terms" listed in your pic I would have the following:

7 groups

1 group = 10 4" x 4" plates or 9 cells

The reactive surface area of the plates after gaskets would be 3.5" x 3.5". Making the reactive surface are of one side of one plate 12.25 sq. in.

So in myold's formula I would have (12.25 X 2)/2 = 12.25
Then 12.25 x 9 cells = 110.25 amps max per group

In Bio's formula I would have (12.25 x 2) x 9 = 220.5
Then 220.5 x 0.25 = 55.125 amps max per group

I understand that this unit can take a lot of amps but I really would like to keep it at the safe and running cool levels so I guess if I want this thing to be able to do 100 amps then I will have to up the plate size a bit or add more cells to each group, right?

OK, first off, they're not "terms"... They're terms, the accepted terms, with the definitions all hammered out a while back. Myself and others here learned the hard way that you don't just coin a phrase to describe established scientific concepts without catching hell about it. Welcome to the club. Ultimately we had to admit that it really was easier when everyone used the same word to describe a particular thing, a lot less confusion that way.
You don't need more plates/area at all. With what you have now you could conceivably push more amps than you can make- Then 220.5 x 0.25 = 55.125 amps max per group x 7 groups = 385.875Amps...

[myoldyourgold]

Bio I could be wrong but I thought his groups were in series and current throughout a series circuit stays the same. Did I miss something here.

[lhazleton]

Myself and others here learned the hard way that you don't just coin a phrase to describe established scientific concepts without catching hell about it.

Why Gus, whatever do you mean???:D:D

[keiththevp]

Keith the difference between my calculation and Bio is that mine is stated as the max. The range is .25 to .50. I use heat to determine where to really run it at. In my reactors it is always under .5 amps per active square inch. In fact you should look for the spot where there is the least heat but the most gas per amp. The right amount of heat is a good thing. To much is very BAD.

The 55.125 amps would be the total amps your system would draw using .25 amps per active square inch of area not per group but all seven groups/sets, each with 10 plates.

Now due to media blasted plates all this might change and is requiring a lot of testing which is going on everyday.

I normally media blast my plates, so I am sure that what ever max amps I come up with will be on the low side. But I would prefer to over build rather then under build. I have learned that expensive lesson too many times. This method also helps to keep the heat down which is one of my main concerns.

There is one thing I seemed to be missing. I was under the impression that all 7 groups would get the same amperage or a total of 110 amps. Not 110 amps per group. So I am looking to pull 100 amps total between all 7 groups therefore I can actually consider reducing the number of cells in my groups. Either way it looks like this setup or even one a little smaller should work just fine for my needs.

[H2OPWR]

I agree with what has been posted here. You should be somewhere between 55 and 110 amps. With that said I beleive we will have to rethink max amp draw with a unipolar design. The lack of current leakage should allow you to draw more amps without excess heat. I was never able to feed the Beastie with enough amps to test the theory.

My problem was always that I couldn't get the heat I wanted.

Larry

[keiththevp]

OK, first off, they're not "terms"... They're terms, the accepted terms, with the definitions all hammered out a while back. Myself and others here learned the hard way that you don't just coin a phrase to describe established scientific concepts without catching hell about it. Welcome to the club. Ultimately we had to admit that it really was easier when everyone used the same word to describe a particular thing, a lot less confusion that way.
You don't need more plates/area at all. With what you have now you could conceivably push more amps than you can make- Then 220.5 x 0.25 = 55.125 amps max per group x 7 groups = 385.875Amps...

Bio, I was not trying to start a flame war nor offend you by quoting terms. If it came across that way then I am sorry. I was just trying to rephrase my description in the format you use so that you could understand my question.

The only thing I will say in regards to using the proper terminology is this. What is excepted on this forum as proper terminology is in fact very different from other forums, yahoo groups, and YouTube communities. I agree that having set terms and definitions makes things MUCH easier. But calling it the scientific gospel, this is far from the case I have viewed first hand in the last 4 years of participating in the various HHO communities. Especially the water car group!! So sometimes I find it hard to keep up with the specific terms that have very different definitions used on many different groups.

[BioFarmer93]

Why Gus, whatever do you mean???:D:D

Hey Lee!
Good to hear from you, man... So, how's your ass growing back? Mine is pretty much filled back in on the right, but that left side's gonna be a while...

[BioFarmer93]

Bio, I was not trying to start a flame war nor offend you by quoting terms. If it came across that way then I am sorry. I was just trying to rephrase my description in the format you use so that you could understand my question.

The only thing I will say in regards to using the proper terminology is this. What is excepted on this forum as proper terminology is in fact very different from other forums, yahoo groups, and YouTube communities. I agree that having set terms and definitions makes things MUCH easier. But calling it the scientific gospel, this is far from the case I have viewed first hand in the last 4 years of participating in the various HHO communities. Especially the water car group!! So sometimes I find it hard to keep up with the specific terms that have very different definitions used on many different groups.

It's cool- no flames, no offense taken. However, you did throw me off with this last post- What is excepted on this forum as proper terminology is in fact very different from other forums -I think you mean 'accepted'... Excepted means to exclude or leave out. OR, did you really mean 'excepted'?
As far as it being gospel, well, ah, yeah, you see it actually is... It was delivered unto us from on high by GROD, (Great Researcher Of Definitions) and was substantiated and confirmed by no less than three unimpeachable sources before it was given to the masses... Amen

[myoldyourgold]

Holy Moly not The GROD again. I better watch out or.............:D I needed a good laugh!!

[BioFarmer93]

Bio I could be wrong but I thought his groups were in series and current throughout a series circuit stays the same. Did I miss something here.

Carter,
I don't think you missed anything... I think I did though. All this time I have based my design on total sq. in., not just that of a single group. Is this possibly a mis-application of accepted current density practice for bipolar e-lyzers? Larry, chime in here buddy- Did you happen to notice any non-linear production at higher current densities? I feel like I just fell off of a cliff... I just don't see how it can work like that, I mean, that current is being used, and used hard by a lot of sq. in. of plate area... You aren't going to get the same production from 55A out of 1080 sq. in. that you will from 240 sq. in., the current density will only be 1/7th of the necessary amount, no?

[myoldyourgold]

Gus as I mentioned earler I am in the process of verifying this in an actual test for other reasons but it will also prove this one way or another. I believe that it will follow standard current flow just like in any other circuit. Don't forget I am a bean counter and require a 10 key to do that with any accuracy. LOL


There is a single path for current in a series circuit. The amount of current is determined by the total resistance of the circuit and the applied voltage.

Using the above I see what you mean. Now I need to do some thinking. Larry where are you?

http://www.wisc-online.com/objects/viewobject.aspx?id=dce302

Go to slide 4 and 5 of this presentation.

[H2OPWR]

Guys, This is really quite simple. Add the total available area of all the unipolar plates in each stack then treat the entire calculation like a normal dry cell.

Larry

[BioFarmer93]

Guys, This is really quite simple. Add the total available area of all the unipolar plates in each stack then treat the entire calculation like a normal dry cell.

Larry

Larry,
That means nothing to me, sorry. I have zero background in what you consider to be normal drycells. There are no plates in a unipolar reactor that are being forced to be pos on one side and neg on the other. I have a great deal of difficulty understanding how the same rules can be made to apply. Not trying to be a pain in your ass, but probably am anyway. My eyes are crossing, see you gents in the a.m. Gus

[keiththevp]

Bio,

Yes I had a brain fart or auto correct issue and meant "accepted"

Larry,

I was thinking along the same lines. This size cell can take 55-100 amps not 300-400 amps. But I will just have to build it, test it, and find out for sure.

MyOld,

If you finish your testing before me then let me know what you find. After reviewing your link, I think I just found a possible downside, or situation that can create confusion to the unipolar design. If any one cell or group has a problem it will replicate this problem on all the other groups. Example: If for some reason some of the plates in a single group were damaged and this lowered that group's ability to draw current then all the other groups would suffer as well.

[BioFarmer93]

Guys, This is really quite simple. Add the total available area of all the unipolar plates in each stack then treat the entire calculation like a normal dry cell.

Larry

Larry,
About the 30th or 40th time I read this, I realized that what is being described here IS every plate. Unless of course, you meant to type "in a stack" rather than "in each stack". In which case I'm back to where I started-LOL!

[keiththevp]

I'll know more when I test mine but as far as I understand it the amperage will be the same in every group. So take the total wet area in all the cells in one group and that is your surface area to divide by .25 to .5 amps per sq. in. Hope that helps Bio.

[BioFarmer93]

Kieth,
I'm pretty sure you mean what I think you thought you said, I just don't know if you think that I mean that you thought that I think you meant what you thought you meant, you know?:D

Yes, I understand, just having a little fun.. I am waiting to hear what Carter finds out about current density in a unipolar configuration. I have an idea that the rules are going to undergo a bit of an update.

[myoldyourgold]

I'll know more when I test mine but as far as I understand it the amperage will be the same in every group. So take the total wet area in all the cells in one group and that is your surface area to divide by .25 to .5 amps per sq. in. Hope that helps Bio.

Keith, I am going to suggest one thing and that is instead of wet area you say active area. Many of us have pockets/gaskets on the last plates which gives the end plates two wet surfaces, but because of no other conductive plate to pair with, one side or these two plates are considered inactive for calculation purposes in a unipolar design. Example, if you have 6 plates in a set, there is 12 wet surfaces but for calculation purposes you only have 10 active areas not 12. I think you knew this but state this to make it clear for everyone else. The end plates are wet on both sides in some designs but only have one active side. I will make this comment here, many have seen small amounts of gas being made on the inactive side of the end plates. The only explanation for this is that there is some current leakage even in unipolar designs but is insignificant. These end pockets help in keeping electrolyte levels even through out the reactor and in cooling.

Gus I am working on it but am slow. Should have some results next week if everything goes like it should. It never does though. LOL

[BioFarmer93]

Take your time Carter, It's not like I don't have anything else to do!

[keiththevp]

Keith, I am going to suggest one thing and that is instead of wet area you say active area. Many of us have pockets/gaskets on the last plates which gives the end plates two wet surfaces, but because of no other conductive plate to pair with, one side or these two plates are considered inactive for calculation purposes in a unipolar design. Example, if you have 6 plates in a set, there is 12 wet surfaces but for calculation purposes you only have 10 active areas not 12. I think you knew this but state this to make it clear for everyone else. The end plates are wet on both sides in some designs but only have one active side. I will make this comment here, many have seen small amounts of gas being made on the inactive side of the end plates. The only explanation for this is that there is some current leakage even in unipolar designs but is insignificant. These end pockets help in keeping electrolyte levels even through out the reactor and in cooling.

Gus I am working on it but am slow. Should have some results next week if everything goes like it should. It never does though. LOL

Sometimes I assume that people would already know certain things, but your right my old I should be a bit more detailed for the newbies. Thanks for elaborating because I am too lazy to be a novelist!!!

Bio, good to see we are on the same page. :D