MMW/Efficiency Calculator
A look at Faraday Efficiency:
.627 Liters per hour per amp is representative of 100% Faraday Efficiency at 32oF and 1 atm pressure.
The ratio of .627 liters per hour per amp is the same thing as .627 divided by 60 minutes to get .01045 liters per minute per amp and then, because there are 1000 milliliters in a liter, multiplied by 1000 to get 10.45 milliliters per minute per amp.
So, let's take another short look at how we can get there.
First, we can perform the calculation for the half reaction for Hydrogen in the electrolysis of water to find it's theoretical volume produced per minute per amp.
Electrical Charge in Coulombs (C) = t X I (60 seconds X 1 Amp) = 60 C
The volume of Hydrogen, or any gas for that matter, per mole is a given value. At standard pressure and temperature, the volume of Hydrogen per mole is 22.414 Liters or 22,414 Milliliters which, by the way, is the Ideal Gas Constant (0.0820574587) multiplied by the Standard Temperature in Kelvins (273.15 K which is equal to 0 C or 32 F). Also, this is the point in the calculation where temperature corrections are made to adjust the volume per mole.
For example, many people will use what is commonly referred to as "room temperature" (25 C = 77 F = 298 K) to make these calculations which makes the volume of gas per mole = Ideal Gas Constant (0.0820574587) multiplied by room temperature in Kelvins (298 K) = 24.4531226926 Liters or 24,453 Milliliters per mole. In order to make this more clear, I will carry out this example throughout these calculations.
Anyway,
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 C (1 Faraday). Since we have two moles of electrons, the electrical charge delivered by one mole of Hydrogen = 2 X 96,485 C or 192,970 C.
Hydrogen volume = Electrical charge in Coulombs (60 C) / (divided by) Electrical charge delivered by one mole of Hydrogen (192,970 C) X (multiplied by) Hydrogen Volume per mole (22,414 milliliters or 24,453 milliliters at room temperature ) =
60 C / 192,970 C = .000311
.000311 X 22,414 = 6.97 milliliters
Or, at room temperature (298 K)
.000311 X 24,453 = 7.60 milliliters
Hydroxy contains 100% more Hydrogen than Oxygen, so we need to add 50% of 6.29 which brings us up to 10.45 milliliters. Okay, let's 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.
60 C / 385,940 C = 0.000155
0.000155 X 22,414 = 3.48 milliliters of Oxygen
Or, at room temperature (298 K):
.000155 X 24,453 = 3.80 milliliters
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 K):
7.60 milliliters Hydrogen + 3.80 milliliters Oxygen = 11.4 milliliters of Hydroxy per minute per amp per cell.
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 that to atmospheres by multiplying the value given in millibars or hectopascals by .0009869233
That's the nub of it!
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re MMW calc
Im curious as to what you're using as the Faraday ref, I was under the impression that it was 7 MMW.
11.4 ish
I think he is probably using 11.4 because of this: http://aquauto.com/content/help-pointing-out-scammers#comment-1048
Re-calculated
This time I keep coming up with 6.27 M/M/W as 100% efficient @ STP. I'm not sure how I kept getting 11.4 last time. I should have written it down.
I suppose It would be good if I posted a step by step breakdown of how I am getting this number so everyone can help me check the steps.
MMW Calculator
Hi Nick;
A step by step breakdown would be a big help.
Suggestion:
The calculator would have far more utility if it were able to solve for Milliliters when Voltage, Current and Seconds are given.
Moved
To save people some reading time, I've moved all the information on how I arrived at what I believe is the correct efficiency factor to the opening post.
Re: Suggestion
Suggestion:
The calculator would have far more utility if it were able to solve for Milliliters when Voltage, Current and Seconds are given.
I thought about this and decided against it.
It would be easy enough to make a calculator that would be predictive for electrolyzers that operate at 100% efficiency at all current densities, but that wouldn't be representative of the real world.
I could make a calculator where a person could enter their own efficiency factor, but efficiency varies in a non-linear fashion with current density in various electrolyzer configurations and I cant think of a good way to overcome that obstacle. Just "playing" with the numbers in the above calculator would have the same effect your looking for, I think.
Edit post
Good idea, I think you should do that for your [blogs/nickstone/rally-el-reno-map-and-info|rally post] also. Just make a bolded "Update 11-9-07" and enter whatever new info you have below it. That way there is a central place to get updated.
Done deal.
Good idea. Consider it a done deal, my friend. All future updates will just be added to that opening post.
Built In?
It'd be cool to build that calculator into the add vehicle or electrolyzer features, so we automatically see an efficiency rating for whichever boosters get uploaded. Might (maybe) see some correlation between mpg gains (or lack thereof) and booster efficiency.
Good Idea
I had thought of that too and will implement that when I redo the way we store electrolyzer data at aquauto. I have a lot of things I want to change but am waiting to do them until I upgrade the cms behind aquauto (which hopefully I will be able to do in the next month or so).
Temperature
Nick,
You stated you used "standard pressure and temperature" for your values, but which STP? The standards differ depending on who you talk too.
I just want to verify.
The widely accepted values
Repeated experiments show that at standard temperature (273 K) and pressure (1 atm or 101325 N/m2 ), one mole (n = 1) of gas occupies a volume of 22,414 milliliters (This is the relevant value). It would take extreme deviations from these values to make a difference of more than one or two-ten thousandths of one liter in my calculations.
Just to clarify, high temperatures coupled with low pressures such as we might find in the pertinent, boosting with HHO applications, would make some, but very little difference in the numbers. It is when you have low temperatures coupled with high pressures, such as you might find in some industrial applications, that the numbers start to make a significant difference.
Of course, with high enough temperatures and low enough pressures, you begin to have tangential issues with water vapor, but that's a different animal that can be tamed.
I've had a realization
While 10.45 is the correct conversion factor per watt, it doesn't, by itself, reflect 100% Faraday efficiency for multiple cells at higher amps. So, I reworked the calculator to take those thing into consideration. It now has a predictive feature that I think Walt might appreciate.
Please let me know if you find any corrections that I need to make. I've included the code at the bottom of the post, for everyone to scrutinize and/or for anyone to copy and paste to their site if they want.
Aquauto moves up to #1
It was interesting to see that Aquauto has made it to the top position of Google's search engine results for "MMW calculator"
Cool
Hey, thats pretty cool. Granted we are only competing with 14,000 other though... rather than millions
.
Evolving Calculator
Over the last couple of days, the calculator has evolved so fast, the explanation of how I arrived at the conversion factors I was using became obsolete. Not because they were incorrect, but because with the addition of greater functionality to the calculator, they became ever more intricate and esoteric.
Also, on the advice of d3, I've incorporated a way to accommodate dry cells with various numbers of cell stacks and still get correct MMW and Faraday Efficiency comparison.
Now, that I've spent way more time on this calculator than I ever intended to, I believe it is the best, most comprehensive, most accurate and most useful calculator, to the booster community, on the internet...bar none.
What I need is feedback to find out how difficult it is to figure out how to use it correctly, so please let me know if you're not getting the results you expected and why you expected something different.
sample calculation
With all that said... I still liked how you had a sample calculation posted. Do you think you could put that back up?
Anything for you, my friend!
It was not only a sample calculation, it was also a proof of Faraday's calculations...a subtle distinguishing aspect that was probably lost on many people, but yeah, I'll put it back when you replace that sterile, nondescript, little white Aquauto lettering in the upper left corner of the home page with something...ANYTHING that doesn't scream we totally lack any creativity around here...until you find a logo you like. LOL! Just kidding...I'll put it back regardless, but wouldn't this do, temporarily, for now?
Forced Honesty
The one thing I'm discovering about this calculator, since I incorporated the gas law constants into the code is that it is very sensitive. A 10th of an amp, 1 or 2 degrees of temperature, just a couple of seconds, or a few milliliters in the input fields can make a big difference. It won't let you fudge the numbers even a little bit. If you are getting results that indicate more than 100% Faraday Efficiency, it is either because your instrumentation is not properly calibrated, you have made a mistake, you are flat out lying OR you have found the holy grail of electrolysis (and that is unlikely). The most likely thing is that your measurements, for whatever reason, are not accurate.
I am working on designing a eudiometer that will automatically start a timer the moment it begins to move and automatically stops the timer when the tube is lifted to a certain height that corresponds to a particular volume of gas. There are a couple of ways to go about this, but If I can get it to work correctly and consistently, it will remove the my eager human nature from the stop watch.
Barometric Pressure
Well, as I've said before, I don't think temperature and pressure corrections are necessary for an MMW/Faraday Efficiency caclulator, but I've had conversations with some people who seem to think are and since I've already gone this far, I decided to go ahead and incorporate corrections for pressure and get it over with. So for all you nit pickers out there, enjoy.
Nick
version 6.2.4 MMW Calc
Hi Nick, Thanks for inviting me to join this forum. As you know I also have a spreadsheet to calculate MMW and efficiency. Well I can’t really call it mine because so many other people have made contributions to the sheet. Luther, Greg and D3 have contributed heavily. I’m just here to take the flack and keep it updated.
I’d simply like to share it. The math for MMW efficiency is all there. Step by step with sources for every Equation, Law and Constant. Follow Greg’s Solution and if it makes sense feel free to believe. I believe the math solution is solid and speaks for itself.
http://www.4shared.com/u/zvmszps/94d1bc4f/Bob_Campbell.html
Updates happen almost daily so keep checking for the most recent version. It's come a long way in just a few weeks.
Bob Campbell
Glad to see you could make
Glad to see you could make it, Bob
We're all on the same team and working toward the same goal. The whole point of communities like this one is to pool our knowledge and hope that, collectively, we can learn and generate new insights together that will benefit our personal financial well being, the hydroxy community at large, and ultimately the planet.
I don't see it as a competition. I believe that we can all learn from each other. Even of it's what not to do, it's still a lesson learned. At least we're doing something. The work you have done might completely change my perspective and send me back to the drawing board and that's okay!
You can upload your file here if, when you are signed in, you "click" on the "Create Content" link on the upper left side and then on the "Upload a File" link. I started to upload it for you, but then I remembered that only I would be able to edit it that way. If you upload it, you can edit it as often as you need to.
Again, welcome
Nick
upload your file here
Thanks Nick,
This spreadsheet is updated quite often. If I start spreading it around, keeping the latest version available would become too complicated.
There is no membership required to download from http://www.4shared.com/u/zvmszps/94d1bc4f/Bob_Campbell.html so the latest version is easily available to everyone.
There are a lot of people using old versions that don't seem to know about the corrections and additions we have made. For this reason I’ve considered making sign in a requirement, but I figure it’s better to keep it quick & simple.
.
You obviously have a strong background in science, math, and java script so I would like your feedback on Greg's solution. I've been watching so many attempts to solve for Ideal MMW. This is the first solution that is completely verifiable and backed up by Laws of science not just assumptions. Laws of science also back up your calculator. It's a really useful tool but it is not relevant to Watts. D3 added the same math that you use. It's very solid and important information, so both efficiency formulas are used in our spreadsheet.
I'm somewhat amused by the modifications this spreadsheet has gone through. It started out as a simple volume adjustment relative to the output temperature of the gas measured. Boy! we have come a long way in a matter of weeks!
So I'd like to hear your response to Greg's solution. Half of the solution is Faraday, which you are already familiar with. You may add it to your calculator if you wish, it's all OPENSOURCE. That the way we all need to keep our research.
Bob
Subscribe notified
There are a lot of people using old versions that don't seem to know about the corrections and additions we have made. For this reason I’ve considered making sign in a requirement, but I figure it’s better to keep it quick & simple.
Bob, if you hosted your file on aquauto.com, users that could "subscribe" to the uploaded file so when its updated they would be notified. They can do this in two ways, First would be to be an aquauto member and just click the subscribe button (this would require users to be members of aquauto). Secondly, anonymous users or registered users can subscribe via RSS to just that file post. Just thought I'd let you know.
I'm not familiar with the
I'm not familiar with the concept of "Ideal MMW". Could you explain how this is different from regular MMW and could you post the relevant math for how it is derived here? It would be nice if I could just scroll up instead of going back and forth off-site to reference it. It would also make it easier for the general readership to follow.
Ideal MMW
Would "No Loss MMW" be a better name for it? It is the Holy Grail. The highly sought after efficiency of H H O. The one, the highest, and most noble of figures too be praised and adorned by all who seek to build an H H O electrolyzer.
What's up with that? You can't say H H O here? Should my name HHO4gas be HYDROXY4gas. I don't think so. Yeah, I heard some guy owns the acronym. Screw him if he doesn't like people using the term. Kinda like trying to put a patent on electrolysis.
Are you asking me to duplicate Greg's Solution from the spreadsheet here in your Blog? It's already on Greg's blog but I think looking at it on the spreadsheet is a lot easier.
By the way the spreadsheet offers an explanation of Faraday calculations too.
Bob
HHO4gas wrote:Would "No Loss
After giving it some thought...
and looking over Bob's spreadsheet and his friend's blog, I've come to the conclusion that what Bob means by "Ideal MMW" is what an electrolyzer would produce if it were consuming power and producing gas in a balanced, 100% efficient maner. If this is indeed the case, then "Ideal MMW" is a perfectly good name for it.
It's a relatively simple thing to calculate and the way to do it has been known for a long time. Had I known it was a highly sought after object that was worthy of praise and adornment, I would have incorporated it into the first version of my calculator.
Now that I have learned that it is comparable to the "Holy Grail", I have added it to my calculator along with a Power Efficiency percentage, but I'm going to keep the code for it stored away, for a bit, while I tripple check my math and put it through it's paces and confirm that it is working correctly.
By the way, it does not use the calculations from Bob's spreadsheet or his friend's blog.
Power efficiency
The calculator now gives 3 resulting measures of efficiency (MMW, Faraday Efficiency and Power Efficiency). The method used to calculate the power efficiency is based on the same methods used by Bob Boyce and Tero Ranta. Also, the calculator, for now, has been moved to a secure socket layer server and is only available here at Aquauto. The source code will be made available to anyone who requests it through submitting their name and e-mail address, which will be encrypted by that server during transmission, via the comment box below the calculator.
Solution to Power Efficiency
Nick,
We all seem to agree on the math for Faraday's equations.
I'd like to see the math behind the answer for Power efficiency.
I watched you and D3 discuss the calculations for Faraday, how long have you been able to post on the EBN forum? I ask because I have registered twice over the past two weeks and still have not received access.
Bob Campbell
btcamp@sbcglobal.net
comment box below the calculator
Sorry I don't see a comment box below the calculator.
I posted my request in a comment and left my e-mail address earlier today in this thread. Did I do it wrong? I have not received anything yet. I'm looking forward to seeing your math and your solution.
I had hoped you would look at Greg's solution and give me some feed back. It seemed to me that you had only glanced at it when you asked what I meant by "Ideal MMW" But I'm guessing that you did not want to offer your comment because I was unwilling to duplicate my work here on your site. I feel the spreadsheet offers a better format for expressing the math than a lot of words and numbers in this thread.
I am really feeling at odds with you. I gave you my phone number and e-mail address and asked for your comments but you continually wanted to steer everything to your blog. I found that controlling. If you would like to start again on a more friendly note I am willing to do that if you stop trying to be in control.
Changing H H O to HHO was and still is insulting. I don't appreciate you editing my comments. KOH is also a chemical abbreviation, why do you allow KOH and not H H O?
Bob Campbell
Sorry you feel that way, Bob.
By the way, I do not have administrative privileges nor do I have the capability to edit or alter anyones posts in any way.
So where is the math?
Show me the math. I just checked my inbox and it's not there.
You and your narcissistic
You and your narcissistic attitude of entitlement annoys me, Bob. You are not entitled to anything and I am not obligated to provide you with anything.
Dont take it personally
Dont take it personally HHO4gas, its not about you, nor is it about freedom of speech, but rather about the Aquauto site. Said owner of the trademarked term referring to Hydroxy isn't interested in going after the the individual but rather the webowner of the site where said trademarked term appears.
Personally I fail to see how something so insignificant can be insulting, but really, how hard can it be to refrain from using the term?
I think the tiny bit of effort is worth the international interchange of info that takes place on sites like these.
On another note, if you had used the scroll bar on the calculator (not browser) you'd have spotted the comment box at the bottom, although by now its probably occurred to you.
Incorrect request to userForms
On another note, if you had used the scroll bar on the calculator (not browser) you'd have spotted the comment box at the bottom, although by now its probably occurred to you.
Thank you for your help. It is obviously not the root of what is upsetting to me. It simply appeared to be one more effort on Nicks part to control his audience.
I tried the comment box below the calculator and get an error message
"There was an error while processing your form input:
Incorrect request to userForms
Please go back and try again. "
Let's be friends
Nick
I’d like to start over. I’ve spent the day thinking negative thoughts about you, and I don’t want to be in that space. I’ll bet it no picnic for you either. We were on a bad start before I came here and there has been something about you that rubs me the wrong way, but in retrospect it’s probably not anything you did on purpose. I’ll pledge to you right now in public that I will no longer try to annoy you.
I’d rather we be on the same side working together to find solutions.
I’d like to be your friend, and I’d like you to publish your math. I’d also appreciate your input on Greg’s solution. We are arriving at different results, and I believe it’s in the best interest of our common cause to find out why.
Let’s work together,
Bob
I tired to make it clear
I tired to make it clear twice before, but let me restate my position one more time just to make sure it is crystal clear to everyone reading this.
My intent, all along, was to have an open and frank disscussion of the math involved.
I asked you to present your understanding of the math here for one reason only; so people wouldn't have to bounce back and forth form this site to an off site reference in order to be able to follow the logical lines of reasoning presented during the course of the discussion. Plain and simple. Cut and dry. There was no other hidden agenda or ulterior motive. Even though you have continually accused me of it, I have no desire to control you or anyone else. Period. Case closed.
For some reason, you were, and from what can I see, still are, reluctant to personally present your understanding of the math you used in your spreadsheet. Apparantly, you think it is good enough to post links to off site references that you claim are self explanatory.
That is tantamount to me posting an off site link to the following explanation of the math I used then simply saying it is speaks for itself:
Where
Q = quantity of electricity or charge in coulombs (C)
I = current in amps (A)
t = time (seconds)
E = electrical energy in joules (J)
J = watt-seconds
V = voltage (or EMF) in volts
F = 96,484 C mol-1
n(e) = moles of electrons
then,
I = Q ÷ t
t = Q ÷ I
Q = I x t
Q = n(e) x F
E = Q x V
n(e) = Q ÷ F
There you have it..."The Holy Grail..."
Bob, having the sort of conversation we've been having is totally unnescessary and an exasperating waste of time. It seems you prefer to spend more time talking about your feelings rather than anything substantive.
Well, I'm finished. I would rather be poked in the eye with a sharp stick than continue this conversation. You can carry on, if you like, but it will be without my participation. I'm done.
Current Efficiency = Faraday Efficiency
Hi Nick, I think HHO4gas just wanted to see the internal formulas/code of your efficiency calculator.
Anyway, there seems to be a lot of confusion out there about electrolysis efficiency with regard to Faraday. Remember: Faraday efficiency only deals with "amps" (coulombs) or quantities of electrons, more specifically moles of electrons, and a volume of gas.not voltage, Joules, Watts, or MMW .
According to Faraday law:
4 moles of electrons are required to create 2 moles of H2 gas, and 1 mole of O2 gas (this might be looked at as: 2 electrons are required to break both bonds within a H2O molecule) at 100 percent efficiency.
So just with this piece of information you can determine what 100 percent Faraday efficiency really is:
1.) First determine how many electrons are in 4 moles: 4 * Avagadros_number = 2.4088567e24 (entities / electrons)
2.) Determine how many coulombs are in 2.4088567e24 electrons: 2.4088567e24 / 1 Coulomb (quantity of electrons) = 385941 Coulombs
3.) Convert 385941 Coulombs to Amp Hours (aH): (Coulombs / 3600 Seconds) = Amp Hours ..... (385941 Coulombs) / (3600 Seconds) = 107.25 Amp Hours
... So now we know 107.25 Amp Hours will create 1 mole of O2 gas and 2 moles of H2 gas, from 2 moles of H2O, at 100 percent Faraday Efficiency.
... Next figure out how much volume is in 2 moles of H2 and 1 moles of O2 gas. Gas temperature is an important factor for this.
4.) According to the ideal gas law 1 mole of O2 or H2 will have a volume of 22.4 L (at a temperature of 0C), but since gas expands at higher temperatures ... It is:: 24.446 Liters of gas per mole at 25C. So, 3 moles of HHO gas at 25C has a volume of 73.338 Liters.
5.) At 100 percent efficiency: 107.25 Amps will create 1.2223 Liters per minute (LPM)
Or.. 87.74 Amps are required to make 1 LPM at 25C. MMW, Watts, and Voltage have nothing to do with Faraday efficiency.
Or.. 1 amp will create 0.0113967 LPM at 100 percent efficiency.
If you want to figure out electrolysis energy efficiency with Watts and Voltage, you then need to look at the "Gibbs energy of formation" of water, which has a value of 237.18 kilo Joules per mole (at 25C) . For 2 moles of water at 25C: (2 * 237.18 kJ) = 474.36 kJ of energy is required to make the above volume of gas (73.338 L).
Convert Joules to Watts: 474360 Joules / 3600 seconds = 131.7666 W
With this number we can now determine the voltage (and MMW or W/LPM) at 100 percent efficiency: 107.25 Amps * X Volts = 131.7666 Watts
Volts = (131.766 Watts) / (107.25 Amps)
Volts = 1.23 V (voltage required at 100 percent efficiency, at 25C)
You can also convert this to MMW, and W/LPM , which comes out to: 107.73 W/LPM and 9.28 MMW (100 percent energy efficiency at 25C)
Thanks, nvtowing.
I believe this is essentially correct, but, as it turns out, with a potential difference of 1.23 V, the reaction will not proceed. To be fair, when making efficiency comparisons, my calculator incorporates an added overpotential of 0.6 V bringing the voltage up to 1.83 V
Where TV = the theoretical volume produced, my calculator makes the comparison like this:
Potential Efficiency = (1.83*amperes)/TV/minutes
User's Efficiency = ((voltage/cells)*amperes)/milliliters/minutes
User's Percentage of Potential Efficiency = (Potential Efficiency/User's Efficiency)*100
As you can see, as the user adds neutral plates (cells), the efficiency increases.
Minimum voltage
This is true for the average cell plate gap, and ohmic resistance. But if you could (in theory) get a plate gap close to the width of a water molecule, then 1.23 V would disassociate the water at 100 percent efficiency.
There are also differing
There are also differing reports on the minimum amount of overpotential required, but I've tried to build my calculator to be useful to the average DIY researcher and their real world electrolyzers.
Neutral Plates
I haven't looked at this too closely, but there seems to be an efficiency "illusion" when using neutral electrodes/plates and Faraday efficiency. According to Faraday law, a certain amount of current into a cell will create a certain amount of gas. But with neutral plates, you'll see (For example) with 2 amps in, at 24 Volts, much more gas output per amp (due to neutral plates/cells) compared to a cell wired in parallel at 2V running at 24A.... Same watts in, and similar output, but a big difference in their Amp_input to gas_out ratios. This can create an illusion of over Faraday output, or increased Faraday efficiency, if you don't consider the current passing through each neutral cell.
Yes, you do need to consider
Yes, you do need to consider the current passing through each neutral cell. oops...edit...reconsidering the rest of my response. I'm not sure I follow your example.
netural plates and faraday efficiency
Hi Nick, Here's what I was meaning to say:
A "sealed series" cell running at 12V and 2A, with 6 neutrals, like this: + n n n n n - Producing .25 LPM.
Might appear to have a higher Faraday efficiency than a parallel cell running at 2V and 12A .... wired like this: + - + - + - + - Producing .25 LPM
Someone might look at the "series cell" and think they're only putting 2 Amps into the cell (and getting a good Faraday/Current Efficiency), when they're actually putting 2 Amps of current through 6 separate cells, which would total 12 Amps of current.
The parallel cell doesn't have this problem, no "illusion" there.
nvtowing wrote:Someone might
Huh? They would be right. The amps do not add up from cell to cell in a series. It is the same two amps flowing through the entire system. The two systems, as you have explained them, if they each consist of 6 cells, would produce exactly the same volume and would consume exactly the same amount of energy (24 watts). There are, however, other advantages to the series of cells configuration. In the 6 cell parallel system drawing a total of 12 amps, each cell would only be drawing 2 amps.
Series cells:
Itotal = Icell 1 = Icell 2 = Icell 3 = Icell 4 = Icell 5 = Icell 6
Parallel cells:
Itotal = Icell 1 + Icell 2 + Icell 3 + Icell 4 + Icell 5 + Icell 6
neutral plate cells
I think you are over analyzing what I'm saying.
From the above example (12V 2A), they are only putting 2 amps of current into the whole electrolyser system. But they have 2 amps going through each individual cell, like you said.
So what you'd actually have are 6 individual cells running at 2 volts, with 2 amps of current for each cells (so the total current is 12 amps going to all 6 cells). The neutral plates just chop the voltage down to 2V ... So you have 6 cells * 2 volts * 2 amps ... still 24 watts. It is not divided up like this: 2 amps into the electrolyzer, so you wouldn't have (2 amps) / (6 cells) of current in all the cells at one time, which would come out to: 6 cells * .333 amps * 2 Volts ....4 watts. The origial name of these "series" cells is really confusing and should just be "neutral plate cell".
I agree. It can be
I agree. It can be confusing, but if our goal is to disseminate correct information, it is just as misleading to give the impression that to find the total current in a series circuit you need to multiply the current by the number of cells as it is to give the impression that you can divide the current by the number of cells. Both of these misunderstanding are equally wrong. I'm not over analyzing it. It is simply wrong to say that "the total current is 12 amps going to all 6 cells". The total current going to all 6 cells in a series of cells is exactly 2 amps, no more, no less. It is not multiplied, divided, added or subtracted in any way.
The task of explaining why that is true might be more difficult, but that is what needs to take place if we are going to purport to be giving correct information.
Personally, I think it should be viewed as a "series of resistors circuit", since the plates are not really neutral at all.
current through cells
Amps or current is just coulombs of electrons moving past a point in a certain amount of time, so you can say 2 amps going through each cell every x seconds or 2 amps going into the whole electrolyzer in x seconds, it doesn't really matter..
But saying the more cells you have in series, the more energy/Faraday efficiency you'll have is truely misleading.
What are you talking about?
A single electrolyzer unit with extra resistor plates is clearly more efficient than a single electrolyzer unit without them.