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Hi,
New to the form and looking to the experts. I am beginning to design my first Electrodes and I am wondering if anyone can tell me if using a plate with a higher surface area would produce more HHO than a flat plate or is the production of HHO really more dependent on the amps and resistance?
I am think about using a plates that are corrugated or honey comb 316 SS.
Thoughts?? :confused:
-Joe
Read this post, it may help.. specifically post #8 "Calculating volume of hho production from a cell design."
http://www.hhoforums.com/showthread.php?t=1625
This articular was excellent in helping me understand how I show approach my design, thank for pointing me in the right direction...
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Default Calculating volume of hho production from a cell design.
It has been deduced from Faraday’s laws that one ampere of current for one hour should produce .0147 cubic feet of hydrogen. (Paraphrased from “The Chemistry and Manufacture of HYDROGEN” by P. Litherland Teed page 131 – LONDON Edward Arnold 1919) this book being obtained from www.knowledgepublications.com. This equates to; amps X .000245 = CFM hydrogen. The equation I received from an electrochemical engineer I’m acquainted with is; amps X .000246 = CFM hydrogen, and amps X .0001229 = CFM oxygen. The accepted unit of measure of gas output that we use for the HHO cells we work with is LPM (liters per minute). To convert our calculated CFM of gas to LPM we multiply by 28.3.
For an example; assume a cell of one anode and one cathode (one gap between them) operating at 12 volts and consuming 12 amperes. Generated hydrogen would be 12amps X .000246 = .002952 CFM, generated oxygen would be 12amps X .0001229 = .0014748 CFM. Add the two together and multiply by 28.3 and we have .1253 LPM or 125.3ml/min HHO. We see that this cell doesn’t produce much HHO and being the plate to plate voltage is 12 we know we have a really good hot water heater.
Practical experience tells us that plate to plate voltage should not be much over 2 to minimize heat gain. To achieve this in our test cell we must add 5 bi-polar (or commonly referred to as neutral) plates for a total of 7 plates having 6 gaps between them. The voltage is now reduced between each plate to 2, 12 volts divided by 6 gaps; however the current remains at 12 amps between each plate. Having 6 gaps at 12 amps each we now plug 72 into our equations; (72 X .000246 = .017712 + 72 X .0001229 = .0088488 = .0265608 X 28.3 = .7516 LPM) three quarters of a liter at 12 amps, not bad and very little heat gain. We can increase our gas volume, along with current consumption, without additional heat gain, by connecting two or more of our seven plate cells electrically in parallel.
When I first came upon these equations I wondered how close they were to the real world. Through empirical testing on the calibrated flow bench, of several different cells, I found that these equations are accurate. Some cells getting closer to calculated output than others, none getting more, due most likely to efficiencies of design.
There are many parameters involved when designing a cell; gas quantity desired, sustained current available, and space required for mounting are primary concerns. We now see how we can calculate gas volume using available current. When designing for space requirements we need to consider how much current will be passing each plate. Heat generation results from a combination of voltage and current. We have seen that voltage can be controlled by the number of plates we use in each cell. We can control current via external means by using a pulse width modulator; there are some very good ones available. However, in the design process of our cell, by juggling the amount of parallel cells, the current to be used and the size of the plates, we are able to get a pretty good handle on the heat gain we will experience. An important consideration is current density on each plate in the cell. A good rule of thumb is to try to achieve a current density of .5 amps per square inch or less.
In our test example above running at 12 amps, in order to achieve our .5 amps /sq. inch, since we have 12 amps flowing through each plate, we’ll need plates equaling 24 sq. inch each. Possibly 3 X 8 inches or maybe 4 X 6 inches. If we have room and can make the plates larger, all the better, it will lower the current density and or allow for the use of more current thus producing more gas.
As we have seen, we are able to closely calculate the expected HHO output of cell designs, albeit there are many factors to consider when starting with a clean sheet of paper.
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All,
I have been working on a Cell Design spreadsheet that would aid in understanding the specifics and the output of a given cell design by plugging in certain specs. This would show the potential HHO output of a cell and allow for changes in the design to be tested before building a cell. I am looking for reviewers to critique the math and the spreadsheet.
Let me know if you are interested.
-Joe