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Thread: Figuring Voltage per plate ????

  1. #11
    redDEVIL Guest

    Figuring voltage and amps per plate to produce high gas yield

    Quote Originally Posted by mileageseeker View Post
    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.
    Based on your explanation, let me give an example of calculating gas production and please correct me if i'm wrong. Prior to using the formula, there are 2 consideration must be made :
    1. Current density, it should be .5 amps per square inch or less.
    2. Voltage per plate, max is 2 volts

    So, we must get these two things right to get the formula works well.

    Now, let' say i have 12 plates with 10 neutrals between + and -, the configuration should be like this :

    Config 1.

    + I I I I I I I I I I -

    Np ( number of plate ) = 12 plates
    Nc ( number of cells ) = 11 cell
    Vs ( supply voltage ) = 22 volt
    Vpp ( Voltage per plate Vs/Nc ) = 2 volt
    I = 4 A
    Plate area = 7.4 sq.cm

    Cd ( current density ) = 0.540540541 ampere/sq.in
    Q hho = ( 4A x 11 cell x 0.00246 ) + ( 4A x 11 cell x 0.0001229 )
    = 0.1136476 CFM x 28.3
    = 3.21622708 LPM

    Config 2.

    + I I I I I - I I I I I +

    Np ( number of plate ) = 7 plates
    Nc ( number of cells ) = 6 cell
    Vs ( supply voltage ) = 12 volt
    Vpp ( Voltage per plate Vs/Nc ) = 2 volt
    I = 4 A
    Plate area = 7.4 sq.cm

    Cd ( current density ) = 0.540540541 ampere/sq.in
    Q hho = ( 4A x 6 cell x 0.00246 ) + ( 4A x 6 cell x 0.0001229 )
    = 0.0619896 CFM x 28.3
    = 1.75430568 LPM

    Because there are two cell setups, we multiply 1.75430568 LPM by 2, and the total HHO flowrate will be 3.50861136 LPM.

    This is what i learn from your explanation. What i'd like to ask again is, what about the gap between each cell. One says that it should range between 3 or 5 mm, others say differently.

  2. #12
    Join Date
    Jul 2008
    Posts
    627
    Quote Originally Posted by redDEVIL View Post
    This is what i learn from your explanation. What i'd like to ask again is, what about the gap between each cell. One says that it should range between 3 or 5 mm, others say differently.
    The gap is something that seems to have as many variations as there are types of cells. I currently use shower pan liner from Lowe's. It is .040 inches thick or about 1mm. Back when I was using my VSPB, I used zip ties as my spacers which were .045 inches. I suspect that if you use a 3-5 mm gap you will start to see a decrease in efficiency.
    --
    Some days I get the sinking feeling that Orwell was an optimist!

  3. #13
    sima.z Guest

    Cool Got some questions

    Quote Originally Posted by Q-Hack! View Post
    The gap is something that seems to have as many variations as there are types of cells. I currently use shower pan liner from Lowe's. It is .040 inches thick or about 1mm. Back when I was using my VSPB, I used zip ties as my spacers which were .045 inches. I suspect that if you use a 3-5 mm gap you will start to see a decrease in efficiency.
    Hello, Great info, thanks to all, this is my configuration after many trials and errors : Never tried +NNN-NNN+ As I understand the - (Negative) produces hidro: so this my own, perhaps I could maked it more eficient.

    -NNN+NNN- Dry cell
    5 square Inch using (10) 4" o-rings and 1/8 gap, I was using baking soda and geting about 1/2 LPM at 8-10 amps, (moving into KOHas I read on this forum it is the best)
    Up to my understanding this will be 1.5 volt per cell??

    I am in Latin America not in USA and I will appreciate a promt responce.
    Cheers and Thanks
    SIMA.z

  4. #14
    mbjhho Guest

    wiring

    I am wondering what is the most consistant current across the cells.... Do you attach thw pos. wire to all the electrodes or can you attach the pos. wire to one electrode and let it travel across the line of cells
    thx mike

  5. #15
    Current density, it should be .5 amps per square inch or less. If we are calculating current density in sq. in lets measure our plates in inches.

    Q hho = ( 4A x 11 cell x 0.00246 ) + ( 4A x 11 cell x 0.0001229 )
    (4 X 11 X .000246 = .010824) + (4 X 11 X .0001229 = .005676) = 0.0165CFM X 28.3 = .46695 LPM

    Q hho = ( 4A x 6 cell x 0.00246 ) + ( 4A x 6 cell x 0.0001229 ) (4 X 6 X .000246 =.005904) + (4 X 6 .0001229 = .0029496) = .0088536CFM X 28.3 = .25055688 LPM
    Since we have 2 cells we end up with .50111376 LPM

    Plate area = 7.4 sq.cm 1 sq centimeter = 0.15500031 sq in so 7.4 sq. cm = 1.147002294 sq. in. (rather small plate), if we feed it 4 amps we have a current density of 3.487351351 amps per sq. in.

    This is what i learn from your explanation. What i'd like to ask again is, what about the gap between each cell. One says that it should range between 3 or 5 mm, others say differently. All I’m able to say regarding this is that I have run plate gaps ranging from .8 mm to 2.3 mm successfully. It’s a matter of working with what is available to you for spacers or gaskets.
    "EXPERIENCE" it's what you get when you don't get what you want!

  6. #16
    -NNN+NNN- Dry cell
    5 square Inch using (10) 4" o-rings and 1/8 gap, I was using baking soda and geting about 1/2 LPM at 8-10 amps, (moving into KOHas I read on this forum it is the best)
    Up to my understanding this will be 1.5 volt per cell??

    I am in Latin America not in USA and I will appreciate a promt responce.
    Cheers and Thanks
    SIMA.z


    You have 3 bi-polar plates (neutrals) between each charged plate (+ & -) so you have 4 gaps, assuming you’re using 12 volts you wind up with 3 volts per gap, you'll generate a fair amount of heat, you might try upping to 5 neutrals. You need to count the gaps between each set of charged plates (+ & -), not the total gaps. Your LPM number calculates from .334 LPM to .418 LPM so your measurements are close, just a little high. As for +nnn-nnn+ or –nnn+nnn- it really doesn’t matter, when you crack water you get 2h’s and 1o you won’t change that ratio.
    "EXPERIENCE" it's what you get when you don't get what you want!

  7. #17
    redDEVIL Guest
    To mileageseeker : Thanks for the correction There're few typos in the calculation. Such as plate area 7.4 sq.cm, it should be typed 7.4 sq.in

    Again, I've read that any excess of 2 voltage per cell will create nothing that heat production. That's why we put neutrals for voltage drop. What about the current ( ampere ) ? Supposed that i apply current density more than 0.5 ampere/sq.in. What will be the effect ?
    As i see from the formula, the ampere will be equal to gas production, which means if i increase the ampere, gas production will be increased. And when the plate area is fixed, increasing ampere will increase the current density too.

    I really need your guidance for this as i'm very new in this forum.

  8. #18
    Join Date
    Jul 2009
    Posts
    9

    voltage/cell

    Quote Originally Posted by redDEVIL View Post
    Hi Dennis,

    I've just joined this forum and just started to read few posts, including yours. What i'm gonna ask is, what about +-+-+-+- configuration? I've read that putting neutrals will decrease heat and increase gas production.
    If i have the +-+- configuration and let's say i put 20 plates in series ( 12 Volts ), it means that there will be 10 cells in series and the voltage of each cell will be 12/10 = 1.2 volts. Am i correct?
    I'm gonna make a HHO generator for my class project. Your inputs will be much appreciated.
    I am not 100% sure. but I was under the impression that you simply divide your voltage by the number of cavities (or spaces between the plates) regardless of their polarity.

  9. #19

  10. #20
    redDEVIL Guest
    Quote Originally Posted by PeteVamped View Post
    Mr reddevil is the man I could not answer this question any better my self great job red devil
    Hi PeteVamped ..
    I guess i should say thanks.. I still have to learn much from the experts

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