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ShowMeHHO
10-07-2008, 03:33 PM
Can someone please clear this up for me ....

I read this in the forum on the thread about " Surface Area and Voltage "

One stated " You voltage is 13.8 thus you plates are ~1.97v/p"

Then someone posted this " in a 7 plate setup, there are 6 cell, so 13.8/6=2.3

I had been believing the first but figuring at 13.7 volts in my design but then when I read that post I wanted to post here and get my facts straight before proceeding any farther ....

H2OPWR
10-16-2008, 06:58 PM
Can someone please clear this up for me ....

I read this in the forum on the thread about " Surface Area and Voltage "

One stated " You voltage is 13.8 thus you plates are ~1.97v/p"

Then someone posted this " in a 7 plate setup, there are 6 cell, so 13.8/6=2.3

I had been believing the first but figuring at 13.7 volts in my design but then when I read that post I wanted to post here and get my facts straight before proceeding any farther ....

The post references a plate stack that actually started with 8 plates. It was recommended as working better to drop one of the n plates and use 7 plates instead. That is the difference between the two figures. You measure the voltage where you are mounting your cell at the wires where your cell will be hooked to. Divide the number of spaces between your plates by the measured voltage and you have your voltage per cell. I personally did not like the 8 plate stack. I prefer 7 or less. I can use lower concentrations of KOH and not deal with the harsh electrolite and worry less about getting KOH in my engine. It eats aluminum.

dennis13030
10-21-2008, 08:25 PM
Can someone please clear this up for me ....

I read this in the forum on the thread about " Surface Area and Voltage "

One stated " You voltage is 13.8 thus you plates are ~1.97v/p"

Then someone posted this " in a 7 plate setup, there are 6 cell, so 13.8/6=2.3

I had been believing the first but figuring at 13.7 volts in my design but then when I read that post I wanted to post here and get my facts straight before proceeding any farther ....

I think the following is the equation you need to determine the plate to plate voltage.

source voltage(battery or other) = Vs
total number of plates = Np
total number of cells = Nc = Np - 1
plate to plate voltage = Vpp = Vs/Nc

This assumes a design like +NNN-.

Understand that a design like +NNN-NNN+ is actually two electrolyzers in parallel.

Lets solve this design(+NNN-).

Np=5
Vs=13.8V
Nc=4
Vpp=13.8V/4 = 3.45V

It is recommended that Vpp be in the range 1.5V to 2.0V because over 2.0V and there is a good deal of efficiency lost to heating effects.

Painless
10-21-2008, 08:46 PM
Hey there, Dennis.

Been quite a while since I've seen you post. Good to see you back.

Russ.

redDEVIL
05-15-2009, 02:39 PM
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.

Q-Hack!
05-15-2009, 04:19 PM
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.

Not quite...

Remember that you measure voltage across each + and - so in your configuration of +-+-+-+- while you do have 7 cells, each one has 12v across it. It's the neutral or unconnected plates that give you the voltage drop.

+- = 1 cell 12v ... +-+-+-
+|- = 2 cells 6v ... +|-|+|-
+||- = 3 cells 4v ... +||-||+||-
+|||- = 4 cells 3v ... +|||-|||+|||-
+||||- = 5 cells 2.4v ... +||||-||||+||||-
+|||||- = 6 cells 2v ... +|||||-|||||+|||||-

Does this help you understand a little better?

redDEVIL
05-15-2009, 09:07 PM
Not quite...

Remember that you measure voltage across each + and - so in your configuration of +-+-+-+- while you do have 7 cells, each one has 12v across it. It's the neutral or unconnected plates that give you the voltage drop.

+- = 1 cell 12v ... +-+-+-
+|- = 2 cells 6v ... +|-|+|-
+||- = 3 cells 4v ... +||-||+||-
+|||- = 4 cells 3v ... +|||-|||+|||-
+||||- = 5 cells 2.4v ... +||||-||||+||||-
+|||||- = 6 cells 2v ... +|||||-|||||+|||||-

Does this help you understand a little better?

Ok. Now i understand :) So if i have 21 plates, i can arrange it into 7 cells set up :

+ І І І І І І - І І І І І І + І І І І І І -

thus i can get 1.7 voltage drop at each cell. However, based on author Patrick J. Kelly ( A Practical Guide to "free energy" device ), the max voltage dop should be 1.24 Volt. Any excess voltage will go heating the electrolyte. So,1.7 volt shouldnt be good enough.

A few questions more :
1. What about the amp? what amp do you use? Is there a maximum limit for amp?
( I'm gonna make a HHO generator for gas production purpose, so it's not gonna be installed in vehicles )
2. What is the suitable cell set up to get a proper HHO gas composition. I'm sure we prefer the hydrogen in higher quantities rather than oxygen. So to make it that way, we need to add extra negative charged electrode within the cells.
3. Can we store the splitted gas of HHO in a confined space? Will it be recombine and become water again?

Thanks for your kind response :)

mileageseeker
05-15-2009, 11:53 PM
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.

redDEVIL
05-16-2009, 02:16 AM
Thanks for your explanation. It definetely makes me understand now. I can now calculate the gas flowrate i wanna achieve before i go constructing the generator. And how about the gap between each plate? i've read that the closer it is, the higher the gas production will be. Moreover, what about the plate thickness, does it count?
Can i store the HHO gas in a vessel? Will it turn in liquid phase when it's static?

Q-Hack!
05-16-2009, 12:44 PM
Can i store the HHO gas in a vessel? Will it turn in liquid phase when it's static?

Yes you can store HHO gas in a vessel and it won't turn back into water over time. However, BE VERY CAREFUL... HHO doesn't like to be compressed beyond a certain pressure. 400 psi seems to be the point at which it self explodes.

http://forum.beawindhog.com/cgi-bin/forum/Blah.pl?m-1218081580/

redDEVIL
05-17-2009, 02:56 AM
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.

Q-Hack!
05-17-2009, 03:23 AM
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.

sima.z
05-20-2009, 09:20 AM
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

mbjhho
05-21-2009, 09:28 AM
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

mileageseeker
05-21-2009, 10:21 PM
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.

mileageseeker
05-21-2009, 10:56 PM
-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.

redDEVIL
06-01-2009, 10:55 AM
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.

soda_pop503
07-28-2009, 03:38 PM
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.

PeteVamped
07-31-2009, 05:13 PM
Mr reddevil is the man I could not answer this question any better my self great job red devil

redDEVIL
08-01-2009, 11:54 AM
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 :o

redDEVIL
08-01-2009, 12:13 PM
Hi everyone..
I just want to share a lil experiment of mine. Recently i did an experiment using :

(1) 3 cells working in parallel, where the cell's configuration was :
-NNNNNN+ -NNNNNN+ -NNNNNN+

(2) A battery 12 Vdc and 45Ah.

(3) Caustic soda ( NaOH ) as the aditive

(4) At least 3 liters of electrolyte solution ( distilled water and NaOH ).

(5) 3mm gap per plate.

When the process began, i read 15 Amps and the voltage was steady at 12 V. Coz i was working with parallel config, it means that each cell will have 5 Amps ( 15 Amps divided by 3 cells ) and 12 Vdc each cell which means the voltage per plate was 1.7 Vdc. I got 0.1 Liter per minute.

What happened was, the temperature was rising so hot ( untill it reached 70 Celcius or 158 F ). The Current then dropped to 12 A and continued to drop untill 10 A. Still i got o.1 LPM.

The question is, What's wrong with my electrolyzer ? It should produce much HHO. My aim should be getting higher gas production ( untill 2 LPM ) and hopefully i can burn the gas safely without the risk of explosion.

Anyone please help me... :confused:

Q-Hack!
08-01-2009, 09:55 PM
Hi everyone..

The question is, What's wrong with my electrolyzer ? It should produce much HHO. My aim should be getting higher gas production ( untill 2 LPM ) and hopefully i can burn the gas safely without the risk of explosion.

Anyone please help me... :confused:

Can you post a pic of your setup?

Mike waterstar
09-25-2009, 09:04 AM
Can someone please clear this up for me ....

I read this in the forum on the thread about " Surface Area and Voltage "

One stated " You voltage is 13.8 thus you plates are ~1.97v/p"

Then someone posted this " in a 7 plate setup, there are 6 cell, so 13.8/6=2.3

I had been believing the first but figuring at 13.7 volts in my design but then when I read that post I wanted to post here and get my facts straight before proceeding any farther ....

The 6 cells and 2.3 V/Cell is correct. You have a bipolar stack. One anode, 5 BiPolar Plates and one Cathode. (The bipolar plates are an anode on one face and a cathode on the other side.) You have a 6 cell stack so 13.8/6 = 2.3V

Popsie
09-30-2009, 11:14 AM
number of cells divided into voltage= voltage per cell

borgdrone
09-30-2009, 09:38 PM
so is there really any benifit to the Smackbooster wet cell design,
+nnn-nnn+
(actually 6 nuetral plates each side, but connected by jam-nut so they are like one plate?)

I was thinking of making mine smaller, and having only 4 or 5 nuetral plates instead of 6 (12).

jerrygoldsmith
10-01-2009, 12:11 PM
so is there really any benifit to the Smackbooster wet cell design,
+nnn-nnn+
(actually 6 nuetral plates each side, but connected by jam-nut so they are like one plate?)

I was thinking of making mine smaller, and having only 4 or 5 nuetral plates instead of 6 (12).

you'd be dropping more voltage per plate by reducing the number of plates.

I don't think one or two plates would make that huge a difference though? Someone else would know that probably.

altern8energy
10-07-2009, 03:34 PM
Anyone got a formula for figuring the LPM of a 24" x 12" system, with 62
plates running at 120VDC with 15-20A??

Plate material will be HIGH MOLYBDENUM-CHROMIUM-NICKEL for excellent non-corrosive properties.

-Altern8Energy

ttsoares
10-14-2009, 08:23 PM
There are news about titanium coated plates... but for now SS is the best solution.

Roland Jacques
10-15-2009, 09:14 AM
Hi everyone..
I just want to share a lil experiment of mine. Recently i did an experiment using :

"(1) 3 cells working in parallel, where the cell's configuration was :
-NNNNNN+ -NNNNNN+ -NNNNNN+

(2) A battery 12 Vdc and 45Ah.

(3) Caustic soda ( NaOH ) as the aditive

(4) At least 3 liters of electrolyte solution ( distilled water and NaOH ).

(5) 3mm gap per plate.

When the process began, i read 15 Amps and the voltage was steady at 12 V. Coz i was working with parallel config, it means that each cell will have 5 Amps ( 15 Amps divided by 3 cells ) and 12 Vdc each cell which means the voltage per plate was 1.7 Vdc. I got 0.1 Liter per minute.

What happened was, the temperature was rising so hot ( untill it reached 70 Celcius or 158 F ). The Current then dropped to 12 A and continued to drop untill 10 A. Still i got o.1 LPM.

The question is, What's wrong with my electrolyzer ? It should produce much HHO. My aim should be getting higher gas production ( untill 2 LPM ) and hopefully i can burn the gas safely without the risk of explosion.

Anyone please help me... :confused:

"(1) 3 cells working in parallel, where the cell's configuration was :
-NNNNNN+ -NNNNNN+ -NNNNNN+"

Your problem is these +- cant be next to each other

try -NNNNNN+NNNNNN-NNNNNN+

HHO_Control
11-17-2009, 08:58 AM
"(1) 3 cells working in parallel, where the cell's configuration was :
-NNNNNN+ -NNNNNN+ -NNNNNN+"

Your problem is these +- cant be next to each other

try -NNNNNN+NNNNNN-NNNNNN+


I'm thinking he was set up as 3 single cells rather than one cell w/ the +- plates next to each other.

SilverHHO
06-30-2010, 07:31 AM
Hi,

I'm brand new to this forum, found it's simply containing awsome resource !
Your post is pretty useful and you seem to get a great HHO knowledge, I have some questions for you :

I have noticed that when talking about plates configuration every body seem to mention this kind : +nnnn-nnnn+
Personnaly I use this : -nnnn+nnnn-
My question is : Is there any difference beetween those two configs ?

I also need some explanation about CFM. My mother language is not english and I simply can't figure out what you're calling CFM. So what is CFM ?

Where is coming from your 0.5 amps per square inch value, how did you determine this ?

Willing to get answers.

Christian.

fredackermann
07-01-2010, 09:30 PM
Hi,


I also need some explanation about CFM. My mother language is not english and I simply can't figure out what you're calling CFM. So what is CFM ?


Christian.
I too am a Noob, but I can answer at least this question. CFM = Cubic Feet per Minute

:cool:

mike915
07-30-2010, 02:45 PM
Yes you can store HHO gas in a vessel and it won't turn back into water over time. However, BE VERY CAREFUL... HHO doesn't like to be compressed beyond a certain pressure. 400 psi seems to be the point at which it self explodes.

http://forum.beawindhog.com/cgi-bin/forum/Blah.pl?m-1218081580/


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.

So if I run the cell from your example at 120VDC at very low amp what will happen???

badog
08-19-2011, 12:33 PM
So I spreadsheeted the formulas.

I have 11 plates, 10.1X10.1 inches running at 13.5 with .05a per sq in.
The output looks way off....

An using the other formulas for Faraday nothing figures out.
What am I doing wrong?

Anyone?:confused:

amps/sq in volts volts/plate
0.05 13.5 1.35

width height sq/in amps plates cells sq in Watts
10.1 10.1 102.01 51.005 11 1020.1 688.5675
CFM 1.317408145
LPM 37.2826505



+ n n n n n n n n n +

Faradays Law
Volts Amps Seconds Factor Output LPM
13.5 51.005 20 3600 3.825375
V*A*S C16 / D16
Efficiency Equation
28871684.55 41314.05 698.834526

MMW
54.14523704

BioFarmer93
08-22-2011, 06:40 PM
I'm thinking you may may be off by a decimal point.. As in the difference between 0.5A per sq. in., and what you're showing - 0.05A per sq. in., as that would be 1/10th of the usually figured current density...:confused:

Mr_The_Carper
09-24-2011, 05:07 AM
Hi,

I'm brand new to this forum, found it's simply containing awsome resource !
Your post is pretty useful and you seem to get a great HHO knowledge, I have some questions for you :

I have noticed that when talking about plates configuration every body seem to mention this kind : +nnnn-nnnn+
Personnaly I use this : -nnnn+nnnn-
My question is : Is there any difference beetween those two configs ?

I also need some explanation about CFM. My mother language is not english and I simply can't figure out what you're calling CFM. So what is CFM ?

Where is coming from your 0.5 amps per square inch value, how did you determine this ?

Willing to get answers.

Christian.

I use -nnnn+nnnn- too...and 3mm seal between the plates

one hole 6mm in the top,3mm hole down in the corner left,next plate right,and again left..
and so to the end, until the last plate

http://img90.imageshack.us/img90/2448/p2204111937.jpg (http://img90.imageshack.us/i/p2204111937.jpg/)

Uploaded with ImageShack.us (http://imageshack.us)

look hole in the corner...another plate on other corner...understand?