Formular to calculate energy per liter of HHO gas

[hartiberlin]

Yes, Stefan

of course, you are right, So your original proposal was not far off anyhow.

But since the experimenters collect their gas generally through bubblers and measure volume, I would say my formula is more straight forward to use for calculating efficiency.

Hi ?ystein,
but your formular for:
3,52 watt hrs pr liter PURE H2
is not so good,
cause all the people working on this mostly directly
produce mixed H2 and O2 gas, so they do not seperate the gases
and to know,
that they just have to break the barrier of 2.15 Watthours of energy is more easy for them,
cause they then can directly measure their HHO gas (mixed H2 and O2 gas) output.
But anyway, many thanks for putting up the right calculations..

Another item; I get warnings from my security software of "spyware" when I open your website, and it states "ads.adbrite.com/adserver/....."

Anything to warry about ?
regards
?ystein

Well, this is just a different text ad provider, but as it seems, that they
can not provide any ads, I will drop them out again.
Nothing to worry.
Regards, Stefan.

[d3adp00l]

I am having a time trying to convert 2.15 watthours per litre into  XXXX (millilitre/minute)/watt. Does anyone want to give it a go for me. I keep getting numbers that are erroneous.

[hartiberlin]

Again:

Energy input in Watthours = used (Volts x amps)  x used time in seconds / 3600 seconds =
used Watts  x used time in seconds / 3600 seconds

So let your HHO gas generator produce 1 Liter of HHO gas= 1000 milliLiters HHO gas and measure the time
in seconds it needs to produce this amount of gas.

Then note the DC Voltage and DC amps down and multiply it to get the Watts.
( e.g. 12 Volts and 10 amps )
Then multiply with the seconds it needed to produce 1 Liter and divide this by 3600.
Then you have the Watthours of energy it needed to produce 1 Liter of HHO gas.

Now if you get less than 2.15 Watthours you have overunity.
If you have more than 2.15 Watthours you have less than 100 % efficiency and are underunity.


So the above example would be, if it takes 60 seconds to fill
up the 1 Liter bottle of HHO gas with 12 Volts at 10 amps :

12 Volts x 10 amps x 60 seconds / 3600 = 2 Watthours per Liter
and you would be overunity, cause it is less than 2.15 Watthours

Regards, Stefan.

[d3adp00l]

Alright taking the time to figure it, it now looks like 7.75 (ml/min)/watt is the goal. Sorry stephan I don't like the kilowatt measurement, just a personal method, right now my best data is good but not good enough.

[hartiberlin]

Yes, you are right,
7.75 milliLiters of HHO per Wattminute, that means for
1 Watt of input power 1 minute long applied to your electrolyzer.

Example:
If you have a 12 Volts DC battery and you power your electrolyzer with it,
you draw 0.0833 amps from it ( 1 Watt) into your electrolyzer for 1 minute constantly,
you must be able to produce more than 7.75 milliLiters of HHO gas.

If you draw more amps, so you have an input power of e.g. 100 Watts, you
must produce in that minute more than 100 x 7.75 mL= 775 mL= 0.775 Liters of HHO gas.

So to reach 1 Liter of HHO gas, you must not use more than 129 Watts of power
1 minute long.
So if you use a 12 Volts battery, you must use less than 10.75 amps for 1 minute long
to produce 1 Liter of HHO in 60 seconds to be overunity.

Or in other words:
if you apply 10.75 amps at 12 Volts  ( 129 Watts) you must be able
to produce 1 Liter HHO gas in
less than 60 seconds to be overunity !
If you need more than 60 seconds for 1 Liter HHO , then you are underunity....

Hope this helps.
Regards, Stefan.