Formular to calculate energy per liter of HHO gas

[CrazyEwok]

lol... you know i hope none of you here with these calculations of thousands of liters per minute actually get a cell or setup that can actually create this... Your gonna blow your entire engine apart... Your calculations are based purly on the petrol numbers... i think you should consider the explosive force and rate at which HHO burns before you start throwing numbers around on what you need to run a car. Lets consider some points first. You will need something in your cylinder besides HHO and atmosphere in order to slow the reaction down (and to cool your exhaust valves) alot of people run steam to do both in high end performance cars. There is now a 3rd item to put into your ratio equation. HHO : Atmosphere : Steam ratio's.
Also being the budding scientists that you are (yes all sarcasm intended) what is the explosive force of HHO compared to petrol??? I hear numbers of around 2.5 times being thrown around (and that is calculated on explosive force) so now to get the same bang per cycle you should be dividing your required amount by 2.5 minimum. Next off the bat is to those of you wanting to run a closed system with HHO... It has been reported that pure HHO when ignited IMPLODES at 4x the power of normal reaction. THATS 4 TIMES THE POWER OF YOUR 2.5 TIMES THE POWER OF PETROL!!! AND IMPLODES i can't see anyone flooding their engine with excessive amounts off straight HHO having it implode mid cycle causing any problems when their cyclinders try to eat their block from the amounts of force they will have to withstand (be aware that engines are designed to run on EXplosive reactions not INplosive) BUT!!! i may be wrong and if you guy do run engines on a closed cycle HHO can you film it and also if/when your engine eats its own cyclinders can you put it on youtube...

Ok so now the sarcasm is out the way some serious thoughts... Off a guestimate from previous experiances i would say you would need a way of distributing your HHO better. From the look of things most peoples cells "might" be able to run an engine (don't get excited) but your delivery system is now where you need to look at things. i would say its now your "air to fuel ratio" that is screwing this up. There is no way you will be able to create enough HHO to feed the vacuum of your air intake to run your car BUT if your not going to use petrol there is now a "free" port that could be used as a controlled delivery system...

[Kator01]

Hi Inventor81,

It was my understanding that the faraday law stated something along the lines of 250KJ per mole of water. This is roughly 250KJ per 1.5 liters of collected gas - i.e. 1 mole of H2 and .5 moles of O2 from one mole of water, or 18 grams

Now if you go back to the hyperphysic-link you can read that
237 kJ are necessaray to create 1 Mol of H2 and 1/2 Mole of Oxygen. All numbers there are based on the SATP-Condition ( Standard Ambient temperature ) that is  298 K which is about 25 Celsius.
1 Mol of hydrogen has a volume of 22,4 Ltr and 1 Mol of Oxygen 11,2 Ltr but these numbers are related to 273 Kelvin or 0 celsius !

So numbers are given in different books based on different conditions which is very confusing.

Now, I am not well versed in thermodynamics and have to research the literature for the correct calculation of the Gas-Volumina of 1 Mol Hydrogen at 25 Celsius.
Based on this :  1 Molvolumen of H - lets make a guess of 28 Ltr â€" will reduce the figures of necessary production-energy for 1 Ltr H-Gas to :
237 Kj / 28 Ltr = 8.46 KJoule per 1 Ltr H-Gas whis is the same as 8460 Watt-sec. 8460 Watt-sec / 3600 sec = 2.35 Watt-Hours.

Now compare this with what Stefan has stated in the first post
in this thread.

Best thing to do a good measurement is to produce a bigger amount of gas ( l would say 150 Ltr HHO ) because the influence of the measurement-failure-rate is bigger with a small volumina produced.

I would produce 150 Ltr of HHO then divide 100 Ltr/ 28 ltr ( 1Mol) = 3.57 Mole and the multiply 237 KJ x 3.57 = 846 KJ.

846 Kj equal 846 000 Wattsec/ 3600 sec = 235 Watt Hours.

With a power-supply of 12 Volt and 10 Ampere this 150 Ltr HHO should be produced in a little less than 2 hours.

You also have to take into account the rising temperature during the process and this is another reason that you should first let the cell run until temperature has leveled and then start Volumen-measurement.

By the way 250 Kj for 1.5 HHO-Gas- as you have stated - is totally wrong.

Regrads

Kator01

[llewgnal]

  Yes the only real way to get a real reading instead of an approximation is to run a real time digitally monitoring program and a lot of sensors.

[newbie123]

Any formula that disregards temperature and pressure will be inaccurate.

In electrolysis,  the change in Gibbs Free Energy (Î"G)  is 237.18 kJ.  This energy  represents the total  electrical input energy required to electrolyze 1 mole of H2O with 100 percent energy efficiency (at 298K with  one atmosphere pressure (or 101.325 kPa))...

1 mole of H2O electrolyzed will produce  one mole of H2 (gas), and a half mole of O2 (gas). 

We can figure out the actual volume with the following:

Using the ideal gas law:  pV = nRT
http://en.wikipedia.org/wiki/Ideal_gas_law

or:     V = (nRT / p)

We know 1 mole of H2 gas has a volume of   24.453 L    (at 298K and 101.325 kPa)      and .5 moles of O2 will have a volume of half of this:  12.226L

The total H2 O2 gas generated would be  36.679 L

So with 237.18 kJ input energy...   or  65.883 Watt-Hours

you should generate (36.68L / 60 Minutes)   .6116 Liters per Minute (LPM)  at 65.883 W

And  generate:

1 LPM for every 107.71 Watts (at 298k / 101.325 kPa)

The MMW would be 9.28



If the temperature of your gas is higher, or the pressure is lower it will appear that you're generating more H2/O2 than you actually are...

Also, the electrolyte temperature will change   Î"G  as well.

[JimH]

Hello... Newbie to the forum... I think    I'm sure I was a member a few years ago but I can't remember my password or user name...    Had a bad accident and it's taken a few years to get back on my feet... fine now though.

I've been using Oystla's formula to calculate for years and it works fine for my general experiments.
I've been using rain water successfully for years without additives... acidic rain, I know...
I've been concentrating on heat reduction... but with reasonable hydrogen production and my latest set-up just might be the ticket... LOL... anyway, it looks good at the minute... I'll be testing it over the next few days to make sure that I haven't messed up.