Formular to calculate energy per liter of HHO gas

[d3adp00l]

The amount of hho to run a vehicle has not been tested in a fashion the tells anything. No one to my knowledge has ran a car on hho and the published the results. So no know one knows if an engine need to consume 100% hho or if it an be mixed with air and still get enough power out of it to run the vehicle. The sad state of affairs with accurate data on hho is that the best info on burnable ratios comes from youtube and he was useing spice jars, and didn't seal it very well. And the test was not done under pressure as a piston in an engine would be. I am working in that direction, but first thing is first and thats to quantify the basics.

[sm0ky2]

WHEN TESTING THE EFFICIENCY OF AN ELECTROLYZER::::::::

The above posted calculations for calculating the efficiency are designed for electrolysis of Pure Water.
Contaminants (even electrolytes) effect this calculation.

Even if you plan to USE electrolytes in your device, efficiency tests should be done with Pure Water, as it s difficult to know the precise energy values of the electrolyte, and dissolution %, ect.

if you use an electrolyte during the efficiency testing, the above calculations may show over unity, but incorrectly so.

[sm0ky2]

recent electric vehicles were reported to consume 200A @ 36v in order to travel 45 MPH.

so     Given 7744 Joules / (200x36) = roughly 1.08 seconds of motor run-time from 1 liter of gas?
that is assuming your fuelcell is converting H2+O into electricity+water at 100% efficiency.

[Omega_0]

As we all know, calculating/measuring the INPUT power is pretty easy. Just hook an ammeter and  voltmeter right after the battery.
OUTPUT measurement is more tricky and whatever was posted above will only give you theoretical output value. It is not necessary that you'd be able to use all that energy.

The only way to get its value is to BURN the gas (thats collected over a specified time (say 1 min)) and let it heat a known amount of water in a calorimeter. The rise in temperature of water will give you the useful energy produced. Of course it means that you need access to a precise calorimeter or a lab.

Another method is to drive a tabletop IC engine and connect a known load on its shaft. Use up the collected gas. The work done on the load is your actual output.

Another method is to use the heat generated by the gas to heat a thermoelectric junction and discharge it through a known resistor and measure the current as usual.

HHO just sitting there in a jar is totally useless. You must get the work out of it, that work is your actual output. Now you can happily divide it by the input and check whether its gives an USEFUL overunity.

I've seen many projects on HHO, that yell OU, but I've never seen anyone do a serious measurement. Whats with that?

[nedh20power]

Hello Stefan,

I'd like to make some clarification regarding your calculations of overunity for electrolysis.

1. Because a liter is a unit of volume, the mass of HHO gas in a liter will vary with ambient pressure and temperature according to the Ideal Gas Law, i.e.  PV = nRT.  Since energy release from burning HHO depends on mass rather than volume of the gas, it's better to assess overunity based on HHO mass produced by electrolysis.

2. Taking into account the rate of energy release from burning hydrogen, which is 142.35 kJ/g H, one can derive the following formula for estimating the expected energy release (Eh, kJ) from burning HHO gas produced via electrolysis of W grams of water:

Eh = 15.817*W   (in kJ)

This formula assumes that the HHO gas is pure and contains no other gases in its volume such as water vapor or air. So, for example, the electrolysis of 1 kg of water (i.e. 1000 g H2O) will produce HHO gas, which after burning will release 15.817*1000 = 15817 kJ of energy.

3. To assess the efficiency of electrolysis and whether it's overunity, one needs to calculate the electric energy (Ee, kJ) consumed to split W mass of water. This can be done using the formula:

Ee = 0.001*V*A*T  (in kJ)

where V is average voltage used (Volts), A is the average current consumed (Amps), and T is the time (in seconds) needed to split W mass of water. The constant 0.001 serves to convert Joules into kilojoules, so that Ee can be compared to Eh.

The efficiency of electrolysis (Eff, %) can then be computed as:

Eff = 100*(Eh/Ee)  (%)

Efficiency over 100% will indicate overunity.

Thank you,
- Ned Nikolov