Formular to calculate energy per liter of HHO gas

[oystla]

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.

OK;

1. Most experimenters I've seen measure the actual gas volume production.

So therefore we should of course propose a simple formula they may use to calculate the efficiency based on actual gas volume produced. Gas volume will here mean the O2 / H2 mixture together with some water vapor.

2. The formula I propose and which is accurate to wathever degree is required here is:

EFFICIENCY (%) = 100 * 7744 * V /(U*I*t)  where

V= Volume H2/O2 gas produced in liters
U = Voltage
I= Amperage
t= Test period in seconds

The above formula is calculated from ideal gas law and Daltons law.

Assumptions included in the formula
- Gas collected holds 20 degC. However, If it is instead 10 degC or 30 degC, it will only have marginal consequences.
- Gas collected holds 1 atmospheric pressure, which is very close to the truth in most cases.
- The collected gas is saturated with water vapor, since most experimenters use a bubbler to collect the gas.

And yes, Higher heating value of Hydrogen is 141,9 KJ/g.

So forget about mass basis calculations. My formula is fit for purpose and is more than accurate enough for any experimenters needs.

regards
?ystein

[shimondoodkin]

there might be an overunity the other way.
lol

for example you make hydrogen and oxygen from water then on the higher place you combine air oxygen with extracted hydrogen.
the extracted oxygen go to the floor and the oxygen from the ceiling is burned with the extracted hydrogen.

it is like moving oxygen up without moving it.
i guess such process should take heat from surrounding

[Tacmatricx]

In that case... Does this look right?

In a collected volume of O2/H2 gas the amount of hydrogen energy will be 7744 Joules pr liter collected gas.

1L = 7744J

1000cm3 = 7744J

1cm3 = 7.744J

1J = 1 / 7.744 = 0.1291cm3

H2 + O2 gas at normal atmospheric pressure has 7.744J/cm3 or 0.1291cm3/J



The Average small home uses 1000kWh per month

1kWh = 3,600,000J

1000kWh * 12 = 12,000kWh per year

12,000kWh / 52 weeks = 230.7692kWh per week

230.7692kWh per week / 7 days = 32.9670kWh per day

32.9670kWh / 24 = 1.3736kWh per hour

1.3736kWh per hour = 4,944,960J per hour

4,944,960J per hour * 0.1291cm3/J = 638,394.3360cm3 of gas per hour or 638.3943l per hour

638.3943l per hour / 60 = 10.6399l per minute to supply a 1000kWh per month house?

(This is assuming 100% efficiency converting it from H2 and O2 to Electricity)

If a fuel cell was 60% efficient then you'd need another 30% of gas to make up the thermal losses.

[magnetrecharge]

Did any of you run the calculations to run a ICE on hydrogen?  You guy's are just to funny.
The amount of oxyhydrogen needed to run an internal combustion engine is spectacular. Idling a small engine (e.g. 5hp) would require 500-1000 LPH (liters per hour), while idling a car engine would probably consume about 3000LPH of oxyhydrogen. Driving down the highway would probably consume 20000-30000 LPH of oxyhydrogen. Have fun.

[ramset]

well I guess Meyer  Boyce Dingle etc etc etc must be comedians or magicians