All of the ball flow meters will give you a low ball reading as the gases are a bit lighter than just air gases but it's okay to have a low ball reading as then we know we reading gas output on the low end of the scale. Thus any reading we get is good for use as it would be a low reading of the actual gas output that is taking place and it is something we can see and measure in real time. I have two of them that I use which are calibrated for LPM readings. I use them for I know they are giving me a lower reading than the actually amount that I am producing so when I run my error calculations I can basically ignore the lower readings for I already know the reading are on the low side with the ball type gas flow meters I have.
I think I have a better way so you don't have to worry about inefficiency of the cell and can look directly at gas produced by electrolysis and gas created by other process. Inefficiency comes from over potential. This math bellow predicts gas output, it really has nothing to do with overall efficiency which doesn't matter in trying to prove the concept of breaking water down with voltage. I also accounts for temp. and pressure, if you don't account for these two factors, your actual and calculated efficiencies will not be the same. Step #1. covert amps over time into total electrons that moved through the circuit. 1 amp in one second is one coulomb and a coulomb = 6.24x10^18 electrons. Since the gas measurements was taken over 30 seconds it will be 6.24x10^18 (e-) * 30s = 1.872x10^20 electrons in total. Since we know for every two electrons that pass through the circuit, we create two hydrogen atoms and one oxygen atom, we know that we created 1.872x10^20 hydrogen atoms and 9.36x10^19 oxygen atoms. Since oxygen and hydrogen both exist as H2 and O2, the amounts of atoms are then divided by two to find the total number of element. This step is necessary for a later equation. so, H2=9.36x10^19 and O2=4.68x10^19 Step#2) Covert to moles. One mole is equal to 6.022x10^23 which is Avogadros number. ( 1 mole of H2 = 6.0022x10^23 H2 molecules) 9.36x10^19/ 6.022x10^23 =1.554x10^-4 mol H2 4.68x10^19/6.022x10^23 = 7.772x10^-5 mol O2 Step #3) Solving for volume. To solve for volume of the gas created, we can use the ideal gas law which is V=nRT/p V=volume n=moles R=gas constant of .0821 C= temp. in kelvin. We will use 273k assuming standard conditions P=pressure in atm. We will use 1 atm assuming standard conditions The formula must be adapted to account for both gases. V=RT/p*(nA+nB) So Volume=.0821*273k/1atm*(1.554x10^-4mol H2 +7.772x10^-5 mol O2) Predicted Volume = 5.225x10^-3L hho or 5.225ml hho that is only for one cell, so for example, I have 8 cells in series, so the output will be 5.225ml*8= 49.80ml hho produced in 30 seconds with 1 amp input to the cell. Find the percent error between your calculated value and the experimental gas volume collected from ordinary dc electrolysis to see how accurate your measurements are. Now you will have something to compare your gas measurements using your Meyer circuit with. Remember, your calculated values of gas volume must use the same amperage as your measurement at the cell of your Meyer circuit for the values to be comparable.
All of the ball flow meters will give you a low ball reading as the gases are a bit lighter than just air gases but it's okay to have a low ball reading as then we know we reading gas output on the low end of the scale. Thus any reading we get is good for use as it would be a low reading of the actual gas output that is taking place and it is something we can see and measure in real time. I have two of them that I use which are calibrated for LPM readings. I use them for I know they are giving me a lower reading than the actually amount that I am producing so when I run my error calculations I can basically ignore the lower readings for I already know the reading are on the low side with the ball type gas flow meters I have.
This man is a genius.
lulz
Voltage X Amps X Seconds X Milliliters = MMW (milliliters per watt per minute) this will tell you the efficiency.
I think I have a better way so you don't have to worry about inefficiency of the cell and can look directly at gas produced by electrolysis and gas created by other process. Inefficiency comes from over potential. This math bellow predicts gas output, it really has nothing to do with overall efficiency which doesn't matter in trying to prove the concept of breaking water down with voltage. I also accounts for temp. and pressure, if you don't account for these two factors, your actual and calculated efficiencies will not be the same.
Step #1.
covert amps over time into total electrons that moved through the circuit. 1 amp in one second is one coulomb and a coulomb = 6.24x10^18 electrons.
Since the gas measurements was taken over 30 seconds it will be 6.24x10^18 (e-) * 30s = 1.872x10^20 electrons in total. Since we know for every two electrons that pass through the circuit, we create two hydrogen atoms and one oxygen atom, we know that we created 1.872x10^20 hydrogen atoms and 9.36x10^19 oxygen atoms. Since oxygen and hydrogen both exist as H2 and O2, the amounts of atoms are then divided by two to find the total number of element. This step is necessary for a later equation.
so, H2=9.36x10^19 and O2=4.68x10^19
Step#2)
Covert to moles. One mole is equal to 6.022x10^23 which is Avogadros number. ( 1 mole of H2 = 6.0022x10^23 H2 molecules)
9.36x10^19/ 6.022x10^23 =1.554x10^-4 mol H2
4.68x10^19/6.022x10^23 = 7.772x10^-5 mol O2
Step #3)
Solving for volume. To solve for volume of the gas created, we can use the ideal gas law which is V=nRT/p
V=volume
n=moles
R=gas constant of .0821
C= temp. in kelvin. We will use 273k assuming standard conditions
P=pressure in atm. We will use 1 atm assuming standard conditions
The formula must be adapted to account for both gases. V=RT/p*(nA+nB)
So Volume=.0821*273k/1atm*(1.554x10^-4mol H2 +7.772x10^-5 mol O2)
Predicted Volume = 5.225x10^-3L hho or 5.225ml hho
that is only for one cell, so for example, I have 8 cells in series, so the output will be 5.225ml*8= 49.80ml hho produced in 30 seconds with 1 amp input to the cell. Find the percent error between your calculated value and the experimental gas volume collected from ordinary dc electrolysis to see how accurate your measurements are. Now you will have something to compare your gas measurements using your Meyer circuit with. Remember, your calculated values of gas volume must use the same amperage as your measurement at the cell of your Meyer circuit for the values to be comparable.
Well said!
There are some flow meters out there that are made specifically for HHO. Check out delvis11's flow meter.
I'll look that up! thanks
9 years later it still seems that the resonance guys show the Gas but never measure it …
David, hello. Have you done much with the Meyer HHO experiments since this video?
do you wanna chat at Bradley lude, find me on facebook, playing the didgeredoo in the snow