Faraday's law and 7.64 mmw

[BoyntonStu]

C&P from another forum:

BoyntonStu
------------------------------------------------------------

Faraday's laws for electrolysis are:

First Law

The quantity of a substance produced by electrolysis is
proportional to the quantity of electricity used.

Second Law

For a given quantity of electricity the quantity of substance
produced is proportional to its weight.

The magic numbers:

Faraday's number is 96,484 Coulombs/mol (often rounded to 96,500.) (see
http://www.ausetute .com.au/faradayl .html)

One liter of hydrogen weighs 0.08988 grams/liter, so 1 liter of hydrogen
is 0.08988 moles. (see http://en.wikipedia .org/wiki/ Hydrogen)

Faraday's laws can be summarized by

m \ = \ \left({ Q \over F }\right)\left( { M \over z }\right)

where

/m/ is the mass of the substance altered at an electrode
/Q/ is the total electric charge passed through the substance
/F/ = 96 485 C mol^-1 is the Faraday constant
<http://en.wikipedia .org/wiki/ Faraday_constant>
/M/ is the molar mass of the substance
/z/ is the valence number of ions <http://en.wikipedia .org/wiki/ Ion>
of the substance (electrons transferred per ion) (see
http://en.wikipedia .org/wiki/ Faraday%27s_ law_of_electroly sis)

Solving for "Q" we get Q=m*F*z/M Hydrogen has a valence of one and a
mass of one, let's put in the numbers:

Q=0.08988 * 96484 * 1 = 8672 (roughly) coulombs per liter of hydrogen.
Since Amperes are Coulombs/second, 1 LPM needs 144 Amperes FOR A SINGLE
CELL. For a series cell, this is divided by the number of cells, so for
a 6 cell system (7 plates), we should need about 24 Amperes. However,
when we produce hydroxy (or whatever name you wish to use), we also
produce 1/2 mol of oxygen per mol of hydrogen.

So, for each mol (gram) of Hydrogen, we produce 1/2 mol (6 grams) of
oxygen. Oxygen has a density of 1.429 g/L, so 4 Liters of Oxygen for
each 11 liters of Hydrogen or 15 Liters of hydroxy, so we really only
need 144 *11/15 or 105.6 Amperes or 17.6 Amperes per liter per minute
for a 6 cell system.

Voltage (and power) do not appear in Faraday's equation as far as I can
tell. However efficiency can be derived from the potential at which
electrolysis starts (1.24 Volts, IIRC) and the current per liter per
minute (105.6 Amperes) givng us 1,000 mL/minute/(105. 6*1.24) or about
7.64 M/M/W. When higher values are claimed, either there is water vapor
in the gas, the temperature correction has not been applied, or
something else is odd. Claims of higher than 7.64 MMW should be
examined for possible error.

Does that help?

--
David G. LeVine

[Davehho1]

Apples and oranges

[Roland Jacques]

Apples and oranges

Why?????????????

[Davehho1]

Why?????????????

Everyone on this site is running H H and O, back in the day ole faraday was doing the math on straight Hydrogen. Also: Everyone is deducting about 33 % by volume for the "O" in the HHO ,While still using the total Watts used to produce complete HHO, further reducing its percieved value. I also believe that the "O" as an oxidizer is of great valve in almost all the applications that we are experimenting with.

While faradays calculations are of use, they are for use with straight H, we need to update them for HHO. Since comparing H to HHO is like comparing apples to oranges.

I, like many others believe that theres two kinds of "H" being produced and I believe that of the two, one type adheres to the "O" better and makes for a rather perfect combination for burning in a torch or burner configuation or as a fuel additive or running an ICE
. If I'm not out on the fringe or crazy, then all the guys that are deducting the 1/3 of the volume of the produced gas are wrong , so they are completely wrong when they do the math on how much HHO is required to either increase fuel economy or to run an ICE off HHO alone.

If I'm correct, then it will take significantly less HHO to run an ICE and HHO on demand is a goal we can obtain.
Thats one of the reasons I"m so interested in the work HHOPWR is doing, since his geographic location gives us important information requarding temp differences compared to volume. I'm betting as the weather warms up in his location , his production go's up and his results improve.

Its all good stuff !

D

[BoyntonStu]

Everyone on this site is running H H and O, back in the day ole faraday was doing the math on straight Hydrogen. Also: Everyone is deducting about 33 % by volume for the "O" in the HHO ,While still using the total Watts used to produce complete HHO, further reducing its percieved value. I also believe that the "O" as an oxidizer is of great valve in almost all the applications that we are experimenting with.

While faradays calculations are of use, they are for use with straight H, we need to update them for HHO. Since comparing H to HHO is like comparing apples to oranges.

I, like many others believe that theres two kinds of "H" being produced and I believe that of the two, one type adheres to the "O" better and makes for a rather perfect combination for burning in a torch or burner configuation or as a fuel additive or running an ICE
. If I'm not out on the fringe or crazy, then all the guys that are deducting the 1/3 of the volume of the produced gas are wrong , so they are completely wrong when they do the math on how much HHO is required to either increase fuel economy or to run an ICE off HHO alone.

If I'm correct, then it will take significantly less HHO to run an ICE and HHO on demand is a goal we can obtain.
Thats one of the reasons I"m so interested in the work HHOPWR is doing, since his geographic location gives us important information requarding temp differences compared to volume. I'm betting as the weather warms up in his location , his production go's up and his results improve.

Its all good stuff !

D

"
While faradays calculations are of use, they are for use with straight H, we need to update them for HHO."

This statement is false!

Electrolysis of water produces hydrogen and oxygen.

The Faraday Constant is for all electrolysis.


In physics and chemistry, the Faraday constant (named after Michael Faraday) is the magnitude of electric charge per mole of electrons.[1] While most uses of the Faraday constant, denoted F, have been replaced by the standard SI unit, the coulomb, the Faraday is still widely used in calculations in electrochemistry. It has the currently accepted value:[2][3]

F = 96 485.339 9(24) C/mol

The constant F has a simple relation to two other physical constants:

F = NAe

where

N_{A} = 6.022 \times 10^{23}mol^{-1}
e = 1.602 \times 10^{-19}C

NA is the Avogadro constant, and e is the elementary charge or the magnitude of the charge of an electron. This relation is true because the amount of charge of a mole of electrons is equal to the amount of charge in one electron, multiplied by the number of electrons in a mole.

The value of F was first determined by weighing the amount of silver deposited in an electrochemical reaction in which a measured current was passed for a measured time, and using Faraday's law of electrolysis.[4] Research is continuing into more accurate ways of determining the interrelated constants F, NA, and e.

This is basic science 101.

What does apples and oranges have to do with Faraday's Constant?


BoyntonStu

[Davehho1]

I did find an example of Faraday and both H and O




Heres an example how people "Deduct" the " O " from their calculations


This ones from energysupply2008 Off one of Smartscarecrows videos

http://www.youtube.com/watch?v=RqvxdBeOGMY&feature=PlayList&p=E936393A0A697565&index=9

( Sic em Boynton , lol )

" HHO, two parts H and one part O so divide 0.01079 by 3 x 2 = energy density of 0.007193 MJ per liter at room temperature.

Typical HHO cell temperature is 120 F. At that temperature, the HHO expands 17.8 percent.

0.007193 divided by 117.8 x 100 is an energy density of 0.006106 MJ per liter."

[Davehho1]

And , if memory serves, Faraday (published 1832) was using about 1.45 v's
Is that correct ? Bet his current (amps ) thru the roof.
I'm using 110vac , running it thru a full wave bridge and hitting my cell with about 500to 900 watts depending on the application..

Does anyone have a quick link handy to the style of cell Faraday was using ?
Including electrolyte, watts , plate number, ga, size, etc etc
Id be curious how we compare by todays standards..

[HHO BLASTER]

And , if memory serves, Faraday (published 1832) was using about 1.45 v's
Is that correct ? Bet his current (amps ) thru the roof.
I'm using 110vac , running it thru a full wave bridge and hitting my cell with about 500to 900 watts depending on the application..

Does anyone have a quick link handy to the style of cell Faraday was using ?
Including electrolyte, watts , plate number, ga, size, etc etc
Id be curious how we compare by todays standards..

Why are you using 120 volts AC through a bridge, what's the current rating of the bridge and are you using a filter cap

[nvtowing]

Hi BoyntonStu,

I've come up with a different MMW number for 100 percent efficiency. I think you might be including the "environmental input energy" as electrical input energy.. I may be wrong, though... I did these calculations quite a while ago.

Here is a reference: http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/electrol.html


According to Faraday Law:

107.205 amps continuous for 1 hour will create 2 moles of H2 and 1 mole of O2 gas (3 moles total), which has a volume of 73.338 Liters (at 25C/101.325 kPa) at 100 percent efficiency.

(Please double check all this)



If we multiply the change in Gibbs Free Energy of formation (or electrical input energy required to create 1.5 moles of gas) by 2, we should have the actual energy (joules) required to create 3 moles of gas (at 100 percent efficiency, in the above conditions).

237.18 kJ * 2 = 474.36 kJ

Convert 474.36 kJ to Watts:

474360 Joules / 3600 seconds = 131.7666 Watts

Now put everything together (amps &voltage)

Since:,
Watts = Amps * Volts

131.7666 Watts = 107.204 Amp * Volts

So, V = (131.7666 W) / (107.204 A)

V = 1.23 Volts

Which is known to be the minimum voltage for electrolysis at 25C, 101.325 kPa. Which is also 100 percent Faraday, and "Gibbs" Efficiency.

Here's how you'd calculate "Ideal" W/LPM (Watts per LPM) at 25C and 101.325 kPa.

73.338 L/Hr / 60 Minutes = 1.223 LPM
Requires 131.76 Watts.

So,

1 LPM / 1.223 LPM = .81766
.81766 * 131.76 Watts = 107.73

At 25C/101.325 kPa it takes 107.73 Watts to generate 1 liter per minute at 100 percent efficiency.

And the MMW (the standard HHO measurement method) would be:

1000/131.76
9.282 MMW, at 25C 101.325 kPa.

[nvtowing]

Hi BoyntonStu,
1000/131.76
9.282 MMW, at 25C 101.325 kPa.

Oops..
(1000 / 107.73 Watts) = 9.28 MMW

[Q-Hack!]

I have seen the term Gibbs law of free energy batted around a few times... everything I remember about Gibbs law of free energy was something like: The potential that is realized while a system reaches equilibrium. Or another way to think about it is as a chemical reaction occurs Gibbs free energy is the math behind the thermal reaction. Not sure how that applies to HHO generation.

Of course I could be completely out of my gourd when it comes to understanding Gibbs law of free energy. (more likely) :rolleyes:

[nvtowing]

The link above "hyper-physics" does seem to imply the "change in gibbs free energy" is the same "electric input energy" (or potential, if nothing else). And my calculations seem to confirm that. No?

[Q-Hack!]

I have had to go back and read up on Gibbs energy of formation and I do think you are correct. Even if I do have a headache now. :p

I know that stu has previously posted a link to a good MMW calculator.
http://dc106.4shared.com/download/73830865/e9076bfd/MMW_V625.xls?tsid=20090525-030111-8d8cab67

The more I look at it I notice that it does account for Gibbs efficiency. And as I sit here and play with the numbers it looks like it does allow a 9.276 MMW without being overunity (according to Gibbs law). However it is over Faraday efficiency... enough to bang ones head against the wall... repeatedly.

5N1
25.00 degrees C
9.88V
10A
60.00 seconds
1000ml
1.000 LPM-uncompensated
.916 LPM-temperature compensated
10.125 MMW-uncompensated
9.276 MMW-temperature compensated
146.12% of Faraday (temp comp)
99.99% Unity - Gibbs Law

[nvtowing]

I have had to go back and read up on Gibbs energy of formation and I do think you are correct. Even if I do have a headache now. :p


I'm thinking now "Faraday Efficiency" is only current efficiency (and voltage is meaningless), and "Gibbs" is voltage potential efficiency (which can vary with temperature and pressure.. I.E 1.23 V)

[nvtowing]

Actually, "Gibbs free energy" efficiency is probably just the actually energy efficiency (in joules).


Lots of confusion about this out there, btw.(even with chemists)