Hydroxy Output vs Faraday Efficiency

Farrah Day

Hydroxy Output vs Faraday Efficiency
« on January 5th, 2014, 10:05 AM »
I thought it might be useful to put up some figures relating to the production of hydroxy from an electrolyser in order to have some base figures to work from.

The maximum gas output (theoretical) produced by standard Faraday electrolysis during the electrolysis of water tends to be the base figure against which all electrolysers and all other hydroxy processes are measured. Everyone compares their gas output with Faraday's maximum in order to evaluate the efficiency of their own electrolyser systems. And rightly so, because Faraday's laws of electrolysis are clear, concise and infallible.

So let's break this down into bite-size pieces.

Now, at the cathode an electrode potential of  -0.41 V is required to produce hydrogen.

At the anode an electrode potential of +0.82 V is required to produce oxygen.

This gives us an overall potential of 1.23 volts.

This is the ideal voltage potential required to initiate electrolysis at the electrodes, and this is the voltage that Faraday's maximum gas output is based upon.

Now it takes 3.952 kW of electrical power to produce 1 mole of hydrogen and half a mole of oxygen per minute, which is 2.016 grams of hydrogen per minute and 16 grams of oxygen per minute.

If we now convert these grams into volumes of gas we get 24.46 litres of hydrogen and 12.23 litres of oxygen. So a total hydroxy gas output of 36.69 litres per minute.

W/V = A

3952 watts/1.23 volts = 3213 amps! This is 100% Faraday efficiency

But here's the thing, at 1.23 volts, minimal current flows through the water and the reality is that electrolysis takes place at a very, very slow pace. And though over time 36.69 litres of gas will be produced from 3.952kW it will take forever to get there as we simply cannot under the given circumstances generate the high current necessary to achieve this in a reasonable time.

In reality we employ voltages between around 1.5 volts and 2.0 volts, but this immediately reduces the Faraday efficiency. So too does adding an electrolyte. An electrolyte will increase the current flow for any given voltage, but it reduces the Faraday efficiency because as a consequence a slightly higher voltage potential is required now to initiate electrolysis.

You also have to consider that the electrodes themselves can play an important part in reducing Faraday efficiency as some will require greater potentials than -0.41 and +0.82. Platinum is generally regarded as the most efficient material to use as electrodes, but is obviously quite expensive.

Anything greater than Faraday's ideal 1.23 volts is known as an over-potential, and as such reduces the Faraday efficiency. And once you start to employ more voltage than you need to reach the potential to initiate electrolysis, you then start to lose efficiency due to energy being lost as heat.  You will also lose efficiency due to side reactions - that brown scum you get means that side reactions have been taking place and any reaction that is unwanted will be detrimental to your gas output.  

It seems very common within the HHO community to employ millilitres per minute per watt (usually denoted as mmw) as opposed to litres per minute (or hour) per kilowatt. I often see people talking about the mmw they need to exceed in order to achieve over-Faraday results.

So what mmw figure is 100% Faraday?

OK, if 100% Faraday produces 36.69 litres of hydroxy per minute at 3.952kW, we can convert this to mmw simply by division.

36.69 litres = 36690 ml

36690/3952 = 9.28 ml/m/w

But the reality of it is that you will be lucky to achieve 70% Faraday efficiency, which would be 6.49 ml/m/w with straight DC in a standard everyday Faraday electrolyser.

You also have to consider that your gas will surely contain some water vapour. Also be aware that the gas figures shown here are based on 1 atm and 25 deg C. The volume of the gases you collect will be directly dependent on atmospheric pressure and temperature.

If you see anything wrong with any of my calculations, please speak out.

 

Matt Watts

RE: Hydroxy Output vs Faraday Efficiency
« Reply #1, on January 5th, 2014, 10:09 AM »Last edited on January 5th, 2014, 10:47 AM by Matt Watts
Thank you for posting this Farrah.  I will stick this thread for future reference by all.

Critical thresholds are:
  • Must exceed:  9.28 mmw (milliliters per minute per watt)
  • Must be less than:  108 wlm (watts per Liter per minute)
If these thresholds are crossed with dry gas at STP, one must look into the production mechanism.  All possible measurement and calculation errors must be ruled out.

adys15

RE: Hydroxy Output vs Faraday Efficiency
« Reply #2, on January 5th, 2014, 06:01 PM »
Farah ,3.952kW is way to high for 30lpm.A regular drycell would output 1lpm per 12v,10A,So 3600w for 30lpm.
Still thinking what actualy heats the water...voltage or amps?Its said if you pass the 1.7v you lose efficency by heat,but when you increase the voltage the amps rushess too,so who is heating the water?I made a little test with rectified AC and rain water with a single tube set and water was heating up very fast(30sec)with an input of 220v 5A.I will try 5v 5A to see if amps are the cause of heat.

geenee

RE: Hydroxy Output vs Faraday Efficiency
« Reply #3, on January 5th, 2014, 06:51 PM »
great thread.

thanks
geenee

Farrah Day

RE: Hydroxy Output vs Faraday Efficiency
« Reply #4, on January 6th, 2014, 03:07 AM »Last edited on January 6th, 2014, 06:48 AM by Farrah Day
Quote from adys15 on January 5th, 2014, 06:01 PM
Farah ,3.952kW is way to high for 30lpm.A regular drycell would output 1lpm per 12v,10A,So 3600w for 30lpm.
Still thinking what actualy heats the water...voltage or amps?Its said if you pass the 1.7v you lose efficency by heat,but when you increase the voltage the amps rushess too,so who is heating the water?I made a little test with rectified AC and rain water with a single tube set and water was heating up very fast(30sec)with an input of 220v 5A.I will try 5v 5A to see if amps are the cause of heat.
Firstly, it is the current that heats the water, but it is increasing the voltage above the threshold to initiate electrolysis that causes the extra current flow... so you decide.

The figures I have given are based on Faraday and science fact, and to the best of my knowledge are correct, and you will note that I stated 36.69 lpm, not 30 lpm!

A dry cell will be producing a lot of water vapour too, nevertheless a regular dry cell producing 1lpm at 120 watts is quite efficient, about 86%.


edxhemphill

RE: Hydroxy Output vs Faraday Efficiency
« Reply #5, on January 7th, 2014, 09:36 AM »Last edited on January 7th, 2014, 09:39 AM by edxhemphill
Hi Gang, I receved a E-book from a friend I want share with all of you.It has a chapter on hho that covers many of the leading experts in the field .Thats chapter 10 . It also has a chapter on under standing electronics ,chapter12 plus many useful circuits and how they work and what they do are covered. It has over 90 over unity devices that may or may not work plus some real insite as to who the enemy is aganist free energy .I haven't finished reading all of it but I found it very interesting so far.I truly think it's worth your time to look into it . The research goes on Ed  you'll find it at web site http;//www.free-energy-info.tuks.nl/  

HMS-776

RE: Hydroxy Output vs Faraday Efficiency
« Reply #6, on January 9th, 2014, 06:38 PM »
Anyone taking measurements should be using a dryer to make sure they are not measuring steam or moisture.

Farrah Day

RE: Hydroxy Output vs Faraday Efficiency
« Reply #7, on January 10th, 2014, 02:01 AM »Last edited on February 4th, 2014, 10:11 AM by Farrah Day
Quote from HMS-776 on January 9th, 2014, 06:38 PM
Anyone taking measurements should be using a dryer to make sure they are not measuring steam or moisture.
Agreed, however as this is not always practical for most amateur experimenters, it would be a good idea simply to assume that the gas will be saturated, as it likely will be, and allow for this in your calculations.

I believe saturated air (ie, 100% humidity) would contain 4% water vapour. The hotter the air the more water vapour it can hold at saturation, but this gives a reasonable ball park figure. Granted this is for water vapour in air and not oxyhydrogen, so this figure could be different, but at least it gives you an initial figure to work from.   

Quote from edxhemphill on January 7th, 2014, 09:36 AM
Hi Gang, I receved a E-book from a friend I want share with all of you.It has a chapter on hho that covers many of the leading experts in the field .Thats chapter 10 . It also has a chapter on under standing electronics ,chapter12 plus many useful circuits and how they work and what they do are covered. It has over 90 over unity devices that may or may not work plus some real insite as to who the enemy is aganist free energy .I haven't finished reading all of it but I found it very interesting so far.I truly think it's worth your time to look into it . The research goes on Ed  you'll find it at web site http;//www.free-energy-info.tuks.nl/
Yes, Patrick Kelly has been compiling this information for years, and it is indeed a good source of info and interesting reading, with everything related to water fuel and OU devices to be found in one place under various sections of his book. And, it is continually being updated as and when more info comes in.

However it has to be borne in mind that Kelly's book is simply a compilation of other peoples work, theories and ideas, many of which have never been categorically substantiated.  So a healthy level of scepticism would be prudent.

HMS-776

Re: Hydroxy Output vs Faraday Efficiency
« Reply #8, on February 3rd, 2014, 05:55 PM »Last edited on February 3rd, 2014, 05:59 PM by HMS-776
Farrah,

In Stan's international independent test evaluation report on page 60 you can see his calculations.

What is interesting is he compares the WFC gas production at 55 watts to electrolysis at only 8.8 watts. He compared the two at the same current values, not the same power. I guess it makes sense since it's the exchange of electrons that produces gas, not voltage. So, he may be correct in his comparison since the same number of electrons are being exchanged.

But at the same time you would think he would compare the two at the same power.
Stan's WFC used 12.5 volts at 4.4 amps which is 55 watts
Perhaps he should compare electrolysis at 2V and 27.5 amps which is also 55 watts.

From my calculations (using the same power) his process is not 1,696% efficient, it's 149% efficient. Which is still great if it can be reproduced and proven.



firepinto

Re: Hydroxy Output vs Faraday Efficiency
« Reply #9, on February 3rd, 2014, 06:02 PM »
Quote from HMS-776 on February 3rd, 2014, 05:55 PM
Farrah,

In Stan's international independent test evaluation report on page 60 you can see his calculations.

What is interesting is he compares the WFC gas production at 55 watts to electrolysis at only 8.8 watts. He compared the two at the same current values, not the same power. I guess it makes sense since it's the exchange of electrons that produces gas, not voltage. So, he may be correct in his comparison since the same number of electrons are being exchanged.

But at the same time you would think he would compare the two at the same power.
Stan's WFC used 12.5 volts at 4.4 amps which is 55 watts
Perhaps he should compare electrolysis at 2V and 27.5 amps which is also 55 watts.

From my calculations (using the same power) his process is not 1,696% efficient, it's 149% efficient. Which is still great if it can be reproduced and proven.
Has anyone been able to run a cell at 2 volts and 27.5 amps with out massive amounts of chemicals in the water?  Comparing apples to apples, the 2 volt faraday cell should also use tap water.

HMS-776

Re: Hydroxy Output vs Faraday Efficiency
« Reply #10, on February 3rd, 2014, 06:41 PM »
Well, I just don't completely agree with the way Meyer made the comparison.

I understand why he did it that way, but at the same time it is a little misleading.
He could have used a series cell with 6-7 cells in it with enough electrolyte to draw 4.4 amps. That way each cell would have about 2v potential and he could have produced more gas and made the comparison at the same power and current levels.



securesupplies

Re: Hydroxy Output vs Faraday Efficiency
« Reply #11, on May 6th, 2014, 01:36 AM »Last edited on May 6th, 2014, 02:18 AM
JUST FYI

Attached there is one more letter I am searching for it for  you

I have read a Letter from a German Group Writing to Stan to Confirm the Measurement of the vs of the 2 methods

doc is out there on web some where I did have a copy so if you find it post it here



Dan

freethisone

Re: Hydroxy Output vs Faraday Efficiency
« Reply #12, on July 8th, 2014, 01:10 PM »
what if oxyclean was added to the water? wold you make a more power full fuel. it contains

Sodium percarbonate is a chemical, an adduct of sodium carbonate and hydrogen peroxide (a perhydrate), with formula 2Na2CO3 · 3H2O2. It is a colorless, crystalline, hygroscopic and water-soluble solid.[1] It is used in some eco-friendly cleaning products and as a laboratory source of anhydrous hydrogen peroxide.

This product contains the carbonate anion, and should not be confused with sodium peroxocarbonate Na2CO4 or peroxodicarbonate Na2C2O6, which contain different anions.