I need a simple formula to calculate the efficiency of an electrolysis device

I need a simple formula to verify my math concerning the efficiency of an electrolysis device. (Actual percentage of electrons used to crack water)
These are the numbers I have to work with: Total amount of electrons sent to the electrolizer from the input power supply is 100 coulombs.(1000 joules if this helps). …Also refer to statement below….
(1 mole of water produces 1 mole of H2 and 0.5 mole of O2 so that would be 24359cc/mole * 1.5mole / 18cc/mole = 2030cc of H2+O2 per cc of water at 75 °F the vapor pressure of water is about .03 atmospheres….)
Using the above information what would be the volume of total gas if the device were “100% efficient”.   Once I am sure of this amount, it will be easy to calculate future experiments
From actual test, the total gas produced (H and O) is 3.5cc for test 1 and 5.75 cc for test 2.

Keep in mind guys (and gals) I’m not an engineer or chemist, just a humble inventor trying to figure this out. So please have mercy on me!!!!!

whiskers1Asked:
Who is Participating?
 
aburrCommented:
grg99 has some good points particularly about the internal resistance of the capacitors.
He also said
If you really want to measure the cell's efficiency, you'll need a high-speed integrating amp meter.  Nothing else will do.

It is possible to get a relatively inexpensive data logger such as one sold by PASCO ($300) which should serve your purpose
0
 
aburrCommented:
It depends on what you mean by efficiency. Note also there is a big difference between 100 coulombs and 100 joules.
I think that you could use a number like total volume of gas/total energy input.
The input energy is easy to measure but the correct total volume of gas is not. You have to consider the temperature of the water, the temperature of the gas, the purity of the gas produced (including consideration of any impurities present or introduced to the water to make it conducting), the pressure in the gas collector, the outside pressure, and perhaps some items I have neglected to put here.
0
 
aburrCommented:
To sim;lify my previous post the formula I am proposing is

figure of merit = total volume of gas produced/total energy input to the device.

Note the difficulties mentioned above of measuring the total volume of gas produced. The method of measuring that quantity MUST be exactly the same from test run to test run.
0
Free Tool: Site Down Detector

Helpful to verify reports of your own downtime, or to double check a downed website you are trying to access.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

 
aburrCommented:
The energy input, by the way, will be the voltage across the input to the device times the current through the device all intergrated over the time the device is in operation.
0
 
whiskers1Author Commented:
1st Thanks Aburr for responding,
 To help you better understand what the heck I’m doing--
The input power supply is a 6F capacitor bank (two 3F caps connected together) charge to 12v and discharged to 8v. From my calculations, this will provide approx. 24 coulombs
 After discharging into the electrolysis device, 3.5 cc total volume of gas is repeatedly produced.

The electrolysis device is simply a test tube with 2 copper 12g wire electrodes, water/baking soda solution as the dielectric. The test tube is configured with a special water reservoir that easily allows for expansion of gas produced (total gas is collected at the closed end of the tube). This is so that its main limitation to gas expansion is atmospheric pressure (approx 100ft above sea level / weather conditions).
Dielectric temperature is approx. 75 deg f.  before and after experiment. (If there is a temp change in the dialectric, it is so slight that I cant detect it.) This is called test A

Test B is identical to Test A- including power supply (total input power) electrolysis device, dielectric etc.  The only difference is that a special circuit is added inline, which repeatedly produces 5.75cc of gas.

From this, I calculated an approx.  65% gain in gas production with test B.
 
Note: I have used many different electrodes and dielectric solutions - Because of this, the amount of gas produced in Test A may vary---But ALL have produced the SAME percentage of increase in gas production in test B

Due to non-liner voltage and current rates from the discharge of the caps, it is very difficult to calculate accurately total input power using a meter. This is one of the reasons I’m working with electrolysis, hoping that this may give me some idea of the actual total power being sent to the load..

Aburr if you have any more ideas concerning my dilemma, I’m all ears!!!!

Thanks Brad
0
 
whiskers1Author Commented:
Aburr

figure of merit = total volume of gas produced/total energy input to the device.

By the way you cant get simpler than this -----GOOD JOB!!!!!!!

Now to figure the total energy?????

Many Thanks Brad
0
 
aburrCommented:
If you use the same capacitors in all tests, you can use the fact that the energy stored in a capacitor is
(1/2) C times  V^2
The energy supplied to all your equipment is the initial energy (V=12) minus the energy remaining after the run (V=8)
There must be no other power at all supplied to the equipment (including your "special circuit")
0
 
whiskers1Author Commented:
"There must be no other power at all supplied to the equipment (including your "special circuit")"

Aburr--
1st-the same components are always used for test A & B
2nd- You have stepped into the big mystery-there is no additional power but I'm still producing more gas with test B..
Note: the special circuit gets its operational power from the input power supply (the caps)
It should be the opposite due to insertion losses from the special circuit- less gas should have been produced, not more !?!?!?!?!

Any thoughts, I would like to hear them...

Thanks Brad
0
 
aburrCommented:
I have answered your questions to the best of my ability. There is no way I can check your work remotely and there is no way that I can offer any useful comments on an unknown "special circuit".
0
 
whiskers1Author Commented:
Aburr
1st,  I going to get back on topic. I  sincerely apologize jumping around with this thing.
Back to business
Lets take another stab at this, if I misunderstood all this, please be gentle... Im here to learn..
                                       
from your last post am I correct to say

part 1...(capacitance) 6f divided by 1/2= 3f x  12v  squared = 422
part 2...(capacitance) 6f divided by 1/2= 3f x  8v squared =    192
part 3... 422 minus 192 = 230 watts total input power provided by the caps

I bet I screwed up, but at least I  tried.

Thanks for all your help and time
Brad

                                                                               
0
 
aburrCommented:
I bet I screwed up

you did but not badly. I get the input power as 432 J. You forgot to carry the 1 from the first mutiplication.
0
 
grg99Commented:
Are you trying to use electrolysis to measure power?

If so, I suggest you give up.   Electrolysis is unlikely to give you power measurements that are within 20%.
Power meters accurate to 1% you can rent for under $200 a month.

Many, Many problems with electrolysis:

(1)  A certain amount of the power goes into just heating the water.  This could be anything from 1% to 30% depending on the current level, electrode spacing, and types of electrodes and elecrolytes.

(2) An unknwable amount of the oxygen and hydrogen are going to dissolve into the electrolyte.

(3)  Using copper electrodes is going to cause some of the energy to go into making copper ions.

(4)  A certain amount of the hydrogen and oxygen are going to recombine.

(5) The amount of gas liberated is proportional to the CURRENT, but only weakly related to the VOLTAGE.  Power per uni ttime is the product of those two numbers.  How can you compute power when you're not measuring one of the parameters?

(6)  Power is voltage times current, measured at the same time and integrated.  There's no way you can measure the instantaneous current and voltage with an electrolytic cell.


0
 
whiskers1Author Commented:
1st thanks aburr for checking my work, now I have an idea of actual input power.
 2nd grg99- I totaly agree that this electroylsis device is very inefficient. But Im trying to find just how inefficient.
ie: efficency would be the "actual"  amount (percentage) of electrons that produced gas by cracking water (2 per molecule)  example: total of 100 electrons are passed to the electroylsis device from my cap bank/ input power supply, only 25 water molecules are cracked and produces a certain volume of gas that is measured--
this would  mean that the device is ,for my purpose, 50% efficent   25molecules x 2 electrons = 50 electrons.


After a lot of thought should my question be "how do I calculate the amount of coulombs of electrons  actually used to produce a know volume of gas."

 Again many thanks for everyones time and effort

Brad
0
 
whiskers1Author Commented:
almost forgot

After calculating how many electrons were used to "make" gas, subtract from know amount of electrons sent by the power supply, from that calculate efficency
0
 
aburrCommented:
your last question is much harder than the original
0
 
ozoCommented:
1  coulombs = 1 amp*second = 6.2415*10^18 electrons
1 mole = 6.0225 * 10^23
But even if every electron liberates a hydrogen atom, that's not very energy efficient if you have to use 1000000 volt electons.
0
 
grg99Commented:
A few more problems:

(7)  Capacitors, especially the "super" kind, have many drawbacks for your application.

(8)  There is a very loose tolerance on the capacitance-- anything from plus or minus 20 percent, up to minus 20 to plus 100 percent for some.

(9) There is a huge variation in capacitance versus temperature-- up to 50% less at low temps.

(10) There is a large  variatioon in capacitance versus voltage, at least 20 percent when varying the voltage from 10 to 100 percent of rating.

(11)  There is a variable series resistance.

(12)  There is a variable parallel leakage resistance.

----------------

BY the time you add up all these loose tolerances, I'd be surprised if you can measure the electrolytic cell efficiencey within plus or minus 30 percent from trial to trial.

-------

If you really want to measure the cell's efficiency, you'll need a high-speed integrating amp meter.  Nothing else will do.

0
 
whiskers1Author Commented:

Thanks to you both for you help. I look forward to future Q & A with you guys on other subjects.

Again its been a pleasure and one heck of an learning experience.

Thanks for living up to the true spirit of this website which is to teach others..

Good job and happy trails

Brad
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.