Author Topic: How to find the efficiency of a fuel cell  (Read 17693 times)

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How to find the efficiency of a fuel cell
« on: April 24, 2008, 02:50:55 am »
 I’ve been testing a new fuel cell design and am wondering if I am figuring the efficiency of it properly. I am getting 1.5 litters every 15 sec. or 6 litters a min or 360 litters an hour. I am at 13.2 volts and 49 amps. I think this is 646.8 watts of power consumed. So if I’m right that’s 646.8 / 360 =  1.80/watts/litters. Is this correct? And how much power will a 100% efficient unit use. Thanks for the help.

hydro

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Re: How to find the efficiency of a fuel cell
« Reply #1 on: April 24, 2008, 05:31:44 am »
first thing first, are you using an additive? this means something to me..

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Re: How to find the efficiency of a fuel cell
« Reply #2 on: April 24, 2008, 13:00:11 pm »
Hi KB,

I made a calculation for you:
Liters: 1,5
Seconds: 15
Volts: 13,2
Amps: 49
LPM: 6
Watts: 646,8
Wh/L: 1,8
mL/W/h: 556,59
Percentage compared to Faraday: 130,9%

That is impressive!
I ll gues you used some additives in your cell.
Can you publish some pictures?

Br
Steve

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Re: How to find the efficiency of a fuel cell
« Reply #3 on: April 24, 2008, 22:29:08 pm »
 I am using 2 tsb. of KOH. The cell is a design unlike anything that I have seen published. Its size is 18" long x 10"wide x 1" tall. It makes very little heat unless you add kOH to raise the amps to 75 amps or higher. I will be going out of town tomorrow for the weekend but will post pictures early next week, but I won’t release the inner working until I try the addition of the pulsed alternator on this design. You guys are going to love how simple this design is.

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Re: How to find the efficiency of a fuel cell
« Reply #4 on: April 24, 2008, 22:32:52 pm »
 To correct an error thats 2 table spoons of KOH.

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Re: How to find the efficiency of a fuel cell
« Reply #5 on: April 24, 2008, 22:36:20 pm »
 And I should add I’m using a gap of .25" between anode and cathode.

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Re: How to find the efficiency of a fuel cell
« Reply #6 on: April 24, 2008, 22:50:49 pm »
I am using 2 tsb. of KOH. The cell is a design unlike anything that I have seen published. Its size is 18" long x 10"wide x 1" tall. It makes very little heat unless you add kOH to raise the amps to 75 amps or higher. I will be going out of town tomorrow for the weekend but will post pictures early next week, but I won’t release the inner working until I try the addition of the pulsed alternator on this design. You guys are going to love how simple this design is.

Ok sir  :D

Looking forward to your pictures!
I like the KISS methode...have a nice weekend!

br
Steve

hydro

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Re: How to find the efficiency of a fuel cell
« Reply #7 on: April 25, 2008, 03:26:38 am »
Hi KB,

I made a calculation for you:
Liters: 1,5
Seconds: 15
Volts: 13,2
Amps: 49
LPM: 6
Watts: 646,8
Wh/L: 1,8
mL/W/h: 556,59
Percentage compared to Faraday: 130,9%

That is impressive!
I ll gues you used some additives in your cell.
Can you publish some pictures?

Br
Steve

you cant go by farraday, comparing your output to what farraday says means nothing, the only way it would mean something is if you used the same fuel cell farraday used, if you use the cell he used then you can use his math, till then you have to use common math and common sense, farraday can produced 1 liter using 2.4 watts, farraday did this with his cell, if he was to use yours things would change.

don't apply farraday with an additive, and don't apply farraday to a cell that farraday did not base his measurements on, also do not apply farraday to Pulsed dc because farradays law with hydrogen is for his cell using pure dc. when you use pulsed things change.

the resistance, conductance, and capacitance of the cell all changes per different cell, farraday cant be applied to some cell out there you have.

if you want to calculate how efficient you are you must turn your gas and oxygen into watts, then you compare the amount of gas in watts you produce to what you put in, this will give you a percentage of what you have done.