Author Topic: How to find the efficiency of a fuel cell  (Read 21261 times)

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Re: How to find the efficiency of a fuel cell
« Reply #64 on: September 21, 2008, 14:53:05 pm »
Hey there, i'm wondering where you all got your 2.4Wh/l from.
After Faraday's law the Voltage has nothing to do with the electrolysis, i am confused a bit.

I calculated it myself after the Faraday equation,
m/M = I * t / z * F

and i get Vh2o2 = I * t * 1,7425 * 10^(-4)  litres/Coulomb, (same parts O2 and H2, which doesn't fit really... hm)

example: 1 Amp; 1 hour (3600s)

0,6273 litres < 2/3 litres


Someone one got more information on how to calculate the efficiency? Or how do i get to 2.4 Wh/l ?


edit: I found a good site for calculcating the efficiency: http://www.overunity.com/index.php?topic=3107.0
« Last Edit: September 29, 2008, 15:12:56 pm by haithar »

sucahyo

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Re: How to find the efficiency of a fuel cell
« Reply #65 on: November 11, 2008, 07:05:34 am »
KB's design is maybe what my ideal cell looks like, thumb up :).

I currently experimenting a cell with similar priciple, although I use up rising bubble to make the water flow.  And I will put them all inside same container full of water (open bath version).

I use the same tube configuration. The negative electrode is small battery carbon. Single cell draw 0.2 Amps at 12 volts brute force electrolysis using tap water producing 2cc/minute gas .

The problem is the positive electrode can not keep up and corrode in hours. Fast corroding make a lot of debris and short circuiting cell, also make the water dirty very fast. I still never try using high grade stainless steel though.

I still looking for other easy to obtain metal that can at least hold for days. I plan to use 2 volts for actual run.