Fuel Cells

[deleted-546552]

Hi, I am doing the Fuel Cells: Fueling the Future project and have a question. What setting on a Innova 3320 Auto-Ranging DMM multimeter do you use for this project and can you give any advice on the calculations?


Thanks,Dodge04

[LeungWilley]

Hi Dodge04,
For the Innova 3220 DMM, you are going to want to use the "DCV" to measure voltage and the "ohm" symbol to measure resistance. In regards to the calculations, please help us understand if there's a particular equation that's giving you trouble.

Thanks and Have fun with the experiment!
Willey

[deleted-546552]

Thanks so much! We tried to use the DCma setting and it didn't work! As far as calculations I have that figured out.

Thanks again, Dodge04

[deleted-546552]

Hi, I am working on the Fuel Cells Fueling the Future project. When calculating the efficiency of the fuel cell, I came up with 138.511%. Is it possible for the result of my project to show 138.511% efficiency?



Thanks, Dodge04
moderator note: I've merged your most recent post in with your previous posts on the same topic. Please keep your posts together so the expert who has been helping you can see that you have a follow-up questions. Thank you and good luck with your project.

[deleted-546552]

Hello again,
With the help of my dad, I have re-checked all of my testing results and calculations for the experiment. Summarized below are the results of my experiment. Would you please review those results and let me know if any of the numbers are outside of the perimeters of what they should be? The electrolysis of the water only took about 2 1/2 minutes and the motor ran for about 8 minutes. I realize that the percentage efficiency figure is too high but I am not sure which numbers in the calculations of the formulas are incorrect.

Equation 1: Power = Average Current x Average Voltage
Power = 1.028 x 1.837 = 1.888
Equation 2: Energy = Power x Time
Energy = 1.888 x 165
Energy = 311.52
Equation 4: Output energy = (Average current of fuel cell) x (Average voltage of fuel cell) x (Running time)
Output energy = 125.518 x 0.719 x 477.667 =
Output energy = 43,108.224
Equation 5: Efficiency = Output energy
Input energy
Efficiency = 43,108.24
311.592
Efficiency = 138.348%

Thanks, Dodge04

[LeungWilley]

HI Dodge04,
Looking at your numbers, can you confirm the units for the output energy calculation please?
The average current of the fuel cell looks really high at 125.518Amps, (Could this be milli-amp?)

Also, please double check your equation for efficiency. It should be Output Energy divided by Input Energy *100. (You should see a number similar to the one above equation 5 in the procedure)

You are very close and doing great!
Good Luck and please post again if you have any other questions.
Willey

[deleted-546552]

Thank you for your help. In answer to your question, those measurements were in amps and were the ones we got. Also i have a question: What does *100 mean in your post? Equation 5 was done as you said and the efficiency was according to those results. To try to make sure we got the right measurement, we tried the DCma setting on our multimeter and it showed overload. Also the general internet research i did showed my fuel cell as being 50-70 percent efficient. Your post suggested that the efficiency may be more like 10%. Do you have an estimated range of efficiency that should be expected for the fuel cell I am using?


Thanks, Dodge04

[LeungWilley]

Hi Dodge04,
I am sorry to say that the 125A still does not make sense to me. According to the spec on your innova meter, it can only read up to 10A. Are you still using this meter for the measurement? If so, can you attach a picture to show how the red / black leads were setup when you took the measurement?

The *100 meant to multiply the result of the fraction by 100 in order to convert it into a percentage.
(In the example, the fraction came out to 0.1, multiply by 100 would make it 10%.)
Good Luck!
Willey