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Calculating the efficency of a hydrogen fuel cell

  1. Nov 3, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm trying to calulate the efficency of a small hydrogen fuell cell that I built. I have all of the measurements, but I'm not coming up with a resonable effiency.

    2. Relevant equations

    [tex]\frac{P OUT}{P IN}[/tex]x 100

    The power to electrolyze the water is the P IN.
    3.25 W =2.5 V x .00021 Amps (.21 mA)

    We measured the electrical power out finding the voltage and amperage.
    P OUT = V x I
    2.01 x 10^-4 W=.96 V x .00021 AMPS (.21 mA)

    I caulcate the force to be 1.421 N


    Then I take the electrical P Out and use it as the P IN for the car

    the P IN for the car is the electrical P Out

    and the P Out for the car is F x V

    To find the % efficiency:

    [tex]\frac{p Out of the Car}{P in Car}[/tex] x 100

    The avg velocity is .0198 m/s


    3. The attempt at a solution


    I'm coming up with an unreasonable calculation
     
  2. jcsd
  3. Nov 3, 2007 #2
    To the Electrical power I'm coming up with:

    P IN = 2.5 V x 1.3 AMPS = 3.25 W

    P Out = .96 x .00021 amps = 2.01 x 10 ^ -4 W

    I'm taking the P Out and using it as the power In for the car.

    For the car:

    P in = 2.01 x 10^-4 W
    P Out = F x V
    2.81 x 10^-2 J = 1.421 N x .0198 m/s

    P out/ p in x 100
    13980 definately isn't reasonable.
    Should my P in and P out for the car both be in Joules?

    Also I'm not sure if i'm calculating the force in th right way. We used a spring scale and my measurement was .160 kg. I'm I just supposed to multiply that by 9.8 m/s^2. I know that this way gives me Newtons, but the car isn't being affected by gravity anyways. If someone could look this over I'd appreciate it.
     
  4. Nov 4, 2007 #3
    can someone just let me know if I'm calculating the force the rigt way?
     
  5. Nov 4, 2007 #4
    correction: it took 1.3 amps to electroylze the water
     
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