# Calculating the efficency of a hydrogen fuel cell

1. Nov 3, 2007

### redsox5

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

$$\frac{P OUT}{P IN}$$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:

$$\frac{p Out of the Car}{P in Car}$$ x 100

The avg velocity is .0198 m/s

3. The attempt at a solution

I'm coming up with an unreasonable calculation

2. Nov 3, 2007

### redsox5

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.

3. Nov 4, 2007

### redsox5

can someone just let me know if I'm calculating the force the rigt way?

4. Nov 4, 2007

### redsox5

correction: it took 1.3 amps to electroylze the water