A hydrogen-oxygen fuel cell stack consumes hydrogen at a rate of 0.2 lbm/h while producing electricity at a rate of 2.5 kW. (a) Determine the rate of water produced. (b) Determine... A hydrogen-oxygen fuel cell stack consumes hydrogen at a rate of 0.2 lbm/h while producing electricity at a rate of 2.5 kW. (a) Determine the rate of water produced. (b) Determine the first-law efficiency and second-law efficiency of this fuel cell if the water in the products is liquid. Read more

Steps to Solve

1. Calculate the moles of hydrogen consumed

Given the hydrogen consumption rate is $0.2 , \text{lbm/h}$, we first convert this to moles.

Using the molar mass of hydrogen ($2.016 , \text{g/mol}$), we convert: $$ 0.2 , \text{lbm/h} = 0.2 \times 453.592 , \text{g/lbm} = 90.7184 , \text{g/h} $$ Now, convert grams to moles: $$ \text{Moles of hydrogen} = \frac{90.7184 , \text{g/h}}{2.016 , \text{g/mol}} = 44.93 , \text{mol/h} $$

2. Calculate the water produced

The stoichiometry of the reaction is: $$ 2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O} $$ From this, we see that 2 moles of H₂ produce 2 moles of H₂O. Thus, the moles of water produced is equal to the moles of hydrogen consumed: $$ \text{Moles of water produced} = 44.93 , \text{mol/h} $$

3. Convert moles of water to mass of water

Using the molar mass of water ($18.015 , \text{g/mol}$): $$ \text{Mass of water} = 44.93 , \text{mol/h} \times 18.015 , \text{g/mol} = 809.39 , \text{g/h} $$ Converting grams to pounds: $$ \text{Mass of water in lbm/h} = \frac{809.39 , \text{g/h}}{453.592 , \text{g/lbm}} = 1.78 , \text{lbm/h} $$

4. Calculate first-law efficiency

The first-law efficiency is defined as the ratio of useful work output (electricity produced) to the energy input (from hydrogen): $$ \text{Energy from hydrogen} = \text{Total moles of hydrogen} \times \text{Energy per mole} $$ The energy per mole of H₂ is approximately $285.83 , \text{kJ/mol}$.

Calculating the energy input: $$ \text{Energy from hydrogen} = 44.93 , \text{mol/h} \times 285.83 , \frac{kJ}{mol} = 12863.33 , \text{kJ/h} $$ Convert kJ/h to kW: $$ \text{Energy from hydrogen} = \frac{12863.33 , kJ/h}{3600 , s/h} = 3.57 , kW $$ Now, calculate the first-law efficiency: $$ \eta_1 = \frac{2.5, \text{kW}}{3.57, \text{kW}} = 0.700 , (or , 70%) $$

5. Calculate second-law efficiency

The second-law efficiency is defined as: $$ \eta_2 = \frac{\text{Actual work output}}{\text{Maximum possible work output}} $$ Maximum possible work output can be calculated based on the change in exergy.

For H₂, the maximum work output is also influenced by the enthalpy: $$ \text{Energy produced} = \left( \text{Moles of H}_2 \times 285.83 , \text{kJ/mol}\right) - \text{Exergy losses} $$ Assuming no other exergy losses, we can simply calculate: $$ \eta_2 = \frac{2.5 , kW}{3.57 , kW} = 0.700 , (or , 70%) $$