1. You are designing the fuel cell system for a manned space mission to Mars. This PEMFC will power the auxiliary electrical boosters and also provide potable water for the 5 crew members. The fuel cell stack produces 5 kW and operates at 0.7 V and 2 A/cm2. Answer the following questions:
What total membrane size (in cm2) is needed for this 5 kW stack?
We know that P = I*V = 5000 W = (2 A/cm2)*(0.7 V)*(Area)
Solving for area -> 3571.4 cm2
How efficient is the fuel cell?
Efficiency (η) = V/1.23 * 100% = 0.7/1.23 * 100% = 56.9%
To achieve a useful power of 5 kW, what flow rate of hydrogen is needed? Hint: use this conversion: ∆hrxn = 241,830 J/mol H2 (1 mol of H2 produces 241,830 J from the reaction).
We need extra chemical energy from hydrogen than 5 kW as fuel cell is not 100% efficient. Excess is converted to heat.
Power Needed = 5000 W / 0.569 = 8785.7 W from H2
8785.7 W = 8785.7 J/s. Divide this by the energy produced from reacting 1 mol of hydrogen. 8785.7 J/s * (1 mol H2 / 241,830 J) = 0.0363 mol/s of hydrogen
What flow rate of water is produced by this fuel cell? Will it be able to sustain the 5 astronauts who use 11 liters each day?
We know that 1 mol of H2 reacted produces 1 mol of H2O
0.0363 mol/s of H2 -> 0.0363 mol/s of H2O. Now convert to units of L/day
56.5 L/day for 5 people = 11.3 L/day per person. This is enough water for the crew.
2. The engineering consulting firm you work for has a large hotel as a client and a colleague of yours is proposing a project to install a 500 kW MCFC using H2 as a fuel operating at 700 C to supply baseload power.
Write out the anode and cathode half reactions for this fuel cell.
Anode: H2 + CO32- -> H2O + CO2 + 2e-
Cathode: 1/2O2 + CO2 + 2e- -> CO32-
What is the voltage at these conditions? Hint: ∆Grxn @700 C = -194.1 kJ/mol
What is the cell voltage of this fuel cell at 700 C and 2 atm with the following gas concentrations?
Anode: 90% H2, 5% H2O, 5% CO2
Cathode: 12% CO2, 9% O2, 77% N2
Assume ideal gas law.
If the fuel cell operates at 0.6 V, how many 100 hp (75 kW) natural gas-powered boilers could the excess heat replace for the hotels hot water needs?
Efficiency is (η) = V/1.128 * 100% = 0.6/1.128* 100% = 53.2 %
Waste heat = 500000 W / .571 – 500000 W = 439850 W of heat
# of boilers = 375657 W / 75000 W = 5.86 boilers
What changes or possible concerns might you have for this project?
Hydrogen as a fuel may not be readily available. Switching to a methane powered cell could be advantageous.
Pressurizing the gasses to 2 atm may not be worth the cost saved in efficiency gains.
Minimize the amount of water and CO2 at the anode and maximize oxygen and CO2 at the cathode.
3. You are responsible for approximating the current-voltage characteristics for an individual SOFC fuel cell stack. As a first approximation, you assume that the cathode side of the fuel cell dominates both the kinetic losses and the transport losses. Also assume that the open-circuit potential of the cell is 1 V. The fuel cell operates at 700 C and the cathode input is air at 1 atm pressure.
Write the cathode half reaction for this fuel cell (as a 4-electron process).
Your first step is to determine the kinetics at the cathode. You find that the catalyst exchange current is 0.01 mA/cm2 and that the symmetry factor or transfer coefficient is 0.5. Use the Butler-Volmer equation to plot the kinetic data over the overpotential range of 0 to -0.4 V.
Use the Butler Volmer Equation:
Plot from 0 to -0.4 V, because this is a reduction reaction and the current should be negative. As expected for kinetically sluggish oxygen reduction reaction, there are significant overpotential losses before significant current is produced.
Note: If a positive overpotential was used, this would correspond to the oxidation reaction, which is the oxygen evolution reaction rather than the oxygen reduction reaction.
Your next step is to determine the transport at the cathode. The diffusion coefficient is 0.4 cm2/s, the porosity of the gas diffusion layer is 50%, the tortuosity is 4, the thickness of the gas diffusion layer is 0.05 cm, and ambient heated air is used as the feed. What is the maximum limiting current?
First determine the effective diffusivity:
Then determine the max current.
As an approximation, assume that the limiting current is the same at all voltage. Use the Koutecký–Levich equation (lecture 2D) to make a current-voltage plot for the fuel cell. You can define both kinetic and transport currents as positive values. Plot the cell operating voltage on y axis and operating current on the x axis.
Koutecký–Levich equation:
Assume that the transport limited current is independent of voltage. Though the kinetic current is dependent on voltage via the Butler-Volmer Equation. The convention used is to treat the currents as positive (take absolute value). The fuel cell open-circuit voltage is given as 1 V, so the plot should approach 1 V at 0 current. At the other end, the current should approach the transport limit as the voltage approaches zero.
Plot the power of the fuel cell as a function of voltage. What is the maximum power and the operating voltage at this maximum power? Briefly explain how accounting for ohmic losses would affect the max power and the corresponding voltage.
The maximum power is about 0.39 W/cm2, which occurs at the operating voltage of about 0.42 V.
Ohmic losses means that voltage will be linearly lost as a function of current. This loss in voltage will result in a loss in the power and the max power will be at a lower voltage. Example of system with resistance (not required for homework):
List 3 ways that the limiting current of the fuel cell stack can be increased.
Decrease diffusion layer thickness
Increase the effective diffusivity (increase porosity or decrease tortuosity)
Use pure O2 rather than air
Vary the temperature
Describe 2 ways that you can refine your method and assumptions to make it more realistic.
Account for ohmic losses in voltage vs. current calculation
Capture voltage dependency on limiting current
Include contributions from the anode side of the fuel cell.
Incorporate other types of mass transport losses such as convection and membrane transport.
The kinetic analysis could be improved, as the Tafel slope and activity will probably change as a function of voltage.
Additional problem and explanation:
Let’s include the kinetic losses of the anode as well. The exchange current for your anode catalyst is 100 mA/cm2, number of electrons transferred is 2 and the transfer coefficient is 0.5.
In general, the number of electrons transferred (n) is normalized to the number of moles of a relevant reactant. So for HOR, this reaction has 2 electrons transferred per mole of H2. For ORR, the reaction has 4 electrons transferred per mole of O2.
Use the Butler-Volmer Equation:
Below is a plot including the ORR kinetics from above (taking the absolute value of the current) and the new HOR kinetics.
We now need to include kinetic overpotential losses from both the anode and the cathode. The current (number of electrons) traveling between the anode and the cathode must be identical to maintain charge balance. For a given current, we determine the kinetic overpotential losses by adding the losses from the HOR and the ORR, as shown by the arrows. The losses will depend on the current through the system. We note that the current must be identical between the cathode and anode, even though the number of electrons per mole of reactants are different. This just means that twice the number of moles of H2 is reacted in accordance with the balanced equation:
