Dissemination of IT for the Promotion of Materials Science (DoITPoMS)

DoITPoMS Teaching & Learning Packages Fuel Cells Fuel utilisation and reaction optimisation
Fuel Cells AimsBefore you startIntroductionThe principleHistory of the technologyTypes of fuel cellsSolid oxide fuel cells (SOFCs)• Electrolyte• Electrode materials• Interconnection• Thermodynamics• System and outlookMolten carbonate fuel cells (MCFCs)• Electrolyte• Electrochemistry• ElectrodesProton exchange membrane fuel cells (PEMFCs)• Membrane• Electrodes and membrane-electrode assembly• Efficiency and reaction conditionsFuelling requirementsCase study: Fuelling practicalitiesSummaryQuestionsGoing further
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Fuel utilisation and reaction optimisation

In practice, not all the fuel supplied to the anode can be reacted; some must pass through as it is. If all the fuel supplied to the electrodes was oxidised, we’d see areas of the anode at which there were lower fuel concentrations at any given time, so as a whole the fuel cell wouldn’t be operating to its full potential. Instead it’s better to have some of the fuel pass through un-reacted. We therefore need to consider the factor of fuel utilisation in our efficiency calculations. The utilisation co-efficient μf is given as:

$${\mu _f} = {{{\rm{Mass\; reacted}}} \over {{\rm{Mass\; input}}}} \approx {\rm{ 95\% }}$$

Typically, only 5% of the fuel supplied should pass through un-reacted. This of course can be re-circulated to the cell. We can put this value of μf into the calculations like this:

Recall that EMF of the hydrogen cell is given by:

$$E = {{ - \Delta {{\overline h }_f}} \over {2F}} = 1.25{\rm{ V}}$$for the lower heating value (steam product)

Using the cell voltage Vc it is possible to estimate the efficiency η:

$$\eta = {{{V_c}} \over {1.25}} \times 100\% $$

and then we take account of the fuel utilisation co-efficient like this:

$$\eta = {\mu _f}\left( {{{{V_c}} \over {1.25}}} \right) \times 100\% $$

Hence we can estimate the efficiency of a cell by measuring its voltage and knowing a little about how it works.


Voltage losses

We’ve seen the limits of efficiency, but due to voltage losses, fuel cells will never operate at these limits. There are 4 main losses to be considered:

  • Fuel crossover – The electrolyte is never a perfectly impermeable material with respect to the fuel gas and also with respect to electron conductivity. It should be noted that an electron pair crossing the electrolyte from the cathode to anode is essentially the same as a molecule of fuel crossing the other way. Both events serve to waste just a little fuel. Fuel crossover is summarised in the animation below:

  • Ohmic losses – These are simply losses due the resistance of the system. These become significant in higher temperature cells such as SOFCs and much effort goes into minimising loss of this kind.

  • Mass transfer losses – This refers to the fact that concentration of reactants at the electrodes affects voltage, as we have seen using the Nernst equations. If the reactant concentrations fall in some area of the electrode for some reason, then the cell does not work to its maximum efficiency. The extent of the concentration changes around the electrodes depends on the current drawn and on how well the reactant gases are circulated.

  • Activation losses – This refers to the slowness of the reaction due to the activation over-potential needed to make it work. Activation over-potential follows a Tafel equation like this:

$$\Delta {V_{act}} = A\ln \left( {{i \over {{i_0}}}} \right)$$

Where i is the current density and i0 is the exchange current density at which we see no net reaction.

Each fuel cell type will be more susceptible to a different kind of voltage loss. These are discussed in more detail in their respective sections.


Optimum pressure

In equations (10) and (11), we introduced the concept of activity as being an important factor determining the EMF of a cell. Activity has a large dependence on both ambient pressure and the partial pressure of the reactant gas. Higher pressure leads to higher power but pressurised systems are heavier and require more energy to run. Therefore determining the optimum pressure for a given application and system is a difficult task, and it’s often solved by a computer program.

Note: If dealing with PEMFCs, a minimum of 2 bar at 80 °C must be maintained to achieve the humidity required for the membrane to conduct.


Optimum humidity

A PEM requires a certain degree of humidity to function, this is often achieved by complicated water management equipment and control systems. The optimum humidity is once more a difficult parameter to calculate. Too dry and the membrane won’t conduct H+, where as too humid and product water won’t dissipate, clogging the electrodes and preventing gases from reaching the catalysts.

 

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