Emf, Maximum Power and Efficiency of Fuel Cells
- 1 June 1993
- journal article
- Published by ASME International in Journal of Energy Resources Technology
- Vol. 115 (2) , 100-104
- https://doi.org/10.1115/1.2905975
Abstract
The ideal voltage of steady-flow fuel cells is usually expressed by Emf = −ΔG°/nF where ΔG° is the “Gibbs free energy of reaction” for the oxidation of the fuel at the supposed temperature of operation of the cell. Furthermore, the ideal power of the cell is expressed as the product of the fuel flow rate with this emf. Such viewpoints are flawed in several respects. While it is true that if a cell operates isothermally, the maximum conceivable electrical work output is equal to the difference between the Gibbs free energy of the incoming reactants and that of the leaving products; nevertheless, even if the cell operates isothermally, the use of the conventional ΔG° of reaction (a) assumes that the products of reaction leave separately from one another (and from any unused fuel); and (b) when ΔS of reaction is positive, it assumes that a free heat source exists at the operating temperature, whereas if ΔS is negative, it neglects the potential power which theoretically could be obtained from the heat released during oxidation. Moveover, (c) the usual cell does not operate isothermally, but (virtually) adiabatically. Comment (a) is often accounted for by employing the Nernst equation to correct for the dilution of reactants and/or products. Nevertheless, comments (b) and (c) remain pertinent. Rather than with emf, the proper starting place is with power output. The ideal power is that which would be obtained if the fuel were oxidized without irreversible entropy generation. Among other factors, this ideal power output depends upon the ratio of oxidant to fuel flow rate (e.g., air-fuel ratio) and the percentage of fuel oxidation. The ideal voltage is deduced from the ideal power, because it is defined as electrical work output per unit of charge delivered. It is a local characteristic which varies with the percent of fuel oxidized. Therefore, (d) ideal power is not equal to the product of emf with current (unless the amount of fuel utilized is infinitesimal). Examples are presented which illustrate such affects and their importance for the evaluation of ideal power and of efficiency.Keywords
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