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Is the Carnot efficiency valid for fuel cells?

  1. Jun 5, 2012 #1
    So, I've read in books and on Wikipedia (see for instance http://en.wikipedia.org/wiki/Carnot...cs)#Applicability_to_fuel_cells_and_batteries) that the Carnot efficiency cannot be applied to a fuel cell because it is not a heat engine that produces work, operating between two heat reservoirs.

    However, I have managed to come across the following article, which claims to "dispel the misconception that an ideal fuel cell is potentially more efficient than an ideal heat engine". (Taken from the abstract of

    "Thermodynamic comparison of fuel cells to the Carnot cycle", International Journal of Hydrogen Energy, Volume 27, Issue 10, October 2002, Pages 1103-1111
    Andrew E. Lutz, Richard S. Larson, Jay O. Keller)

    I don't know if copying and pasting from a research journal is allowed here, and I don't think I'd do justice by trying to summarize what it says, so I hope the people that are interested in discussing this topic have access to the article (it's available at ScienceDirect, if you have access to that.)

    So, is the article right? If it is, can the Carnot efficiency be applied to any physical system that produces work, and not just fuel cells?

    I'd appreciate to hear your thoughts, thanks. :biggrin:
  2. jcsd
  3. Jun 5, 2012 #2


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    Staff: Mentor

    The title doesn't imply to me that they applied a Carnot efficiency calculation to a fuel cell - and I'm not sure how you would (just look at the equation!). No, carnot efficiency applies to heat engines only.
  4. Jun 6, 2012 #3
    Hey, thanks for answering. :smile:

    Well, your particular question is addressed in section 4 of the article. They state:

    The first thing they do to answer the question is to consider the high temperature reservoir:

    They then introduce a modified heat engine, where the high temperature reservoir is replaced by a combustion reactor.

    In the associated sketch, the high temperature reservoir has H2+O2 going in it, a ΔH_R being produced where the high temperature reservoir is, and H2O coming out. The rest of the engine is a typical heat engine, with a cycle producing work and a low temperature (cold) reservoir receiving heat from the cycle.

    They then state that to obtain the maximum Carnot efficiency, we want the highest temperature possible in the reactor. This will happen when the change in Gibbs free energy vanishes.

    [tex]\Delta G_R = 0 = \Delta H_R - T \Delta S_R[/tex]​

    With this in mind, they then define a combustion temperature T_C, that is

    [tex]T_C = \left. \frac{\Delta G_R}{\Delta S_R} \right |_{T_C}[/tex]​

    Now with this temperature in mind, they apply the First law to the reactor to arrive at

    [tex]Q_H = -\Delta H_R (T_C)[/tex]​

    Juggling around with the Second law, they arrive at the other additional result:

    [tex]Q_L = -T_L \Delta S_R(T_C)[/tex]​

    where [tex]T_L[/tex] is the temperature of the cold reservoir.


    [tex]\eta_{\text{Carnot}} = \frac{Q_H - Q_L}{Q_H} = \frac{\Delta H_R (T_C) - T_L \Delta S_R(T_C)}{\Delta H_R (T_C)}[/tex]​

    To finish showing the equivalence, the authors use an approximation.


    [tex]\Delta H_R (T_C) \approx \Delta H_R (T_L),[/tex]

    [tex]\Delta S_R (T_C) \approx \Delta S_R (T_L).[/tex]​


    [tex]\eta_{\text{Carnot}} = \frac{\Delta H_R (T_C) - T_L \Delta S_R(T_C)}{\Delta H_R (T_C)} \approx \frac{\Delta H_R (T_L) - T_L \Delta S_R(T_L)}{\Delta H_R (T_L)} = \left.\left( \frac{\Delta G_R}{\Delta H_R} \right)\right|_{T_L}[/tex]​

    Now, if we go back a bit, any process that takes in energy to produce work has a First law efficiency defined by

    [tex]\eta = \frac{W_\text{out}}{Q_\text{in}}[/tex]​

    For a fuel cell, both terms need to be evaluated with the help of the First law of Thermodynamics. Assuming a constant pressure process that is done with reversible heat transfer, and using the Gibbs free energy formula of H - TS, the authors arrive at

    [tex]W_\text{out} = - \Delta G_R,[/tex]​


    [tex]\Delta G_R = G(T, \text{Products}) - G(T, \text{Reactants})[/tex]​

    Which is presumably T_L from before.

    The Q_in term is evaluated as the change in enthalpy for the reaction,

    [tex]Q_\text{in} = - \Delta H_R[/tex]​

    So we have

    [tex]\eta_\text{fuel cell} = \frac{W_\text{out}}{Q_\text{in}} = \frac{\Delta G_R(T_L)}{\Delta H_R(T_L)} = \left.\left( \frac{\Delta G_R}{\Delta H_R} \right)\right|_{T_L}[/tex]​

    Having finished their proof in section 4, the authors state

    They then go on to analyze some example reactions to show a good agreement between their derivation and what the actual calculations yield.

    So, is this derivation any good, or are there any flaws?

    I hope this helped, and I hope I didn't copy and paste too much to warrant a copyright infringement! :tongue:
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