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3.4 Thermodynamics of Combustion System                         83

              Since a typical flue gas or engine exhaust is at temperature greater than 100 °C,
            where water exists as vapor, LHVs are more relevant than HHVs to fuel com-
            bustion analysis. The heating values of typical fuels are available in the literature,
            for example, Fundamentals of Combustion Processes by McAllister et al. [12].
              Table A.5 shows the HHV and LHV of some typical fuels.
              The LHV can be related to the enthalpy of formation of a fuel. Enthalpy of
            formation is available at certain temperature, typically T = 298 K.

                                "                     #
                              1  X      o     X    o
                 LHV 298KÞ ¼         n i h f ;i     n i h f ;i  ¼ DH R 298Kð  Þ  ð3:59Þ
                     ð
                              n f
                                  R           P
            where the enthalpy of formation of the fuel is used for the calculation of the total
            enthalpy of the reactants.
            Example 3.9: Heating values
            Consider stoichiometric combustion of methane with pure oxygen proceeds as
            CH 4 þ 2O 2 ! CO 2 þ 2H 2 O. The mass-based HHV of methane is 55.65 MJ/kg at
            298 K and 1 atm, estimate (a) LHV of CH 4 , (b) enthalpy of formation of methane.
            Solution
            The conversion from HHV (MJ/kg) to HHV (MJ/kmole) is needed by considering
            the molar weight of the fuel given based on mass, and it can be converted by

                        HHVðMJ=kmoleÞ¼ HHVðMJ=kgÞ  M fuel
                                       ¼ 55:65ðMJ=kgÞ  16 ðkg=kmoleÞ
                                       ¼ 890:4MJ=kmole at 298KÞ
                                                      ð


            (a) The LHV can be determined by considering the latent heat of vaporization of
                water at 298 K, which is h fg ¼ 43:92MJ=kmole.

                                           n H 2 O
                             LHV ¼ HHV         h fg
                                           n fuel
                                  ¼ 890:4   2   43:92  ð MJ=kmoleÞ
                                                                         ð3:60Þ
                                  ¼ 802:6 ð MJ=kmoleÞ
                                  ¼ 802:6 ð MJ=kmoleÞ or 50:1MJ=kg

            (b) The combustion CH 4 þ 2O 2 ! CO 2 þ 2H 2 O gives


                               h o  þ n H 2 O h o      h o       h o     ð3:61Þ
                                           f ;H 2 O    n CH 4 f ;CH 4  þ n O 2 f ;O 2
                   LHV ¼ n CO 2 f ;CO 2
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