Page 224 - Advanced Thermodynamics for Engineers, Second Edition
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212    CHAPTER 10 THERMODYNAMICS OF COMBUSTION




             nitrogen). It is normal to assume, at this level, that the nitrogen is an inert gas and takes no part in the
             process. Combustion of methane with air is given by


                                     79                                          79
                   CH 4  þ   2 O 2 þ   N 2   /     CO 2   þ     2H 2 O    þ  2     N 2
                                     21                                          21
                  1 kmol        9:52 kmol         1 kmol        2 kmol       7:52 kmol    (10.3)
                  12 þ 4     2  ð32 þ 105:2Þ     12 þ 32     2  ð2 þ 16Þ     7:52   28
                  16 kg         274:4kg           44kg          36kg         210:67 kg



             10.2.1 STOICHIOMETRY
             There is a clearly defined, and fixed, ratio of the masses of air and fuel that will result in complete
             combustion of the fuel. This mixture is known as a stoichiometric one and the ratio is referred to as the
             stoichiometric air-fuel ratio. The stoichiometric air-fuel ratio, ε stoic , for methane can be evaluated from
             the chemical equation (Eqn (10.3)). This gives
                                        mass of air  2  ð32 þ 105:33Þ
                                                                    ¼ 17:17
                                 ε stoic ¼        ¼
                                        mass of fuel       16
                This means that to obtain complete combustion of 1 kg CH 4 it is necessary to provide 17.17 kg of
             air. If the quantity of air is less than 17.17 kg then complete combustion will not occur and the mixture
             is known as rich. If the quantity of air is greater than that required by the stoichiometric ratio then the
             mixture is weak.

             10.2.2 COMBUSTION WITH WEAK MIXTURES

             Aweak mixture occurs when the quantity of air available for combustion is greater than the chemically
             correct quantity for complete oxidation of the fuel; this means that there is excess air available. In this
             simple analysis, neglecting reaction rates and dissociation etc., this excess air passes through the
             process without taking part in it. However, even though it does not react chemically, it has an effect on
             the combustion process simply because it lowers the temperatures achieved due to its capacity to
             absorb energy. The equation for combustion of a weak mixture is

                                 2                              1   f      7:52
                           CH 4 þ ðO 2 þ 3:76N 2 Þ/CO 2 þ 2H 2 O þ 2  O 2      N 2        (10.4)
                                 f                                f         f
             where f is called the equivalence ratio, and
                                               actual fuel   air ratio
                                                                                          (10.5)
                                        f ¼
                                            stoichiometric fuel   air ratio
                For a weak mixture f is less than unity. Consider a weak mixture with f ¼ 0.8, then

                              CH 4 þ 2:5ðO 2 þ 3:76N 2 Þ/CO 2 þ 2H 2 O þ 0:50 2   9:4N 2 :  (10.6)
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