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206                          7 Combustion Process and Air Emission Formation

            7.5.2 Char Oxidation

            Similar to the volatile oxidation analysis above, a simple model [13, p. 150] for the
            oxidation of char or other similar solid carbon-based fuels is
                                           1   k f1
                                      C s þ O 2 ! CO                     ð7:21Þ
                                           2

                                               k f2
                                      C s þ CO 2 ! CO                    ð7:22Þ
            where C s is in solid phase carbon. The oxidizers are not limited to O 2 and CO 2 ;
            other species such as OH, O, and H 2 O can also be effective oxidizers.
              The oxidation takes place at the surface of the solid char and the corresponding
            rate of reaction is governed by the diffusion of oxidizers to the surface of the char.
              The apparent rate of char oxidation can be estimated using the equation given by
            Flagan and Seinfeld [13]

                                        E     n           2

                            r ¼ Aexp        P        kg=m   s            ð7:23Þ
                                       RT    O 2
            where P n  is the partial pressure of the oxidier, with a unit of atm. A, E/R, and n are
                  O 2
            constants for specific solid fuels and they can be determined experimentally.
              The formation of carbon monoxide and hydrogen carbon is driven by local fuel
            rich combustion in the flame and imperfect mixing between fuel and oxygen.
              The level of CO in the exhaust or flue gas depends on the design of the com-
            bustion system. Oxidation of CO to CO 2 takes place in the luminous zone by the
            following mechanisms [26].

                               CO þ OH ! CO 2 þ H
                                                                         ð7:24Þ

                                                    3
                          k f1 ¼ 4:4T  1:5  exp 372=TÞ  m mol   s
                                       ð
                               H þ CO 2 ; ! CO þ O þ H
                                                                         ð7:25Þ
                                 k f2 ffi 0
                              CO þ O 2 ! CO 2 þ O
                                                                         ð7:26Þ
                                      6                   3
                          k f3 ¼ 2:5   10 exp  24;060=Tð  Þ  m mol   s
                              CO þ O þ M ! CO 2 þ M
                                                                         ð7:27Þ
                                                  6

                          k f4 ¼ 53exp 2;285=Tð  Þ  m mol   s
                             CO þ HO 2 ! CO 2 þ OH
                                                                         ð7:28Þ
                                8                   3
                    k f5 ¼ 1:5   10 exp  11;900=Tð  Þ  m mol   s:
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