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Sec. 6.3   Algorithm for  Solution to Complex Reactions        309
                                     3.  Stoichiometry   u  = uo                            (ECi-7.8)

                                                        FH  = vOCH                          (E6-7.9)

                                                       FM  = VOCM                          (E6-7.10)

                                                        Fx  =v&                            (E6-7.11)

                                                        FT  = VOCT = (FMO - FM) - Fx       (E6-7.12)

                                                       FMe  = UOCMe = vO (CHO - cH  )      (E6-7.13)
                                     4. Combining and letting z = V/u, (space-time) yields:

                                                    c~0-c~ =(ki(C:’CM  + k2C;’Cx)~         (E6-7.14)


                                                    CM, - CM = (k, ci’cc,) T               (E6-’7.15)
                                                         CX =(klc, cM-k~c~’Cx)‘T
                                                                 112
                                                                                           (E6’7.16)
                                  Next, we put these equations in a form such that they can be readily  solved using
                                  POIXMATH.


                                                      0
                                                                                           (E6-’7.17)
                                               f(c~) = ~H-~HO+(~~C:~CM+~~C~’C~)T
                                                    =
                                              f(cM)  = o= c~-cp,,jo+k,C;~CMT               (E6-’7.18)
                                               f(cx) = 0 = (k,CFCM-k,C[’Cx)z-  c,          (E6-’1.19)
                                  The POLYMATH program and  solution are  shown  in  Table E6-7.1.  The problem
                                  was solved for different values of  z and the results are plotted in Figure E6-7.1.
                                       For a space-time of  z  = 0.5, the exiting concentrations are CH = 0.0089,
                                  CM = 0.0029, and Cx = 0.0033. The overall conversion is
                                       Hydrogen:  X,  = - -                           -
                                                      FHO - FH  CHO - CH - 0.021 - 0.0089 - o.58
                                                                        -
                                                              =
                                                        FHO       cHO        0.02 1
                                       Mesitylene:  XM = -              -              -
                                                      FMO- FM = CMO - CM  0.0105 - 0.0029 - o.72
                                                                        -
                                                        FMO                   0.0 105

                                               TABLE E6-7.1.  POLYMATH how AND SOLUTION
                                                                                      ~~~
                                                    Equations                          Initial Values
                                   fC~ch~=ch-.0rM+~55.2W~mWchWW.5+30.2X~xXch~.5~~tau     0.006
                                   f <cml=cm-.  OIO59CS5. PXcmXchW. 5)Xtau               0.0033
                                   fC1:x>=~55.2Wcm#~M~.5-30.2XoxXchXX.S~Xtau-cx          0.005
                                   I tacJ=O.  5
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