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306  Chapter 12: Batch Reactors (BR)

                           values of  C,  for A, R, and S are, respectively, 185.6, 104.7, and 80.9 J  mol-’  K-l.  The
                           enthalpy of reaction is -6280 J (mol A)-l,  and the reaction is first-order with respect to A,
                           with   kA  =  1()‘4e-‘0,000/T   h-l.   D t        and  T  versus  t,  if the process
                                                      e ermine the profiles of  fA
                           is adiabatic, and T and t for fA  = 0.99.

      SOLUTION

                           We first obtain the appropriate forms of the material-balance and energy-balance equations
                           (12.3-3 1 and 12-3-29, respectively) for use in the above algorithm.
                             From the given rate law, we have

                                            (-YA)  = kAcA  = k,n,lV  = k,n,,(l   -  fA)IV


                           so that equation 12.3-3 1 becomes
                                                       t=         d f*                           (4


                                                               k,(l   -  fA)
                           where

                                                       kA  =  1014e-10,000/T
                                                                                                  (B)
                             For equation 12.3-29, we may not be able to use the simplified result in equation 12.3-
                           30, since  n,Cp   may not be constant but depend on fA;  to investigate this, we form

                                               = n, C xiCpi  =  n, 2 (niln,)Cpi
                                          %CP
                                               =  2  nicpj   =  nACpA   +  nRCpR  + nscpS
                                               = nA,(l-f~)C~~ $- nAofACPR +nAofACPS
                                               = nA,[CPA + ccPR  + cPS - cPA)fA]
                                               =  1856n,,

                           Since  ntCp   is  independent of fA,  we  cm  use equation 12.3-30:

                                              T = 300 + ;;;e2A;  fA =  300 + 33.8fA               0


                           Equations (A), (B) and (C) are used in the algorithm to obtain the information required.
                           Step (3) is used to calculate  kA   from equation (B), and step (4) is not required. Results are
                           summarized in Table 12.2, for  the  arbitrary step-size in  fA  indicated; G = ll[k,(  1 - fA)],
                           and  G*  represents the average of two consecutive values of G. The last column lists the
                           time required to achieve the corresponding conversion in the second column. These times
                           were obtained as approximations for the value of the integral in equation (A) by means of
                           the trapezoidal rule:

                                              tj  =  tj-1  +  0*5(G,   +  Gj-l)(fAj  -  fA,j-1)
                           The estimated time required to achieve a fractional conversion of 0.99 is 1.80 h, and  the
                           temperature at this time is 333.5 K, if the reactor operates adiabatically. The  fA(t)  profile
                           is given by the values listed in the second and last columns; the  T(t)  profile is given by
                           the third and last columns.
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