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Reactor Design                                                  389


                 Then, for 95% conversion the reaction time,

            t R = (1 / 0.02102) hi [1 / (1 -  0.95)] = 142.5 min (2.375 h)

                From Equation 7.8.5,  calculate  the  batch time,  t&. Because  we  do  not  have
            values for t, t, 1c, and tc for this reaction, we will have to make use of the times
                     F  H     B
            given in Table 7.10 for polymerization reactions. Except for the charging time and
            cooling times,  select the worst case  from  Table 7.10. We will assume that it takes
            the same time to cool the reactor as it does to heat the reactor.  We have some con-
            trol over the time it takes to charge the reactor.  By adjusting  a control valve, as-
            sume  that  we  can  charge  the  reactor  in  1.5  hours.  Thus,  the  batch  time,  from
            Equation 7.8.5,

            t B = 1.5 + 2.0 + 2.375 +  2.0 + 1.0 = 8.875 h

                 Now,  calculate the reaction volume from Equation 7.8.6.

                                                       3
            V r =  1000 (8.875) / 56.2 = 157.9 ft 3  (1181 gal, 4.47 m )
                 Next,  select  a  standard  (4.54 m3)  reactor  size,  from  Equation 7.8.7.  From
            Table 7.3, we  find  that  there  is  a  1200 gal  standard  reactor.  To  allow  for  some
            flexibility  select a  1500 gal (5.68 m3) reactor. Even if the production rate requires
            1181  gal, the reactor will be filled to  1500 gal, which increases the production rate.
                Now,  we have to decide on how to remove the enthalpy of reaction -  using a
            jacket,  a  coil,  a coil  and  a jacket  or  an  external  heat  exchanger.  First,  check  if  a
            jacket  will  suffice.  Because  the  reaction  is  an  unsteady-state  process,  the  heat
            transfer  will vary with time. Initially, the reaction rate will be a maximum because
            the  concentration  of  acetylated  castor  oil  (AO) is  at its  maximum value.  As the
            reaction  proceeds,  the  concentration  of  acetylated  oil  will  decrease,  as  will  the
            heat-transfer  rate.  In  this problem Ah R is  given. Thus,  we do not  need  Equations
            7.8.2  and 7.8.18 to  7.8.20. From Equation 7.8.4,  calculate the initial rate of reac-
            tion.

                               1     min       Ib acid     Ib AO       Ib acid
            r Ao = kc Ao = 0.02102  —— 60  —— 0.156  ——— 56.20  ———=11.06 ———
                                                                         3
                              min     h        IbAO         ft 3       ft -h
                Now,  from  Equation  7.8.1 calculate  Q R.  The  reaction  volume  now  equals
            1500  gal.
                450.0  Btu   11.06 Ibacid  1500 gal  1   ft 3
            Q  =  ______________  _____________  ___________  ________
                                    3
                   1   Ibacid  1   ft -h  1       7.481 gal




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