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382                                                       Chapter 7

                                                                      3
           maximum  capacity,  and  maintain  the  volumetric  flow  rate  at  5600  ft /h  (21.2
           m3/h). In this case, the residence time increases because  of the  increased reaction
           volume,  increasing  the  conversion.  Choosing  the  third  option,  substitute  1069  ft3
                          3
           (8000 gal, 30.3 m )  for the reaction volume and the  same values for x 1A, V v, m 1A,
           and k  into the equations above for n =  1, and  solve  for x 2A.  Using POLYMATH
            [22],  the  conversion is  now  0.4182.  This  design will  give  some  flexibility.  If the
           demand  for product  increases,  the  feed  rate  can be  increased,  but  the  conversion
           will decrease. The original required conversion is 0.37.
                The mole balance can now be completed for one CSTR. The inlet molar flow
           rate  for propylene  oxide is  calculated above. The  inlet molar  flow  rate  of metha-
           nol,

                  800
           m 1M  = ——— 0.7914 (62.43) =  1,234 Ibmol/h (560 kgmol/h)
                 32.04
           and the inlet molar flow rate of water,

                 4000
               = ——— 0.9941 (62.43) = 13,780 Ibmol/h (6240 kg/mol/h)
           m  w
                 18.02
           The inlet  flow  rates (stream  1) are entered into Table 7.1.2. For x = 0.4182, the
                                                                2A
           outlet flow rates are also entered into Table 7.1.2.
                Next, select a heat exchanger and calculate the heat transfer  area. First, cal-
           culate the required heat transfer,  Q, from  an energy balance. Obtain the enthalpy
                                        n
           of reaction from  Equation  7.4.19  and the  standard  enthalpies  of reaction  listed  in
           Table 7.1.1.
           AH° R = - 222,600 - (-  123, 000 -  66,600)

           AH° R = -33,000 Btu/lbmol (-76,760 kJ/kgmol)

                The  enthalpy flowing  into  and out of the reactor for each component is cal-
           culated relative to  25 °C (77  °F), using  heat capacities  from  Table  7.1.1.  The  re-
           sults are contained in Table 7.1.3.
                Solve for the required heat transferred,  Q n, using the energy equation, Equa-
           tion 7.4.2 in Table 7.4.  Substituting the enthalpy of reaction, the enthalpy into the
           reactor, and the enthalpy out of the reactor, obtained from Table 7.1.3, we find  that
           Q n = -  2,085,000 -  33,000 (308.9) +  9,465,000)

                                      s
                        6
           Q n = -2.814xl0  Btu/h (-2.97xlO  kJ/h)



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