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44                                  Conversion and Reactor Sizing   Chap. 2


                                                           dm.s
                                           V = 0.867  [ 800 mol (OX)]  = 554.9 dm3
                                                    S
                             The CSTR volume necessary to achieve 80% conversion at the specified temper&-
                            ture and pressure is 555 dm3.



                            Example 2-3  Sizing a PFR

                            The reaction described by  the data in Tables 2-1  and 2-2 is to be carried out in a
                            PFR. The entering molar flow rate of A is 0.867 mol/s. Calculate the reactor volume
                            necessary to achieve 80% conversion in a PFR (a) First, use one of  the integration
                            formulas given in Appendix A.4  to  determine the PFR reactor volume.  (b) Next,
                             shade the area in Figure 2-1 which when multiplied by FA0 would give the PFR vol-
                             ume.  (c) Make a qualitative sketch of  the conversion, X, and the rate of  reaction,
                             -rA,  down the length (volume) of the reactor.

                            Solution
                            (a) For the PFR, the differential form of the mole balance is

                                                                                        (2-15)


                            Rearranging and integrating gives

                                                                                        (2-16)


                            For 80% conversion, we will use the five-point quadratic formula with AX = 0.2.





                                               1    +-+-   4      2        4
                                                                        -rA(0.6)
                                           -rA(X  = 0)  -rA(0.2)   -rA(0.4)  f-+-   -rA(0.8)
                            Using values of  1 / - r,  in Table 2-2 yields

                                                                                    s.dm3
                               V = (0.867 mol/s)(0.2/3)[189  + 4(200) + Z(250) + 4(400) + (800)J mol
                                 = (0.867 mol/s)(259.3 s  dm3/m01)
                                 = 225 dm3
                            (b) The integral in Equation (2-16) can be evaluated for the area under the curve of
                            a plot of (l/-rA) versus X.


                                  -!!-  % = area under the curve between X  = 0 and X = 0.8
                                     =
                                  F*O            (see appropriate shaded area in Figure E2-3.1)
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