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Sec. 2 3   Applications of the Design Equations for  Continuous-Flow Reactors   45






















                                                            0.2        0.4       0.6       0.8

                                                                Conversion, X
                                                      Figure E2-3.1  Levenspiel PFR plot.

                                   The: product of this area and FAO will give the tubular reactor volume necessary to
                                   achieve  the  specified  conversion  of  A.  For  80% conversion,  the  shaded  area  is
                                   roughly equal to 260 dm3*(s/mol). The tubular reactor volume can be deterimined
                                   by  multiplying this  area  [in  dm3.(s/mol)] by  FA,, (mol/s).  Consequently,  for  an
                                   entering molar flow rate of 0.867 mol/s the PFR volume necessary to achieve 80%
                                   conversion is 225 dm3.
                                   (c)  Sketch -rA and X down the length of the reactor. We know that as we proceed
                                   down the reactor and more and more of  the reactant is consumed, the concentration
                                    of reactant decreases, as does the rate of  disappearance of A. However, the conver-
                                    sion increases as more and more reactant is converted to product. For X  = 0.2 we
                                   cak!ulate the corresponding reactor volume using Simpson’s rule with AX  = 0.1.






                                                       [I89 + 4(192) + 2001   ~
                                                                        mol * s
                                                   L                   1 dm3
                                         = 33.4 dm3
                                    For X = 0.4, we can again use Simpson’s rule with AX  = 0.2:


                                             v = FAO -                 4      +
                                                     3
                                                    “I -rA(x  = 0) + -TA(X  = 0.2)  -rA(X = 0.4)
                                                       L
                                                          0.2 { 189 + 4(200) + 2.501


                                               = 71.6 dm3
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