Page 64 - Applied Process Design For Chemical And Petrochemical Plants Volume II
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Distillation                                          53


             Example 8-15: Batch Distillation, Vapor Boil-up Rate for
             Fixed Trays (used by permission of Treybal[129];
             clarification added by this author)

               Distill a small quantity each day to obtain relatively pure
             *xylene  from a mixture of ortho and para xylene, having
             boiling points of  142.7"C and  138.4"C, respectively. The
             feed is 15 lbmols  (about 225 gallons) per batch, at 0.20
             mol fraction para. The desired residue product is 0.020 in
             the kettle, while the distillate is to be 0.400 mol fraction   N,i,   +1=-- 1'514  -24.6
             para. A  distillation column  equivalent to  50  theoretical   0.06145
             plate is to be used.
               The time requirement is  to  complete the distillation/   Nmin (in column) + 1 (kettle) = 24.6
             recovery in  six  hours,  allowing an  extra  two hours for
             charging, emptying, and cleaning. What is  the constant   The results indicate that 25 theoretical plates are mini-
             rate that the distillation must be carried out?       mum; then by assuming an efficiency of 50%, total actual
                                                                   trays of  50 should be adequate. Choose values of k (see
             F = 15 lb-mols/hr                                     nomenclature) and solve for b and xs by:

             Xf  = 0.20                                            b = (ak/c) - k

             XD  = 0.400                                           c = 1 + (a - 1)k

             x,  = 0.02                                            AX:  + Bx, + C = 0
             The material balance:                                   The tabulated results are:


                                                                             k             b             xs
             D = 15 (0.2 - 0.02)/(0.400  - 0.020) = 7.105 mols             0.0200       0.00297        0.01899
                                                                           0.0500       0.00716       0.04842
             Then, F = D - W                                               0.0750       0.01043        0.07301
                                                                           0.1000       0.01347        0.09728
             W = 15 - 7.105 = '7.893 mols                                  0.1250       0.0 163 1      0.12109
                                                                           0.1500       0.01895        0.14470
             el = D/G                                                      0.1750       0.02137        0.17145
                                                                           0.2000       0.02360        0.18133
             and, 0 = 7.105/G
                                                                     Graphical integration shows the area under the curve,
                    (1 - xsi la   a-1  (1 - Xf 1"                  Figure 8-38A, to be 15.764. Applying this to:
                      Xsi
                                                                   02 = (W/G)r (dx, / b)
             (1 - 0.18330)1.'52           (1 - 0.2)1.'52                      xw
                 0.18530                     0.2
                                                                   Then, 02 = 7.895 (15.764)/G = 124.46/G
                        4.26 = 4.26
                                                                   81 + 02 = 6 hr = 7.105/G + 124.46/G
               At  138.4"C, the vapor pressures of  ortho and para are
             660 and  '760 mm  Hg, respectively. Because Raoult's  law   G = 21.93 Ib-mols vapor/hr
             applies:
                                                                     This is  the boilup rate, which is approximately 3.3 ft3
             CI  = 760/660  = 1.132                                vapor/sec. An  approximately 1 ft 0 in. diameter column
                                                                   can handle this rate; however, because it is in the usual size
               Solving the equation by  trial and error shows that xsi =   for a packed tower (or cartridge trays), the diameter must
             0.18330.  Solving  for  the  minimum  number  of  plates   be checked using the packed tower calculations in Chap-
             required:                                             ter 9 of this volume.
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