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05_chap_wang.qxd  05/05/2004  3:46 pm  Page 207
                    Wet and Dry Scrubbing                                                     207
































                                    Fig. 2. Flooding correction in randomly packed towers.




                       The actual determination of the tower diameter is as follows:
                       Calculate the abscissa (ABS):
                                                ABS = (L / G)(D / D ) 0.5                      (6)
                                                              G   L
                    where L is the solvent flow rate (lb/hr) = (MW  )(L  ), G is the gas flow rate (lb/hr) =
                                                            solvent  mol
                                                                3
                    (MW )(G    ), D is the density of the liquid (lb/ft ), and D is the density of the gas or
                         e   mol   L                                    G
                                       3
                    emission stream (lb/ft ).
                       D can be approximated as
                        G
                                                     D = PM/RT                                (6a)
                                                       G
                    where P is the pressure (atm) (note, normally 1), M is the molecular weight of gas (lb/lb-
                                                     3
                    mol), R is the gas constant (0.7302 ft atm/lb-mol ºR), and T is the temperature (ºR).
                       The values for L and G are determined by multiplying L  and G  by their respec-
                                                                        mol     mol
                    tive molecular weights. Next, the flooding line in Fig. 2 is used to read the ordinate
                    (ORD). Now, the ordinate expression for G  at flooding is
                                                           area, f
                                                     2
                                                          3
                                      ORD = [(G     ) (a/e )(µ  0.2 )] / [(D D g )]            (7)
                                                area, f       L        G  L  c
                    Solving for G    , we have
                                area, f
                                                                   3
                                        G     = {(ORD D D g ) / [(a/e )(µ  0.2 )]} 0.5         (8)
                                          area, f       G  L c        L
                    where G     is the gas stream flow rate based on tower cross-sectional area at the flood-
                            area,f
                                  2
                    ing point (lb/ ft -s), µ is the viscosity of the scrubbing (cP) (1 when using water), and
                                       L
                                                        2
                    g is the gravitational constant (32.2 ft/s ). Note that a and e are packing factors (11).
                     c
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