Page 249 - Distillation theory
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P1: JPJ/FFX  P2: JMT/FFX  QC: FCH/FFX  T1: FCH
            0521820928c07  CB644-Petlyuk-v1                                                      June 11, 2004  20:18





                                7.2 Distillation Trajectories of Finite Columns                   223

                                                                                2
                                                                            1
                                                                                     +
                                  Because the dimensionality of region Reg sh,R  (S − S − N ) is larger by one
                                                                      sep,r
                                                                           r
                                                                                     r
                                                                                r
                                                                       2
                                than that of its boundary element Reg min,R (S − N ), the dimensionality of the
                                                                            +
                                                                 sep,r  r   r
                                                 sh
                                set of points {x f −1 }  at given reflux larger than minimum should be larger by
                                                 lin
                                one than the dimensionality at minimum reflux. As far as at minimum reflux, the
                                                     sh
                                dimensionality of {x f −1 }  is equal to zero then at reflux larger than minimum
                                                     lin
                                                        sh                                         sh
                                the dimensionality of {x f −1 }  is equal to one; that is, the set of points {x f −1 }
                                                        lin                                        lin
                                                                            2
                                                                       1
                                is a segment lying in linear region Reg sh,R (S − S − N ). Similarly, the set of
                                                                                 +
                                                                  sep,r  r  r   r
                                         sh                                 sh,R  1   2
                                                                                           +
                                points {x f }  is a segment lying in linear region Reg  (S − S − N ). There is
                                         lin                                sep,s  s  s    s
                                a correspondence between each point of segment [x f −1 ] sh  and certain point of
                                                                                 lin
                                segment [x f ] sh  that is connected with the first one by Eq. (5.18). The greater the
                                           lin
                                                               sh
                                length of segments [x f −1 ] sh  and [x f ] , the greater the reflux number.
                                                     lin       lin
                                  One of the ends of segment [x f −1 ] sh  that we designate (x ∞  ) sh  should lie at
                                                                lin                 f −1 lin
                                                               2
                                linear boundary element Reg min,R (S − N ) lin . The section trajectory starting in
                                                                    +
                                                         sep,r  r   r
                                                                                          1
                                                                                   2
                                point (x ∞  ) sh  should pass through two stationary points S and S , that is, for
                                       f −1 lin                                    r      r
                                point (x  ∞  ) sh  the top section is infinite not only at sharp, but also at quasisharp
                                       f −1 lin
                                separation (Fig. 7.1b). The other end of the segment [x f −1 ] sh  that we designate
                                                                                   lin
                                                                         2
                                                                     1
                                    )
                                                                              +
                                (x min sh  should lie inside region Reg sh,R (S − S − N ) lin , the farthest possible
                                  f −1 lin                      sep,r  r  r   r
                                                              2
                                from boundary element Reg min,R (S − N ) lin . The section trajectory starting in
                                                                   +
                                                         sep,r  r  r
                                                                                 1
                                point (x min sh  should pass only through stationary point S , the farthest one from
                                         )
                                       f −1 lin                                  r
                                               2
                                stationary point S . Therefore, in this case, at quasisharp separation the top section
                                              r
                                has the smallest number of trays.
                                                                                             ∞ sh
                                  Similarly, for the bottom section, the ends of segment [x f ] sh  are (x )  and
                                                                                     lin     f  lin
                                                                                      ∞ sh
                                                                          sh
                                    ) .Point(x
                                                ) correspondstopoint(x
                                (x min sh     min sh                  ∞  ) andpoint(x ) corresponds
                                  f  lin      f  lin                   f −1 lin       f  lin
                                to point (x min sh  (Fig. 7.1c). In the first case at quasisharp separation, there is an
                                            )
                                         f −1 lin
                                infinite number of trays in the top section and the smallest number in the bottom
                                section, and vice versa in the second case.
                                  The smallest summary number of trays of two sections at quasisharp sepa-
                                                                                              sh
                                ration corresponds to some middle location of points (x f −1 ) sh  and (x f ) . Such
                                                                                    lin       lin
                                compositions in the feed cross-section are optimal.
                        7.2.2. Possible Compositions in Feed Cross Section
                                The coordinates of segments [x f −1 ] sh  and [x f ] sh  can be determined from purely
                                                              lin       lin
                                geometric considerations from the known coordinates of the stationary points and
                                of point x F .
                                  While solving this task, we act the way we did when we determined points x f −1
                                and x f in the mode of minimum reflux (see Section 5.6).
                                  It follows from the condition of material balance in the feed cross-section (Eq.
                                [5.18]) that segments [x f −1 ] sh  and [x f ] sh  should be parallel to each other and to
                                                       lin       lin
                                                                                    1
                                                                                         2
                                the line of intersection of surfaces or hypersurfaces Reg sh,R  (S − S − N ) lin and
                                                                                              +
                                                                                    r
                                                                               sep,r
                                                                                         r
                                                                                             r
                                             2
                                        1
                                                  +
                                                                            ) , (x )
                                Reg sh,R (S − S − N ) lin . We examine points (x min sh  ∞ sh  and x F for which
                                   sep,s  s  s   s                        f −1 lin  f  lin
                                                                                      ∞ sh
                                Eq. (5.18) should be valid. Therefore, straight line (x min sh  − (x )  − x F should
                                                                               )
                                                                             f −1 lin  f  lin
                                                                                    1
                                                                                        2
                                                                                             +
                                be the intersection line for linear manifolds Reg sh,R  ≡ (S − S − N ) lin and
                                                                            sep,r
                                                                                        r
                                                                                             r
                                                                                    r
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