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2.3 Gas–Liquid Interfacial Behavior                             55

            Fig. 2.10 Interfacial mass                        Driving
            transfer diving force                 y           force    Equilibrium
                                                                       line
                                                      Driving
                                             Gas      force
                                            phase
                                             mole  y*
                                            ratio, y



                                                            x      x*
                                                       Liquid phase molar ratio, x

                                   0
                                  n ¼ K y y   y Þ  gas phase             ð2:83Þ
                                        ð
                                  0
                                 n ¼ K x x   xð  Þ liquid phase          ð2:84Þ
            where K x , K y = Overall mass transfer coefficients for the liquid and gas phases
                            2
            respectively (mol/m s). x ; y ¼ Hypothetical mole fraction corresponding to y, x


            in the bulk fluids, and they are determined by the Henry’s law as
                                          y ¼ Hx                         ð2:85Þ

                                          y ¼ Hx                         ð2:86Þ
              The difference between the hypothetical mole fraction and the corresponding
            actual mole fraction results in the driving force illustrated in Fig. 2.10.
              If we consider gas phase only in Eq. (2.83) above, the total mass transfer rate
                  2
            (mole/m s) is described as
                             0
                                              ð
                                             ½
                            n ¼ K y y   yð  Þ ¼ K y y   y i Þ þ y i   y ފ  ð2:87Þ
                                                      ð
              Dividing both sides by K y gives,
                                   n 0
                                       ð
                                     ¼ y   y i Þ þ Hx i   xÞ             ð2:88Þ
                                                 ð
                                  K y
              The single-phase mass transfer Eqs. (2.81) and (2.84) above also give
                                 n 0             n 0
                                   ¼ y   y i  and  ¼ x i   x             ð2:89Þ
                                k y              k x

              Substitute Eq. (2.89) into (2.88) and dividing both sides with n , one can get
                                                                  0
                                        1    1   H
                                          ¼    þ                         ð2:90Þ
                                        K y  k y  k x
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