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13.4  Separation of Nanoaerosol from the Air                    413

            properties of a glass fiber filter for NaCl nanoparticles 1–100 nm when filter fiber
            diameter d f ¼ 5 lm.
            Solution
            For salt particles and glass fiber filter

                                      H p ¼ 7   10  20  J

                                     H f ¼ 8:5   10  20  J

                                                    m 2
                                                  11
                                    K p ¼ 0:75   10
                                                    N
                                                    m 2
                                                  11
                                   K f ¼ 0:443   10
                                                     N
              Then the Hamaker constant
                                     p
                                 1=2   ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
                                                  20             20
                       H ¼ H p H f  ¼  7   8:5   10  J ¼ 7:71   10  J
              And the specific adhesion energy is

                                       7:71   10  20  J
                              A H                                 2
                        Dc ¼      ¼                   ¼ 0:0128 J m
                                                9 2
                             12pZ e 2  12p 0:4   10 Þ m 2
                                       ð
              The composite Young’s modulus of bodies with the mechanical constants of K p
            and K f

                          4     1       4              1

                    Y ¼               ¼              11           11
                         3p K p þ K f   3p 0:75   10   þ 0:443   10
                                 11
                       ¼ 3:51   10 ðPaÞ
              The characteristic radius of two bodies is

                                                     1
                                         1  1    1

                                     R ¼      þ
                                         2 d p  d f
              The impact contact area is determined by

                                                    1=3

                                         R

                                    a ¼    ð 2DcpR Þ
                                         Y
              Then the adhesion energy is calculated using
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