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282                                                          APPENDIX A












           Fig. A.10. Changes in contact angle as a result of movement of water–oil interface. y is the contact angle
           at static position, y a is the contact angle when oil is displaced by water (advancing angle), and y b is the
           contact angle when water is displaced by oil (receding angle).














             Fig. A.11. Flow through a two-branch capillary and trapping of oil in a small-diameter capillary.

           and
                          2
               n ¼ q=A ¼ d Dp =32 mL                                         (A.18)
                             t
                                             3
           where q is the volumetric rate of flow (cm /sec), d is the diameter of capillary (cm), Dp t
                                                                  2
           is the total pressure drop (dyn/cm), A is the cross-sectional area (cm ), m is the viscosity
           (cP), L is the flow path length (cm), and n is the velocity (cm/s).
             The total pressure drop, Dp , is equal to
                                     t
               Dp ¼ Dp þ P c .                                               (A.19)
                  t    i
           where P c is the capillary pressure (see Eq. A.1) and Dp is the applied pressure (dyn/
                                                         i
             2
           cm ). Solving for n in each capillary, by combining Eqs. A.1, A.18, and A.19 gives
                     2
               n 1 ¼ d =32 m L 1 ðDp þ 4s cos y=d 1 Þ                        (A.20)
                     1    1     i
           and
                     2
               n 2 ¼ d =32 m L 2 ðDp þ 4s cos y=d 2 Þ                        (A.21)
                                i
                          2
                     2
             Setting L 1 ¼ L 2 and m ¼ m , and dividing Eq. A.20 by Eq. A.21 gives the
                                 1
                                      2
           following relationship:
                        2                 2
               n 1 =n 2 ¼ ðd Dp þ 4s cos yd 1 Þ=ðd Dp þ 4s cos yd 2 Þ        (A.22)
                                          2
                                             i
                           i
                        1
             Therefore, when Dp   P c ,
                              i
                       2
               n 1 =n 2   d =d 2                                             (A.23)
                       1  2
           and when Dp   P c ,
                      i
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