Page 79 - Gas Wettability of Reservoir Rock Surfaces with Porous Media
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Evaluation Methods and Influencing Factors               CHAPTER 2           63




              Table 2.10  Surface Energy of Measuring Liquids
                                                                         2
                                     p
                                                           2
                                            2
                                                                                 P
                                                   d
              Measuring Liquids     γ (mJ/m )     γ (mJ/m )      γ (mJ/m )     γ =γ d L  Polarity
                                                                  L
                                     L
                                                                                 L
                                                   L
              Water                 51            21.8           72.8          2.36      Polarity
              N-hexadecane          0             27.6           27.6          0         Nonpolarity
              γ 5 γ 1 γ P                                                   (2.20)
                   d
               L   L   L
                             d d 1=2    p p 1=2
              γ ð1 1 cosθÞ 5 2ðγ γ Þ  1 2ðγ γ Þ                             (2.21)
               L
                                          L
                                        s
                             s
                               L
              In the equation, the free energy γ of the solid surface can be divided into dis-
                                           s
                                            p
                          D
              persion part γ and polarity part γ s .
                          s
              γ  Free energy of the liquid surface can also be divided into dispersion part
               L
                                p
               d
              γ and polarity part γ .
               L                L
              From the Eq. (2.21) it can be seen that if the surface free energy γ , dispersion
                                                                       L
                                      p
                   d
              part γ , and polarity part γ of liquid are known and contact angle θ of the
                                      L
                   L
              liquid on solid surface is measured, it still results in two unknown values, γ d
                   p                                                             s
              and γ s , in the equation. Two equations are needed to obtain these unknown
              values. Thus two kinds of measuring liquids are adopted, and the equations
              are established as follows:
                               d d  1=2   P P  1=2
              γ ð1 1 cosθ 1 Þ 5 2ðγ γ Þ  1 2ðγ γ Þ                          (2.22)
               L1
                                          s
                                 L1
                                            L1
                               s
                               d d  1=2   P P  1=2
              γ ð1 1 cosθ 2 Þ 5 2ðγ γ Þ  1 2ðγ γ Þ                          (2.23)
               L2
                               s
                                            L2
                                 L2
                                          s
              The data in Table 2.10 is used in Eqs. (2.19) (2.23)
                                         d d
                                                      P P
              Water: γ  ð1 1 cosθ water Þ 5 2ðγ γ  1=2  1 2ðγ γ  1=2        (2.24)
                     water               s  water  Þ  s  water  Þ
              Inn-hexadecane:γ inn2hexadecane  ð1 1 cosθ inn2hexadecane Þ
                                                                            (2.25)
                   d d
              5 2ðγ γ          1=2  1 2ðγ γ        1=2
                                       P P
                   s  inn2hexadecane Þ  s  inn2hexadecane Þ
              Following are acquired by using the polarity part and dispersion part of water
              and n-hexadecane into Eqs. (2.24) and (2.25):
                                p ffiffiffiffiffiffi q ffiffiffiffiffi  p ffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffi
                                       P
              72:8ð1 1 cosθ water Þ 5 2 51  γ 1 2 21:8  γ d                 (2.26)
                                       S            S
                                       p ffiffiffi q ffiffiffiffiffi  p ffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffi
                                             P
              27:6ð1 1 cosθ inn2hexadecane Þ 5 2 0  γ 1 2 27:6  γ d         (2.27)
                                             S            S
              The following are acquired by working out the Eqs. (2.26) and (2.27):
              q ffiffiffiffiffi  126:8 1 191:23cosθ water 2 64:43cosθ inn2hexadecane
                 P
                γ 5                                                         (2.28)
                 S
                                      37:52
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