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108                                     NATURAL GASES AND CONDENSATES

             In practice, however, the Clapeyron equation with the compressibility factor Z
           (due to intermolecular attraction) is used:
               pV ¼ ZmRT                                                      (6.7)
           where m is the mass of gas, R ¼ the gas constant, T ¼ the absolute temperature, and
           Z ¼ the compressibility of the natural gas (which is a function of pressure, temper-
           ature, and gas composition). For ideal gases, Z ¼ 1. The compressibility of a gas
           mixture may be computed as a weighted average of compressibility of all components.

           6.3.4. Deviation of pressure at bottom of gas column

             If point 2 (in a gas column) lies at distance DL below point 1 ( p 1 in psi), the
           pressure at point 2 ( p 2 in psi) is equal to
               p ¼ p þ ðg DLÞ=144                                             (6.8)
                2    1
           where g is the specific weight of gas in lb/cu ft.
             On using the equation of state and considering 1 lb of gas,
               pv ¼ ZNRT                                                      (6.9)
           where p is the absolute pressure in lb/sq ft, v ¼ the specific volume in cu ft/lb,
           Z ¼ the compressibility factor, N ¼ the number of moles of gas, R ¼ the universal
           gas constant, and T ¼ the absolute temperature in 1R.
             Thus,
               v ¼ ZNRT=p                                                    (6.10)
             In as much as
               v ¼ 1=g                                                       (6.11)
           and
               g ¼ p=ZNRT                                                    (6.12)
             On substituting Eq. 6.12 into Eq. 6.8,
               p ¼ p þ ð p=144ZNRTÞDL                                        (6.13)
                2    1
             Rearranging,
               p   p ¼ ð p=144ZNRTÞDL                                        (6.14)
                2    1
           and
               dp=p ¼ dL=144ZNRT                                             (6.15)
             In integral form, Eq. 6.15 becomes
                            L
               Z          Z
                 p bc
                    dp=p ¼   ð1=144ZNRTÞ dL                                  (6.16)
                           0
                p s
             Thus,
               lnð p =p Þ ¼ L=144Z av NRT av                                 (6.17)
                   bc
                      s
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