Page 172 - Air and gas Drilling Field Guide 3rd Edition
P. 172

6.5 Air and Gas Drilling Model    163




                                          ð        ð  H
                                           P ai  dP
                                                 ¼   dh;                       (6-68)
                                             B i ðPÞ
                                          P in      0
              where
                                              8                       9
                                                                    32
                                2            3         2
                                                          P g  T av
                                              >                       >
                                              >                       >
                                              >                   Q g  >
                                              <                       =
                                       _ w g                  T g
                                6            7      f  6  P         7
                                6            7  1      6            7  :
                          B i ðPÞ¼                             2
                                   P g  T av  >    2gD i       i      >
                                4            5         4     p  D   5
                                              >              4        >
                                              >                       >
                                   P    T g  Q g :                    ;
              Using Equations (6-6), (6-7), and (6-13), Equation (6-68) can be rearranged to give
                                  PdP      a i
                            ð                ð  H
                             P ai
                                         ¼      dh;                            (6-69)
                                2
                                      2
                               ðP   b i T Þ  T av
                            P in      av      0
              where
                                           S g
                                       a i ¼  R e                              (6-70)
                                                    2  2
                                            f          _ w g
                                       b i ¼    R e       :                    (6-71)
                                                     p 2

                                           2gD i  S g    4
                                                        D
                                                     4   i
              In the form just described, both sides of Equation (6-69) can be integrated. Using
              the constants in Equations (6-70) and (6-71), the solution to Equation (6-69) is
                                                   P
                                                   ai
                                     1    2           a i  H
                                        lnðP   b i T Þ  ¼  jhj :               (6-72)
                                               2
                                       2       av     T av  0
                                                  P in
              Evaluating Equation (6-72) at the limits and rearranging the results give
                                            2    2
                                          P   b i T   2a i
                                       ln  ai    av  ¼  H:                     (6-73)
                                                 2
                                           2
                                          P   b i T av  T av
                                           in
              Raising both sides of Equation (6-73) to the natural exponent gives
                                                 2
                                           2
                                          P   b i T av  ¼ e T av :             (6-74)
                                                     2a i H
                                           ai
                                                 2
                                           2
                                          P   b i T av
                                           in
              Equation (6-74) can be rearranged and a solution obtained for P in . This is
                                                           30:5

                                         2          2a i H
                                           2
                                          P þ b i T  2  e T av   1
                                           ai    av
                                    P in ¼  4             5  :                 (6-75)
                                                 2a i H
                                                 e Tav
              The von Karman empirical correlation can be used to determine the friction fac-
              tor in Equation (6-71)[1]. This empirical expression is
                                          2              3 2
                                                 1
                                          6              7
                                       f ¼  6            7  :                  (6-76)
                                          4              5
                                                D i
                                           2 log   þ 1:14
                                                e
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