Page 190 - Air and gas Drilling Field Guide 3rd Edition
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7.5 Air and Gas Drilling Model    181




              where
                                           8                             9
                                                                       32
                              2           3               2
                                                             P g  T av
                                           >                             >
                                           >                             >
                                                                     Q g
                                           >                             >
                                           <                             =
                              6     _ w g  7         f    6  P   T g   7
                              6           7                               :
                       B a PðÞ ¼             1             6           7
                                P g  T av  >    2gD h   D p    D   D     >
                              4           5               4  p  2   2  5
                                           >                    h   p    >
                                           >                4            >
                                P    T g  Q g :                          ;
              Using Equations (7-6), (7-7), and (7-14), Equation (7-56) can be rearranged to give
                                     ð                 ð H
                                            PdP      a a
                                      P bh
                                                    ¼     dh;                  (7-57)

                                          2
                                         P   b a T  2
                                      P in      av   T av  0
              where
                                                  S
                                              a a ¼  R e                       (7-58)
                                        f           2     _ w  2 g
                                b a ¼           R e              :             (7-59)
                                                                2
                                                     p 2
                                    2gD h   D p  S       D   D 2
                                                          2
                                                     4    h    p
              In this form, both sides of Equation (7-57) can be integrated. Using Equations
              (7-58) and (7-59), the solution to Equation (7-57) is
                                                     P
                                                     bh
                                     1       2         a a  H
                                       ln P   b a T       h jj :               (7-60)
                                                2    ¼
                                     2                 T av
                                                av          0
                                                   P in
              Evaluating Equation (7-60) at the limits and rearranging the results give
                                              2     2
                                            P    b a T   2a a
                                        ln   bh     av  ¼   H:                 (7-61)
                                             2
                                            P   b a T  2  T av
                                             in     av
              Raising both sides of Equation (7-63) to the natural exponential exponent gives
                                                     2a a H
                                                2
                                        P 2 bh    b a T av  T av :
                                         2
                                        P   b a T  2 av  ¼ e                   (7-62)
                                         in
              Equation (7-62) can be rearranged and a solution obtained for P in . This is
                                                          13 0:5
                                                 0
                                                   2a i H
                                       2
                                         2
                                       6       2 B e      C7
                                               av
                                         ai
                                        P þ b i T @  T av   1A7
                                       6
                                       6                    7
                                       6                    7
                                  P in ¼  6                 7  :               (7-63)
                                       6        2a i H      7
                                       6                    7
                                                e  T av
                                       6                    7
                                       4                    5
                 The von Karman empirical correlation can be used to determine the friction
              factor given in Equation (7-59)[1]. This empirical expression is
                                            2          3 2
                                                 1
                                        f ¼  4          5  ;                   (7-64)
                                             2 log  D h  D p
                                                   e
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