Page 440 - Air and Gas Drilling Manual
P. 440

9-44    Air and Gas Drilling Manual
                                                      (.
                                             001)
                                            (.
                                                                           0 375)
                                                                0 00015)
                                                      0 656)
                                                   4
                                      e
                                       av3
                                                                     0 375)
                                                        0 656)
                                                       (.  2  2  (.  (.  4  2  (.  2
                                                    4            4
                                             .
                                      e av3  0 0076 ft
                               Equation 6-77 must  be solved by trial and error.  A value for f a3 must  be  selected
                               that will satisfy the Colebrook equation.  This value is
                                      f a3  0 065
                                             .
                               The Colebrook equation is
                                                           0 0076
                                                            .
                                                                 .
                                                         .
                                                                             .
                                         1              0 656   0 375       251
                                                 2 log
                                        0 065                37         2 688  0 065
                                                                         ,
                                                              .
                                        .
                                                                                .
                               The right and left sides of the above yield
                                      392    392
                                       .
                                              .
                                   Equation  6-74  for  the  third  increment  in  the  annulus  can  be  solved  for  the
                               pressure at the bottom  of  the  increment,  P a3.    This  involves  selecting  this  upper
                               limit by  a trial and error procedure.   The magnitude of the upper limit  pressure on
                               the left side of the equation is  selected to  allow the left side  integral  to  equal  the
                               right side integral.  The integration can be carried out  on the computer using  one of
                               the commercial analytic software programs.    The  trial  and  error  magnitude  of  the
                               upper limit pressure is
                                      p a3  3 094 3 psia
                                             ,
                                                .
                                      P a3  p a3  144
                                                      2
                                      P a3  445 579 lb/ft abs
                                               ,
                                   Substituting the values of  ˙ w , Q g, Q m, P g, P a2, P a3, T g, T av3, D h, D 1, f a3, and g
                                                       t
                               into the left side of Equation 6-74 gives
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