Page 235 - Air and Gas Drilling Manual
P. 235

Chapter 5: Shallow Well Drilling Applications    5-77
                                                             2
                                                            
                                                 f
                                                          R
                                         =
                                      b
                                                         
                                                            
                                             gD −
                                                               π
                                       a
                                                          S 
                                                               
                                                         
                                           2
                                               (
                                                     D )
                                                 h
                                                      p
                                                           g
                                                                           2 2
                                                                 2  ( D w ˙ 2 h g 2  −  D )  (5-18)
                                                                           p
                                                                4
                                                                 2
                                                                
                                      f =       D − 1                              (5-19)
                                                       D 
                                                
                                           2 log.               
                                                   h
                                                        p
                                                           + 1 14
                                                  e         
                                                    av          
                                   The outer wall absolute surface roughness for the limestone sequence is  assumed
                               to be 0.005 ft (see Table 8-1) [2].   The inner surface of the borehole annulus is  the
                               outer surface of the drill  pipe.   The absolute surface roughness of commercial pipe,
                               e p, is given as 0.00015 ft [2].  The average absolute surface roughness of the annulus
                               is  approximated  by  the  calculating  the  surface  area  average  of  these  surface
                               roughness.  Thus, the value for e av is
                                                π
                                                              π
                                                
                                            eDH +         e     DH
                                                                  2
                                                    2
                                             oh     h    p     p
                                                              4
                                                4
                                      e  =
                                                           π
                                               π
                                       av
                                                 DH +      2
                                                    2
                                                  h        DH
                                                               p
                                                4
                                                           4
                               The above reduces to
                                                            π
                                                π
                                                          
                                            e      D 2  +  e   D 2
                                             oh     h  p     p
                                                            4
                                                4
                                      e  =                                             (5-20)
                                                         π
                                                π
                                                  2  +     2
                                       av
                                                  D h     D p
                                                4
                                                         4
                               Thus, e av, is
                                                                        π
                                                   π
                                                         2                  2
                                            0
                                            (.005 )     ( .021 )  + (.0 00015 )     (.458 )
                                                      1
                                                                           0
                                      e  =         4                    4
                                                                 π
                                                    π
                                       av                 2          2
                                                      (. 1 021 )  +     ( . 0 458 )
                                                    4
                                                                 4
                                             .
                                      e av  = 0 0042 ft
                               Equation 5-19 becomes
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