Page 200 - Air and Gas Drilling Manual
P. 200

5-42    Air and Gas Drilling Manual
                                                                        π
                                                   π
                                                   
                                                       0
                                            (.005
                                            0
                                                 )
                                                                            0
                                                   
                                                                        
                                                                        4
                                                   4
                                         =
                                      e
                                                    π
                                                                 π
                                                    
                                                                 
                                       av
                                                              +
                                                                    (. 0 240
                                                                         )
                                                       (. 0 521
                                                                 
                                                     (.521 )  2 ) 2  + (.0 00015 )    2  (.240 )  2
                                                                 4
                                                    4
                                             .
                                      e av  = 0 0042 ft
                               Equation 5-5 becomes
                                                                    2
                                                                   
                                      f =            1             
                                                  .
                                                          .
                                                0 521  − 0 240   
                                                                  .
                                           2 log           + 114 
                                                             
                                                      .
                                                    0 0042         
                                           .
                                      f = 0 043
                               Equations 5-2 and 5-3 become, respectively,
                                      a  =   10 .    1  +   0 897.   
                                       a                   
                                              .
                                            53 36     1 599.  
                                            .
                                      a  = 0 029
                                       a
                               and
                                                                               1 599)
                                                   .
                                     b a  =       0 043        53 36.     2  2  (.  2
                                                                       π
                                               .
                                           2 32 2) ( 0 521  − 0 240.  )  10.     [  2  2 ] 2
                                                   .
                                            (
                                                                            0 521)
                                                                                    (.
                                                                         (.     − 0 240)
                                                                       4
                                      b  = 617 6
                                              .
                                       a
                                   The  bottomhole  temperature  is  determined  from  the  surface  geothermal
                               temperature rock formations at the surface, the geothermal gradient, and the depth of
                               the borehole.  The geothermal temperature of the rock formations near the surface is
                               assumed to be represented by the temperatures given in Table 4-1 (also see Appendix
                               D).  Thus, for a 4,000 ft surface location this  temperature is  assumed to  be 44.7˚F.
                               For this  example the geothermal gradient is  approximated  to  be  0.01˚F/ft.    Thus,
                               T bh, is
                                              o
                                      t r  = 44 7 F
                                             .
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