Page 339 - Air and Gas Drilling Manual
P. 339

8-22    Air and Gas Drilling Manual
                                   Illustrative Example 8.3b  Determine  the  bottomhole  and  injection  pressures
                               while  drilling  at  10,000  ft  of  depth  for  the  drilling  operations  described  in
                               Illustrative Examples 8.2 and 8.3a.  Compare the injection pressure to the maximum
                               pressure capability of the selected semi-trailer mounted Dresser Clark Model CFB-4,
                               four-stage,  reciprocating  piston  compressor.    Also,  determine  the  horsepower
                               required by the compressor and the maximum  derated horsepower of the Caterpillar
                               Model D398 prime mover.  Compare the compressor  required  horsepower  and  the
                               maximum derated horsepower from the prime mover.
                                 General Data
                                   Table 4-1 gives an average atmospheric pressure of 12.685 psia for a surface
                               location of 4,000 ft above sea level (mid latitudes North America) (also see
                               Appendix D).  The actual atmospheric pressure for the air at the drilling location
                               (that will be utilized by the compressor), P at, is
                                             .
                                      p at  = 12 685 psia
                                      P at  =  p 144
                                             at
                                                   2
                                            ,
                                      P at  = 1 827 lb/ft abs
                               The actual atmospheric temperature of the air at the drilling location (that will be
                               used by the compressor), T at, is
                                             o
                                      t at  = 60 F
                                                   .
                                      T  =  t  + 459 67
                                       at   at
                                                 o
                                      T at  = 519 67 R
                                              .
                               Thus, P g and T g become

                                      P =  P at  = 1 827 lb/ft abs
                                                         2
                                                 ,
                                       g
                                                      o
                                      T =  T at  = 519 67 R
                                                   .
                                       g
                               Using Equation 4-11, the specific weight of the gas entering the compressor is
                                             (1 827,  ) (1 0.  )
                                      γ  =
                                       g
                                            (53 36.  ) (519 67.  )
                                      γ g  = 0 0659 lb/ft 3
                                            .
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