Page 239 - Air and Gas Drilling Manual
P. 239

Chapter 5: Shallow Well Drilling Applications    5-81
                               psig  at API standard conditions.   When the drilling  rig  is  moved  to  the  6,000  ft
                               surface location, the volumetric flow rate from  the  compressor  is  840  acfm.    The
                               compressor  volumetric  flow  rate  is  greater  than  the  required  adjusted  minimum
                               volumetric  flow  rate  of  280  acfm  (factor  of  safety  of  3.00).    Therefore,  this
                               compressor  unit  is  capable  of  producing  the  volumetric  flow  rate  to  drill  the
                               borehole.
                                   A comparison must also be made between the injection pressure required to  drill
                               this  borehole and the pressure capability of this  compressor system.    The injection
                               pressure  is  obtained  by  calculating  through  a  sequence  of  calculation  steps  that
                               determine the bottomhole pressure in the annulus, the pressure at the bottom  of the
                               inside  of  the  drill  string,  and  then  the  injection  pressure  (upstream  through  the
                               circulation system).    In  these  calculations,  the  actual  volumetric  flow  rate  of  the
                               compressor  is  used,  the  840  acfm.    Also,  in  these  calculations,  the  actual
                               atmospheric conditions at the drilling  location will  be used  (i.e.,  11.769  psia  and
                               40˚F).  The atmospheric pressure of the air, P at, entering the compressor is
                                             .
                                      p at  = 11 769 psia
                                      P  =  p 144
                                       at    at
                                                   2
                                      P at  = 1 695 lb/ft abs
                                            ,
                               The atmospheric temperature of the air, T at, entering the compressor is
                                             o
                                      t at  = 40 F
                                      T  =  t  + 459 67
                                                   .
                                       at   at
                                                 o
                                              .
                                      T at  = 499 67 R
                               Thus, P g and T g become

                                      P =  P at  = 1 695 lb/ft abs
                                                         2
                                                 ,
                                       g
                                      T =  T at  = 499 67 R
                                                      o
                                                   .
                                       g
                               Using Equation 4-11, the specific weight of the gas entering the compressor is
                                             (1 695,  ) (1 0.  )
                                      γ  =
                                       g
                                            (53 36.  ) (499 67.  )
                                      γ  = 0 0636 lb/ft 3
                                            .
                                       g
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