Page 188 - Air and Gas Drilling Manual
P. 188

5-30    Air and Gas Drilling Manual
                                                                         503
                                                                           .27
                                                                . )
                                                                                    
                                                                                   
                                      P in  =      (, 5 456 ) 2  + ( , 7 610 ) (503 27  2     e  )  2  (. 0 019 ) ( , 1 200 )  −  1       . 05
                                                            2
                                                                ) ( , 1 200
                                                             (. 0 019
                                                          e   503 .27              
                                                                                   
                                                                                   
                                                    2
                                      P in  = 13 860 lb/ft abs
                                             ,
                                            P in
                                      p in  =
                                            144
                                              .
                                      p in  = 96 2 psia
                                   The above  pressure  is  the  approximate  injection  pressure  into  the  top  of  the
                               inside of the drill  string.    These calculations  have  been  carried  out  neglecting  the
                               drill  collar outside and inside diameters.    Also,  the  calculations  have  ignored  the
                               existence of a blooey line type structure  (most  shallow  drilling  operations  do  not
                               have  blooey  lines).    These  minor  losses  are  not  important  in  shallow  drilling
                               operations  (Chapter  8  calculation  examples  will  consider  these  additional  minor
                               losses).  The above pressure slightly underestimates the actual pressure that is seen at
                               the pressure gauge just downstream of the compressor.   In order for this  compressor
                               to be used for this drilling operation, the above injection pressure must  be less than
                               the derated fixed pressure of the rotary screw Sullair Model 840 compressor.
                                   The fixed pressure capability of the Sullair Model 840 compressor has a pressure
                               output of 340 psig.   However, this  output  must  be derated when the compressor is
                               placed at a surface drilling location above sea level.   To determine the derated fixed
                               pressure capability of this  compressor for the 6,000  ft surface drilling  location, the
                               design  fixed  ratio  of  the  compressor  must  be  determined.    As  demonstrated  in
                               Chapter 4, the fixed pressure ratio is  referenced to  sea level conditions (usually API
                               standard conditions).  Thus, the assumed design input  pressure, p 1 (air flowing into
                               the compressor from the atmosphere), is
                                      p =  14 696 psia
                                             .
                                       1
                               The output pressure, p 2, is
                                      p 2  =  340 +  14 696 =  354 696 psia
                                                   .
                                                            .
                               The total fixed compression ratio across the two stages of this compressor, r c, is

                                           p
                                      r =   2
                                       c
                                            p 1
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