Page 243 - Air and Gas Drilling Manual
P. 243

= 0 333 ft
                                      D
                                       io
                               Equation 5-19 becomes
                                                                     2
                                            .         Chapter 5: Shallow Well Drilling Applications    5-85
                                                                    
                                                                   
                                      f =            1            
                                                          .
                                                  .
                                                
                                          2 log   0 333  − 0 229  + 114 
                                                                  .
                                                 0 00015         
                                                     .
                                      f = 0 021
                                           .
                               Equations 5-17 and 5-18 become, respectively,
                                             .
                                      a a  =  10
                                             .
                                           53 36
                                      a a  = 0 019
                                            .
                               and
                                                                               0 890)
                                                  .
                                     b  =         0 021        53.36   2    (.    2
                                     a                                   2
                                                                        π
                                                  .
                                          2 32 2) ( 0 333  − 0 229.  )   1 0.     [  2  2 ] 2
                                              .
                                           (
                                                                            0 333)
                                                                          (.     − 0 229(.  )
                                                                        4
                                      b a  = 3 409
                                            ,
                               Equation 5-16 becomes
                                                                      2  (. 0 019 ) ( , 1 200 )     . 05
                                             (,960 ) 2  + ( , 3 409 ) (503 27  2  e   503 .27  −  1  
                                             10
                                                                . )
                                                                                  
                                                                                    
                                      P  =                                          
                                                           2  (. 0 019  ) ( , 1 200  )  
                                       in
                                                          e   503 .27               
                                                                                    
                                                                                    
                                                    2
                                      P  = 13 550 lb/ft abs
                                             ,
                                       in
                                            P in
                                      p in  =
                                            144
                                              .
                                      p  = 94 1 psia
                                       in
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