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RADIAL LAMINAR FLOW OF GAS             45 1


                              For  other  well locations inside  closed (or  bounded)  drainage  area
                            shapes (f  = 0), the general form of Equation 7.105 is:


                                       WP2 - P$)
                            qsc =                                                      (7.107)
                                  1,422pgzT[ln(rG/r,)  + SI

                            where r  is given by Equation 7.64~.

                            EXAMPLE

                              A  well  is  producing  275  MSCFD  from  a  gas  reservoir under  the
                            influence  of  a  partial  water  drive  with  an  index  of  0.5.  Calculate
                            the wellbore pressure and the pressure at the drainage boundary. The
                            following reservoir and fluid properties are known:

                            pg = 0.035 CP            k=5mD                Swi  = 17%
                             z = 0.95                h=35ft                 @= 12%
                            T = 130°F                re = 2,640 ft

                             p = 2,720 psia         r,  = 0.5 ft
                             s=o

                            SOLUTION

                              The wellbore pressure can be obtained from Equation 7.105:
                            pw = [p'  - m(Ln (')   - 0.75 + 0.25f)l  0.5

                                               r,

                           where:


                            m=      qscPgZT
                                 0.702 x lO-3kh

                            Substituting the values of the fluid and reservoir properties gives:

                                 (275)(0.035)(0.95)(460  + 130)
                            m=                                = 43,895
                                     (0.702 x  10-3)(5)(35)

                            and:
                                                    ( ('I:)
                                   2,720'  - (43,895)  In  - - - + 0.25)]0.5 -  = 2,655 psia
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