Page 32 - Fundamentals of Reservoir Engineering
P. 32

CONTENTS                                      XXXII


               d(HCPV) = − (c w V w + c f V f ) ∆p                                       (1.38)     26

                G p           (c S wc  +  c ) p    E
                                           ∆
                                         f
                                w
                G   =  1−     1 −  1 S wc          E i                                   (1.39)     27
                                    −
                Production   =  GIIP −   Unproduced Gas
                   (sc)         (sc)          (sc)                                       (1.40)     29
                                             G
                   G p       =  G     −       −  W E
                                                  e
                                            E i
                p   p     G p         WE                                                 (1.41)     29
                                          i
                                        e
                     i
                Z  =  Z i     1−  G          1−  G
                         G p
                G =                                                                      (1.42)     31
                 a
                      1 −  E /E i
                     G —  W e  E
                      p
                G =                                                                      (1.43)     31
                      1E /E  i
                        −
                               WE
                G =   G +        e                                                       (1.44)     31
                 a
                            1 −  E /E i
                               dp
               p =  p g GWC  −       dD     GWC  × ∆ D                                   (1.45)     34


                               p
                Z      =                                                                 (1.46)     40
                 2 phase
                  −
                          p     G ' p
                           i
                          Z i     1−  G
                         scf        rb
               (R R )          × B g       =  (R R ) B g       (rb. free gas / stb)       (2.1)     48
                                                                   −
                  −
                                            −
                     s
                                               s
                          stb       scf
               (Underground withdrawal)/stb = B o + (R − R s) B g     (rb/stb)            (2.2)     48
                       rb       1
                 B g        =                                                             (2.3)     48
                       scf    5.615E
                        y
                x(B + (   −  R )B ) rb / day                                              (2.4)     50
                                 g
                   o
                              s
                        x
                                       v      rb
               v o (rb/ rb b)     B o   B o  =  o                                         (2.5)     63
                                       c
                                         f b     stb
                                            5.615 F rb

               F (stb/rb b)     R s   R s  =  R si f  −  c b f      stb                   (2.6)     63


                                       1        rb
               E (scf/rcf)     B g   B =                                                  (2.7)     63
                                g
                                    5.612 E scf
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