Page 48 - Fundamentals of Reservoir Engineering
P. 48

CONTENTS                                     XLVIII



                         t
                     m D  ( D n  −  t D j 1 )  =  m D  () =  2 t π ′ DA  +  1 2  ln  4t′ D  −  1 2  m DMBH )  ( DA  )
                                                                                t′
                                          t′
                                           D
                                                                           (
                                                               γ
                                 −
                     ∆ Q =  Q − Q j 1                                                    (8.40)     260
                             j
                        j
                                  −
                and
                     S′ = S DQ  n
                           +
                      n
                  kh  (          (         2    n  Q m   t    t  )  Q S                  (8.41)     260
                         () m p
                                      −
                                             =
                           i
                1422T  mp −        wf n  ) FQ n )   ∆  j  D  ( D n  −  D j 1  +  n
                                                               −
                                               j1
                                                =
                       2
                Q =  C p − p 2 wf  ) n                                                   (8.42)     262
                     ( i
                                1422QT          4A
                  () mp
               mp −      (  wf  ) =  kh       1 2 ln  γ C r 2  + S    + FQ 2             (8.43)     263
                   i

                                                  Aw
               mp  i     (p wf  ) BQ FQ=  +  2                                           (8.44)     263
                  () m−
                mp  i    (  wf n )
                  () m p−
                      Q n       versus Q n                                               (8.45)     263
                m(p ) m(p wf n  ) FQ 2 n  2.359T  n  Q j   1422T    1  ln  4A    S       (8.46)     264
                              −
                     −
                   i
                                                         j
                                                 j1 Q
                                       ( ) c Ah
                        Q n         =  φµ   i        n  t ∆ +  kh       2  γ C r 2  +
                                                 =
                                                                            A w
                m(p ) m(p wf n  ) FQ 2 n    n  Q j
                     −
                              −
                   i
                        Q n         versus      n  t ∆  j                                (8.47)     264
                                            j1 Q
                                            =
                                                    2
                                               6
               m(p) = (0.3457p — 414.76) × 10 psia /cp                                   (8.48)     265
                  () ( )
                mp −   mp  wf n  − FQ 2 n  n  ∆ Q j                   k
                    i
                                                                                 +

                                                  ( n
                                        j1 Q
                                                                    ( ) c r
                        Q n         = m      n  log t − t j 1 ) m log+         φµ  i  w 2  − 3.23 .87S  (8.49)  273
                                                        −

                                        =
                m(p ) m(p wf n  ) FQ 2 n  1422T  n  ∆ Q j          1422T
                              −
                     −
                   i
                                                        t
                                             j1 Q
                        Q n         =   kh       n  m D  ( D n  − t D − j 1 )  +  kh  S  (8.50)     275
                                             =
               m(t ) ′ =  α −  1 2  m D(MBH) (t )                                        (8.51)     275
                                       ′
                    D
                 D
                                       DA
                   kh                              t ) m ( t )                           (8.52)     279
                                        =
                               −
                                           D
                                    ws
                             i
                                              D
                1422Q T  (m(p ) m(p )) m (t + ∆    D  −  D  ∆  D
                                              1
                      1
                   kh   (m(p ) m(p       )) 1.151 log  t +∆ t          1  ln 4t D 1      (8.53)     279
                                                       1
                                                                D
                                                                  D
                1422Q T      i  −   ws(LIN)  =           t ∆  +  m (t ) −  2  γ
                                                                   1
                      1
                    1637Q T
               m =        1                                                              (8.54)     280
                       kh
                                      (m(p ws(LIN)1 hr  ) m(p ))  k
                                                 −
                                                      wf
                S′ =  S DQ = 1.151            −           − log       2  + 3.23          (8.55)     280
                      +
                           1
                 1
                                                                ( c)r
                                              m                φµ   iw
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