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REAL GAS FLOW: GAS WELL TESTING                              283


                     2)    From the flow tests determine k, S′, S′  and hence D, or F.
                                                                2
                                                            1
                     EXERCISE 8.3 SOLUTION

               1)    Buildup Analysis

                     For a flowing time of 3 hours, the data necessary to draw the Horner buildup plot are
                     listed in table 8.12.

                                      t +∆ t                                   t +∆ t
                                      1
                                                                                1
                         ∆t       log             m(p ws)         ∆t       log             m(p ws)
                                                                                               2
                                                     2
                         hrs             t ∆     psia /cp        hrs              t ∆      psia /cp
                         .5         .845      1002.61×10  6      3.5         .269       1062.07×10 6
                       1.0          .602      1056.19    "       4.0         .243       1062.76   "
                       1.5          .477      1058.96    "       5.0         .204       1063.45   "

                       2.0          .398      1060.34    "       6.0         .176       1063.80   "
                       2.5          .342      1061.03    "       7.0         .155       1064.14   "

                       3.0          .301      1061.72    "       8.0         .138       1064.49   "

                                                                 (9.0)       (.125)    (1064.65) "
                                                          TABLE 8.12


                     The corresponding buildup plot is shown as fig. 8.15 (a), from which the slope has
                     been determined as

                                                                   1637Q T
                                                2
                                          6
                           m 16.17 ×    10 psia    / cp /log cycle =     1
                             =
                                                                      kh
                     which for a fully penetrating well gives
                                               3
                              1637 ×   40 ×  10 ×   660
                           k =               6          = 53.5 mD
                                  16.17 ×  10 ×   50

                     and the extrapolation to ∆t = ∞ gives

                                                 2
                                            6
                     m(p i) =   1066.7 × 10  psia /cp
                     p i  =     4285 psi
                                                                                              6
                                                                                                   2
                     The value of  ( )  1hr taken from the extrapolated linear trend is 1057 ×10  psia /cp
                                  mp′
                                      ws
                                         −
                     and therefore, using equ. (8.55)
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