Page 372 - Fundamentals of Reservoir Engineering
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NATURAL WATER INFLUX                                   307

                     EXERCISE 9.1 APPLICATION OF THE CONSTANT TERMINAL PRESSURE
                                      SOLUTION

                     A reservoir-aquifer system has the geometry and dimensions as shown in fig. 9.8.







                                                        80 O
                                                            Oil

                                                                 Water
                                                     r = 5000’
                                                      o

                                                            r = 15000’
                                                             e
                     Fig. 9.8   Water influx from a segment of a radial aquifer

                     If the aquifer properties are as follows

                           h    =   50 ft       µ   =   0.4 cp
                                                               -6
                           φ    =   0.25        c w =   3.0×10  /psi
                                                               -6
                           k    =   50 mD       c f  =  6.0×10  /psi
               1)    Calculate the water influx at times t = 0.5, 1, 1.5, 2 and 3 years after an instantaneous
                     pressure drop of ∆p = 100 psi, at the oil water contact, at time t=0.

               2)    What would be the corresponding water influx if it is assumed that the same pressure
                     drop is transmitted instantaneously throughout the aquifer?


                     EXERCISE 9.1 SOLUTION

               1)    Since t is measured in years, then

                               2.309kt         2.309 50 t
                                                     ×
                                                         ×
                           t =         =               6         2
                            D
                                                       −
                                             ×
                                                   ×
                               φµ cr  2 0  .25 .4x9 10 ×  (5000)
                                   t D  =  5.131t
                     The encroachment angle is 80°, therefore f = 80°/360° = 0.222 and
                                        2
                     W e  = 1.119 fφhcr ∆pW D(t D)
                                        0
                                                              2
                                                     -6
                          = 1.119×.222×.25×50×9×10 ×(5000) ×100W D (t D)
                     W e  = 69868 W D (t D) bbls
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