Page 95 - A Practical Companion to Reservoir Stimulation
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PRACTICAL COMPANION TO RESERVOIR STIMULATION




           EXAMPLE F-5                                           EXAMPLE F-6

           Fracture Face Damage                                  Apparent Fracture Face Damage

            Clay swelling from a water-base fracturing fluid resulted in a   Long-term production from a very tight, heavy gas-conden-
            permeability impairment of  95% (k/k,=20).  If  the fracture   sate  well  (1;,=0.328  ft)  results  in  a  skin  effect  during
            half-length is 500 ft and the penetration of  damage, b,s, is   pseudoradial flow equal to -6.4.  If the liquid gas condensate
            2 in., estimate the impact of this damage on production.   extends  20  ft  into  the  reservoir  (normal  to  the  fracture),
                                                                 calculate the  apparent reservoir permeability-to-gas reduc-
            Solution (Ref. Section 11-4)                         tion. The fracture half-length, measured early in the life of the
            From Eq.  11-20,                                     well, is 1000 ft.
                       (3.14)(2/ 12)( 19)                        Solution (Ref. Section 1-4)
                  Sf.7  =             = 9.9~           (F- 18)
                          (2)  (500)                             The variables on the ordinate of Fig. 1 1-2 1 can be calculated:
            and the reciprocal  s,:;’   = 100.                          r,#,,  r,.e-.’   (0.328)e+6.4
                                                                           -
                                                                           -
              From  Fig.  11-21 it  can be  determined that  r:/xf  = 0.5;   --    -            = 0.2.      (F-19)
                                                                                   -
            i.e., there is no impact on the well performance as a result of   Xf   Xr   1000
            this damage.                                         Therefore, from Fig. 11-21, q-’ = 1.2, and thus, sf., = 0.83.
                                                                   Using Eq. 1 1-20 and rearranging,
                                                                         k    (2) ( 1000) (0.83)
                                                                            -
                                                                            -
                                                                         -                 + 1 = 27.4.      (F-20)
                                                                         k,s    (3.14)(20)
                                                                 Therefore, the apparent permeability impairment is
                                                                               (  1 - - 0.96,               (F-21)
                                                                                        =
                                                                                   2i.4)
                                                                 or 96% reduction.






































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