Page 222 - Fundamentals of Enhanced Oil and Gas Recovery
P. 222

210                                                                    Mohammad Ali Ahmadi





                                                 α

                Figure 7.2 An inclined linear reservoir bed.

                   Darcy’s equation for oil and water flow rate with respect to dip angle is given as

                                              kk ro A @P o
                                       q o 52           1 P o g sin α                 (7.11)
                                               μ     @x
                                                 o

                                              kk rw A @P w
                                      q w 52             1 P w g sin α                (7.12)
                                               μ w   @x
                   Replacing water pressure by P w 5 P o 2 P cow so that

                                              kk rw A @P w
                                      q w 52             1 P w g sin α                (7.13)
                                               μ     @x
                                                w
                   After rearranging, the equations may be written as
                                              μ
                                                o    @P o
                                         2 q o    5      1 P o g sin α                (7.14)
                                             kk ro A  @x
                                          μ
                                            w    @P o  @P cow  1 P w g sin α
                                    2 q w      5     2                                (7.15)
                                         kk rw A  @x     @x
                   Subtracting the first equation from the second one, we get


                                    1      μ      μ        @P cow
                                 2      q w  w  2 q o  o  52    1 P w g sin α         (7.16)
                                    kA    k rw    k ro      @x
                   Substituting for

                                                q t 5 q w 1 q o                       (7.17)
                and

                                                       q w
                                                  f w 5                               (7.18)
                                                       q t
                   Also, solving for a fraction of water flowing, we obtain the expression for a frac-
                tion of water flowing:
                                     1 1 kk ro A=q t μ        @P cow =@x 2 Δρg sin α


                                f w 5             o                                   (7.19)
                                                1 1 k ro μ =μ k rw
                                                        w  o
   217   218   219   220   221   222   223   224   225   226   227