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Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 36 3.1.2007 8:30pm Compositor Name: SJoearun




               3/36  PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS

                Forapartialtwo-phasereservoir,modelconstantJ inthe  Well B:
               generalizedVogelequationmustbedeterminedbasedonthe

               range of tested flowing bottom-hole pressure. If the tested  J ¼       q 1            2
               flowing bottom-hole pressure is greater than bubble-point  (  p   p b ) þ  p b  1   0:2  p wf 1    0:8  p wf 1
                                                              p

               pressure, the model constant J should be determined by  1:8   p b    p b

               J ¼   q 1  :                       (3:30)  ¼                   900
                   p
                   (  p   p wf 1 )                                                             2
                                                            (5,000   3,000) þ  3,000  1   0:2  2,000    0:8  2,000
                                                                                  3,000
                                                                                          3,000
                                                                         1:8
               If the tested flowing bottom-hole pressure is less than

               bubble-point pressure, the model constant J should be  ¼ 0:3156 stb=day-psi
               determined using Eq. (3.28), that is,
                                                         Calculated points are
                                                   !

               J ¼           "    q 1             2 # :
                           p b      p wf 1   p wf 1
                    p
                    (  p   p b ) þ  1   0:2    0:8                p wf (psia)  q (stb/day)
                           1:8       p b      p b
                                                                     0           1,157
                                                  (3:31)           500           1,128
               Example Problem 3.4 Construct IPR of two wells in an  1,000       1,075
               undersaturated oil reservoir using the generalized Vogel  1,500    999
               equation. The following data are given:            2,000           900
                                                                  2,500           777
                  Reservoir pressure:        p p ¼ 5,000 psia     3,000           631
                  Bubble point pressure:  p b ¼ 3,000 psia        5,000             0
                  Tested flowing bottom-hole
                   pressure in Well A:   p wf 1 ¼ 4,000 psia  The IPR curve is plotted in Fig. 3.13.
                  Tested production rate                  For a two-phase (saturated oil) reservoir, if the Vogel
                   from Well A:           q 1 ¼ 300 stb=day  equation, Eq. (3.20), is used for constructing the IPR
                  Tested flowing bottom hole             curve, the model constant q max can be determined by
                   pressure in Well B:   p wf 1 ¼ 2,000 psia          q 1
                  Tested production rate                 q max ¼     p wf 1      p wf 1   2  :  (3:32)
                   from Well B:           q 1 ¼ 900 stb=day   1   0:2    p p    0:8    p p
                                                         The productivity index at and above bubble-point pres-
               Solution
                                                         sure, if desired, can then be estimated by
               Well A:


                          J ¼    q 1                     J ¼  1:8q max  :                   (3:33)
                              (  p   p wf 1 )                    p p
                               p
                                  300                    If Fetkovich’s equation, Eq. (3.22), is used, two test points
                            ¼
                              (5,000   4,000)            are required for determining the values of the two model
                            ¼ 0:3000 stb=day-psi         constant, that is,

               Calculated points are                         log  q 1 q 2
                                                         n ¼                                (3:34)
                                                                 p p 2  p 2
                                                            log  wf 1
                        p wf (psia)  q (stb/day)                 p p 2  p 2
                                                                 wf 2
                           0           1,100             and
                         500           1,072
                        1,000          1,022             C ¼  2  q 1  2  n  ,               (3:35)
                                                             p
                        1,500           950                 (  p   p wf 1 )
                        2,000           856              where q 1 and q 2 are the tested production rates at tested
                        2,500           739              flowing bottom-hole pressures p wf 1 and p wf 1 , respectively.
                        3,000           600
                        5,000             0              Example Problem 3.5 Construct IPR of a well in a
                                                         saturated oil reservoir using both Vogel’s equation and
                The IPR curve is plotted in Fig. 3.12.   Fetkovich’s equation. The following data are given:
                  6,000
                                                           6,000
                  5,000
                                                           5,000
                  4,000                                    4,000
                p wf  (psia)  3,000                       p wf  (psia)  3,000
                  2,000
                                                           2,000
                  1,000
                                                           1,000
                     0
                      0   200   400  600  800  1,000 1,200    0
                                   q (stb/day)                 0   200   400  600   800  1,000 1,200
                                                                             q (stb/day)
               Figure 3.12 IPR curves for Example Problem 3.4,
               Well A.                                   Figure 3.13 IPR curves for Example Problem 3.4, Well B.
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