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OILWELL TESTING                                    211

                                                                                       5
                     The original paper on the subject was presented by Odeh and Jones  in which the
                     analysis technique is precisely as described above except that the p  functions in
                                                                                       D
                     equ. (7.69) were evaluated for transient flow as

                                        4t
                           p D  () =  1 2 ln  D                                                     (7.23)
                              t
                               D
                                        γ
                     This leads to the test analysis equation (with t in hours)

                                           p −
                                      kh  ( i  p wf  )       n  ∆ q j                k
                           7.08 10 − 3           n  = 1.151       log t −  t j 1 ) log  2  −  3.23 0.87S
                                ×
                                                                                               +
                                                                             +
                                                                     ( n
                                                                           −
                                      µ B o  q n             j1 q n               φµ cr w
                                                            =
                                                                                                    (7.72)
                     which, providing the assumption of transient flow is appropriate for the test, will give a
                     linear plot of (p i−p wf)/q n versus Σ ∆q j/q n log(t n−t j-1), with slope m = 162.6µB o/kh and
                                           2
                     intercept m(log(k/φµcr ) −3.23 + .87S), from which k and S can be calculated.
                                           w
                     It is frequently stated in the literature that the separate flow periods should be of short
                     duration so that transient flow conditions will prevail at each rate. While this condition is
                     necessary, it is insufficient for the valid application of transient analysis to the test.
                     Instead, the entire test, from start to finish, should be sufficiently short so that
                     transience is assured throughout the whole test period. The reason for this restriction is
                     that the largest value of the dimensionless time argument, for which the p D functions in
                     equ. (7.69) must be evaluated, is equal to the total duration of the test. This point is
                     illustrated in fig. 7.27 (b), which again demonstrates the basic principle of superposition
                     and shows that in evaluating the flowing pressure at the very end of the test there is
                     still a component of the pressure response due to the first flow rate to be included in
                     the superposed constant terminal rate solution. The following example will illustrate the
                     magnitude of the error that can be made by automatically assuming that a multi-rate
                     flow test can be interpreted using transient analysis techniques.


                     EXERCISE 7.8 MULTI-RATE FLOW TEST ANALYSIS

                     An initial test in a discovery well is conducted by flowing the well at four different rates
                     over a period of 12 hours as detailed in table 7.10.

                             Flowing time                   Oil rate                       p wf
                                (hours)                      (stb/d)                      (psia)

                                   0                             0                       3000(p i)
                                   3                           500                        2892
                                   6                         1000                         2778
                                   9                         1500                         2660
                                  12                         2000                         2538
                                                          TABLE 7.10
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