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Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 199  3.1.2007 9:07pm Compositor Name: SJoearun




                                                                                           GAS LIFT  13/199
                                      Pressure                     where
                                 p                  p
                              p hf  hf,d       p c,s  k,s
                                                                     p cd2 ¼ design injection pressure at valve 2, psig
                                                                     G fd ¼ design unloading gradient, psi/ft.
                                                      Kick-off     Applying Eq. (13.9) gives
                                        G s
                               Gf  G f,d              Pressure                      D 2
                                                                 p cd2 ¼ p c,s   Dp cm  1 þ  :      (13:52)
                                                                                   40,000
                                                                 Solving Eqs. (13.51) and (13.52) yields

                                         G s
                                                      ∆p             p c,s   Dp cm   p hf,d þ G s   G fd D 1
                                 ∆p tm                 km        D 2 ¼      G s    p c   Dp cm :    (13:53)
                          Depth           G s     ∆p cm          Similarly, the depth to the third valve is
                                                                                40,000

                                           G s
                                                                     p c,s   Dp cm   p hf,d þ G s   G fd D 2
                                                                 D 3 ¼                              (13:54)
                                                                               p c   Dp cm
                                                                            G s       :
                                            G s
                                                                                40,000
                             Operating Tubing  G s  Injection Operating  Thus, a general equation for depth of valve i is
                                Pressure             Pressure
                                                                     p c,s   Dp cm   p hf,d þ G s   G fd D i 1
                                                                 D i ¼                              (13:55)
                                             G s
                                                                                p c   Dp cm
                                                                            G s       :
                                                                                 40,000
                        Figure 13.18 A flow diagram to illustrate procedure of  Depths of all valves can be calculated in a similar manner
                        valve spacing.                           until the minimum valve spacing (   400 ft) is reached.
                                                                 Example Problem 13.5 Only 1 MMscf/day of lift gas
                                                                 is available for the well described in the Example
                       5. Starting from available kickoff surface pressure p k,s ,  Problem 13.1. If 1,000 psia is available to kick off the
                         establish kickoff casing pressure line. This can be done  well and then a steady injection pressure of 800 psia is
                         using Eq. (13.7) or Eq. (13.9).         maintained for gas lift operation against a wellhead
                       6. Starting from p k   Dp km at surface, where the kickoff  pressure of 130 psia, design locations of unloading
                         pressure margin Dp km can be taken as 50 psi, establish a  and operating valves. Assume a casing pressure margin
                         design kickoff line parallel to the kickoff casing pres-  of 50 psi.
                         sure line. Pressures in this line, denoted p kd , represent
                         kickoff pressure after adjustment for kickoff pressure  Solution The hydrostatic pressure of well fluid (26 8API
                         margin. Set Dp km ¼ 0 if kickoff casing pressure margin  oil) is (0.39 psi/ft) (5,200 ft), or 2,028 psig, which is greater
                         is not required.                        than the given reservoir pressure of 2,000 psia. Therefore,
                       7. Calculate depth of the first valve. Based on the fact  the well does not flow naturally. The static liquid level
                         that p hf þ G s D 1 ¼ p kd1 , the depth of the top valve is  depth is estimated to be
                         expressed as
                                                                        5,200   (2,000   14:7)=(0:39) ¼ 110 ft:
                                                                 Depth of the top valve is calculated with Eq. (13.49):
                           p kd1   p hf
                       D 1 ¼      ,                       (13:47)
                              G s                                          1,000   50   130
                                                                      D 1 ¼            ¼ 2,245 ft > 110 ft
                       where                                              0:39    1,000   50
                           p kd1 ¼ kickoff pressure opposite the first valve (psia)  40,000
                            G s ¼ static (dead liquid) gradient; psi/ft  Tubing pressure margin at surface is (0.25)(800), or
                                                                 200 psi. The modified Hagedorn–Brown correlation gives
                          Applying Eq. (13.9) gives              tubing pressure of 591 psia at depth of 5,000 ft. The design
                                                                 tubing flowing gradient is G fd ¼ [591   (130 þ 200)]=
                                          D 1
                       p kd1 ¼ p k,s   Dp km  1 þ  :      (13:48)  (5,000) or 0.052 psi/ft. Depth of the second valve is calcu-
                                         40,000                  lated with Eq. (13.53):
                       Solving Eqs. (13.47) and (13.48) yields       1,000   50   330 þ 0:39   0:052Þ(2,245)
                                                                                   ð
                                                                  D 2 ¼                           ¼ 3,004 ft
                           p k,s   Dp km   p hf                                   1,000   50
                       D 1 ¼                              (13:49)            0:39
                               p k   Dp km                                         40,000
                            G s       :
                                40,000                           Similarly,
                         When the static liquid level is below the depth calculated  1,000   50   330 þ 0:39   0:052Þ(3,004)
                                                                                   ð
                         by use of Eq. (13.49), the first valve is placed at a depth  D 3 ¼  1,000   50  ¼ 3,676 ft
                         slightly deeper than the static level. If the static liquid  0:39
                         level is known, then                                      40,000
                                                                                   ð
                       D 1 ¼ D s þ S 1 ,                  (13:50)  D 4 ¼  1,000   50   330 þ 0:39   0:052Þ(3,676)  ¼ 4,269 ft
                                                                                  1,000   50
                         where D s is the static level and S 1 is the submergence of  0:39    40,000
                         the valve below the static level.
                                                                                   ð
                                                                     1,000   50   330 þ 0:39   0:052Þ(4,269)
                       8. Calculate the depths to other valves. Based on the fact  D 5 ¼  1,000   50  ¼ 4,792 ft,
                         that p hf þ G fd D 2 þ G s (D 2   D 1 ) ¼ p cd2 , the depth of  0:39    40,000
                         valve 2 is expressed as
                                                                 which is the depth of the operating valve.
                           p cd2   G fd D 1   p hf
                       D 2 ¼            þ D 1 ,           (13:51)  Similar problems can be quickly solved with the com-
                                 G s                             puter spreadsheet GasLiftValveSpacing.xls.
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