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Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 240 22.12.2006 6:14pm




               15/240  PRODUCTION ENHANCEMENT
                                12,000

                                10,000
                               Test Flow Rate (Mcf/D)  8,000       Unloaded



                                 6,000

                                 4,000
                                 2,000                      ?      Nearly loaded up
                                                                   Loaded up
                                                                   Questionable
                                    0
                                     0     2,000  4,000  6,000  8,000  10,000  12,000
                                            Calculated Minimum Flow Rate (Mcf/D)
                      Figure 15.19 Calculated minimum flow rates with the Turner et al. model and test flow rates.

                    cde
               m ¼      ,                         (15:41)  Substituting Eq. (15.44) into Eq. (15.34) results in
                      2
                   1 þ d e
                                                                              b
                                                                1   2bm    m þ n   bm 2
               and                                       144ba 1 þ    ln a 2    c p ffiffiffi
                                                                  2             n           (15:45)
                     2
                    c e
               n ¼       ,                        (15:42)       tan  1  b 1   tan  1  b 2 ¼ g,
                  ð 1 þ d eÞ 2
                      2
               where                                     where
                  A ¼ cross-sectional area of conduit, ft 2              2
                 D h ¼ hydraulic diameter, in.                     5  S g T bh Q gm
                                                         a 1 ¼ 9:3   10      p hf ,         (15:46)
                                                                      2
                 f M ¼ Moody friction factor                         A E km
                                                                      i
                  g ¼ gravitational acceleration, 32:17 ft=s 2           2     2
                  L ¼ conduit length, ft                     1:34   10  2 S g T bh Q gm  þ m  þ n
                                                                      A 2 E km
                  p ¼ pressure, psia                     a 2 ¼         i    2     ,         (15:47)
                 p hf ¼ wellhead flowing pressure, psia           144p hf þ m þ n
                 Q G ¼ gas production rate, Mscf/day        1:34   10  2 S g T bh Q 2 gm  þ m
                 Q o ¼ oil production rate, bbl/day      b 1 ¼     p ffiffiffi A 2 E km  ,        (15:48)
                                                                      i
                                      3
                 Q s ¼ solid production rate, ft =day                n
                 Q w ¼ water production rate, bbl/day       144p hf þ m
                 S g ¼ specific gravity of gas, air ¼ 1  b 2 ¼  p ffiffiffi n  ,                  (15:49)
                 S o ¼ specific gravity of produced oil,
                      freshwater ¼ 1                     and
                 S w ¼ specific gravity of produced water,
                                                                 2
                      freshwater ¼ 1                     g ¼ a 1 þ d e L:                   (15:50)
                 S s ¼ specific gravity of produced solid,
                      freshwater ¼ 1                     All the parameter values should be evaluated at Q gm .
                 T av ¼ the average temperature in the butting, 8R  The minimum required gas flow rate Q gm can be solved
                  0
                  « ¼ pipe wall roughness, in.           from Eq. (15.45) with a trial-and-error or numerical
                  u ¼ inclination angle, degrees.        method such as the Bisection method. It can be shown
                                                         that Eq. (15.45) is a one-to-one function of Q gm for
               15.5.2.3 Minimum Required Gas Production Rate  Q gm values greater than zero. Therefore, the Newton–
               A logical procedure for predicting the minimum required  Raphson iteration technique can also be used for solving
               gas flow rate Q gm involves calculating gas density r g , gas  Q gm . Commercial software packages such as MS Excel can
               velocity v g , and gas kinetic energy E k at bottom-hole con-  be used as solvers. In fact, the Goal Seek function built
               dition using an assumed gas flow rate Q G , and compare  into MS Excel was used for generating solutions presented
               the E k with E km . If the E k is greater than E km , the Q G is  in this chapter. The spreadsheet program is named
               higher than the Q gm . The value of Q G should be reduced  GasWellLoading.xls.
               and the calculation should be repeated until the E k is very
               close to E km . Because this procedure is tedious, a simple
               equation was derived by Guo et al. for predicting the  Example Problem 15.2 To demonstrate how to use
               minimum required gas flow rate in this section. Under  Eq. (15.45) for predicting the minimum unloading gas
               the minimum unloaded condition (the last point of the  flow rate, consider a vertical gas well producing 0.70
               mist flow regime), Eq. (15.33) becomes    specific gravity gas and 50 bbl/day condensate through
                                                         a 2.441-in. inside diameter (ID) tubing against a
                                2
               E km ¼ 9:3   10  5  S g T bh Q gm ,  (15:43)  wellhead pressure of 900 psia. Suppose the tubing string
                              2
                             A p                         is set at a depth of 10,000 ft, and other data are given in
                              i
                                                         Table 15.1.
               which gives
                              2
               p ¼ 9:3   10  5  S g T bh Q gm :   (15:44)  Solution The solution given by the spreadsheet program
                           2
                          A E km                         GasWellLoading.xls is shown in Table 15.2.
                           i
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