Page 410 - Characterization and Properties of Petroleum Fractions - M.R. Riazi
P. 410

QC: IML/FFX
              P2: IML/FFX
  P1: IML/FFX
  AT029-09
                                           June 22, 2007
            AT029-Manual
         390 CHARACTERIZATION AND PROPERTIES OF PETROLEUM FRACTIONS
                 300    AT029-Manual-v7.cls  T1: IML    14:25  and at 285 K,     y i /K i VS  = 1.073. Finally at T = 285.3877 K,
                                                                    VS
                                 29 wt%  16 wt%  0 wt%           y i /K i  = 1.00000, which is the correct answer. Thus the
                                                                                                       ◦
                                             MEOH             HFT for this gas at 30 bar is 285.4 K or 12.2 C. At this
                 250      Experimental
                                                              temperature from Eq. (9.45), K 1 = 2.222, K 2 = 0.7603, and
                          Calculated                          K 3 = 0.113. Composition of hydrocarbons in a water-free
                Pressure, atm__  150                          base hydrate is calculated as x = y i K i  , which gives x =
                 200
                                                                                        S
                                                                                                              S
                                                                                               VS

                                                                                        i
                                                                                                              1
                                                              0.36, x = 0.197, and x = 0.443. (b) At 414 bars Eq. (9.46)
                                                                    S
                                                                                  S
                                                                    2
                                                                                  3
                                                              with coefficients in Table 9.12 should be used. At this pres-
                                                              sure HFT is calculated as 303.2 K or HFT = 30 C. (c) To de-
                                                                                                      ◦
                 100
                                                              crease HFT at 30 bars an inhibitor solution that can cause
                                                              depression of  T = 12.2 − 5 = 7.2 C is needed. Rearrang-
                                                                                            ◦
                  50                                          ing Eq. (9.47): wt% = 100[M T/(A + M T)], where wt% is
                                                              the weight percent of inhibitor in aqueous solution. From
                   0                                          Table 9.13 for methanol, A = 1297.2 and M = 32. Thus with
                   260     270     280     290    300          T = 7.2, wt% = 15.1. Since calculated wt% of methanol
                               Temperature, K                 is less than 20% use of Eq. (9.47) is justified. For pressure
                                                              of 414 bars Eq. (9.48) should be used for methanol where
                FIG. 9.24—Depression of hydrate formation
              temperature in methane by methanol-calculated   upon rearrangement one can get x MeOH = 1 − exp(− T/72).
                                                                                ◦
              versus measured values. Lines represent coex-   At  T = 30 − 5 = 25 C we get x MeOH = 0.293. For an aque-
              istence curves for methane, hydrate, and aque-  ous solution (M H 2 O = 18) and from Eq. (1.15), the wt% of
              ous solutions of MeOH. Taken with permission    methanol (M = 32) can be calculated as: wt% = 42.4.
              from Ref. [55].
                                                              9.6 APPLICATIONS: ENHANCED OIL
         for some common inhibitors are given in Table 9.13. Values  RECOVERY—EVALUATION OF GAS
         of A are corrected values as given in Ref. [14].     INJECTION PROJECTS
          Equation (9.47) is recommended for sweet natural gases
         (H 2 S content of less than 4 ppm on volume basis, also see  In this section another application of some of the methods
         Section 1.7.15) with inhibitor concentrations of less than  presented in this book is shown for the evaluation of gas in-
         20 mol%. For concentrated methanol solutions, like those  jection projects. Gas is injected into oil reservoirs for differ-
         used to free a plugged-up tubing string in a high-pressure well,  ent purposes: storage of gas, maintenance of reservoir pres-
         Whitson [1] suggests a modified form of Hammerschmidt  sure, and enhanced recovery of hydrocarbons. In the last case,
         equation:
                                                              understanding and modeling of the diffusion process is of
        (9.48)           T =−72 ln (1 − x MeOH)               importance to the planning and evaluation of gas injection
                                                              projects. Gases such as natural gas, methane, ethane, lique-
                                      ◦
         where  T is the decrease in HFT in C (or in kelvin) and x MeOH  fied petroleum gas (LPG), or carbon dioxide are used as misci-
         is the mole fraction of methanol in the aqueous solution.  ble gas flooding in EOR techniques. Upon injection of a gas, it
                                                              is dissolved into oil under reservoir conditions and increases
         Example 9.4—Composition of a natural gas in terms of mol%  the mobility of oil due to decrease in its viscosity. To reach a
         is as follows: 85% C 1 , 10% C 2 , and 5% C 3 . Calculate
                                                              certain mobility limit a certain gas concentration is required.
         a. HFT at 30 bars and composition of hydrate formed.  For planning and evaluation of such projects, it is desired to
         b. HFT at 414 bars.                                  predict the amount of gas and duration of its injection in an
         c. wt% of methanol solution needed to decrease HFT to 5 C  oil reservoir. In such calculations, properties such as density,
                                                         ◦
           for each case.                                     viscosity, diffusivity and phase behavior of oil and gas are
                                                              needed. The purpose of this section is to show how to apply
         Solution—(a) At 30 bar pressure (<70 bar) the HFT can  methods presented in this book to obtain desired information
         be calculated from Eqs. (9.44) and (9.45) by trial-and-error  for such projects. This application is shown through modeling
         method. Assuming HFT of 280 K, the sum in Eq. (9.44)  of fractured reservoirs for a North Sea reservoir for the study
         is     y i /K VS  = 2.848 since it is greater than 1, temperature  of nitrogen injection. Laboratory experimental data are used
                 i
         should be increased in order to decrease K VS  values. At T =  to evaluate model predictions as discussed by Riazi et al. [56].
         300 K,     y i /K i VS  = 0.308;, at T = 290 K,  i    y i /K i VS  = 0.504;  An idealized matrix–fracture system is shown in Fig. 9.25,
                                                              where matrix blocks are assumed to be rectangular cubes. Di-
                                                              mensions of matrix blocks may vary from 30 to 300 cm, and
            TABLE 9.13—Constants in Eq. (9.47) for hydrate formation  the thickness of fractures is about 10 –10 −4  cm. When a gas
                                                                                             −2
                              inhibitors.                     is injected into a fractured reservoir, the gas flows through
         Hydrate formation inhibitor  Formula  M        A     the fracture channels in horizontal and vertical directions.
         Methanol                CH 3 OH      32      1297.2
         Ethanol                 C 2 H 5 OH   46      1297.2  Therefore, all surfaces of a matrix block come into contact
         Ethylene glycol         C 2 H 6 O 2  62      1500    with the surrounding gas in the fracture. The injected gas
         Diethylene glycol       C 4 H 10 O 3  106    2222.2  comes into contact with oil in the matrix block at the matrix–
         Triethylene glycol      C 6 H 14 O 4  150    3000    fracture interface. The gas begins to diffuse into oil and light



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