Page 189 - Fundamentals of Enhanced Oil and Gas Recovery
P. 189

177
             Thermal Recovery Processes

                      Rate of growth for the heated zone is

                                           dA h        _ m s H s
                                               5                 G 1 ðt D Þ             (5.124)
                                           dt     M R T s 2 T R Þh
                                                     ð
                                                              p
                                                                ffiffiffiffiffi
                                                        t D
                                               G 1 t D 5 e erfc  t D                    (5.125)
                                                 ðÞ
                   Example 5.3: Radius of steam zone for constant injection rate
                   The steam with the quality of 80% at pressure of 500 psig is injected to the reservoir
                   at rate of 500 BWPD CWE. The reservoir has 25% porosity. The initial saturations of
                   oil and water is S oi 5 0.2 and S wi 5 0.8. The volume occupied by the steam is 40% of
                   the PV.
                                                      2
                      k h 5 1.5 Btu/h ft F,  α 5 0.0482 ft /h,  M R 5 32.74 Btu/ft 3   F,  M(5k h /α) 5


                   31.12 Btu/ft 3   F, T R 5 80 F.
                      Determine the heated area radius while 14 days is passed from injection [36].
                      At 500 psig, T s 5 470.9 F

                      H wTR 5 77 Btu/lb m (at 80 F)

                      H wTs 5 452.9 Btu/lb m (at 470.9 F)

                      λ s 5 751.4 Btu/lb m
                      m s 5 (500 bbl/d)(350 lb m /bbl)(24 h/d) 5 7292 lb m /h
                                                     _ m s H s M R h
                                            A h 5               Gðt D Þ
                                                 4 T s 2 T R ÞαM 2
                                                   ð

                                                     T s
                                              H s 5 H 2 H  T R
                                                     w     w  1 f s λ s
                                                        M   2

                                                              αt
                                                t D 5 4
                                                       M R    h 2
                                                          k h
                                                     M 5
                                                          α
                                               2   !
                                                  αt
                                     t D 5 4  M
                                            M R   h 2
                                                 ! 2                        !
                                                     ð0:0482Þ 14d 3 24 h=d
                                            31:12
                                       5 4
                                            32:74             20 2
                                       5 0:146

                      G(t D ) 5 0.113
                      H s 5 (452.9 2 77) 1 0.8 3 (751.4) 5 977 Btu/lb m
   184   185   186   187   188   189   190   191   192   193   194