PEGN 513 Reservoir Simulation I Fall 2009 Homework #3 Solution The simplest form for waterflood of a linear core initially saturated with oil and at residual water saturation, without gravity or capillary effects, is represented by the 1-dimensional Buckley-Leverett material balance equation, xfutSwtw∂∂−=∂∂ (1) Explicit solution Considering the space discretization shown in Figure 1, the explicit solution of this equation is given by, xffutSSniwniwtniwniw∆−−=∆−−+1,,,1, (2) To improve the computation process and avoid using an imaginary grid block at the left of the first node, the following equation is used for the first node, 221,1,1,11,xffutSSnwnwtnwnw∆−−=∆−+ (3) These equations (Eqs. 2 and 3) can be solved explicitly for water saturation at the new time step (n+1). Figure 1. 1-dimensional core space discretization 1 2 3 i-1 i i+1 IMAX-1 IMAX i-1/2 i+1/2 IMAX-1/2 1 1/2 1/2 ∆xiPEGN 513 Reservoir Simulation I Fall 2009 2/8 Implicit solution Considering the space discretization shown in Figure 1, the implicit solution of equation (1) is given by, xffutSSniwniwtniwniw∆−−=∆−+−+++ 11,1,1,1, (4) The fractional flow at time n+1 can be related to the known fractional flow at time n using the Taylor polynomial of first order, ( )niwniwniwwniwniwSSSfff,1,,1,−∂∂+=++ (5) Substituting (5) into (4) and reordering, an equation with two unknowns (Sw’s at n+1) is obtained, niwniwniwniwwniwtiniwwniwtiniwwniwniwwffSSfStuxSfStuxSfSSf1,,1,1,1,11,1−−−++−−−+∂∂+∆∆+∂∂−=∆∆+∂∂−∂∂ (6) For node 1, this equation can be solve explicitly as follows, ∆∆+∂∂−∂∂−−+∂∂+∆∆+∂∂−=++tuxSfSSfffSSfStuxSfStnwwnwnwwnwnwnwnwwnwtnwwnw2/2/1112/1,2/12/1,1,2/1,2/11,1111, (7) where, rwnwSS =2/1, 12/1,=nwf 02/1=∂∂nwwSf A back-substitution procedure can be applied to sequentially solve for the rest of the node saturations, using equation (6). Both, the explicit and implicit formulations were coded in MATLAB to solve the 1-dimensional Buckley-Leverett problem. The procedure to solve analytically for the derivative of fractional flow with respect to water saturation is shown at the end of this report as an addendum.PEGN 513 Reservoir Simulation I Fall 2009 3/8 Solution for Case 1: Explicit solution, ∆∆∆∆t = 0.1 day, IMAX = 10, ∆∆∆∆x = 10 ft Table 1 shows the values of water saturation for the 10 grid blocks at 10 days. A plot of water saturation profile as a function of distance along path of flow at every five time steps is shown in Figure 2. The initial condition (at t = 0) corresponds to the horizontal line of constant water saturation equal 0.25 (i.e., Swr). The figure shows an increase of water saturation with time and how the water injection front propagates along the 1-dimensional core. The spread of the curves is an indicative of numerical dispersion. Table 1. Water saturation values for the 10 grid blocks at 10 days Grid block index 1 2 3 4 5 6 7 8 9 10 Sw (fraction) 0.68313 0.66975 0.66058 0.65302 0.64641 0.64045 0.63497 0.62986 0.62503 0.62045 0 20 40 60 80 10000.10.20.30.40.50.60.70.80.91Distance along path of flow (ft)Water Saturation (fraction) Figure 2. Water saturation profile along the path of flow at every five time stepsPEGN 513 Reservoir Simulation I Fall 2009 4/8 Solution for Case 2: Explicit solution, ∆∆∆∆t = 0.1 day, IMAX = 20, ∆∆∆∆x = 5 ft Table 2 shows the values of water saturation for the 20 grid blocks at 10 days for Case 2. A plot of water saturation profile as a function of distance along path of flow at every five time steps is show in Figure 3. This case uses a small ∆x than Case 1 and more nodes; it can be observe a sharp water injection front in comparison with the spread curves of Case 1, which means a decrease of numerical dispersion with a decrease of grid block size in the x direction. Table 2. Water saturation values for the 10 grid blocks at 10 days. Case 2 Grid block index 1 2 3 4 5 6 7 8 9 10 Sw (fraction) 0.6894 0.6807 0.6746 0.6695 0.6650 0.6609 0.6572 0.6537 0.6504 0.6473 Grid block index 11 12 13 14 15 16 17 18 19 20 Sw (fraction) 0.6444 0.6415 0.6388 0.6362 0.6336 0.6312 0.6288 0.6264 0.6242 0.6220 0 20 40 60 80 10000.10.20.30.40.50.60.70.80.91Distance along path of flow (ft)Water Saturation (fraction) Figure 3. Water saturation profile along the path of flow at every five time steps. Case 2PEGN 513 Reservoir Simulation I Fall 2009 5/8 Solution for Case 3: Implicit solution, ∆∆∆∆t = 0.1 day, IMAX = 10, ∆∆∆∆x = 10 ft Table 3 shows the values of water saturation for the 10 grid blocks at 10 days of Case 3. A plot of water saturation profile as a function of distance along path of flow at every five time steps is show in Figure 4. It can be observe the water injection front propagation with time along the 1-dimensional core in the form of spread saturation curves, because of numerical dispersion. Table 3. Water saturation values for the 10 grid blocks at 10 days. Case 3 Grid block index 1 2 3 4 5 6 7 8 9 10 Sw (fraction) 0.6823 0.6684 0.6589 0.6511 0.6442 0.6381 0.6324 0.6271 0.6221 0.6173 0 20 40 60 80 10000.10.20.30.40.50.60.70.80.91Distance along path of flow (ft)Water Saturation (fraction) Figure 4. Water saturation profile along the path of flow at every five time steps. Case 3PEGN 513 Reservoir Simulation I Fall 2009 6/8 Solution for Case 4: Implicit solution, ∆∆∆∆t = 0.1 day, IMAX = 20, ∆∆∆∆x = 5 ft Table 4 shows the values of water saturation for the 20 grid blocks at 10 days for Case 4. A plot of water saturation profile as a function of distance along path of flow at every five time steps is show in Figure 5. This case uses a small ∆x than Case 3 and more nodes, equal to Case 2; it can be observe a sharp water injection front, but not that sharp as in Case 3, which
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