Degenerate Two-Phase Flow in Porous Media including Dynamic Effects

Transcription

1 Degenerate Two-Phase Flow in Porous Media including Dynamic Effects Xiulei Cao Supervisor: Iuliu Sorin Pop Center for Analysis, Scientific Computing and Applications (CASA) April 9, 2014 Where innovation starts

2 Outline 2/37 Motivation Mathematical model Existence, non-degenerate case Existence, degenerate case Numerical results Conclusions and future work

3 Outline 3/37 Motivation Mathematical model Existence, non-degenerate case Existence, degenerate case Numerical results Conclusions and future work

4 Motivation 4/37 Non-monotone (p n p w )(s) curves measured under non-equilibrium condition: Water saturation [-] Pn-Pw, [kpa] time [s] time [s] Figure 1.1 1: Dynamic pressure *S. Bottero, Advances in the Theory of Capillarity in Porous Media.

5 5/37 Figure 1.2 2: The saturation measured under non-equilibrium condition *David A. DiCarlo, Experimental measurements of saturation overshoot on infiltration, Water

6 Outline 6/37 Motivation Mathematical model Existence, non-degenerate case Existence, degenerate case Numerical results Conclusions and future work

7 Mathematical model 7/37 Darcy Mass balance Gives Note: v α = k µ α k rα p α, α = w, n. t (φs α) + v α = 0, α = w, n. t (φs α) ( k k rα µ α p α ) = 0, α = w, n. s w + s n = 1.

8 Dynamic effects in capillary pressure 8/37 Standard model: p n p w = p c (s w ). Dynamic effects: Constant τ with τ > 0. Non-linear τ p n p w = p c (s w ) τ s w t, p n p w = p c (s w ) τ(s w ) s w t, with τ(s w ) > 0. *S.M. Hassanizadeh & W. Gray, Water Resour. Res. 29 (1993),

9 9/37 The model becomes t s w + (k w p w ) = 0, with ((k w + k n ) p w ) + (k n p c ) + (k n (τ t s w )) = 0, k α = k rα µ α, α = w, n.

10 k w, k n 10/37 Figure 2.1 1: k w, k n

11 Outline 11/37 Motivation Mathematical model Existence, non-degenerate case Existence, degenerate case Numerical results Conclusions and future work

12 Global pressure and complementary pressure Define the global pressure p 12/37 p = p w sw and the complementary pressure θ 0 k w k (z)p c (z)dz p c(s w ) + τ(s w ) t s w, with θ(s w ) = sw 0 k n k w k (z)p c (z)dz, k = k w + k n. * S. N. Antontsev (1972) and G. Chavent and J. Jaffré (1978).

13 13/37 With u = s w, the model becomes Existing results: t u + (k n p) θ(u) = 0, (k p) + (k w (τ t u)) = 0. Z. Chen (1997): Existence, uniqueness, τ = 0. A. Mikelić (2011): Existence, τ Constant, given total flow. B. Schweizer (2011): Alternative formulation - non-degenerate case.

14 Basic setting 14/37 Given: R d - bounded and connected, - Lipschitz. T i > 0, Q = (0, T i ]. Initial and boundary conditions: u(0, ) = u 0 in, u = u D at for all t > 0. Define and V := u D + W 1,2 0 ( ), W := {p W 1,2 ( ) p dz = 0}, T(u) = u u D τ(z)dz. Assume: T(u 0 ) W 1,2 ( ) and 0 u 0 1, 0 < u D < 1.

15 Weak solution 15/37 Problem P: Find u W 1,2 (0, T i ; V), p L 2 (0, T i ; W) such that u(0, ) = u 0, k w (u) t T(u) L 2 (0, T i ; L 2 ( ) d ), and ( t u, φ) (k n p, φ) + ( θ(u), φ) = 0, for all φ, ψ L 2 (0, T i ; V). (k p, ψ) + (k w t T(u), ψ) = 0, Note: k w (0) = k n (1) = 0, the problem becomes degenerate. First step: regularization (avoid "0").

16 Regularization 16/37 k wδ k w Figure 3.1 1: The regularization for k w

17 Regularization 17/37 k nδ k n Figure 3.2 2: The regularization for k n

18 Non-degenerate case 18/37 Probelm P δ : Find u δ W 1,2 (0, T; V), p δ L 2 (0, T; W), such that u δ (0, ) = u 0, t T δ (u δ ) L 2 (0, T; L 2 ( ) d ) and ( t u δ, φ) (k nδ p δ, φ) + ( θ δ, φ) = 0, (k δ p δ, ψ) + (k wδ t T δ (u δ ), ψ) = 0, for all φ, ψ L 2 (0, T; W 1,2 0 ( )).

19 19/37 Define: G, Ɣ : R R {± }, G(u) = G δ (u) = u u u D u D τk k w k n (z)dz, and Ɣ(u) = τ δ k δ k wδ k nδ (z)dz, and Ɣ δ (u) = u u D G(z)dz. u u D G δ (z)dz.

20 20/37 (a) Ɣ(u) function (b) Ɣ δ (u) function Figure 3.3 3: The profiles of Ɣ and Ɣ δ

21 21/37 Assume: Idea: show that Ɣ δ (u)dx C, This gives (δ 0): Ɣ(u 0 )dx < +. for all t (C is δ independent). 0 u 1.

22 Time discretization 22/37 With N N, let h = T i /N and consider the Euler implicit discretization Problem Pδ n : Given un 1 δ V(n = 1, 2...N), find uδ n V and pδ n W, such that ( un δ un 1 δ, φ) (k nδ (uδ n h ) pn δ, φ) + ( θ δ(uδ n ), φ) = 0, (k δ (uδ n ) pn δ, ψ) + (k wδ(uδ n ) T δ(uδ n) T δ(uδ n 1 ), ψ) = 0, h for any φ, ψ W 1,2 0 ( ). Lemma Problem Pδ n has a solution.

23 Interpolation in time 23/37 For t (t n 1, t n ], define and T δn (t) = T δ (u n 1 δ U δn (t) = u n 1 δ ) + t t n 1 h (T δ (u n δ ) T δ(u n 1 δ )), + t t n 1 (uδ n h un 1 δ ), T δn (t) = T δ (u n δ ), Ū N (t) = u n δ, P N (t) = p n δ.

24 24/37 Lemma With C not dependent on h, we have U N (t, x) 2 L 2 (0,T i ;L 2 ( )) C U N (t, x) 2 L 2 (0,T i ;L 2 ( )) C t U N (t, x) 2 L 2 (0,T i ;L 2 ( )) C ŪN (t, x) 2 L 2 (0,T i ;L 2 ( )) C T δ N(t) 2 L 2 (0,T i ;L 2 ( )) C T δn (t) 2 L 2 (0,T i ;L 2 ( )) C T δn (t) 2 L 2 (0,T i ;L 2 ( )) C P N 2 L 2 (0,T i ;L 2 ( )) C δ t T δn (t) 2 L 2 (0,T i ;L 2 ( )) C t T δn (t) 2 δ L 2 (0,T i ;L 2 ( )) C δ Ɣ δ (Ū N (t, x))dx C.

25 Existence 25/37 Theorem There exists a subsequence (U N, P N ) N N, which converges weakly to a solution pair (u δ, p δ ) of Problem P δ.

26 Outline 26/37 Motivation Mathematical model Existence, non-degenerate case Existence, degenerate case Numerical results Conclusions and future work

27 Existence for degenerate case 27/37 Assumptions: k w (v) = v α p c (v) = v λ k n (v) = (1 v) β τ(v) = (1 v) ω

28 Estimates for Problem P δ 28/37 ( u δ + δ) 2 α L (0,T i ;L 1 ( )) C ( 1 u δ + δ) 2 β ω L (0,T i ;L 1 ( )) C t u δ 2 L 2 (0,T i ;L 2 ( )) C u δ 2 L (0,T i ;L 2 ( )) C T δ (u δ ) 2 L (0,T i ;L 2 ( )) C p cδ τ δ u δ 2 L 2 (0,T i ;L 2 ( )) C k nδ p δ 2 L 2 (0,T i ;L 2 ( )) C 1 τδ (u δ ) tt δ (u δ ) 2 L 2 (0,T i ;L 2 ( )) C k wδ k nδ t T δ (u δ ) 2 k L 2 (0,T δ i ;L 2 ( )) C

29 Estimates for Problem P δ 29/37 Theorem Let d 3, if α λ > 4, ω > 2, β > 2, for d = 1, 2. α λ > 5, and ω > 5/2, β > 5/2, for d = 3. Then we have the following estimates: t T δ (u δ ) L r 1 (Q) C, with r 1 (1, 2), p δ L r 2 (Q) C, with r 2 (1, 2), t T δ (u δ ) L r 3(Q) C, with r 3 (1, 2).

30 Theorem Let d 3, if α λ > 4, ω > 2, β > 2, for d = 1, 2. α λ > 5, and ω > 5/2, β > 5/2, for d = 3. Then Problem P has a solution pair (u, p), and 0 u 1 a.e. in for all t. 30/37

31 Outline 31/37 Motivation Mathematical model Existence, non-degenerate case Existence, degenerate case Numerical results Conclusions and future work

32 Numerical results 32/37 (a) τ = 0 (b) τ = 0 Figure 5.4 4: Saturation τ = 0 (monotone) and τ = 0 (over shoot)

33 Outline 33/37 Motivation Mathematical model Existence, non-degenerate case Existence, degenerate case Numerical results Conclusions and future work

34 Conclusions and future work 34/37 Conclusions: Proved the existence of weak solutions for the 2-phase model including dynamic effects in the capillary pressure (degenerate case). Future work: Uniqueness? Heterogeneities? Acknowledgements: China Scholarship Council

35 One equation model (total flow given constant) 35/37 φ s t q k w x k ( k k wk n x k x {p c(s, x) + τ s ) t } = 0. Capillary pressure p c p c II I entry 0 s s 1 Figure 6.5 5: Capillary pressure p c

36 Heterogeneous results 36/37 S entry S I S II = 0 (a) τ = 0 (b) τ = 0 Figure 6.6 6: Oil saturation τ = 0 (trapping) and τ = 0 (oil flow)

37 Thank you for your attention! 37/37