Water Artificial Circulation For Eutrophication Control

Transcription

1 Water Artificial Circulation For Eutrophication Control Francisco Javier Fernández Centro Universitario de la Defensa, Escuela Naval Militar, Marn, Spain Joint work with Lino J. Álvarez-Vázquez and Aurea Martínez Depto. Matemática Aplicada II, Universidade de Vigo, E.I. Telecomunicación Vigo, Spain

2 Table of contents. The environmental problem Mathematical formulation Mathematical analysis Numerical resolution Numerical tests

3 The environmental problem

4 The environmental problem Eutrophication (environmental problem) Eutrophication is one of the most important problems of large masses of water and it is caused by high levels of pollutants that reach the waters. This pollutants come mainly from human activities and can cause an excessive phytoplankton growth that lead to undesirable effects like algal blooms. 1 Artificial circulation (main idea) Its is a management technique for oxygenating eutrophic water bodies subject to quality problems, such as loss of oxygen, sediment accumulation and algal blooms. It disrupts stratification and minimizes the development of stagnant zones that may be subject to water quality problems (low levels of oxygen). In our case we are interested in increase the dissolved oxygen content in the bottom layers.

5 The environmental problem 1 Artificial circulation (how it works) A flow pump takes water from the well aerated upper layers by means of a collector and injecting it into the poorly oxygenated bottom layers, setting up a circulation pattern that prevents stratification. Oxygen-poor water from the bottom is circulated to the surface, where oxygenation from the atmosphere and photosynthesis can occur. Objective We want to optimize the volume of pumped water in order to minimize the energy cost and ensure that the concentration level of dissolved oxygen in the bottom layer is over a threshold.

6 Mathematical formulation

7 Mathematical formulation 1 N C 1 C 2 C T 2 T 1 S T 3 The control variables C 3 C 4 T 4 N N The physical domain R 3 (for instance a lake). C, control domain. N CT number of pairs collector-injector. = Γ S Γ N Γ C Γ T, where: Γ S, surface of the lake, Γ N, bottom and walls, Γ C, collectors, Γ C = N CT k=1 C k, Γ T, injectors, Γ T = N CT k=1 T k. g(t) = (g 1 (t),..., g N CT (t)), where g k (t) is the volume of water displaced by pump k. The state equations The hydrodynamic model: modified Navier-Stokes equations (Smagorinsky). ( ) v/ t + vv (2ν + 2ν tur [ɛ(v) : ɛ(v)] 1/2 )ɛ(v) + p = F in (0, T ), v = 0 in (0, T ), v = 0 on (Γ S Γ N ) (0, T ), v = +g k (t)/µ(c k ) n on C k (0, T ), for k = 1,..., N CT, v = g k (t)/µ(t k ) n on T k (0, T ), for k = 1,..., N CT, v(0) = v 0 in,

8 Mathematical formulation The eutrophication model: Michaelis-Menten kinetics. u 1 generic nutrient (usually, nitrogen and/or phosphorus), u 2 phytoplankton u 3 zooplankton, u 4 organic detritus, and u 5 dissolved oxygen. Interactions between five above species can be modelled by the following system: u i / t + v u i (µ i u i ) = A i (u) + f i in (0, T ), u i n = 0 on (Γ S Γ N Γ C ) (0, T ), u i = C k u i dγ/µ(c k ) on T k (0, T ), k = 1,..., N CT, u i (0) = u i 0 in, for i = 1,..., 5, where the reaction term A = (A i ) : (0, T ) [R +] 5 R 5 is: ) u C nc (L(x, 1 t) K N + u 1 u2 K r u 2 + C nc K rd Θ θ(x,t) 20 u 4 ( ) u 1 L(x, t) K N + u 1 u2 K r u 2 K mf u 2 u 2 K z K F + u 2 u3 A(x, t, u) = u 2 C fz K z K F + u 2 u3 K mz u 3 K mf u 2 + K mz u 3 K rd Θ θ(x,t) 20 u 4 ) u C oc (L(x, 1 t) K N + u 1 u2 K r u 2 C oc K rd Θ θ(x,t) 20 u 4

9 Mathematical formulation The optimal control problem where: (P) mín{j(g) : g U ad and 1 u(t) dx K C, t [0, T ]}, µ( C ) C U ad = {g U : 0 < c 1 g k (t) c 2, t [0, T ], k = 1,..., N CT }, with U a suitable functional space and the constants c 1, c 2 related to technological constraints on the pumps. We will have to redefine it in order to perform the mathematical analysis of the problem. J(g) is the cost functional given by: J(g) = 1 N CT T g k (t) 2 dt, 2 k=1 0 representing the energy saving related to pumps operation. K C R 5 is the rectangle: K C = [λ m 1, λm 1 ] [λm 5, λm 5 ], with λ m k and λm k the minimal and maximal mean concentrations allowed for the species in C. (v, u) are the solutions of the state system associated to control g. We have a optimal control problem with nonlinear state constraints.

10 Mathematical analysis

11 Mathematical analysis The space for the control variables The state equations U ad = { g H s 1/2 ( (0, T )) L 2 (0, T ; H 1 ( )) : c 1 g(x, t) 0, a.e. (x, t) T k (0, T ), k = 1,..., N CT, +0 g(x, t) c 2, a.e. (x, t) C k (0, T ), k = 1,..., N CT, g(x, t) = 0 a.e. (x, t) (Γ N Γ S ) (0, T ), k = 1,..., N CT, T k C k g(t) dγ = 0, a.e. t (0, T ), k = 1,..., N CT, g H s 1/2 ( (0,T )) L 2 (0,T ; H1 ( )) c 3 } The hydrodynamic model. If s 7/2, there exists an unique solution: v W 1,,2 (0, T ; W, [L 2 ()] 3 ) C([0, T ]; W), v(0) = v 0 a.e. in. v = g n, a.e. (x, t) (0, T ). v verifies the variational formulation: v t z dx + vv z dx + β(ɛ(v))ɛ(v) : ɛ(z) dx = F z dx, a.e. t (0, T ), z V. where: W = {v [W 1,3 ()] 3 : v = 0 and v (ΓN Γ S ) = 0}, V = {v [W 1,3 ()] 3 : v = 0 and v = 0} W 1,p,q (I ; X, X ) = { u L p (I ; X ) : u/ t L q (I ; X ) }.

12 Mathematical analysis The eutrophication model. If v W 1,,2 (0, T ; W, [L 2 ()] 3 ) C([0, T ]; W) is the solution of the hydrodynamic model, then there exists a unique solution for the eutrophication model (we also need compatibility conditions for oefficients and data): u W 1,2,2 (0, T ; [H 1 ()] 5, [H 1 ()] 5 ), u i (0) = u i 0 a.e. x, u i T k = 1 u i dγ a.e. (x, t) T k (0, T ), µ(c k ) C k u i verifies the variational formulation: u i t η dx + v u i η dx + µ i u i η dx = f i η dx + A i (u)η dx, η X, a.e. t (0, T ). The control problem Existence. Assuming that the coefficients and data of the problem satisfy the necessary conditions for the existence of a solution of the state systems, there exists, at least, a solution of the optimal control problem (P). Optimality conditions. We can prove the existence of first order necessary conditions that must be verified by the optimal control.

13 Numerical resolution

14 Numerical resolution Space-Time discretization A partition {t 0, t 1,..., t N } of the time interval [0, T ] for a time step t = T /N, where t n = n t for n = 0, 1,..., N, will be considered. We will use the Characteristics Method for the discretization of the material derivatives. We will consider a regular triangular mesh τ h of a polygonal domain and we will use the Finite Element Method for the space discretization, with: W h /V h (P 1b FEM space) for the water velocity (u), M h (P 1 FEM space) for the water pressure (p), Zh/X h (P 1 FEM space) for the species concentration of the eutrophication model (u). Software development The space and time discretizations will be implemented in the scientific software Freefem++. The numerical resolution of the constrained optimization problem (obtained from the space-time discretization of the continuous control problem) has been developed with the help of the interior point algorithm IPOPT, interfaced with Freefem++

15 Numerical resolution The discretized control problem After the space-time discretization of the continuous problem (P), we obtain the following constrained nonlinear minimization problem: where: (Q) mín{j(g) : g U ad }, Discretized control: We will consider the following set of admissible discrete controls: U ad = {g R N N CT : c 1 g k,n c 2, k = 1,..., N CT, n = 1,..., N}, with c 1 and c 2 are bounds related to pumps technical characteristics. Discretized cost functional: We consider the following discretized cost function: J(g) = 1 N N CT (g k,n ) 2. 2 n=1 k=1 Discretized state constraints: We define the real-valued mapping: G = (G 1,..., G N ) : U ad R 5N g G(g), where, for each n = 1,..., N, G n 1 (g) = u n+1 dx, µ( C ) C with u n+1 Z h the solution of the discretized eutrophication model. Thus: U ad = {g U ad : G n (g) K C, n = 1,..., N}.

16 Numerical resolution Discretized state equations: Given an admissible control g = (g 1,1, g 2,1,..., g N CT,1,..., g 1,N, g 2,N,..., g N CT,N ) U ad, }{{}}{{} g 1 g N The hydrodynamic model: For n = 0, 1,..., N 1, (v n+1, p n+1 ) W h M h, with v n+1 T k = g k,n+1 /µ(t k ) n, v n+1 C k = g k,n+1 /µ(c k ) n, k = 1,..., N CT, is the solution of: α v n+1 z dx + 2ν ɛ(v n+1 ) : ɛ(z) dx +2ν tur [ɛ(v n+1 ) : ɛ(v n+1 )] 1/2 ɛ(v n+1 ) : ɛ(z) dx p n+1 z dx v n+1 q dx λ p n+1 q dx = α (v n X n ) z dx + F n+1 z dx, z V h, q M h. The eutrophication model: For n = 0,..., N, u n+1 Z h, with u n+1 T k = u n+1 dγ/µ(c k ), k = 1,..., N CT, C k is the solution of: α u n+1 η dx + Λ µ u n+1 : η dx = α (u n X n ) η dx + f n+1 η dx + A n+1 (u n+1 ) η dx, η X h.

17 Numerical resolution Remark. It should be noted that we employ a different number of time steps in both models: N for the hydrodynamic model and N + 1 for the species model. This shift is motivated by the following dependency scheme v 0 u 0 g 1 v 1 u 1 g 2 v 2 u 2 G 1 (g) = G 1 (g 1 ) g 3 v 3 u 3 G 2 (g) = G 2 (g 1, g 2 ).... g N v N u N G N 1 (g) = G N 1 (g 1, g 2,..., g N 1 ) u N+1 G N (g) = G N (g 1, g 2,..., g N 1, g N ) The Jacobian associated to state constraints is a block lower triangular matrix: δ g1 G 1 (g) δ g1 G 2 (g) δ g2 G 2 (g) δ gg(g) = δ g1 G N 1 (g) δ g2 G N 1 (g)... δ gn 1 G N 1 (g) 0 δ g1 G N (g) δ g2 G N (g)... δ gn 1 G N (g) δ gn G N (g).

18 Numerical resolution Computing the Jacobian associated to state constraints In order to use the IPOPT algorithm in the numerical resolution of the problem (Q), we need a function that evaluates the Jacobian associated to the state constraints. We can use the linearized state equations or the adjoint state equations. The choice of one method or another depends on the relationship between the size of the control and the constraints. If we want to use the linearized equations, we must solve N CT N times this equations. If we employ the adjoint state equations, we must solve 5 N times this equations. We have that if N CT < 5 it is more convenient to use the linearized equations, whereas if N CT > 5, it is more advantageous to employ the adjoint state. In the example presented in the next section, we have considered N CT = 4, therefore, we will use the linearized equations for computing the Jacobian of the discretized state constraints: δg n 1 (g)(δg) = δu n+1 dx, n = 1,..., N, µ( C ) C where δv 0 = 0, δu 0 = 0 and, given a direction δg, the linearized state is obtained by:

19 Numerical resolution The linearized hydrodynamic model: For n = 0,..., N 1, (δv n+1 h, δp n+1 h ) W h M h, with δv n+1 k,n+1 δg T k = µ(t k ) n, k,n+1 δg δvn+1 Ck = µ(c k ) n, k = 1,..., N CT, is the solution of: α δv n+1 z dx + 2ν ɛ(δv n+1 ) : ɛ(z) dx +2ν tur [ɛ(v n+1 ) : ɛ(v n+1 )] 1/2 ɛ(δv n+1 ) : ɛ(v n+1 ) ɛ(v n+1 ) : ɛ(z) dx +2ν tur [ɛ(v n+1 ) : ɛ(v n+1 )] 1/2 ɛ(δv n+1 ) : ɛ(z) dx δp n+1 z dx δv n+1 q dx λ δp n+1 q dx = α (δv n X n ) z dx ( v n X n )δv n z dx, v V h, q M h. The linearized eutrophication model: For n = 0,..., N, the element δu n+1 Z h, with δu n+1 T k = 1 δu n+1 dγ, k = 1,..., N CT, µ(c k ) C k is the solution of: α δu n+1 η dx + Λ µ u n+1 : η dx = α (δu n X n ) η dx ( u n X n )δv n η dx + δ ua n+1 (u n+1 )(δu n+1 ) z dx, η X h.

20 Numerical tests

21 Numerical tests The following data have been considered for testing the methodology proposed (2D example): Computation domain Time discretization (4 scenarios) Test 1: 12h, t = 3600s. Test 2: 12h, t = 1800s. C 1 C 2 S C 3 C 4 Test 3: 24h, t = 3600s. Test 4: 48h, t = 3600s. N N Space discretization = [0, 14] [0, 16] (meters). N CT = 4 pairs collector-injector. C T 2 T 1 T 3 T 4 N C = [0, 14] [0, 2]. We will consider a regular mesh of 3075 vertices.

22 Numerical tests Test 1 (12h, t = 3600s) Test 2 (12h, t = 1800s)

23 Numerical tests Test 1 (12h, t = 3600s) Test 2 (12h, t = 1800s) Dissolved Oxygen, time step 12 IsoValue Dissolved Oxygen, time step 25 IsoValue Vec Value 0 Vec Value e e e e e

24 Numerical tests Test 3 (24h, t = 3600s) Test 4 (48h, t = 3600s)

25 Numerical tests Test 3 (24h, t = 3600s) Test 4 (48h, t = 3600s) Dissolved Oxygen, time step 25 IsoValue Dissolved Oxygen, time step 49 IsoValue Vec Value 0 Vec Value e e e e e e e e e e e e e e e e e

26 Thank you for your attention. Any Questions?