Commit 87d5c1f6 by Jonathan Lambrechts

### pffff

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 \section{Finite Element Formulation} We present a finite element formulation for the simulation of a viscous incompressible flow that is an intermediate between a free flow (Stokes flow) and porous media flow (Darcy flow). Consider our problem in a domain $\Omega$ with homogeneous Dirichlet boundary conditions on its boundary $\Gamma$: \begin{eqnarray} \nabla p + {\mu \over K} \mvu - \bar{\mu} \nabla^2 \mvu &=& {\mathbf 0}~~\mbox{in}~~ \Omega\nonumber \\ -\nabla \cdot (\mvu + \mvu_s) &=& 0~~\mbox{in}~~ \Omega\nonumber \\ \mvu + \mvu_s& =& {\mathbf 0} ~~\mbox{on}~~ \Gamma\nonumber \end{eqnarray} We use the following notations: The variational formulation of the problem is the following. Find $(\mvu,p) \in [H^1_0(\Omega)]^3 \times L^2_0(\Omega)$ such that \begin{eqnarray} \label{eq:var} &&\left(\nabla p, \mvv\right)_{\Omega} + \left(\mvu, \nabla q\right) _{\Omega} + \left(\bar{\mu} \nabla \mvu, \nabla \mvv \right) _{\Omega} + \left({\mu \over K} \mvu, \mvv \right) _{\Omega} = \left(\nabla \cdot \mvu_s, q\right) _{\Omega}\nonumber \\ &&\forall (\mvv,q) \in [H^1_0(\Omega)]^3 \times L^2_0(\Omega). \end{eqnarray} Equation \eqref{eq:var} can be written in the usual abstract form: find $(\mvu,p) \in [H^1_0(\Omega)]^3 \times L^2_0(\Omega)$ such that $$B(\mvu,p;\mvv,q) = L(\mvv,q),~~~~\forall (\mvv,q) \in [H^1_0(\Omega)]^3 \times L^2_0(\Omega).$$ This continuous problem has a unique solution. We assume a partitioning ${\mathcal T}$ of the domain $\Omega$. We denote $T$ an element of the partitioning, and its size of $h_T$. Let us first define standard continuous finite element spaces $${\mathbf V}_h =\left\{\mvv \in [H^1_0(\Omega) \cap C(\Omega)]^3 ~|~\mvv|_T \in [P_k(K)]^3,~~\forall T \in {\mathcal T} \right\},$$ $${ Q}_h =\left\{p \in L^2(\Omega) \cap C(\Omega)~|~p|_T \in P_k(K),~~\forall T \in {\mathcal T} \right\}.$$ We use a stabilized method for ensuring the stability of the discrete formulation: find $(\mvu_h, p_h) \in {\mathbf V}_h \times Q_h$ such that $$B_h(\mvu_h,p_h;\mvv,q)=L_h(\mvv,q)~~ \forall (\mvv_h, q_h) \in {\mathbf V}_h \times Q_h.$$ For that, we define a the \emph{stabilized bilinear form} \cite {Juntunen} \begin{eqnarray} &&B_h(\mvu_h,p_h;\mvv,q) = B(\mvu_h,p_h;\mvv,q) - \nonumber \\ &&\alpha \sum_{T \in {\mathcal T}} {h_T^2 \over K+h_T^2} \left ( \nabla p_h + {\mu \over K} \mvu_h - \mu \nabla^2 \mvu_h, {K\over\mu} \nabla q + \mvv - K \nabla^2 \mvv \right)_T \end{eqnarray} and \begin{eqnarray} L_h(\mvv,q) = L(\mvv,q) - \alpha \sum_{T \in {\mathcal T}} {h_T^2 \over K+h_T^2} \left (\nabla \cdot \mvu_s,q \right)_T \end{eqnarray} Coefficient $\alpha$ has to be positive. If linear finite elements are used, this simplifies to \begin{eqnarray} &&B_h(\mvu_h,p_h;\mvv,q) = B(\mvu_h,p_h;\mvv,q) - \nonumber \\ &&\alpha \sum_{T \in {\mathcal T}} {h_T^2 \over K+h_T^2} \left[ \left ({K\over\mu} \nabla p_h,\nabla q\right)_T + \left ({\mu\over K} \mvu_h,\mvv\right)_T + \left (\nabla p_h,\mvv\right)_T + \left (\mvu_h, \nabla q\right)_T \right].\nonumber \end{eqnarray} \subsection {FE formulation ($v^β = 0$)} \begin{eqnarray*} «φv_\alpha ·ṉ» - ‹\nabla φ·v_\alpha › &=& 0 \\ «\mu ψn · \nabla v_\alpha » -‹\mu \big(\nabla ψ·\nabla v_\alpha + ψ\nabla v^\alpha ·\nabla c_\alpha \big)› - ‹ψc_\alpha \nabla p^\alpha › -‹\frac{\mu (1-c_\alpha )²}{kc²_\alpha}v_\alpha › &=& 0 \end{eqnarray*} where $\nabla v^\alpha = \frac{\nabla v_\alpha}{c_\alpha} - \frac{v_\alpha \nabla c_\alpha}{c²_\alpha}$

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 \section{Avancement des particules} \section{Couplage temporel} \subsection{Avancement des particules} Pour le moment, on résout en déplacement (si j'ai bien compris ce qu'on entendait par là). C'est à dire que: \begin{align*} a^i &= a(x^i, v^i)\\ ... ... @@ -22,7 +23,7 @@ x^{i+1} &= x^i + v^{i+1}\, Δt } \section{Accélération partiellement implicite} \subsection{Accélération partiellement implicite} \begin{align*} \acc(x, v) &= \af(x, v) + \aext\\ \af(x, v) &= -β∇p -α(v-\vf)\,(v-\vf) ... ... @@ -46,7 +47,7 @@ telle que $\tia(x, v, v) = a(x, v)$ et $v$ est traité implicitement dans l'approximation $\aim$ et explicitement dans la correction $\aex$. Comme $\tiv$, on prend la vitesse au début du pas de temps. \nb{\item si on applique RK à $x$, il faut aussi introduire un $\tilde x$ pour la partie implicite} \section{Formalisme IMEX} \subsection{Formalisme IMEX} Je suis ici les notations d'\cite{ascher_implicit-explicit_1995}: $u$ est le vecteur inconnu, $g$ la partie implicite et $f$ la partie explicite. Pour un pas de temps, on a: \begin{description} ... ... @@ -64,10 +65,10 @@ $u^n_i = u^n + Δt \sum_{j=1}^i\left(a_{i,j}\,K_j+\hat a_{i+1,j}\,\hat K_j\righ NB: Lorsque$\hat b_{s+1} = 0$, l'évaluation de$\hat K_{s+1}$n'est pas nécessaire. \section{Application} \subsection{Application} Dans notre cas, vu que la linéarisation se fait autour de$v^n$et que$x$ne rentre pas dans le schéma RK, on a$\hat K_1 = 0$, ce qui réduit de 1 le nombre d'évaluation de la partie explicite. Le$u^{n+1}$désigne en fait$\vfree^{n+1}$et la seule résolution des contacts a lieu, à la fin de l'itération. \subsection{Forward-backward Euler (1,1,1)} \subsubsection{Forward-backward Euler (1,1,1)} $\text{implicite: }\begin{array}{c|cc} 0 & 0 & 0 \\ ... ... @@ -96,7 +97,7 @@ On résout le problème de Stokes en utilisant: = \frac{-β∇p-α(v^n-\vf)(v^n - \vf + Δt\,\aext)}{1 + Δt\,α(v^n-\vf)}$ Et on obtient la vitesse libre: $\vfree^{n+1} = v^{n} + Δt\, (\af + \aext)$ \subsection{Forward-backward Euler (1,2,1)} \subsubsection{Forward-backward Euler (1,2,1)} $\text{implicite: }\begin{array}{c|cc} 0 & 0 & 0 \\ ... ... @@ -130,7 +131,7 @@ où \haf{2} est obtenu en résolvant le problème de Stokes en utilisant v = \[\haf{2} = -β∇p-α(v_2-\vf)(v_2 - \vf)$ On trouve alors$\vfree^{n+1}$à partir du {$\haf{2}} explicite : $\vfree^{n+1} = v^n + Δt\,({\af}_2 + \aext + \haf{2} - {\af}_2) = v^n + Δt\,(\haf{2} + \aex)$ \subsection{Implicit-Explicit midpoint (1,2,2)} \subsubsection{Implicit-Explicit midpoint (1,2,2)} \text{implicite: }\begin{array}{c|cc} 0 & 0 & 0 \\ ... ... @@ -152,5 +153,5 @@ Même coût que le précédent mais second ordre de précision. \end{description} Les calculs sont identiques, il faut juste utiliser \frac{Δt}{2} au lieu de Δt pour le calcul de {\af}_2 et v_2. \subsection{plus haut ordre de précision} \subsubsection{plus haut ordre de précision} On va déjà essayer les précédents ...  ... ... @@ -56,10 +56,12 @@ \newpage \input{equations.tex} \input{particle-fluid-interaction.tex} \input{fe.tex} \input{stability.tex} \input{stability2.tex} \input{imex.tex} \input{scontact.tex} \input{validation.tex} \newpage \bibliography{zotero} \end{document}  \section{Particle-fluid interaction forces} Assume a porous medium formed of particles of average diameter d_p and a fluid characterized by its dynamic viscosity \mu [\mbox{Pa} \cdot \mbox{sec}]. Assume that the representative elementary volume (REV) is caracterized by a porosity \phi defined as \phi = {\mbox{Volume of fluid in the REV} \over \mbox{Volume of the REV}}. Let us call \mvu = (u_1,u_2,u_3) and p the average fluid velocity and the average fluid pressure over a REV. Assume the pressure gradient to be proportional both to the velocity and its Laplacian (Brinkman \cite{brinkman}, Forchheimer \cite{forchheimer}): \nabla p = -{\mu \over K} \mvu -{C_E \over {\sqrt{K}}} \|\mvu\| \mvu + \bar{\mu} \nabla^2 \mvu. where K [\mbox{m}^2] is the permeability of the REV, where C_E [\mbox{m}^{-1}] is the Ergun coefficient that accounts for inertial (kinetic) effects and where \bar{\mu} is the effective viscosity. The Ergun coefficient is strongly dependent on the flow regime. For slow flows, C_E is very small and the quadratic terme can be neglected. The effective viscosity can be computed using Einstein’s formula \cite{einstein} for the effective viscosity of a dilute suspension of rigid spherical particles in an ambient fluid: \bar{\mu} \approx \mu \left(1+{5\over 2} (1-\phi)\right). The permeability is interpreted in terms of spatial parameters \phi and d_p as K = {d^2_p \phi^3 \over 150 (1-\phi)^2}. Let us first neglect inertial effects (C_E = 0). Assume a domain of charachetistic length L \gg d_p. The following adimensional number (called the Fidel Castro Number \mbox{Fc}) accounts for the relative importance of the large scale viscous effects (Stokes term) and the small scale effects (Darcy term) \mbox{Fc} = {\mbox{Stokes} \over \mbox{Darcy}} = {K \over L^2} = {d^2_p \phi^3\over 150 L^2 (1-\phi)^2}. If a significative amount of particles are present in the REV (\phi is not very close to 1) , \mbox{Fc} \ll 1 and large scale effects viscous effects can be neglected. Yet, for \phi \simeq 1, stokes effects can become proeminant and this is clearly the case when no particles are present in the REV. More precisely, if (1-\phi) L\gg d_p, then large scale viscous effects are negligible. \subsection*{New} NB : all those papers deal with gaz-particle mixture in general and fluidized bed in particular. I'm not sure this can be applied to subsurface water. While other forces can be seen in the literature \citep{zhu_discrete_2007}, the principal forces present in the gas-particle interactions are the drag force G and the pressure gradient force F. \subsubsection*{Pressure gradient force} ... ...  \section*{TODO} \subsection*{For the Hydro-Quebec report} \subsection*{1 Hydro-Quebec report {\small (Hopefully before July 3)}} \begin{itemize} \item coupling with lmgc \item doc python \item build with python instead of (or in addition of*) cmake \item coupling with lmgc for hydroc-quebec 2d and 3d test cases \item zone of prescribed porosity \item doc python of hydroc-quebec 2d and 3d test cases \item build with python instead of (or in addition of) cmake \item doc scontactplot ? \end{itemize} \subsection*{For the model} \subsection*{2 Model} \begin{enumerate} \item imex : include d/dt in the fluid equations, increase the order, find why there is a delay when we increase dt \item friction in scontact \item free surface/ale \end{enumerate} \subsection*{For the paper} \subsection*{3 Paper} \begin{itemize} \item introduction \item notation consistency \item introduction + state of the art (partially done in my FNRS proposal) \item conclusion \item result section \item result section (which testcases ?) \item references (some can be found in my FNRS proposal) \item polish/structure \end{itemize}  \section{Validation} \begin{figure} \center \includegraphics[width=0.7\textwidth]{fig/darcy/darcy_aligned_mesh.png}\\ \includegraphics[width=0.7\textwidth]{fig/darcy/darcy_aligned_pressure.png}\\ \includegraphics[width=0.7\textwidth]{fig/darcy/darcy_aligned_velocity.png} \caption{aligned, c_β = 0.4, d=0.0128} \end{figure} \begin{figure} \center \includegraphics[width=0.7\textwidth]{fig/darcy/darcy_random_mesh.png}\\ \includegraphics[width=0.7\textwidth]{fig/darcy/darcy_random_pressure.png}\\ \includegraphics[width=0.7\textwidth]{fig/darcy/darcy_random_velocity.png} \caption{random, c_β = 0.4, d=0.0128} \end{figure} \begin{figure} \center \begin{tabular}{cc} \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_aligned_01.pdf}& \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_aligned_02.pdf}\\ \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_aligned_03.pdf}& \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_aligned_04.pdf}\\ \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_aligned_05.pdf}& \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_aligned_06.pdf}\\ \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_aligned_07.pdf}& \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_aligned_08.pdf} \end{tabular} \caption{Aligned particles} \end{figure} \begin{figure} \center \begin{tabular}{cc} \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_random_01.pdf}& \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_random_02.pdf}\\ \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_random_03.pdf}& \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_random_04.pdf}\\ \includegraphics[width=0.4\textwidth]{fig/darcy/pressure_random_045.pdf} \end{tabular} \caption{Random particles} \end{figure} \begin{figure} \center \includegraphics[width=0.7\textwidth]{fig/darcy/K.pdf}\\ \includegraphics[width=0.7\textwidth]{fig/darcy/coeff.pdf} \caption{Permeability} \end{figure} \section{Densité linéaire de particules} d' est une densité constante dans les coordonées (x', y'). d(x, y) = \left(\frac{∂x}{∂x'} \frac{∂y}{∂y'} - \frac{∂x}{∂y'}\frac{∂y}{∂x'}\right)^{-1} d' une densité fonction de x peut donc être obtenue en choisissant : \frac{∂x}{∂x'} = \frac{∂y}{∂y'} = \left(\frac {d'}{d(x)}\right)^{\frac 1 2} \hspace*{1cm} \frac{∂x}{∂y'} \frac{∂y}{∂x'} = 0\hspace{1cm} \mbox{avec }L' = \frac{αL}{1-α} une densité constante en y et linéaire (de 1 en 0 à α en L) en x, s'obtient donc en posant : \frac{∂x}{∂x'} = \frac{∂y}{∂y'} = \left(1 - \frac{x}{L'}\right)^{-\frac 1 2} \hspace*{1cm} \frac{∂x}{∂y'} \frac{∂y}{∂x'} = 0\hspace{1cm} \mbox{avec }L' = \frac{L}{1-α} d'où \begin{align*} x &= L - L\Big(1 ± \frac{3x'}{2L}\Big)^{\frac 2 3} %x &= \Big[\Big(1 - \frac{3x'}{2L'}\Big)^\frac 2 3 - 1\Big] L' & y &= \Big(1 - \frac{3x'}{2L'}\Big)^{-\frac 1 3}y' \end{align*} \subsection{solution théorique} Seulement Darcy : \begin{align*} ∇p &= -\frac{μ}{K(x)}u \dx & K = \frac{d²_pφ(x)³}{150(1-φ(x))²}\\ Δp &= ∫_0^L -\frac{150(1-φ(x))²}{d_p²φ(x)³} μu\dx & φ(x)= φ_0 (1 + \frac x L (α - 1)) \\ &= -\frac{150 μuL}{d^2_pφ_0(α-1)}\left[-\frac{1}{2φ²} + \frac{2}{φ} + \log{φ}\right]_{φ_0}^{αφ_0}\\ &= -\frac{150 μuL}{d^2_p}\left(\frac{α+1}{2α²φ_0³} - \frac{2}{αφ_0²} + \frac{\log(α)}{(α - 1)φ_0}\right) \end{align*} Complete (avec u = \text{cst}) : \begin{align*} ∇p &= \left(\frac{|∇φ|²}{φ³}-\frac{1}{K(x)}\right)μu\\ \end{align*} Le rapport entre les deux est: \[ \frac{1}{150}\left(\frac{|∇φ|d_p}{1-φ}\right)² Autrement dit, le terme supplémentaire est significatif quand|∇φ| ≥ \frac{10}{d_p(1-φ)}ce qui est impossiple. On a donc un terme supplémentaire : \begin{align*} ∫_0^L\frac{φ_0² (α-1)²}{L²φ³}\dx &=-\left[\frac{φ_0 (α-1)}{2Lφ^2}\right]_{φ_0}^{αφ_0}\\ &=-\frac{φ_0(α-1)(1-α^2)}{2Lα^2φ_0^2}\\ &=\frac{(1-α)^2(1+α)}{2Lα^2φ_0} \end{align*} %\subsection{flow around a porous cylinder} %\begin{figure} %\includegraphics[width=\textwidth]{fig/complete/porosity.png}\\ %\end{figure} %\begin{figure} %\includegraphics[width=\textwidth]{fig/complete/velocity.png}\\ %\includegraphics[width=\textwidth]{fig/complete/pressure.png} %\caption {complete term with-\tilde P^\alpha\nabla c_\alpha$} %\end{figure} %\begin{figure} %\includegraphics[width=\textwidth]{fig/incomplete/velocity.png}\\ %\includegraphics[width=\textwidth]{fig/incomplete/pressure.png} %\caption {incomplete term without$-\tilde P^\alpha\nabla c_\alpha\$} %\end{figure} %\begin{figure} %\includegraphics[width=\textwidth]{fig/orig/velocity.png}\\ %\includegraphics[width=\textwidth]{fig/orig/pressure.png} %\caption {previous formulation} %\end{figure} \newpage