Difference between revisions of "Team:Paris Saclay/Modeling"

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n(r) = -\frac{\alpha}{r} + \beta
 
n(r) = -\frac{\alpha}{r} + \beta
 
\end{equation}
 
\end{equation}
\alpha and \beta being constants to be determined.
+
$\alpha$ and $\beta$ being constants to be determined.
 
+
One of the boundary conditions is determined by the density of the external medium, when the distance from the center of the bead is far enough compared to the size of a bead.
 +
When $r\rightarrow + \infty \, n=n_\infty $.
  
  
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%%%%Start of LATEX
 
%%%%Start of LATEX
  
Diffusion de particules - Loi de Fick
 
 
 
 
Equation de continuité -> conservation nombre de particules
 
 
 
Eq. \ref{eq:fick} + Eq. \ref{eq:continuity_eq} : $\partial_t n = D \, \triangle n \, (+\sigma)$
 
 
Expression du laplacien en coordonnées sphériques
 
\begin{equation}
 
\triangle a = \frac{1}{r^2} \frac{\partial}{\partial r} \Big(r^2 \frac{\partial a}{\partial r} \Big) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \Big(\sin \theta \frac{\partial a}{\partial \theta} \Big) + \frac{1}{r^2 \sin \theta} \frac{\partial^2 \varphi}{\partial \varphi^2}
 
\end{equation}
 
 
 
 
Let's suppose that in the beads $n(r,\theta,\varphi,t) = n(r,t)$ $\rightarrow$ spherical symmetry, no dependence on angles
 
 
Steady diffusion
 
 
\begin{equation}
 
\frac{1}{r^2} \frac{\partial}{\partial r} \Big[ r^2 \frac{\partial n}{\partial r} \Big] =0 \rightarrow 2r\frac{\mathrm{d}n}{\mathrm{d}r} + r^2 \frac{\mathrm{d}^2n}{\mathrm{d}r^2} = 0
 
\end{equation}
 
Let's consider $m=\frac{\mathrm{d}n}{\mathrm{d}r}$ $\rightarrow  2rm + r^2 \frac{\mathrm{d}m}{\mathrm{d}t}=
 
2 r +r^2 \frac{\mathring{d}(\log m)}{\mathrm{d}r} =0$
 
  
$\rightarrow \frac{\mathrm{d}(\log m)}{\mathrm{d}r} = -\frac{2}{r} \rightarrow \log m = -2 \log r+ cte \rightarrow m\frac{alpha}{r^2}=\frac{\mathrm{d}n}{\mathrm{d}r}$
 
  
\begin{equation}
 
n(r) = -\frac{\alpha}{r} + \beta
 
\label{steady_eq}
 
\end{equation}
 
When $r\rightarrow + \infty \, n=n_\infty $ densité de particule loin du centre de la bille / densité du milieu ext.
 
  
 
\ref{steady_eq} $\rightarrow n(+\infty) = \beta = n_\infty \rightarrow n(r) = -\frac{\alpha}{r} + n_\infty$
 
\ref{steady_eq} $\rightarrow n(+\infty) = \beta = n_\infty \rightarrow n(r) = -\frac{\alpha}{r} + n_\infty$

Revision as of 03:08, 19 September 2015

Modeling

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The diffusion of particules is based on Fick's first law which is given by: \begin{equation} \textbf{j} = -D \, \bf{\nabla} n \label{eq:fick} \end{equation} In this equation, $\textbf{j}$ is the diffusion flux, $D$ the diffusion coefficient and $n$ the concentration of particles. This equation can be coupled with the continuity equation $\partial_t n = \mathbf{\nabla} \cdot \mathrm{j} \quad (+\sigma)$ expressing the conservation of the total number of diffusing particles. $\sigma$ is the net particle production rate. The beads being spherical, it is more interesting to work with spherical coordinates. The Laplace operator is then defined by : \begin{equation} \triangle a = \frac{1}{r^2} \frac{\partial}{\partial r} \Big(r^2 \frac{\partial a}{\partial r} \Big) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \Big(\sin \theta \frac{\partial a}{\partial \theta} \Big) + \frac{1}{r^2 \sin \theta} \frac{\partial^2 \varphi}{\partial \varphi^2} \end{equation} Let's suppose that there is no dependence on angles in the beads, i.e. there is a spherical symmetry. We can write $n(r,\theta,\varphi,t) = n(r,t)$.

Steady diffusion

In steady diffusion, the equation $(eq number)$ is simpler as there is no dependence in time. \begin{equation} \frac{1}{r^2} \frac{\partial}{\partial r} \Big[ r^2 \frac{\partial n}{\partial r} \Big] =0 \end{equation} The equation above leads to the following differential equation : \begin{equation} 2r\frac{\mathrm{d}n}{\mathrm{d}r} + r^2 \frac{\mathrm{d}^2n}{\mathrm{d}r^2} = 0 \end{equation} Let's consider a variable m defined by $m=\frac{\mathrm{d}n}{\mathrm{d}r}$. The equation $eq number$ can be rewritten $ 2rm + r^2 \frac{\mathrm{d}m}{\mathrm{d}t}= 2 r +r^2 \frac{\mathrm{d}(\log m)}{\mathrm{d}r} =0$. We can show that the differential equation is then $m\frac{\alpha}{r^2}=\frac{\mathrm{d}n}{\mathrm{d}r}$. The concentration of particles is then : \begin{equation} n(r) = -\frac{\alpha}{r} + \beta \end{equation} $\alpha$ and $\beta$ being constants to be determined. One of the boundary conditions is determined by the density of the external medium, when the distance from the center of the bead is far enough compared to the size of a bead. When $r\rightarrow + \infty \, n=n_\infty $.