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

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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}=
 
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$.
 
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.
  
  

Revision as of 03:03, 19 September 2015

Modeling

Note

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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.