Team:Paris Saclay/Modeling

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(Modeling)
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In equation after Fick's law above, we need to know the parameter $D$. We then searched in scientific literature and article '''[1]'''  describes a method to get it. Referring to this article,  the diffusion coefficient of oxygen in agarose is $ D = 0{,}256 \times 10^{-8} m^2 s^{-1} $.
In equation after Fick's law above, we need to know the parameter $D$. We then searched in scientific literature and article '''[1]'''  describes a method to get it. Referring to this article,  the diffusion coefficient of oxygen in agarose is $ D = 0{,}256 \times 10^{-8} m^2 s^{-1} $.
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The phenomenon we are facing is quite difficult to tackle if taken as a whole and We must emit a number of intuitively acceptable hypothesis.
   
   
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* To simplify the problem, we consider that the diffusion of oxygen particle occurs only in one direction. So $\overrightarrow{J}(M,t) = J(x,t) \overrightarrow{e}_x $.
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* To simplify the problem, we consider that the diffusion of oxygen particle occurs only in one direction. So $\overrightarrow{J}(M,t) = J(x,t) \overrightarrow{e}_x $. This hypothesis is credible because we seek the maximum penetrance of oxygen. This is why we consider that the diffusion occurs in the line of greatest slope.
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* Spatial variations in the density of particles are connected to spatial variations of the vector $\overrightarrow{J}(M,t)$ by '''the material's equation of conservation''' in presence of volume distribution of particle source $\sigma (x,t)$ (device which injects or subtracted particles to the system) :  
* Spatial variations in the density of particles are connected to spatial variations of the vector $\overrightarrow{J}(M,t)$ by '''the material's equation of conservation''' in presence of volume distribution of particle source $\sigma (x,t)$ (device which injects or subtracted particles to the system) :  
\[ \frac{\partial n}{\partial t} (x,t) = - \frac{\partial J}{\partial x} (x,t) + \sigma (x,t)  \qquad (2) \]
\[ \frac{\partial n}{\partial t} (x,t) = - \frac{\partial J}{\partial x} (x,t) + \sigma (x,t)  \qquad (2) \]
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By replacing $(2)$ in $(1)$, we obtain the following '''equation of diffusion'''
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By replacing $(2)$ in $(1)$, we obtain the following '''equation of diffusion''' :
\[ \forall t, \forall x,  \bigg(  \frac{\partial}{\partial t} - D \frac{\partial^2}{\partial x^2} \bigg) n(x,t) =  \sigma (x,t)  \qquad (3) .\]
\[ \forall t, \forall x,  \bigg(  \frac{\partial}{\partial t} - D \frac{\partial^2}{\partial x^2} \bigg) n(x,t) =  \sigma (x,t)  \qquad (3) .\]
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As our lemon is exposed to the ambient air, we stay in steady state where the source $ \sigma (x,t) $ is equal to $N_0$ the quantity of $O_2$  in the air.
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As our lemon is exposed to the ambient air, we stay in steady state where the source $ \sigma (x,t) $ is equal to $N_0$ the quantity of $O_2$  in the air.  
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To solve this equation, we use Fourier's analysis (+ d'explications)
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To solve this equation, we use Fourier's analysis and it is kwown (classical solution of the heat equation) that
\[ \forall x, \forall t>0, n(x,t) = \frac{N_0}{\sqrt{4 \pi D t}} exp \bigg(- \frac{x^2}{4 D t} \bigg)  + \int_{0}^{t} \underbrace{N_0 * exp \bigg(- \frac{|x|^2}{4 D \tau} \bigg)}_{= 0 \text{ by symmetry of the gaussian distribution }} \frac{d\tau}{\sqrt{4 \pi D \tau} }  \]
\[ \forall x, \forall t>0, n(x,t) = \frac{N_0}{\sqrt{4 \pi D t}} exp \bigg(- \frac{x^2}{4 D t} \bigg)  + \int_{0}^{t} \underbrace{N_0 * exp \bigg(- \frac{|x|^2}{4 D \tau} \bigg)}_{= 0 \text{ by symmetry of the gaussian distribution }} \frac{d\tau}{\sqrt{4 \pi D \tau} }  \]
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The average dispersion particle is given by the variance $\Delta x = \sqrt{2Dt}$. Using this formula, we deduct that oxygen will penetrate $3 \times 10^{-3} m$ in $1956.522 s = 32.6082 $ minutes.
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Whence, \[ \forall x, \forall t>0, n(x,t) = \frac{N_0}{\sqrt{4 \pi D t}} exp \bigg(- \frac{x^2}{4 D t} \bigg) /]
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 +
Thus we deduce easily that the average dispersion particle is given by the variance $\Delta x = \sqrt{2Dt}$. Using this formula, we deduct that, for example, oxygen will penetrate $3 \times 10^{-3} m$ in $1956.522 s = 32.6082 $ minutes.
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 +
The charts below give us a very clear idea of ​​penetrance evaluated over time.
=== Graphical visualization ===
=== Graphical visualization ===

Revision as of 17:44, 30 September 2014


Modeling

To realise our artwork, we use an agarose gel to obtain the shape of a lemon. To push the resemblance to the extreme, we wish to have a crust in the edge of the lemon when we seperate it. In fact, we build bacteria who produce yellow/green color in presence of oxygen. Thus we must evaluate the penetration of the oxygen in the gel !

Oxygen penetrates into the gel by diffusion phenomenon that we will study below. We first use the following phenomenological law suggest by Adolphe Fick in 1855:



In an homogeneous and isotropic environment, containing particles distributed inhomogeneously,appears spontaneously a volumetric flow density vector particle $\overrightarrow{J}(M,t) $. In any point $M$ in space, this vector is proportional to the gradient of the particle density $n(M,t)$. Mathematicaly, this relationship take the form: \[ \overrightarrow{J}(M,t) = - D \times \nabla n(M,t) \qquad (1) \] where $D$ is the diffusion coefficient and $\nabla$ is the gradient vector.


In equation after Fick's law above, we need to know the parameter $D$. We then searched in scientific literature and article [1] describes a method to get it. Referring to this article, the diffusion coefficient of oxygen in agarose is $ D = 0{,}256 \times 10^{-8} m^2 s^{-1} $.

The phenomenon we are facing is quite difficult to tackle if taken as a whole and We must emit a number of intuitively acceptable hypothesis.

  • To simplify the problem, we consider that the diffusion of oxygen particle occurs only in one direction. So $\overrightarrow{J}(M,t) = J(x,t) \overrightarrow{e}_x $. This hypothesis is credible because we seek the maximum penetrance of oxygen. This is why we consider that the diffusion occurs in the line of greatest slope.
  • Spatial variations in the density of particles are connected to spatial variations of the vector $\overrightarrow{J}(M,t)$ by the material's equation of conservation in presence of volume distribution of particle source $\sigma (x,t)$ (device which injects or subtracted particles to the system) :

\[ \frac{\partial n}{\partial t} (x,t) = - \frac{\partial J}{\partial x} (x,t) + \sigma (x,t) \qquad (2) \]

By replacing $(2)$ in $(1)$, we obtain the following equation of diffusion : \[ \forall t, \forall x, \bigg( \frac{\partial}{\partial t} - D \frac{\partial^2}{\partial x^2} \bigg) n(x,t) = \sigma (x,t) \qquad (3) .\]

As our lemon is exposed to the ambient air, we stay in steady state where the source $ \sigma (x,t) $ is equal to $N_0$ the quantity of $O_2$ in the air.

To solve this equation, we use Fourier's analysis and it is kwown (classical solution of the heat equation) that

\[ \forall x, \forall t>0, n(x,t) = \frac{N_0}{\sqrt{4 \pi D t}} exp \bigg(- \frac{x^2}{4 D t} \bigg) + \int_{0}^{t} \underbrace{N_0 * exp \bigg(- \frac{|x|^2}{4 D \tau} \bigg)}_{= 0 \text{ by symmetry of the gaussian distribution }} \frac{d\tau}{\sqrt{4 \pi D \tau} } \]

Whence, \[ \forall x, \forall t>0, n(x,t) = \frac{N_0}{\sqrt{4 \pi D t}} exp \bigg(- \frac{x^2}{4 D t} \bigg) /]

Thus we deduce easily that the average dispersion particle is given by the variance $\Delta x = \sqrt{2Dt}$. Using this formula, we deduct that, for example, oxygen will penetrate $3 \times 10^{-3} m$ in $1956.522 s = 32.6082 $ minutes.

The charts below give us a very clear idea of ​​penetrance evaluated over time.

Graphical visualization

Paris Saclay oxygenGraph.jpeg



References:

[1] A.C. Hulst, H.J.H. Hens, R.M. Buitelaar and J. Tramper, Determination of the effective diffusion coefficient of oxygen in gel materials in relation to gel concentration, Biotechnology Techniques Vol 3 No 3 199-204 (1989).

[2] Vincent Renvoizé, Physique PC-PC*, Cap Prepas, Pearson Education, 2010.