$5\mathrm{.}3$ Chemical Reaction and Diffusion in Pore

The differential equation for diffusion and reaction in a pore is given by

$\frac{d^{2}C}{d{x}^2}-\frac{k}{D}C=0$

Let us define

$m=\sqrt[\phantom{2}]{\frac{k}{D}}$

Thus the differential equation can be written as

$\frac{d^{2}C}{d{x}^2}-m^{2}C=0$

The boundary conditions are

$atx=0$

$C=C_S$

and at $x=L$

$\frac{dC}{dx}=0$

where $C_S$ is the concentration at the surface of the pore.

EXAMPLE $5\mathrm{.}8$ Diffusion and reaction take place in a pore of $1mm$ in length. The rate constant of the reaction, $k={10}^{-3}{s}^{-1}$ and effective diffusivity of species, $D={10}^{-9}\frac{m^2}{s}.$ Make $100$ parts of the pore and determine the concentration at $x=0\mathrm{.}5mm.$ The concentration at the surface of the mouth of the pore is $1mo\frac{l}{m^3}.$

Solution We have

$m=\sqrt[\phantom{2}]{\frac{k}{D}}=\sqrt[\phantom{2}]{\frac{{10}^{-3}}{{10}^{-9}}}=1000m^{-1}$

The pore length is $1mm={10}^{-3}m$ and $100$ parts are made, thus $x=\frac{{10}^{-3}}{100}={10}^{-5}m.$ The schematic diagram of the pore is shown in Fig. $5\mathrm{.}10.$ The first node is labelled $0$ and therefore the last node is the ${100}^{th}$ node as $100$ parts are made.

$C=$

$-0$

The boundary conditions are

At $x=0$:$,C=1$

At $x={10}^{-3}m$:$,\frac{dC}{dx}=0$

Discretizing the differential equation at node $i,$ we get

$\frac{C_{1+1}+C_{1-1}-2C_1}{x^2}-{10}^6{C}_1=0$

At node $1$

$\frac{C_2+C_0-2C_1}{x^2}-{10}^6{C}_1=0$

It is given that $C_0=1,$ therefore

$\frac{C_2+1-2C_1}{x^2}-{10}^6{C}_1=0$

Substituting the value of $x={10}^{-5}m,$ we get

$C_2+1-2C_1-{10}^6{10}^{-10}{C}_1=0$

Simplifying, we get

$-2\mathrm{.}0001C_1+C_2=-1$

At node $2$

$\frac{C_3+C_1-2C_2}{x^2}-{10}^6{C}_2=0$

$C_3+C_1-2C_2-{10}^{-4}{C}_2=0$

$C_1-2\mathrm{.}0001C_2+C_3=0$

At node $99$

$\frac{C_{100}+C_{98}-2C_{99}}{x^2}-{10}^6{C}_{99}=0$

$C_{100}+C_{98}-2C_{99}-{10}^{-4}{C}_{99}=0$

$C_{98}-2\mathrm{.}0001C_{99}+C_{100}=0$

At node $100$

$\frac{C_{101}+C_{99}-2C_{100}}{x^2}-{10}^6{C}_{100}=0$

$C_{101}+C_{99}-2C_{100}-{10}^{-4}{C}_{100}=0$

But $\frac{dC}{dx}=0$; thus $\frac{C_{101}-C_{99}}{2x}=0$ and therefore $C_{101}=C_{99}.$

Substituting $C_{101}=C_{99}$ in the previous equation, we get

$2C_{99}-2\mathrm{.}0001C_{100}=0$

The $100$ equations can be written in tridiagonal form as

The solution can be obtained by modifying the parameters to $N$ to $N$ and to $N-1$ and to $N$ in Program $1\mathrm{.}1$ given in the Appendix. From the results, we get $C=0\mathrm{.}73$ at node

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