Consider unsteady diffusion across a membrane

of thickness L$.$ Initially, the concentration of solute in the

membrane is zero. At time zero, the concentration at

$x=0$ is raised to $C_0$ and the concentration at $x=L$ is

maintained at zero.

$($a$)$ Use the following dimensionless variables:

$=\frac{x}{L},=\frac{C_m}{C_0},=\frac{t{D}_m}{L^2}.$

Show that the solution for $$ can be rewritten as

$(,)=()+(,)$

where $()$ is the steady$-$state solution. State

boundary conditions for the problem. Show that

$()=1-.$

$($b$)$ Show that $(,)$ can be solved by separation of

variables, and obtain the following solution for

$(,)$:

$=1--\frac{2}{}{\displaystyle \underset{n=1}{\overset{}{}}}\frac{sin(n)}{n}$exp$(-n^2{}^2).$

$($c$)$ Use Equation $(6\mathrm{.}12\mathrm{.}10)$ to assess the time required

for the membrane to reach steady state. Compare

this result with the result obtained from the quasi$-$

steady$-$state analysis.

$($d$)$ The following relation is derived from a mass balance:

$V\frac{d{C}_1}{dt}=A{D}_m\frac{del{C}_m}{\text{d}\text{e}\text{l}\text{x}}{|}_x=0,$

where $C_1$ is the concentration on side $1$ of the

membrane, as shown in Figure $6\mathrm{.}24.$ At $t=0,$

$C_1=C_0.$ Show that the solution for $C_1$ is

The problem pertains to one side of

the membrane. You may assume that

volume to be $V.$

please write clearly and label all parts of the solution