# Separation Process Principles

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Exercise. This is the solution to exercise 3.1 in the book.

Solution. Assume that the concentrations of AC and AL in air are zero. Since the components evaporate independently then the molar fluxes according to Fick’s law are:

$\begin{array}{rcll}{J}_{AC}& \propto & {D}_{AC}{y}_{AC}& \text{}\\ {J}_{AL}& \propto & {D}_{AL}{y}_{AL}& \text{}\end{array}$

where ${y}_{i}$ is the vapor concentration of each component at the surface of the liquid. The amount that evaporates is also proportional to the flux, hence:

$\begin{array}{rcll}{n}_{AC}^{evap}& \propto & {D}_{AC}{y}_{AC}& \text{}\\ {n}_{AL}^{evap}& \propto & {D}_{AL}{y}_{AL}& \text{}\end{array}$

The proportionality constant is the same and depdends on geometry and elapsed time. Thus:

$\frac{{n}_{AC}^{evap}}{{n}_{AL}^{evap}}=\frac{{D}_{AC}{y}_{AC}}{{D}_{AL}{y}_{AL}}$

or

$\frac{{n}_{0,AC}-{n}_{AC}}{{n}_{0,AL}-{n}_{AL}}=\frac{{D}_{AC}{y}_{AC}}{{D}_{AL}{y}_{AL}}$

When,

${n}_{AL}={n}_{AL}^{evap}=\frac{{n}_{0,AL}}{2}$

then

${n}_{0,AC}-{n}_{AC}=\frac{{n}_{0,AL}}{2}\frac{{D}_{AC}{y}_{AC}}{{D}_{AL}{y}_{AL}}$

${n}_{AC}={n}_{0,AC}-\frac{{n}_{0,AL}}{2}\frac{{D}_{AC}{y}_{AC}}{{D}_{AL}{y}_{AL}}$

But also:

${n}_{0,AC}={n}_{0,AL}$

thus

${n}_{AC}={n}_{0,AL}-\frac{{n}_{0,AL}}{2}\frac{{D}_{AC}{y}_{AC}}{{D}_{AL}{y}_{AL}}$

and with:

${n}_{AL}=\frac{{n}_{0,AL}}{2}$

Taking the ratio:

$\frac{{x}_{AC}}{{x}_{AL}}=2-\frac{{D}_{AC}{y}_{AC}}{{D}_{AL}{y}_{AL}}$

Using Raoul’s law:

$\frac{{y}_{AC}}{{y}_{AL}}=\frac{{P}_{AC}{x}_{AC}}{{P}_{AL}{x}_{AL}}$

Thus:

$\frac{{x}_{AC}}{{x}_{AL}}\left(1+\frac{{D}_{AC}{P}_{AC}}{{D}_{AL}{P}_{AL}}\right)=2$

Using the values provided:

$\frac{{D}_{AC}{P}_{AC}}{{D}_{AL}{P}_{AL}}=1.3843$

Thus:

$\begin{array}{rcll}{x}_{AL}+{x}_{AC}& =& 1& \text{}\\ \frac{{x}_{AC}}{{x}_{AL}}& =& 0.83882& \text{}\end{array}$

and

${x}_{AC}=0.456\phantom{\rule{1em}{0ex}}{x}_{AL}=0.544$