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:

JAC DACyAC JAL DALyAL

where yi is the vapor concentration of each component at the surface of the liquid. The amount that evaporates is also proportional to the flux, hence:

nACevap D ACyAC nALevap D ALyAL

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

nACevap nALevap = DACyAC DALyAL

or

n0,AC nAC n0,AL nAL = DACyAC DALyAL

When,

nAL = nALevap = n0,AL 2

then

n0,AC nAC = n0,AL 2 DACyAC DALyAL

nAC = n0,AC n0,AL 2 DACyAC DALyAL

But also:

n0,AC = n0,AL

thus

nAC = n0,AL n0,AL 2 DACyAC DALyAL

and with:

nAL = n0,AL 2

Taking the ratio:

xAC xAL = 2 DACyAC DALyAL

Using Raoul’s law:

yAC yAL = PACxAC PALxAL

Thus:

xAC xAL 1 + DACPAC DALPAL = 2

Using the values provided:

DACPAC DALPAL = 1.3843

Thus:

xAL + xAC = 1 xAC xAL = 0.83882

and

xAC = 0.456xAL = 0.544