Example $8\mathrm{.}18$ : A feed containing $60$ mole $\%$ hexane and $40$ mole $\%$ octane is fed to a pipe still through a pressure reducing valve into a flash separator. The vapour and liquid leaving the separator are assumed to be in equilibrium. If $50$ mole $\%$ of the feed is vaporised, find the composition of the top and bottom products. The equilibrium data is given below.

$\backslash$table$[[x,$ mole fraction of hexane in liquid,$1\mathrm{.}0,0\mathrm{.}69,0\mathrm{.}40,0\mathrm{.}192,0\mathrm{.}045,0],[y,$ mole fraction of hexane in vapour,$1\mathrm{.}0,0\mathrm{.}932,0\mathrm{.}78,0\mathrm{.}538,0\mathrm{.}1775,0]]$

Solution : Basis : $1$kmol of feed.

Operating line for flash distillation is

$y=\frac{-(1-f)}{f}x+\frac{x_F}{f}$

slope $=-\frac{1-f}{f}$

Draw the equilibrium curve and draw the operating line with slope equal to $-1$ passing through $(0\mathrm{.}6,0\mathrm{.}6)$ on the diagonal. It intersects the equilibrium curve at $P$ which gives us the equilibrium liquid and vapour compositions as $0\mathrm{.}41$ and $0\mathrm{.}79$ molefraction hexane respectively.