= Distillation material balance
{title2=$F=D+B,\quad Fz=Dx_D+Bx_B$}
Consider a nonreacting two-component <mixture> processed into just two products, with no losses or unaccounted inventory. Let $F,D,B$ be feed, distillate, and bottoms <amount of substance> values in <moles>, and let $z,x_D,x_B$ be their component-A <mole fractions>. These are whole-stream compositions; $x_D$ is not automatically the instantaneous equilibrium <vapor> composition $y$ above the pot. Total and component balances give
$$
F=D+B,\qquad Fz=Dx_D+Bx_B.
$$
The component-B balance follows by subtracting the second equation from the first. Eliminating $B$ gives, for $x_D\ne x_B$,
$$
\frac DF=\frac{z-x_B}{x_D-x_B},\qquad
\frac BF=\frac{x_D-z}{x_D-x_B}.
$$
Nonnegative product amounts require $z$ between $x_B$ and $x_D$. For an original stipulated feed $F=1\,\mathrm{mol}$, $z=0.35$, $x_D=0.80$, and $x_B=0.10$,
$$
D=\frac5{14}\,\mathrm{mol},\qquad B=\frac9{14}\,\mathrm{mol}.
$$
Check the component inventories:
$$
Dx_D=\frac27\,\mathrm{mol},\qquad Bx_B=\frac9{140}\,\mathrm{mol},\qquad
\frac27+\frac9{140}=\frac7{20}=0.35.
$$
The distillate contains a fraction $(Dx_D)/(Fz)=40/49\approx0.816327$ of the original component A, despite its higher <mole fraction> $x_D=0.80$. This distinction between composition and recovery is useful when deciding how much enriched product to collect. These balances determine neither <temperatures>, <heat> input, nor the number of <equilibrium stages>; they do not prove that the specified products are attainable in a particular <fractionating column>.
Back to article page
