A liquid feed of $F=10$mole$\frac{s}{h}r$ enters a distiller with a microbubble sparge rate of

$B=60$mole$\frac{s}{h}r.$ The liquid product produced $L=0\mathrm{.}5$ moles $\frac{?}{hr}$ exits from the distiller.

An azeotropic mixture with mole fractions, $x_w^F=0\mathrm{.}087,x_{\text{t}\text{i}\text{l}\text{d}\text{e}(e)}^F={0\mathrm{.}913}_{{?}_{?}},$ makes up the

composition of the liquid feed $F.$ The outlet liquid stream composition is nearly all

water: $x_x^L=0\mathrm{.}999,x^L={0\mathrm{.}001}_{?}.$ It is assumed that the sparge stream B is only air, and

there is no air dissolved in either of the liquid streams. Assume the vapour $V$ is an

ideal gas, so that the moles in the vapour stream are proportional to the volumetric

flow rates of the individual species.

$($i$)$ Draw a process diagram. List all your assumptions necessary to write a

material balance.

$($ii$)$ List your unknowns.

$($iii$)$ Demonstrate a complete set of material balances in symbolic form.

$($iv$)$ Solve your system of equations and state the values of the unknown variables.

Is the exit stream more or less acidic than the feed stream?

$($v$)$ Calculate the ratio $\frac{x_e^V}{x_{\text{t}\text{i}\text{l}\text{d}\text{e}(w)}^V+x_e^V}$ tor the vapour stream coming trom the aistiller, which

would be the ethanol mole fraction from the condensate of a total condenser.

Would you expect this to be greater or less than the azeotropic value? Why?