An ideal equimolar mixture of hexane and decane is sent to a flash drum, which

operates at equilibrium at $760mmHg(1$atm$)$ and $120C.$ The saturation pressure $($in $mm$

$Hg)$ of each component can be calculated using the following Antoine equations:

$lo{g}_{10}(p_{\text{h}\text{e}\text{x}\text{a}\text{n}\text{e}})=6\mathrm{.}91058-\frac{1189\mathrm{.}64}{T(C)+226\mathrm{.}28}$

$lo{g}_{10}(p_{\text{d}\text{e}\text{c}\text{a}\text{n}\text{e}})=7\mathrm{.}44-\frac{1843\mathrm{.}12}{T(C)+230\mathrm{.}22}$

solved: x$1=0\mathrm{.}215$ x$2=0/785$ y$1=0\mathrm{.}845$ y$2=0\mathrm{.}155$

b$.$What is the fraction of the outlet $($effluent$)$ that is vapor $($i$.$e$.$ the vapor outlet

flow rate over the inlet flow rate$).$

$($Hint: Assuming that the inlet flow rate is equal to $1mo\frac{l}{h}$ will make your calculations easier.$)$ Step $4$:

$($b$)$ The boiling of both the component is given as$,$

$T_{b{p}_1}=69C$

$T_{b{p}_2}=174C$

But at $120C,$ there is only chance of hexane has to be vaporized.

On the basis of feed is $1mo\frac{l}{h}r.$

Someones answerr

but where did we get xf from?? and then later after calculating V and doiing the fraction of v$/$f they use as V value$/0\mathrm{.}5$ but didnt we just establish F as $1$ mol$/$hr

$F=1molh{r}^{-1}$

$x_F=0\mathrm{.}5$

$x_1=0\mathrm{.}215$

$y_1=0\mathrm{.}845$

Overall Balance:

$F=V+L$

$1=V+L$

Component Balance: $F{x}_F=V_y+L{x}_1$