Question

A flash distillation chamber operating at 101.3 kpa is separating an ethanol water mixture the feed mixture contains z weight ethanol and a molar flow rate of a feed is

Answer 1

Answer:

The answer is $[\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454$ mol/hr

$[\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454$mol/hr

Explanation:

For flash distillation

F = V+L

$\frac{V}{F} + \frac{L}{F} = 1$

$\frac{F}{V} -\frac{L}{V} = 1$

Fz = Vy+Lx

Y = $\frac{F}{V}\times Z - \frac{L}{V}\times X$                  let, $\frac{V}{F} = F$

$y = \frac{Z}{F} -[ \frac{1}{F} -1]\times X$

Highlighted reading

F = 299;  $\frac{V}{F}$ = 0.85 ; z = 0.36

y = $\frac{0.36}{0.85} - (-0.15)\times X$

 = 0.423 + 0.15x ------------(i)

$y^{*}$ = -43.99713$x^{6}$ + 148.27274$x^{5}$ - 195.46$x^{4}$+127.99$x^{3}$-43.3$x^{2}$+ 7.469$x^{}$+ 0.02011

At equilibrium, $y^{*}$ = y

0.423+0.15$x^{}$ = $y^{*}$

-43.99713$x^{6}$+ 148.27274$x^{5}$ - 195.46$x^{4}$+127.99$x^{3}$-43.3$x^{2}$+ 7.319$x^{}$-0.403

F(x) for Newton's Law

Let $x_{0} = 0$

     $x_{1}$     = $\frac{0-[{-0.403}]}{7.319}$

             = 0.055

     $x_{2}$      = $\frac{{0.055}-{f(0.055)} {{{{{{{}}}}}}}}{f^{'} (0.055)}$

             = $\frac{{(0.055)}-(-0.11)}{3.59}$

             = 0.085

    $x^{3}$    = $\frac{{0.085}-(0.024)}{2.289}$

           = 0.095

   $x^{4}$     = $\frac{{0.095}-(-0.0353)}{-1.410}$

            = 0.07

From This x and y are found from equation (i) and L and V are obtained from $\frac{V}{F}$  and F values

$[\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454$ mol/hr

$[\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454$mol/hr