Answer:
The answer is [tex][\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454[/tex] mol/hr
[tex][\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454[/tex]mol/hr
Explanation:
For flash distillation
F = V+L
[tex]\frac{V}{F} + \frac{L}{F} = 1[/tex]
[tex]\frac{F}{V} -\frac{L}{V} = 1[/tex]
Fz = Vy+Lx
Y = [tex]\frac{F}{V}\times Z - \frac{L}{V}\times X[/tex] let, [tex]\frac{V}{F} = F[/tex]
[tex]y = \frac{Z}{F} -[ \frac{1}{F} -1]\times X[/tex]
Highlighted reading
F = 299; [tex]\frac{V}{F}[/tex] = 0.85 ; z = 0.36
y = [tex]\frac{0.36}{0.85} - (-0.15)\times X[/tex]
= 0.423 + 0.15x ------------(i)
[tex]y^{*}[/tex] = -43.99713[tex]x^{6}[/tex] + 148.27274[tex]x^{5}[/tex] - 195.46[tex]x^{4}[/tex]+127.99[tex]x^{3}[/tex]-43.3[tex]x^{2}[/tex]+ 7.469[tex]x^{}[/tex]+ 0.02011
At equilibrium, [tex]y^{*}[/tex] = y
0.423+0.15[tex]x^{}[/tex] = [tex]y^{*}[/tex]
-43.99713[tex]x^{6}[/tex]+ 148.27274[tex]x^{5}[/tex] - 195.46[tex]x^{4}[/tex]+127.99[tex]x^{3}[/tex]-43.3[tex]x^{2}[/tex]+ 7.319[tex]x^{}[/tex]-0.403
F(x) for Newton's Law
Let [tex]x_{0} = 0[/tex]
[tex]x_{1}[/tex] = [tex]\frac{0-[{-0.403}]}{7.319}[/tex]
= 0.055
[tex]x_{2}[/tex] = [tex]\frac{{0.055}-{f(0.055)} {{{{{{{}}}}}}}}{f^{'} (0.055)}[/tex]
= [tex]\frac{{(0.055)}-(-0.11)}{3.59}[/tex]
= 0.085
[tex]x^{3}[/tex] = [tex]\frac{{0.085}-(0.024)}{2.289}[/tex]
= 0.095
[tex]x^{4}[/tex] = [tex]\frac{{0.095}-(-0.0353)}{-1.410}[/tex]
= 0.07
From This x and y are found from equation (i) and L and V are obtained from [tex]\frac{V}{F}[/tex] and F values
[tex][\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454[/tex] mol/hr
[tex][\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454[/tex]mol/hr
