Answer:
Explanation:
From the given information:
The percentage wieght of mixture [tex]x_w[/tex] = 0.4
The distillate = 70
From equilibrium data in Exercise 7.33
The Feed W = 100
The Liquid residue F = Feed(W) - Distillate (D)
= 100 - 70 = 30
By applying rayleigh equation;
[tex]In \bigg (\dfrac{F}{W}\bigg)= \int^{xF}_{x_w}\dfrac{1}{y-x} \ dx[/tex]
From the plot of the graph of [tex]\dfrac{1}{y-x} \ vs \ x[/tex]; the area under the curve is being calculated between the point {[tex]x_1 = 0.4 \ and \ x_2[/tex] }.
Such that; the area [tex]= In \bigg( \dfrac{100}{30}\bigg)[/tex] = 1.209
Similarly, the value of xF = 0.067
∴
[tex]y_D = \dfrac{F_{xF} - W_{xW}}{D}[/tex]
[tex]y_D = \dfrac{30(0.067)-100(0.4)}{70}[/tex]
[tex]y_D = 0.543[/tex]
