Answer:
Explanation:
Given that:
The distillation is carried out at a pressure of 1 atm
The feed harbors 25% mole of methanol (z)
The total moles of feed is usually 100 moles
In the system, we have both methanol and water
Using the total mole balance for the distillation column.
Fz = Lx + Vy
where;
F = amount of feed
z = mole fraction of ethanol (in feed)
L = amount of liquid product out of the column
V = amount of vapor product out of the column
x = mole fraction of methanol out of the liquid
y = mole fraction of methanol out of the vapor.
SO;
(a)
If all the feed is vaporized, then the vapor will likely have the same composition as the feed.
(b)
If no vaporization of the feed takes place, then the bottoms moving out of the column contains the same composition as the feed.
(c)
If 1/3 of the feed is vaporized; then 2/3 of the feed is at the bottom.
The balance equation would be:
[tex]Fz = (\dfrac{2}{3}F) x + (\dfrac{1}{3}F)y \\ \\ z = \dfrac{2}{3}x+\dfrac{1}{3}y[/tex]
Replacing z = 0.25; we have:
[tex]0.25 = \dfrac{2}{3}x+\dfrac{1}{3}y[/tex]
0.75 = 2x + y
(d)
If 2/3 of the feed is vaporized;
Then:
[tex]Fz = (\dfrac{1}{3}F) x + (\dfrac{2}{3}F)y \\ \\ z = \dfrac{1}{3}x+\dfrac{2}{3}y[/tex]
replacing z = 0.25
[tex]0.25 = \dfrac{1}{3}x+\dfrac{2}{3}y[/tex]
0.75 = x + 2y
