Question

A small still is separating propane and butane at 135 °C, and initially contains 10 kg moles of a mixture whose composition is x = 0.3 (x = mole fraction butane). Additional mixture (x = 0.3) is fed at the rate of 5 kg mole/hr. The total volume of the liquid in the still is constant, and the concentration of the vapor from the still (xp) is related to x, as follows: Xp = How long will it take for X, to change from 0.3 to 0.35.​

Answer 1

Answer:

Hello the needed relation is missing below is the required relation

$X_{p} = \frac{x_{s} }{1+x_{s} }$   composition : propane = 0.70, butane = 0.3

Answer : ≈ 5.75 hrs

Explanation:

Applying the data given in regards to the material balance

Butane balance input into the still = 5 mole feed/hr | 0.30 mol butane/molfeed

since the total volume of the liquid in the still is constant

The output from the still is = 5mol condensed/hr | x$_{p}$ mol butane/mol condensed

unsteady state equation = $\frac{dx_{s} }{dt}$ = 0.15 - $0.5X_{p}$

note : to reduce the equation a single dependent variable we have to substitute for $x_{p}$

$\frac{dx_{s} }{dt}$ $= 0.15 + x_{s} / 1 + (0.5)x_{s}$

In order to find the time it will take for X to change from 0.3 to 0.35

integrate the above equation using the limits : t = 0, x$_{s}$ = 0.3 and t = Ф,

x$_{s} = 0.35$

= $[ - (x_{s} /0.35 - (1/(0.35)^2)* In(0.15 - 0.35x_{s} ) ]_{0.3} ^{0.35}$

hence t = Ф ≈ 5.75 hrs