# Flash Vaptorization of Benzene-Toluene

A liquid mixture containing 50 mole % each of benzene and toluene at 40^{o}C is to be continuously flash vaporized to vaporize 60 mole % of the feed. The residual liquid product contains 35 mole % benzene. If the enthalpies per mole of feed, the liquid product and the vapor product are respectively 2,5 and 30 kJ/mole,

- calculate the heat added in kJ per mole of vapor product
- represent the process on a H-x-y diagram

Data:

*Basis*: 100 mole of feed

Feed composition, mole fraction of more volatile component (*z*_{F}) = 0.5

Feed (F) = 100 mole

Since 60% of the feed is vaporized,

Vapor (D) = 60 mole

Liquid (W) = 40 mole

Mole fraction of more volatile component in the liquid (*x*_{W}) = 0.35

Formulae:

F = D + W à 1

F*z*_{F} = D*x*_{D} + W*x*_{W }à 2

FH_{F} + Q = DH_{D} + WH_{W} à 3

Calculations:

From equations, 1 and 2

100 x 0.5 = 60*x*_{D} + 40 x 0.35

Therefore, *x*_{D} = 0.6

From equation 3,

100 x 2 + Q = 60 x 30 + 40 x 5

Therefore, Q = 1800 kJ

**Heat to be added per mole of vapor product =** Q/D = 1800/60 = **300 kJ**

The H-x-y diagram for the above process is given below: