Question

Liquid enters a valve at Tin and exits at Tout, Pout = 1 atm. Determine Tin and Tout to vaporize 20% of the feed if the liquid is 1/3 benzene and 2/3 toluene.

Answer 1

Given the mole fraction of benzene in liquid phase, X_b = 0.333

Mole fraction of toluene in liquid phase, X_t = 0.667

The out side pressure Pt = 1atm = 760 mm Hg

Here let's consider the out side temperature be normal temperature i.e 25 DegC

Suppose for the inside temperature T, 20 % of the liquid is vaporized

Vapor pressure of benzene and toluene at temperature T can be calculated from Antoine equation which is

logP⁰ = A - B / (C+T)

For benzene, logP⁰_b = 6.90565 - 1211.033 / (220.790 + T)

=> P⁰_b = 10^[6.90565 - 1211.033 / (220.790 + T)]

For toluene, logP⁰_t = 6.95464 - 1344.800 / (219.482+T)

=> P⁰_t = 10^[6.95464 - 1344.800 / (219.482+T)]

Now the vapor phase mole fractions can be calculated from the following formulae.

For benzene the mole fraction in vapor phase, Y_b = X_bxP⁰_b / Pt

For toluene the mole fraction in vapor phase, Y_t = X_txP⁰_t / Pt

Given Y_b + Y_t = 0.2

=>  X_bxP⁰_b / Pt +  X_txP⁰_t / Pt = 0.2

=> X_bxP⁰_b +  X_txP⁰_t = 0.2xPt = 0.2 x 760 mmHg = 152 mm Hg

=> 0.333x10^[6.90565 - 1211.033 / (220.790 + T)] + 0.667x10^[6.95464 - 1344.800 / (219.482+T)] = 152

=> 10^[6.90565 - 1211.033 / (220.790 + T)] + 2x10^[6.95464 - 1344.800 / (219.482+T)] = 152x3 = 456

Now we can find the value of T by solving the above equation