Question

Question 3 of question on vapour pressure

(c) If the solution is fed to a flash tank operating at 220°F, what percentage (by moles) of the mixture is vaporised? What will be the liquid and vapour compositions?

Vapour pressure data for n-hexane and n-octane are given in the following table. Vapour Pressure (mmHg) n-Hexane n-Octane Temperature (°F) 155.7 760 121 175 1025 173 200 1480 278 225 2130 434 250 3000 654 258.2 3420 760 Fig. Q1 (a) Construct a T-xy plot for the system at 1atm (760 mmHg) pressure. Also, plot y versus x. Base the plots on the hexane. State any assumptions clearly. Graph paper is attached, but you can also use any relevant software package to illustrate the graph.

Answer 1

(c)

We are given a flash tank that operates at
220°F
. We can see the graph of the process as,

270 250 Tvs y 230 L M N T=220 F 210 190 TVS X 170 150 O 0.1 0.3 0.4 0.5 X,Y 0.6 0.7 0.8 0.9 1 0.2 x=0.225 X=0.35 y=0.595

The solution will reach a point M in the flash tank and will separate into liquid represented by L and a vapour represented by N.

Using Inverse lever arm rule we can write,

(moles of vapour) L(moles of liquid) line LM line NM

Similarily we can write,

fraction vapourised N L+N line LM line NM +LM line LM line LN

From the given graph we can calculate this as,

fraction vapourised = 0.35 – 0.225 0.595 – 0.225 0.3378

So the percentage of mixture vapourised at
220°F
is 33.78%

The liquid contains 22.5% of Hexane

The vapour contains 59.5% of Hexane