Partial Equilibrium Vaporization Calculation and its Relation to Separation Processes

A liquid mixture of 50 mol % n-pentane and 50 mol % n-heptane, initially at a low temperature, is partially vaporized by heating at a constant pressure of 1.013 bar (1 atm). Find the equilibrium vapor and liquid compositions and the equilibrium temperature as a function of the fraction that is vaporized.

Question

Partial Equilibrium Vaporization Calculation and its Relation to Separation Processes

A liquid mixture of 50 mol % n-pentane and 50 mol % n-heptane, initially at a low temperature, is partially vaporized by heating at a constant pressure of 1.013 bar (1 atm). Find the equilibrium vapor and liquid compositions and the equilibrium temperature as a function of the fraction that is vaporized.

Accepted Answer

Here we have for the equilibrium conditions

[latex]y_{ i }=x_{ i } \frac{P_{ i }^{ vap }(T)}{P\left(x_{5}, x_{7}, T ight)}[/latex]

where [latex]P_{ i }^{ vap }(T)[/latex] is obtained from the Antoine equation data in Illustration 10.1-1. From Raoult’s law, the total pressure is

[latex]P\left(x_{5}, x_{7}, T ight)=x_{5} P_{5}^{ ext {vap }}(T)+x_{7} P_{7}^{ ext {vap }}(T)[/latex]

We also have, from Eqs. 10.1-5–10.1-8,

[latex]\sum_{i} x_{i}=1[/latex] (10.1-5)

[latex]\sum_{i} y_{i}=1[/latex] (10.1-6)

[latex]x_{ i } L+y_{ i } V=z_{ i , F } \quad i =1,2, \ldots, C[/latex] (10.1-7)

[latex]L+V=1[/latex] (10.1-8)

[latex]\begin{array}{l}x_{5}+x_{7}=1 \y_{5}+y_{7}=1\end{array}[/latex]

and

[latex]\begin{aligned}x_{5} L+y_{5} V &=0.50 \x_{7} L+y_{7} V &=0.50 \L+V &=1.0\end{aligned}[/latex]

With pressure fixed, for each value of the fraction of liquid L, the equilibrium temperature and the compositions in vapor and liquid phases can be computed by iteration. The results are given below and shown in Fig. 10.1-7. In this diagram each of the horizontal tie lines shown connecting the vapor and liquid compositions is labeled with its equilibrium temperature. Note that the first bubble of vapor occurs at 327.8 K and has an n-pentane mole fraction of 0.888. As the temperature increases, more of the liquid phase evaporates, and each of the phases becomes increasing more concentrated in n-heptane and less concentrated in n-pentane. Of course, when all the liquid has evaporated (L = 0), the vapor will be of the same composition as the initial liquid. Also, the end points of this figure at L = 0 and L = 1 in fact can be computed somewhat more easily from bubble point and dew point calculations, respectively.

[Using Aspen Plus[latex]^R[/latex] and the folder Aspen Illustrations>Chapter 10.1>10.1-5 on the Wiley website for this book the calculation is done using the two-phase flash block and varying the fraction that is liquid from 1 (the bubble point of the feed) to 0 (the dew point of the feed). The results are given in the file Illustration 10.1-5.xlsx in that folder, which are agreement with the results above.]

Question 10.1.5: Partial Equilibrium Vaporization Calculation and its Relatio...

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