Partial Vaporization Calculation

A liquid mixture of 25 mol % n-pentane, 45 mol % n-hexane, and 30 mol % n-heptane, initially at 69°C and a high pressure, is partially vaporized by isothermally lowering the pressure to 1.013 bar (1 atm). Find the relative amounts of vapor and liquid in equilibrium and their compositions.

Question

Partial Vaporization Calculation

A liquid mixture of 25 mol % n-pentane, 45 mol % n-hexane, and 30 mol % n-heptane, initially at 69°C and a high pressure, is partially vaporized by isothermally lowering the pressure to 1.013 bar (1 atm). Find the relative amounts of vapor and liquid in equilibrium and their compositions.

Accepted Answer

From the Antoine equation data in Illustration 10.1-2, we have

[latex]P_{5}^{ ext {vap }}=2.721 ext { bar } \quad P_{6}^{ ext {vap }}=1.024 ext { bar } \quad ext { and } \quad P_{7}^{ ext {vap }}=0.389 bar[/latex]

Also, using the simplifications for this system introduced in the previous illustrations, the equilibrium relation [latex]\bar{f}_{ i }^{ L }=\bar{f}_{ i }^{ V }[/latex] reduces to [latex]x_{ i } P_{ i }^{ vap }=y_{ i } P[/latex], or

[latex]\frac{y_{ i }}{x_{ i }}=\frac{P_{ i }^{ vap }}{P}[/latex]

For convenience, we will use the K factor, defined by the relation [latex]y_{ i }=K_{ i } x_{ i }[/latex], in the calculations; for the ideal mixtures considered here [latex]K_{ i }=P_{ i }^{ vap }(T) / P[/latex]. Thus we obtain the following three equations:

[latex]\begin{array}{l}y_{5}=x_{5} K_{5} & ext { where } & K_{5}=2.7406 \quad (1)\y_{6}=x_{6} K_{6} & &K_{6}=1.0109\quad (2)\y_{7}=x_{7} K_{7} && K_{7}=0.3844\quad(3)\end{array}[/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}{r}x_{5}+x_{6}+x_{7}=1\quad (4) \y_{5}+y_{6}+y_{7}=1\quad(5)\end{array}[/latex]

and

[latex]\begin{array}{r}x_{5} L+y_{5} V=0.25\quad (6)\x_{6} L+y_{6} V=0.45\quad (7)\x_{7} L+y_{7} V=0.30\quad (8)\L+V=1.0\quad (9)\end{array}[/latex]

Thus we have eight independent equations for eight unknowns [latex]\left(x_{5}, x_{6}, x_{7}, y_{5}, y_{6}, y_{7}, L, ight. ext { and },V)[/latex], and any numerical procedure for solving algebraic equations may be used to solve this set of equations. One method is to use Eqs. 1, 2, and 3 to eliminate the vapor-phase mole fractions and the overall mass balance, Eq. 9, to eliminate the amount of vapor. In this way the eight algebraic equations are reduced to five linear algebraic equations:

[latex]\begin{aligned}x_{5}+x_{6}+x_{7} &=1\quad (4) \K_{5} x_{5}+K_{6} x_{6}+K_{7} x_{7} &=1\quad(5') \x_{5}\left[L+K_{5}(1-L) ight]=x_{5}\left[L\left(1-K_{5} ight)+K_{5} ight] &=0.25\quad(6') \x_{6}\left[L\left(1-K_{6} ight)+K_{6} ight] &=0.45\quad (7')\x_{7}\left[L\left(1-K_{7} ight)+K_{7} ight] &=0.30\quad(8')\end{aligned}[/latex]

These equations are most easily solved by trial and error. In particular, a value of L is guessed and then used in Eqs. 6'–8' to compute [latex]x_{5}, x_{6}, ext { and } x_{7}[/latex]. These trial values of the liquid-phase mole fractions are then tested in Eqs. 4 and 5'. If those equations are satisfied, the guessed value of L and the computed values of the [latex]x_{i}[/latex]'s are correct, and the vapor-phase mole fractions can be computed from Eqs. 1–3. If Eqs. 4 and 5' are not satisfied, a new guess for L is made, and the procedure repeated. Alternatively, one can use an equation-solving computer program such as MATHCAD, MATLAB or Aspen Plus[latex]^R[/latex] and the file in the folder Aspen Illustrations>Chapter 10.1>10.1-4 on the Wiley website for this book.

The solution is

[latex]\begin{array}{rlrl}L & =0.564 & V & =0.436 \x_{5} & =0.142 & y_{5} & =0.390 \x_{6} & =0.448 & y_{6} & =0.453 \x_{7} & =0.410 & y_{7} & =0.158\end{array}[/latex]

[Using Aspen Plus[latex]^R[/latex] and the folder Aspen Illustrations>Chapter 10.1>10.1-4 on the Wiley website for this book the following results are obtained

[latex]\begin{array}{rlrl}L & =0.569 & V & =0.431 \x_{5} & =0.144 & y_{5} & =0.391 \x_{6} & =0.448 & y_{6} & =0.452 \x_{7} & =0.408 & y_{7} & =0.157\end{array}[/latex]

The slight differences between those obtained using Aspen Plus[latex]^R[/latex] and previous is a result of the different vapor pressures in the Aspen databank.]

Comments

1. The K-factor formulation introduced in this calculation is frequently useful in solving vapor-liquid equilibrium problems. The procedure is easily generalized to nonideal liquid and vapor phases as follows:

[latex]\frac{y_{ i }}{x_{ i }} \equiv K_{ i }=\frac{\gamma_{ i } P_{ i }^{ vap }}{\bar{\phi}_{ i } P}[/latex]

In this case [latex]K_{ i }[/latex] is a nonlinear function of the liquid-phase mole fraction through the activity coefficient [latex]\gamma_{ i }[/latex], and also a function of the vapor-phase composition through the fugacity coefficient [latex]\bar{\phi}_{ i }[/latex]. This makes solving the equations much more difficult.

2. It was not necessary to assume ideal solution behavior to solve this problem. One could, for example, assume that the solution is regular, in which case [latex]\gamma_{i}\left( ext { and } K_{i} ight)[/latex] would be a nonlinear function of the mole fractions. The calculation of the vapor- and liquid-phase mole fractions is then more complicated than was the case here; however, the basis of the calculation, the equality of the fugacity of each species in both phases, remains unchanged.

3. In some cases it may not be possible to find an acceptable solution to the algebraic equations. Here by acceptable we mean a solution such that each mole fraction, L, and V are each greater than 0 and less than 1. This difficulty occurs when the vapor and liquid phases cannot coexist in the equilibrium state at the specified conditions. For example, if the flash vaporization temperature were sufficiently high or the total pressure so low that all the [latex]K_{ i }[/latex]’s were greater than 1, there would be no set of mole fractions that satisfies both Eqs. 4 and 5'. In this case only the vapor is present. Similarly, if all the [latex]K_{ i }[/latex]’s are less than 1 (which occurs at low temperatures or high pressures), only the liquid is present, and again there is no acceptable solution to the equations. It is also possible that there will not be an acceptable solution even with a distribution of values of the K factors if some, but not all, their values are much greater than or much less than unity.

4. For this system C = 3, P = 2, and M= 0, so that, from the Gibbs phase rule, the number of degrees of freedom is

[latex]F =3-0-2+2=3[/latex]

Since the equilibrium temperature and pressure were specified, one degree of freedom remains. If no further information about the system were given, that is, if one were asked to determine the equilibrium compositions of vapor and liquid for a pentane–hexane–heptane mixture at 69°C and 1.013 bar with no other restrictions, many different vapor and liquid compositions would be solutions to the problem. A problem that does not have a unique solution is said to be ill posed. With the initial liquid composition given, however, the species mass balances (Eqs. 6, 7, and 8) provide the additional equations that must be satisfied to ensure that there will be no more than one solution to the problem.

Question 10.1.4: Partial Vaporization Calculation A liquid mixture of 25 mol...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions