1.1.4 Continuous closed distillation binary ideal mixture

Points: 9 No answer

This task is supposed to make you familiar with the computation of a continuous closed distillation of an ideal binary mixture.

A sketch of the continuous closed distillation is provided in the figure below.

 

The feed $$F ̇$$ is a binary mixture of benzene (1) and p-Xylene (2). I chose benzene and p-xylene, as this binary mixture can be taken as an ideal mixture. This simplifies the equations that follow, as we can set all the parameters accounting for non-idealities to 1.
The composition of the feed is $$z_1=0.6$$.

The table below shows vapour pressures of benzene and p-xylene as a function of $$T$$. It also specifies the compositions of the coexisting phases for a total pressure of $$p_t=0.1013MPa$$. I generated these vapour pressures using the Antoine equation with the Antoine parameters for benzene and p-xylene, which are provided by the NIST services (just in case you want to reproduce them).

Temperature in $$K$$        $$p_1^{sat}$$       in  $$MPA$$     $$p_2^{sat}$$     
in $$MPA$$
$$x_1$$ $$y_1$$     $$L ̇/G ̇$$
3530.101300.015551.00 1.00 -1.00
3600.124280.020200.78 0.96 
3700.165290.028790.53 0.87 3.88
3800.216050.040150.35 0.74 
3900.277970.054870.21 0.57 -0.07
4000.352540.073660.10 0.34 -0.51
4100.441320.097240.01 0.05 -0.93
4120.460930.101300.00 0.00 

 

From the lecture we know the operation equation for the continuous closed distillation. It is 

$$y_i=-L ̇/G ̇  x_i+(1+L ̇/G ̇ ) z_i$$

The distillation kettle is one separation stage. The streams leaving the kettle are in thermodynamic equilibrium. This implies that they feature the same temperature, the same pressure and that chemical equilibrium is obtained. When you look at the table above, you can see that the ratio of the liquid and vapour phase composition $$y_i/x_i=K_i$$ is a function of the temperature $$T$$. In other words, for various temperatures $$T$$, you will have to consider different combinations of $$x_i$$ and $$y_i$$ in the operation equation and as a consequence different ratios of $$L ̇/G ̇$$ will be obtained.

TASK 1: $$L ̇/G ̇$$-ratio

Compute for the different temperatures $$T$$ specified in the table above the ratios $$L ̇/G ̇$$ and add the missing values into the respective column.
When you look at the sketch above, you can see that both streams $$L ̇$$ and $$G ̇$$ leave the distillation kettle. Therefore, they should feature the same sign and thus their ratio should be positive. But, some of the $$L ̇/G ̇$$-values you compouted are negative!

Let us first have a look at the rows in the table, which state positive $$L ̇/G ̇$$-values. 
The positive $$L ̇/G ̇$$-values are the meaningful ones. We obtain meaningful $$L ̇/G ̇$$-values, whenever the composition of the feed is in between the compositions of the liquid and the vapour phase streams $$x_i<z_i<y_i$$.

Now let us consider the rows in the table, which state negative $$L ̇/G ̇$$-values.
This are the non-meaningful $$L ̇/G ̇$$-values (non-meaningful with respect to the sketched distillation). These non-meaningful $$L ̇/G ̇$$-values occur whenever $$z_i$$ is not in the range of compositions between $$x_i$$ and $$y_i$$. 

Theoretically the substance balance still works even for negative  $$L ̇/G ̇$$-values. Let us look at the first row in the table, where $$L ̇/G ̇=-1$$. We change the direction of $$L ̇$$ and let it stream into the distillation kettle with the composition $$x_1=1$$ (pure compound 1). We do NOT change the direction of  $$G ̇$$. $$G ̇$$ still strems out of the distillation kettle. As  $$L ̇$$ is going into and $$G ̇$$ is going out of the kettle, the ratio $$L ̇/G ̇$$ is negative. The ratio $$L ̇/G ̇=-1$$, as the magnitudes of both streams are the same. This (under stationary conditions) is possible only, if  $$F ̇=0$$. Under these theoretical conditions it is totally obvious that the stream $$G ̇$$ is pure compound 1 ($$y_1=1$$) as we feed via the stream $$L ̇$$ only pure compound 1 ($$y_1=1$$). The temperature T inside the kettle would then be the boiling temperature of pure compound 1.

TASK 2: Graphical considerations of the $$L ̇/G ̇$$-ratio

The figure below shows the $$yx$$-diagram of the binary mixture of benzene (1) and p-Xylene (2). The feed composition is indicated by the cross in the diagram. A vertical and a horizontal line pass through the feed composition point. These lines represent the two extreme meaningful operation lines. Only those operation lines, which are in between these two extremes are meaningful, too. Operation lines out of this range would result from non-meaningful negative $$L ̇/G ̇$$-values.

What is the larges $$x_i$$= you can subtract with the stream $$L ̇$$ from the distillation kettle? What is the ratio $$L ̇/F ̇$$= then? (Attention $$L ̇/F ̇$$, NOT $$L ̇/G ̇$$ and pay attention to the differently directed streams). Please estimate (by linear interpolation) the temperature inside the distillation kettle for this condition $$T$$= . (This is the boiling $$T$$ of the feed)

What is the smallest $$y_i$$= you can subtract with the stream $$G ̇$$ from the distillation kettle? What is the ratio $$G ̇/F ̇$$= then? Please estimate (by linear interpolation) the temperature inside the distillation kettle for this condition $$T$$= . (This is not the boiling $$T$$ of the feed, this is the dew $$T$$ of the feed)