Page 219 - Distillation theory

P1: FCH/FFX P2: FCH/FFX QC: FCH/FFX T1: FCH
0521820928c06 CB644-Petlyuk-v1 June 11, 2004 20:17

6.6 Determination of Minimum Flow Rate of Entrainer 193

To determine the values of (L/V) min and (L/V) min , the general algorithm of
m1 m2
calculation of the minimum reflux mode for two-section columns at various splits
is used.
In particular, for the most widespread split with one-component entrainer and
one-component top product (m m = 2, m r = 1), the joining of intermediate section
trajectories with the trajectories of the top and the bottom sections goes on the
way it is at direct split in two-section columns. This uses the simplest modification
of the algorithm of calculation of the minimum reflux mode.

6.6. Determination of Minimum Flow Rate of Entrainer
The second important parameter, besides the parameter (L/V) min , at designing
m
of sharp extractive distillation columns with two feeds is the parameter (E/D) min .
The theory of trajectory tear-off easily determines this parameter at any splits in
an extractive distillation column.
We express the value of the parameter (E/D) through limit conditions when
the point of tear-off of reversible distillation trajectory coincides with the saddle
point of trajectory bundle of the intermediate section (e.g., x t = S m at Fig.
rev,e
2
6.8, x t = S at Fig. 6.9a or x t = S m at Fig. 6.9b,c). The conditions in the
rev,e m rev,e
tear-off point of this trajectory establish connections between coordinates of the
t
tear-off point x rev,e and of the pseudoproduct point x the way it was done earlier

D
for the connection between coordinates of the tear-off point and of the product
point in two-section columns (see Eq. 4.20). The corresponding equation for the
intermediate section looks the same way as Eq. 4.20:
t
t
t
t
x = x (K − K )/(1 − K ) (6.13)
D,i i i j j
Here, j is the component not entering into the number of components of the
top product and the entrainer.
We now express the parameter E/D through x with the help of material
D,i
balance equation of the intermediate section (Eq. [6.3]). After transformations,
we get the following:
(1 − E/D)x (6.14)
D,i = x D,i − (E/D)x E,i
If, for example, the top product is component 1 and the entrainer does not
contain component 1, that is, x D , 1 = 1 and x E , 1 = 0, then we get the following from
Eq. (6.14):
E/D = 1 − 1/x D,i (6.15)

After substitution into Eq. (6.13), we get:
t
t
t
t
E/D = 1 − (1 − K )/x (K − K ) (6.16)
j
i
j
i
For the mixture acetone(1)-water(2)-methanol(3) at side 1-2, at which there is
t
an intermediate section trajectory tear-off segment Reg , the dependence of phase
e