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Unformatted text preview: Fig. 8.26a. Minimum extract reflux
ratio = 3.49. Equilibrium data in Exercise 8.22.
Find: With the Maloney-Schubert method on a Y - X Janecke diagram:
(a) Raffinate reflux ratio.
(b) Amount of aniline that must be removed by solvent removal at the top
of the extractor, as shown in Fig. 8.26a.
(c) Amount of solvent that must be added to the mixer at the bottom of the
extractor, as shown in Fig. 8.26a.
Analysis: The liquid-liquid phase equilibrium data are given, for each phase, as mass % MCH
on an aniline-free basis, and mass of aniline per mass of aniline-free mixture. Thus, the data can
be plotted directly to obtain the Janecke diagram, which is shown on the next page.
Referring to Fig. 8.26a, the given compositions of the feed, F, the solvent-free extract, D,
and the solvent-free raffinate, B, are plotted on the diagram. By construction on the Janecke
diagram, on the previous page, the compositions of the solvent-containing extract, VN , and the
solvent-containing raffinate, L1 , are determined as intersections of vertical lines (drawn up from
the solvent-free points to the extract and raffinate equilibrium curves. The wt% compositions of
these points are read from the diagram with the aid of the tabulated equilibrium data given in the
exercise statementto be:
Extract, VN Raffinate, L1
Take a basis of 1,000 kg/h of feed and make material balances for the minimum extract reflux
condition. Because the feed is 50 wt% MCH and 50 wt% C, and the two solvent-free product
compositions are symmetrical with respect to MCH and C, the solvent-free material balance is
quickly determined to be:
Feed, F Extract, D Raffinate, B
500 Exercise 8.27 (continued)
Analysis: (continued) Exercise 8.27 (continued)
Analysis: (continued) (b) Assume the given minimum extract reflux ratio = 3.49 = LR/D in Fig. 8.36a.
Then, LR = 3.49(500) = 1,745 kg/h. Therefore, the flow rate of solvent-free extract in stream VN
= 1,745 + 500 = 2,245 kg/h. But, from above, this stream contains 84 wt% aniline solvent.
Therefore, the flow rate of aniline in stream VN = (84/16)(2,245) = 11,790 kg/h. This is the flow
rate of aniline that must be removed by the separator at the top of the cascade.
(c) From above, the raffinate L1 leaving at the bottom of the cascade is 7.5 wt% aniline.
Therefore, aniline flow rate in B is (7.5/92.5)(500) = 40 kg/h and B = 540 kg/h
The entering solvent rate SB is, therefore, 40 + 11,790 = 11,830 kg/h.
This is the amount of fresh solvent that must be added to the mixer at the bottom of the column.
(a) Referring to Fig. 8.36a, define the raffinate reflux ratio as (L1 - B)/B. Because the cascade is
operating at minimum reflux, assume infinite stages so that VB is in equilibrium with L1. From
the Janecke diagram on the previous page, the composition of VB is obtained by the dashed tie
line from B, which gives 14.5 kg anili...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land