This preview shows page 1. Sign up to view the full content.
Unformatted text preview: entering the extractor. On the following page, the equilibrium stages are stepped
on another right triangle diagram, as in Fig. 8.17, by determining the operating point P from
extensions of lines drawn through points F and E, and S and R, followed by alternating between
operating lines and equilibrium tie lines. The result is 5 equilibrium stages. Analysis: Exercise 8.15 (continued) Case of 80oC: The given liquidliquid equilibrium data are plotted in the righttriangle
diagram below. Included on the diagram are composition points F for the feed, R for the
raffinate on the equilibrium curve, and S for the solvent. A straight line extends from point F to
point S. Because the mass flow rates of the feed and solvent are equal, the mixing point, M, is
located at the midpoint of this line. Another straight line extends from point R to point M, and
then to an intersection with the equilibrium curve at point E, which is the final extract. Using the
inverse leverarm rule on line RME, the mass ratio of R to E is 0.383. Combining this with an
overall material balance: F + S = 500 + 500 = 1,000 = R +E
gives R = 277 kg/h and E = 723 kg/h. From the diagram, the mass fraction of DPH in the
extract is 0.271. Therefore, the DPH in the extract is 0.271(723) = 195.9 kg/h, which is 93.3% of
the total DPH entering the extractor. On the following page, the equilibrium stages are stepped
on another right triangle diagram, as in Fig. 8.17, by determining the operating point P from
extensions of lines drawn through points F and E, and S and R, followed by alternating between
operating lines and equilibrium tie lines. The result is 4+ equilibrium stages. o Exercise 8.15 (continued) Analysis: (case of 80 C continued) Exercise 8.16
Subject: Selection of extraction method.
Given: Four ternary systems in Fig. 8.43. Diagrams 1, 2, 4 are Type I. Diagram 3 is Type II.
Find: Method for most economical process for each system. Methods are:
(a) Countercurrent extraction (CE).
(b) CE with extract reflux (ER).
(c) CE with raffinate reflux (RR).
(d) CE with ER and RR.
Analysis: Note that y1 is the composition of the extract.
Raffinate reflux (RR) is of little value, so don't use it.
For Type I diagrams, extract reflux is rarely useful.
All three Type I diagrams exhibit solutropy, making it almost impossible to obtain a good
separation.
The Type II diagram uses a poor solvent, making the use of extract reflux questionable.
Summary: Use CE for all four systems. However, better solvents should be sought for all four
systems. Exercise 8.17
Subject: Stage requirements for extraction of acetone (A) from two feeds, of different
composition of acetone (A) and water (C), with a solvent of 1,1,2  trichloroethane (S).
Given: Feed F of 7,500 kg/h containing 50 wt% A in C. Second feed F' of 7,500 kg/h
containing 25 wt% A in C. Solvent of 5,000 kg/h of S. Raffinate to contain 10 wt% A. Liquidliquid equilibrium data from Exercise 8.11.
Find: Number of equilibrium stages and feed locations, using a righttriangle...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details