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Unformatted text preview: following results:
acetic acid (A)
ethyl acetate (S) 0.7059 0.9073
Define KD = mass fraction in extract phase/mass fraction in raffinate phase.
(KA)D = 0.0595/0.0564 = 1.055
(KC)D = 0.0332/0.9242 = 0.036
(KS)D = 0.9073/0.0194 = 46.8
βAC = (KA)D /(KC)D = 1.055/0.036 = 29.3
(a) The value of βAC is high, indicating a high selectivity.
(b) The value of (KA)D is not high, indicating a solvent capacity that is not high.
(c) Since (KS)D is high and (KC)D is low, recovery of solvent is relatively easy.
A better solvent should be sought because ethyl acetate does not have a high capacity. From the
results of Exercise 8.7, four other solvents are selected. Again using CHEMCAD, the following
results are obtained for distribution coefficients and relative selectivity:
No solvent meets all four criteria. Ethyl acetate may be the compromise. No solvent has a high
capacity. Exercise 8.10
Subject: Estimation of interfacial tension. Given: Composition of a ternary mixture.
Find: Method for estimating the interfacial tension from the values of surface tension in air for
each component in the mixture.
Analysis: Antonoff's rule can be used to compute the interfacial tension for two liquid phases, I
and II, in equilibrium,
σi = ( σII )in air − ( σI )in air
This equation, which is discussed in,
1. J. Russ. Phys. Chem. Soc., 39, 342 (1907).
2. Hart, Duga, Res./Devel., 16 (9), 46 (1965).
requires values of the multicomponent surface tensions of each phase [see Reid, Prausnitz, and
Poling, "The Properties of Liquids and Gases", 4th ed. (1986)]. Exercise 8.11
Subject: Extraction of acetone (A) from water (C) by 1,1,2-trichloroethane (S) at 25oC
Given: 1,000 kg/h of 45 wt% A in C. Liquid-liquid equilibrium data. 10 wt% A in raffinate.
Find: Using Hunter-Nash method with an equilateral triangle diagram.
(a) Minimum flow rate of S.
(b) Number of equilibrium stages for solvent rate of 1.5 times minimum.
(c) Flow rate and composition of each stream leaving each stage.
Analysis: Using the given equilibrium data in weight fractions, the triangular diagram is shown
on the next page, where a solid line is used for the equilibrium curve and dashed lines are used
for the tie lines, only three of which are given. Additional tie-line locations can be made using
either of the two techniques illustrated in Fig. 8.16.
(a) The minimum solvent flow rate corresponds to an infinite number of equilibrium
stages. To determine this minimum: (1) Points are plotted on the triangular diagram for the
given compositions of the solvent, S (pure S), feed, F (45 wt% A, 55 wt% C), and raffinate, RN
(10 wt% A on the equilibrium curve), followed by drawing an operating line through the points S
and RN and extending it only...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land