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raffinate, R, and the extract, E. By total material balance, R + E = F + S = 45 + 45 = 90. By the
inverse lever arm rule, R/E = 0.61. Combining these equations gives: E = 55.9 kg, R = 34.1 kg.
From the diagram, the composition of the extract is: 19 wt% B, 76 wt % C, and 5 wt% A.
Therefore, the extract contains (0.19)(55.9) = 10.62 kg B. The composition of the raffinate is 8
wt% B, 84 wt% A, and 8 wt% C.
The % extraction of glycol (B) = 10.62/[(0.30)(45)] x 100% = 78.7%.
The amount of solventfree extract = (55.9)(1  0.76) = 13.4 kg.
The amount of solventfree raffinate = (34.1)(1  0.08) = 31.4 kg. Exercise 4.46 (continued)
Analysis: (continued)
(d) The maximum possible glycol purity in the extract occurs when the minimum
amount of solvent is used, as discussed in part (a), giving raffinate R1. In this case, have say just
one drop of extract, corresponding to point E1, which connects to R1 by a tie line in the diagram
below. At point E1, have 48 wt% glycol.
The maximum purity of water in the raffinate occurs when the maximum amount of
solvent is used, as discussed in part (b), giving extract E2. In this case, have say just one drop of
raffinate, corresponding to point R2, which connects to E2 by a tie line in the diagram below. At
point R2, have 90 wt% water. Exercise 4.47
Subject: Representation of the composition of a ternary mixture on a triangular diagram.
Given: A triangular diagram where each vertex represents a pure component.
Assumptions: Consider the case of an equilateral triangular diagram.
Prove: The composition of any point inside the triangle is proportional to the length of the
respective perpendicular drawn from the point to the side of the triangle opposite the vertex in
question.
Analysis: In the triangular diagram below, the pure components are E, R, and S. A mixture of
these components is represented by the point M. For the equilateral triangle shown, the sum of
the lengths of the three perpendiculars drawn from an interior point, such as M, to the three sides
equals the altitude, is the same length from each side. Therefore, divide each altitude into 100
divisions and number these divisions starting with 0 at the base to 100 at the apex. Thus, if the
divisions represent wt% (or mol%), the sum of the perpendiculars equals 100%. In the diagram
below, each division is 10%. The perpendicular s measures 28% and is the composition of S.
The perpendicular r measures 18% and is the composition of R. The perpendicular e measures
54% and is the composition of E. This proof is extended to a triangle of any shape on pages 11
to 18 of "Chemical Process Principles, Part I" by Hougen, Watson, and Ragatz. Exercise 4.48
Subject: Liquidliquid extraction of acetic acid from chloroform by water at 18oC and 1 atm. Given: Equilibrium data for the ternary mixture
Assumptions: Equilibrium stages.
Find: (a) Compositions and weights of raffinate and extract when 45 kg of a 35 wt%
chloroform (C) and 65 wt% acetic acid (A) feed mixture (F) is extracted wi...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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