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Unformatted text preview: is results in an impossible negative QSC .
The data are inconsistent, at least for the Hf/Zr ratio in the final raffinate. This error has a
significant effect on the volumetric flow rate of the raffinate, which then affects other flow rates.
(c) The extractor as shown removes almost all of the ZrO2 and is, therefore, effective. However,
the Hf/Zr ratio in the aqueous phase, as shown in the Stagewise Analyses table, decreases, rather
than increases, in Stages 13 and 14. This indicates the need for more nitric acid in the recycle
solvent. Also, Stages 1 and 2 appear to do little extraction and might be eliminated. Exercise 8.22
Subject: Extraction of diphenylhexane (DPH) from docosane (C) with furfural (U) at 45oC in a
continuous, countercurrent, multistage liquidliquid extraction system.
Given: Feed, F, of 5,000 kg/h of 65 wt% DPH, 28 wt% C, and 7 wt% U. Solvent is pure U.
Raffinate to contain 12 wt% DPH. Liquidliquid equilibrium data from Exercise 8.15.
Find: (a) Minimum solvent flow rate.
(b) Flow rate and composition of extract at minimum solvent rate.
(c) Number of equilibrium stages at a solvent rate of 1.5 times the minimum.
Analysis: The given liquidliquid equilibrium data at 45oC are plotted in the righttriangle
diagram below. Included on the diagram are composition points F for the feed, RN for the
raffinate on the equilibrium curve, and S for the solvent.
(a) To find the minimum solvent flow rate, Smin , corresponding to an infinite number
of equilibrium stages, A straight line is drawn on the diagram below that joins points RN and S
and extends to either side of the diagram. Similar lines are extended from the tie lines to where
they intersect the first line. The construction is similar to Figure 8.18. For this exercise the
intersection points are closest to the raffinate side as they are in Figure 8.18. Therefore the
intersection farthest away corresponds to Smin. and locates the composition of the final extract, E1.
From either the inverse leverarm rule or a mass balance, the minimum solvent rate = 2,190 kg/h.
b) From the diagram, the composition of the extract is:
2 wt% docosane
28 wt% DPH
70 wt% furfural
The composition of the raffinate is:
83 wt% docosane
12 wt% DPH
5 wt% furfural
The total flow rate of extract + raffinate = feed + min. solvent = 5,000 + 2,190 = 7,190 kg/h
Using the inverse leverarm rule or mass balances, E1 = 3,355 kg/h and RN = 3,835 kg/h
(c) The solvent rate for finite stages = 1.5 Smin = 1.5(2,190) = 3,285 kg/h. The mixing
point for this rate is at a composition of:
DPH: 0.28(5,000)/(5,000 + 3,285) = 0.169 or 16.9 wt%
Furfural: [0.07(5,000) + 3,285]/(5,000 + 3,285) = 0.439 or 43.9 wt%
The second figure below shows the new extract point for this solvent rate and a mixing point.
The operating point, P, is the intersection of the two lines, from through RN and S; the other
through F and E1. The point is not shown because far to the left off the page. However, using
that point and alternating between operating lines and equilibrium tie lines...
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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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