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Unformatted text preview: , as in Figure 8.17,
only 2 equilibrium stages are required to move from E1 to RN . Analysis: (a) and (b) Exercise 8.22 (continued)
S E1 F Pmin is further down RN Analysis: (a) and (b) Exercise 8.22 (continued)
S E1 F P is further down RN Exercise 8.23
Subject: Extraction of diphenylhexane (DPH) from docosane (C) with furfural (U) at 45oC in a
single-stage and a continuous, countercurrent, multistage liquid-liquid extraction system.
Given: Feed, F, of 1,000 kg/h of 20 wt% DPH and 80 wt% docosane. Solvent is pure furfural. .
Liquid-liquid equilibrium data from Exercise 8.15.
Find: (a) For a single equilibrium stage, the compositions and flow rates of extract and raffinate
for solvent rates of 100, 1,000, and 10,000 kg/h.
(b) Minimum solvent flow rate to form two liquid phases.
(c) Maximum solvent flow rate to form two liquid phases.
(d) For two countercurrent-flow equilibrium stages, the compositions and flow rates of
extract and raffinate for a solvent flow rate of 2,000 kg/h.
(a) First, calculate the composition of the mixing point for each solvent flow rate, noting
that the feed is 800 kg/h of docosane and 200 kg/h of DPH.
Solvent flow rate, kg/h
Feed + Solvent, kg/h
Wt% Furfural 100
0.910 These three mixing points can be plotted on the triangular phase diagram shown below. Then a
tie line is drawn through each mixing point. The intersections of the tie lines with the bimodal
curve give the following compositions from a single equilibrium stage, in the manner of Figure
4.16, for each of the three cases:
Solvent flow rate, kg/h
Raffinate composition, wt%:
Extract composition, wt%
Docosane 100 1,000 10,000 18.2
0.1 Analysis: (a) continued: Exercise 8.23 (continued) The following diagram shows the construction for a solvent rate of 1,000 kg/h. R R F
R M E S Analysis: (a) continued: Exercise 8.23 (continued) Next, mass balances can be used to compute the flow rates of the raffinate and the extract for
each of the three cases, as follows.
Case 1. Solvent flow rate = 100 kg/h
DPH mass balance: 0.182 R + 0.180 E = 200
Total mass balance: R + E = F + S = 1,000 + 100 = 1,100
Solving, R = 1,000 kg/h and E = 100 kg/h
Case 2 Solvent flow rate = 1,000 kg/h
DPH mass balance: 0.101 R + 0.099 E = 200
Total mass balance: R + E = F + S = 1,000 + 1,000 = 2,000
Solving, R = 1,000 kg/h and E = 1,000 kg/h
Case 3. Solvent flow rate = 10,000 kg/h
Docosane mass balance:
0.937 R + 0.000 E = 800
Total mass balance: R + E = F + S = 1,000 + 10,000 = 11,000
Solving, R = 850 kg/h and E = 10,150 kg/h
(b) and (c)
The minimum and maximum solvent flow rates for the formation of two liquid phases is
obtained from the intersection with the bimodal curve of a straight line between the feed and
solvent compositions. The construction is shown on a separate page below. The solvent rates
are obtained using the inverse lever-arm rule with that diagram, resulting in the following results: Smin = 64 kg/h and Smax = 142,000 kg/h (d) The determination of the raffinate and extract obtained from a two-equilibrium-stage,
countercurrent-flow extraction system requires a trial-and-error procedure. One approach is to
assume a sequence of equilibrium raffinate compositions and step off the corresponding stages
until 2 stages result. A good way to start the start sequence is to solve the single-s...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land