Unformatted text preview: tage problem
in a manner similar to that in Part (a) of this exercise. That construction is shown below and
results in a raffinate containing 6.7 wt% DPH. For two stages, the wt% DPH will be less. In this
manner, it is found that the raffinate will contain 2.5 wt% DPH for 2 stages. The construction is
shown below and the compositions and flow rates of the corresponding raffinate and extract are
as follows for the solvent flow rate of 2,000 kg/h:
Flow rate, kg/h
DPH, wt %
Furfural, wt% Raffinate
91.5 Analysis: (b, c) continued: Exercise 8.23 (continued) The following diagram shows the construction to determine minimum and maximum
solvent rates to produce two liquid phases. R R F
S min S max
S Analysis: (d) continued: Exercise 8.23 (continued) The diagram bellows shows construction for an assumed raffinate containing 2.5 wt%
DPH and two stages. The mixing point, M, corresponds to 200 kg/h of DPH and 2,000 kg/h of
furfural in a total of 3,000 kg/h sent to the extraction unit. This gives 6.7 wt% DPH and 66.7
wt% furfural. R R Analysis: (d) continued Exercise 8.23 (continued) Exercise 8.24
Subject: Extraction of acetone (A) from water (C) by 1,1,2-trichloroethane (S) at 25oC
Given: Liquid mixture of 27 wt% A and 73 wt% C. Liquid-liquid equilibrium data. Raffinate
essentially of acetone Find: (a) Minimum solvent-to-feed ratio.
(b) Composition of the extract at the minimum solvent-to-feed ratio.
(c) Composition of the extract stream leaving Stage 2 from the feed end. Analysis: Using the given equilibrium data in weight fractions, the right-triangle diagram is shown
below, 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 (27 wt% A, 73 wt% C), and raffinate, RN (0 wt% A
on the equilibrium curve), followed by drawing an operating line through the points S and RN and
extending it only to the right because the tie lines slope down from left to right; (2) Because the
pinch region, where equilibrium stages crowd together, may occur anywhere, additional potential
operating lines may be drawn through the tie lines extended to the right until they cross the operating
line at the solvent end, drawn in (1) The intersections are not shown in the diagram because they are
far off the page. (3) The intersection point farthest from the triangular diagram is the one used to
determine the minimum solvent rate. For this exercise, that point is quite sensitive. Assume the
controlling line is the one through F, corresponding to a pinch point at the feed end of the cascade;
(4) An operating line is drawn from Pmin through F to the dete...
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