Unformatted text preview: using just the feed and one portion of the solvent, i.e.
100 ml of organic solvent and 100 ml of water containing 100 mg each of A and B, the
resulting products are: Extract containing 100 ml solvent, 66.7 mg A, and 33.3 mg B.
Raffinate containing 100 ml water, 33.3 mg A, and 66.7 mg B.
If all 4 extracts and all 4 raffinates for the Fig. 5.19 are combined, we find:
Extract containing 400 ml solvent, 66.7 mg A, and 33.3 mg B.
Raffinate containing 400 ml water, 33.3 mg A, and 66.7 mg B.
The process in Fig. 5.19 does produce extracts of varying purity, but overall, it is not
(d) The cascade above in Part (a) can be modified to give a more countercurrent process
by reversing the flows of the solvents as shown in the following diagram. However, it is
not possible to carry this out batchwise. Exercise 5.3
Two-stage membrane cascades for removal of nitrogen from methane by
Given: Glassy polymer membrane that is selective for nitrogen. Assumptions: Desired degree of separation can not be achieved with one stage.
Find: Two different two-stage cascades.
Analysis: The two cascades are shown below. In both cases, the retentate is methanerich and the permeate is nitrogen-rich. In the first cascade, the permeate from Stage 2 is
recycled to the feed to obtain a more pure methane retentate. In the second cascade, the
retentate from Stage 2 is recycled to the feed to obtain a more pure nitrogen permeate. Exercise 5. 4
cascades. Leaching of oil from soybeans with benzene using countercurrent-flow Given: Soybean meal containing 2,500 kg/h of oil = L0 and 2,500 kg/h of insoluble
solids. Benzene solvent as in Example 4.9.
Assumptions: Equilibrium leaching stages where, for each stage, the weight fraction of
oil in the overflow equals that in the underflow liquid. No solids in the overflows.
Underflows contain 65 wt% solids.
Find: Percent extraction of oil if:
(a) Two stages are used with 5,000 kg/h of benzene = S.
(b) Three stages are used with 5,000 kg/h of benzene = S.
and (c) Number of stages to extract 98% of the oil with two times the min. benzene rate.
Analysis: Let xi = mass fraction of oil in the underflow liquid from Stage i and yi =
mass fraction of oil in the overflow liquid from Stage i. Let Li = underflow liquid leaving
Stage i and Vi = overflow liquid leaving Stage i. At any stage, yi = xi.
solids entering a stage leave in the underflow from that stage and the total underflow is
65 wt% solids with 2,500 kg/h of solids, Li = (35/65)2,500 = 1,346 kg/h
(a) The two-section cascade is shown below. Soybean meal enters Stage 1 at
5,000 kg/h and the benzene solvent enters Stage 2 at 5,000 kg/h. Underflow liquids are
L1 = L2 = 1,346 kg/h. Total liquid material balance around Stage 2:
S + L1 = V2 + L2 or 5,000 + 1,346 = V2 + 1,346. Therefore, V2 = 5,000 kg/h.
Total liquid material balance around Stage 1:
V2 + L0 = V1 + L1 or 5,000 + 2,500 = V1 + 1,346. Therefore, V1 = 6,154 kg/h.
Oil material balance around Stage 1:...
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