We wish to remove acetic acid from water using pure isopropyl ether as solvent. The operation is at \(293 \mathrm{~K}\) and \(1 \mathrm{~atm}\) (see Table 7.2). The feed is \(45 \mathrm{wt} \%\) acetic acid and \(55 \mathrm{wt} \%\) water. The feed flow rate is \(2000 \mathrm{~kg} / \mathrm{h}\). A multistage countercurrent extraction cascade is used to produce a final extract that is \(20 \mathrm{wt} \%\) acetic acid and a final raffinate that is also \(20 \mathrm{wt} \%\) acetic acid. Calculate how much solvent and how many equilibrium stages are required.

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Table 7.2 Water-Isopropyl Ether-Acetic Acid Equilibrium Dataa Isopropyl ether phase, wt% Water phase, wt% Water Ether Acetic acid Water Ether Acetic acid 98.10 1.2 0.69 0.5 99.3 0.18 97.10 1.5 1.41 0.7 98.9 0.37 95.50 1.6 2.89 0.8 98.4 0.79 91.70 1.9 6.42 1.0 97.1 1.93 84.40 2.3 13.30 1.9 93.3 4.82 71.10 3.4 25.30 3.9 84.7 11.40 58.90 4.4 36.70 6.9 71.5 21.60 45.10 10.6 44.30 10.8 58.1 31.10 37.10 16.5 46.40 15.1 48.7 36.20 "Water and ether phases on the same line are in equilibrium with each other. Source: Data from Treybal (1980) at 293 K and 1 atm.