Separation Process Principles- 2n - Seader & Henley - Solutions Manual

Assume ideal liquid solutions component lbh acetic

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Unformatted text preview: ne/ kg solvent-free extract and 5.7 wt% MCH in the extract. This gives 14.5/15.5 x 100% = 93.5 wt% aniline. A total material balance around the bottom Mixer gives:SB - B = 11,830 - 540 = 11, 290 = VB - L1 . An aniline balance around the bottom Mixer gives:11,830 - 40 = 0.935VB - 0.075L1. Solving these two equations gives L1 = 1,435 kg/h. Therefore, the raffinate reflux ratio = (1,435 - 540)/540 = 1.66. Exercise 8.28 Subject: Design of a mixer-settler unit for extraction of acetic acid (A) from water (C) by isopropyl ether (S). Given: 12,400 lb/h of 3 wt% A in C. 24,000 lb/h of S. 1.5 minutes residence time in mixer. 4 gal/min-ft2 in settling vessel. Find: (a) (b) (c) (d) Diameter, DT , and height, H, of mixing vessel if H/DT = 1. Agitator Hp for mixing vessel. Diameter, DT , and length, L, of settling vessel, if L/DT =4. Residence time in settling vessel in minutes. Analysis: First compute total volumetric flows into mixer vessel. Because feed is dilute in A, assume density of feed is that of water = 62.4 lb/ft3. From Perry's Handbook, specific gravity of S = 0.725 or a density of 0.725(62.4) = 45.2 lb/ft3. Therefore, total flow rate of feed plus solvent = Q = 12,400/62.4 + 24,000/45.2 = 730 ft3/h or 12.2 ft3/min. (a) Vessel volume = V = Qtres = 12.2(1.5) = 18.3 ft3 For a cylindrical vessel of H/DT = 1, V = πDT2H/4 = πDT3/4 Solving, DT3 = 4V/π = 4(18.3)/3.14 = 23.3 ft3. Therefore, DT = H = 2.85 ft (b) Assume 4 Hp/1,000 gallons for mixer agitator power. Gallons in vessel = 7.48V = 7.48(18.3) = 137 gal Therefore, need 4(137)/1,000 = 0.55 Hp. (c) Assume volumetric flow into mixer = volumetric flow leaving settler Therefore, Q = 12.2 ft3/min or 7.48(12.2) = 91 gpm The ft2 disengaging area needed = 91/4 = 22.8 ft2 = DTL = DT(4DT) = 4DT2 Solving, DT = 2.4 ft and L = 4DT = 4(2.4) = 9.6 ft. (d) Settler volume = V = πDT2L/4 = 3.14(2.4)2(9.6)/4 = 43.4 ft3 Residence time of phases in settler = tres = V/Q = 43.4/12.2 = 3.6 minutes Exercise 8.29 Subject: Extraction of acetic acid (A) from water (C) by ethyl acetate (S) in available extractor. Given: Cascade of 6 mixer-settler units. Each mixer is 10-ft diameter by 10-ft high with 20-Hp agitator. Each settler is 10-ft diameter by 40-ft long. Feed and solvent compositions , and solvent-to-feed mass ratio, are those in Fig. 8.1. Find: Allowable feed flow rate. Analysis: From Fig. 8.1, the total entering component flow rates are as follows, with liquid densities from Perry's Handbook. Assume ideal liquid solutions. Component lb/h Acetic acid Water Ethyl acetate 6,660 26,100 68,600 density, lb/gal 8.74 8.33 7.51 gpm 12.7 52.2 152.2 Therefore, total flow rate = Q = 12.7 + 52.2 + 152.2 = 217.1 gpm The volume of one mixer = V = πDT2H/4 = 3.14(10)2(10)/4 = 785 ft3 or 5,872 gal. First consider the mixer residence time factor. With the flow rates in Fig. 8.1, residence time in mixer = tres = V/Q = 5,872/217 = 27 minutes. Since the liquid phase viscosities are each less than 5 cP (actually less than 1 cP) and the de...
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