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

00 1200 differential plot of predicted screen

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Unformatted text preview: e magma, so we can compute the residence time in the crystallizer. The flow rate of solution in the magma = 10,000 – 1,777 = 8,223 kg/h From Perry’s Handbook, the density of BaCl2 2H 2 O crystals = 3,097 kg/m3, and the density of the solution in the magma = 1,279 kg/m3 The volumetric flow rate of the magma is, 8,223/1,279 + 1,777/3,097 = 7.00 m3/h Average residence time, τ, of crystals in the crystallizer = 2.0/7.0 = 0.286 h = 17 minutes Growth rate, G = 4.0 x 10-7 m/s = 0.0024 cm/min From (17-49), the predominant crystal size is given by, Lpd = 3Gτ = 3(4.0 x 10-7)[17(60)] = 0.00122 m = 0.122 cm = 1.22 mm Exercise 17.24 (continued) (c) Now estimate the mass fraction of crystals in the size range from 10 to 25 mesh. From Table 17.4, 20 mesh has an aperture of 0.850 mm = 0.085 cm 25 mesh has an aperture of 0.710 mm = 0.071 cm Calculate the dimensionless crystal size from (17-48), z = L/Gτ For 20 mesh, z = 0.085/[(0.0024)(17)] = 2.08 For 25 mesh, z = 0.071/[(0.0024)(17)] = 1.74 From Table 17.9, the cumulative mass fraction of crystals smaller than size L, is: z 2 z3 xm = 1 − 1 + z + + exp(-z) 26 For L = 0.085 cm, z = 2.08. Substitution into (2), gives, xm = 1 − 1 + 2.08 + 2.082 2.083 + exp(−2.08) = 0.158 2 6 For L = 0.071 cm, z = 1.74. Substitution into (2), gives, xm = 1 − 1 + 1.74 + 1.74 2 1.743 + exp(−1.74) = 0.099 2 6 Therefore, we have 15.8 wt% crystals smaller than 0.085 cm, and 9.9 wt% crystals smaller than 0.071 cm Therefore, the mass fraction of crystals between 20 and 25 U.S. mesh = 0.158 = 0.099 = 0.059 or 5.9 wt% (2) Exercise 17.25 Subject: Operation of an MSMPR crystallizer Given: Feed of 5,000 kg/h of 40 wt% sodium acetate in water. Monoclinic crystals of the trihydrate are formed. 20% of the water in the feed is evaporated at 40oC. Crystal growth rate, G, is 0.0002 m/h. Predominant crystal size, Lpd, is 20 U.S. standard mesh. Solubility of sodium acetate at 40oC = 65.5 g/100 g water. Crystal density = 1.45 g/cm3. Mother liquid density = 1.20 g/cm3. Assumptions: Equilibrium and applicability of the MSMPR model. Find: (a) The kg/h of crystals in the exiting magma. (b) The kg/h of mother liquor in the exiting magma. (c) The volume in m3 of magma in the crystallizer. Analysis: (a) and (b) Compute a mass balance around the crystallizer. The feed contains 0.4(5,000) = 2,000 kg/h of sodium acetate and 3,000 kg/h of water. MW of sodium acetate = 82.04. MW of sodium acetate trihydrate = 136.09. Let x = kg/h of water in the crystals. Sodium acetate mass balance: 65.5 82.04 2, 000 = ( 3, 000 − 600 − x ) + x 100 136.09 − 82.04 Solving, x = 496 kg/h of water in the crystals Therefore, water in the mother liquor = 3,000 – 600 – 496 = 1,904 kg/h Sodium acetate in the mother liquor = 0.655(1,904) = 1,247 kg/h Sodium acetate in the crystals = 2,000 – 1,247 = 753 kg/h Therefore, the flow rate of crystals in the exiting magma = 496 + 753 = 1,249 kg/h The flow rate of mother liquor in the exiting magma = 1,904 + 1,247 = 3,151 kg/h (c) The t...
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