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

# Analysis use cgs units assume a thermodynamic path of

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Unformatted text preview: H 2 O crystals formed Analysis: At 100oC, have a solution of 3,750 lb/h of FeCl3 and 3,750 lb/h of water. Note that this amount of FeCl3 is far below its solubility at that temperature. The molecular weights are: 162.22 for FeCl3 and 270.32 for FeCl3 6 H 2 O The final solution will contain less water than the initial solution because of the water of hydration in the crystals. Let x = lb/h of water in the final aqueous solution. Then, 3,750 – x = water in crystals. 162.22 The crystals are (100%) = 60.0% FeCl3 and 40.0% water 270.32 A mass balance for FeCl3 at 20oC gives, 3, 750 = 91.8 162.22 x + (3, 750 − x) 100 270.32 − 162.22 (1) Solving (1), gives x = 3,222 lb/h of water in solution and, therefore, 3,750 – 3,222 = 528 lb/h of water in crystals. Therefore, the FeCl3 still in the solution = 0.918(3,222) = 2,958 lb/h Crystals of FeCl3 6 H 2 O = 3,750 – 2,958 + 528 = 1,320 lb/h Exercise 17.11 Subject: Continuous vacuum crystallization of MgSO4. Given: Feed to the crystallizer consisting of a mixture of 5,870 lb/h of aqueous 35 wt% MgSO4 at 180oF and 25 psia as a concentrate from an evaporator, and 10,500 lb/h of saturated aqueous recycle filtrate of MgSO4 at 80oF and 25 psia. The vacuum crystallizer operates at 85oF and 0.58 psia to produce steam and a magma of 25 wt% crystals and 75 wt% saturated solution. Assumptions: Equilibrium in the crystallizer. Find: Evaporation rate of water in the crystallizer in lb/h and the production rate of crystals in tons/day (dry basis for 2,000 lb/ton) Analysis: First compute the total feed to the crystallizer. Concentrate: 0.35(5,870) = 2,055 lb/h MgSO4 5,870 – 2,055 = 3,815 lb/h water At 189oF, from Fig. 17.2, solubility of MgSO4 = 39 wt%. So, all of the sulfate is in solution. Filtrate: From Fig. 17.2, the solubility of MgSO4 in water at 80oF = 27.5 wt% 0.275(10,500) = 2,888 lb/h MgSO4 10,500 – 2,888 = 7,612 lb/h water Total feed: 2,055 + 2,888 = 4,943 lb/h MgSO4 3,815 + 7,612 = 11,427 lb/h water Compute the mass balance around the crystallizer. From Fig. 17.2, the solubility of MgSO4 in water = 28 wt% and MgSO4 is crystallized as the heptahydrate. Let: L = lb/h of solution in the magma V = lb/h of water vaporized in the crystallizer S = lb/h of crystals in the magma MW of MgSO4 = 120.4. MW of MgSO 4 7H 2 O = 246.4 Crystals are 120.4/246.4 = 0.4886 mass fraction MgSO4 and 0.5114 mass fraction water. Magma is 25 wt% crystals and 75 wt% of a solution of 28 wt% MgSO4. MgSO4 mass balance around the crystallizer: 4,943 = 0.28 L + 0.4886 S But, S = 0.25(S + L) Solving these two equations, L = 11,160 lb/h and S = 3,720 lb/h Therefore, the tons/day of crystals = 3,720(24)/2,000 = 44.6 tons/day Total mass balance around the crystallizer: 4,943 + 11,427 = V + L + S = V + 11,160 + 3,720 Solving, V = 1,490 lb/h of water vapor produced in the crystallizer. Exercise 17.12 Subject: Crystallization of urea from an aqueous solution by cooling and evaporation. Given: Feed is a 90% saturated solution of urea in water at 100...
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