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

10 a2 find time to leach 95 of the sucrose analysis a

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Unformatted text preview: 40 kg/h. Therefore, xN = 200/59,940 = 0.00334 is the target. Feed to the system is 199,800 kg/h TiO2, (20/50)199,800 = 79,920 kg/h of solutes, and (30/50)199,800 = 119,880 kg/h of water. Fresh water flow rate = total feed flow rate = 199,800 + 79,920 + 119,880 = 399,600 kg/h Therefore, by overall mass balances, Water in final extract = 119,880 + 399,600 – 59,740 = 439,660 kg/h Solutes in final extract = 79,920 - 200 = 79,720 kg/h Summary of overall mass balance in kg/h: Component Feed Wash Water Final Underflow Final Extract TiO2 199,800 199,800 Solutes 79,920 200 79,720 Water 119,880 399,600 59,740 459,740 Total 399,600 399,600 259,740 539,460 Refer to Figure16.7. Calculate the leaching stage. xL = yL = 79,720/539,460 = 0.148. A solute mass balance gives: 79,920 + y1V1 = 79,720 + xL LL or, y1V1 = -200 + 0.148LL (2) Exercise 16.7 (continued) A solution mass balance gives: V1 + 79,920 + 119,800 = LL + 79,720 + 459,460 or, V1 = LL + 339,460 We have 2 equations in 3 unknowns. However, from (1), RL = LL/199,800 = 0.2 xL + 0.3 = 0.2(0.148) + 0.3 = 0.3296 or, LL = 65,850 kg/h From (3), V1 = 65,850 + 339,460 = 405,310 kg/h From (2), y1 = [-200 + 0.148(65,850)]/405,310 = 0.0236 = x1 (3) Now compute Stage 1: From (1), R1 = L1/199,800 = 0.2x1 + 0.3 = 0.2(0.0236) + 0.3 = 0.3047 Therefore, L1 = 60,880 kg/h From a liquid flow balance around Stage 1, LL + V2 = L1 + V1 or, V2 = 60,880 + 405,310 – 65,850 = 400,340 kg/h From a solute balance around Stage 1, xLLL + y2V2 = x1L1 + y1V1 or, y2 = [0.0236(60,880) + 0.0236(405,310) – 0.148(65,850)]/400,340 = 0.00314 Therefore, x2 = y2 = 0.00314 But, x2 is already below the target of 0.00334. Therefore, 2 washing stages is sufficient together with the leaching stage or a total of 3 stages. Exercise 16.8 Subject: Rate of leaching of solute from a flake when the internal resistance to mass transfer is negligible, such that mass transfer is controlled by the external resistance of the solvent. Find: Derive Equation (16-20) Analysis: For diffusion from either side of a flake when the rate of mass transfer is controlled by the external resistance, (16-13), gives for mass-transfer from the surface of the solid to the solvent surrounding the solid, ni = kc A ( Yi ) s − (Yi )b (1) where, ni = rate of mass transfer of solute from the flake to the solvent, mass/time kc = external mass-transfer coefficient on a concentration driving force basis, A = area for mass transfer at the surface of the flake (Yi )s = concentration of solute in the solvent at the surface of the flake, mass solute/volume of solvent (Yi )b = concentration of solute in the bulk of the solvent, mass solute/volume of solvent Also, let, ( X i ) = concentration of solute in the solid, mass solute/volume of flake ( X i )o = initial concentration of solute in the solid, mass solute/volume of flake Because the mass-transfer resistance of the solid is negligible, values of ( X i ) stay uniform throughout the solid, but decrease with time, t. Let the equilibrium ratio for the solute concentration between t...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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