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

# Find derive equation 16 24 analysis treat the

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Unformatted text preview: derflow depends on concentration of solutes/kg solution as follows (note the linear relationship): Solute Concentration, Retention of Solution, kg solutes/kg solution kg solution/kg TiO2 0.0 0.30 0.2 0.34 0.4 0.38 0.6 0.42 Analysis: Because the underflow basis is on total liquid (not just the water), use mass fractions to express the solutes composition in the overflow and in the liquid in the underflow. Let: yi = mass fraction of solutes in the overflow from Stage i xi = mass fraction of solutes in the underflow from Stage i Vi = kg/h of liquid in the overflow from Stage i Li = kg/h of liquid in the underflow from Stage i (a) First compute the overall mass balance: Final underflow from Stage N is 0.999(200,000) = 199,800 kg/h of TiO2 and 200 kg/h of solutes. Solution in the final underflow = 0.4(199,800) = 79,920 kg/h. Water in the final underflow = 79,920 – 200 = 79,720 kg/h. Therefore, xN = 200/79,920 = 0.00250. 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 – 79,720 = 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 79,720 439,760 Total 399,600 399,600 279,720 519,480 Exercise 16.7 (continued) Referring to Figure 16.7, all underflow liquid flow rates = L = 79,920 kg/h. All overflow liquid flow rates, except for the leaching stage = V = 399,600 kg/h. xL = yL = 79,720/519,480 = 0.153. Calculate the solute mass balance for just the leaching stage: 79,920 + y1V1 = 79,720 + xL (79,920) or, 79,920 + y1(399,600) = 79,720 + 0.153(79,920) Solving, y1 = 0.0301 x − y N +1 0.00250 − 0 log N log yL − y1 0.153 − 0.0301 = = 2.4 or 3 washing stages From (16-8), N= L 79,920 log log V 399, 600 Therefore, need 1 ideal leaching stage and 3 ideal washing stages. (b) Because the flow rate of liquid in the underflow depends on the solute concentration and, therefore, varies from stage to stage, we can not apply the McCabe-Thiele algebraic method used in Part (a). Instead, we must set up solute mass balances for each washing stage. The relationship between the retention of solution on the solids = Ri in kg solution/kg TiO2 and the mass fraction, xi, of solutes/kg solution is linear and we can develop the following equation from the given data in the above table: Ri = 0.2 xi + 0.30 (1) First compute the overall mass balance: Final underflow from Stage N is 0.999(200,000) = 199,800 kg/h of TiO2 and 200 kg/h of solutes. From Part (a), the final underflow will likely have a xN < 0.01. Therefore, let RN = 0.3. Therefore, solution in the final underflow = 0.3(199,800) = 59,940 kg/h. Water in the final underflow = 59,940 – 200 = 59,7...
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