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

Total na2so4 in the crystals 500 220 720 kg therefore

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Unformatted text preview: h. This leaves 4,000 - 634 = 3,366 lb/h of water and 1,000 - 500 = 500 lb/h of sulfate in the mother liquor. The total flow rate of mother liquor is 3,366 + 500 = 3,866 lb/h. Thus, the concentration of sulfate in the mother liquor = 500/3,866 x 100% = 12.9 wt% Na2SO4. From the diagram below, the required temperature for this concentration at C in the mother liquor is 18oC. 500 Exercise 4.59 Subject: Dissolving crystals of Na2SO4 with water. Given: 500 kg of Na 2SO 4 ⋅ 10H 2 O crystals and 500 kg of Na2SO4 crystals at 20oC. Assumptions: Equilibrium according to phase diagram of Fig. 4.24. Find: Amount of water to dissolve the crystals at 20oC. Analysis: From Fig. 4.24, the solubility of Na2SO4 in water at 20oC is 15 wt% Na2SO4. Molecular weight of Na2SO4 = 142. Molecular weight of Na 2SO 4 ⋅ 10H 2 O = 322. Therefore, the kg of Na2SO4 in Na 2SO 4 ⋅ 10H 2 O = 500(142/322) = 220 kg. The water in Na 2SO 4 ⋅ 10H 2 O = 500 - 220 = 280 kg. Total Na2SO4 in the crystals = 500 + 220 = 720 kg. Therefore, need a total of 720(85/15) = 4,080 kg water. Additional water needed = 4,080 - 280 = 3,800 kg. Exercise 4.60 Subject: Adsorption of phenol (B) from an aqueous solution at 20oC with activated carbon. Given: One liter of aqueous solution containing 0.01 mol phenol. Freundlich isotherm equation for adsorption of phenol from aqueous solution by activated carbon at 20oC. Assumptions: Attainment of equilibrium. Find: Grams of activated carbon for (a) 75%, (b) 90%, and (c) 98% adsorption of phenol * 0.233 Analysis: At equilibrium, Eq. (1) in Example 4.12 is: q B = 2.16cB (1) * where, q B = mmol phenol adsorbed/g carbon cB = mmol phenol in solution/liter solution For each case, cB = 0.01(1,000 mmol/mol)(1-fraction adsorbed) mmol adsorbed = 0.01(1,000 mmol/mol)(1 liter)(fraction adsorbed) * g activated carbon needed = mmol adsorbed/ q B Using these equations, the results are as follows: * Case g activated cB at equilib, mmol phenol qB , adsorbed carbon mmol/L mmol/g (a) 75% adsorbed 2.5 7.5 2.67 2.81 (b) 90% adsorbed 1.0 9.0 2.16 4.17 (c) 98% adsorbed 0.2 9.8 1.48 6.62 Exercise 4.61 Subject: Adsorption of a colored substance (B) from an oil by clay particles at 25oC. Given: Oil with a color index of 200 units/100 kg oil. Adsorption equilibrium data Assumptions: One adsorption equilibrium contact. Find: (a) Freundlich equation for the adsorption equilibrium data. (b) kg clay to reduce color index to 20 units/100 kg oil for 500 kg of oil * 1/ Analysis: (a) From Eq. (4-30), the Freundlich equation is q B = AcB n , where here, * q B = color units/100 kg clay and cB = color units/100 kg oil By nonlinear regression of the data, A = 0.6733 and (1/n) = 0.5090 * The linearized form of the Freundlich equation is log q B = log A + (1 / n) log cB By linear regression of the data with the linearized form, A = 0.6853, (1/n) = 0.5050 The two results are close and both fit the data quite well. Use the nonlinear regression result: * 0.509 qB = 0.6733cB (b) (1) ( Q = 500 kg oil S = kg clay cBF ) = 200 units/kg oil Need, at equilibrium, cB = 20 units/kg oil * From Eq. (1), qB = 0.6733(20) 0.509 = 3.09 units/kg clay Apply Eq. (4-29 for a material balance on the color units: * q B = 3.09 = − Q Q 500 500 ( cB + cBF ) = − (20) + (200) S S S S Solving Eq. (2), S = 29,100 kg of clay (2) Exercise 4.62 Subject: Absorption o...
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