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

# Total na2so4 in the crystals 500 220 720 kg therefore

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online