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

# Assumptions crossflow find permeance for water and

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Unformatted text preview: .3 mol / h Therefore, the material balance in kmol/h is: Component Feed Sweep Retentate Permeate HCl 60 0 6.0 54.0 NaCl 30 0 6.7 23.3 H2 O 16,650 16,650 16,650.0 16,650.0 Total 16,740 16,650 16,662.7 16,727.3 Case 2 (95% transfer of HCl: Transfer of HCl to the permeate = 0.95(60) = 57 mol/h Neglect any transfer of water. Then the concentrations of HCl are: cHCl P = 57 / 300,000 = 0.00019 mol / cm3 = 3 / 300,000 = 0.00001 mol / cm3 cHCl R cHCl Sweep = 0.0 mol / cm3 Note that because the volumetric flow rates of feed and sweep are equal to each other and to the permeate and retentate, by material balance, the concentration driving forces for HCl and NaCl are equal at both ends and equal to ci F − ci P Therefore, ∆cHCl LM = cHCl F − cHCl P = 0.0002 - 0.00019 = 0.00001 mol/cm3 nHCl (57 / 60) = = 1, 730, 000 cm 2 or 173.0 m 2 K HCl ( ∆cHCl )LM 0.055(0.00001) Now determine the rate of permeation of NaCl. From Eq. (14.57, nNaCl = KNaCl AM ∆cNaCl LM = 0.021(1,730,000) ∆cNaCl LM = 36,300 ∆cNaCl LM (1) From Eq. (14-57), AM = But, nNaCl = cNaCl ∆cNaCl LM P = cNaCl 300,000 = 5,000 cNaCl P 60 − cNaCl P = 0.0001 - cNaCl F (2) P (3) Exercise 14.14 (continued) Analysis: Case 2 (continued) Substituting Eqs. (2) and (3) into (1), 5,000 cNaCl P = 36,300 0.0001 − cNaCl P Solving, cNaCl P = 0.0000879 and the transfer of NaCl = 30(0.0000879 / 0.0001) = 26.4 mol / h Therefore, the material balance in kmol/h is: Component Feed Sweep Retentate Permeate HCl 60 0 3.0 57.0 NaCl 30 0 3.6 26.4 H2 O 16,650 16,650 16,650.0 16,650.0 Total 16,740 16,650 16,656.6 16,733.4 Case 3 (98% transfer of HCl: Transfer of HCl to the permeate = 0.98(60) = 58.8 mol/h Neglect any transfer of water. Then the concentrations of HCl are: cHCl P = 58.8 / 300,000 = 0.000196 mol / cm3 = 12 / 300,000 = 0.000004 mol / cm3 . cHCl R cHCl Sweep = 0.0 mol / cm3 Note that because the volumetric flow rates of feed and sweep are equal to each other and to the permeate and retentate, by material balance, the concentration driving forces for HCl and NaCl are equal at both ends and equal to ci F − ci P Therefore, ∆cHCl LM = cHCl F − cHCl P = 0.0002 - 0.000196 = 0.000004 mol/cm3 nHCl (58.8 / 60) = = 4, 450, 000 cm 2 or 445.0 m 2 K HCl ( ∆cHCl )LM 0.055(0.000004) Now determine the rate of permeation of NaCl. From Eq. (14.57), nNaCl = KNaCl AM ∆cNaCl LM = 0.021(4,450,000) ∆cNaCl LM = 93,500 ∆cNaCl LM (1) From Eq. (14-57), AM = 300,000 = 5,000 cNaCl P P 60 = cNaCl F − cNaCl P = 0.0001 - cNaCl LM But, nNaCl = cNaCl ∆cNaCl Substituting Eqs. (2) and (3) into (1), 5,000 cNaCl P = 93,500 0.0001 − cNaCl (2) P (3) P Solving, cNaCl P = 0.0000949 and the transfer of NaCl = 30(0.0000949 / 0.0001) = 28.5 mol / h Therefore, the material balance in kmol/h is: Component Feed Sweep Retentate Permeate HCl 60 0 1.2 58.8 NaCl 30 0 1.5 28.5 H2 O 16,650 16,650 16,650.0 16,650.0 Total 16,740 16,650 16,652.7 16,737.3 Exercise 14.15 Subject: Desalinization of an aqueous solution by electrodialysis. Given: 86,000 gal/day of an aqueous solution of 3,000...
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