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

0002 0000196 0000004 molcm3 nhcl 588 60 4 450

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Unformatted text preview: From a modification of Eq. (14-55), AM = 90.4 − 65.0 − 22.9 − 0.0 = 24.2 kg / m3 90.4 − 65.0 ln 22.9 − 0.0 mNa 2SO 4 KNa 2SO 4 ∆cNa 2SO 4 = 56.5(0.4536) = 692 m2 −3 153 × 10 24.2 . Because the membrane area for Na2SO4 is larger than for A, the % transfer of Na2SO4 is less than the assumed 75%. Repeat the calculations assuming 70% transfer of Na2SO4, but use the same liquid densities. Now the permeate contains 0.7(75.3) = 52.7 lb/h Na2SO4 and the retentate contains 22.6 lb/h Na2SO4. The new concentrations are: Concentrations in kg/m3: Component Feed Sweep Retentate Permeate Na2SO4 90.4 0.0 28.1 65.0 A 67.8 0.0 63.1 7.0 3 -3 ∆c of Na2SO4 = 26.8 kg/m and membrane area = 52.7(0.4536)/[1.53x10 (26.8)] = 583 m2 ∆c of Na2SO4 = 62.0 kg/m3 and membrane area = 5.7(0.4536)/[7.63x10-5 (62.0)] = 547 m2 By extrapolation, membrane area = 545 m2 and the % transfer of Na2SO4 = 65%. Exercise 14.14 Subject: Removal of HCl from an aqueous solution by dialysis. Given: 300 L/h of an aqueous solution of 0.1 M NaCl and 0.1 M HCl. Sweep of 300 L/h of H2O. Microporous membrane for which lab experiments give the following overall masstransfer coefficients: Component K, cm/min H2 O 0.0025 NaCl 0.021 HCl 0.055 Assumptions: Log mean concentration driving force. Negligible change in volume flow rate of feed and sweep with permeation. Find: Membrane area in m2 and material balance for 90, 95, and 98% transfer of HCl. Analysis: Determine feed conditions, noting that it is very dilute in NaCl and HCl. Therefore, assume that it contains 55.5 mol of H2O per liter. For 300 L/h, nNaCl F = 0.1(300) = 30 mol / h nHCl F nH 2 O F = 0.2(300) = 60 mol / h = 55.5(300) = 16,700 mol / h The feed concentrations are: cNaCl F = 0.1 / 1000 = 0.0001 mol / cm3 cHCl F cH 2 O F = 0.2 / 1000 = 0.0002 mol / cm3 = 55.5 / 1000 = 0.0555 mol / cm3 The sweep flow rate = 55.5(300) = 16,700 mol/h of pure water. Case 1 (90% transfer of HCl: Transfer of HCl to the permeate = 0.9(60) = 54 mol/h Neglect any transfer of water. Then the concentrations of HCl are: cHCl P = 54 / 300,000 = 0.00018 mol / cm3 = 6 / 300,000 = 0.00002 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 - 0.00018 = 0.00002 mol/cm3 Exercise 14.14 (continued) Analysis: Case 1 (continued) nHCl (54 / 60) = From Eq. (14-57), AM = = 818, 000 cm 2 or 81.8 m 2 K HCl ( ∆cHCl )LM 0.055(0.00002) Now determine the rate of permeation of NaCl. From Eq. (14.57, nNaCl = KNaCl AM ∆cNaCl LM = 0.021(818,000) ∆cNaCl LM = 17,200 ∆cNaCl LM (1) 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 = 17,200 0.0001 − cNaCl (2) P (3) P Solving, cNaCl P = 0.0000775 and the transfer of NaCl = 30(0.0000775 / 0.0001) = 23...
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