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

00018 mol cm3 6 300000 000002 mol cm3 chcl r chcl

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Unformatted text preview: -flow dialyzer, using a pure water sweep of 100 gal/h. Membrane of microporous cellophane with ε = 0.5, wet thickness, lM = 0.0051 cm, τ = 4.1, and pore diameter = dp = 31 Angstroms. Properties of Na2SO4 and A: Component Na2SO4 A Molecular weight 142 1000 Molecular diameter, Angstroms 5.5 15.0 Molecular diffusivity, cm2/s 0.77 x 10-5 0.25 x 10-5 Assumptions: Log-mean driving forces. Mass-transfer resistances on each side of the membrane are each 25% of the total mass-transfer resistance for each component. No transfer of water through the membrane. Find: Membrane area in m2 for a 10% transfer of A through the membrane. Corresponding percent transfer of Na2SO4. Analysis: For a concentration driving force, using Eqs. (14-13), (14-14), and (14-15) and taking into account restrictive diffusion because of the small pore size, the permeances are given by: P Mi For Na2SO4, PM Na SO = 2 4 εD d = i 1− M lM τ dp 0.5 0.77 × 10−5 0.0051(4.1) . 55 1− 31 4 (1) 4 = 8.43 × 10−5 cm / s 4 0.5(0.25 × 10 −5 ) 15.0 1− = 0.424 × 10 −5 cm / s 0.0051(4.1) 310 . Using Eq. (14-56), with the two film resistances together being equal to half the total resistance, the overall mass transfer coefficients are one-half of the permeances. Therefore, KNa 2SO 4 = 4.22 × 10 −5 cm / s or 1.52 × 10-3 m / h For A, PM A = KA = 0.212 × 10 −5 cm / s or 7.63 × 10-5 m / h Assume that the density of the feed is that of 14 wt% aqueous Na2SO4 at 20oC. From Perry's Handbook, the density = 1.13 g/cm3 or 1130 kg/m3 or 9.41 lb/gal. Therefore, the mass flowrate of the feed = 100(9.41) = 941 lb/h, comprised of 0.08(941) = 75.3 lb/h of Na2SO4, 56.5 lb/h of A, and 941 - 75.3 - 56.5 = 809.2 lb/h of water. The density of the sweep is 8.33 lb/gal. Exercise 14.13 (continued) Analysis: (continued) The sweep liquid enters at a flow rate of 100(8.33) = 833 lb/h of water. The permeate contains 10% of A in the feed or 0.10(56.5) = 5.7 lb/h. To obtain the concentrations, assume that 75% of the Na2SO4 in the feed is transferred through the membrane. Thus, the permeate contains 0.75(75.3) = 56.5 lb/h of Na2SO4. The retentate contains 56.5 - 5.7 = 50.8 lb/h of A and 75.3 - 56.5 = 18.8 lb/h of Na2SO4. Compute the concentrations in kg/m3. For example, the feed is 941 lb/h or 427 kg/h. Its volume is 427/1130 = 0.378 m3/h. Therefore, the concentration of Na2SO4 in the feed = 0.08(427)/0.378 = 90.4 kg/m3. Assume the density of the permeate is 1.07 g/cm3 or 1070 kg/m3. Assume the density of the retentate is 1.07 g/cm3 or 1070 kg/m3. The total permeate = 895.2 lb/h or 406.1 kg/h with a volume of 0.394 m3/h. The total retentate = 878.8 lb/h or 398.6 kg/h with a volume of 0.373 m3/h. The results of the concentration calculations are: Concentrations in kg/m3: Component Feed Sweep Retentate Permeate Na2SO4 90.4 0.0 22.9 65.0 A 67.8 0.0 61.8 6.6 67.8 − 6.6 − 618 − 0.0 . = 615 kg / m3 . 67.8 − 6.6 ln 618 − 0.0 . mA 5.7(0.4536) From a modification of Eq. (14-55), AM = = = 551 m2 −5 KA ∆cA 7.63 × 10 615 . The log-mean driving force for A is ∆cA = The log-mean driving force for Na2SO4 is ∆cA =...
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