13.2-1. |
Diffusion Through Liquids and a Membrane. A membrane process is being designed to recover solute A from a dilute solution where c1 = 2.0 × 10-2 kg mol A/m3 by dialysis through a membrane to a solution where c2 = 0.3 × 10-2. The membrane thickness is 1.59 × 10-5 m, the distribution coefficient K' = 0.75, DAB = 3.5 × 10-11 m2/s in the membrane, the mass-transfer coefficient in the dilute solution is kc1 = 3.5 × 10-5 m/s, and kc2 = 2.1 × 10-5.
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A1: | Ans. (a) Total resistance = 6.823 × 105 s/m, 11.17% resistance; (b) NA = 2.492 × 10-8 kg mol A/s · m2, area = 111.5 m2 |
13.2-2. |
Suitability of a Membrane for Hemodialysis. Experiments are being conducted to determine the suitability of a cellophane membrane 0.029 mm thick for use in an artificial-kidney device. In an experiment at 37°C using NaCl as the diffusing solute, the membrane separates two components containing stirred aqueous solutions of NaCl, where c1 = 1.0 × 10-4 g mol/cm3 (100 g mol/m3) and c2 = 5.0 × 10-7. The mass-transfer coefficients on either side of the membrane have been estimated as kc1 = kc2 = 5.24 × 10-5 m/s. Experimental data obtained gave a flux NA = 8.11 × 10-4 g mol NaCl/s · m2 at pseudo-steady-state conditions.
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13.3-1. |
Gas-Permeation Membrane for Oxygenation. To determine the suitability of silicone rubber for its use as a membrane for a heart-lung machine to oxygenate blood, an experimental value of the permeability at 30°C of oxygen was obtained, where PM = 6.50 × 10-7 cm3 O2 (STP)/(s · cm2 · cm Hg/mm).
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A3: | Ans. (b) 1.953 m2 |
13.4-1. |
Derivation of Equation for Permeate Concentration. Derive Eq. (13.4-11) for Case 2 for complete mixing. Note that xo from Eq. (13.4-9) must first be substituted into Eq. (13.4-6) before multiplying out the equation and solving for yp. |
13.4-2. |
Use of Complete-Mixing Model for Membrane Design. A membrane having a thickness of 2 × 10-3 cm, permeability = 400 × 10-10 cm3 (STP) · cm/(s · cm2 · cm Hg), and = 10 is to be used to separate a gas mixture of A and B. The feed flow rate is Lf = 2 × 103 cm3 (STP)/s and its composition is xf = 0.413. The feed-side pressure is 80 cm Hg and the permeate-side pressure is 20 cm Hg. The reject composition is to be xo = 0.30. Using the complete-mixing model, calculate the permeate composition, fraction of feed permeated, and membrane area. |
A5: | Ans. yp = 0.678 |
13.4-3. |
Design Using Complete-Mixing Model. A gaseous feed stream having a composition xf = 0.50 and a flow rate of 2 × 103 cm3 (STP)/s is to be separated in a membrane unit. The feed-side pressure is 40 cm Hg and the permeate is 10 cm Hg. The membrane has a thickness of 1.5 × 10-3 cm, permeability = 40 × 10-10 cm3 (STP) cm/(s · cm2 · cm Hg), and = 10. The fraction of feed permeated is 0.529.
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A6: | Ans. (a) Am = 5.153 × 107 cm2 (c) xoM = 0.2478 |
13.4-4. |
Effect of Permeabilities on Minimum Reject Concentration. For the conditions of Problem 13.4-2, xf = 0.413, = 10, pl = 20 cm Hg, ph = 80 cm Hg, and xo = 0.30. Calculate the minimum reject concentration for the following cases:
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13.4-5. |
Minimum Reject Concentration and Pressure Effect. For Example 13.4-2 for separation of air, do as follows:
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A8: | Ans. (b) xoM = 0.0624 |
13.5-1. |
Separation of Multicomponent Gas Mixtures. Using the same feed composition and flow rate, pressures, and membrane as in Example 13.5-1, do the following, using the complete-mixing model:
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13.5-2. |
Separation of Helium from Natural Gas. A typical composition of a natural gas (S1) is 0.5% He (A), 17.0% N2 (B), 76.5% CH4 (C), and 6.0% higher hydrocarbons (D). The membrane proposed to separate helium has a thickness of 2.54 × 10-3 cm, and the permeabilities are = 60 × 10-10 cm3 (STP) · cm/(s · cm2 · cm Hg), = 3.0 × 10-10, and = 1.5 × 10-10. It is assumed that the higher hydrocarbons are essentially nonpermeable ( ≅ 0). The feed flow rate is 2.0 × 105 cm3 (STP)/s. The feed pressure ph = 500 cm Hg and the permeate pressure pl = 20 cm Hg.
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13.6-1. |
Design Using Cross-Flow Model for Membrane. Use the same conditions for the separation of an air stream as given in Example 13.6-1. The given values are xf = 0.209, = 10, ph = 190 cm Hg, pl = 19 cm Hg, Lf = 1 × 106 cm3 (STP)/s, = 500 × 10-10 cm3 (STP) · cm/(s · cm2 · cm Hg), and t = 2.54 × 10-3 cm. Do as follows using the cross-flow model:
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A11: | Ans. (a) yp = 0.452, xo = 0.0303, Am = 6.94 × 108 cm2 (S6); (b) yp = 0.655, xo = 0.209 |
13.7-1. |
Equations for Cocurrent and Countercurrent Flow Models. Derive the equations for the following cases:
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13.7-2. |
Design Using Countercurrent-Flow Model for Membrane. Use the same conditions as given in Example 13.6-1 for the separation of an air stream. The given values are xf = 0.209, = 10, ph = 190 cm Hg, pl = 19 cm Hg, Lf = 1 × 106 cm3 (STP)/s, = 500 × 10-10 cm3 (STP) · cm/(s · cm2 · cm Hg), and t = 2.54 × 10-3 cm. Using the countercurrent-flow model, calculate yp, xo, and Am for θ = 0.40. (Note that this problem involves a trial-and-error procedure along with the numerical solution of two differential equations.) |
13.8-1. |
Effect of Pressure Drop on Asymmetric Membrane Calculations. Using the same conditions as in Example 13.8-1, do as follows:
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A14: | Ans. (a) NT = 1.267 × 105, (z2 − z1) = 1.413 m, Δph = 20.74 kPa; (b) Am = 54.10 m2; (c) Δph = 82.96 kPa |
13.8-2. |
Spreadsheet Calculation for Asymmetric Membrane. Using the conditions for Example 13.8-1, write the detailed spreadsheet and calculate the results. Compare these results with those in Table 13.8-2. |
13.8-3. |
Comparison of Experimental and Predicted Spreadsheet Results in a Pilot-Unit Asymmetric Membrane. A pilot-size membrane used to separate air in order to obtain nitrogen has the following dimensions with the feed inside the tubes: total NT = 3.8 × 104 fibers, ID = 95 μm, OD = 135 μm, length of fibers = 19 in. An experimental run gave the following results. Feed rate Lf = 1.086 m3/h, Lo = 0.654 m3/h, Vp = 0.450 m3/h, xo = 0.067 mole fraction oxygen, yp = 0.413, ph = 703.81 kPa, pl = 98.10 kPa, and T = 25°C. The permeances determined previously are (P'O2/t) = 1.50 × 10-10 m3 (STP)/(s · m2 · Pa), (P'N2/t) = 2.47 × 10-11. Flow is counter current. Do as follows:
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A16: | Ans. (a) Δph = 4.455 kPa; (b) xo = 0.079, Vp = 0.442 m3/hr, yp = 0.398 |
13.9-1. |
Osmotic Pressure of Salt and Sugar Solutions. Calculate the osmotic pressure of the following solutions at 25°C and compare with the experimental values:
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A17: | Ans. (a) π = 24.39 atm; (b) π = 0.0713 atm; (c) π = 0.768 atm |
13.9-2. |
Determination of Permeability Constants for Reverse Osmosis. A cellulose-acetate membrane with an area of 4.0 × 10-3 m2 is used at 25°C to determine the permeability constants for reverse osmosis of a feed salt solution containing 12.0 kg NaCl/m3 (ρ = 1005.5 kg/m3). The product solution has a concentration of 0.468 kg NaCl/m3 (ρ = 997.3 kg/m3). The measured product flow rate is 3.84 × 10-8 m3/s and the pressure difference used is 56.0 atm. Calculate the permeability constants and the solute rejection R. |
A18: | Ans. Aw = 2.013 × 10-4 kg solvent/s · m2 · atm, R = 0.961 |
13.9-3. |
Performance of a Laboratory Reverse-Osmosis Unit. A feed solution at 25°C contains 3500 mg NaCl/L (ρ = 999.5 kg/m3). The permeability constant Aw = 3.50 × 10-4 kg solvent/s · m2 · atm and As = 2.50 × 10-7 m/s. Using a ΔP = 35.50 atm, calculate the fluxes, solute rejection R, and product solution concentration in mg NaCl/L. Repeat, but using a feed solution of 3500 mg BaCl2/L. Use the same value of Aw, but As = 1.00 × 10-7 m/s (A1). |
13.10-1. |
Effect of Pressure on Performance of Reverse-Osmosis Unit. Using the same conditions and permeability constants as in Example 13.10-1, calculate the fluxes, solute rejection R, and product concentration c2 for ΔP pressures of 17.20, 27.20, and 37.20 atm. (Note: The values for 27.20 atm have already been calculated.) Plot the fluxes, R, and c2 versus the pressure. |
13.10-2. |
Effect of Concentration Polarization on Reverse Osmosis. Repeat Example 13.10-1 but use a concentration polarization of 1.5. (Note: The flux equations and the solute rejection R should be calculated using this new value of c1.) |
A21: | Ans. Nw = 1.171 × 10-2 kg solvent/s · m2, c2 = 0.1361 kg NaCl/m3 |
13.10-3. |
Performance of a Complete-Mixing Model for Reverse Osmosis. Use the same feed conditions and pressures given in Example 13.10-1. Assume that the cut or fraction recovered of the solvent water will be 0.10 instead of the very low water recovery assumed in Example 13.10-1. Hence, the concentrations of the entering feed solution and the exit feed will not be the same. The flow rate q2 of the permeate water solution is 100 gal/h. Calculate c1 and c2 in kg NaCl/m3 and the membrane area in m2. |
A22: | Ans. c1 = 2.767 kg/m3, c2 = 0.0973 kg/m3, area = 8.68 m2 |
13.11-1. |
Flux for Ultrafiltration. A solution containing 0.9 wt % protein is to undergo ultrafiltration using a pressure difference of 5 psi. The membrane permeability is Aw = 1.37 × 10-2 kg/s · m2 · atm. Assuming no effects of polarization, predict the flux in kg/s · m2 and in units of gal/ft2 · day, which are often used in industry. |
A23: | Ans. 9.88 gal/ft2 · day |
13.11-2. |
Time for Ultrafiltration Using Recirculation. It is desired to use ultrafiltration for 800 kg of a solution containing 0.05 wt % of a protein to obtain a solution of 1.10 wt %. The feed is recirculated past the membrane with a surface area of 9.90 m2. The permeability of the membrane is Aw = 2.50 × 10-2 kg/s · m2 · atm. Neglecting the effects of concentration polarization, if any, calculate the final amount of solution and the time to achieve this, using a pressure difference of 0.50 atm. |
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