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

3300001 00416 lbmin and a concentration difference in

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Unformatted text preview: ppm of NaCl. Desalinization to 400 ppm by electrodialysis with a 40% conversion. 4 stages with 3 stacks of 150 cell pairs in each stage. Fractional desalinization is the same in each stage. Expected current efficiency = 90%. Applied voltage for first stage = 220 v. Each cell pair has an area = 1160 cm2. Find: Current density in mA/cm2, current in amperes, and power in kW for the first stage. Analysis: For the four stages, the overall fractional demineralization is: f = (3,000 - 400)/3,000 = 0.8667 For equal fractional demineralization in each stage, (1- f) = (1 - fs )4 Therefore, (1 - 0.8667) = (1 - fs )4 . Solving, (1 - fs ) = 0.6043 Thus, the salt concentration changes from stage to stage as follows: Stage Feed leaving 1 leaving 2 leaving 3 leaving 4 NaCl concentration, ppm 3,000 1,813 1,096 662 400 For a 50% conversion, the volumetric feed flow rate = Q = 80,000 (0.5)/[24(3,600)] = 0.463 gal/s or 0.00175 m3/s. Take the solution density = 1,000,000 g/m3. Then for Stage 1, the NaCl concentrations are: Feed: 3,000 ppm (assume by weight) or c = 3,000(1,000,000)/1,000,000 = 3,000 g/m3 Leaving Stage 1: 1,813 ppm or c = 1,813(1,000,000)/1,000,000 = 1,813 g/cm3 MW of NaCl = 58.5. Therefore, in mol, ∆c = (3,000 - 1,813)/58.5 = 20.3 mol/m3 or equiv/m3 Membrane area for Stage 1 = AM = (1,160 cm2/cell pair)(150 cell pairs/stack)(3 stacks/stage) = 522,000 cm2 or 52.2 m2 Use a rearrangement of Eq. (14-58) to compute the current density: i= zFQ (∆c) (1)(96, 520)(0.00175)(20.3) = = 73 amp/m 2 AM ξ (52.2)(0.9) Therefore, the current flow through a cell pair = I = iAM/n = (73)(52.5)/150 = 25.6 amp From Eq. (14-59), for a voltage of 220 volts across all cell pairs, the power = P = IE = 25.6(220) = 5,630 W = 5.6 kW Exercise 14.16 Subject: Reverse osmosis of sea water. Given: 30,000,000 gal/day of sea water at 20oC, containing 3.5 wt% dissolved solids. Permeate is 10,000,000 gal/day of water containing 500 ppm of dissolved solids. Retentate contains 5.25 wt% dissolved solids. Feed-side pressure = 2000 psia constant, and permeate pressure = 50 psia constant. One stage of spiral-wound membranes with total membrane area of 2,000,000 ft2. Assumptions: Crossflow. Find: Permeance for water, and salt passage for the dissolved solids. Analysis: First make an approximate calculation assuming perfect mixing on the permeate side and an arithmetic average of feed and retentate on the feed side. Rearranging Eq. (14-69), nH 2 O 10,000,000 PM H O = = (1) 2 AM ∆P − ∆π avg 2,000,000 ∆P − ∆π avg ∆P = 2,000 - 50 = 1,950 psi To estimate osmotic pressure, assume dissolved solids are NaCl (M = 58.5). Salt concentrations are approximately: Feed: 3.5(1,000)/[58.5(96.5)] = 0.620 mol/L Retentate: 5.25(1,000)/[58.5(94.75)] = 0.947 mol/L 500 (1,000z 0 Permeate: (100) = 0.00855 mol / L 1,000,000 58.5(99.95) From Eq. (14-68), for T = 293 K and mi = 2 mNaCl Feed: π = 1.12(293)(2)(0.620) = 407 psia Retentate: π = 1.12(293)(2)(0.947) = 622 psia Permeate: π = 1.12(293)(2)(0.00855) = 5.6 psia 407...
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