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

02 105 0002 c k l a cout c 4 the quantity c is the

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Unformatted text preview: volume of liquid in the tank = 16,300 m3. Therefore, the solids volume = (2.5/97.5)(16,300) =418 m3. Assume a particle density = 850 kg/m3. Therefore, S = 418 (850) = 355,000 kg = 3.55 x 108 g and Eq. (8) becomes, dq = 176 × 10−6 − 3.08 × 10 −10 q 1.773 . (9) dt Now, compute the run time to achieve a cumulative cout = 0.01 mg/L, where, t ccum = 0.01 = cout dt (10) t Eq. (9) represents an initial value problem in ODEs, which can be solved by the Euler method using a spreadsheet, if a small time step is taken. The Euler form, with j as the iteration index, is, 0 q ( j +1) = q ( j ) + ( ∆t ) [1.76 × 10 −6 − 3.08 × 10−10 q 1.773 ( j) (11) The total run time is specified to be not less than 10 times the average liquid residence time; thus, 10(86,100) = 861,000 minutes. Try a ∆t of 100,000 minutes. From the spreadsheet, the first 5 and last 5 values are: t, min. q, mg/g c*, mg/L c out c cum 0 0 0 0.00473 0 100,000 0.176 2.66E-05 0.00476 0.00474 200,000 0.352 9.09E-05 0.00482 0.00477 300,000 0.528 0.000187 0.00492 0.00480 400,000 0.704 0.000311 0.00504 0.00484 500,000 0.880 0.000462 0.00519 0.00490 3,200,000 3,300,000 3,400,000 3,500,000 3,600,000 5.625 5.800 5.975 6.151 6.326 0.012377 0.013069 0.013778 0.014502 0.015243 0.01709 0.01778 0.01849 0.01921 0.01995 0.00919 0.00944 0.00970 0.00996 0.01023 Thus, the run time is just more over 3,500,000 minutes (2,430 days), which is greater than the above 861,000 minutes. Unfortunately, the vessel size is enormous. Exercise 15.22 Subject: Comparison of batch, continuous, and semicontinuous modes of slurry adsorption for removal of benzene (B) and m-xylene (X) from water with activated carbon. Given: Feed of water at 25oC containing 0.324 mg/L of B and 0.630 mg/L of X. Activated carbon with an average particle diameter of 1.5 mm = 0.15 cm and the following Freundlich adsorption isotherms at 25oC: q =32 c0.428 for B and q =125 c0.333 for X, where q = mg adsorbed/g carbon and c = mg in solution/L solution. Sherwood number = 30. Particle external surface area = 5 m2/kg. Assumptions: Mass-transfer resistance in the pores is negligible compared to the external resistance. Find: For an effluent of 0.002 mg each of B and X, (a) Minimum amount of adsorbent needed per liter of feed solution. (b) For batch mode with twice the minimum amount of adsorbent, the contact time. (c) For continuous mode with twice the minimum amount of adsorbent, the required residence time. (d) For semicontinuous mode at a feed rate of 50 gpm and a liquid residence time of 1.5 times that in Part (c), the amount of activated carbon to give a reasonable vol% solids in the tank and run time of not less than 10 times the liquid residence time. Analysis: (a) The minimum amount of adsorbent corresponds to equilibrium with a solution of the effluent concentration of 0.002 mg of B or X, with the other solute at a lower concentration.. Must determine which solute controls. From the Freundlich equation for B, the equilibrium loading is given by: q = 32(0.002)0.428 = 2.24 mg B/g carbon. By mater...
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