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

Exit humidity of water vapor to be 00009 lb h2olb dry

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Unformatted text preview: 2q 1.773 (8) . . . dt Let the volume of solids in the tank = 2.5 vol%. The solids-free volume of liquid in the tank = 2,200 m3. Therefore, the solids volume = (2.5/97.5)(2,200) =56.4 m3. Assume a particle density = 850 kg/m3. Therefore, S = 56.4 (850) = 48,000 kg = 4.8 x 107 g and Eq. (8) becomes, dq = 8.58 × 10 −7 − 6.5 × 10 −8 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 ) [ 8.58 × 10−7 − 6.5 × 10−8 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(7,770) = 77,700 minutes. Try a ∆t of 25,000 minutes. From the spreadsheet, the first 6 and last 6 values are: t, min. q, mg/g c*, mg/L c out c cum 0 0 0 0.00485 0 25,000 0.0215 1.86E-05 0.00487 0.00486 50,000 0.0429 6.34E-05 0.00491 0.00487 75,000 0.0643 0.000130 0.00498 0.00490 100,000 0.0858 0.000217 0.00506 0.00493 125,000 0.1072 0.000322 0.00516 0.00497 975,000 1,000,000 1,025,000 1,050,000 1,075,000 1,100,000 0.8208 0.8411 0.8614 0.8816 0.9017 0.9218 0.011887 0.012413 0.012948 0.013491 0.014043 0.014602 0.01648 0.01699 0.01752 0.01805 0.01859 0.01913 0.00910 0.00929 0.00949 0.00968 0.00988 0.01009 By interpolation, the run time is 1,090,000 minutes (757 days), which is greater than the above 77,700 minutes. Unfortunately, the vessel size is very large. Exercise 15.24 Subject: Use of three beds in a cycle to adsorb 1,2-dichloroethane from water, with application of the MTZ concept. Given: Three fixed-bed adsorbers, each containing 10,000 lb of activated carbon of ρb = 30 lb/ft3, to be used to reduce the concentration of 1,2-dichloroethane (D) in 250 gpm of water from 4.6 mg/L to 0.001 mg/L. Each bed has an H/D = 2. Two beds in series for adsorption (a lead bed and a trailing bed). One bed being regenerated by replacing the spent carbon. When the lead bed is saturated, it is regenerated, the trailing bed becomes the lead bed, and the regenerated bed becomes the trailing bed. The adsorption equilibrium isotherm is q = 8 c0.57, with q in mg/g and c in mg/L. Assumptions: Constant pattern front so that the width of the MTZ is a constant. Find: How often must the carbon in a bed be replaced. The maximum width of the MTZ to allow saturation of the lead bed. Analysis: Volume of each bed = V = mass of carbon/ρb = 10,000/30 = 333 ft3 For H = 2D, V = 333 = πD3/2. Solving, D = 6.0 ft and H = 2(6) = 12.0 ft. For ideal equilibrium adsorption from Eq. (15-92), the time for the ideal wave front to move through one of the three beds is tb = qF S QF cF (1) From the adsorption isotherm, the loading in equilbrium with the feed = qF = 8 cF0.57 = 8(4.6)0.57 = 19.1 mg/g. The grams of adsorbent in each bed = S = 10,000(453.6) = 4,536,000 g The volumetric...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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