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

0277 atm we have 01264100 0001264 lb h2oft3 gas

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Unformatted text preview: feed rate = QF = 250(3.785) = 946 L/min, and cF = 4.6 mg/L Therefore, from Eq. (1), tb = 19.1(4,536,000) = 19,900 minutes = 332 hours = 13.8 days 946(4.6) The lead bed must be replaced every 13.8 days. The width of the MTZ zone can not exceed the height of a bed or 12 ft. Otherwise, breakthrough will occur before the trailing bed becomes the lead bed. Exercise 15.25 Subject: Use of three or more beds in a cycle to adsorb benzene and m-xylene 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 remove 0.185 mg/L of benzene (B) and 0.583 mg/L of m-xylene (X) from 250 gpm of water. Each bed has an H/D = 2. Two or more beds in series for adsorption (at least 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 isotherms are: qB = 32 cB0.428 and qX = 125 cX0.333, with q in mg/g and c in mg/L. Widths of the mass-transfer zones are MTZB = 2.5 ft and MTZX = 4.8 ft. Assumptions: Constant pattern front so that the widths of the MTZs are constant. Find: How often must the carbon in a bed be replaced. 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 qS through one of the three beds for one of the solutes is: tb = F (1) QF cF From the adsorption isotherm of B. the loading in equilbrium with the feed = qF = 32 cF0.428 = 32(0.185)0.428 = 15.5 mg/g. From the adsorption isotherm of X. the loading in equilbrium with the feed = qF = 125 cF0.333 = 125(0.583)0.333 = 104.4 mg/g. The grams of adsorbent in each bed = S = 10,000(453.6) = 4,536,000 g The volumetric feed rate = QF = 250(3.785) = 946 L/min. For benzene, with cF = 0.185 mg/L, and 15.5(4,536,000) from Eq. (1), tb = = 402,000 minutes = 6,700 hours = 279 days 946(0185) . For m-xylene, with cF = 0.583 mg/L, and 104.4(4,536,000) from Eq. (1), tb = = 859,000 minutes = 14,300 hours = 596 days 946(0.583) So, benzene controls. A satisfactory bed arrangement is to use two beds in series for adsorption with a time of 279 days to change the beds. Because the MTZs are less than the bed height of 12 ft, no breakthrough will occur. The lead bed will become saturated with benzene but not with mxylene. Exercise 15.26 Subject: Comparison of equilibrium model to mass-transfer model for fixed-bed adsorption of water vapor from air with silica gel. Given: Feed of air with a relative humidity of 80% at 80oF and 1 atm, with a superficial velocity of 100 ft/min in a fixed bed of 2.8-mm-diameter spherical particles of silica gel (ρb = 39 lb/ft3). Bed height = H = 5 ft. Linear adsorption isotherm for water vapor = q = 15.9 p, where q is in lb H2O/lb gel, a...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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