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

D fry chem eng education fall 1990 pp 204 207 the

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Unformatted text preview: eight of carbon bed for a time-to-breakthrough of 8 h, for the same conditions of flow rate, temperature, and pressure as the lab data. Analysis: Use a continuity equation, similar to Eq. (6-43) to compute the bed diameter: πD 2 Q (1) m = u s A ρ = Qρ o r A = = 4 us At operating conditions, Q = 12,000(560/520) = 12,900 ft3/min. us = 60 ft/min From Eq. (1), A = 12,900/60 = 215 ft2. D = [4(215)/3.14]1/2 = 16.5 ft. Use Eq. (15-122) to compute the ideal bed length, LES: cQt c (12,900)(8 × 60) c LES (in ft) = F F B = F = 960 F (2) q F ρb A q F (30)(215) qF Want cF in lb EA/ft3 inlet gas. MW of EA = 88.1. One scf of gas contains 379 lbmol. Therefore, have 12,000/379 = 31.66 lbmol/min. The mole fraction of EA = 0.005. Concentration of EA in the entering gas of 0.005(31.66)(88.1)/12,900 = 0.00108 lb EA/ft3 gas. From the given equilibrium data, for pEA = 0.005 atm, qF = 0.270 lb EA/lb C. From Eq. (2), 0.00108 LES = 960 = 385 ft . 0.270 From Eq. (15-120), the required bed length taking into account the MTZ is, LB = LES + LUB, where LUB is given by Eq. (15-121). The breakthrough data for Le = 2 ft are plotted on the next page as yEA out against time in minutes. The data are almost straight in the mole fraction range of 0.00475 (c/cF = 0.95), where time = te = 160 minutes, to 0.00025 (c/cF = 0.05), where time = tb = 75 minutes. Therefore, the midpoint time, ts = (160 + 75)/2 = 117.5 minutes. Alternatively, ts can be evaluated by integration, using Eq. (15-123). From Eq. (15-121), LUB = (ts - tb)Le/ts = (117.5 - 75)2/117.5 = 0.72 ft. Therefore, LB = 3.85 + 0.72 = 4.57 ft. So the bed is 16.5 ft in diameter by 4.57 ft high. This is a poor bed height-to-diameter ratio. Might increase the breakthrough time by a factor of 4 to get a bed height approximately equal to the bed diameter. Exercise 15.28 (continued) Analysis: (continued) 0.0050 Mole fraction of EA in gas effluent 0.0045 Lab Breakthrough Data 0.0040 te 0.0035 0.0030 0.0025 0.0020 0.0015 ts tb 0.0010 0.0005 0.0000 0 20 40 60 80 100 Time, minutes 120 140 160 180 Exercise 15.29 Subject: Desorption of benzene from silica gel using air under isothermal, isobaric conditions. Given: Values are given in a table below for dimensionless benzene concentration, φ, and loading, ψ, profiles in a bed following adsorption. Bed is 2 feet in diameter and 30 ft high. Bed contains silica gel adsorbent. Desorption by air at 1 atm and 145oF at an interstitial velocity, u, of 98.5 ft/min. Bed porosity = εb = 0.5. During desorption, mass-transfer coefficient = k = 0.206 min-1 and Henry's law equilibrium adsorption constant = 1,000 in Eq. (1) on p. 837. Assumptions: Plug flow with a constant interstitial velocity equal to the given inlet value. Negligible axial dispersion. Desorption air is passed up the bed, opposite in direction to the initial flow of benzene-air feed mixture during adsorption. Find: Desorption time required to remove 90% of the benzene from the bed. Analysis: First, compute the initial amount of benzene adsorbed in the bed. Bed volume = V...
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