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

# Analysis use a continuity equation similar to eq 6 43

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Unformatted text preview: 47. Therefore, the particle density 39 ρ from a rearrangement of Eq. (15-4) is ρ p = b = = 73.6 lb gel/ft3 of particles. 1 − ε b (1 − 0.47) When the partial pressure of H2O = 0.0277 atm, we have 0.1264/100 = 0.001264 lb H2O/ft3 gas. Therefore, K = 15(73.6)(0.0277)/0.001264 = 24,200. The equilibrium adsorption isotherm is now, q in lb H2O/ft3 silica gel particles = 24,200 c in lb H2O/ft3 gas From above, k = 1/[0.0292 K] = 1/[0.0292(24,200)] = 1.42 x 10-3 s-1 or 0.0849 min-1 From Eq. (2), k c = The Klinkenberg Eq. (15-114) is: where, from Eq. (15-115), ξ = c1 ≈ 1+ cF 2 kKz 1 − ε b u εb τ− ξ+ 1 8τ + 1 (3) 8ξ (4) u = actual (interstitial velocity) = superficial velocity/εb = 100/0.47 = 213 ft/min = 1.08 m/s 1 z 1 − 0.47 0.0292 From Eq. (4), ξ = (5) = 35.8z (in meters) 108 . 0.47 Exercise 15.26 (continued) Analysis: (continued) From Eq. (15-116), z z τ=k t− = 0.0849 t − = 0.0849t (in minutes) − 0.00131z (in meters) (6) u 1.08(60) The maximum value of z, the distance through the bed, is the bed height = 5 ft = 1.524 m. At breakthrough, c/cF = 0.05 at z = 1.524 m and, therefore, ξ = 35.8(1.524) = 54.6 and τ = 0.0849t - 0.00131(1.524) = 0.0849t (in minutes) - 0.00200 Must solve Eq. (1) for t at c/cF = 0.05 and ξ= 54.6. The time will be less than the ideal equilibrium time above of 679 minutes. Can do this by trial and error with a spreadsheet or with a nonlinear solver that can handle the error function, erf. From a spreadsheet, breakthrough time = 451 minutes. To obtain the width of the MTZ, solve Eq. (1) for z at c/cF = 0.95 and t = 451 minutes. By trial and error from a spreadsheet, at breakthrough, the MTZ extends from z = 0.718 m to z =1.524 m. Therefore, the width of the MTZ = 1.524 - 0.718 = 0.81 m = 2.6 ft. Breakthrough curves similar to Figs. 15.30 and 15.31, but in terms of distance, z, through the bed and time, instead of the dimensionless groups, ξ and τ , are shown on the next two pages. They are computed with a spreadsheet from Eq. (3) above for gas concentration and Eq. (15-119) for adsorbent loading, together with Eqs. (5) and (6) to convert from ξ and τ to z and t, respectively. With the spreadsheet, the error function, erf, must be used carefully, noting that erf(-x) = -erf(x). Thus, when the argument of the error function is negative, the argument should be made positive and the result should be multiplied by -1. To compute the average loading of the bed at the breakthrough time of 451 minutes, use the equation in Example 15.11: 1.524 qavg = qdz , where z is in meters. (7) 1.524 A spreadsheet is used with Eq. (15-119) to compute q as a function of z for t = 451 minutes 0 (breakthrough time) and q * = 0.440 lb H2O/lb silica gel from above (value in equilibrium with F the feed. The results are: q , lb/lb 0.440 0.440 0.433 0.382 0.260 0.125 0.042 0.018 z, meters 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.524 Solving Eq. (7) numerically by the trapezoid method, qavg = [0.440(0.2) + 0.440(0.2) + 0.4365(0.2) + 0.4075(0.2) + 0.321(0.2) + 0.1925(0.2) + 0.0835(0.2)...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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