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

Exercise 1530 subject pressure swing adsorption

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Unformatted text preview: = πD2L/4 = 3.14(2)2(30)/4 = 94.2 ft3 Volume of SG particles = (1-εb)V = (1-0.5)(94.2 = 47.1 ft3 During adsorption at 70oF, from Example 15.10, the concentration of benzene in the entering airbenzene mixture = 0.00181 lb benzene/ft3 = cF. The adsorbent loading in equilibrium with the feed, as given in Example 15.10, is qF* = 5,120 cF = 5,120(0.00181) = 9.27 lb benzene/ft3 of silica gel particles. If the bed were completely saturated with the feed, the benzene in the bed would be 9.27(47.1) = 437 lb benzene. If the breakthrough occurred for c just greater than zero at the given 641 minutes for adsorption (i.e. all of the benzene in the feed were adsorbed), then with a feed flow rate, from Example 15.10, of 310 ft3/min, the benzene loading would be: 310(641)(0.00181) = 360 lb benzene, which, as expected, is less than complete saturation. However, the breakthrough at 641 minutes corresponds to φ = c/cF = 0.05 and not at 0. Therefore, it is best to integrate the given loading data at the end of adsorption to obtain a more accurate value of benzene loading. Using the given data on the next page, compute the average value of ψ by numerical integration. Note that because the flow of desorption agent is opposite to the feed gas, the bed distances below are given as z' = 30 ft - z. ψ avg = 0.5ψ z '= 0 + 29 z ' =1 ψ at z ' + 0.5ψ z '= 30 30 = 0.022 + 24.079 + 0.500 = 0.820 30 Therefore, the amount of benzene on the adsorbent in the bed at the end of the adsorption period is: 0.820(9.27)(47.1) = 358.0 lb benzene, which is very close to the above rough estimate of 360. The desorption period must be long enough to reduce the loading to (1-0.90)(358) = 35.8 lb benzene. Exercise 15.29 (continued) Analysis: (continued) Given conditions at the beginning of the desorption period: z, ft z' = 30 - z, ft φ = c/cF ψ = q/ qF* 30 0 0.050 0.044 29 1 0.090 0.081 28 2 0.150 0.137 27 3 0.235 0.217 26 4 0.343 0.321 25 5 0.468 0.444 24 6 0.599 0.575 23 7 0.722 0.701 22 8 0.825 0.808 21 9 0.901 0.890 20 10 0.951 0.944 19 11 0.978 0.975 18 12 0.992 0.990 17 13 0.997 0.997 16 14 0.999 0.999 15 15 1.000 1.000 14 to 0 16 to 30 1.000 1.000 The desorption calculations utilize the following forms of Eqs. (15-123) and (15-124): 1 − εb 1 − 0.5 ∂φ ∂φ ∂φ = −u − (0.206)(1,000) φ − ψ kK φ − ψ = −98.5 − 0.5 ∂t ∂z ' εb ∂z ' = −98.5 ∂φ − 206 φ − ψ ∂z ' ∂ψ = 0.206 φ − ψ ∂t (1) (2) Conditions for φ and ψ at t = 0 are given as a function of z' in the above table. Condition for φ at z' = 0 is 0 because the desorption gas contains no benzene. Equations (1) and (2) can be solved for φ{t,z'} and ψ{t,z'} for z' = 0 to 30 ft and t > 0 by various means, including: (1) A FORTRAN code similar to that on for Example 15-13, with the method of lines and a stiff integrator in connection with Eqs. (15-127) to (15-133). (2) Use of the windows code PDESOL (www.pdesol.com), which also utilizes the method lines ∂φ in connection with a number of different integrators an...
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