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

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Unformatted text preview: drop at the end of stage 1 = 25 psia = 172,400 Pa Rearrange (14-103) to compute the time, t, in seconds to reach the pressure drop of 25 psia, using SI units. t= ∆P Rm 172, 400 1.43 × 1010 − = − J 2µK 2 cF K 2 JcF ( 9.63 × 10−5 )2 ( 0.001) ( 3.78 × 1012 ) ( 4.3) ( 3.78 × 1012 )( 9.63 ×10 −5 ) ( 4.3) = 1,138 – 9 = 1,129 s = 18.82 min During stage 1, the cumulative volume, mL, as a function of time in minutes is given by, V (in mL) = 10 t (in minutes) up to 18.82 minutes Exercise 14.27 (continued) Cumulative Permeate Volume, mL Analysis: (continued) For the stage 2 with operation at constant pressure drop of 25 psia, (14-110) is used to compute the cumulative permeate volume V, starting from VCF = volume at the end of stage 1 = 10(18.82) = 188.2 mL and tCF = 18.82 minutes = 1,129 s. The decrease in permeate flux, J, in stage 2 is obtained from (14-111). The calculations are best carried out with a spreadsheet. The results shown in the following two plots show that stage 2 ends at 2,850 seconds or 47.5 minutes. 400 350 300 250 200 150 100 50 0 0 10 20 30 40 50 Time, minutes Permeate Flux, mL/min-sq cm 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00 0 5 10 15 20 25 30 Time, minutes 35 40 45 50 Exercise 15.1 Subject: Estimation of adsorbent characteristics. Given: Porous particles of activated alumina. BET area = 310 m2/g = Sg. Particle porosity = 0.48 = εp. Particle density = 1.30 g/cm3 = ρp. Assumptions: Straight pores of circular cross-section with uniform diameter. Find: (a) Vp = specific pore volume in cm3/g. (b) ρs = true solid density, g/cm3. (c) dp = average pore diameter, angstoms. Analysis: (a) From Eq. (15-3), Vp = εp/ρp = 0.48/1.30 = 0.369 cm3/g (b) From a rearrangement of Eq. (15-5), ρs = ρp/(1 - εp) = 1.30/(1 - 0.48) = 2.50 g/cm3 (c) From a rearrangement of Eq. (15-2), dp = 4 εp/Sgρg = 4(0.48)/[310 x 104(1.30)] = 4.76 x 10-7 cm or 47.6 angstroms. Subject: Exercise 15.2 Estimation of surface area of two forms of molecular sieves. Given: Form A with Vp = 0.18 cm3/g and average dp = 5 angstroms. Form B with Vp = 0.38 cm3/g and average dp = 2.0 microns. Assumptions: Straight pores of circular cross-section with uniform diameter. Find: Surface area, Sg, of each form. Analysis: By substitution of Eq. (15-3) into (15-2), Sg = 4 Vg/dp (1) Form A: From Eq. (1), Sg = 4(0.18)/(5 x 10-8) = 14.4 x 106 cm2/g or 1,440 m2/g Form B: From Eq. (1), Sg = 4(0.38)/(2 x 10-4) = 7,600 cm2/g or 0.760 m2/g This is a very large difference in surface area. Exercise 15.3 Subject: Consistency of the characteristics of a small-pore silica gel. Given: Small-pore silica gel. Pore diameter = 24 angstroms = dp. Particle porosity = 0.47 = εp. Particle density = 1.09 g/cm3 = ρp. Specific surface area = 800 m2/g = Sg. Assumption: Straight pores of circular cross-section with uniform diameter. Find: (a) Reasonableness of the above values. (b) Fraction of a monolayer adsorbed if the adsorption capacity for water vapor at 25oC and 6 mmHg partial pressure is 18% by weight. Anal...
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