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

3 for stage 2 0 x1 x2 t2y2 t2x2 20 eq 4 for

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Unformatted text preview: a spreadsheet with the results shown in the table on the next page. The perfectly mixed case is compared to the crossflow case in the following plot of the mole fraction of propylene in the permeate as a function of the cut. The results are similar to those shown in Fig. 14.10. For example at a cut of 60%, the mole fraction of propylene in the permeate is 0.75 for crossflow, while only 0.70 for perfect mixing. Analysis: (b) (continued) Results of crossflow: Exercise 14.10 (continued) Exercise 14.11 Subject: Material balance and membrane area for the separation of propylene and propane by gas permeation using two-stage membrane systems. Given: Gas feed of 100 lbmol/h of 60 mol% propylene (C3=) and 40 mol% propane (C3) at 25oC and 300 psia. Polyvinyltrimethylsilane polymer membrane with a 0.1µm skin in spiralwound modules. From Table 14.9, PM for C3= is 9 barrer, and 2.8 barrer for C3. Pressure on the permeate side = 15 psia. Assumptions: Perfect mixing on both sides of the membrane. No pressure drop on feed side or on the permeate side. Negligible film resistances on either side of the membrane. Equal membrane areas for the two stages. Find: Material balances and membrane areas in m2 for 40 lbmol/h of final retentate for: Case 1: Two-stage stripping cascade, as shown in Fig. 14.12a. Case 2: Two-stage enriching cascade, as shown in Fig. 14.12b. Analysis: Assume that the permeate-side pressure is 15 psia for each stage for each case. PM C3= 9 Ideal separation factor from Eq. (14-40) = α * = = 3.21 C3=,C3 = PM C3 2.8 Pressure ratio = PPiPF = 15/300 = 0.05. From Eq. (14-1) and the definition of a barrer in Section 14.1,, −10 PM C3= 9 1 × 10 PM C3= = = = 9 × 10 −5 cm3 (STP) / cm2 − s − cmHg lM 01 × 10−4 . αx y= Rearranging Eq. (14-36), 1 + (α − 1) x where, y and x refer to C3= and α = α C3=,C3 . (1) , x (α − 1) + 1 − rα x (α − 1) + 1 − 0.05α = 3.21 (2) x (α − 1) + 1 − r x (α − 1) + 1 − 0.05 For a given value of x, Eq. (2) is solved for α. Substituting this α into Eq. (1), y is obtained. From Eq. (14-43), α = α * 3=,C3 C Case 1: On the feed side, xR = x. On the permeate side, yP = y. From rearrangements of Eqs. (3) and (5) of Example 14.5, for each stage, x −x for C3=, θ = F (3) y−x yθnF and the membrane area for each stage is given by, AM = PM C3= xPF − yPP where, nF = combined feed, F, is in lbmol/h times (453.6)(22,400)/3600 to give cm3 (STP)/s Exercise 14.11 (continued) Analysis: Case 1 (continued) Pressures are PF = 300(76/14.7) = 1,551 cmHg, and PP = 15(76/14.7) = 77.6 cmHg yθnF (2820) (4) 9.0 × 10 x (1,551) − y (76.6) (10,000) Assume that the membrane areas for Stages 1 and 2 are equal. Let 1 refer to Stage 1 (Memb 1) and 2 refer to Stage 2 (Memb 2) in Fig. 14.12a. Let F = feed rate, R = retentate rate, and P = permeate rate, all in lbmol/h, for a stage. Let T = θ and xF1 refer to the combined feed for Stage 1. For x = xF = 0.60, Eq. (2) gives α = 3.05. Assume this value is constant for both stages. The governing equations then are as follows where R2 = 40 lbm...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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