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

Either h2s or co2 will control the initial value of n

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Unformatted text preview: i Therefore, the permeances are: PM O = 150,000 1.70 × 10-10 = 2.55 × 10-5 lbmol / ft 2 - h - psi 2 Using the given selectivity, PM N = 7.29 × 10-6 lbmol / ft 2 - h - psi 2 Assume that α = αO2-N2 = a constant = 3.28 from above. Then, for crossflow with a binary mixture, we have the following relationships between oxygen mole fractions in the permeate and retentate as a function of the cut, θ = nP/nF. x − x R (1 − θ) 0.21 − x R (1 − θ) From material balance Eq. (1), Example 14.6, y P = F = (5) θ θ From Eq. (14-49), where the y and x pertain to O2, yP = x 1 1− α R 1− θ θ 1 − xR α α −1 xF 1 − xF α α −1 − xR Combine this equation with Eq. (5) to obtain: 0.21 1− θ = − x R + x R0.4386 0.1486 1 − x R α α −1 1.439 − = x R0.4386 1− θ 01486 1 − x R . θ 1.439 − x1.439 R (6) − x 1.439 R For O2, xR begins at 0.21 at the feed end and decreases toward the retentate end. Use Eq. (6) to compute values of θ as xR is stepped toward 0. Use Eq. (5) to compute the corresponding value of yP. For a differential molar transfer rate, dn, the differential membrane area required is given by Eq. (14-50), which is also given in incremental form for numerical calculation. If we base this calculation on O2, d AM = PM O 2 ydn , xPF − yPP ∆AM = 2.55 × 10 −5 y∆n x (500) − y (20) (9) where y is the local value given by Eq. (3) and ∆n is the incremental decrease in the feed flow. Exercise 14.20 (continued) Analysis: Company A (continued) Using the above equations, the calculations are carried out on a spreadsheet with results below. From the spreadsheet results, the membrane area is, by interpolation, 2,388 ft2 when the oxygen mole fraction in the permeate = 0.40. However, at that condition, the calculations show a retentate with an oxygen mole fraction of 0.128 and, therefore, a nitrogen mole fraction of 0.872. Thus, the ideal of 90 mol% N2 is not quite achieved. At that value, the oxygen content of the permeate drops to 37.4 mol% and the membrane area increases to 3,262 ft2. Exercise 14.21 Subject: Removal of CO2 and H2S from high-pressure sour natural gas by gas permeation to produce a pipeline gas. Given: Feed and product conditions: Stream Feed gas Pipeline gas Pressure, psia 1000 980 Composition, mol%: C H4 70 97.96 H2 S 10 0.04 C O2 20 2.00 Hollow-fiber membrane of 0.5 µm skin thickness with a permeability for CO2 of 13.3 barrer, and selectivities of CO2 to CH4 and of H2S to CH4 , both of 50. Assumptions: T = 60oF and permeate-side pressure = 20 psia. Crossflow. Find: Membrane area for feed flow rate of 10,000 scfm (0oC, and 1 atm). Analysis: The feed molar flow rate = 10,000 scfm/[(379 scf/lbmol)(60 s/min)] = 0.44 lbmol/s The membrane thickness = lM = 0.5 microns = 0.5 x 10-4 cm From Eq. (14-1), PM 13.3 PM CO2 = CO2 = = 266,000 barrer / cm lM 0.5 × 10 −4 = 2.66 × 10−5 cm3 (STP) / cm2 - s - cmHg Convert permeance to American Engineering units of lbmol/ft2-s-psi: 1 barrer/cm = 1 x 10-10 (30.48 cm/ft)2 (76 cmHg/14.696...
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