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

# Assumptions assume membrane is 1 micron thick and

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Unformatted text preview: Pa: From Section 14.1, 1 barrer/cm = 10-10 cm3(STP)/cm2-s-cmHg Conversion factors: 22.42 x 106 cm3(STP)/kmol, 10-4 m2/cm2, 3600 s/h, and 1.333 kPa/cmHg 1 × 10−10 (3600) kmol Therefore, 1 barrer / cm = = 1205 × 10−10 2 . 6 −4 m - h - kPa 22.42 × 10 10 1333 . kmol N2 m - h - kPa kmol PM = 10,000(1.205 × 10 −10 ) = 1.21 × 10 −6 2 CH 4 m - h - kPa Let x = kmol/h of CH4 in permeate. Then 800 - x = kmol/h of CH4 in retentate. nN 180 N N2 = 2 = = 6.03 × 10−6 ∆PN 2 (1) avg AM A M PM = 50,000(1205 × 10 −10 ) = 6.03 × 10−6 . N CH 4 = nCH 4 AM = x = 121 × 10 −6 ∆PCH 4 . AM 0.2(5500) + 5450 ∆PN 2 avg = 2 20 820 − x − 100 2 0.8(5500) + 5450 (2) avg 800 − x 820 − x 180 180 + x (3) x (4) avg 2 180 + x Combining Eqs. (1) and (3), and (2) and (4) and solving the resulting two nonlinear equations, x = 281 kmol/h of CH4 in the permeate and the membrane area = 48,800 m2. ∆PN 2 = 612 kPa and ∆PCH 4 = 4760 kPa ∆PCH 4 = avg avg − 100 Exercise 14.4 Subject: Hollow-fiber module Given: Hollow fibers, each 42 microns i.d., 85 microns o.d., and 1.2 m long in a membrane module with 4,000 ft2 of membrane surface area based on the i.d. Assumptions: Module is a right circular cylinder in shape. Find: (a) Number of hollow fibers in the module. (b) Diameter of the module if fibers on a square spacing of 120 microns center to center. (c) Membrane surface area per unit volume of module (packing density) in m2/m3. Compare results to Table 14.4. Analysis: (a) For one fiber, the inside surface area = Ai = πDiL = 3.14(42/106)(1.2) = 0.000158 m2 or 0.000158/(9.29 x 10-2) = 0.001704 ft2. Therefore, number of fibers = N = 4000/0.001704 = 2,350,000 fibers. (b) For square spacing of 120 microns center to center, each fiber occupies a crosssectional area of 120 by 120 microns or 120(120)/1012 = 14.4 x 10-9 m2. For 2,350,000 fibers, cross-sectional area of the membrane module = 2,350,000(14.4 x 10-9) = 0.0338 m2 = 338 cm2. If the area is circular, the diameter of the module = [4(338)/3.14]1/2 = 20.8 cm. (c) The volume of the module for a length of 1.2 m = 0.0338(1.2) = 0.0406 m3. The packing density = membrane surface area based on the inside diameter/unit module volume. Thus, the packing density = 0.000158(2,350,000)/0.0406 = 9,150 m2/m3. Table 14.4 gives a range of 500 to 9,000 m2/m3 for a hollow-fiber module. Exercise 14.5 Subject: Geometry of a spiral-wound membrane module. Given: Module of 0.3 m diameter (D) and 3 m length (L). Packing density = 500 m2/m3. Assumptions: Collection tube of 1 cm in diameter. Find: Center to center spacing of membrane in the spiral. Analysis: Membrane volume = V = πD2L/4 = 3.14(0.3)23/4 = 0.212 m3. Membrane area = membrane volume x packing density = 0.212(500) = 106 m2. Length around the spiral = 106/3 = 35.3 m Circumference of outer circle = πD = 3.14(0.3) = 0.943 m Circumference of inner circle = 3.14(0.01) = 0.0314 m Average circumference = (0.943 + 0.0314)/2 = 0.487 m Number of turns of the spiral = length around spiral/aver. cir...
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