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

Assumptions assume membrane is 1 micron thick and

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online