{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

# For each step in x eq 4 is used to calculate use eq 6

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: eously gives, R1 = 599.5 lb/h and P2 = 400.5 lb/h Dissolved solids in purified water, P2, equals [300/(1,000,000 + 300)](400.5) = 0.12 lb/h Dissolved solids in the concentrate (retentate from Stage 1) = 150 - 0.12 = 149.88 lb/h Compute material balance for Stage 2: Let P1 = permeate flow rate from Stage 1 = feed flow rate to Stage 2. Dissolved solids material balance: 0.004 P1 = 0.12 + 0.026 (P1 - 400.5) (3) Solving Eq. (3), P1 = 467.9 lb/h, containing 0.004(467.9) = 1.87 lb/h dissolved solids Retentate flow rate from Stage 2 = 467.9 - 400.5 = 67.4 lb/h, containing 1.87 - 0.12 = 1.75 lb/h dissolved solids The complete material balance is as follows: Component Dissolved solids Water Total Flow rate, lb/h: Fresh Combined feed feed to Stage 1 150 151.75 850 1,000 915.65 1,067.40 Retentate from Stage 1 149.88 Permeate from Stage 1 1.87 Retentate from Stage 2 1.75 Permeate from Stage 2 0.12 449.62 599.50 466.03 467.90 65.65 67.40 400.38 400.50 Exercise 14.18 (continued) Analysis: (continued) Now calculate the membrane area of each stage using Eq. (14-69). Assume osmotic pressure = π = C (wt% solids), with π = 1,700 psia for 25 wt% solids. Therefore, C = 1,700/25 = 68. For the feed, π = 68(151.75/1,067.40)100 = 967 psia For the retentate from Stage 1, π = 68(25) = 1,700 psia For the permeate from Stage 1, π = 68(0.4) = 27.2 psia For the retentate from Stage 2, π = 68(2.6) = 176.8 psia For the permeate from Stage 2, π = 68(0.12/400.50)100 = 2.0 psia nH 2 O nH 2 O From Eq. (14-69), AM = = PM H O ∆P − ∆π avg 0.0134 ∆P − ∆π avg 2 Stage 1: ∆P − ∆π avg = (1,756 - 15) - [(967 + 1,700)/2 - 27.2] = 1,741 - 1,306 = 435 psi AM = Stage 2: ∆P − ∆π avg 466.03 = 80 ft 2 0.0134(435) = (518 - 15) - [(27.2 + 176.8)/2 - 2.0] = 503 - 100 = 403 psi AM = 400.38 = 74 ft 2 0.0134(403) Exercise 14.19 Subject: Recovery of VOCs from air by gas permeation. Given: Feed gas of 1500 scfm (0oC, 1 atm) of air at 40oC and 1.2 atm, containing 0.5 mol% acetone (A). Pressure of permeate side = 5 cmHg. Membrane of thin-composite with 2µm-thick silicone rubber skin gives 4 barrer for air and 20,000 barrer for acetone. Assumptions: Crossflow. Find: Membrane area for retentate containing 0.05 mol% acetone and permeate containing 5 mol% acetone. Analysis: First complete the overall material balance. Feed flow rate = 1,500(60)/359 = 250.7 lbmol/h Acetone in the feed = 0.005(250.7) = 1.25 lbmol/h Air in the feed = 250.7 - 1.25 = 249.45 lbmol/h Overall total molar balance: F = 250.7 = R + P (1) Overall acetone molar balance: 1.25 = 0.0005 R + 0.05 P (2) Solving Eqs. (1) and (2) simultaneously, R = 228.0 lbmol/h and P = 22.7 lbmol/h The resulting component material balance is as follows: lbmol/hr: Component: Feed Retentate Permeate Acetone 1.25 0.11 1.14 Air 249.45 227.89 21.56 Total 250.70 228.00 22.70 Convert the given permeabilities in barrer to permeances in American Engineering units. 20,000 PM A = = 10 × 108 barrer / cm . −4 2 × 10 = 10 × 10 −2 cm3 (STP...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern