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

Separation Process Principles 2n Seader& Henley Solutions Manual

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: the first increment of n, which takes n to 0.44 - 0.005 = 0.435, with an navg = (0.44 + 0.435)/2 =0.4375 Because the permeate side pressure is very low compared to the feed-side pressure, the initial composition of the permeate is given by simplifications of Eqs. (4) and (5): yA/yC = 50(xA/xC) = 50(0.1/0.7) = 7.14 and yB/yC = 50(xB/xC) = 50(0.2/0.7) = 14.29 (0) (0) (0) Substitution into Eqs. (6) to (8) gives yA = 0.3183, yB = 0.6372, yC = 0.0445 From Eqs. (1), (2), and (3), ( ( xA1) = xA0) + x (1) B =x ( 0) B ( ( yA0) − xA0) 0.3183 − 010 . ∆n = 0.10 + (-0.005) = 0.0975 navg 0.4375 ( ( yB0) − xB0) 0.6372 − 0.20 + ∆n = 0.20 + (-0.005) = 0.1950 navg 0.4375 ( ( ( xC1) = 1 − xA1) − xB1) = 1- 0.0975 - 0.1950 = 0.7075 Substitution into Eqs. (4) and (5) gives: yA yC yB yC (1) = α* AC ( ( xA1) PF − yA0) PP 0.0975(1000) - 0.3183(20) = 50 = 6.449 (1) ( 0) xC PF − yC PP 0.7075(1000) - 0.0445(20) =α ( ( xB1) PF − yB0) PP 0.1950(1000) - 0.6372(20) = 50 = 12.897 (1) ( 0) xC PF − yC PP 0.7075(1000) - 0.0445(20) (1) * BC Using Eqs. (6) to (8), 1 = 0.0491 1 + 6.449 + 12.897 = 0.04915(6.449) = 0.3170 ( yC1) = ( yA1) ( yB1) = 0.04915(12.897) = 0.6339 Exercise 14.21 (continued) Analysis: (continued) From Eq. (9), ∆AM = PM A ( yA1) ∆n 0.3170(0.005) = = 1,380 ft 2 -8 (1) (1) 1.26 × 10 [0.0975(1000) − 0.3170(20)] xA PF − yA PP Calculations for succeeding increments of n are given below from spreadsheet calculations. From the spreadsheet results, it is seen that when the mole fraction of H2S in the retentate reaches 0.0004, the mole fraction of CO2 is well below the specified 0.02. Thus, H2S controls. The mole fractions of the permeate in the spreadsheet table are the local values. The average composition of the final combined permeate is obtained by material balance from the feed gas and calculated pipeline gas (final retentate from the spreadsheet) with the following results: Component mol%: C H4 H2 S C O2 Total flow,lbmol/s Feed gas Pipeline gas 70 10 20 0.440 99.89 0.04 0.07 0.255 Permeate 28.80 23.73 47.47 0.185 From 10,000 scfm of feed gas, 1,000(0.255/0.44) = 580 scfm of pipeline gas is obtained. From the spreadsheet table, the final retentate pressure is very close to the required 980 psia. The required membrane area = 197,000 ft2. If each module contained 4,000 ft2, as mentioned for one case in the industrial example at the beginning of this chapter, then the number of modules in parallel = 197,000/4,000 = say 50 modules. This is not unreasonable. Analysis: (continued) Exercise 14.21 (continued) Exercise 14.22 Subject: Separation of ethyl acetate (EA) from water by pervaporation. Given: 100,000 gal/day of water containing 2.0 wt% EA at 30oC and 20 psia. Dense polydimethylsiloxane with a 1 µm-thick skin in a spiral-wound module. Permeate pressure = 3 cmHg. Membrane flux = 1.0 L/m2-h with a separation factor of 100 for EA with respect to H2O. Assumptions: Crossflow. Find: Membrane area in m2 and temperature drop of feed for a retentate of 0.2 wt% EA and a permeate of 45.7 wt% EA as a vapor at 3 cmHg pressure. Analysis: Use the Chemcad process simulator with the Wilson...
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