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

004 0002 000402 0008 0004 000807 0014 0006 00142 0017

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: f acetone (A) from air by water at 1 atm. Given: y-x data for the distribution of acetone between air and water. Assumptions: Only acetone undergoes mass transfer so that there is no water in the gas phase and no air in the liquid phase. Find: (a) Plots of the equilibrium data as (1) Y-X , (2) p-X' '(in mass units) , and (3) y-x. (b) Compositions of the exiting streams when 20 moles of gas containing 0.015 mole fraction of acetone is brought to equilibrium with 15 moles of water. Analysis: (a) Equilibrium data are given in acetone mole fractions (y and x). To convert to acetone mole ratios (Y and X), acetone partial pressure (p), and acetone mass ratio in the liquid (X'), the following equations are used, with molecular weight of acetone = 58: y x 58 ' YA = A , XA = A , pA = yA P , XA = XA 1 − yA 1 − xA 18 Using these four equations, the following values are obtained from the given data: yA xA YA, mol A/mol air XA, mol A/mol water 0.004 0.002 0.00402 0.008 0.004 0.00807 0.014 0.006 0.0142 0.017 0.008 0.0173 0.019 0.010 0.0194 0.020 0.012 0.0204 Plots are given on the next page. 0.002004 0.00402 0.00604 0.00807 0.0101 0.01215 pA, atm 0.004 0.008 0.014 0.017 0.019 0.020 ' X A , g A/g water 0.00645 0.01294 0.01945 0.02599 0.0325 0.0391 (b) Solve by the intersection of the acetone material balance line with the equilibrium curve on the Y-X plot. Feed gas contains (0.015)(20) = 0.30 moles acetone and 20 - 0.3 = 19.7 moles air. The acetone material balance is: 0.30 = 19.7YA + 15 X A or by rearrangement , YA = −0.761 X A + 0.01523 (1) This equation is plotted on the Y-X plot on the next page and gives an intersection at YA = 0.0114 moles acetone/mole air and XA = 0.0051 moles acetone/mole water. The material balance is: Component: Air Water Acetone Gas in, moles 19.7 0.0 0.3 Liquid in, moles 0.0 15.0 0.0 Gas out, moles 19.7 0.0 0.224 Liquid out, moles 0.0 15.0 0.076 Analysis: (a) (continued) Exercise 4.62 (continued) Exercise 4.63 Subject: Separation of air into O2 and N2 by absorption into water followed by desorption. Given: Air containing 79 mol% N2 and 21 mol O2. Pressures ranging from 101.3 kPa (1 atm) to 10,130 kPa (100 atm), and temperatures ranging from 0 to 100oC. Assumptions: Applicability of Henry's law for solubility of H2 and O2 in water. Find: (a) A workable process. (b) Number of batch absorption steps to make 90 mol% pure O2 and the corresponding yield of O2. Analysis: Henry's law constants are given in Fig. 4.27 as (1/H) as function of temperature, where from Eq. (4-32), xi = (1/Hi)yiP with pressure in atm. Absorption is most efficient at high pressures, while desorption is best at low pressures. From Henry's law in this form, Ki = yi/xi = 1/[(1/Hi)P]. Using Fig. 4.27, the following are calculated: P = 100 atm, 25oC: P = 1 atm, 100oC: -1 -1 Ki Ki Component 1/Hi, atm αN2-O2 1/Hi, atm αN2-O2 -5 -6 Nitrogen 1.1 x 10 910 2.02 7.5 x 10 133,000 1.60 Oxygen 2.2 x 10-5 450 1.2 x 10-5 83,000 (a) A possible continuous scheme is the following, where air is compressed to 100 atm, followed by...
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