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

Raoults law is badly in error when compared to the

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Unformatted text preview: the given vapor pressure data are fitted to Antoine equations, we obtain: xA = Exercise 4.15 (continued) Analysis: (e) continued PAs = exp 18.4854 − 392196 . T + 230.91 (7) PBs = exp 25.0173 − 8010.6 T + 353.238 (8) Where vapor pressure is in torr and temperature is in oC. Solving, Eqs. (1) to (8), T, oC 82.5 84.0 86.0 88.0 90.0 92.0 94.0 96.0 98.0 100.0 Ps o f B Ps o f A KB KA xB yB αB-A 760.0 392.1 1.0000 0.5159 1.000 1.000 1.938 809.5 416.2 1.0651 0.5476 0.874 0.931 1.945 879.9 450.2 1.1578 0.5924 0.721 0.835 1.954 955.7 486.6 1.2575 0.6402 0.583 0.733 1.964 1037.3 525.3 1.3649 0.6912 0.458 0.626 1.975 1125.0 566.6 1.4803 0.7456 0.346 0.513 1.985 1219.3 610.6 1.6043 0.8035 0.245 0.394 1.997 1320.5 657.4 1.7375 0.8650 0.155 0.269 2.009 1429.1 707.2 1.8804 0.9305 0.073 0.138 2.021 1545.6 760.0 2.0336 1.0000 0.000 0.000 2.034 These results are plotted below. Raoult’s law is badly in error when compared to the experimental data. For part (b), Raoult's law predicts a bubble-point vapor with an isopropanol mole fraction of 0.56. By coincidence, this compares well with the result determined with the experimental data. For part (c), however, Raoult's law predicts isopropanol mole fractions of 0.28 for the liquid and 0.43 for the vapor. These are drastically different from the values of 0.14 and 0.50, respectively from the experimental data. Raoult's law can not be used for the isopropanol-water system, for which it also fails to predict an azeotrope. Exercise 4.15 (continued) Analysis: (e) continued Exercise 4.16 Subject: Vaporization of mixtures of n-hexane (H) and n-octane (C) at 1 atm Given: T-x-y diagram in Fig. 4.3, and y-x diagram in Fig. 4.4. 100 kmol mixture. Find: Temperature, kmol of vapor, mole fractions of H in liquid and vapor at equilibrium for various flash conditions. Analysis: Let zH = mole fraction of n-hexane in the feed and Ψ = V/F. Use inverse lever-arm rule as displayed by Line DEF in Fig. 4.3. The results for parts (a) through (f) are as follows: Given (a) zH = 0.5, Ψ = 0.2 (b) zH = 0.4, yH = 0.6 (c) zH = 0.6, xC = 0.7 (d) zH = 0.5, Ψ = 0.0 (e) zH = 0.5, Ψ = 1.0 (f) zH = 0.5, T = 200oF T, o F 196 220 210 188 230 200 V, kmol 20 48.6 73.7 0.0 100 31 yH 0.80 0.60 0.70 0.84 0.50 0.77 xH 0.43 0.21 0.30 0.50 0.14 0.38 Exercise 4.17 Subject: Derivation of equilibrium flash equations for a binary mixture (1, 2). Given: Eqs. (5), (6), and (3) of Table 4.4. Find: Derive given equations for x1, x2, y1, y2, and Ψ = V/F. Analysis: First derive the equation for Ψ = V/F. From Eq. (3), Table 4.4, z1 1 − K1 1 − z1 1 − K2 + =0 1 + Ψ K1 − 1 1 + Ψ K2 − 1 (1) Solving Eq. (1) for Ψ, and simplifying, Ψ= z1 K1 − K2 / 1 − K2 − 1 − z1 1 − K1 − 1 − z1 1 − K2 = z1 1 − K1 K2 − 1 + 1 − z1 1 − K2 K1 − 1 K1 − 1 (3) Substituting Eq. (3) into Eq. (5) of Table 4.4 and simplifying gives the required equation for x1. Then use y1 = K1x1 and simplify, followed by x2 = 1 - x1 and y2 = 1- y1 . Exercise 4.18 Subject: Conditions for R...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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