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

They are of the redlichkister form see walas s m phase

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: mptions: Modified Raoult's law, Eq. (4), Table 2.3 applies. Find: Dew point, bubble point, and equilibrium vapor and liquid at a temperature halfway between the bubble and dew points. Analysis: The Rachford-Rice flash equations can be used from Table 4.4: f {Ψ} = zi 1 − Ki =0 i =1 1 + Ψ Ki − 1 C zi Ki 1 + Ψ Ki − 1 zi xi = 1 + Ψ Ki − 1 yi = (1) (2) (3) where, zW = 0.5 and zA = 0.5 and the modified Raoult's law is: γ iL Pi s Ki = P (4) Antoine vapor pressure equations are given in Perry's Handbook for water and acetic acid: 1730.63 T ( C ) + 233.426 1936.01 log PAs = 8.02100 − o T ( C ) + 258.451 s log PW = 8.07131 − o (5) (6) The equations for the liquid-phase activity coefficients, given in the Chemical Engineering Science article of 1967 by Sebastiani and Lacquaniti, are incorrect. They are of the RedlichKister form (see Walas, S. M., "Phase Equilibria in Chemical Engineering", Butterworth, 1985, page 184) and should be: 2 log γ W = xA B + C 4 xW − 1 + C xW − xA 6 xW − 1 (7) 2 log γ A = xW B + C 4 xW − 3 + C xW − xA 6 xW − 5 (8) Exercise 4.33 (continued) Analysis: (continued) where, 64.24 T ( K) 43.27 B = 01735 − . T ( K) C = 01081 . A = 0.1182 + (9) (10) (11) Since P = 1 atm and the normal boiling points of water and acetic acid are 100oC and 118.1oC, respectively, it might be expected that the dew and bubble point of the mixture would be in the vicinity of 100oC, unless the liquid-phase activity coefficients are much different from 1. To check this, activity coefficients are computed from Eqs. (7) and (8) with a spread sheet at 100oC, with the following result as a plot. It is seen that in the vicinity of mole fractions equal to 0.5, the coefficients are not large, but are about 1.2. Exercise 4.33 (continued) Analysis: (continued) At the bubble point, Ψ = V/F = 0, and Eq. (1), combined with (4), becomes: f {T} = zW 1 − s γ W PW {T} γ P s {T} + zA 1 − A A =0 P P (12) Also, at the bubble point, xi = zi = 0.5. Then, the only unknown in Eq. (12) is T. Solving nonlinear Eq. (12), with Eqs. (5) to (11), by trial and error with a spreadsheet, starting from a guess of T = 100oC, quickly leads to a bubble-point temperature of 101.6oC. The composition of the vapor bubble is obtained from Eq. (2), which at the bubble point reduces to yi = xiKi = ziKi.. The K-values at the bubble point are computed to be KW = 1.364 and KA = 0.636, giving yW = 0.682 and yA = 0.318. At the dew point, Ψ = V/F =1, and Eq. (1), combined with (4), becomes: f{xW, xW, T} = zW P P − 1 + zA −1 = 0 s γ W {xW , xA , T} PW {T} γ A {xW , xA , T} PAs {T} where because yi = zi = 0.5, T , xW , and xA = (1 - xW) are left as unknowns. The liquid phase mole fractions are from Eq. (3), xi = zi /Ki . Solving these equations by trial and error with a spread sheet, starting from T = 105oC, xW = 0.4 and xA = 0.6, quickly leads to a dew-point temperature of 105.8oC. The K-values at the dew point are computed to be KW = 1.578 and KA = 0.732, with xW = 0.3169 and xA = 0.6832. The equilibrium flash calculation is carried out at T = (101.6 + 10...
View Full Document

Ask a homework question - tutors are online