Prausnitz Thermodynamics Notes 26

Click here to load reader

  • date post

    27-Oct-2014
  • Category

    Documents

  • view

    71
  • download

    6

Embed Size (px)

Transcript of Prausnitz Thermodynamics Notes 26

Advanced Chemical Engineering Thermodynamics

Class 25 Group Contribution Methods Chapters 6-7 and Appendix F, Prausnitz

GE and Activity Coefficients We demonstrated that

giE = RT ln [ i ] Now we need to remember thatGE = i ni giE

Thus

GE = RT i ni ln i

These are the most important relations for the chapter

Expressions for GEgE = 0 gE = 0 when when x1 = 0 x2 = 0

The two-suffix Margules equations were derived from: gE = A x1 x2 A more complex form for gE as a function of composition can be: gE = x1 x2 [ A + B (x1 - x2) + C (x1 - x2)2 + D (x1 - x2)3 + ] This is the Redlich-Kister expansion

Two-Suffix MargulesThe two-suffix Margules equations were derived from: gE = A x1 x2

ln 1 = A/RT x22 ln 2 = A/RT x12The activity coefficients at infinite dilution are:

ln 1 = A/RT ln 2 = A/RT

Wohls ExpansiongE / [RT (x1q1 + x2q2)] = 2 a12 z1z2 + 3 a112 z12z2 + 3 a122 z1z22 + 4 a1112 z12z2 + 4 a1222 z1z23 + 6 a1122 z12z22 + z1 = x1 q1 / [ x1 q1 + x2 q2 ] z2 = x2 q2 / [ x1 q1 + x2 q2 ]

The parameters q are effective volumes or cross sections of the molecules The parameters a are interaction parameters The contribution to gE of the interaction parameters a is weighted by the products that contain the factors z

Van Laar EquationgE / RT = 2 a12 x1 x2 q1 q2 / [ x1 q1 + x2 q2 ]We can calculate the fugacity coefficients of the two components

ln 1 = A / [ 1 + A/B x1/x2 ]2 ln 2 = B / [ 1 + B/A x2/x1 ]2And, at infinite dilution, we obtain

A = 2 q1 a12 B = 2 q2 a12

ln 1 = A ln 2 = B

Three-Suffix Margules EquationsgE / [RT (x1q1 + x2q2)] = 2 a12 z1z2 + 3 a112 z12z2 + 3 a122 z1z22 + 4 a1112 z12z2 + 4 a1222 z1z23 z1 = x1 q1 / [ x1 q1 + x2 q2 ] z2 = x2 q2 / [ x1 q1 + x2 q2 ]

ln 1 = A x22 + B x23 ln 2 = (A + 3/2 B) x12 - B x13 At infinite dilution ln 1 = A + B ln 2 = A + 1/2 B

UNIQUACgE / RT = [ gE / RT ]combinatorial + [ gE / RT ]residual For a binary mixture UNIQUAC states that: [gE/RT]combinatorial = x1 ln [1*/x1] + x2 ln [2*/x2] + z/2 { x1q1 ln [1/1*] + x2q2 ln [2/2*]} [gE/RT]residual = - x1 q1 ln [ 1 + 2 21 ] - x2 q2 ln [ 2 + 1 12 ] * is the segment fraction , and are area fractions

z is the coordination number In UNIQUAC z=10

UNIQUACSegment fractions 1* = x1 r1 / [ x1 r1 + x2 r2 ] 2* = x2 r2 / [ x1 r1 + x2 r2 ]

The parameters r are pure-component molecular structure constants

Area fractions 1 = x1 q1 / [ x1 q1 + x2 q2 ] 1 = x1 q1 / [ x1 q1 + x2 q2 ] 2 = x2 q2 / [ x1 q1 + x2 q2 ] 2 = x2 q2 / [ x1 q1 + x2 q2 ]

For fluids other than water and lower alcohols q = q

Energy parameters 12 = exp ( -u12 / RT ) exp ( -a12 / RT ) 21 = exp ( -u21 / RT ) exp ( -a21 / RT )These parameters are given in terms of characteristic energies

Parameters, from Table 6.9Compound CCl4 CH3OH C2H5OH H2O Toluene N-Hexane r 3.33 1.43 2.11 0.92 3.92 4.50 q 2.82 1.43 1.97 1.40 2.97 3.86 q 2.82 0.96 0.92 1.00 2.97 3.86 * * *

Binary Parameters, from Table 6.10System Acetone/Chloroform Acetone/Water Acetone/Methanol Ethanol/n-octane Ethanol/n-heptane Ethanol/Benzene Ethanol/CCl4 Formic acid/Water N-Hexane/Nitromethane T/K 323 331-368 323 348 323 350-369 340-351 374-380 318 a12/K -171.71 530.99 379.31 -123.57 -105.23 -75.13 -138.90 924.01 230.64 a21/K 93.93 -100.71 -108.42 1354.92 1380.30 242.53 947.20 -525.85 -5.86

UNQUAC Activity CoefficientsgE / RT = [ gE / RT ]combinatorial + [ gE / RT ]residual ln 1 = ln (1*/x1) + z/2 q1 ln (1/1*) + 2* (l1 - l2 r1/r2 ) - q1 ln ( 1 + 2 21 ) + 2q1 [ 21 / (1 + 2 21) - 12 / (2+112)] ln 2 = ln (2*/x2) + z/2 q2 ln (2/2*) + 1* (l2 - l1 r2/r1 ) - q2 ln ( 2 + 1 12 ) + 1q2 [ 12 / (2 + 1 12) - 21 / (1+212)]

l1 = z/2 (r1 - q1) - (r1 - 1)

l2 = z/2 (r2 - q2) - (r2 - 1)

CommentsIn all the methods we discussed, it is necessary to have some experimental data to obtain the fitting parameters for any binary mixture of interest

When experimental data are not available, an alternative approach is to look at the molecular structure of the compounds and to try obtaining the fitting parameters from the molecular structure of the interacting molecules

ExampleLets suppose we have one mixture of acetone and toluene

ExampleLets suppose we have one mixture of acetone and toluene To obtain the interaction parameters, we first look at the chemical structure of both compounds: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3

ExampleLets suppose we have one mixture of acetone and toluene Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3

ExampleLets suppose we have one mixture of acetone and toluene Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3

ExampleLets suppose we have one mixture of acetone and toluene Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3

ExampleLets suppose we have one mixture of acetone and toluene Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3

ExampleLets suppose we have one mixture of acetone and toluene Then we identify a number of groups, within these molecules, that we think are responsible for molecule-molecule interactions: Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3

ExampleLets suppose we have one mixture of acetone and toluene Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3

The basic assumption is that each group will behave in any mixture independently on the molecule it is part of

ExampleLets suppose we have one mixture of acetone and toluene Acetone CH3 C=O CH3 CH CH CH Toluene CH CH C CH3

The basic assumption is that each group will behave in any mixture independently on the molecule it is part of We obtain this contribution from available experimental data, and we predict the behavior of untested mixtures

UNIFACThe Universal Functional Activity Coefficient theory is the best known theory that uses this approach First introduced by Fredenslund, Jones, and Prausnitz in 1975 The main advantage of the procedure is that in typical mixtures of non electrolytes the number of group-group interactions is much less than the possible number of molecule-molecule pairs Another advantage is that correlations are always easier than the experiments they try to substitute Remember, however, that when no experimental data are used to improve the estimation of the interaction parameters, UNIFAC works, at best, as a good first approximation in VLE, but is often poor in LLE predictions

UNIFAC: The Methodln i = ln iC + ln iR

UNIFAC: The Methodln i = ln iC + ln iR

This is the combinatorial part, calculated as in the case of UNIQUAC

This is the residual contribution, which depends on group-group interactions

UNIFAC: The Methodln i = ln iC + ln iR

This is the combinatorial part, calculated as in the case of UNIQUAC ln iC = FC (x, , ) ln iR = FR (X, Q, T, amn)

This is the residual contribution, which depends on group-group interactions As defined for UNIQUAC

X = group mole fraction Q = group external surface area amn = interaction energy between groups n and m

Example: Fig. F4

methanol + water @ 50C Such good theory-experiment agreement is not common

Example: Fig. F5

n-hexane + methyl ethyl ketone @ 65 C This is more typical

Appendix FMany binary systems have been studied and the corresponding data have been published DIPPR-AIChE provides a continuously-improving compilation Unfortunately, many other systems have not been reported UNIFAC, or other group-contribution methods, are in general not very accurate n-hexane + methyl ethyl ketone P/bar UNIFAC Experiments Fig. F-5 L

V x/n-hexane

Appendix FLets suppose we can obtain only 2 experimental data point Which ones shall we get? P/bar L

V x/n-hexane

Appendix FLets suppose we can obtain only 2 experimental data point Which ones shall we get? P/bar L

V x/n-hexane For all binary mixtures, regardless of whether or not they form an azeotrope, binary parameters can be calculated from activity coefficients at infinite dilution These often provide the most valuable experimental information

Appendix F: ExampleLets suppose we decide to use any functional form for gE of the mixture of interest we like: gE = F (x, A, B) x = composition A,B the two parameters that depend on the specific F we like

Appendix F: ExampleLets suppose we decide to use any functional form for gE of the mixture of interest we like: gE = F (x, A, B) x = composition A,B the two parameters that depend on the specific F we like We saw that, depending on F, we can express the activity coefficient for both compounds as RT ln 1 = F1 (x, A, B) RT ln 2 = F2 (x, A, B) The specific F1 and F2 will depend on our choice for F

Appendix F: ExampleWe saw that, depending on F, we can express the activity coefficient for both compounds as RT ln 1 = F1 (x, A, B) RT ln 2 = F2 (x, A, B) A,B are the two parameters we need to implement F If we measure 1 and 2, thenx10

lim [F1 (x, A, B) / RT] = ln 1 lim [F2 (x, A, B) / RT] = ln 2 Which is a system of 2 equations in 2 unknowns

x20

Appendix F: ExampleVan Laars equations

ln 1 = A ln 2 = B

Appendix F: ExampleVan Laars equations

ln 1 = A ln 2 = BAnd we can predict the activity coefficients of both components at any composition:

ln 1 = A / [ 1