Prausnitz Thermodynamics Notes 26

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Transcript of Prausnitz Thermodynamics Notes 26
Advanced Chemical Engineering Thermodynamics
Class 25
Group Contribution MethodsChapters 67 and Appendix F, Prausnitz
• We demonstrated that
• Now we need to remember that
• Thus
• These are the most important relations for the chapter
GE and Activity Coefficients
giE = RT ln [ γi ]
GE = ∑i ni giE
GE = RT ∑i ni ln γi
Expressions for GE
gE = 0 when x1 = 0gE = 0 when x2 = 0
The twosuffix 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 RedlichKister expansion
TwoSuffix MargulesThe twosuffix Margules equations were derived from:
gE = A x1 x2
ln γ1 = A/RT x22
ln γ2 = A/RT x12
ln γ1∞ = A/RT
ln γ2∞ = A/RT
The activity coefficients at infinite dilution are:
Wohl’s ExpansiongE / [RT (x1q1 + x2q2)] = 2 a12 z1z2
+ 3 a112 z12z2 + 3 a122 z1z2
2
+ 4 a1112 z12z2 + 4 a1222 z1z2
3
+ 6 a1122 z12z2
2 + …
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 Equation
gE / RT = 2 a12 x1 x2 q1 q2 / [ x1 q1 + x2 q2 ]
We can calculate the fugacity coefficients of the two components
ln γ2 = B’ / [ 1 + B’/A’ x2/x1 ]2ln γ1 = A’ / [ 1 + A’/B’ x1/x2 ]2
And, at infinite dilution, we obtain
A’ = 2 q1 a12
B’ = 2 q2 a12
ln γ2∞ = B’
ln γ1∞ = A’
ThreeSuffix Margules EquationsgE / [RT (x1q1 + x2q2)] = 2 a12 z1z2
+ 3 a112 z12z2 + 3 a122 z1z2
2
+ 4 a1112 z12z2 + 4 a1222 z1z2
3
z1 = x1 q1 / [ x1 q1 + x2 q2 ] z2 = x2 q2 / [ x1 q1 + x2 q2 ]
ln γ1 = A’ x22 + B’ x2
3
ln γ2 = (A’ + 3/2 B’) x12  B’ x1
3
At infinite dilutionln γ1
∞ = A’ + B’
ln γ2∞ = A’ + 1/2 B’
UNIQUAC
For a binary mixture UNIQUAC states that:
[gE/RT]combinatorial = x1 ln [φ1*/x1] + x2 ln [φ2
*/x2] +
z/2 { x1q1 ln [θ1/φ1*] + x2q2 ln [θ2/φ2
*]}
gE / RT = [ gE / RT ]combinatorial + [ gE / RT ]residual
[gE/RT]residual =  x1 q1’ ln [ θ1
’ + θ2’ τ21 ]  x2 q2
’ ln [ θ2’ + θ1
’ τ12 ]
z is the coordination number
In UNIQUAC z=10
φ* is the segment fraction
θ, and θ’ are area fractions
UNIQUAC
φ1* = x1 r1 / [ x1 r1 + x2 r2 ]
Segment fractions
φ2* = x2 r2 / [ x1 r1 + x2 r2 ]
The parameters ‘r’ are purecomponent molecular structure constants
θ1 = x1 q1 / [ x1 q1 + x2 q2 ]Area fractions
For fluids other than water and lower alcohols q = q’
θ2 = x2 q2 / [ x1 q1 + x2 q2 ]
θ1’ = x1 q1
’ / [ x1 q1’ + x2 q2
’ ] θ2’ = x2 q2
’ / [ x1 q1’ + x2 q2
’ ]
Energy parameters
These parameters are given in terms of characteristic energies
τ12 = exp ( Δu12 / RT ) ≡ exp ( a12 / RT )
τ21 = exp ( Δu21 / RT ) ≡ exp ( a21 / RT )
Parameters, from Table 6.9
Compound r q q’
CCl4 3.33 2.82 2.82
CH3OH 1.43 1.43 0.96
C2H5OH 2.11 1.97 0.92
H2O 0.92 1.40 1.00
Toluene 3.92 2.97 2.97
NHexane 4.50 3.86 3.86
*
*
*
Binary Parameters,from Table 6.10
System T/K a12/K a21/K
Acetone/Chloroform 323 171.71 93.93Acetone/Water 331368 530.99 100.71Acetone/Methanol 323 379.31 108.42
Ethanol/noctane 348 123.57 1354.92Ethanol/nheptane 323 105.23 1380.30Ethanol/Benzene 350369 75.13 242.53Ethanol/CCl4 340351 138.90 947.20
Formic acid/Water 374380 924.01 525.85NHexane/Nitromethane 318 230.64 5.86
UNQUAC Activity CoefficientsgE / RT = [ gE / RT ]combinatorial + [ gE / RT ]residual
ln γ2 = ln (φ2*/x2) + z/2 q2 ln (θ2/φ2
*) + φ1* (l2  l1 r2/r1 )
 q2’ ln ( θ2
’ + θ1’ τ12 )
+ θ1’q2
’ [ τ12 / (θ2’ + θ1
’ τ12)  τ21 / (θ1’+θ2
’τ12)]
ln γ1 = ln (φ1*/x1) + z/2 q1 ln (θ1/φ1
*) + φ2* (l1  l2 r1/r2 )
 q1’ ln ( θ1
’ + θ2’ τ21 )
+ θ2’q1
’ [ τ21 / (θ1’ + θ2
’ τ21)  τ12 / (θ2’+θ1
’τ12)]
l1 = z/2 (r1  q1)  (r1  1) l2 = z/2 (r2  q2)  (r2  1)
Comments
In 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
ExampleLet’s suppose we have one mixture of acetone and toluene
ExampleLet’s suppose we have one mixture of acetone and toluene
To obtain the interaction parameters, we first look at the chemical structure of both compounds:
CH3
CH3
C=O
Acetone Toluene
CH3CCH
CH CH
CH CH
ExampleLet’s 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 moleculemolecule interactions:
CH3
CH3
C=O
Acetone Toluene
CH3CCH
CH CH
CH CH
ExampleLet’s 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 moleculemolecule interactions:
CH3
CH3
C=O
Acetone Toluene
CH3CCH
CH CH
CH CH
ExampleLet’s 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 moleculemolecule interactions:
CH3
CH3
C=O
Acetone Toluene
CH3CCH
CH CH
CH CH
ExampleLet’s 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 moleculemolecule interactions:
CH3
CH3
C=O
Acetone Toluene
CH3CCH
CH CH
CH CH
ExampleLet’s 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 moleculemolecule interactions:
CH3
CH3
C=O
Acetone Toluene
CH3CCH
CH CH
CH CH
ExampleLet’s suppose we have one mixture of acetone and toluene
The basic assumption is that each group will behave in any mixture independently on the molecule it is part of
CH3
CH3
C=O
Acetone Toluene
CH3CCH
CH CH
CH CH
ExampleLet’s suppose we have one mixture of acetone and toluene
The basic assumption is that each group will behave in any mixture independently on the molecule it is part ofWe obtain this contribution from available experimental data, and we ‘predict’ the behavior of untested mixtures
CH3
CH3
C=O
Acetone Toluene
CH3CCH
CH CH
CH CH
UNIFACThe Universal Functional Activity Coefficient theory is the best known theory that uses this approachFirst 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 groupgroup interactions is much less than the possible number of moleculemolecule 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 Method
ln γi = ln γiC + ln γi
R
UNIFAC: The Method
ln γi = ln γiC + ln γi
R
This is the ‘combinatorial’part, calculated as in the case of UNIQUAC
This is the ‘residual’contribution, which depends on groupgroup interactions
UNIFAC: The Method
ln γi = ln γiC + ln γi
R
This is the ‘combinatorial’part, calculated as in the case of UNIQUAC
This is the ‘residual’contribution, which depends on groupgroup interactions
ln γiC = FC (x, φ, θ) As defined for UNIQUAC
ln γiR = FR (X, Q, T, amn)
X = group mole fractionQ = group external surface areaamn = interaction energy between groups n and m
Example: Fig. F4
methanol + water @ 50C
Such good theoryexperiment agreement is not common
Example: Fig. F5
nhexane + methyl ethyl ketone @ 65 C
This is more typical
Appendix FMany binary systems have been studied and the corresponding data have been publishedDIPPRAIChE provides a continuouslyimproving compilation
Unfortunately, many other systems have not been reported
UNIFAC, or other groupcontribution methods, are in general not very accurate
P/bar
x/nhexane
L
VUNIFAC
Fig. F5
Experiments
nhexane + methyl ethyl ketone
Appendix FLet’s suppose we can obtain only 2 experimental data point
Which ones shall we get?
P/bar
x/nhexane
L
V
Appendix FLet’s suppose we can obtain only 2 experimental data point
Which ones shall we get?
P/bar
x/nhexane
L
V
For all binary mixtures, regardless of whether or not they form an azeotrope, binary parameters can be calculated from activity coefficients at infinite dilutionThese often provide the most valuable experimental information
Appendix F: ExampleLet’s suppose we decide to use any functional form for gE of the mixture of interest we like:
gE = F (x, A, B)
x = compositionA,B the two parameters that depend on the specific F we like
Appendix F: ExampleLet’s suppose we decide to use any functional form for gE of the mixture of interest we like:
gE = F (x, A, B)
x = compositionA,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 γ1 = F1 (x, A, B)RT ln γ2 = 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 γ1 = F1 (x, A, B)RT ln γ2 = F2 (x, A, B)
A,B are the two parameters we need to implement FIf we measure γ1
∞ and γ2∞, then
lim [F1 (x, A, B) / RT] = ln γ1∞
lim [F2 (x, A, B) / RT] = ln γ2∞
Which is a system of 2 equations in 2 unknowns
x1→0
x2→0
Appendix F: ExampleVan Laar’s equations
ln γ2∞ = B’
ln γ1∞ = A’
Appendix F: ExampleVan Laar’s equations
ln γ2 = B’ / [ 1 + B’/A’ x2/x1 ]2ln γ1 = A’ / [ 1 + A’/B’ x1/x2 ]2
And we can predict the activity coefficients of both components at any composition:
ln γ2∞ = B’
ln γ1∞ = A’
Appendix F: ExampleThreeSuffixMargules equations
ln γ1∞ = A’ + B’
ln γ2∞ = A’ + 1/2 B’
Appendix F: ExampleThreeSuffixMargules equations
And we can predict the activity coefficients of both components at any composition:
ln γ1 = A’ x22 + B’ x2
3
ln γ2 = (A’ + 3/2 B’) x12  B’ x1
3
ln γ1∞ = A’ + B’
ln γ2∞ = A’ + 1/2 B’