De La Salle University College of Engineering Chemical Engineering Department
Problem Set 5
Advanced Thermodynamics (CHE618M) Dr. Nathaniel Dugos
Dizon, Clarissa S. - 10993770
Term 3, AY-2009-2010
12.1
The following is a set of VLE data for the system methanol(l)/water(2) at 333.15 K (60°C) (extracted from K. Kurihara et al., J. Chem. Eng. Data, vol. 40, pp. 679-684, 1995): P/kPa
x1
y1
P/kPa
x1
y1
19.953
0.0000
0.0000
60.614
0.5282
0.8085
39.223
0.1686
0.5714
63.998
0.6044
0.8383
42.984
0.2167
0.6268
67.924
0.6804
0.8733
48.852
0.3039
0.6943
70.229
0.7255
0.8922
52.784
0.3681
0.7345
72.832
0.7776
0.9141
56.652
0.4461
0.7742
84.562
1.0000
1.0000
a. Basing calculations on Eq. (12. I), find parameter values for the Margules equation that provide the best fit of GE/RT to the data, and prepare a Pxy diagram that compares the experimental points with curves determined from the correlation. b. Repeat (a) for the van Laar equation. c. Repeat (a) for the Wilson equation. d. Using Barker's method, find parameter values for the Margules equation that provide the best fit of the P-x1 data. Prepare a diagram showing the residuals δP and δy 1 plotted vs. x1. e. Repeat (d) for the van Laar equation. f. Repeat (d) for the Wilson equation. Solution: a. Parameter values for the Margules Equation, Pxy diagram To get A12 and A21, GE/(RTx1x2) was plotted versus x. Linearizing, A12 would correspond to the intercept and A21 would correspond to the sum of the slope and A12. Using equation 12.6, GE/RT was obtained. G
E
RT
x1
ln 1
x 2 ln 2
γ is calculated using Equation 12.1. i
yi P xi Pi
sat
The table below shows the computed and given values P/kPa
x1
y1
x2
y2
γ1
γ2
GE/RT
GE/(RTx1x2)
39.223
0.1686
0.5714
0.8314
0.4286
1.5720
1.0134
0.0873
0.6229
42.984
0.2167
0.6268
0.7833
0.3732
1.4703
1.0264
0.1039
0.6123
48.852
0.3039
0.6943
0.6961
0.3057
1.3198
1.0752
0.1348
0.6373
52.784
0.3681
0.7345
0.6319
0.2655
1.2455
1.1115
0.1476
0.6346
56.652
0.4461
0.7742
0.5539
0.2258
1.1627
1.1574
0.1482
0.5999
60.614
0.5282
0.8085
0.4718
0.1915
1.0972
1.2330
0.1478
0.5932
63.998
0.6044
0.8383
0.3956
0.1617
1.0497
1.3110
0.1365
0.5707
67.924
0.6804
0.8733
0.3196
0.1267
1.0310
1.3495
0.1166
0.5360
70.229
0.7255
0.8922
0.2745
0.1078
1.0213
1.3822
0.1042
0.5231
72.832
0.7776
0.9141
0.2224
0.0859
1.0125
1.4098
0.0860
0.4975
Plot: 0.7000 0.6500
y = -0.2075x + 0.6828 R² = 0.828
) 2 x 10.6000 x T R ( /0.5500 E G
0.5000 0.4500 0.0000
0.2000
0.4000
0.6000
0.8000
1.0000
x1
The linearized equation of the line is y = -0.2075x + 0.6828. With this, A12 = Intercept = 0.6828 and A21 = Slope + A12 = -0.2075 + 0.6828 = 0.4753 Equations 12.10a and 12.10b computes for the gamma values given the parameter values of Margules ln 1
x 2
2
A
12
2 A21
A12
x 1
ln 2
x1
2
A
21
2 A12
A21
x 2
Using Modified Raoult’s Law, the new final pressure and final composition of the vapor phase is computed.
x1
γ1’
γ2’
Pcalc
y1
y2
0.1686
1.5275
1.0236
38.7575
0.5619
0.4381
0.2167
1.4387
1.0383
42.5919
0.6190
0.3810
0.3039
1.3096
1.0731
48.5603
0.6931
0.3069
0.3681
1.2357
1.1051
52.3976
0.7341
0.2659
0.4461
1.1650
1.1507
56.6630
0.7756
0.2244
0.5282
1.1087
1.2059
60.8733
0.8135
0.1865
0.6044
1.0699
1.2631
64.6544
0.8458
0.1542
0.6804
1.0418
1.3250
68.3878
0.8764
0.1236
0.7255
1.0292
1.3636
70.6086
0.8942
0.1058
0.7776
1.0180
1.4095
73.1916
0.9145
0.0855
Plot of P(x,y) diagram: 90 80 70 60
e r u 50 s s e r 40 P
P-x Margules P-y Margules
30
P-x exp
20
P-y exp
10 0 0.0000
0.5000
1.0000
1.5000
x1,y1
b. Parameter values for the Van Laar Equation, Pxy diagram y vs x was plotted where y is equal to x1x2/(GE/RT). The parameters a12 and a21 are the reciprocal of the intercept and reciprocal of the sum of the slope and intercept respectively. Plot: 2.5 2 ) T R / 1.5 E G ( / 2 1 x 1 x
y = 0.6414x + 1.4185 R² = 0.8148
0.5 0 0.0000
0.2000
0.4000
0.6000 x1
0.8000
1.0000
Linearizing, the slope is 0.6414 and the intercept is 1.4185. Then, a12 = (1/1.4185) a12 = 0.705
a21 = 1/(0.6414 + 1.4185) a21 = 0.485
Equations 12.17a and 12.17b computes for the gamma values given the parameter values of Van Laar
ln 1
a x a12 1 12 1 a21 x2
2
ln 2
a x a 21 1 21 2 a12 x 1
2
Using Modified Raoult’s Law, the new final pressure and final composition of the vapor phase is computed.
x1
γ1’
γ2’
Pcalc
y1
y2
0.1686
1.5230
1.0254
38.7251
0.5607
0.4393
0.2167
1.4316
1.0407
42.4983
0.6173
0.3827
0.3039
1.3022
1.0758
48.4064
0.6913
0.3087
0.3681
1.2298
1.1073
52.2433
0.7328
0.2672
0.4461
1.1616
1.1515
56.5445
0.7749
0.2251
0.5282
1.1077
1.2045
60.8137
0.8135
0.1865
0.6044
1.0704
1.2594
64.6496
0.8462
0.1538
0.6804
1.0430
1.3194
68.4243
0.8770
0.1230
0.7255
1.0306
1.3573
70.6598
0.8948
0.1052
0.7776
1.0193
1.4033
73.2499
0.9150
0.0850
Plot of P(x,y) diagram: 90.0000 80.0000 70.0000 60.0000 e r 50.0000 u s s e r 40.0000 P
30.0000 20.0000 10.0000 0.0000 0.0000 0.2000 0.4000x,y0.6000 0.8000 1.0000
P-x Van Laar P-y Van Laar P-x exp P-y exp
c.
Parameter values for the Wilson Equation, Pxy diagram
The values of the parameters Λ12 and Λ21 are obtained by non-linear regression with the use of equations 12.19a and 12.19b.
12 21 x x x x 2 1 21 1 2 12
ln 1 ln( x 1 x 2 12 ) x 2
12 21 x x x x 2 1 21 1 2 12
ln 2 ln( x2 x1 21 ) x1 The parameters were solved in MS EXCEL. γ1 W
γ2 W
γ1 expt
γ2 expt
γ1W/γ1expt
γ2W/γ2expt
(γ1W/γ1expt)*(γ2W/γ2expt)
1.550
1.029
1.572
1.013
0.986
1.016
1.001
1.448
1.046
1.470
1.026
0.985
1.019
1.004
1.308
1.084
1.320
1.075
0.991
1.008
0.999
1.232
1.117
1.246
1.112
0.989
1.005
0.994
1.162
1.163
1.163
1.157
0.999
1.005
1.004
1.107
1.217
1.097
1.233
1.009
0.987
0.996
1.070
1.273
1.050
1.311
1.019
0.971
0.990
1.043
1.333
1.031
1.350
1.011
0.988
0.999
1.030
1.371
1.021
1.382
1.009
0.992
1.001
1.019
1.417
1.012
1.410
1.007
1.005
1.012
The parameters obtained are: Λ12 =
0.4253
Λ21 =
1.0839
Equations 12.19a and 12.19b computes for the gamma values given the parameter values of Wilson.
12 21 x1 x2 12 x2 x1 21
ln 1 ln( x1 x2 12 ) x2
12 21 x x x x 2 1 21 1 2 12
ln 2 ln( x2 x1 21 ) x1
Using Modified Raoult’s Law, the new final pressure and final composition of the vapor phase is computed.
x1
γ1’
γ2’
Pcalc
y1
y2
0.1686
1.550
1.029
39.1712
0.5641
0.4359
0.2167
1.448
1.046
42.8809
0.6187
0.3813
0.3039
1.308
1.084
48.6710
0.6907
0.3093
0.3681
1.232
1.117
52.4398
0.7314
0.2686
0.4461
1.162
1.163
56.6834
0.7732
0.2268
0.5282
1.107
1.217
60.9153
0.8119
0.1881
0.6044
1.070
1.273
64.7301
0.8448
0.1552
0.6804
1.043
1.333
68.4902
0.8759
0.1241
0.7255
1.030
1.371
70.7182
0.8938
0.1062
0.7776
1.019
1.417
73.2996
0.9142
0.0858
Plot of P(x,y) diagram: 90.0000 80.0000 70.0000 60.0000 e r u50.0000 s s e r40.0000 P 30.0000 20.0000 10.0000 0.0000 0.000
P-x Wilson P-y Wilson P-x exp P-y exp 0.500
1.000
1.500
x,y
d. Parameter values for the Margules Equation using Barker’s method, plot of δP and δy1 vs. x1. Using excel solver, the following parameters are obtained. The minimum value for Pi
exp
i xi Pi
sat
obtained is 1.293.
A12 = 0.749
A21 = 0.432
Residuals of Pressure and Vapor Composition as a function of x1 using Margules: 0.5 0.4 s l a u d i s e R
0.3 P exp - P calc
0.2
y exp - y calc
0.1
0 0.0000 -0.1
0.5000 x1
1.0000
e. Parameter values for the Van Laar Equation using Barker’s method, plot of δP and δy1 vs. x1. Using excel solver, the following parameters are obtained. The minimum value for Pi
exp
i xi Pi
sat
obtained is 2.430.
a12 = 0.8132
a21 = 0.4689
Residuals of Pressure and Vapor Composition as a function of x1 using Van Laar: 0.6 0.5 0.4
s l a 0.3 u d i s 0.2 e R
P exp - P calc y exp - y calc
0.1 0 -0.1
0
0.5
1
x1
f.
Parameter values for the Wilson Equation using Barker’s method, plot of δP and δy1 vs. x1.
Using excel solver, the following parameters are obtained. The minimum value for Pi
exp
i xi Pi
obtained is 2.606. Λ12 =
0.3588
Λ 21 =
1.1847
The values for the vapor composition and total pressure are solved using the modified Raoult’s law.
Residuals of Pressure and Vapor Composition as a function of x1 using Wilson Equation: 0.600 0.500 0.400 s l a u d i s e R
0.300
P exp - P calc
0.200
y exp - y calc
0.100 0.000 0.000 -0.100
0.200
0.400
0.600 x1
0.800
1.000
sat
12.3
The following is a set of VLE data for the system acetone(l)/methanol(2) at 328.15 K (55°C) (extracted from D. C. Freshwater and K. A. Pike, J. Chem. Eng. Data, vol. 12, pp. 179-183 ,1967): P/kPa
x1
y1
P/kPa
x1
y1
68.728
0.0000
0.0000
97.646
0.5052
0.5844
72.278
0.0287
0.0647
98.462
0.5432
0.6174
75.279
0.0570
0.1295
99.811
0.6332
0.6772
77.524
0.0858
0.1848
99.950
0.6605
0.6926
78.951
0.1046
0.2190
100.278
0.6945
0.7124
82.528
0.1452
0.2694
100.467
0.7327
0.7383
86.762
0.2173
0.3633
100.999
0.7752
0.7729
90.088
0.2787
0.4184
101.059
0.7922
0.7876
93.206
0.3579
0.4779
99.877
0.9080
0.8959
95.017
0.4050
0.5135
99.799
0.9448
0.9336
96.365
0.4480
0.5512
96.885
1.0000
1.0000
a. Basing calculations on Eq. (12. I), find parameter values for the Margules equation that provide the best fit of GE/RT to the data, and prepare a Pxy diagram that compares the experimental points with curves determined from the correlation. b. Repeat (a) for the van Laar equation. c. Repeat (a) for the Wilson equation. d. Using Barker's method, find parameter values for the Margules equation that provide the best fit of the P-x1 data. Prepare a diagram showing the residuals δP and δy 1 plotted vs. x1. e. Repeat (d) for the van Laar equation. f. Repeat (d) for the Wilson equation. Solution: a. Parameter values for the Margules Equation, Pxy diagram To get A12 and A21, GE/(RTx1x2) was plotted versus x. Linearizing, A12 would correspond to the intercept and A21 would correspond to the sum of the slope and A12. Using equation 12.6, GE/RT was obtained. G
E
RT
x1
ln 1
x 2 ln 2
γ is calculated using Equation 12.1. i
yi P xi Pi
sat
The table below shows the computed and given values P/kPa x1 y1 x2 y2
γ1
γ2
GE/RT
GE/(RTx1x2)
72.278
0.0287
0.0647
0.9713
0.9353
1.6818
1.0127
0.0272
0.9741
75.279
0.0570
0.1295
0.9430
0.8705
1.7653
1.0111
0.0428
0.7964
77.524
0.0858
0.1848
0.9142
0.8152
1.7234
1.0058
0.0520
0.6632
78.951
0.1046
0.2190
0.8954
0.7810
1.7061
1.0020
0.0576
0.6155
82.528
0.1452
0.2694
0.8548
0.7306
1.5804
1.0263
0.0887
0.7144
86.762
0.2173
0.3633
0.7827
0.6367
1.4972
1.0269
0.1085
0.6379
90.088
0.2787
0.4184
0.7213
0.5816
1.3959
1.0569
0.1329
0.6611
93.206
0.3579
0.4779
0.6421
0.5221
1.2846
1.1027
0.1524
0.6632
95.017
0.4050
0.5135
0.5950
0.4865
1.2435
1.1304
0.1612
0.6689
96.365
0.5512 0.5844
0.5520
0.4488
1.2238
1.1400
0.1628
0.6582
97.646
0.4480 0.5052
0.4948
0.4156
1.1659
1.1933
0.1650
0.6600
98.462
0.5432
0.6174
0.4568
0.3826
1.1551
1.1999
0.1616
0.6512
99.811
0.6332
0.6772
0.3668
0.3228
1.1018
1.2781
0.1514
0.6517
99.950
0.6605
0.6926
0.3395
0.3074
1.0818
1.3168
0.1453
0.6482
100.278
0.6945
0.2876
1.0617
1.3736
0.1385
0.6530
100.467
0.7327
0.7124 0.3055 0.7383 0.2673
0.2617
1.0449
1.4312
0.1280
0.6536
100.999
0.7752
0.7729
0.2248
0.2271
1.0394
1.4846
0.1188
0.6815
101.059
0.7922
0.7876
0.2078
0.2124
1.0370
1.5030
0.1135
0.6893
99.877
0.9080
0.8959
0.092
0.1041
1.0171
1.6444
0.0612
0.7325
99.799
0.9448
0.9336
0.0552
0.0664
1.0179
1.7467
0.0475
0.9111
Plot: 1.0500
y = -0.0182x + 0.7077 R² = 0.0034
0.9500 ) 20.8500 x 1 x T0.7500 R ( / E0.6500 G
0.5500 0.4500 0.0000
0.2000
0.4000 x1 0.6000
0.8000
1.0000
The equation of the line is given as y = -0.0182x + 0.7077. A12 = Intercept = 0.7077 A21 = Slope + A12 = -0.0182 + 0.7077 = 0.6895 And Equations 12.10a and 12.10b computes for the gamma values given the parameter values of Margules
ln 1 x2
2
A12 2 A21 A12 x1
ln 2
x1
2
A
21
2 A12
A21
x 2
Using Modified Raoult’s Law, the new final pressure and final composition of the vapor phase is computed.
x1 0.0287
γ1’
γ2’
Pcalc
y1
y2
1.9478
1.0006
72.2113
0.0750
0.9250
0.0570
1.8729
1.0024
75.3060
0.1373
0.8627
0.0858
1.8019
1.0053
78.1453
0.1917
0.8083
0.1046
1.7583
1.0079
79.8459
0.2232
0.7768
0.1452
1.6707
1.0153
83.1510
0.2827
0.7173
0.2173
1.5353
1.0345
87.9706
0.3674
0.6326
0.2787
1.4375
1.0572
91.2233
0.4255
0.5745
0.3579
1.3316
1.0956
94.5243
0.4885
0.5115
0.4050
1.2780
1.1237
96.1007
0.5218
0.4782
0.4480 0.5052
1.2345
1.1531
97.3276
0.5505
0.4495
1.1838
1.1979
98.6813
0.5872
0.4128
0.5432
1.1544
1.2316
99.4192
0.6111
0.3889
0.6332
1.0965
1.3255
100.6826
0.6681
0.3319
0.6605
1.0820
1.3582
100.9312
0.6860
0.3140
0.6945
1.0658
1.4020
101.1494
0.7090
0.2910
0.7327
1.0499
1.4555
101.2671
0.7359
0.2641
0.7752
1.0349
1.5208
101.2260
0.7679
0.2321
0.7922
1.0297
1.5488
101.1546
0.7813
0.2187
0.9080
1.0057
1.7704
99.6698
0.8877
0.1123
0.9448
1.0021
1.8539
98.7581
0.9288
0.0712
Plot of P(x,y) diagram: 105.0000 100.0000 95.0000 90.0000
P-x Margules
e r u85.0000 s s e r80.0000 P
P-y Margules P-x exp
75.0000
P-y exp
70.0000 65.0000 60.0000 0.0000
0.5000 x1,y1
1.0000
b. Parameter values for the Van Laar Equation, Pxy diagram
Plot: 1.8 1.6 1.4
) T1.2 R / E 1 G ( /0.8 2 x 10.6 x
y = 0.015x + 1.4424 R² = 0.0008
0.4 0.2 0 0.0000
0.2000
0.4000 x1 0.6000
0.8000
1.0000
Linearizing, the slope is 0. 0.015 and the intercept is 1.4424. Then, a12 = (1/1.4424) a12 = 0.6933
a21 = 1/(0.015 + 1.4424) a21 = 0.6862
Equations 12.17a and 12.17b computes for the gamma values given the parameter values of Van Laar
ln 1
a x a12 1 12 1 a21 x2
2
ln 2
a x a 21 1 21 2 a12 x 1
2
Using Modified Raoult’s Law, the new final pressure and final composition of the vapor phase is computed.
x1
γ1’
γ2’
Pcalc
y1
y2
0.0287
1.9226
1.0006
72.1400
0.0741
0.9259
0.0570
1.8511
1.0023
75.1806
0.1360
0.8640
0.0858
1.7832
1.0052
77.9784
0.1901
0.8099
0.1046
1.7413
1.0077
79.6581
0.2215
0.7785
0.1452
1.6571
1.0148
82.9312
0.2811
0.7189
0.2173
1.5262
1.0335
87.7264
0.3663
0.6337
0.2787
1.4314
1.0556
90.9782
0.4248
0.5752
0.3579
1.3281
1.0932
94.2918
0.4884
0.5116
0.4050
1.2756
1.1207
95.8793
0.5220
0.4780
0.4480
1.2328
1.1495
97.1174
0.5510
0.4490
0.5052
1.1829
1.1935
98.4866
0.5879
0.4121
0.5432
1.1538
1.2268
99.2353
0.6119
0.3881
0.6332
1.0964
1.3194
100.5254
0.6691
0.3309
0.6605
1.0820
1.3518
100.7827
0.6870
0.3130
0.6945
1.0659
1.3952
101.0124
0.7100
0.2900
0.7327
1.0500
1.4483
101.1438
0.7369
0.2631
0.7752
1.0351
1.5132
101.1194
0.7688
0.2312
0.7922
1.0299
1.5411
101.0550
0.7822
0.2178
0.9080
1.0058
1.7626
99.6244
0.8881
0.1119
0.9448
1.0021
1.8463
98.7313
0.9291
0.0709
Plot of P(x,y) diagram: 105 100 95 e r u s s e r P
90
P-x exp
85
P-y exp
80
P-x Van Laar
75
P-y Van Laar
70 65 60 0.0000
c.
0.5000 x,y
1.0000
Parameter values for the Wilson Equation, Pxy diagram
The values of the parameters Λ12 and Λ21 are obtained by non-linear regression with the use of equations 12.19a and 12.19b.
12 21 x 1 x 2 12 x 2 x 1 21 12 21 ln( x2 x1 21 ) x1 x x x x 2 1 21 1 2 12
ln 1 ln( x 1 x 2 12 ) x 2 ln 2
The parameters were solved in MS EXCEL. γ1 W
γ2 W
γ1 expt
γ2 expt
γ1W/γ1expt
γ2W/γ2expt
(γ1W/γ1expt)*(γ2W/γ2expt)
1.833
1.001
1.6818
1.0127
1.090
0.988
1.077
1.769
1.002
1.7653
1.0111
1.002
0.991
0.993
1.709
1.005
1.7234
1.0058
0.992
0.999
0.991
1.672
1.007
1.7061
1.0020
0.980
1.005
0.985
1.597
1.014
1.5804
1.0263
1.011
0.988
0.998
1.481
1.031
1.4972
1.0269
0.989
1.004
0.993
1.397
1.051
1.3959
1.0569
1.001
0.994
0.995
1.304
1.085
1.2846
1.1027
1.015
0.984
0.999
1.257
1.110
1.2435
1.1304
1.011
0.982
0.993
1.218
1.136
1.2238
1.1400
0.995
0.997
0.992
1.173
1.176
1.1659
1.1933
1.006
0.986
0.992
1.146
1.207
1.1551
1.1999
0.992
1.006
0.998
1.093
1.292
1.1018
1.2781
0.992
1.011
1.002
1.079
1.321
1.0818
1.3168
0.998
1.003
1.001
1.064
1.361
1.0617
1.3736
1.002
0.991
0.993
1.049
1.411
1.0449
1.4312
1.004
0.986
0.989
1.035
1.471
1.0394
1.4846
0.995
0.991
0.987
1.029
1.498
1.0370
1.5030
0.993
0.996
0.989
1.006
1.709
1.0171
1.6444
0.989
1.039
1.028
1.002
1.791
1.0179
1.7467
0.985
1.025
1.009
The parameters obtained are: Λ12 =
0.7199
Λ21 =
0.6859
Equations 12.19a and 12.19b computes for the gamma values given the parameter values of Wilson.
12 21 x x x x 2 1 21 1 2 12
ln 1 ln( x 1 x 2 12 ) x 2
12 21 x x x x 2 1 21 1 2 12
ln 2 ln( x2 x1 21 ) x1
Using Modified Raoult’s Law, the new final pressure and final composition of the vapor phase is computed.
x1
γ1’
γ2’
Pcalc
y1
y2
0.0287
1.833
1.001
71.8871
0.0709
0.9291
0.0570
1.769
1.002
74.7180
0.1308
0.8692
0.0858
1.709
1.005
77.3388
0.1837
0.8163
0.1046
1.672
1.007
78.9207
0.2147
0.7853
0.1452
1.597
1.014
82.0244
0.2740
0.7260
0.2173
1.481
1.031
86.6338
0.3600
0.6400
0.2787
1.397
1.051
89.8134
0.4199
0.5801
0.3579
1.304
1.085
93.1109
0.4857
0.5143
0.4050
1.257
1.110
94.7162
0.5207
0.4793
0.4480
1.218
1.136
95.9824
0.5509
0.4491
0.5052
1.173
1.176
97.4019
0.5893
0.4107
0.5432
1.146
1.207
98.1902
0.6142
0.3858
0.6332
1.093
1.292
99.5920
0.6731
0.3269
0.6605
1.079
1.321
99.8887
0.6913
0.3087
0.6945
1.064
1.361
100.1719
0.7146
0.2854
0.7327
1.049
1.411
100.3703
0.7418
0.2582
0.7752
1.035
1.471
100.4306
0.7736
0.2264
0.7922
1.029
1.498
100.4037
0.7870
0.2130
0.9080
1.006
1.709
99.2909
0.8912
0.1088
0.9448
1.002
1.791
98.5235
0.9310
0.0690
Plot of P(x,y) diagram: 105 100 95 e r u s s e r P
90
P-x exp
85 80
P-y exp
75 70 65 60 0.0000
0.5000
x,y
1.0000
1.5000
d. Parameter values for the Margules Equation using Barker’s method, plot of δP and δy1 vs. x1. Using excel solver, the following parameters are obtained. The minimum value for Pi
exp
i xi Pi
sat
obtained is 4.8254.
A12 = 0.6492
A21 = 0.6599
Residuals of Pressure and Vapor Composition as a function of x1 using Margules: 1.4000 1.2000 1.0000 s 0.8000 l a u d i 0.6000 s e R 0.4000
P exp - P calc y exp - y calc
0.2000 0.0000 0.0000 -0.2000
0.2000
0.4000
0.6000
0.8000
1.0000
x1
e. Parameter values for the Van Laar Equation using Barker’s method, plot of δP and δy1 vs. x1. Using excel solver, the following parameters are obtained. The minimum value for Pi
exp
i xi Pi
sat
obtained is 4.8241.
a12 = 0.6493
a21 = 0.6599
Residuals of Pressure and Vapor Composition as a function of x1 using Van Laar: 1.4 1.2 1 0.8
s l a u d i s e R
P exp - P calc
0.6
y exp - y calc
0.4 0.2 0 0.0000 -0.2
f.
0.2000
0.4000
0.6000
0.8000
1.0000
x1
Parameter values for the Wilson Equation using Barker’s method, plot of δP and δy1 vs. x1.
Using excel solver, the following parameters are obtained. The minimum value for
Pi
exp
i xi Pi
obtained is 6.337. Λ12 =
0.7042
Λ 21 =
0.6918
The values for the vapor composition and total pressure are solved using the modified Raoult’s law.
Residuals of Pressure and Vapor Composition as a function of x1 using Wilson Equation:
sat
1.800 1.600 1.400 1.200 s l a u d i s e R
1.000 0.800
P exp - P calc
0.600
y exp - y calc
0.400 0.200 0.000 0.0000 -0.200
0.5000
1.0000
x1
12.13
For one of the binary systems listed in the preceding table(see table on book), based on Eq. (10.5) and the Wilson equation prepare a t-x-y diagram for P = 101.33 kPa. Solution: The data points of the t-x-y diagram were determined using a matlab program. The procedure done was trial and error. The t-x-y diagram:
375 370 365
K , e r 360 u t a r 355 e p m350 e T
T-x1 T-y1
345 340 335 0
0.2
0.4
0.6 x1,y1
0.8
1
1.2
12.15
For one of the binary systems listed in the preceding table(see table on book), based on Eq. (10.5) and the NRTL equation prepare a t-x-y diagram for P = 101.33 kPa. Solution:
The data points of the t-x-y diagram were determined using a matlab program. The procedure done was trial and error. The t-x-y diagram: 375 370 K 365 , e r 360 u t a r 355 e p 350 m e T 345
T-x1 T-y1
340 335 0
0.2
0.4
0.6
0.8
1
1.2
x1,y1
Code for 12.13 and 12.15 %CHE618M Problem Set 6 disp('Solution for Problem 12.13 and 12.15') disp('\nWilson and NRTL Equation given a Binary system') x1 = input('Input x1: ');
% for NRTL % PARAMETERS - System dependent b12 = -253.88; % cal/mol b21 = 845.21; % cal/mol alpha = 0.2994; % Initial temperature assumption (Kelvin) T = 300; error = 100; counter = 1;
% difference in pressure by 0.1% allowed; counter is to limit iterations while error>0.001||counter==5000 % Sensitivity to 0.01K T=T+0.01;
x2 = 1 - x1; t12 = b12/(R*T); t21 = b21/(R*T); g12 = exp(-alpha*t12); g21 = exp(-alpha*t21); ft1 = (t21*(g21/(x1 + x2*g21))^2+g12*t12/(x2 + x1*g12)^2); gam1 = exp(ft1*x2^2); ft2 = (t12*(g12/(x2 + x1*g12))^2+g21*t21/(x1 + x2*g21)^2); gam2 = exp(ft2*x1^2); % VAPOR PRESSURE COMPUTATION (iterative) for i=1:comp; Psat(i)=exp(C1(i)-(C2(i)/((T-273.15)+C3(i)))); end % for P = 101.33 kPa Ptotal = Psat(1)*gam1*x1 + Psat(2)*gam2*x2; error = abs((101.33-Ptotal)/101.33); counter = counter + 1; TnewN = T; end % for y = vapor composition y1N = x1*gam1*Psat(1)/Ptotal % For temperature TnewN
% for Wilson % PARAMETERS - System dependent R = 1.9872; %cal/mol-K V1 = 40.73; %cm3/mol V2 = 18.07; %cm3/mol comp = 2; %composition % Data obtained from Smith and Van Ness for VAPOR PRESSURE C1 = [16.5785 16.3872]; %A C2 = [3638.27 3885.70]; %B C3 = [239.500 230.170]; %C % for Wilson a12 = 107.38; a21 = 469.55;
% cal/mol % cal/mol
% Initial temperature assumption (Kelvin) T = 300; error = 100; counter = 1;
% difference in pressure by 0.1% allowed;counter is to limit Iterations while error>0.001||counter==5000
% Sensitivity to 0.01K T=T+0.01; x2 = 1 - x1; A12 = (V2/V1)*exp(-a12/(R*T)); A21 = (V1/V2)*exp(-a21/(R*T)); ft1 = -log(x1 + x2*A12); st1 = x2*((A12/(x1 + x2*A12))-(A21/(x2 + x1*A21))); gam1 = exp(ft1 + st1); ft2 = -log(x2 + x1*A21); st2 = x1*((A12/(x1 + x2*A12))-(A21/(x2 + x1*A21))); gam2 = exp(ft2 - st2); % VAPOR PRESSURE COMPUTATION (iterative) for i=1:comp; Psat(i)=exp(C1(i)-(C2(i)/((T-273.15)+C3(i)))); end % for P = 101.33 kPa Ptotal = Psat(1)*gam1*x1 + Psat(2)*gam2*x2; error = abs((101.33-Ptotal)/101.33); counter = counter + 1; Tnew = T; end
% for y = vapor composition y1 = x1*gam1*Psat(1)/Ptotal % For temperature Tnew
12.26
At 298.15 K (25°C) and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by the equation:
V x1 x 2 45 x1 25 x2 3
where ΔV is in cm mol . At these conditions, V1 = 110 and V2 = 90 cm3 mol-1. Determine the partial
molar volumes
V 1
-1
and V 2 in a mixture containing 40 mol-% of species 1 at the given conditions.
Solution: From equation 12.27, E
V
V xiV i
We know that ΔV = VE,
For the partial molar volumes,
V 1
and V 2 , express V (total volume) as a function of x1
V = ΔV + x1V1 + x2V2 Since x1 = 1 – x2 for binary systems, ΔV = x1(1 - x1)(45x1 + 25(1- x1)) ΔV = -20x13 -5x12 + 25x1
Simplifying, V = -20x13 -5x12 + 25x1 + 110x1 + 90x2 In terms of x1,
V = -20x13 -5x12 + 25x1 + 45x1 + 90 From equations 11.15 and 11.16, V 1
V x2
dV dx1
and
V 2
V
x1
dV dx1
Deriving V with respect to x1 and simplifying: dV dx
2
60 x1
10 x1
45
`
Therefore the expressions for the partial molar volumes are V 1
And
3
20 x1
V 2
20 x1
3
2
5 x1
5 x1
2
2
25 x1 45 x1 90 (1 x1 ) 60 x1
25 x1
45 x1
90 x1
60 x1
2
10 x1 45
10 x1
Since x1 is 0.4, V 1
124.76
cm
3
mol
V 2
93.36
cm
3
mol
45
12.32
A 20-mol-% LiCl/H20 solution at 298.15 K (25°C) is made by mixing a 25-mol-% LiCI/H20 solution at 298.15 K (25°C) with chilled water at 278.15 K (5°C). What is the heat effect in J mol-1 of final solution? Solution: The feed is 75% mol water. With 1 mole basis of feed, LiCl has 0.25 moles and water has 0.75 moles. The product is 80% mol water. With stoichiometric ratios, LiCl has 0.20 moles and water has 0.80 moles.
The heat effect is given by: ΔH = ΔH1 + ΔH2 + ΔH3 ~ Using figure 12.14, and n
3 (3 moles of water per mole of LiCl), ΔH1 is obtained.
ΔH1 = 20.6 KJ/mol
Multiplying equation 4.7 by R, ΔH2 is obtained. T
H R T o
C p R
dT
Where:
C p
A BT CT
2
DT
2
R From Appendix C, the constants for water were gathered to solve for the heat capacity. A B C
8.712 1.25x10-3 -0.18x10-6
T = 298.15 K To = 278.15 K ΔH2 = 1.506 KJ/mol
Using figure 12.14, and
n
~
4 (4 moles of water per mole of LiCl), ΔH3 is obtained.
ΔH3 = -25.5 KJ/mol Therefore the heat effect of the final solution is, ΔH = 20.6 KJ/mol + 1.506 KJ/mol - 25.5 KJ/mol ΔH = 20.6 KJ/mol Since the LiCl in the product stream has 0.2 moles, ΔH = (20.6 KJ/mol)*(0.2 mol) ΔH = -0.679 KJ