P-x,y Diagram from Margules Equation fit to GE/RT data.
0 0.2 0.4 0.6 0.8
36
38
Pi
kPa
x1iy1i
,X1j
,y1calcj
,
(d) Consistency Test: δGERTiGeRT x1ix2i
,
()
GERTi
−:=
γ1x
1ix2i
,
()
⎛
⎞
γ1i
⎛
⎞
0
0.004
δGERTi
0
δlnγ1γ2i
414
PcalcjX1jγ1X1jX2j
,A12
,A21
,C,
()
⋅Psat1
⋅
X2jγ2X1jX2j
,A12
,A21
,C,
()
⋅Psat2
⋅+
…:=
Plot P-x,y diagram for Margules Equation with parameters from Barker’s
Method.
C 0.2:=A21 0.5−:=A12 0.3−:=
Guesses:
Minimize sum of the squared errors using the Mathcad Minimize function.
⎢
⎥
⎢
⎥
γ1x1 x2,A12
,A21
,C,
()
exp x2()
2A12 2A
21 A12
−C−
()
⋅x1⋅+
3C⋅x12
⋅+
…
⎡
⎢
⎣
⎤
⎥
⎦
⋅
⎡
⎢
⎣
⎤
⎥
⎦
:=
Barker’s Method by non-linear least squares:
Margules Equation
(e)
Calculate mean absolute deviation of residuals
415
0 0.2 0.4 0.6 0.8
26
34
38
P-x data
P-y data
P-x calculated
P-y calculated
Pi
kPa
x1iy1i
,X1j
,y1calcj
,
Pcalcix1iγ1x1ix2i
,A12
,A21
,C,
()
⋅Psat1
⋅
…:=
416
Plot of P and y1 residuals.
0 0.5 1
0.4
0.2
PiPcalci
−
kPa
()
RMS deviations in P:
417
γ1x1x2,T,()
exp x2 Λ12 T()
x1 x2 Λ12 T()⋅+
Λ21 T()
x2 x1 Λ21 T()⋅+
−
⎛
⎜
⎝
⎞
⎠
⋅
⎡
⎢
⎣
⎤
⎥
⎦
x1 x2 Λ12 T()⋅+
()
:=
V2 18.07 cm3
mol
⋅:=V1 75.14 cm3
mol
⋅:=
Parameters for the Wilson equation:
C2 230.170 K⋅:=B2 3885.70 K⋅:=A2 16.3872:=
Water:
C1 205.807 K⋅:=B1 3483.67 K⋅:=A1 16.1154:=
1-Propanol:
Antoine coefficients:
It is impractical to provide solutions for all of the systems listed in the
table on Page 474 we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the appropriate
parameter values substituted for those given. The file WILSON.mcd
reproduces the table of Wilson parameters on Page 474 and includes the
necessary Antoine coefficients.
12.12
418
⎜
⎢
⎥
P-x,y diagram at T 60 273.15+()K⋅:=
Guess: P 70 kPa⋅:=
Peq x()
kPa
20.007
30.009
30.97
31.331
31.496
31.467
31.148
30.355
28.759
25.769
=
x
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
=yeq x()
0
0.363
0.395
0.413
0.431
0.453
0.483
0.526
0.594
0.718
=
419
C1 205.807 K⋅:=B1 3483.67 K⋅:=A1 16.1154:=
1-Propanol:
Antoine coefficients:
It is impractical to provide solutions for all of the systems listed in the
table on Page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the
appropriate parameter values substituted for those given. The file
WILSON.mcd reproduces the table of Wilson parameters on Page 474
and includes the necessary Antoine coefficients.
12.13
22
24
28
32
Peq x()
x yeq x(),
T 333.15 K=
P,x,y Diagram at
420
T 90 273.15+()K⋅:=
Guess:
P 101.33 kPa⋅:=
T-x,y diagram at
a21 1351.90 cal
mol
⋅:=a12 775.48 cal
mol
⋅:=
V2 18.07 cm3
mol
⋅:=V1 75.14 cm3
mol
⋅:=
Parameters for the Wilson equation:
421
Teq x()
K
373.149
364.159
361.836
361.264
360.985
360.881
360.99
361.418
362.364
364.195
367.626
370.349
=
x
0
0.05
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
1
=yeq x()
0
0.304
0.381
0.407
0.429
0.453
0.484
0.527
0.589
0.686
0.858
1
=
T,x,y Diagram at P 101.33 kPa⋅:=
0 0.2 0.4 0.6 0.8 1
360
375
γ2x1x2,T,( ) exp x12τ12 T() G12 T()
⎛
⎜
⎞
2
⋅
…
⎡
⎢
⎤
⎥
⋅
⎡
⎢
⎤
⎥
:=
⎜
⎢
⎢
⎥
⎥
⎢
⎢
⎥
⎥
G21 T( ) exp α−τ21 T()⋅
()
:=G12 T( ) exp α−τ12 T()⋅
()
:=
τ21 T() b21
RT⋅
:=
τ12 T() b12
RT⋅
:=
α0.5081:=b21 1636.57 cal
mol
⋅:=b12 500.40 cal
mol
⋅:=
Parameters for the NRTL equation:
⎢
⎥
⎢
⎥
C2 230.170 K⋅:=B2 3885.70 K⋅:=A2 16.3872:=
Water:
C1 205.807 K⋅:=B1 3483.67 K⋅:=A1 16.1154:=
1-Propanol:
Antoine coefficients:
It is impractical to provide solutions for all of the systems listed in the
table on Page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the
appropriate parameter values substituted for those given. The file
NRTL.mcd reproduces the table of NRTL parameters on Page 474 and
includes the necessary Antoine coefficients.
12.14
⎢
⎥
⎢
⎥
P-x,y diagram at T 60 273.15+()K⋅:=
Guess: P 70 kPa⋅:=
Peq x()
kPa
20.007
28.892
30.783
30.959
31.127
31.163
30.922
30.271
29.03
26.868
23.124
20.275
=
x
0
0.05
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
1
=yeq x()
0
0.33
0.382
0.39
0.404
0.427
0.459
0.503
0.564
0.659
0.836
1
=
424
α0.5081:=
b21 1636.57 cal
mol
⋅:=b12 500.40 cal
mol
⋅:=
Parameters for the NRTL equation:
C2 230.170 K⋅:=B2 3885.70 K⋅:=A2 16.3872:=
Water:
C1 205.807 K⋅:=B1 3483.67 K⋅:=A1 16.1154:=
1-Propanol:
Antoine coefficients:
It is impractical to provide solutions for all of the systems listed in the
table on Page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the
appropriate parameter values substituted for those given. The file
NRTL.mcd reproduces the table of NRTL parameters on Page 474 and
includes the necessary Antoine coefficients.
12.15
0 0.2 0.4 0.6 0.8
35
x yeq x(),
T 333.15 K=
P,x,y Diagram at
425
τ12 T() b12
RT⋅
:= τ21 T() b21
RT⋅
:=
G12 T( ) exp α−τ12 T()⋅
()
:= G21 T( ) exp α−τ21 T()⋅
()
:=
T-x,y diagram at P 101.33 kPa⋅:=
Guess: T 90 273.15+()K⋅:=
426
x 0 0.05,1.0..:=
Teq x()
K
373.149
361.745
361.066
360.843
360.697
360.709
360.985
361.66
362.974
365.442
367.449
370.349
=
x
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.95
1
=yeq x()
0
0.377
0.402
0.415
0.434
0.462
0.5
0.552
0.629
0.754
0.853
1
=
0 0.2 0.4 0.6 0.8 1
360
375
Λ21 T() V1
V2 exp a21−
RT⋅
⎛
⎜
⎝
⎞
⎠
⋅:=Λ12 T() V2
V1 exp a12−
RT⋅
⎛
⎜
⎝
⎞
⎠
⋅:=
V2 18.07 cm3
mol
⋅:=V1 75.14 cm3
mol
⋅:=
Parameters for the Wilson equation:
⎢
⎥
⎢
⎥
C2 230.170 K⋅:=B2 3885.70 K⋅:=A2 16.3872:=
Water:
C1 205.807 K⋅:=B1 3483.67 K⋅:=A1 16.1154:=
1-Propanol:
Antoine coefficients:
It is impractical to provide solutions for all of the systems listed in the
table on Page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the appropriate
parameter values substituted for those given. The file WILSON.mcd
reproduces the table of Wilson parameters on Page 474 and includes the
necessary Antoine coefficients.
12.16
428
(c) P,T-flash Calculation
PPdew Pbubl
+
2
:= T 60 273.15+()K⋅:= z1 0.3:=
x1 0.1:= x2 1 y1−:=
Guess: V 0.5:=
y1 0.1:= y2 1 x1−:=
(a) BUBL P: T 60 273.15+()K⋅:= x1 0.3:= x2 1 x1−:=
Guess: P 101.33 kPa⋅:= y1 0.4:= y2 1 y1−:=
Given y1 P⋅x1 γ1x1x2,T,()⋅Psat1T()⋅=
(b) DEW P: T 60 273.15+()K⋅:= y1 0.3:= y2 1 y1−:=
Guess: P 101.33 kPa⋅:= x1 0.1:= x2 1 x1−:=
429
Guess: P 101.33 kPa⋅:= x1 0.3:= x2 1 y1−:=
y1 0.3:= y2 1 x1−:=
x1
x2
⎛
⎜
⎜
⎞
⎟
x1 1 V−()⋅y1 V⋅+ z1=Eq. (10.15)
x1
x2
⎛
⎜
⎞
(d) Azeotrope Calculation
Test for azeotrope at: T 60 273.15+()K⋅:=
γ101,T,( ) 21.296= γ210,T,( ) 4.683=
Since one of these values is >1 and the other is <1, an azeotrope exists.
See Ex. 10.3(e)
430
γ1x1x2,T,( ) exp x22τ21 T() G21 T()
x1 x2 G21 T()⋅+
⎛
⎜
⎝
⎞
⎠
2
⋅
G12 T()τ12 T()⋅
…
⎡
⎢
⎢
⎢
⎤
⎥
⎥
⎥
⋅
⎡
⎢
⎢
⎢
⎤
⎥
⎥
⎥
:=
α0.5081:=
b21 1636.57 cal
mol
⋅:=b12 500.40 cal
mol
⋅:=
Parameters for the NRTL equation:
⎢
⎥
⎢
⎥
C2 230.170 K⋅:=B2 3885.70 K⋅:=A2 16.3872:=
Water:
C1 205.807 K⋅:=B1 3483.67 K⋅:=A1 16.1154:=
1-Propanol:
Antoine coefficients:
It is impractical to provide solutions for all of the systems listed in the
table on Page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the
appropriate parameter values substituted for those given. The file
NRTL.mcd reproduces the table of NRTL parameters on Page 474 and
includes the necessary Antoine coefficients.
12.17
431
⎜
⎢
⎢
⎥
⎥
⎢
⎢
⎥
⎥
⎢
⎥
⎢
⎥
y1 P⋅x1 γ1x1x2,T,()⋅Psat1T()⋅=
(c) P,T-flash Calculation
PPdew Pbubl
+
2
:= T 60 273.15+()K⋅:= z1 0.3:=
x1 0.1:= x2 1 y1−:=
Guess: V 0.5:=
y1 0.1:= y2 1 x1−:=
Given y1 x1 γ1x1x2,T,()⋅Psat1T()⋅
P
=x1 x2+1=
(a) BUBL P: T 60 273.15+()K⋅:= x1 0.3:= x2 1 x1−:=
Guess: P 101.33 kPa⋅:= y1 0.4:= y2 1 y1−:=
(b) DEW P: T 60 273.15+()K⋅:= y1 0.3:= y2 1 y1−:=
Guess: P 101.33 kPa⋅:= x1 0.1:= x2 1 x1−:=
Given
432
P 101.33 kPa⋅:= x1 0.3:= x2 1 x1−:=
y1 0.3:= y2 1 x1−:=
x1
x2
⎛
⎜
⎜
⎞
⎟
x1
x2
⎛
⎜
⎞
(d) Azeotrope Calculation
Test for azeotrope at: T 60 273.15+()K⋅:=
γ101,T,( ) 19.863= γ210,T,( ) 4.307=
Since one of these values is >1 and the other is <1, an azeotrope exists.
See Ex. 10.3(e).
Guess:
433