Λ21 T() V1
V2 exp a21−
RT⋅
⎛
⎜
⎝
⎞
⎠
⋅:=Λ12 T() V2
V1 exp a12−
RT⋅
⎛
⎜
⎝
⎞
⎠
⋅:=
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:
Psat2T( ) exp A2 B2
T 273.15 K⋅−()C2+
−
⎡
⎢
⎣
⎤
⎥
⎦kPa⋅:=
Psat1T( ) exp A1 B1
T 273.15 K⋅−()C1+
−
⎡
⎢
⎣
⎤
⎥
⎦kPa⋅:=
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.
12.18
434
y1 P⋅x1 γ1x1x2,T,()⋅Psat1T()⋅=
(c) P,T-flash Calculation
TTdew Tbubl
+
2
:= P 101.33 kPa⋅:= z1 0.3:=
x1 0.1:= x2 1 y1−:=
Guess: V 0.5:=
y1 0.1:= y2 1 x1−:=
(a) BUBL T: P 101.33 kPa⋅:= x1 0.3:= x2 1 x1−:=
Guess: T 60 273.15+()K⋅:= y1 0.3:= y2 1 y1−:=
(b) DEW T: P 101.33 kPa⋅:= y1 0.3:= y2 1 x1−:=
Guess: T 60 273.15+()K⋅:= x1 0.1:= x2 1 y1−:=
Given
435
x1 y1=y1 y2+1=y2 P⋅x2 γ2x1x2,T,()⋅Psat2T()⋅=
x1 x2+1=y1 P⋅x1 γ1x1x2,T,()⋅Psat1T()⋅=Given
y2 1 x1−:=y1 0.4:=x2 1 y1−:=x1 0.4:=T 60 273.15+()K⋅:=
Since one of these values is >1 and the other is <1, an azeotrope exists.
See Ex. 10.3(e). Guesses:
⎜
(d) Azeotrope Calculation
Test for azeotrope at: P 101.33 kPa⋅:=
Tb1 B1
A1 ln P
kPa
⎛
⎜
⎝
⎞
⎠
−
C1−
⎛
⎜
⎜
⎝
⎞
⎠
273.15 K⋅+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
:= Tb1 370.349 K=
Tb2 B2
A2 ln P
kPa
⎛
⎜
⎝
⎞
⎠
−
C2−
⎛
⎜
⎜
⎝
⎞
⎠
273.15 K⋅+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
:= Tb2 373.149 K=
436
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:
Psat2T( ) exp A2 B2
T 273.15 K⋅−()C2+
−
⎡
⎢
⎣
⎤
⎥
⎦kPa⋅:=
Psat1T( ) exp A1 B1
T 273.15 K⋅−()C1+
−
⎡
⎢
⎣
⎤
⎥
⎦kPa⋅:=
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.
12.19
437
(b) DEW T: P 101.33 kPa⋅:= y1 0.3:= y2 1 x1−:=
Guess: T 90 273.15+()K⋅:= x1 0.05:= x2 1 y1−:=
Given y1 P⋅x1 γ1x1x2,T,()⋅Psat1T()⋅=
x1 x2+1=
y2 P⋅x2 γ2x1x2,T,()⋅Psat2T()⋅=
⎜
γ1x1x2,T,( ) exp x22τ21 T() G21 T()
x1 x2 G21 T()⋅+
⎛
⎜
⎝
⎞
⎠
2
⋅
G12 T()τ12 T()⋅
x2 x1 G12 T()⋅+()
2
+
…
⎡
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎦
⋅
⎡
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎦
:=
γ2x1x2,T,( ) exp x12τ12 T() G12 T()
x2 x1 G12 T()⋅+
⎛
⎜
⎝
⎞
⎠
2
⋅
G21 T()τ21 T()⋅
x1 x2 G21 T()⋅+()
2
+
…
⎡
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎦
⋅
⎡
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎦
:=
(a) BUBL T: P 101.33 kPa⋅:= x1 0.3:= x2 1 x1−:=
Guess: T 60 273.15+()K⋅:= y1 0.3:= y2 1 y1−:=
Given y1 P⋅x1 γ1x1x2,T,()⋅Psat1T()⋅=
y1 y2+1=
y2 P⋅x2 γ2x1x2,T,()⋅Psat2T()⋅=
438
⎜
(d) Azeotrope Calculation
Test for azeotrope at: P 101.33 kPa⋅:=
Tb1 B1
A1 ln P
kPa
⎛
⎜
⎝
⎞
⎠
−
C1−
⎛
⎜
⎜
⎝
⎞
⎠
273.15 K⋅+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
:= Tb1 370.349 K=
⎜
⎜
⎜
⎢
⎢
⎥
⎥
(c) P,T-flash Calculation
TTdew Tbubl
+
2
:= P 101.33 kPa⋅:= 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=
y2 x2 γ2x1x2,T,()⋅Psat2T()⋅
P
=y1 y2+1=
439
⎜
0
161.88−
291.27
⎛
⎞
⎢
⎥
C
228.060
239.500
230.170
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
B
2756.22
3638.27
3885.70
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
A
14.3145
16.5785
16.3872
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
V
74.05
40.73
18.07
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
Molar volumes & Antoine coefficients:12.20
⎜
Since one of these values is >1 and the other is <1, an azeotrope exists.
See Ex. 10.3(e). Guesses:
T 90 273.15+()K⋅:= x1 0.4:= x2 1 y1−:= y1 0.4:= y2 1 x1−:=
Given y1 P⋅x1 γ1x1x2,T,()⋅Psat1T()⋅=x1 x2+1=
y2 P⋅x2 γ2x1x2,T,()⋅Psat2T()⋅=y1 y2+1=x1 y1=
440
⎜
⎜
Given
PP
bubl
:=x31x
1
−x2
−:=x20.2:=x10.05:=
Guess:
y31y
1
−y2
−:=y20.4:=y10.3:=
DEW P calculation:
(a) BUBL P calculation: No iteration required.
x10.3:= x20.4:= x31x
1
−x2
−:=
γix,T,( ) exp 1 ln
j
xjΛij,T,()⋅
()
∑
⎡
⎢
⎣
⎤
⎥
⎦
p
xpΛpi,T,()⋅
j
xjΛpj,T,()⋅
()
∑
∑
+
…
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎦
−
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎦
:=
⎜
(b)
441
⎜
x1
x2
x3
y2
y3
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎞
⎟
⎟
⎟
⎟
⎟
x11V−()⋅y1V⋅+ z1
=Py
1
⋅x1γ1x,T,()⋅Psat 1 T,()⋅=
Given
Use x from DEW P and y from BUBL P as initial
guess.
V 0.5:=
Guess:
z31z
1
−z2
−:=z20.4:=z10.3:=
T 338.15 K=PPdew Pbubl
+
2
:=
P,T-flash calculation:(c)
⎜
442
⎜
⎜
γix,T,( ) exp j
τji,Gji,
⋅xj
⋅
()
∑
Gli,xl
⋅
()
∑
…
⎡
⎢
⎢
⎢
⎤
⎥
⎥
⎥
:=
x31x
1
−x2
−:=x20.4:=x10.3:=
BUBL P calculation: No iteration required.(a)
b
0
222.64
1197.41
184.70
0
845.21
631.05
253.88−
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
cal
mol
⋅:=α
0
0.3084
0.5343
0.3084
0
0.2994
0.5343
0.2994
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
NRTL parameters:
Psat i T,( ) exp Ai
Bi
T
K273.15−
⎛
⎜
⎝
⎞
⎠Ci
+
−
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
kPa⋅:=T 65 273.15+()K:=
C
228.060
239.500
230.170
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
B
2756.22
3638.27
3885.70
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
A
14.3145
16.5785
16.3872
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
V
74.05
40.73
18.07
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
Antoine coefficients:
Molar volumes & Antoine coefficients:12.21
443
⎜
⎜
⎢
⎥
⎢
⎥
(c) P,T-flash calculation: PPdew Pbubl
+
2
:= T 338.15 K=
z10.3:= z20.4:= z31z
1
−z2
−:=
Guess: V 0.5:= Use x from DEW P and y from BUBL P as initial
guess.
(b) DEW P calculation:
y10.3:= y20.4:= y31y
1
−y2
−:=
Guess: x10.05:= x20.2:= x31x
1
−x2
−:= PP
bubl
:=
Given
Py
1
⋅x1γ1x,T,()⋅Psat 1 T,()⋅=Py
2
⋅x2γ2x,T,()⋅Psat 2 T,()⋅=
Py
3
⋅x3γ3x,T,()⋅Psat 3 T,()⋅=
xi
∑1=
444
x31x
1
−x2
−:=x20.4:=x10.3:=
BUBL T calculation: (a)
p13..:=j13..:=i13..:=Λ ij,T,()
Vj
Vi
exp aij,
−
RT⋅
⎛
⎜
⎝
⎞
⎠
⋅:=
a
0
583.11
1448.01
161.88−
0
469.55
291.27
107.38
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
cal
mol
⋅:=
Wilson parameters:
P 101.33kPa:=Psat i T,( ) exp Ai
Bi
T
K273.15−
⎛
⎜
⎝
⎞
⎠Ci
+
−
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
kPa⋅:=
C
228.060
239.500
230.170
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
B
2756.22
3638.27
3885.70
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
A
14.3145
16.5785
16.3872
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
V
74.05
40.73
18.07
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
Molar volumes & Antoine coefficients:12.22
x1
x2
⎛
⎜
⎜
⎞
⎟
445
⎜
⎜
⎜
⎜
y31y
1
−y2
−:=y20.4:=y10.3:=
DEW T calculation:(b)
⎜
⎜
γix,T,( ) exp 1 ln
j
xjΛij,T,()⋅
()
∑
⎡
⎢
⎣
⎤
⎥
⎦
p
xpΛpi,T,()⋅
j
xjΛpj,T,()⋅
()
∑
∑
+
…
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎦
−
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎦
:=
446
x1
x2
x3
⎛
⎜
⎜
⎜
⎞
⎟
⎟
x21V−()⋅y2V⋅+ z2
=Py
2
⋅x2γ2x,T,()⋅Psat 2 T,()⋅=
x11V−()⋅y1V⋅+ z1
=Py
1
⋅x1γ1x,T,()⋅Psat 1 T,()⋅=Given
Use x from DEW P and y from BUBL P as initial
guess.
V 0.5:=
Guess:
z31z
1
−z2
−:=z20.2:=z10.3:=
T 340.75 K=TTdew Tbubl
+
2
:=
P,T-flash calculation:(c)
⎜
⎜
447
τji,T,()Gji,T,()⋅xj
⋅
()
∑
⎡
⎤
x31x
1
−x2
−:=x20.4:=x10.3:=
BUBL T calculation:(a)
⎜
⎜
0
184.70
631.05
⎛
⎜
⎞
0
0.3084
0.5343
⎛
⎜
⎞
NRTL parameters:
Psat i T,( ) exp Ai
Bi
T
K273.15−
⎛
⎜
⎝
⎞
⎠Ci
+
−
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
kPa⋅:=P 101.33kPa:=
C
228.060
239.500
230.170
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
B
2756.22
3638.27
3885.70
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
A
14.3145
16.5785
16.3872
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
V
74.05
40.73
18.07
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
:=
Antoine coefficients:
Molar volumes & Antoine coefficients:12.23
⎜
⎜
⎜
⎜
448
⎢
⎥
⎢
⎥
⎢
⎥
y20.4:= y31y
1
−y2
−:=
Guess: x10.05:= x20.2:= x31x
1
−x2
−:= TT
bubl
:=
Given
Py
1
⋅x1γ1x,T,()⋅Psat 1 T,()⋅=Py
2
⋅x2γ2x,T,()⋅Psat 2 T,()⋅=
Py
3
⋅x3γ3x,T,()⋅Psat 3 T,()⋅=
i
xi
∑1=
x1
⎛
⎞
Guess: T 300K:= y10.3:= y20.3:= y31y
1
−y2
−:=
Given
i
∑
y1
⎛
⎞
(b) DEW T calculation:
y10.3:=
449
x1
x2
⎛
⎜
⎜
⎜
⎞
⎟
⎟
x11V−()⋅y1V⋅+ z1
=Py
1
⋅x1γ1x,T,()⋅Psat 1 T,()⋅=Given
Use x from DEW P and y from BUBL P as initial
guess.
V 0.5:=
Guess:
z31z
1
−z2
−:=z20.2:=z10.3:=
T 341.011 K=TTdew Tbubl
+
2
:=
P,T-flash calculation:(c)
450
⎜
⎜
12.27 V158.63 cm3
mol
⋅:= V2118.46 cm3
mol
⋅:=
moles1
750 cm3
⋅
V1
:= moles2
1500 cm3
⋅
V2
:=
moles moles1moles2
+:= moles 25.455 mol=
x1
moles1
moles
:= x10.503=x21x
1
−:=
VE x1x2
⋅1.026−0.220 x1x2
−
()
⋅+
⎡
⎣
⎤
⎦
⋅cm3
mol
⋅:= VE 0.256−cm3
mol
=
By Eq. (12.27), VVEx
1V1
⋅+ x2V2
⋅+:= V 88.136 cm3
mol
=
12.26 x10.4:= x21x
1
−:= V1110 cm3
mol
:= V290 cm3
mol
:=
VE x1x2
,
()
x1x2
⋅45 x1
⋅25 x2
⋅+
()
⋅cm3
mol
:= VE x1x2
,
()
7.92 cm3
mol
=
By Eq. (12.27): Vx
1x2
,
()
VE x1x2
,
()
x1V1
⋅+ x2V2
⋅+:=
Vx
1x2
,
()
105.92 cm3
mol
=
451
Since there are 11 moles of solution per mole of solute, the result on the
basis of 1 mol of solution is
2(HCl + 2.25 H2O —–> HCl(2.25 H2O)) (1)
HCl(4.5 H2O) —–> HCl + 4.5 H2O (2)
———————————————-
HCl(4.5 H2O) + HCl —–> 2 HCl(2.25 H2O)
12.29
12.28 LiCl.2H2O —> Li + 1/2 Cl2 + 2 H2 + O2 (1)
Li + 1/2 Cl2 + 10 H2O —> LiCl(10 H2O) (2)
2(H2 + 1/2 O2 —> H2O) (3)
——————————————————————–
LiCl.2H2O + 8 H2O(l) —> LiCl(10 H2O)
452
Assume 3 steps in the process:
1. Heat M1 moles of water from 10 C to 25 C
2. Unmix 1 mole (0.8 moles water + 0.2 moles LiCl) of 20 % LiCl solution
3. Mix (M1 + 0.8) moles of water and 0.2 moles of LiCl
Basis: 1 mole of 20% LiCl solution entering the process. 12.31
nLiCl n’LiCl
+0.7667 kmol=
nH2O
nLiCl n’LiCl
+8.15=
Mole ratio, final solution:
nH2O
nLiCl
21.18=
Mole ratio, original solution:
nH2O
0.9 125⋅
18.015 kmol⋅:=nLiCl
0.1 125⋅
42.39 kmol⋅:=
Calculate moles of LiCl and H2O in original solution:12.30
453