To find the maximum, set dVE/dx1 = 0 and solve for x1. Then use x1 to
find VEmax.
(b)
⎡
⎣
⎤
⎦
⎡
⎣
⎤
⎦
VEx1x2
⋅abx
1
⋅+ cx
1
()
2
⋅+
⎡
⎣
⎤
⎦
⋅=
By definition of the excess properties
600
x1ix1,
c 250:=b 3000−:=a 3000−:=
Guess:(a)
358
y21y
1
−:=y10.5:=P 2 bar⋅:=T 75 273.15+()K⋅:=
Propane = 1; n-Pentane = 211.33
Discussion:
⎡
⎣
⎤
⎦
x1 x1,
VEbar1x1() 1x1−()
2a2b⋅x1⋅+ 3c⋅x1()
2
⋅+
⎡
⎣
⎤
⎦
⋅:=
(c)
359
dBdT
ij
yiyj
⋅dBdTij,
⋅
()
∑∑
:= dBdT 3.55 cm3
mol K⋅
=
Use a spline fit of B as a function of T to
find derivatives:
b11
331−
276−
235−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
cm3
mol
⋅:= b22
980−
809−
684−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
cm3
mol
⋅:= b12
558−
466−
399−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
cm3
mol
⋅:=
t
50
75
100
⎛
⎜
⎜
⎜
⎜
⎞
⎟
273.15+
⎡
⎢
⎢
⎢
⎤
⎥
⎥
⎥
K⋅:= t
323.15
348.15
373.15
⎛
⎜
⎜
⎜
⎞
⎟
K=
Differentiate Eq. (11.61):
360
0 0.2 0.4 0.6 0.8
0.94
0.96
0.98
1
⎢
⎥
φhat1 y1( ) exp P
RT⋅B11,1y1−()
2δ12,
⋅+
⎡
⎣
⎤
⎦
⋅
⎡
⎢
⎣
⎤
⎥
⎦
:=
By Eqs. (11.63a) and (11.63b):
δij,2B
ij,
⋅Bii,
−Bjj,
−:=
j1n..:=
B276−
466−
466−
809−
⎛
⎜
⎝
⎞
⎠
cm3
mol
⋅:= i1n..:=n2:=
y21y
1
−:=y10.5:=P 2 bar⋅:=T 75 273.15+()K⋅:=
Propane = 1; n-Pentane = 211.34
0
x1ix1,
⎜
⎜
⎜
⎢
⎥
c 0.01:=b 100−:=a 500−:=
Guess:(a)
()
45.7−
66.5−
118.2−
176.6−
204.2−
174.1−
116.8−
43.5−
22.6−
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
0.0817
0.1177
0.2107
0.3472
0.5163
0.6810
0.8181
0.9276
0.9624
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
11.36
362
0 0.2 0.4 0.6 0.8
1000
500
x1
HEbar1x1( ) HE x1() 1x1−()
x1
HE x1()
d
d
⋅+:=
(c)
HE x1( ) x1 1 x1−()⋅abx1⋅+ cx1
2
⋅+
()
⋅:=x1 0.5:=
Guess:
To find the minimum, set dHE/dx1 = 0 and solve for x1. Then use x1 to
find HEmin.
(b)
⎡
⎣
⎤
⎦
⎡
⎣
⎤
⎦
HEx1x2
⋅abx
1
⋅+ cx
1
()
2
⋅+
⎡
⎣
⎤
⎦
⋅=
By definition of the excess properties
363
0.307
0.2485
0.082
⎛
⎝
⎞
⎠
⎜
508.2
464.851
⎛
⎝
⎞
⎠
⎜
0.233
0.25
⎝
⎠
⎜
1
3Vcj
1
3
⎡
⎢
⎤
⎥
3
209
214.65
⎜
47.104
45.013
⎛
⎞
11.37 (a) (1) = Acetone (2) = 1,3-butadiene
364
⎜
B 598.524−cm3
mol
=B
1
n
i1
n
j
yiyj
⋅Bij,
⋅
()
∑
=
∑
=
:=
Eq. (11.61)
Bij,RTc
ij,
⋅
Pcij,B0ij,ωij,B1ij,
⋅+
()
⋅:=
Eq. (11.69a) + (11.69b)
⎜
0.874−
0.558−
0.098
⎛
⎞
B1ij,B1Trij,
()
:=
Eq. (3.66)
0.74636−
0.6361−
0.16178−
⎛
⎜
⎞
B0ij,B0Trij,
()
:=
Eq. (3.65)
⎜
Prij,P
Pcij,
:=Trij,T
Tcij,
:=
365
Differentiating Eq. (11.61) and using Eq. (11.69a) + (11.69b)
dBdT
1
n
i1
n
j
yiyj
⋅R
Pcij,dB0dTrij,ωij,dB1dTrij,
⋅+
()
⋅
⎡
⎢
⎣
⎤
⎥
⎦
⋅
⎡
⎢
⎣
⎤
⎥
⎦
∑
=
∑
=
:=
366
z1:=
Guess:
q
4.559
3.234
4.77
3.998
4.504
4.691
3.847
2.473
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
Eq. (3.54)
qΨ
ΩTr1.5
⋅
⎛
⎜
⎝
⎞
⎠
→
⎯
⎯⎯
⎯
:=β
0.02
0.133
0.069
0.036
0.081
0.028
0.04
0.121
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
Eq. (3.53)
βΩ
Pr
Tr
⋅
⎛
⎜
⎝
⎞
⎠
→
⎯
⎯
⎯
:=
Ψ0.42748:=Ω 0.08664:=
Redlich/Kwong Equation:11.38
Pr
0.244
2.042
0.817
0.474
0.992
0.331
0.544
2.206
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
Pr P
Pc
→
⎯
:=Tr
1.054
1.325
1.023
1.151
1.063
1.034
1.18
1.585
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
Tr T
Tc
→
⎯
:=
ω
.187
.000
.210
.224
.087
.301
.012
.038
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
Pc
61.39
48.98
48.98
73.83
50.40
30.25
45.99
34.00
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
Tc
308.3
150.9
562.2
304.2
282.3
507.6
190.6
126.2
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
P
15
100
40
35
50
10
25
75
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
T
325
200
575
350
300
525
225
200
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
Data for Problems 11.38 – 11.40
367
z1:=
Guess:
α1c1Tr
0.5
−
()
⋅+
⎡
⎤
2
→
⎯
⎯⎯⎯⎯⎯⎯
⎯
:=c 0.480 1.574 ω⋅+ 0.176 ω2
⋅−
()
→
⎯
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
⎯
:=
Ψ0.42748:=Ω 0.08664:=
Soave/Redlich/Kwong Equation11.39
Eq. (6.65)
Iiln Zβiqi
,
()
βi
+
Zβiqi
,
()
⎛
⎜
⎝
⎞
⎠
:=
i18..:=
⋅
368
⎯
⎯
⎯
⎯
q
5.383
3.946
5.658
4.598
5.359
5.527
4.646
2.924
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
Eq.(3.54)
qΨα⋅
ΩTr⋅
⎛
⎜
⎝
⎞
⎠
→
⎯
⎯
⎯
:=β
0.018
0.12
0.062
0.032
0.073
0.025
0.036
0.108
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
Eq.(3.53)
βΩ
Pr
Tr
⋅
⎛
⎜
⎝
⎞
⎠
→
⎯
⎯
⎯
:=
α1c1Tr
0.5
−
()
⋅+
⎡
⎣
⎤
⎦
2
→
⎯
⎯⎯⎯⎯⎯⎯
⎯
:=
c 0.37464 1.54226 ω⋅+ 0.26992 ω2
⋅−
()
→
⎯
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
⎯
:=
Ψ0.45724:=Ω 0.07779:=ε 12−:=σ 12+:=
Peng/Robinson Equation11.40
fiφiPi
⋅:=
Eq. (11.37)
φiexp Z βiqi
,
()
1−ln Z βiqi
,
()
βi
−
()
−qiIi
⋅−
()
:=
Eq. (6.65)
Iiln Zβiqi
,
()
βi
+
Zβiqi
,
()
⎛
⎜
⎝
⎞
⎠
:=
i18..:=
Zβq,
()
Find z():=
Eq. (3.52)
z1β+ qβ⋅ zβ−
zz β+
()
⋅
⋅−=Given
369
Eq. (3.66)
B1 B1Tr()
→
⎯
⎯
:=
Eq. (3.65)
B0 B0Tr()
→
⎯
⎯
:=
Evaluation of φ:
Pr P
Pc
→
⎯
:=Tr T
Tc
→
⎯
:=
ω
.187
.224
.301
.012
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:=
Pc
61.39
73.83
30.25
45.99
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:=
P
15
35
10
25
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:=
Tc
308.3
304.2
507.6
190.6
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:=
T
325
350
525
225
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:=
φ BY GENERALIZED CORRELATIONS
Parts (a), (d), (f), and (g) — Virial equation:
Eq. (6.65)
Ii
1
22
⋅ln Zβiqi
,
()
σβ
i
⋅+
Zβiqi
,
()
εβ
i
⋅+
⎛
⎜
⎝
⎞
⎠
⋅:=
i18..:=
Zβq,
()
Find z():=
Eq. (3.52)
z1β+ qβ⋅ zβ−
zεβ⋅+
()
zσβ⋅+
()
⋅
⋅−=Given
z1:=
Guess:
370
Assume an ideal solution since n-octane and iso-octane are non-polar and
very similar in chemical structure. For an ideal solution, there is no heat of
mixing therefore the heat transfer rate is zero.
a)
x20.667=x21x
1
−:=x10.333=x1
ndot1
ndot3
:=
ndot3ndot1ndot2
+:=ndot24kmol
hr
:=ndot12kmol
hr
:=
11.43
⎜
⎜
⎯
⎯
DB0 0.675
Tr2.6
→
⎯
⎯
:= Eq. (6.89) DB1 0.722
Tr5.2
→
⎯
⎯
:= Eq. (6.90)
⎯
⎯
⎜
⎜
Parts (b), (c), (e), and (h) — Lee/Kesler correlation:
Interpolate in Tables E.13 – E.16:
φ0
.7454
.7517
.7316
.8554
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:= φ1
1.1842
0.9634
0.9883
1.2071
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:= ω
0.000
0.210
0.087
0.038
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:=
371
c) To calculate the entropy change, treat the process in two steps:
1. Demix the air to O2 and N2
2. Mix the N2 and combined O2 to produce the enhanced air
11.44 For air entering the process: xO21 0.21:= xN21 0.79:=
For the enhanced air leaving the process: xO22 0.5:= xN22 0.5:=
ndot250 mol
sec
:=
a) Apply mole balances to find rate of air and O2 fed to process
Guess: ndotair 40 mol
sec
:= ndotO2 10 mol
sec
:=
Given
372
b 13.549−J
mol K⋅
=b1
3
i
Bi
∑
=
3
:=
Use averaged b value
⎜
⎜
⎜
⎜
⎝
⎠
Rearrange to find b using estimated a and c values along with GE and T data.
GE a−Tln T
K
⎛
⎜
⎝
⎞
⎠
⋅T−
⎛
⎜
⎝
⎞
⎠
⋅bT⋅+ c+=
GE is of the form:
c 1.544 103
×J
mol
=c intercept T HE,():=
a 2.155−J
mol K⋅
=a slope T HE,():=
Find a and c using the given HE and T values.
HE c a T⋅+=
Assume Cp is constant. Then HE is of the form:
HE
932.1
893.4
845.9
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
J
mol
:=GE
544.0
513.0
494.2
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
J
mol
:=T
10
30
50
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
K 273.15K+:=
11.50
373
⎜
⎜