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b)For water as solvent:
M
s
18.015
gm
mol
:=
For CO2 in H2O:
k
i
0.034
mol
kg bar
⋅
:=
By Eq. (5):
H
i
1
M
s
k
i
⋅
:=
H
i
1633bar
=
Ans.
The value is Table 10.1 is 1670 bar.
The values agree within about 2%.
10.36
Acetone:
Psat
1
T
() e
14.3145
2756.22
T
degC
228.060
+
−
kPa
⋅
:=
Acetonitrile
Psat
2
T
14.8950
3413.10
T
degC
250.523
+
−
kPa
⋅
:=
a) Find BUBL P and DEW P values
T
50degC
:=
x
1
0.5
:=
y
1
0.5
:=
10.35 a) The equation from NIST is:
M
i
k
i
y
i
⋅
P
⋅
=
Eq. (1)
x
i
H
i
⋅
y
i
P
⋅
=
Eq. (2)
Solving to eliminate P gives:
H
i
M
i
k
i
x
i
⋅
=
Eq. (3)
By definition:
M
i
n
i
n
s
M
s
⋅
=
where M is the molar mass and the
subscript s refers to the solvent.
Dividing by the toal number of moles gives:
M
i
x
i
x
s
M
s
⋅
=
Eq. (4)
Combining Eqs. (3) and (4) gives:
H
i
1
x
s
M
s
⋅
k
i
⋅
=
If x
i
is small, then x
s
is approximately equal to 1 and:
H
i
1
M
s
k
i
⋅
=
Eq. (5)
333
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1
Psat
1
T
()
⋅
y
1
P
⋅
=
1x
1
−
Psat
2
T
⋅
1y
1
−
P
⋅
=
x
1
DEWT
⎛
⎜
⎝
⎞
⎠
Find x
1
T
,
:=
DEWT
51.238degC
=
Ans.
At P = 0.5 atm, two
phases will form between T = 46.3 C and 51.2 C
10.37 Calculate x and y at T = 90 C and P = 75 kPa
Benzene:
Psat
1
T
() e
13.7819
2726.81
T
degC
217.572
+
−
kPa
⋅
:=
Toluene:
Psat
2
T
13.9320
3056.96
T
degC
217.625
+
−
kPa
⋅
:=
a) Calculate the equilibrium composition of the liquid and vapor at the flash T and P
T
90degC
:=
P
75kPa
:=
Guess:
x
1
0.5
:=
y
1
0.5
:=
BUBLP
x
1
Psat
1
T
⋅
1
−
Psat
2
T
⋅
+
:=
BUBLP
0.573atm
=
Ans.
DEWP
1
y
1
Psat
1
T
1
−
Psat
2
T
+
:=
DEWP
0.478atm
=
Ans.
At T = 50 C two
phases will form between P = 0.478 atm and 0.573 atm
b)Find BUBL T and DEW T values
P
0.5atm
:=
x
1
0.5
:=
y
1
0.5
:=
Guess:
T
50degC
:=
Given
x
1
Psat
1
T
⋅
1
−
Psat
2
T
⋅
+
P
=
BUBLT
Find T
:=
BUBLT
46.316degC
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 Spring '11
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