APPLICATIONS Applications of Raoult’s law Qualitative description of phase diagrams for mixtures.
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Transcript of APPLICATIONS Applications of Raoult’s law Qualitative description of phase diagrams for mixtures.
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APPLICATIONS
• Applications of Raoult’s law
• Qualitative description of phase diagrams for mixtures
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Raoult’s law
• Model the vapor phase as a mixture of ideal gases:
• Model the liquid phase as an ideal solution
ivi Pyf ˆ
isati
li xPf ˆ
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VLE according to Raoult’s law:
222
111
xPPy
xPPysat
sat
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Acetonitrile (1)/nitromethane (2)
• Antoine equations for saturation pressures:
209/
64.972,22043.14/ln
224/
47.945,22724.14/ln
2
1
CTkPaP
CTkPaP
osat
osat
Calculate P vs. x1 and P vs. y1 at 75 oC
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66.72
0.75
Bubble line
Dew line
Diagram is at constant T
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Calculate the P-x-y diagram
satsatsatsatsat
sat
sat
PxPPxPxPP
:Summing
xPPy
xPPy
21211211
222
111
)()1(
12
111
1 yyP
Pxy
sat
Bubble pressure calculations
Knowing T and x1, calculate P and y1
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59.74
0.43
Diagram is at constant T
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Knowing T and y1, get P and x1
satsat
satsat
satsat
Py
Py
P
summing
xP
PyxPPy
xP
PyxPPy
2
2
1
1
22
2222
11
1111
1
Dew point calculation
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In this diagram, the pressureis constant
78oC
0.51 0.67
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Calculate a T-x1-y1 diagram
ii
isati
sat
sat
CPA
BT
xTPPy
xTPPy
ln
)(
)(
222
111
get the two saturation temperaturesThen select a temperature from the range between T1
sat and T2sat
At the selected T,summing (1) and (2) solve for x1
(1)
(2)
Why is this temperaturea reasonable guess?
Given P and y1 solvefor T and x1
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Given P and x1, get T and y1
212
1221
2
1
2
2211
222
111
xxPP
PPxx
P
P
P
P
xPxPP
xPPy
xPPy
sat
satsat
sat
sat
sat
satsat
sat
sat
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Iterate to find T, then calculate y1
209/
64.972,22043.14/ln
224/
47.945,22724.14/ln
2
1
CTkPaP
CTkPaP
osat
osat
209
64.972,2
224
47.945,20681.0ln
2
1
TTP
Psat
sat
212
12
xxPP
PP
sat
satsat
Estimate P1
sat/P2sat using a guess T
Then calculate P2sat from (III)
Then get T from (I)Compare calculated T with guessed T
(I)
(II)
(III)
Finally, y1 = P1sat x1/P and y2 = 1-y2
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In this diagram, the pressureis constant
78oC
0.51
76.4
0.75
Dew pointsBubblepoints
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Knowing P and y, get T and x
• Start from point c last slide (70 kPa and y1= 0.6)
sat
satsat
satsat
satsat
satsat
P
PyyPP
P
y
P
yP
P
PyxxPPy
P
PyxxPPy
2
1211
2
2
1
1
2
22222
1
11111
1
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Iterate to find T, and then calculate x
sat
satsat
P
PyyPP
2
1211
209/
64.972,22043.14/ln
224/
47.945,22724.14/ln
2
1
CTkPaP
CTkPaP
osat
osat
209
64.972,2
224
47.945,20681.0ln
2
1
TTP
Psat
sat
Estimate P1sat/P2
sat using a guess TThen calculate P1
sat from (III)And then get T from (I)
(I)
(II)
(III)
x1= Py1/P1sat
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79.6
0.44
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Ki = yi/xi
Ki = Pisat/P
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ReadExamples 10.4, 10.5, 10.6
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Flash Problem
T and P
1 mol ofL-V mixtureoverallcomposition {zi}
V, {yi}
L, {xi}
mass balance:
L + V =1
mass balance component i
zi = xi L + yi V for i = 1, 2, …n
zi = xi (1-V) + yi V
Using Ki values, Ki = yi/xi
xi= yi /Ki;
yi = zi Ki/[1 + V(Ki -1)]
read and work examples 10.5 and 10.6 1
11
i i
ii
ii KV
Kzy
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Flash calculations
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F=2-+N
For a binary
F=4-
For one phase:P, T, x (or y)
Subcooled-liquidabove the upper surface
Superheated-vaporbelow the under surface
L is a bubble point
W is a dew point
LV is a tie-line
Line of critical points
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Each interior loop represents the PTbehavior of a mixture of fixed composition
In a pure component, the bubble and dewlines coincide
What happens at points A and B?
Critical point of a mixture is the point wherethe nose of a loop is tangent to the envelopecurve
Tc and Pc are functions of composition, and do not necessarilycoincide with the highest T and P
How do we calculate a P-T envelope?
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Fraction of theoverall systemthat is liquid
At the left of C, reductionof P leads to vaporization
At F, reduction in P leads tocondensation and then vaporization (retrograde condensation)
Important in the operation ofdeep natural-gas wells
At constant pressure, retrograde vaporization may occur
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Class exercise
• From Figure 10.5, take P = 800 psia and generate a table T, x1, y1. We call ethane component 1 and heptane component 2. In the table complete all the T, x1, y1 entries that you can based on Figure 10.5. For example, at T= 150 F, x1 = 0.771, we don’t know y1 (leave it empty for now). Continue for all the points at P = 800 psia. Once the table is complete, graph T vs. x1, y1. Also fill in the empty cells in the table reading the values from the graph.
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Minimum and maximum ofthe more volatile speciesobtainable by distillation at this pressure(these are mixture CPs)
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This is a mixture of very dissimilarcomponents
azeotrope
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The P-x curve in (a) lies belowRaoult’s law; in this case there are strongerintermolecular attractions between unlikethan between like molecular pairs
This behavior may result in a minimumpoint as in (b), where x1=y1 Is called an azeotrope
The P-x curve in (c) lies above Raoult’s law; in this case there are weakerintermolecular attractions between unlikethan between like molecular pairs; it could end as L-L immiscibility
This behavior may result in a maximumpoint as in (d), where x1=y1, it is alsoan azeotrope
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Usually distillation is carriedout at constant P
Minimum-P azeotrope is amaximum-T (maximum boiling)Point (case b)
Maximum-P azeotrope is a minimum-T (minimum boiling)Point (case d)
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Limitations of Raoult’s lawWhen a component critical temperature is < T, the saturation pressure is not defined.
Example: air + liquid water; what is in the vapor phase? And in the liquid?
Calculate the mole fraction of air in water at 25oC and 1 atm
Tc air << 25oC
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Henry’s law
iii HxPy
For a species present at infinite dilution in the liquid phase,
The partial pressure of that species in the vapor phase is directly proportional to the liquid mole fraction
Henry’s constant
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Calculate the mole fraction of air in water at 25oC and 1 atm.
First calculate y2 (for water, assuming that air does not dissolve in water)
Then calculate x1 (for air, applying Henry’s law)
See also Example 10.2
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Modified Raoult’s law
2222
1111
xPPy
xPPysat
sat
Fugacity vapor
Fugacity liquid
is the activity coefficient, a function of composition andtemperature
It corrects for non-idealities in the Liquid phase