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
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
VLE according to Raoult’s law:
222
111
xPPy
xPPysat
sat
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
66.72
0.75
Bubble line
Dew line
Diagram is at constant T
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
59.74
0.43
Diagram is at constant T
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
In this diagram, the pressureis constant
78oC
0.51 0.67
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
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
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
In this diagram, the pressureis constant
78oC
0.51
76.4
0.75
Dew pointsBubblepoints
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
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
79.6
0.44
Ki = yi/xi
Ki = Pisat/P
ReadExamples 10.4, 10.5, 10.6
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
Flash calculations
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
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?
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
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.
Minimum and maximum ofthe more volatile speciesobtainable by distillation at this pressure(these are mixture CPs)
This is a mixture of very dissimilarcomponents
azeotrope
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
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)
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
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
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
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
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