Transcript Class 2. VLE in its simplest form

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
ˆf v  Py
i
i
ˆf l  P sat x
i
i
i
VLE according to Raoult’s law:
Py1  P x
sat
1
1
Py 2  P x2
sat
2
Acetonitrile (1)/nitromethane (2)
• Antoine equations for saturation pressures:
sat
1
ln P
ln P2sat
2,945.47
/ kPa  14.2724 o
T / C  224
2,972.64
/ kPa  14.2043 o
T / C  209
Calculate P vs. x1 and P vs. y1 at 75 oC
Diagram is at constant T
Bubble line
66.72
Dew line
0.75
Calculate the P-x-y diagram
Knowing T and x1, calculate P and y1
Py1  P1sat x1
Py2  P2sat x2
Sum m ing:
P  P1sat x1  P2sat (1  x1 )  ( P1sat  P2sat ) x1  P2sat
Bubble pressure calculations
x1P1sat
y1 
P
y2  1  y1
Diagram is at constant T
59.74
0.43
Knowing T and y1, get P and x1
Py1
Py1  P x  sat  x1
P1
sat
1
1
Py2
Py2  P x2  sat  x2
P2
sat
2
sum m ing
P
1
y1
y2
 sat
sat
P1
P2
Dew point calculation
In this diagram, the pressure
is constant
78oC
0.51
0.67
Calculate a T-x1-y1 diagram
Py1  P (T ) x1
sat
1
(1)
Py2  P (T ) x2
sat
2
Ti
sat
Given P and y1 solve
for T and x1
(2)
Bi

 Ci
Ai  ln P
Why is this temperature
a reasonable guess?
get the two saturation temperatures
Then select a temperature from the range between
T1sat and T2sat
At the selected T,
summing (1) and (2) solve for x1
Given P and x1, get T and y1
Py1  P x
sat
1
1
Py2  P x2
sat
2
P  P x  P x2
sat
1
1
sat
1
sat
2
P
P

sat
P2
P
sat
2
x1  x2  P
sat
2

P
P1sat
x  x2
sat 1
P2
Iterate to find T, then calculate y1
2,945.47
T / oC  224
2,972.64
/ kPa  14.2043 o
T / C  209
ln P1sat / kPa  14.2724
ln P2sat
(I)
P1sat
2,945.47 2,972.64
ln sat  0.0681

P2
T  224 T  209
sat
2
P

P
sat
1
sat
2
P
P
x1  x2
(III)
(II)
Estimate P1sat/P2sat using a guess T
Then calculate P2sat from (III)
Then get T from (I)
Compare calculated T with guessed T
Finally, y1 = P1sat x1/P and y2 = 1-y2
In this diagram, the pressure
is constant
Dew points
Bubble
points
78oC
76.4
0.51
0.75
Knowing P and y, get T and x
• Start from point c last slide (70 kPa and y1= 0.6)
Py1
Py1  P x  x1  sat
P1
sat
1
1
Py2
Py2  P x2  x2  sat
P2
sat
2
 y1
y2
1  P sat  sat
P2
 P1
sat


P
  P1sat  P y1  y2 1sat
P2





Iterate to find T, and then calculate x
sat
1
ln P
ln P2sat
2,945.47
/ kPa  14.2724 o
T / C  224
2,972.64
/ kPa  14.2043 o
T / C  209
P1sat
2,945.47 2,972.64
ln sat  0.0681

P2
T  224 T  209
sat
1
P
sat

P1
 P y1  y2 sat
P2




(I)
Estimate P1sat/P2sat using a
guess T
Then calculate P1sat from (III)
And then get T from (I)
(II)
(III)
x1= Py1/P1sat
79.6
0.44
Ki = yi/xi
Ki = Pisat/P
Read
Examples
10.4, 10.5, 10.6
Flash Problem
mass balance:
V, {yi}
L + V =1
mass balance component i
zi = xi L + yi V
T and P
for i = 1, 2, …n
zi = xi (1-V) + yi V
1 mol of
L-V mixture
overall
composition {zi}
Using Ki values, Ki = yi/xi
L, {xi}
xi= yi /Ki;
yi = zi Ki/[1 + V(Ki -1)]
read and work examples 10.5 and 10.6
zi K i
i yi  i 1  V K  1  1
i
Flash calculations
F=2-p+N
Line of critical points
For a binary
F=4-p
For one phase:
P, T, x (or y)
Subcooled-liquid
above the upper
surface
Superheated-vapor
below the under surface
L is a bubble point
W is a dew point
LV is a tie-line
Each interior loop represents the PT
behavior of a mixture of fixed composition
In a pure component, the bubble and dew
lines coincide
What happens at points A and B?
Critical point of a mixture is the point where
the nose of a loop is tangent to the envelope
curve
Tc and Pc are functions of
composition, and do not necessarily
coincide with the highest T and P
How do we calculate a P-T envelope?
At the left of C, reduction
of P leads to vaporization
At F, reduction in P leads to
condensation and then
vaporization (retrograde
condensation)
Important in the operation of
deep natural-gas wells
At constant pressure,
retrograde vaporization
may occur
Fraction of the
overall system
that is liquid
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 of
the more volatile species
obtainable by distillation
at this pressure
(these are mixture CPs)
azeotrope
This is a mixture of very dissimilar
components
The P-x curve in (a) lies below
Raoult’s law; in this case there are stronger
intermolecular attractions between unlike
than between like molecular pairs
This behavior may result in a minimum
point 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 weaker
intermolecular attractions between unlike
than between like molecular pairs; it could
end as L-L immiscibility
This behavior may result in a maximum
point as in (d), where x1=y1, it is also
an azeotrope
Usually distillation is carried
out at constant P
Minimum-P azeotrope is a
maximum-T (maximum boiling)
Point (case b)
Maximum-P azeotrope is a
minimum-T (minimum boiling)
Point (case d)
Limitations of Raoult’s law
When 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
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
yi P  xi H i
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
Fugacity vapor
Py1  P  x
sat
1
1 1
Py2  P  2 x2
sat
2
Fugacity liquid
 is the activity coefficient, a function of composition and
temperature
It corrects for non-idealities in the Liquid phase