Transcript Solution Thermodynamics: Applications

Solution Thermodynamics:
Applications
Chapter 12-Part III
Other models for GE/RT
GE
 A  B ( x1  x2 )  C ( x1  x2 ) 2  ....
x1 x2 RT
A  B  C  ..  0
B  C  ...  0
Obtain activity coefficients from the
one-parameter Margules equation
E
G
n1 n2
 Ax1 x2  A
RT
n n
GE
n1
n
 A n2
RT
n
  (nG E

)
An2 1 n1

RT 

(  2)


n1
n n



T , P , n2
 Ax2 (1  x1 )  Ax  ln  1
2
2
problem
• For methanol(1)/methyl acetate(2), the 1parameter Margules equation gives a
reasonable prediction of the activity
coefficients, with A = 2.771 -0.00523 T.
• Vapor pressures as functions of T are
known. T are in Kelvin.
• a) Calculate P and {yi} for T = 318.15K and
x1 =0.25
Calculate P and {yi}
given T = 318.15K and x1 =0.25
VLE, BUBL P calculation
yi P  xi i Pi
sat
For i =1, 2
Calculate 1 and 2 at T and x1 =0.25 using Margules 1parameter
A(T) = 1.107
1 = 1.864
2 = 1.072
and calculate P = 73.5 kPa and y1 = 0.282
Calculate P and {xi}
given T = 318.15K and y1 =0.60
VLE, DEW P calculation
P1sat, P2sat, and A are the same as in the first part
But, we don’t know xi, and 1, 2 are functions of x1, x2
For good initial guesses, solve the problem with Raoult’s law
1
1) P 
Solution:
y1
y2

sat
 1 P1
 2 P2sat
P = 62.89 kPa
y1 P
x1 = 0.817
x1 
sat
 1 P1
1 = 1.038
Evaluate 1, 2, and return to 1)
2 = 2.094
until P converges
Calculate T and {yi}
given P = 101.33 kPa and x1 =0.85
VLE, BUBL T calculation
To obtain an initial T, get the saturation temperatures of both
components (from Antoine)
Ti sat
Bi

 Ci
Ai  ln P
T1sat = 337.71; T2sat = 330.08 K
Use a mole-fraction weighted average of these values to get T
(1)
For that T calculate A, 1, 2 and a =P1sat/P2sat
Then calculate
sat
1
P
B1
T
 C1
sat
A1  ln P1
P

x1 1  x2 2 / a
Get T from Antoine
and return to (1)
Once T converges,
calculate y1
Calculate T and {xi}
given P = 101.33 kPa and y1 =0.40
VLE, DEW T calculation
Same P as in BUBLT calculation, saturation temperatures are the same, get weighted
mole fraction average for initial T = 333.13 K
Since we don’t know {xi} use Raoult’s law to initialize {i}
(1) At the initial T, evaluate A, P1sat, P2sat, a
Calculate x1 = y1P/1 P1sat
sat
1
P
 y1 y2 
 P  a 
 1  2 
Calculate 1, 2
New value of T from Antoine
and return to (1)
T
Once T converges,
calculate x1
B1
 C1
sat
A1  ln P1
Find the azeotropic pressure and the
azeotropic composition for T = 318.15 K
Define the relative volatility
y1
a12 
y2
x1
x2
How much is a12 at the azeotrope?
Get a12 from the VLE equations
From the one-parameter Margules
equation
ln  1  Ax2 ; ln  2  Ax1
2
a12 x 0
1
2
P1sat exp(A)
P1sat

; a12 x1 1  sat
sat
P2
P2 exp(A)
Calculate these values from the data at T = 318.15K
a12 x 0  2.052; a12 x 1  0.224
1
1
This means that a12 is =1 at some point
between x1 = 0 and x1 = 1
Double azeotrope
At the azeotrope, a12 = 1
 1 P1sat
a12 
;
sat
 2 P2
1
P
 az 
 1.47
P
2
1
ln  A(1  2 x1 )
2
az
sat
2
sat
1
1
az
ln az  A(1  2 x1 )  0.388
2
az
 x1  y1az  0.325;
az
P   1 P1sat  73.76kPa
az
az
The Van-Laar equation
'
GE
A12' A21
 '
x1 x2 RT A12 x1  A21' x2

A x 

ln  1  A 1 
 A x 
'
12
'
12 1
'
21 2
2
 A x 

ln  2  A 1 
A x 

'
21
'
21 2
'
12 1
2