solution thermodynamics: applications chapter 12-part iii

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Solution Thermodynamics: Applications Chapter 12-Part III

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Page 1: Solution Thermodynamics: Applications Chapter 12-Part III

Solution Thermodynamics: Applications

Chapter 12-Part III

Page 2: Solution Thermodynamics: Applications Chapter 12-Part III
Page 3: Solution Thermodynamics: Applications Chapter 12-Part III

Other models for GE/RT

0...

0..

....)()( 22121

21

CB

CBA

xxCxxBARTxx

G E

Page 4: Solution Thermodynamics: Applications Chapter 12-Part III

Obtain activity coefficients from the one-parameter Margules equation

12212

212

,,1

21

2121

ln)1(

)1(

)(

2

AxxAx

n

n

n

An

nRT

nG

nn

nA

RT

Gn

n

n

n

nAxAx

RT

G

nPT

E

E

E

Page 5: Solution Thermodynamics: Applications Chapter 12-Part III

problem

• For methanol(1)/methyl acetate(2), the 1-parameter 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

Page 6: Solution Thermodynamics: Applications Chapter 12-Part III

VLE, BUBL P calculation

satiiii PxPy For i =1, 2

Calculate 1 and 2 at T and x1 =0.25 using Margules 1-parameter

A(T) = 1.107

1 = 1.8642 = 1.072

and calculate P = 73.5 kPa and y1 = 0.282

Calculate P and {yi} given T = 318.15K and x1 =0.25

Page 7: Solution Thermodynamics: Applications Chapter 12-Part III

Calculate P and {xi} given T = 318.15K and y1 =0.60

VLE, DEW P calculationP1

sat, 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

sat

satsat

P

Pyx

Py

Py

P

11

11

22

2

11

1

1

Evaluate 1, 2, and return to 1)until P converges

1)Solution:

P = 62.89 kPax1 = 0.8171 = 1.0382 = 2.094

Page 8: Solution Thermodynamics: Applications Chapter 12-Part III

Calculate T and {yi} given P = 101.33 kPa and x1 =0.85

ii

isati C

PA

BT

ln

VLE, BUBL T calculationTo obtain an initial T, get the saturation temperatures of both components (from Antoine)

Use a mole-fraction weighted average of these values to get T

T1sat = 337.71; T2

sat = 330.08 K

For that T calculate A, 1, 2 and =P1sat/P2

sat

/22111 xx

PP sat

Then calculate

Get T from Antoine and return to (1)

(1)

111

1

lnC

PA

BT

sat

Once T converges,calculate y1

Page 9: Solution Thermodynamics: Applications Chapter 12-Part III

Calculate T and {xi} given P = 101.33 kPa and y1 =0.40

VLE, DEW T calculationSame 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, P2

sat,

Calculate x1 = y1P/1 P1sat Calculate 1, 2

2

2

1

11

yyPP sat

New value of T from Antoine and return to (1)

111

1

lnC

PA

BT

sat

Once T converges,calculate x1

Page 10: Solution Thermodynamics: Applications Chapter 12-Part III

Find the azeotropic pressure and the azeotropic composition for T = 318.15 K

Define the relative volatility

2

2

1

1

12

xyx

y

How much is 12 at the azeotrope?

Get 12 from the VLE equations

Page 11: Solution Thermodynamics: Applications Chapter 12-Part III

From the one-parameter Margules equation

212

221 ln;ln AxAx

)exp(

;)exp(

2

1112

2

1012 11 AP

P

P

APsat

sat

xsat

sat

x

Calculate these values from the data at T = 318.15K

224.0;052.2 112012 11 xx

This means that 12 is =1 at some point between x1 = 0 and x1 = 1

Page 12: Solution Thermodynamics: Applications Chapter 12-Part III

Double azeotrope

Page 13: Solution Thermodynamics: Applications Chapter 12-Part III

At the azeotrope, 12 = 1

kPaPP

yx

xA

xA

P

P

P

P

satazaz

azaz

az

az

az

sat

sat

az

az

sat

sat

76.73

;325.0

388.0)21(ln

)21(ln

47.1

;

11

11

1

2

1

12

1

1

2

2

1

22

1112

Page 14: Solution Thermodynamics: Applications Chapter 12-Part III

The Van-Laar equation

2

1'12

2'21'

212

2

2'21

1'12'

121

2'211

'12

'21

'12

21

1ln

1ln

xA

xAA

xA

xAA

xAxA

AA

RTxx

GE