VLE Pada Tekanan Moderat dan Rendah

Referensi:

1) Smith Van Ness. 2001. Introduction to Chemical Engineering Thermodynamic,

6th ed.

2) Sandler. 2006. Chemical, Biochemical adn Engineering Thermodynamics, 4th

ed.

VLE Pada Tekanan Moderat dan Rendah

L

i

V

i

f

f



0

L

V





(

i = 1, 2, . . ., N

)

(1)

(2)

0

i

L

i

i

V

i

i

φ

P

x

f

y





(2)

o i

(4)

1





V

i

sat

i

L

i

i

i

P

x

P

y





(5)

(3)

1





L

i

sat

i

i

i

P

x

P

o

o

γ

o

γ

Perhitungan VLE

1

x₁ y P

20 25 30 35 40

P

/k

P

a

P-x₁(RL)

P-x₁

P-y₁ T= 45o_C

0 5 10 15 20

0,00 0,50 1,00

P

fraksi mol isopropanol

P₂ P₁

P₁(RL) P2(RL)

1 x₁ y P 1 P₂ P P 1 P₂ P

γ

γ





















_sat i i i i

P

x

P

y

ln

ln

4059

0

3669

0

138

18

x

4753

0

928

35

x

3463

0

2 1

,

ln

,

,

,

,

,

ln

P

x

P

y

ln

ln

sat 1 1 1

















































_{i i 1 1 2 2}

E

γ

ln

x

γ

ln

x

γ

ln

x

RT

G

RT

x

/x

γ γ

1

x₁ y

1,500 2,000 2,500

Dua parameter Persamaan Margules

E

γ γ

0,000 0,500 1,000

0,00 0,50 1,00

x1 GE_/x

1x2RT

ln γ2

lnγ1

Contoh 2

. Menentukan parameter Van Laar untuk kesetimbangan uap-cair

Data eksperimen Nilai prediksi

x₁ Pex_{(mmHg) P}calc ∆_{P y}calc

L i V i

f

f



sat i L i i V i

i

φ

P

x

P

y





sat

P

x

P

y





φ

sat i i i i

P

x

P

y





sat 2 2 2 1 2 2 sat 1 2 2 1 2 1

P

x

x

x

exp

x

P

x

x

x

exp

x

P

















































































21 12 12 21 21 12 12 12

A

A

A

A

A

A

A

A

P

P

P



y

₁



y

₂

,1

a

T

a

a

P

log

3 2,1 1,1 sat

1







2 3 2,2 1,2 sat 2 , a T a a P log   



A



₂₁

A

₁₂







x

,

,

P

₁

1

x₂ xcalc

Pexp

Pcalc

PPexp



_{calc calc}



2

P

P 

γ1 γ2 γ1 γ2

1

x₂ x₁

2



1

₂ 1

y₂ y

γ1 γ2

P-x

P-y

P

₂sat

x₁, y₁

P-y

γ γ

ln

224

47

,

945

.

2

2724

,

14

n

l







C

T

/kPa

P

o sat

1

209

47

,

972

.

2

2043

,

14

n

l







C

T

/kPa

P

₂sat _o

209



C

T

o

2

sat

sat 1 1

1

P

x

P

y



sat 2 2

2

P

x

P

y



y

₁

+ y

₂

=

1

:

y

₁

+ y

₂

=

1

:

sat 2 2 sat

1

1

P

x

P

x

P





x

₂

= 1-x

₁

sat 2 1 sat

1

1

P

x

P

x

kPa

98

41

,

P

₂sat



kPa

21

83

,

P

₁sat



P

= 41,98 + (83,21 – 41,98) (0,6) = 66,72 kPa

P

= 41,98 + (83,21 – 41,98) (0,6) = 66,72 kPa

P

P

x

y

sat 1 1 1



 



7483

,

0

72

,

66

21

,

83

6

,

0



t =75o_C

60 80 100

/k

P

a b'

c'

b a

c

P₁sat_=83,21

cairan subcoolid

40 60

P

/k c' c

d

P₂sat_=41,98

P-y1 P-x1

1 1

1 sat

1

C

P

A

B

T







n

l

sat 2 sat 1

sat 2 1

P

P

P

P

x







1

P

5156

0

84

,

46

76

,

91

84

,

46

70

,

x

₁















6759

0

70

76

,

91

5156

,

0

,

P

P

x

y

sat 1 1

1







P =70 kPa

80 85 90 C c' c d

t₂sat_=89,58

t-y 1 t-x 1 uap superjenuh 70 75 t/ o C b' b a

t₁sat_=89,58

cairan subcoolid

2 2 1

Ax

γ

ln



T

A



2

,

771



0

,

00523

2 1 2

Ax

γ

ln



424 , 33

31 , 643 . 3 59158 ,

16 ln

  

T P₁sat

424 , 53

54 , 665 . 2 25326 , 14

  

T P

kPa

51

44

,

P

₁sat



P

₂sat



65

,

64

kPa

A= 2,771 –(0,00523) (318,15) = 1,107

 

Ax

e

xp





1

,

107



0

,

75



2



1

,

864

exp

γ

2

2

1







   

282 0 50

73

51 44 864 1 25 0

, ,

, , ,

y₁  

 

Ax

e

xp





1

,

107



0

,

75





1

,

864

exp

γ

1



2





 

Ax

e

xp





1

,

107



0

,

25



2



1

,

072

exp

γ

2

1

2







sat 2 2 2 sat

1 1

1

γ

P

x

γ

P

x

γ1 =γ2= 1

γ1dan γ2

o

sat 2 2 2 sat 1 1

1

γ

P

y

γ

P

y

1

P





sat 1 1

1 1

P

γ

P

y

x



o

o

1





₁

2

x

x

60 70 80

p-x

P-y

50 60 70 80

p-x

P-y

X₁=0,25 X1=0,0322

P = 67,404 kPa

P₂sat

20 30 40 50

P

/kP

a

20 30 40 50

P

/kP

a _P

i i

i sat

i C

P ln A

B

T 

 

sat 2 2 2 1 1 sat

1

P

γ

x

γ

x

P

P





o

T

_o

= T

₁sat

.x

1

+ T

2sat

.x

2

o

o

P

o

1 sat 1 1

C P

A B

T 

 

ln

o

o o

   

 



 α

γ

y

γ

y P P

2 2

1 1 sat

1

o

1

C B

T   ₁

sat 1 1

C P

A B

T 

 

ln

o

332 334 336 338 340

T

/K

Tx

Ty

332 334 336 338 340

T

/K

Tx

Ty

324 326 328 330

0 0,2 0,4 0,6 0,8 1

324 326 328 330