Termodinamika Lanjut (PTK 213 ) (Advance Thermodynamics)

(1)

Termodinamika Lanjut

(PTK 213 )

(Advance Thermodynamics)

Dr. Istadi, ST, MT

Ir. Danny Soetrisnanto, MEng

Year 2010-2011

Master Program in Chemical Engineering,

Diponegoro University

LITERATURES

●

Credit : 3 credits/SKS

●

Evaluations:



●

References/Textbook:



Smith, J.M., Van Ness, H.C., and Abbott, M.M. (2001).

Introduction to

Chemical Engineering Thermodynamics

. 6th Edition. New York: McGraw

Book Co.



Elliot, J. R. and Lira, C.T., (1999),

Introductory Chemical Engineering

Thermodynamics

, Prentice Hall PTR.

(2)

Outlines for 2

nd

_{Stage Course}

1. Introduction to

Multicomponents VLE Systems

2. VLE Calculation in Mixtures

by an Equation of State

3. Activity Models



Modified Raoult's laws



Margules Equation



Van Laar Equation



Regular-Solution Theory



_{Wilson's Equation}



UNIQUAC



UNIFAC

4. Fitting Activity Models

to Experimental Data

Solving Problem with: EXCEL, MATLAB, CHEMCAD, and/or

HYSYS

Introduction to

Multicomponents VLE Systems

Click to add text

(3)

Phase Diagrams (T-xy & P-xy)

● Click to add an outline

Mass Balance

●

F = L + V (over all)



F (initial mole number), L (moles of liquid), V (moles of vapor)

●

==> 1 = L/F + V/F

●

z

A

F = y

A

V + x

A

L (z

A

= overall mole fraction)

●

==> z

A

= y

A

.V/F + x

A

.L/F

●

Percentage of liquid: L/F = (z

A

-y

A

) / (x

A

-y

A

) ==>

●

Percentage of vapor: V/F = (x

A

-z

A

) / (x

A

-y

A

) ==>

●

Remember that: L/F + V/F = 1

de

ce cd

ce

(4)

Activity, Activity Coefficient,

Fugacity Coefficient

● Click to add an outline

Fugacity for Gas Mixtures

• The simplest type of mixture bevavior is

IDEAL GAS

BEHAVIOR

• A

component fugacity coefficient

is to quantify the

deviations from component behavior in ideal-gas

mixtures.

• Fugacity of a vapor-phase component in real solutions:

• IDEAL SOLUTIONS are intermediate between ideal

gases and real mixtures.

For non-ideal Liquid ===?



i

=



f

_i

y

_i

P



f

_i

=y

_i

P



f

_i

=y

_i

_

_i

P

(5)

Fugacity of Non-Ideal Liquid

Mixtures

●

For ideal gases ==>

γ

i

= 1 and f

i

o

_=P

●

For LIQUID:



Activity: ==>



Activity Coefficient ==>

GAMMA APPROACH

●

Remember:

●

Fugacity of component i in LIQUID:

●

For low to moderate pressure, f

i

o

≈

P

i

sat

a

_i

= 

f

_i

/

f

i

o

γ

_i

= 

f

_i

/

x

_i

f

i

o



f

_i

=y

_i

P

f

i

o

=

i

sat

P

i

sat

exp



V

i

L



P

i

−

P

i

sat



RT





f

i

L

=

γ

i

x

i

P

i

sat



f

i

L

=

γ

i

x

i

f

i

o

=

γ

i

x

i



i

sat

P

i

sat

exp



V

i

L



P

i

−

P

i

sat



RT



Summary for Component

Fugacities

(6)

Ideal Solutions

●

_{Ideal Solutions}

:



No synergistic effect of the components in mixture



each component operates independently



no energy change for mixing



no volume change



●

_{LEWIS/RANDALL}

Rule:



f

_i

is

f

_i

=x

i

 

f

i

is

_=x

i

f

i

VLE in Ideal Solutions

• Bagaimana menghitung

K

i

≡

y

i

/x

i

• Equilibrium constraint:

• In ideal solutions:

• Fugacity of the liquid:

• Combining the equations:

• Dinyatakan dalam rasio

K

i

:

• Pada tekanan rendah:



Hukum Raoult





f

_i

V

= 

f

_i

L

y

_i

f

_i

V

=

x

_i

f

_i

L

f

i

L

=

i

sat

P

i

sat

exp



V

i

L



P

i

−

P

i

sat



RT



y

_i



i

V

P=x

_i



i

sat

P

i

sat

exp



V

i

L



P

i

−

P

i

sat



RT



K

_i

=

y

_x

i i

=

P

isat

P





isat

exp

[

V

_iL



P

i

−

P

_isat



/

RT

]



iV





i sat



i

=

1, dan exp

[

V

iL



P

i

−

P

isat



/

RT

]

K

_i

=

P

i sat

P

_i

atau

y

_i

P

₌

x

_i

P

_i

sat

(7)

System of Raoult’s Law



binary system

● Click to add an outline

Click to add title

(8)

Shortcut Estimation of VLE K-ratios

K

_i

₌

P

i

sat

P

≈

P

_c,i

10

7

3 

1 

ω





1 −

1 T

_{r, i}



P

VLE CALCULATIONS

● Jenis-jenis Perhitungan VLE:



Bubble-point Pressure (BP)



Dew-point Pressure (DP)



_{Bubble-point Temperature (BT)}



Dew-point Temperature (DT)

(9)

Jenis-jenis Perhitungan

Kesetimbangan Fase

Paling sulit

x

i

, y

i

, L/F

P, T, z

i

FL

Sulit

T, x

_i

P, y

_i

=z

_i

DT

Sulit

T, y

_i

P, x

_i

=z

_i

BT

Mudah

P, x

_i

T, y

_i

=z

_i

DP

Paling

mudah

P, y

_i

T, x

_i

=z

_i

BP

Konvergensi

Kriteria

Dihitung

Diketahui

Tipe

∑

i

y

_i

=

∑

i

K

_i

x

_i

=

1 ∑

i

x

_i

=

∑

i

y

_i

K

_i

=

1 ∑

i

y

_i

=

∑

i

K

_i

x

_i

=

1 ∑

i

x

_i

₌

∑

i

y

_i

K

_i

=

1 ∑

i

z

_i

_

1 −

K

_i





K

_i

_

L

_/

F

_

1 −

K

_i



Perhitungan Kesetimbangan

Fasa untuk Hukum Raoult Biner



Bubble Pressure Calculation

:



Tidak diperlukan iterasi, karena temperature dan

tekanan uap diketahui.



Hk Raoult



linear bubble pressure line (P-x,y)

∑

i

y

_i

=

1, or

∑

i

K

_i

x

_i

=

1 ,

∑

i

P

_isat

P

x

i

=

1 P

₁

sat

P

x

1 

P

₂

sat

P

x

2 =

1 P

₌

x

₁

P

₁

sat

_

x

₂

P

₂

sat

x

₂

₌

1 −

x

₁

(10)

Hukum Raoult Biner …..(2)



Dew-Pressure Calculation:



Diselesaikan tanpa iterasi, sebab tekanan uap adalah

tertentu pada temperatur yang ditentukan, sehingga:

∑

i

x

_i

=

1, or

∑

i

y

_i

K

_i

=

1 y

₁

P

₁

sat



y

₂

P

₂

sat

=

1 P

=

1 y

₁

P

₁

sat



y

₂

P

₂

sat

Hukum Raoult Biner …..(3)



Bubble-Temperature Calculation:



Diselesaikan dengan iterasi

Temperatur

(yang

mengubah

P

_i

sat

),

hingga tekanan sama dengan tekanan

yang diketahui

.

∑

i

y

_i

=

1, or

∑

i

K

_i

x

_i

=

1 _P

₌

_x

1 P

1 sat



x

₂

P

₂

sat

(11)

Hukum Raoult Biner …..(4)



Dew-Temperature Calculation:



Diselesaikan dengan iterasi

Temperatur

(yang

mengubah

P

_i

sat

),

hingga tekanan sama dengan tekanan

yang diketahui

.

∑

i

x

_i

=

1, or

∑

i

y

_i

K

_i

=

1 P

=

_y

1

1 P

₁

sat



y

₂

P

₂

sat

Hukum Raoult Biner …..(5)



Flash-drum Calculation:



Feed: liquid/cairan (vaporized after entering flash drum)



Feed composition =

z

_i

dan L/F = liquid-to-feed ratio



V/F =

1 -L/F,

Component balance:

==>

y

_i

=K

_i

x

_i

==>

z

_i

=

x

_i

L

F



y

i

V

F

x

_i

=

z

i

K

_i



L

F



1 −

K

i



y

_i

=

z

i

K

i

K

_i



L

F



1 −

K

i



(12)

Binary Flash Calculation ....(6)

● Dalam perhitungan flash, L/F harus diiterasi

hingga

Σ

x

_i

=1

,

● Tetapi dalam flash, kita juga harus

menyelesaikan

Σ

y

_i

=1

● Untuk penyelesaian uap dan cairan, maka

secara simultan: (

Σ

x

_i

-

Σ

y

_i

)=0

==> fungsi

objective

● Note that: 0<L/F<1

● Kasus-kasus flash:



flashing liquid



_{partial condensation}

● z

i

, feed flow rate, P, T ==> diketahui

Multicomponent VLE

Calculations

● Bubble Calculation:

● Dew Calculation:

● Rules:



bubble- & dew-pressure calculation ==> no iteration

required



bubble- & dew-temperature calculation ==> iteration

required

∑

i

y

_i

=

1,

atau

∑

i

x

_i

K

_i

=

∑

i

x

_i

P

_i

sat

P

=

1 ∑

i

x

_i

₌

1,

atau

∑

i

y

_i

K

_i

=

P

∑

i

y

_i

P

_i

sat

=

1

(13)

Multicomponent VLE Calculations...(2)

●

Tebakan awal Temperatur ==> scr. kasar

●

General formula for

_{ISOTHERMAL FLASH}

Calculation:

T

=

∑

i

x

_i

T

_i

sat

atau T

=

∑

i

y

_i

T

_{r ,i}

T

_i

sat

∑

i

y

_i

T

_{c, i}

∑

i

x

_i

−

∑

i

y

_i

=

∑

i

z

_i

_

1 −

K

_i



K

_i

_

L

_/

F

_

1 −

K

_i



=

D

i

=

0 Contoh Perhitungan VLE dgn

MS Excel

Produk atas suatu kolom distilasi (seperti pada gambar)

mempunyai komposisi (

z

_i

) sebagai berikut: 23% propane, 67%

isobutane, dan 10% n-butane. Jika dianggap kolom ideal, uap

yang meninggalkan tray dalam keadaan keseimbangan fasa

dengan cairan yang meninggalkan tray tersebut. Dalam kasus

partial condenser

maka uap dan cairan meninggalkan

condensor

dalam keadaan kesetimbangan fasa.

a)

Hitung temperatur kondensor agar uap dari kolom distilasi bisa

terkondensasi semua pada tekanan 8 bar.

b)

Jika diasumsikan bahwa produk atas kolom distilasi

berkeseimbangan dengan cairan di tray paling atas, hitunglah

temperatur produk uap dan komposisi cairan di tray tersebut jika

dioperasikan pada tekanan 8 bar.

c)

Berapakah fraksi cairan hasil kondensasi, jika uap

terkondensasi dlm sebuah kondensor parsial pada 8 bar dan

320 K

(14)

Penyelesaian

(a)

Temperatur dimana semua uap terkondensasi ==>

bubble point temperatur

Dengan MS Excel:

Dengan Interpolasi:

T = 310 + ((1,000-0,827)/(1,061-0,827))(320-310)=317 K*

Tekanan (bar) =

8 Tebak T (K) =

310 Tebak T (K) =

320 zi

Pci (bar) Tci (K)

ω

Ki

yi

Ki

yi

0,23

42,48

369,8

0,152

1,609

0,370

2,027

0,47

0,67

36,48

408,1

0,181

0,612

0,410

0,795

0,53

0,1

37,96

425,1

0,200

0,433

0,043

0,571

0,06

1 0,82

1,06

● Click to add an outline

1 1 2 3 Flash Feed Vapor Liquid Stream No. 1 2 3 Name Feed Vapor Produc Liquid Produ Overall

-Molar flow kmol/h 1.0000 0.0000 1.0000 Mass flow kg/h 54.8968 0.0000 54.8968 Temp K 100.0000 0.0000 319.7421 Pres bar 8.0000 0.0000 8.0000 Vapor mole fraction 0.0000 0.0000 0.0000 Vapor mass fraction 0.0000 0.0000 0.0000 Enth MJ/h -166.77 0.00000 -142.68 Heating values (60 F)

Gross J/kmol 2.722E+009 2.722E+009 Net J/kmol 2.511E+009 2.511E+009 Actual vol m3/h 0.0743 0.0000 0.1072 Std liq m3/h 0.0991 0.0000 0.0991 Std vap 0 C m3/h 22.4136 0.0000 22.4136 Component mole fractions

Propane 0.230000 0.000000 0.230000 I-Butane 0.670000 0.000000 0.670000 N-Butane 0.100000 0.000000 0.100000

(15)

(b). Dew point Temperature

(b)

uap kesetimbangan dgn cairan ==> Uap jenuh ==>

dew

point temperatur

Dengan MS Excel:

Dengan interpolasi:

T = 325 + ((1,00-0,994)/(1,123-0,994))(320-325) = 324,8 K*

Tekanan (bar) =

8 Tebak T (K) =

325 Tebak T (K) =

320 zi

Pci (bar) Tci (K)

ω

Ki

xi

Ki

xi

0,23

42,48

369,8

0,152

2,262

0,102

2,027

0,11

0,67

36,48

408,1

0,181

0,900

0,744

0,795

0,84

0,1

37,96

425,1

0,200

0,651

0,154

0,571

0,18

1,000

0,999

1,13

● Click to add an outline

(b). ChemCAD 2

-Molar flow kmol/h 1.0000 1.0000 0.0000 Mass flow kg/h 54.8968 54.8968 0.0000 Temp K 100.0000 324.9329 0.0000 Pres bar 8.0000 8.0000 0.0000 Vapor mole fraction 0.0000 1.000 0.0000 Enth MJ/h -166.77 -125.26 0.00000 Heating values (60 F)

Gross J/kmol 2.722E+009 2.722E+009 Net J/kmol 2.511E+009 2.511E+009 Actual vol m3/h 0.0743 2.8142 0.0000 Std liq m3/h 0.0991 0.0991 0.0000 Std vap 0 C m3/h 22.4136 22.4136 0.0000 Component mole fractions

(16)

(c). Isothermal Flash Calculation

(c) Partial condenser

==> Flash to Liquid and Vapor

Dengan MS Excel:

Dengan interpolasi:

L/F = 0,7684 K

Tekanan (bar)

8 Temperature (K)=

320 Tebak L/F

0,5

Tebak L/F =

0,6

zi

Pci (bar) Tci (K)

ω

Ki

Di

0,23

42,48

369,8

0,152

2,027

-0,1560

-0,1674

0,67

36,48

408,1

0,181

0,795

0,1531

0,1497

0,1

37,96

425,1

0,200

0,571

0,0546

0,0518

1,000

0,0517

0,0341

Tebak L/F =

0,77

Ki

Di

2,027

-0,1910

0,795

0,1442

0,571

0,0476

0,0008

x

_i

=

z

i

K

_i



L

F



1 −

K

i



y

_i

=

z

i

K

i

K

i



L

F



1 −

K

i



∑

i

x

i

−

∑

i

y

i

=

∑

i

z

i



1 −

K

i



K

_i



L

/

F



1 −

K

_i



=

D

i

=

0 Click to add title

● Click to add an outline

-Molar flow kmol/h 1.0000 0.0356 0.9644 Mass flow kg/h 54.8968 1.8823 53.0145 Temp K 100.0000 320.0000 320.0000 Pres bar 8.0000 8.0000 8.0000 Vapor mole fraction 0.0000 1.000 0.0000 Enth MJ/h -166.77 -4.3367 -137.72 Heating values (60 F)

Gross J/kmol 2.722E+009 2.626E+009 2.726E+009 Net J/kmol 2.511E+009 2.421E+009 2.514E+009 Average mol wt 54.8968 52.8252 54.9733 Actual vol m3/h 0.0743 0.0992 0.1035 Std liq m3/h 0.0991 0.0034 0.0956 Std vap 0 C m3/h 22.4136 0.7987 21.6150 Component mole fractions