Termodinamika Lanjut (PTK 213 )

(Advance Thermodynamics)

Dr. Istadi, ST, MTIr. 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.

etc

Outlines for 2nd Stage Course

1. Introduction to Multicomponents VLE Systems2. VLE Calculation in Mixtures by an Equation of State3. 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

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Phase Diagrams (T-xy & P-xy)

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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

AF = y

AV + x

AL (z

A = overall mole fraction)

● ==> zA = y

A.V/F + x

A.L/F

● Percentage of liquid: L/F = (zA-y

A) / (x

A-y

A) ==>

● Percentage of vapor: V/F = (xA-z

A) / (x

A-y

A) ==>

● Remember that: L/F + V/F = 1

dece cd

ce

Activity, Activity Coefficient, Fugacity Coefficient

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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

yi P

f i =yi P

f i =yi i P

Fugacity of Non-Ideal Liquid Mixtures

● For ideal gases ==> γi = 1 and f

io =P

● For LIQUID: Activity: ==> Activity Coefficient ==>

GAMMA APPROACH● Remember:● Fugacity of component i in LIQUID:

● For low to moderate pressure, fio ≈P

isat

ai= f i/ fio

γi= f i /xi fio

f i =yi P

fio=

isat P

isat exp V

iL Pi−P

isat

RT

fiL =γi xi P

isat

fiL =γi xi f

io =γi xi

isat P

isat expV

iLPi−P

isat

RT

Summary for Component Fugacities

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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 iis

f i

=xi f iis =xi f i

VLE in Ideal Solutions

• Bagaimana menghitung Ki ≡ yi /xi

• Equilibrium constraint:• In ideal solutions:

• Fugacity of the liquid:• Combining the equations:

• Dinyatakan dalam rasio Ki :

• Pada tekanan rendah:

Hukum Raoult

f iV= f i

L

yi f iV=xi f i

L

fiL=

isat P

isatexp V

iLPi−P

isat

RT yii

V P=xiisat P

isat expV

iLPi−P

isat

RT Ki=

yi

xi

=P

isat

P isat exp [V i

LPi−Pisat /RT ]

iV

isat

i

=1, dan exp[V iLPi−P

isat /RT ]

Ki=Pi

sat

Pi

atauyi P= xi Pi

sat

System of Raoult’s Law binary system

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Shortcut Estimation of VLE K-ratios

Ki=Pi

sat

P≈

Pc,i 10

731ω1− 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) Isothermal Flash (FL)

Jenis-jenis Perhitungan Kesetimbangan Fase

Paling sulitxi, yi, L/FP, T, ziFL

SulitT, xiP, yi=ziDT

SulitT, yiP, xi=ziBT

MudahP, xiT, yi=ziDP

Paling mudah

P, yiT, xi=ziBP

KonvergensiKriteriaDihitungDiketahuiTipe

∑i

yi=∑i

K i xi=1

∑i

xi=∑i

yi

Ki

=1

∑i

yi=∑i

K i xi=1

∑i

xi=∑i

yi

Ki

=1

∑i

zi1−K i

KiL /F 1−Ki

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

yi=1, or ∑i

K i xi=1 ,∑i

Pisat

Pxi=1

P1sat

Px1

P2sat

Px2=1

P=x1 P1satx2 P2

sat

x2=1− x1

P=x1 P1sat1−x1 P2

sat=x1 P1sat−P2

sat P2sat

Hukum Raoult Biner …..(2)

Dew-Pressure Calculation:

Diselesaikan tanpa iterasi, sebab tekanan uap adalah tertentu pada temperatur yang ditentukan, sehingga:

∑i

xi=1, or ∑i

yi

Ki

=1y1 P

P1sat

y2 P

P2sat=1

P=1

y1

P1sat

y2

P2sat

Hukum Raoult Biner …..(3)

Bubble-Temperature Calculation:

Diselesaikan dengan iterasi Temperatur (yang mengubah Pi

sat), hingga tekanan sama dengan tekanan yang diketahui.

∑i

yi=1, or ∑i

K i xi=1 P=x1 P1satx2 P2

sat

Hukum Raoult Biner …..(4)

Dew-Temperature Calculation:

Diselesaikan dengan iterasi Temperatur (yang mengubah Pi

sat), hingga tekanan sama dengan tekanan yang diketahui.

∑i

xi=1, or ∑i

yi

Ki

=1 P=1

y1

P1sat

y2

P2sat

Hukum Raoult Biner …..(5)

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

Feed composition = zi dan L/F = liquid-to-feed ratio

V/F = 1-L/F, Component balance:

==>

==> yi=K

ix

i ==>

zi=xi

LF yi

VF

xi=zi

K iLF1−Ki

yi=zi Ki

K iLF1−Ki

Binary Flash Calculation ....(6)

● Dalam perhitungan flash, L/F harus diiterasi hingga Σx

i=1,

● Tetapi dalam flash, kita juga harus menyelesaikan Σy

i=1

● Untuk penyelesaian uap dan cairan, maka secara simultan: (Σx

i-Σy

i)=0 ==> fungsi

objective● Note that: 0<L/F<1● Kasus-kasus flash:

flashing liquid partial condensation

● zi, 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

yi=1, atau ∑i

xi K i =∑

i

xi Pisat

P= 1

∑i

xi=1, atau ∑i

yi

Ki

= P∑i

yi

Pisat= 1

Multicomponent VLE Calculations.....(2)

● Tebakan awal Temperatur ==> scr. kasar

● General formula for ISOTHERMAL FLASH Calculation:

T =∑i

xiT isat atau T =

∑i

yiT r ,i T isat

∑i

yiT c, i

∑i

xi−∑i

yi =∑i

zi1−KiKiL /F 1−K i

= Di = 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

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) = 8Tebak T (K) = 310 Tebak T (K) = 320

zi Pci (bar) Tci (K) ω Ki yi Ki yi0,23 42,48 369,8 0,152 1,609 0,370 2,027 0,470,67 36,48 408,1 0,181 0,612 0,410 0,795 0,530,1 37,96 425,1 0,200 0,433 0,043 0,571 0,06

1 0,82 1,06

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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

(a) with ChemCAD

(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) = 8Tebak T (K) = 325 Tebak T (K) = 320

zi Pci (bar) Tci (K) ω Ki xi Ki xi0,23 42,48 369,8 0,152 2,262 0,102 2,027 0,110,67 36,48 408,1 0,181 0,900 0,744 0,795 0,840,1 37,96 425,1 0,200 0,651 0,154 0,571 0,18

1,000 0,999 1,13

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(b). ChemCAD 21

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

(c). Isothermal Flash Calculation

(c) Partial condenser ==> Flash to Liquid and VaporDengan MS Excel:

Dengan interpolasi: L/F = 0,7684 K

Tekanan (bar) 8 Temperature (K)= 320Tebak L/F 0,5 Tebak L/F = 0,6

zi Pci (bar) Tci (K) ω Ki Di Di0,23 42,48 369,8 0,152 2,027 -0,1560 -0,16740,67 36,48 408,1 0,181 0,795 0,1531 0,1497

0,1 37,96 425,1 0,200 0,571 0,0546 0,05181,000 0,0517 0,0341

Tebak L/F = 0,77Ki Di

2,027 -0,19100,795 0,14420,571 0,0476

0,0008

xi=zi

K iLF1−Ki

yi=zi Ki

K iLF1−Ki

∑i

xi−∑i

yi =∑i

zi1−KiKiL /F 1−K i

= Di = 0

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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 0.0356 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

Propane 0.230000 0.377689 0.224543

I-Butane 0.670000 0.557528 0.674156

N-Butane 0.100000 0.064784 0.101301