Post on 22-Aug-2019
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