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:
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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. 2. 3.
Introduction to Multicomponents VLE Systems VLE Calculation in Mixtures by an Equation of State Activity Models
4.
Modified Raoult's laws Margules Equation Van Laar Equation Regular-Solution Theory Wilson's Equation UNIQUAC UNIFAC
Fitting Activity Models to Experimental Data
Solving Problem with: EXCEL, MATLAB, CHEMCAD, and/or HYSYS
Introduction to Multicomponents VLE Systems Click to add text
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 zAF = yAV + xAL (zA = overall mole fraction) ==> zA = yA.V/F + xA.L/F de Percentage of liquid: L/F = (zA-yA) / (xA-yA) ==> ce cd Percentage of vapor: V/F = (xA-zA) / (xA-yA) ==> ce Remember that: L/F + V/F = 1
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 ===?
f =y P i i
i =
f i yi P
f =y P i i i
Fugacity of Non-Ideal Liquid Mixtures ● ●
● ●
f =y P i i
For ideal gases ==> γi = 1 and fio =P For LIQUID: Activity: ==> Activity Coefficient ==>
a i = f i / f
o i
GAMMA APPROACH V L Pi −P sat i i Remember: Fugacity of component i in LIQUID: f io = i sat P isat exp RT
i
i
i
i
For low to moderate pressure, fio ≈Pisat
V
i
L
P −P i
RT
f
i
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i
sat
L =γi x i P sat i
Summary for Component Fugacities ●
o
γ i = f i /x i f
f =γ x f =γ x P exp L o sat sat i i i i ●
i
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 is i fi
is =xi f i =xi f i
VLE in Ideal Solutions • • • • •
f V = f L i i
Bagaimana menghitung Ki ≡ yi /xi Equilibrium constraint: In ideal solutions: Fugacity of the liquid: Combining the equations:
f L = sat P i
i
i
sat exp
V
V
y i i P=xi sat P sat exp i
• Dinyatakan dalam rasio Ki : Ki=
• Pada tekanan rendah:
yi xi
P =
sat
i
i
Hukum Raoult
P
atau
Pi
Pi
L
P −P i
i
sat
RT V L P i −P i
i
sat
RT
L
[ P −P / RT ]
isat exp V
=1, dan
sat
Ki=
i
sat
i
i
V
y i f i =x i f i
i
L
i
i
sat
V
i
[ P −P / RT ]
exp V
i
L
i
y i P= xi Pisat
i
sat
System of Raoult’s Law binary system ●
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Click to add title ●
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Shortcut Estimation of VLE K-ratios
Ki=
sat Pi
P
≈
Pc,i 10
7 1ω 1− 1 3 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 Tipe
Diketahui
Dihitung
Konvergensi
∑ y i =∑ K i xi =1
P, yi
T, xi=zi
BP
Kriteria
i
i
y
DP
T, yi=zi
P, xi
BT
P, xi=zi
T, yi
∑ y i =∑ K i xi =1
P, yi=zi
T, xi
∑ x i =∑ Ki =1
∑ x i =∑ Ki =1 i
DT
P, T, zi
i
i
i
FL
Paling mudah Mudah
xi, yi, L/F
i
i
i
Sulit
y
Sulit
i
∑ z i 1−K i
Paling sulit
i
K i L /F 1−K i
Perhitungan Kesetimbangan Fasa untuk Hukum Raoult Biner Bubble Pressure Calculation: ∑ y i =1, or i
∑ K i x i=1 , ∑ i
i
sat
sat
Pi
P
P1
xi =1 sat
P
sat
x1
P2
P
x 2 =1
sat
P=x1 P 1 x 2 P2
Tidak diperlukan iterasi, karena temperature dan tekanan uap diketahui. Hk Raoult linear bubble pressure line (P-x,y) x 2 =1− x1
sat sat sat sat P=x1 P sat 1 1−x 1 P 2 =x 1 P 1 −P 2 P 2
Hukum Raoult Biner …..(2) Dew-Pressure Calculation: y y1 P y 2 P sat =1 ∑ x i =1, or ∑ Ki =1 sat i
i
P1
i
P2
Diselesaikan tanpa iterasi, sebab tekanan uap adalah tertentu pada temperatur yang ditentukan, sehingga:
P=
1 y1
y2
P1
P sat 2
sat
Hukum Raoult Biner …..(3) Bubble-Temperature Calculation: sat sat ∑ yi =1, or ∑ K i xi=1 P=x1 P 1 x 2 P2 i
i
Diselesaikan dengan iterasi Temperatur (yang mengubah Pisat), hingga tekanan sama dengan tekanan yang diketahui.
Hukum Raoult Biner …..(4) Dew-Temperature Calculation: y
∑ x i =1, or ∑ Ki =1 i
i
1
P=
i
y1 P1sat
y2 P sat 2
Diselesaikan dengan iterasi Temperatur (yang mengubah Pisat), hingga tekanan sama dengan tekanan yang diketahui.
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 L V V/F = 1-L/F, Component balance:
z i =x i
==>
==> yi=Kixi ==>
xi=
zi K i
F
yi
L 1−K i F yi=
zi K i L K i 1−K i F
F
Binary Flash Calculation ....(6) ●
●
●
● ●
Dalam perhitungan flash, L/F harus diiterasi hingga Σxi=1, Tetapi dalam flash, kita juga harus menyelesaikan Σyi=1 Untuk penyelesaian uap dan cairan, maka secara simultan: (Σxi-Σyi)=0 ==> fungsi objective Note that: 0
●
flashing liquid partial condensation
zi, feed flow rate, P, T ==> diketahui
Multicomponent VLE Calculations ●
Bubble Calculation:
∑ y i=1,
atau
i
●
i
P
i
=1
Dew Calculation:
∑ x i=1, i
●
∑ xi K i =
∑ xi Psati
Rules:
atau
y
∑ Ki i
i
= P∑ i
yi Pisat
=1
bubble- & dew-pressure calculation ==> no iteration required bubble- & dew-temperature calculation ==> iteration required
Multicomponent VLE Calculations.....(2) ●
Tebakan awal Temperatur ==> scr. kasar
T = ∑ xi T i
sat
atau T =
i
∑ y i T r ,i T sati i
∑ y i T c, i i
●
General formula for ISOTHERMAL FLASH Calculation:
z 1−K
i = Di = 0 ∑ x i−∑ yi = ∑ K Li / F 1−K i
i
i
i
i
Contoh Perhitungan VLE dgn MS Excel Produk atas suatu kolom distilasi (seperti pada gambar) mempunyai komposisi (zi) 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: zi
Pci (bar) Tci (K) 0,23 42,48 369,8 0,67 36,48 408,1 0,1 37,96 425,1 1
Tekanan (bar) = Tebak T (K) = ω Ki 0,152 1,609 0,181 0,612 0,200 0,433
8 310 Tebak T (K) = yi Ki 0,370 2,027 0,410 0,795 0,043 0,571 0,82
320 yi 0,47 0,53 0,06 1,06
Dengan Interpolasi: T = 310 + ((1,000-0,827)/(1,061-0,827))*(320-310)=317 K
(a) with ChemCAD 2
1
Vapor
●
Click to add an outline 1
Flash
Feed
3 Liquid
Stream No. Name - - Overall - Molar flow kmol/h Mass flow kg/h Temp K Pres bar Vapor mole fraction Vapor mass fraction Enth MJ/h Heating values (60 F) Gross J/kmol Net J/kmol Actual vol m3/h Std liq m3/h Std vap 0 C m3/h Component mole fractions Propane I-Butane N-Butane
1 Feed
2 Vapor Produc
3 Liquid Produ
1.0000 54.8968 100.0000 8.0000 0.0000 0.0000 -166.77
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.00000
1.0000 54.8968 319.7421 8.0000 0.0000 0.0000 -142.68
2.722E+009 2.511E+009 0.0743 0.0991 22.4136
0.0000 0.0000 0.0000
2.722E+009 2.511E+009 0.1072 0.0991 22.4136
0.230000 0.670000 0.100000
0.000000 0.000000 0.000000
0.230000 0.670000 0.100000
(b). Dew point Temperature (b) uap kesetimbangan dgn cairan ==> Uap jenuh ==>
dew
point temperatur Dengan MS Excel:
zi
Tekanan (bar) = Tebak T (K) = ω Ki 0,152 2,262 0,181 0,900 0,200 0,651
Pci (bar) Tci (K) 0,23 42,48 369,8 0,67 36,48 408,1 0,1 37,96 425,1 1,000
8 325 Tebak T (K) = xi Ki 0,102 2,027 0,744 0,795 0,154 0,571 0,999
320 xi 0,11 0,84 0,18 1,13
Dengan interpolasi: T = 325 + ((1,00-0,994)/(1,123-0,994))*(320-325) = 324,8 K
(b). ChemCAD 2 2
1
Vapor 1
●
Flash
Click to add an outline Feed
3 Liquid
Stream No. Name - - Overall - Molar flow kmol/h Mass flow kg/h Temp K Pres bar Vapor mole fraction Enth MJ/h Heating values (60 F) Gross J/kmol Net J/kmol Actual vol m3/h Std liq m3/h Std vap 0 C m3/h Component mole fractions Propane I-Butane N-Butane
1 Feed
2 Vapor Produc
3 Liquid Produ
1.0000 54.8968 100.0000 8.0000 0.0000 -166.77
1.0000 54.8968 324.9329 8.0000 1.000 -125.26
0.0000 0.0000 0.0000 0.0000 0.0000 0.00000
2.722E+009 2.511E+009 0.0743 0.0991 22.4136
2.722E+009 2.511E+009 2.8142 0.0991 22.4136
0.0000 0.0000 0.0000
0.230000 0.670000 0.100000
0.230000 0.670000 0.100000
0.000000 0.000000 0.000000
(c). Isothermal Flash Calculation (c) Partial condenser ==>
Flash to Liquid and Vapor
Dengan MS Excel: Tekanan (bar) 8 Temperature (K)= 320 Tebak L/F 0,5 Tebak L/F = Pci (bar) Tci (K) ω Ki Di 0,23 42,48 369,8 0,152 2,027 -0,1560 0,67 36,48 408,1 0,181 0,795 0,1531 0,1 37,96 425,1 0,200 0,571 0,0546 1,000 0,0517
zi
Tebak L/F = Ki 2,027 0,795 0,571
z 1−K
i = Di = 0 ∑ x i−∑ yi = ∑ K Li /F 1−K i
i
i
i
i
Dengan interpolasi: L/F = 0,7684 K
xi=
0,77 Di -0,1910 0,1442 0,0476 0,0008
zi K i
0,6 Di -0,1674 0,1497 0,0518 0,0341
yi =
L 1−K i F
zi K i L K i 1−K i F
2
1
Vapor
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Flash
Feed
●
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Liquid Stream No.
1
2
3
Feed
Vapor Produc
Liquid Produ
1.0000 54.8968 100.0000
0.0356 1.8823 320.0000
0.9644 53.0145 320.0000
8.0000 0.0000 -166.77
8.0000 1.000 -4.3367
8.0000 0.0000 -137.72
2.722E+009 2.511E+009
2.626E+009 2.421E+009
2.726E+009 2.514E+009
Average mol wt Actual vol m3/h Std liq m3/h
54.8968 0.0743 0.0991
52.8252 0.0992 0.0034
54.9733 0.1035 0.0956
Std vap 0 C m3/h Component mole fractions
22.4136
0.7987
21.6150
0.230000 0.670000 0.100000
0.377689 0.557528 0.064784
0.224543 0.674156 0.101301
Name - - Overall - Molar flow kmol/h Mass flow kg/h Temp K Pres bar Vapor mole fraction Enth MJ/h Heating values (60 F) Gross J/kmol Net J/kmol
Propane I-Butane N-Butane
