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2/3/2009
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Department of Chemical EngineeringFaculty of Industrial TechnologyParahyangan Catholic University
ITK 226 Termodinamika Teknik KimiaITK 226 Termodinamika Teknik Kimia
3.0 Volumetric Properties of Pure Fluids
Author: Dr. Ir. Buana Girisuta
Contents
• PVT Behavior of Pure Substances• Virial Equations of State• The Ideal Gas• Application of the Virial Equations• Cubic Equations of State• Generalized Correlations for Gases• Generalized Correlations for Liquids
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3.1 PVT Behavior of Pure Substances
• Phase diagram (PT diagram)
Ftp = ?Ftp = 0
Fvap. curve = ?
Fliq = ?Fliq = 2(divariant)
F = 1
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• Liquid if vaporization results from P reduction at constant T• Gas if condensation results from T reduction at constant P• The region where T > Tc is termed supercritical fluid
tp(invariant)
Fvap. curve 1(univariant)
3.1 PVT Behavior of Pure Substances
• In a PV diagram, the phase boundaries becomes areas/regions
Triplepoint
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• T1 and T2 are subcritical temperatures consist of three segments• Line BC singlephase (saturated) liquids at boiling temperature• CD singlephase (saturate) vapor at condensation temperature
point
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3.1 PVT Behavior of Pure Substances
• Singlephase region• A relation connecting P, V, and T for any pure homogeneous fluid in
equilibrium states may be expressed through an equation of state:f (P V T) 0f (P, V, T) = 0
• The simplest equation is for an ideal gas, PV = RT• For example, if V = V (T, P), then:
• Volume expansivity:
• Isothermal compressibility:
dPPVdT
TVdV
TP
⎟⎠⎞
⎜⎝⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
=
PTV
V⎟⎠⎞
⎜⎝⎛∂∂
≡1β
V⎟⎞
⎜⎛ ∂−≡
1κ
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• Combining these equations yield:
• The isotherms for liquid phase in PV diagram are very steep due to the low values β and κ. For incompressible fluid β and κ are zero.
TPV⎟⎠
⎜⎝ ∂
≡κ
dPdTVdV κβ −=
3.2 Virial Equations of State
• PVT behavior of a pure substance is complex difficult to describe by an equation.
H f th i l l ti l i l ti ft
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• However, for the gas region alone relatively simple equations often suffice.
• For example:• Where a, B’, C’, etc., are constants for a given temperature and a given
chemical species.
( )K++++= 32 '''1 PDPCPBaPV
3.2 Virial Equations of State
• Idealgas temperatures• Homework: read this topic in the textbook.
• Universal gas constant (R)
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3.2 Virial Equations of State
• By defining the compressibility factor:
RTPVZ ≡
• The virial equation can be rewritten in two forms as follow:• Pressure function:
• Volume function:
• The parameters are called virial coefficients B’ and B are secondvirial coefficients, C’ and C are third virial coefficients, etc.
K++++= 32 '''1 PDPCPBZ
K++++= 321VD
VC
VBZ
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• The two sets of coefficients are related as follow:
RTBB ='
( )22
'RT
BCC −=
( )3323'
RTBBCDD +−
=
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3.3 The Ideal Gas
• The term B/V arises on account of interactions between pairs of molecules; the C/V2 term, on account of threebody interactions.• As the pressure of real gas is reduced at constant temperature V
i d th t ib ti f th t B/V C/V2 t dincreases and the contribution of the terms B/V, C/V2, etc., decrease.• As P approaches zero, V becomes infinite, then Z approaches unity:
or (ideal gas)
• Internal energy of a real gas is a function of P and T• P dependency is the result of forces between the molecules.• If such forces did not exist (ideal gas behavior) the internal energy of
d d t t l !
1=Z RTPV =
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a gas depends on temperature only!(ideal gas))(TUU =
3.3 The Ideal Gas
• Implied property relations for an ideal gas:• Heat capacity at constant V (CV) is a function of temperature only
( )TCdUUC VV ==⎟⎞
⎜⎛ ∂≡ dTCdU V= ∫=Δ dTCU V
• Enthalpy (H) is a function of temperature only
• Heat capacity at constant P (CP) is a function of temperature only
• Relation between CP and CV
( )TCdTT
C VV
V ⎟⎠
⎜⎝ ∂
≡
( ) ( )THRTTUPVUH =+=+≡
( )TCdTdH
THC P
PP ==⎟
⎠⎞
⎜⎝⎛∂∂
≡
dTCdU V ∫Δ dTCU V
dTCdH P= ∫=Δ dTCH P
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This equation does not imply that CP and CV are themselves constant for an ideal gas, but only that they vary with temperature in such a way that their difference is equal to R.
RCRdTdU
dTdHC VP +=+==
3.3 The Ideal Gas
• Equations for process calculations:• For a unit mass or a mole of ideal gas in any mechanically reversible
closedsystem process:
dTCdWdQ V=+
PdVdW −=PdVdTCdQ V +=
VRTP =
PRTV =
RPVT =
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VdVRTdTCdQ V +=
VdVRTdW −=
PdPRTdTCdQ P −=
PdPRTRdTdW +−=
PdVRCVdP
RCdQ PV +=
PdVdW −=
3.3 The Ideal Gas
• Isothermal process: 0=Δ=Δ HU
1
2
1
2 lnlnPPRT
VVRTWQ −==−=
• Isobaric process:
• Isochoric process:
∫=Δ dTCU V
( )12 TTRW −−=
∫=Δ dTCH P
∫=Δ= dTCHQ P
∫=Δ dTCU V ∫=Δ dTCH P
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0=W∫=Δ= dTCUQ V
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3.3 The Ideal Gas
• Adiabatic process (constant heat capacities):
( ) constantconstant
1
1
=
=−
−
γγ
γ
TPTV
V
P
CC
≡γ
constantconstant
=
=γPV
TP V
TCdTCUW VV Δ==Δ= ∫
11112222
−−
=−−
=γγ
VPVPRTRTW
( ) ( )
⎥⎤
⎢⎡
−⎟⎟⎞
⎜⎜⎛
=⎥⎤
⎢⎡
−⎟⎟⎞
⎜⎜⎛
=−−
111
21
1
211
γγγγPRTPVPW
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The process should be mechanically reversible!How about for irreversible process?
⎥⎥⎦⎢
⎢⎣
⎟⎟⎠
⎜⎜⎝−
=⎥⎥⎦⎢
⎢⎣
⎟⎟⎠
⎜⎜⎝−
= 11
11 11 γγ PP
W
3.3 The Ideal Gas
• Irreversible process• Those equations that relate changes in state functions only are valid for
ideal gases regardless of the process reversible and irreversibleprocesses in both open and closed systemprocesses in both open and closed system.
• On the other hand, Q and W is specific to the process considered in its derivation.
• Calculation of work for an irreversible process:• 1st step: determine W for a mechanically reversible process that
accomplishes the same change of state as the actual irreversible process maximum work obtained or minimum work required.
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p q• 2nd step: Multiply or divide this result by an efficiency to give the actual
work multiply by an efficiency if the process produces work and divide by an efficiency if the process required work.
Exercise 31
• One kmol of an ideal gas is taken through a fourstep cyclic process:• A to B: the gas is subjected to an isothermal expansion at 600 K from 5
to 4 bar.B t C di b ti i t 3 b• B to C: an adiabatic expansion to 3 bar.
• C to D: a constantpressure cooling.• D to A: a constantvolume heating.
• All processes are assumed reversible. For these processes, it is reasonable to assume CP is constant and equal to 30 kJ kmol1 K1.
• Calculate Q, W, ΔU, and ΔH for each step and for the entire process.
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Homework
• Please read and redo the following examples from the textbook:• Smith, J.M., Van Ness, H.C., and Abbot M. M. Introduction to Chemical
Engineering Thermodynamics, 6th ed., McGraw Hill, 2001.E l 3 2 3 6 78 87• Examples 3.2 – 3.6 on pp. 7887.
• Derive the equations to calculate Q, W, ΔU, and ΔH for an ideal gas that goes trough an adiabatic process.
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3.4 Application of the Virial Equations
• General virial equations are infinite series
K++++= 32 '''1 PDPCPBZ K++++= 321VD
VC
VBZ
• For engineering purposes two or three terms are sufficient! (for gases and vapors at low to moderate pressures)
PBZ '1+=
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3.4 Application of the Virial Equations
• Truncated virial equations:• Two terms
RTBP
RTPVZ +== 1
PVT behavior of vapors at subcritical T and P < 5 bar.
• Three terms
Suitable for 5 bar < P < Pcrit
• Extended virial equationsBenedict/Webb/R bin eq ation
2000 abRTTCARTBRT −−−
21VC
VB
RTPVZ ++==
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• Benedict/Webb/Rubin equation:
• Used in the petroleum and naturalgas industries for light hydrocarbons and a few other commonly encountered gases.
22236
32000
exp1VVTV
cVa
VabRT
VTCARTB
VRTP
γγα −⎟⎠⎞
⎜⎝⎛ +++
++=
Exercise 32
• Reported values for the virial coefficients of isopropanol vapor at 200 °C are:
B = 388 cm3 mol1 C = 26000 cm6 mol2
Calculate V and Z for isopropanol at 200 °C and 10 bar by:• The idealgas equation• Truncated virial equation to two terms• Truncated virial equation to three terms
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3.5 Cubic Equations of State
• The van der Waals EOS• Proposed by Johannes Diderik van der Waals (1837 – 1923)
aRTP = 2VbVP −
−=
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3.5 Cubic Equations of State
• Determination of EOS parameters• Suitable estimates of EOS parameters are usually found from values for
the critical constants Tc and Pc critical isotherm exhibits a horizontal inflection at the critical point!inflection at the critical point!
• More straightforward procedure p. 94 from the textbook!• Estimates for the van der Waals EOS:
02
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
=⎟⎠⎞
⎜⎝⎛∂∂
cc TT VP
VP
c
cc P
RTV83
=c
c
PTRa
22
6427
=c
c
PRTb
81
=
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• Substitution of Vc into the vdW EOS gives:
• Unfortunately, each chemical species has its own value of Zc! see Table B.1 of App. B from the textbook.
83
=≡c
ccc RT
VPZ
3.5 Cubic Equations of State
• Generic Cubic EOS( )
( )( )bVbVTa
bVRTP
σε ++−
−=
• Constants ε and σ are pure numbers, the same for all substances.• Constants a(T) and b are substance dependent.• The temperature dependence of a(T) is specific to each EOS.
• Parameter estimates
( ) ( )c
cr
PTRTTa
22αΨ=
c
c
PRTb Ω=
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• Constants Ω and Ψ are pure numbers determined for a particular EOS from the values assigned to ε and σ.
• Function α(Tr) is an empirical expression specific to a particular EOS
cP c
3.5 Cubic Equations of State
• Theorem of corresponding statesAll fluids, when compared at the same reduced temperature (Tr) and reduced pressure (Pr), have approximately the same compressibility factor, and all deviate from the idealgas behavior to about the same degree.and all deviate from the ideal gas behavior to about the same degree.
• This theorem is very nearly exact for the simple fluids (Ar, Kr, Xe).• However, systematic deviations are observed for more complex fluids.
Therefore an acentric factor (ω) is introduced as follow:
Th l f b d t i d f fl id f T P d i l
cr T
TT ≡c
r PPP ≡
( ) 7.0log0.1 =−−≡rT
satrPω
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• The value of ω can be determined for any fluid from Tc, Pc, and a single vaporpressure measurement made at Tr = 0.7.
• App. B lists the values of ω and the critical constants Tc, Pc, and Vc for a number of fluids.
3.5 Cubic Equations of State
• Vapor and vaporlike roots of the generic cubic EOS
( )( )σβεββββ++
−−+=
ZZZqZ 1
• Iterative solution starts with the value of Z = 1.
• Liquid and liquidlike roots of the generic cubic EOS
• Iterative solution starts with the value of Z = β.
( )( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛ −++++=
ββσβεββq
ZZZZ 1
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• For both equations, the following equations apply:( )
r
r
TTq
ΩΨ
=α
r
r
TP
Ω=β
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3.5 Cubic Equations of State
• Parameter assignments for EOS
EOS α(Tr) σ ε Ω Ψ Zc
vdW (1873) 1 0 0 1/8 27/64 3/8
RK (1949) 1 0 0.08664 0.42748 1/3
SRK (1972) αSRK(Tr; ω)† 1 0 0.08664 0.42748 1/3
PR (1976) αPR(Tr; ω)‡ 0.07779 0.42748 0.3074021+ 21−
21−rT
( ) ( )( )[ ]2212 1176.0574.1480.01; rrSRK TT −−++= ωωωα
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( ) ( )( )[ ]2212 126992.054226.137464.01; rrPR TT −−++= ωωωα
3.6 Generalized Correlations for Gases
• Pitzer correlations for the compressibility factor:
• Z0 and Z1 are functions of both Tr and Pr.
10 ZZZ ω+=
• The commonly used Pitzertype correlation was developed by Lee and Kesler, which is given in table form in App. E.
• The Lee/Kesler correlation provides reliable results for gases which are nonpolar or slightly polar (errors of no more than 2 or 3%).
• The correlation for quantum gases (H2, He, Ne) is accommodated by use of temperaturedependent effective critical parameters.
F h d
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• For hydrogen:
• T is absolute temperature in kelvins.
( )T
KTc
016.28.211
6.43
+= ( )
T
barPc
016.22.441
5.20
+= ( )
T
Vc
016.291.91
5.51mol cm 13
−=
3.6 Generalized Correlations for Gases
• Pitzer correlations for the second virial coefficient:
r
r
r
r
TPB
TPBZ 101 ω++=
6.10 422.0083.0
rTB −= 2.4
1 172.0139.0rT
B −=
10 ZZZ ω+=
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Exercise 33
• Determine the molar volume of nbutane at 510 K and 25 bar by each the following:• The idealgas equation• The generalized compressibilityfactor correlation• The generalized virialcoefficient correlation
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3.7 Generalized Correlations for Liquids
• The simplest generalized equation for the estimation of molar volume of saturated liquids, proposed by Rackett:
( ) 2857.01sat rTccZVV −=
• Lydersen, Greenkorn, and Hougen developed a twoparameter correspondingstates correlation for estimation of liquid volumes:
• ρr1 ≡ reduced density• ρc ≡ density at the critical point
2
112
r
rVVρρ
=VVc
cr =≡
ρρρ
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• V2 ≡ required volume• V1 ≡ known volume• ρr1 , ρr2 ≡ reduced densities read from Fig. 3.17(p. 109)
Thank you … Questions or comments?
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