chapter 3. volumetric properties of pure fluids
TRANSCRIPT
Chapter 3. Volumetric Properties of
Pure Fluids
Introduction
Thermodynamic properties (U, H and thus Q, W) are
calculated from PVT data
PVT data are important for sizing vessels and pipelines
Subjects
PVT behavior of pure fluids
Ideal gas behavior
Real gas behavior
Generalized correlation when experimental data are lacking
3.1 PVT Behavior of Pure Substances
P-T Phase Diagram
P
T
Solid
Phase (subcooled)
Liquid
Phase
(superheated)
Vapor
Phase
Pc
Tc
critical point
triple
point
Gas
Super-critical (fluid) phase
Vapor vs. Gas?
- Gas : noncondensable (cannot become liquid)
- Vapor : condensable
3.1 PVT Behavior of Pure Substances
P-T Phase Diagram
P
T
Solid
Phase
Liquid
Phase
Vapor
Phase
T1
pvap
vapor pressure curve
Vaporization curve
• Component =1
• Phase = 2 (V and L)
• F = 1
• Vapor pressure at T1
3.1 PVT Behavior of Pure Substances
P
T
Solid
PhaseLiquid
Phase
Vapor
Phase
Pc
Tc
triple
point
Gas
sublimation pressure
sublimation
point
freezing pressure
freezing
point
melting curve
fusion curve
sublimation curve
P-T Phase Diagram
3.1 PVT Behavior of Pure Substances
P
T
Solid
Phase
Liquid
Phase
Vapor
Phase
Pc
Tc
triple
point
Gas
3.1 PVT Behavior of Pure Substances
P
T
Solid
PhaseLiquid
Phase
Vapor
Phase
Pc
Tc
triple
point
Gas
3.1 PVT Behavior of Pure Substances
PV Phase Diagram
Compare with PT diagram
- triple point?
- vapor pressure line?
3.1 PVT Behavior of Pure Substances
PV diagram vs. PT diagram
heating heating
3.1 PVT Behavior of Pure Substances
Single Phase Region
Equation relating P, V, T Equation of State
Volume expansivity
Isothermal compressibility
0),,( TVPf
dPP
VdT
T
VdV
TP
PT
V
V
1
TP
V
V
1
dPdTV
dV
3.2 Virial Equation of State
PVT behavior is complex cannot be represented by simple
equation
Gas phase alone can be correlated with simple equation
PV term is expanded as a function of P by a power series
...2 cPbPaPV
...)'''1( 32 PDPCPBaPV
At low pressure, truncation after two terms usually provide satisfactory result
Ideal-Gas Temperature : Universal Gas Constant
B’, C’, D’, …. depends on component, temperature
a same for all species, function of temperature only
...)'''1( 32 PDPCPBaPV
aPVP ,0
)()( * TfaPV
Useful for defining the
temperature scale
(PV)*, the limiting value of PV as P 0, is
independent of the gas
Ideal-Gas Temperature : Universal Gas Constant
Ideal Gas Temperature Scale
Make (PV)* proportional to T, (R: proportionality constant)
Assign the value 273.16 K to triple point of water
RTaPV *)(
KRPV t 16.273)( *
K
KT
PV
PV
t 16.273
/
)(
)(*
*
*
*
)(
)(16.273/
tPV
PVKT
Kelvin scale temperature
Ideal-Gas Temperature : Universal Gas Constant
P 0
Molecular volume becomes smaller
Molecules are separated infinite distances
Intermolecular forces approaches zero
Universal Gas Constant
K.
)PV(R
*
t
16273
mol.R.psia/lbft 10.73R
mol.R.atm/lbft 0.7302R
mol.RBtu/lb 1.987R
cal/mol.K 1.987R
.atm/mol.Kcm 82.06R
kPa/mol.Kcm 8314R
bar/mol.Kcm 83.14R
Pa/mol.Km 8.314R
J/mol.K 8.314R
3
3
3
3
3
3
(PV)t* = 22,711.8 m3bar/molK
Two forms of the Virial Equation
Compressibility
B, B’ : the second virial coefficient
C, C’ : the third virial coefficient
RT
PVZ
...'''1 32 PDPCPBZ
...132
V
D
V
C
V
BZ
Virial expression
RT
BB '
2
2
)('
RT
BCC
3
323
)RT(
BBCD'D
Importance of the Virial Equation
Many form of equation of state have been proposed
Virial Equation has a firm basis in theory
Statistical mechanics
B/V : interaction between two molecules (two-body interaction)
C/V2 : three-body interaction
...132
V
D
V
C
V
BZ
3.3 The Ideal Gas
No interaction between molecules (B/V, C/V2 ,…=0)
Internal energy of gases
Real gas : function of T and P
Ideal gas : function of T only
- Pressure dependency resulting from intermolecular force
- No intermolecular force in ideal gas no dependency in P
- The Equation of State
- Internal Energy
RTPVZ or 1 Ideal gas
RTPVZ or 1
)(TUU
Implied Property Relations for an Ideal Gas
Heat capacity at const. volume is a function of temperature only
Enthalpy is also a function of temperature only
Heat capacity at const. pressure is a function of temperature only
Heat Capacity Relationship
)(TCdT
dU
dT
dUC v
V
v
)(THRTUPVUH
)(TCdT
dH
dT
dHC P
P
P
RCdT
RTUd
dT
PVUd
dT
dHC VP
)()(Caution:
Cv and Cp are not constant,
they vary with T while
keeping Cp = Cv + R
Equations for Process Calculation for Ideal Gases
From the first law of thermodynamics
Combine ideal gas law
PdVdQdWdQdTCdU V
VRTP / RCC VP
V
dVRTdTCdQ V
V
dVRTdW
P
dPRTdTCdQ P
P
dPRTRdTdW
T, V
T, P
PdVR
CVdP
R
CdQ PV PdVdW P,V
Equations for Process Calculation for Ideal Gases
- (1) Isothermal Process
PdVdQdWdQdTCdU V
sP dWdQdTCdH
0U
0H
dWdQ
1
2
1
2 lnlnP
PRT
V
VRTQ
1
2
1
2 lnlnP
PRT
V
VRTW
DO NOT memorize the equation !
Memorize the sequences of derivation !
T= const.
V
dVRTdTCdQ V
P
dPRTdTCdQ P
V
dVRTdW
P
dPRTRdTdW
Equations for Process Calculation for Ideal Gases
- (2) Isobaric Process
PdVdQdWdQdTCdU V dTCU V
dTCH P
QH
dTCHQ P
)( 12 TTRW
P= const.
sP dWdQdTCdH
V
dVRTdW
P
dPRTRdTdW
PdVdW
Equations for Process Calculation for Ideal Gases
- (3) Isochoric Process
PdVdQdWdQdTCdU V dTCU V
dTCH P
QU
dTCUQ V
0W
V= const.
sP dWdQdTCdH
V
dVRTdW
P
dPRTRdTdW
PdVdW
Equations for Process Calculation for Ideal Gases
- (4) Adiabatic Process
PdVdQdWdQdTCdU V dTCU V
dTCH P
0Q dQ=0
sP dWdQdTCdH
V
dV
C
R
T
dT
V
integration
VCR
V
V
T
T/
2
1
1
2
0
V
dVRTdTCdQ V
0P
dPRTdTCdQ P
0 PdVR
CVdP
R
CdQ PV
P
dP
C
R
T
dT
P
integration
PCR
P
P
T
T/
1
2
1
2
V
dV
C
C
P
dP
V
P integrationVP CC
V
V
P
P/
2
1
1
2
Equations for Process Calculation for Ideal Gases
- (4) Adiabatic Process
VCR
V
V
T
T/
2
1
1
2
PCR
P
P
T
T/
1
2
1
2
VP CC
V
V
P
P/
2
1
1
2
VVP CCC
V
V
T
T/)(
2
1
1
2
PVP CCC
P
P
T
T/)(
1
2
1
2
.1 constTV
./)1( constTP
.constPV
VP CC /
0Q
TCdTCW VV
These equations are restricted
to const. heat capacities and
reversible, adiabatic processes
Equations for Process Calculation for Ideal Gases
- (4) Adiabatic Process
Other expression for WORK calculation
TCdTCW VV
1
RCV
1
TRTCW V
11
111
)(/)1(
1
21
/)1(
1
21112
P
PRT
P
PVPTTRTCW V
Use P instead of T1 or T2
Equations for Process Calculation for Ideal Gases
- (4) Adiabatic Process
Values of Cp/Cv
Monatomic gases : 1.67 (He, Ne, Kr,…)
Diatomic gases : 1.4 (H2, N2, O2, …)
Simple polyatomic gases : 1.3 (CO2, SO2, NH3, CH4,…)
Equations for Process Calculation for Ideal Gases
- (5) Polytropic Process
Polytropic : “Turning many ways”
.constPV
0 constP
1 constRTPV
constPV
constV
isobaric
isothermal
adiabatic
isochoric
Can be used to represent
a in-between processes
Equations for Process Calculation for Ideal Gases
- (5) Polytropic Process
Using ideal gas equation,
.constPV
VRTP /
.1 constTV
.constTP /)1(
1P
P
1
RTPdVW
/)1(
1
21
1P
P
)1)(1(
RT)(Q
/)1(
1
21
constant heat capacity
Irreversible Processes
Equations in (1)-(5)
Only valid for mechanically reversible, closed system for ideal
gases
Properties changes (U, H) are also same regardless of the
process State Properties
Heat and Work amount depends on the nature of process
(reversible/irreversible, closed/open) Path function
For irreversible processes, the following procedures are
commonly employed
W is determined
For reversible process
Efficiency (h)
is multiplied
Example 3.2
Air is compressed from an initial state of 1 bar and 25 oC to a final state of
5 bar and 25 oC by three different mechanically reversible processes in a
closed system.
(a) Heating at constant volume followed by cooling at constant pressure.
(b) Isothermal compression.
(c) Adiabatic compression followed by cooling
at constant volume.
Assume air to be an ideal gas with the constant
heat capacities, Cv = 5/2R and Cp = 7/2R.
Calculate W, Q, U, and H.
Example 3.2 – solution
Choose the system as 1 mol of air (basis)
Cv = 20.785 Cp = 29.099 J/molK
Since T is same, U = H = 0
Also, U = Q + W = 0 Q = -W
(although the absolute value of Q & W can be different –
path function)
P1 = 1 bar
T1 = 298.15 K
V1 = 0.02479 m3
P2 = 5 bar
T2 = 298.15 K
V2 = 0.04958 m3
Example 3.3
An ideal gas undergoes the following sequence of mechanically reversible
processes in a closed system.
(a) From an initial state of 70 oC and 1 bar, it is compressed adiabatically to
150 oC.
(b) It is then cooled from 150 to 70 oC at constant P.
(c) Finally, it is expanded isothermally to
its original state.
Calculate W, Q, U, and H for each of
the three processes and for the entire cycle.
Assume air to be an ideal gas with the constant
heat capacities, Cv = 3/2R and Cp = 5/2R.
Example 3.3 – solution
Choose the system as 1 mol of gas (basis), R = 8.314 J/molK
Cv = 12.471 Cp = 20.785 J/molK = Cp/Cv = 5/3
(a) Adiabatic compression, Q = 0
U = W = Cv T = 12.471(150 - 70) = 998 J
H = Cp T = 20.785(150 - 70) = 1663 J
P2 from PT/(-1) = const P2 = 1.689 bar
(b) Const P process
Q = H = Cp T = 20.785(70 - 150) = -1663 J
U = Cv T = 12.471(70 - 150) = -998 J
W = U – Q = -998 – (-1663) = 665 J or W = -P V
(c) Isothermal process
U = H = 0 const T
Q = -W = RTln(P3/P1) = RTln(P2/P1) = (8.314)(343.15)ln(1.689/1) = 1495 J
P1 = 1 bar
P2
3.4 Application of the Virial Equations
Virial Equation
Infinite Series
Only useful for engineering purpose when convergence is rapid
- Two or three terms
Derivatives of compressibility
Compressibility-factor graph for methane
...'''1 32 PDPCPBZ
...'3'2' 2
PDPCB
P
Z
T
'0;
BP
Z
PT
RT
BP
RT
PVZ 1
Truncation equation
Virial Equation for Engineering Purpose
Truncated two terms
Truncated three terms
Extended Virial equation
(Benedict-Webb-Rubin Equation)
V
BZ 1PBZ '1
21
V
C
V
B
RT
PVZ 2''1 PCPBZ
similar accuracy
but convenient
more accurate
32
2
000 /
V
abRT
V
TCARTB
V
RTP
22236exp1
VVTV
c
V
a
eight parameters
,,,,,,, 000 cbaCBA
Example 3.8
Reported values for the virial coefficients of isopropanol vapor at
200 oC are:
B = -388 cm3/mol C = -26,000 cm6/mol2
Calculate V and Z for isopropanol vapor at 200 oC and 10 bar by:
(a) the ideal-gas equation
(b) Second virial coefficient, B
(c) Second virial coefficient, B, and third virial coefficient, C
Example 3.8 - solution
T = 473.15 K and R = 83.14 cm3bar/mol K
(a) For ideal gas, Z = 1
(b) Using B,
(c) with B and C, use the iteration or solver to solve for V
mol/cm,).)(.(
P
RTV 39343
10
154731483
mol/cm,,BP
RTV 354633889343
21
V
C
V
B
RT
PVZ
3.5 Cubic Equations of State
PVT behavior of liquid & vapor over wide range
Equation must not be so complex
Polynomial equation at least cubic equation
Cubic EOS are simplest equation
P
V
P
Vreal behavior cubic EOS representation
The van der Waals Equation of State
J. D. Van der Waals
First proposed in 1873
Won Novel Prize in 1910
2V
a
bV
RTP
2V
a
bV
RTP
V
RTP
. .
. .
.
.
Ideal gas : no volume, no interactionVan der Walls gas
volume + interaction
The van der Waals Equation of State
T > Tc ; monotonically
decreasing function
T = Tc ; Critical Region
Saturated liquid and vapor
T < Tc ; Liquid region
T < Tc ; Unrealistic Behavior
T < Tc ; Vapor region
T < Tc ; Real behavior : two phase region
The van der Waals Equation of State
2V
a
bV
RTP
b
Molar volume cannot be smaller than b
b often called “hard sphere volume”
At T = Tc and T > Tc, only single root
exists
At T<Tc, three roots exist
Smaller : liquid-like volume
Middle : no significance
Larger : vapor-like volume
Inaccurate critical point prediction
Inaccurate vapor pressure prediction
a parameters are not optimized to fit vapor pressure
Only historical interest
Limitation of Van der Waals EOS
375.0c
ccc
RT
VPZ For all fluids
Zc = 0.24 to 0.29 for real fluid
(mainly hydrocarbons)
Cubic Equation of State - History
Improvement over Van der Waals EOS
Redlich and Kwong (1949)
- Slightly different volume dependence in attractive term
- Improved critical compressibility (0.3333) and Second Virial Coefficient
- Vapor pressure and liquid density are still inaccurate
Wilson (1964)
- First introduced T dependency of a parameter
)]11
)(62.157.1(1[ r
rT
T )70at ( 1ln c
c
T.TP
P
Acentric factor
2V
a
bV
RTP
)bV(VT
a
bV
RTP
/
21
)bV(V
)T(a
)bV(
RTP c
a(T) = ac(T)
Cubic Equation of State - History
Soave (1972)
More refined temperature dependency of a parameter
Improved vapor-pressure calculation
Also called Soave-Redlich-Kwong (SRK) EOS
Peng-Ronbinson (1976)
)bV(V
a)T(
bV
RTP c
)bV(VT
a
bV
RTP
/
21
2
2
156130551711485090
11
...m
)T/T(m c
)bV(b)bV(V
)T(a
)bV(
RTP c
Cubic Equation of State - History
Equation of State with more parameters
Scmidt and Wenzel (1980)
Hamens and Knapp (1980)
Patel and Teja (1982)
22 331 bbv)(v
)T(a
)bv(
RTP
22 1 b)c(cbvv
)T(a
)bv(
RTP
bcv)cb(v
)T(a
)bv(
RTP
2
Results can be
improved …
But meaning of the
third parameter is not
clear…
It is not easy to write a
meaningful mixing
rule for the third
parameter
Note:
a(T) = ac(T)
Cubic Equation of State - History
a(T) law improvement
Stryjek-Vera (1986)
Soave (1984)
Carrier (1988)
Mathias and Copeman (1983)
Numerous functional forms have been proposed …
)1/1()1(1 21 rr TCTC
25.0)1(1 rT
)1()1(1 21 rr TCTC
23
3
2
211 ccc rT1
)7.0)(1(5.0
10 rr TT
32
0
0196554.017131848.0
4897153.1378893.0
Cubic Equation of State - History
Ref) J. O. Valderrama, Ind. Eng. Chem. Res., 42,1603 (2003)
- Reference in textbook, page 92
A Generic Cubic Equation of State
Several hundred cubic equation of state have been proposed
Most EOS can be reduced to a generic form;
))((
)(
)( 2
h
VVbV
V
bV
RTP
))((
)(
)( bVbV
Ta
bV
RTP
and : pure number (same for all substance)
a(T) and b : substance dependent (a(T) = f(Tr, Tc, Pc) and b = f(Tc, Pc))
e.g., for van der Waals EOS,
a(T) = a, = = 0
A Generic Cubic Equation of State
Determination of Equation-of-State Parameters
How to determine a and b parameters ?
An EOS should represent PVT behavior of pure fluid
Two conditions at critical point
At critical point ; T = Tc, P = Pc, V = Vc
0;
crTV
P0
;
2
2
crTV
P
c
c
P
TRa
64
27 22
c
c
P
RTb
8
93-94 page derivation (Van der Waals EOS)
375.0c
ccc
RT
VPZ
Determination of Equation-of-State Parameters
For other EOS (RK, SRK, PR), similar equation can be obtained
c
c
P
TRa
64
27 22
c
c
P
RTb
8
c
c
P
TRa
22
c
c
P
RTb
Determination of Equation-of-State Parameters
For better representation of vapor pressure, a parameters are
assumed to be temperature dependent
c
c
P
TRa
22
)()(
)(22
rc
c
cr TaP
TRTTa
Theorem of Corresponding State: Acentric Factor
a(T) law can be fitted using vapor pressure data
Is there any method not using vapor pressure ?
Corresponding state theorem acentric factor ( )
)()(
)(22
rc
c
cr TaP
TRTTa
)1/1()1(1 21 rr TCTC
Theorem of Corresponding State: Acentric Factor
Reduced Properties : How far from critical point?
Theorem of Corresponding State (Simplest Form)
All fluids, when compared at the same reduced
temperature and pressure, have approximately the
same compressibility factor and all deviate from ideal
gas behavior to about the same degree.
Only valid for simple fluid (argon, krypton and xenon)
Systematic deviations are observed
cr TTT / cr PPP /
Theorem of Corresponding State: Acentric Factor
Acentric Factor
Introduced by K. S. Pitzer and coworkers
Plot of log (Prsat) vs. (1/Tr)
STd
Pd
r
sat
r )/1(
log
Slope varies depending on
the chemical species
Theorem of Corresponding State: Acentric Factor
Acentric Factor
Basis : = 0 for Ar, Kr, Xe
This parameter represent the degree how far from simple gases
Example: CH4 : 0.012, C2H5: 0.100, C3H8: 0.152, C4H10: 0.200, …., C10H22: 0.492
See Table B.1 for , Tc, Pc, Vc, and Zc for many chemical species
7.0)log(0.1 rT
sat
rP
(1/Tr)
log(1/Prsat)
S=-2.3 (for Ar, Kr, Xe)
Slope varies depending on
the chemical species
(1/0.7)
-1
Theorem of Corresponding State: Acentric Factor
Three-Parameter Theorem of Corresponding State
All fluids having the same value of , when compared at the same
Tr and Pr, have about the same value of Z, and all deviate from
ideal gas behavior to about the same degree.
Use of acentric factor in Cubic EOS
How to Solve Cubic EOS ?
Solution Technique
Root formula for Cubic equation
Iterative calculation
- Newton-Raphson iteration or Secant iteration
- Successive Substitution
Use the solver function
- Engineering calculator
- Computer program: Mathematica, Matlab, etc
Initial guess
- For gas phase : Ideal Gas Root (V = RT/P)
- For liquid phase : Hard Sphere Volume (V = b)
Solution Method (3) Successive Substitution
Vapor or Vapor-Like Roots
Start with Ideal Gas Root :V=RT/P
Liquid and Liquid-Like Roots
Start with : V=b
))((
)(
bVbV
bV
P
Tab
P
RTV
)())((
Ta
VPbPRTbVbVbV
Example 3.9
Given that the vapor pressure of n-butane at 350 K is 9.4573 bar, find
the molar volume of (a) saturated-vapor and (b) saturated-liquid n-
butane using the Redlich/Kwong equation
Example 3.9 - solution
Tc and Pc from Table B.1, Tc = 425.1 K and Pc = 37.96 bar
From Table 3.1,
(a) For vapor-like root, start with V = RT/P = 83.14×350/9.4573 = 3157.5 cm3/mol
Using solver with initial value of V = 3157.5 cm3/mol, then V = 2555 cm3/mol
(b) For liquid-like root, start with b = 80.667 cm3/mol
Using solver with initial value of V = 80.667 cm3/mol, then V = 133.3 cm3/mol
823301425
350.
.T
TT
c
r 249109637
45739.
.
.
P
PP
c
r
26722212221
105519637
1425148382330427480 mol/barcm.
.
....
P
TRTa
/
c
c
/
r
mol/cm667.8096.37
1.42514.8308664.0
P
RTb 3
c
c
).V(V
.V
.
..
.
.
)bV)(bV(
bV
P
)T(ab
P
RTV
66780
66780
45739
1055166780
45739
3501483 7
3.6 Generalized Correlations for Gases
Pitzer correlations for the Compressibility factor
Z0 and Z1 are given as a generalized function of Tr and Pr
- They are tabulated
- See Appendix E (Table E.1 – E.4)
Lee-Kesler Method (1975)
- Given by chart (Z0 and Z1) : Table E.1 – E.4
- Basis : Benedict/Webb/Rubin Equation
For quantum gases (hydrogen, helium, neon)
- Small molecules
- Do not conform corresponding behavior
- Use effective critical parameters (3.58-3.60)
Errors
- 2-3 percent for nonpolar or slightly polar small molecules (mainly hydrocarbons)
- Larger errors for polar/associating components, large molecules
10 ZZZ
3.6 Generalized Correlations for Gases
Pitzer correlations for the Second Virial Coefficient
Not recommended
Not accurate for highly polar or associating molecules
r
r
T
PB
RT
BP
RT
PVZ ˆ11
10ˆ BBB
6.1
0 422.0083.0
rTB 2.4
1 172.0139.0
rTB
10 ZZZ
3.7 Generalized Correlation for Liquids
Cubic EOS
Not accurate for liquid phase molar volume (5-10 % error)
Lee-Kesler correlation
Accurate liquid volume but not accurate for polar species
Saturated molar volume
Rackett equation (1978)
7/2)1( rT
cc
sat ZVV
])1(1[ 7/2
rT
c
r
rsat ZT
PZ
Example 3.10
Determine the molar volume of n-butane at 510 K and 25 bar by
each of the following:
(a) The ideal-gas equation
(b) The generalized compressibility-factor correlation
Example 3.10 - solution
(a) For ideal gas,
(b) Tc and Pc from Table B.1, Tc = 425.1 K and Pc = 37.96 bar
from Table E.1 and E.2 for Z0 and Z1,
Z0 = 0.865, Z1 = 0.038, and = 0.200 (Table B.1)
Z = Z0 + Z1 = 0.865 + (0.200)(0.038) = 0.873
mol/cm.,))(.(
P
RTV 316961
25
5101483
20011425
510.
.T
TT
c
r 65909637
25.
.P
PP
c
r
mol/cm.,))(.)(.(
P
ZRTV 374801
25
51014838730
Homework
Homework
3.8, 3.10, 3.14, 3.20, 3.32, 3.44, 3.68
Due:
Other Recommend Problems
3.2, 3.9, 3.12, 3.15, 3.16, 3.17, 3.18, 3.21, 3.22, 3.23, 3.24, 3.27, 3.36