Chapter 6. Thermodynamic Properties
of Fluids
Introduction
Numerical values of thermodynamic properties are most important for heat and work calculations
Enthalpy, Internal Energy, Entropy, …..
Measurable Quantities : PVT data and heat capacity
Properties relations are important to calculate one properties from the other.
Generalized correlations are used when experimental values are not available.
6.1 Properties Relations for Homogeneous Phases
From the 1
stlaw and 2
ndlaw of Thermodynamics,
rev
rev dW
dQ )
nU (
d dWrev Pd(nV) ) nS ( Td dQrev
) nV ( Pd )
nS ( Td )
nU (
d
Although this eqn. has been derived for the special case of reversible process, it only contain state properties of the system.
Can be applied to any process
from one equilibrium state to another
The only requirements are that the system be closed and that the change occur between equilibrium states.
6.1 Properties Relations for Homogeneous Phases
PdV TdS
dU
When the equations are written in molar units;
PV U
H TS U
A TS H
G
VdP PdV
PdV TdS
) PV ( d dU
dH dH TdSVdP
SdT TdS
PdV TdS
) TS ( d dU
dA dA PdVSdT
SdT TdS
VdP TdS
) TS ( d dH
dG dG VdP SdT
6.1 Properties Relations for Homogeneous Phases
PdV TdS
) V , S (
dU
Fundamental Properties Relations
VdP TdS
) P , S (
dH
SdT PdV
) T , V (
dA
SdT VdP
) P , T (
dG
These fundamental properties relations are general equation for homogeneous fluid of constant composition
Exact Differential Equation
If a function F is a property (not depending on the path)
The total differentiation
6.1 Properties Relations for Homogeneous Phases
) y , x ( F F
y dy dx F
x dF F
y x
Ndy Mdx
dF
y y x
N F x ,
M F
6.1 Properties Relations for Homogeneous Phases
Exact Differential Equation
Condition for exactness
y dy dx F
x dF F
y x
y x
F y
M 2
x
Ndy Mdx
dF
y x
F x
N 2
y
x x y
N y
M
6.1 Properties Relations for Homogeneous Phases
Maxwell’s Relation
PdV TdS
) V , S (
dU
VdP TdS
) P , S (
dH
SdT PdV
) T , V (
dA
SdT VdP
) P , T (
dG
V
S S
P V
T
P
S S
V P
T
T
V V
S T
P
T
P P
S T
V
x x y
N y
M
Maxwell Relation
S H P
U G
V A T
PdV TdS
dU
VdP TdS
dH
SdT PdV
dA SdT VdP
dG
V
S S
P V
T
P
S S
V P
T
T
V V
S T
P
T
P P
S T
V
Enthalpy as a Function of T and P
T and P : easily measurable
VdP TdS
) P , S (
dH
P P
T C
H
dP
? dT
? ) P , T (
dH
T V T V
P V T S
P H
P T
T
T dP T V
V dT
C dH
P
P
Entropy as a Function of T and P
T and P : easily measurable
P P
T C
H
dP
? dT
? ) P , T (
dS
P P
P T
T S T
C H
T dP V T
C dT dS
P
P
VdP TdS
) P , S (
dH
P
T T
V P
S
Internal Energy as a Function of T and P
T and P : easily measurable
PV H
U
dP
? dT
? ) P , T (
dU
P P
P T
P V T
H T
U
P dP P V
T T V
T dT P V
C dU
T P
P
P
P V P V
P H P
U
T T
T
Ideal-Gas State
T dP T V
V dT
C dH
P
P
dP
T V T
C dT dS
P
P
P R T
) P / RT ( T
V
P P
ig
dT C dH P
P R dP T
C dT
dS P
Alternative Form for Liquid
For liquids,
T V V
P
T dP T V
V dT
C dH
P
P
dP
T V T
C dT dS
P
P
1 T
VdPdT C
dH P VdP
T C dT
dS P
Internal Energy as a Function of T and V
Sometimes T and V are more useful than T and P
Equation of States are expressed as P = F(V,T) PdV
TdS )
V , S (
dU
V V
T C
U
dV
? dT
? ) V , T (
dU
T P T P
V P T S
V U
V T
T
dV T P
T P dT
C dU
V
V
Entropy as a Function of T and V
Sometimes T and V are more useful than T and P
Equation of States are expressed as P = F(V,T) PdV
TdS )
V , S (
dU
V V
V T
T S T C
U
dV
? dT
? ) V , T (
dS
V
T T
P V
S
T dV P T
C dT dS
V
V
Gibbs Free Energy as a Generating Function
G = G(T,P)
T, P Canonical Variables for Gibbs energy
Conforms to a general rule that is both simple and clear
SdT VdP
) P , T (
dG
RT dT dP H
RT V
RT dT dT S
RT dT H
RT dP S
RT V
dT TS RT H
SdT 1 RT VdP
1
RT dT dG G
RT 1 RT
d G
2
2 2 2
Gibbs Free Energy as a Generating Function
RT dT dP H
RT V RT
d G 2
Fundamental Properties Relationship
P T
RT / G RT
V
T P
RT / T G
RT
H
Gibbs energy, when given as a function of T and P, serves as a generating function for other thermodynamic properties, and
implicitly represents complete property information
Example 6.1
Determine the enthalpy and entropy changes of liquid water for a change of state from 1 bar and 25 oC to 1,000 bar and 50 oC. Data for water are given in the following table.
T (oC) P (bar) Cp (J/molK) V (cm3/mol) (K-1)
25 1 75.305 18.071 256×10-6
25 1,000 …… 10.012 366×10-6
50 1 75.314 18.234 458×10-6
50 1,000 …… 18.174 568×10-6
Example 6.1
1 T
VdPdT C
dH P T VdP
C dT
dS P
) P P ( V ) T 1
( ) T T ( C
H P 2 1 2 2 1
) P P ( T V
ln T C
S 2 1
1 2
P
Use mean averaged values from Table
K mol / J 310 . 2 75
314 . 75 305 .
CP 75
For P = 1 bar,
For T = 50 oC,
mol / cm 204 . 2 18
174 . 18 234 .
V 18 3
1 6
6 513 10 K
2 10 568
458
mol / J 400 , 3 ) P P ( V ) T 1
( ) T T ( C
H P 2 1 2 2 1
K mol / J 13 . 5 ) P P ( T V
ln T C
S 2 1
1 2
P
6.2 Residual Properties
No experimental method to measure G as a function of T and P
Residual Property
Departure function from ideal gas value
M is any molar property : V, U, H, S, A or G
id
R M M
M
id
R G G
G
P
V ZRT )
1 Z P ( RT P
V RT V
V
VR id
RT dT dP H
RT V RT
d G 2
dT
RT dP H
RT V RT
d G 2
R R
R
6.2 Residual Properties
Fundamental residual-property relation RT dT dP H
RT V RT
d G 2
R R
R
T R
R
P RT / G RT
V
P R
R
T RT / T G
RT
H
P ) dP 1 Z ( J
RTdP V RT
G RT
G P
0 P
0 R
0 P R
R
P dP T
T Z RT
H
P P
0
R
P ) dP 1 Z P (
dP T
T Z R
S P
P 0 P
0
R
(const T)
(const T)
(const T)
Derive yourself!
(Page 209 – 210)
Residual Properties as P 0
A gas become ideal as P 0
Then, all residual properties are zero???
- Not generally true…
ig 0
P R
0
P
U 0 , as lim U U
lim
0 ) PV ( lim U
lim H
lim
R0 P R
0 P R
0
P
0 T P 0
P R
0
P P
lim Z P RT
1 lim Z
RT V
lim
RT J G RT
lim G T
RT / T G
RT H
0 P R R
0 P P
R
R
See Figure 3.9 (Page 87)
Enthalpy and Entropy from Residual Properties
R
ig H
H
H S Sig SR dT
C H
H T
T ig P ig
0 ig
0
0 T
T ig P ig
0 ig
P ln P T R
C dT S
S
0
T R T
ig P ig
0 C dT H
H H
0
R0 T
T ig P ig
0 S
P ln P T R
C dT S
S
0
Residual Properties
Advantages of using Residual Properties
Generating Function Normally function of T, P
Ideal Gas Property
- P = RT/V
- Cp = Cp (T) = dH/dT , Cv = Cv(T)=dU/dT, Function of T only
If we can use other method to evaluate residual properties, we can calculate all the other properties
- Equation of State
- Generalized Correlation
Example 6.3
Calculate the enthalpy and entropy of saturated isobutane vapor at 360 K from the following information.
1. Table 6.1 gives compressibility-factor data (Z) for isobutane vapor.
2. The vapor pressure of isobutane vapor at 360 K is 15.41 bar.
3. Set for the ideal-gas reference state at 300 K and 1 bar.
4. The ideal gas heat capacity of isobutene vapor is K mol J
S and mol J
H0ig 18,115 / 0ig 295.976 /
) Kelvin in
T ( T 10 037 . 33 7765 . R 1
Cp 3
Example 6.3
T R T
ig P ig
0 C dT H
H H
0
R0 T
T ig P ig
0 S
P ln P T R
C dT S
S
0
P dP T
T Z RT
H
P P
0
R
P )dP 1 Z P (
dP T
T Z R
S P
P 0 P
0
R
P )dP 1 Z ( P and
dP T
evalute Z to
Need P
P 0 P
0
From the data in Table 6.1,
Fit the data Z vs. T at different P…
2596 . P 0
)dP 1 Z (
K 10 37 . P 26
dP T
Z
P 0
1 4 P
P 0
Then, you can perform the rest of calculation!
6.3 Residual Properties by Equations of State
To evaluate above expression, equation of state is to be expressed as a function of P at const T.
Volume explicit: Z (or V) = f(P)
Pressure explicit: Z (or V) = f(P)
However, many cases are pressure explicit!
P dP T
T Z RT
H
P P
0
R
SR T TZ dPP 0P(Z 1)dPP PP 0
R
(const T)
1 B'P C'P2 Z
2
V C V
1 B Z
6.3 Residual Properties by Equations of State
For two-term virial equation,
RT 1 BP Z
RT BP P
)dP 1 Z RT (
G P
0
R
2P P
0 R
T B dT
dB T
1 R
P P
dP T
T Z RT
H
dT dB R
P P
)dP 1 Z P (
dP T
T Z R
S P
P 0 P
0
R
6.3 Residual Properties by Equations of State
Pressure-explicit EOS
In general, EOS can be expressed as Z = PV/RT
By replacing V by r (molar density) P = ZrRT
Differentiation gives
Z dZ d
P ) dP
dZ Zd
( RT
dP
r
r
r
r
Z ln 1 d Z
) 1 Z P (
) dP 1 Z RT (
G
0 P
0
R
r
r
r1 d Z
T T Z
RT H
0
R
r
r
r
rr
r r
r
rr
r d
) 1 Z d (
T T Z
Z R ln
S
0 0
R
Residual Properties as a function of T and P
From the supplement, we can also derive the following alternative expression.
P dP T
T Z T dP
T V V P
T H P
T H P
T H
P P
P
P ig
R( , ) ( , ) ( , )
0
0 P Z dP
P dP T
T Z P dP
R T
P V T S P T S P
T
S P
P P
P
P ig
R( , ) ( , ) ( , )
0
0
0 ( 1)P Z dP
V dP V RT
P T G P
T G P
T
GR( , ) ( , ) ig( , )
0P
0P( 1)Residual Properties as a function of T,V
However, the application of the previous expression is not straightforward, especially for complicate E.O.S.
From the supplement, we derived the alternative expressions, which is much more useful to use for any pressure explicit E.O.S.
) , ( )
1 (
) ,
( P dV RT Z H T P0
T T P V
T
H V ig
V
) , ( ln
ln )
,
( 0 S T P0
P R P
Z R V dV
R T
V P T
S V ig
V
) , ( ln
) 1 (
ln
) , ( )
, ( )
, (
0
0 G T P
P RT P
Z RT Z
RT V dV
P RT
V T S T V
T H V
T G
V ig
6.4 Two-Phase Systems
Vapor-Liquid Transition
Specific Volume : large change
U, H, S : also large change
G ?
At constant T and P
P
T
Solid Phase
Liquid Phase
Vapor Phase Pc
Tc triple
point
Gas Vapor-Liquid Boundary
SdT VdP
) P , T (
dG
0 ) nG (
d
G G
6.4 Two-Phase Systems
Derivation of Clapeyron Equation
Phase Transition
dG dG
SdT VdP
) P , T (
dG
dT S dP
V dT
S dP
V sat sat
V S V
V
S S
dT dPsat
H T S
V T
H dT
dPsat
6.4 Two-Phase Systems
Derivation of Clapeyron Equation
Especially for vapor liquid transition
vl vl sat
V T
H dT
dP
vl sat
vl Z
P
V RT
vl 2
vl sat
Z RT
H dT
P ln d
vlvl sat
Z R
H T
/ 1 d
P ln d
Relation between
Vapor pressure and heat of vaporization
Example 6.5
The Clapeyron equation for vaporization may be simplified by introduction of reasonable approximations, namely, that the vapor phase is an ideal gas and that the molar volume of the liquid is negligible compared with the molar volume of the vapor. How do these assumptions alter the Clapeyron equation?
vlvl sat
Z R
H T
/ 1 d
P ln d
From the assumption,
1 Z P or
V RT
Vvl v sat lv
The Clapeyron equation becomes
1/Td P ln R d H
sat vl
the Calusius/Calpeyron equation
T A B
P
ln sat
6.4 Two-Phase Systems
Temperature Dependence of Vapor Pressure
Empirical equation for P vs. T
vlvl sat
Z R
H T
/ 1 d
P ln d
T A B
P
ln sat Simple Expression
C T A B P
ln sat
Antoine Equation
1
D C
B P A
ln
6 3
5 . 1 sat
Wagner Equation
Tr
1
Specific Temperature Range
6.4 Two-Phase Systems
Two phase Liquid/Vapor System
When system consist of saturated liquid and saturated vapor
Other extensive thermodynamic properties, M (M=U, H, S, A, G,..)
v v l
lV n V
n
nV
v v l
lV x V
x
V
v v l
v)V x V
x 1 (
V
v v l
v)M x M
x 1 (
M
lv v
l x M
M
M
6.5 Thermodynamic Diagrams
Graph showing for a particular substance a set of properties (T, P, V, H, S,…)
TS Diagram
PH Diagram (ln P vs. H)
HS Diagram (Mollier Diagram)
6.6 Tables of Thermodynamic Properties
U, H, S, V,… Thermodynamic function
Tables of Thermodynamic Data
Tabulation of values of thermodynamic functions (U, H, V,..) at various condition (T
and P)
It is impossible to know the absolute values of U , H for process materials Only
changes are important ( U, H,…)
Reference state
- Choose a Tand Pas a reference state and measure changes of Uand Hfrom this reference state
tabulation
Steam Tables
Compilation of physical properties of water (H, U, V)
Reference state: liquid water at triple point (0.01 oC, 0.00611 bar)
Properties of Superheated Steam
Example 6.7
Superheated steam originally at P1 and T1 expands through a nozzle to an exhaust pressure P2. Assuming the process is reversible and adiabatic, determine the downstream state of the steam and H for the following condition.
P1 = 1,000 kPa, T1 = 250 oC, and P2 = 200 kPa
Solution
Since it is adiabatic process, S = 0 S1 = S2 To calculate H, find H1 and H2.
At the initial state, find the H1 and S1 at P1 = 1,000 kPa and T1 = 250 oC
From steam table, interpolation between 240 and 260 oC gives H1 = 2942 kJ/kg, S1 = 6.9252 kJ/kgK = S2
Since the entropy of saturated vapor at 200 kPa is greater than S2, the final state is the two phase liquid/vapor region!
From page 731
Steam Table (page 731)
sat.
liq.
sat.
vap.
175 (448.15)
200 (473.15)
220 (493.15)
240 (513.15)
260 (533.15)
280 (553.15)
300 (573.15)
325 (598.15)
Steam Table (page 725)
sat.
liq.
sat.
vap.
300 (573.15)
350 (623.15)
400 (673.15)
450 (723.15)
500 (773.15)
550 (823.15)
600 (873.15)
650 (923.15)
Example 6.7
From the steam table (page 725), the final T2 is the saturation temperature at 200 kPa,
T2 = 120.23 oC At T2 = 120.23 oC,
Since the entropy of saturated vapor at 200 kPa is greater than S2, the final state is the two phase liquid/vapor region!
, S x S ) x 1 ( S
From v l v v
kg / kJ 3 . 706 , 2 H
, kg / kJ 7 . 504 H
K kg / kJ 1268 .
7 S
, K kg / kJ 5301 .
1 S
v 2 l
2
v 2 l
2
9640 .
0 x
1268 .
7 x 5301 .
1 ) x 1 ( 9252 .
6
v
v v
kg / kJ 9 . 315 9
. 942 , 2 0 . 627 , 2 H
H H
kg / kJ 0 . 627 , 2 3 . 706 , 2 9640 .
0 7 . 504 )
9640 .
0 1 ( H
x H
) x 1 ( H
1 2
v 2 v l
2 v 2
From page 725
Calculation of Enthalpy and Entropy for Real Fluids
Using residual properties, HR and SR, and ideal gas heat capacities, we can now calculate the enthalpy and entropy for real fluids at any T and P!
For a change from state 1 to state 2,
R 1 T
T ig P ig
0
1 H C dT H
H
0
TT R2ig P ig
0
2 H C dT H
H
0
R 1 R
2 T
T
ig P 1
2 H C dT H H
H
H 2
1
R 1 R
2 1
T 2 T
ig
P S S
P ln P T R
C dT S 2
1
Calculation of Enthalpy and Entropy for Real Fluids
Step 1 1ig
Step 1ig 2ig
Step 2ig 2
R 1 1
ig H H
H1 Sig S1 S1R
1
dT C H
H
H 2
1 1 2
T T
ig P ig
ig
ig
1 T 2
T ig P ig
1 ig
2 ig
P ln P T R
C dT S
S
S 2
1
R 2 ig
2
2 H H
H S2 Sig2 SR2
Homework
Problems
6.3, 6.11, 6.23, 6.48, 6.54, 6.74
For 6.54
- use the heat capacity of n-butane from Table C.1.
- For HRand SR, use Pitzer correlations for the Compressibility factor (Z).
Due:
Recommend Problems
6.2, 6.4, 6.9, 6.12, 6.17, 6.30, 6.31, 6.49, 6.57, 6.58, 6.71, 6.72, 6.76, 6.80, 6.92, 6.93, 6.94
