Advanced Thermodynamics
Thermodynamic Properties of Fluids
Property relations for homogeneous phases
First law for a closed system
d (nU )  dQ  dW
a special reversible process
d (nU )  dQrev  dWrev
dWrev   Pd (nV )
dQrev  Td (nS )
d (nU )  Td (nS )  Pd (nV )
Only properties of system are involved:
It can be applied to any process in a closed system (not necessarily
reversible processes).
The change occurs between equilibrium states.
d (nU )  Td (nS )  Pd (nV )
The primary thermodynamic properties: P, V , T , U , and S
The enthalpy:
The Helmholtz energy:
H  U  PV
The Gibbs energy:
G  H  TS
A  U  TS
For one mol of homogeneous fluid of constant composition:
dU  TdS  PdV
 T 
 P 

   
 V  S
 S V
dH  TdS  VdP
 T   V 

 


P

S

S 
P
dA  PdV  SdT
 P 
 S 

 

 T V  V T
dG  VdP  SdT
 V 
 S 




 

T

P
 P T
Maxwell’s equations
Enthalpy, entropy and internal energy change calculations f (P,T)
 H 
 H 
dH  
dT



 dP
 T  P
 P T
 H 

  CP

T

P
dH  TdS  VdP
 H 
 S 

T


  V
 P T
 P T
 H 
 S 

T




 T  P
 T  P
  S 

dH  CP dT   T    V dP
  P T

 V 
 S 




 
 T  P
 P T

 V  
dH  CP dT  V  T 
 dP
 T  P 

 S 
 S 
dS  
 dT    dP
 T  P
 P T
C
 S 

  P
 T  P T
 V 
 S 

   
 T  P
 P T
dT  V 
dS  CP

 dP
T  T  P
H  U  PV
 H   U 
 V 

 
  P
 V
 P T  P T
 P T
 U 
 V 
 V 

  T 
  P

 P T
 P T
 P T
Determine the enthalpy and entropy changes of liquid water for a
change of state from 1 bar and 25°C to 1000 bar and 50°C. The data for
water are given.
H1 and S1 at 1 bar, 25°C
H2 and S2 at 1 bar, 50°C
H3 and S3 at 1000 bar, 50°C
 V 

  V

T

P

 V  
dH  CP dT  V  T 
 dP
volume expansivity
 T  P 

dH  CP dT  1  T VdP
H  CP (T2  T1 )  1   T2  V ( P2  P1 )
1  (513 10
H  75.310(323.15  298.15) 
dT  V 
dS  CP

 dP
T  T  P
6

)(323.15) (18.204)(1000  1)
 3400 J / mol
10
S  C P ln


T2
  V ( P2  P1 )
T1
323.15 513 106 (18.204)(1000  1)
S  75.310 ln

 5.13J / mol
298.15
10
Internal energy and entropy change calculations f (V,T)
 U 
 U 
dU  
 dT  
 dV
 T V
 V T
 U 
 U 
 S 

  CV 
  T
 P

T

V

V

V

T

T
  S 

dV
dU  CV dT   T 

P

  V T

 S 
 P 






 V T
 T V
  P 


dU  CV dT   T 
  P dV
  T V

 S 
 S 
dS  
 dT  
 dV
 T V
 V T
 U 
 S   S 
 P 
  


  T
 
 T V
 T V
 T V  V T
CV
 S 




T

V T
dT  P 
dS  CV

 dV
T  T V
Gibbs energy G = G (P,T)
dG  VdP  SdT
Thermodynamic property of great potential utility
1
G
 G 
d
dG 
dT

2
RT
 RT  RT
G  H  TS dG  VdP  SdT
 G
d
 RT
H
 V

dP

dT

2
RT
 RT
 G   (G / RT ) 
 (G / RT ) 
d

 dP  
 dT
P
T
 RT  
T

P
V
 (G / RT ) 


RT 
P
T
H
 (G / RT ) 
 T 

RT
T

P
G/RT = g (P,T)
The Gibbs energy serves as a generating function
for the other thermodynamic properties.
Residual properties
• The definition for the generic residual property:
M R  M  M ig
– M and Mig are the actual and ideal-gas properties,
respectively.
– M is the molar value of any extensive thermodynamic
properties, e.g., V, U, H, S, or G.
• The residual Gibbs energy serves as a generating
function for the other residual properties:
 GR
d 
 RT
 VR
HR
 
dP 
dT
2
RT
 RT
 GR
d 
 RT
 VR
HR
 
dP 
dT
2
RT
 RT
const T
 GR
d 
 RT
 VR
 
dP
 RT
R
PV
GR

dP
0
RT
RT
V R  V  V ig 
P  Z  dP
HR
 T  

0
RT
 T  P P
S R H R GR


R RT RT
H
 (G / RT ) 
 T 

RT

T

P
P
GR
dP
  ( Z  1)
0
RT
P
ZRT RT

P
P
const T
P  Z  dP
P
SR
dP
 T  
  ( Z  1)

0
0
R
P
 T  P P
const T
Z = PV/RT: experimental measurement .Given PVT data or an appropriate equation
of state, we can evaluate HR and SR and hence all other residual properties.


P  Z  dP 

2
H  H  H  H   C dT   RT  


T0
0

T
P


P


P  Z  dP 

ig
2
 H 0  R  ICPH (T0 , T ; A, B, C , D)  RT  


0

T
P

P


ig

R
T
ig
0
ig
P

P  Z  dP 

2
T  T0    RT 0  
 H  C


T
P

P


P  Z  dP 

ig
2
 H 0  R  MCPH (T0 , T ; A, B, C , D)T  T0   RT  


0

T
P

P


ig
0
ig
P H
T
P  Z  dP
P


dT
P
dP 
S  S ig  S R  S 0ig   C Pig
 R ln   R  T  

(
Z

1
)



T0
0
0
T
P

T
P
P

P
0



P  Z  dP
P
 ig

P
dP 
 S 0  R  ICPS (T0 , T ; A, B, C , D)  R ln   R  T  
  ( Z  1) 

0
0
P0 
P
 T  P P


P  Z  dP
P


T
P
dP 
 S 0ig  C Pig ln  R ln   R  T  

(
Z

1
)



S
0
0
T
P

T
P
P

P
0
0



P  Z  dP
P
 ig

T
P
dP 
 S 0  R  MCPS (T0 , T ; A, B, C , D) ln  R ln   R  T  

(
Z

1
)



0
0
T
P

T
P
P


P
0
0



Calculate the enthalpy and entropy of saturated isobutane vapor at 360
K from the following information: (1) compressibility-factor for
isobutane vapor; (2) the vapor pressure of isobutane at 360 K is 15.41
bar; (3) at 300K and 1 bar, H 0ig  18115 J / mol S0ig  295.976 J / mol  K (4) the
ideal-gas heat capacity of isobutane vapor: CPig / R  1.7765  33.037 103 T
P  Z  dP
P  Z  dP
P
HR
SR
dP
 T  


T

(
Z

1
)




0
0
0
RT

T
P
R

T
P

P

P P
 Z 
 / P and (Z-1)/P vs. P.
 T  P
Graphical integration requires plots of 
P  Z  dP
HR
 T  
 (360)( 26.37 10 4 )  0.9493

0
RT
 T  P P
H R  2841.3J / mol
P  Z  dP
P
SR
dP
 T  
  ( Z  1)
 0.9493  (0.02596)  0.6897

0
0
R

T
P
P

P


S R  5.734 J / mol  K
H  H 0ig  R  ICPH (300,360;1.7765,33.037 E  3,0.0,0.0)  H R  21598.5


S  S 0ig  R  ICPS (300,360;1.7765,33.037 E  3,0.0,0.0)  R ln
J
mol
15.41
J
 S R  286.676
1
mol  K
Residual properties by equation
of state
• If Z = f (P,T):
P  Z  dP
HR
 T  

0
RT
 T  P P
P  Z  dP
P
SR
dP
 T  
  ( Z  1)

0
0
R
P
 T  P P
• The calculation of residual properties for gases
and vapors through use of the virial equations and
cubic equation of state.
Use of the virial equation of state
BP
Z
1

RT
dP two-term virial equation
P
GR
  ( Z  1)
0
RT
P
P  Z  dP
HR
 T  

0
RT
 T  P P
Z 1 
BP
RT
G R BP

RT RT
P 
HR
BP  dP
 T  
(1 
)
0
RT
RT  P P
 T
P   P  1 dB
HR
B   dP
 T    
 2  
0
RT
  R  T dT T   P P
H R P  B dB 
  

RT R  T dT 
S R H R GR


R RT RT
SR
P dB

RT
R dT
Three-term virial equation
Z  1  B  C 2
GR
3
 2 B  C 2  ln Z
RT
2
P
GR
dP
  ( Z  1)
0
RT
P
P  Z  dP
HR
 T  

0
RT
 T  P P
P  ZRT
 B dB 
HR
 C 1 dC  2 
 T  
   
 
RT
 T 2 dT  
 T dT 
S R H R GR


R RT RT
Application up to moderate pressure
Use of the cubic equation of state
• The generic cubic equation of state:
RT
a(T )
P

V  b (V  b)(V  b)
 d ln  (Tr ) 
H
 Z 1  
 1 qI
RT
 d ln Tr

q
a (T )
bRT
R
 d ln  (Tr ) 
SR
 ln Z     
 1 qI
R
 d ln Tr

bP

RT
1
b
Z
q
1  b
(1  b)(1  b)




0
0
( Z  1)
d

  ln( 1  b)  qI
dq
 Z  d

I


dT
 T   
I 

0
d ( b)
(1  b)(1  b)
const T
Find values for the residual enthalpy HR and the residual entropy SR
for n-butane gas at 500 K and 50 bar as given by Redlich/Kwong
equation.
P
 (Tr )
500
50
 3.8689
 1.317    r  0.09703 q 
 1.176 Pr 

T
T
r
37.96
425.1
r
Tr 
Z 
Z  1    q
( Z   )( Z   )
I 

0
d ( b)
(1  b)(1  b)
 0
Z  0.685
 1
const T
I
 Z   
1
  0.13247
ln 
    Z   
 d ln  (Tr ) 
HR
 Z 1  
 1 qI
RT
 d ln Tr

H R  4505
J
mol
 d ln  (Tr ) 
SR
 ln Z     
 1 qI
R
 d ln Tr

S R  6.546
J
mol  K
Two-phase systems
• Whenever a phase transition at constant
temperature and pressure occurs,
– The molar or specific volume, internal energy, enthalpy,
and entropy changes abruptly. The exception is the
molar or specific Gibbs energy, which for a pure
species does not change during a phase transition.
– For two phases α and β of a pure species coexisting at
equilibrium:
G  G 
where Gα and Gβ are the molar or specific Gibbs energies of the individual phases
dG  dG 
G  G 
dG  VdP  SdT

V dP
sat


 S dT  V dP
dH  TdS  VdP
sat

 S dT
dP sat S   S 
S 
 


dT
V V
V 
H   TS 
The latent heat of phase transition
dP sat H 

The Clapeyron equation dT
TV 
V   V v 
RT
P sat
Ideal gas, and Vl << Vg
d ln P sat
H   R
d (1 / T )
The Clausius/Clapeyron equation
lv
dP sat H 

dT
TV 
The Clapeyron equation: a vital connection between the properties of different phases.
P  f (T )
ln P sat  A 
ln P sat  A 
ln Prsat
B
T
For the entire temperature range from the triple point
to the critical point
B
The Antoine equation, for a specific temperature range
T C
A  B 1.5  C 3  D 6 The Wagner equation, over a wide

temperature range.
1 
 1  Tr
Two-phase liquid/vapor systems
• When a system consists of saturated-liquid and
saturated-vapor phases coexisting in equilibrium,
the total value of any extensive property of the
two-phase system is the sum of the total properties
of the phases: M  (1  xv )M l  xv M v
– M represents V, U, H, S, etc.
– e.g., V  (1  x v )V v  x vV v
Thermodynamic diagrams and tables
• A thermodynamic diagram represents the temperature,
pressure, volume, enthalpy, and entropy of a substance on
a single plot. Common diagrams are:
– TS diagram
– PH diagram (ln P vs. H)
– HS diagram (Mollier diagram)
• In many instances, thermodynamic properties are reported
in tables. The steam tables are the most thorough
compilation of properties for a single material.
Fig. 6.2
Fig. 6.3
Fig. 6.4
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 ΔH for the following conditions:
(a) P1 = 1000 kPa, T1 = 250°C, and P2 = 200 kPa.
(b) P1 = 150 psia, T1 = 500°F, and P2 = 50 psia.
Since the process is both reversible and adiabatic, the entropy change
of the steam is zero.
(a) From the steam stable and the use of interpolation,
At P1 = 1000 kPa, T1 = 250°C:
H1 = 2942.9 kJ/kg, S1 = 6.9252 kJ/kg.K
At P2 = 200 kPa,
S2 = S1= 6.9252 kJ/kg.K < Ssaturated vapor,
the final state is in the two-phase liquid-vapor region:
S 2  (1  x2v ) S 2l  x2v S 2v
6.9252  1.5301(1  x2v )  7.1268 x2v
x2v  0.9640
H  H 2  H1  315.9
kJ
kg
H 2  (1  0.964)(504.7)  (0.964)(2706.3)  2627.0
(a) From the steam stable and the use of interpolation,
At P1 = 150 psia, T1 = 500°F:
H1 = 1274.3 Btu/lbm, S1 = 1.6602 Btu/lbm.R
At P2 = 50 psia,
S2 = S1= 1.6602 Btu/lbm.K > Ssaturated vapor,
the final state is in the superheat region:
H 2  1175.3 Btu / lbm
T2  283.28 F
H  H 2  H1  99.0
Btu
lbm
A 1.5 m3 tank contains 500 kg of liquid water in equilibrium with pure water
vapor, which fills the remainder of the tank. The temperature and pressure are
100°C, and 101.33 kPa. From a water line at a constant temperature of 70°C and
a constant pressure somewhat above 101.33 kPa, 750 kg of liquid is bled into the
tank. If the temperature and pressure in the tank are not to change as a result of
the process, how much energy as heat must be transferred to the tank?
Energy balance:
d (mU ) cv
 Q  H m 
dt
d (mU ) cv
dm
 Q  H  cv
dt
dt
Q  (mU )cv  H mcv
H  U  PV
Q  (mH )cv  ( PmV )cv  H mcv
Q  (m2 H 2 )cv  (m1H1 )cv  H mcv
At the end of the process, the tank still contains saturated liquid and
saturated vapor in equilibrium at 100°C and 101.33 kPa.
Q  (m2 H 2 )cv  (m1H1 )cv  H mcv
H   293.0
kJ
kg
saturated liquid @ 70 C
kJ
kg
saturated liquid @100 C
H cvl  419.1
H cvv  2676.0
kJ
kg
saturated vapor @100 C
From the steam table, the specific volumes of saturated liquid and saturated
vapor at 100°C are 0.001044 and 1.673 m3/kg , respectively.
(m1 H1 ) cv  m1l H1l  m1v H1v  500(419.1) 
m2  500 
1.5  (500)(0.001044)
(2676.0)  211616kJ
1.673
1.5  (500)(0.001044)
 750  m2l  m2v 1.5  1.673m2v  0.001044m2l
1.673
(m2 H 2 ) cv  m2l H 2l  m2v H 2v  524458kJ
Q  (m2 H 2 )cv  (m1H1 )cv  H mcv  524458  211616  (750)(293.0)  93092kJ
R
H
 T 
0
RT
P
 Z  dP



T

P P
P  Pc Pr
P  Z  dP
P
SR
dP
 T  

(
Z

1
)

0
0
R
P
 T  P P
T  TcTr
Pr  Z
HR
 Tr2  
0
RTc
 Tr
Pr  Z
SR
 Tr  
0
R
 Tr
 dPr

 Pr Pr
Pr
 dPr
dP

  ( Z  1) r
0
Pr
 Pr Pr
Z  Z 0  Z 1
1

Pr  Z 
H R  2 Pr  Z 0  dPr 
dPr 
2


    Tr  

  Tr  
0
0
RTc 

 Tr  Pr Pr 
 Tr  Pr Pr 

0
1

Pr  Z 
Pr
Pr  Z 
Pr
SR 
dPr
dPr 
dPr
dPr 
0
1


    Tr  

  Tr  
  ( Z  1)
  ( Z  1)
0
0
0
0
R 

T
P
P

T
P
P

r 
r 
 r  Pr r
 r  Pr r



0
1

H R  H R 
HR 


  HRB (TR, PR, OMEGA)
RTc  RTc
RTc 
0
1

S R  S R 
SR 


  SRB (TR, PR, OMEGA)
R  R
R 
Table E5 ~ E12
T2
H 2  H   C Pig dT H 2R
ig
0
T0
T1
H1  H   C dT H
ig
0
T0
ig
P
R
1
T2
H   C Pig dT H 2R  H1R
T1
H  CPig
T2
S 2  S   C Pig
ig
0
T0
dT
P
 R ln 2 S 2R
T
P0
dT
P1
S1  S   C
 R ln S1R
T0
T
P0
ig
0
T1
H
(T2  T1 )  H 2R  H1R
S   C Pig
dT
P
 R ln 2 S 2R  S1R
T
P1
S  C Pig
ln
T2
T1
ig
P
S
P2
 S 2R  S1R
P1
Estimate V, U, H, and S for 1-butene vapor at 200°C and 70 bar if H and S are
set equal to zero for saturated liquid at 0°C. Assume that only data available are:
Tc  420 K
Pc  40.43 bar
  0.191
Tn  266.9K (nomal boiling pt.)
CPig / R  1.967  31.630 103 T  9.837 106 T 2
Tr  1.127
Pr  1.731
Z  Z 0  Z 1
 0.485  (0.191)(0.142)
 0.512
Assuming four steps:
(a) Vaporization at T1 and P1 = Psat
(b) Transition to the ideal-gas state at (T1, P1)
(c) Change to (T2, P2) in the ideal-gas state
(d) Transition to the actual final state at (T2, P2)
ZRT
cm3
V
 287.8
P
mol
Fig 6.7
Fig 6.7
Step (a)
ln P sat  A 
B
T
B
ln 1.0133  A 
266.9
B
ln 40.43  A 
420
For = 273.15K,
Psat = 1.2771 bar
H nlv 1.092(ln Pc  1.013)

 9.979
The latent heat at normal boiling point:
RTn
0.930  Trn
 1  Tr
H

lv
H n  1  Trn
lv
The latent heat at 273.15 K:
J
lv
S  79.84
mol  K
Step (b)
H   TS 
H lv  21810




0.38
J
mol
Tr  0.650 Pr  0.0316
0
1
J

H R  H R 
HR 
R


  HRB (TR, PR, OMEGA)  0.0985  H 1  344
RTc  RTc
RTc 
mol
0
1
J
R

S R  S R 
SR 


  SRB (TR, PR, OMEGA)  0.1063  S1  0.88
mol  K
R  R
R 
Step (c)
J
mol
70
S ig  8.314  ICPS (273.15,473.15;1.967,31.630 E  3,9.837 E  6,0.0)  8.314 ln
1.2771
J
 22.18
mol  K
H ig  8.314  ICPH (273.15,473.15;1.967,31.630 E  3,9.837 E  6,0.0)  20564
Step (d)
Total
Tr  1.127 Pr  1.731
0
1

H R  H R 
HR 


  2.430
RTc  RTc
RTc 
H 2R  8485
0
1

S R  S R 
SR 


  1.705
R  R
R 
J
S  14.18
mol  K
J
mol
R
2
J
mol
J
S  S  79.84  (0.88)  22.18  14.18  88.72
mol  K
J
U  H  PV  34233  (70)( 287.8) / 10  32218
mol
H  H  21810  (344)  20564  8485  34233
Gas mixtures
• Simple linear mixing rules
   yii TPc   yiTci PPc   yi Pci
i
i
i
Estimate V, HR, and SR for an equimolar mixture of carbon dioxide and propane
at 450K and 140 bar
TPc   yiTci  (0.5)(304.2)  (0.5)(369.8)  337 K
i
PPc   yi Pci  (0.5)(73.83)  (0.5)( 42.48)  58.15bar
i
   yii  (0.5)(0.224)  (0.5)(0.152)  0.188
i
T pr  1.335
Z 0  0.697
Ppr  2.41
Z 1  0.205
Z  Z 0  Z 1  0.736
0
1
1
ZRT
cm3 H R  H R 0

S R  S R 
SR 

HR 
V
 196.7




  1.029
  1.762
R
R
R
RT
RT
RT
P
mol


pc
pc
pc 


H R  4937 J / mol
S R  8.56 J / mol  K