# THERMODYNAMIC PROPERTIES OF FLUIDS by hcj

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Chapter 3                                                   Thermodynamic Properties of Fluids

CHAPTER 3

THERMODYNAMIC PROPERTIES OF FLUIDS

3.1 Introduction

There are two (2) main objectives of this chapter; the first is to develop the overall
relation between thermodynamic properties and second, to apply the relations so
numerical and meaningful values are acquired based on actual processes.

3.2 Thermodynamic Properties Function Relations

Primary thermodynamic properties are P, V, T, U and S, while H, A and G are secondary
properties given by the following relations;

Enthalpy        H  U  PV                                                          (3.1)
Helmholtz       A  U  TS                                                          (3.2)
Gibbs          G  H  TS                                                           (3.3)

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From the first and second law of thermodynamics, for a closed system with changes
occur only between equilibrium states, an equation is derived for reversible process

d nU   Td nS   Pd nV                                                          (3.4)

Based on the above equations, for 1 mole of a homogeneous fluid at constant
composition, the fundamental property relations are as follows;

dU  TdS  PdV                                                                        (3.5)
dH  TdS  VdP                                                                        (3.6)
dA   PdV  SdT                                                                      (3.7)
dG  VdP  SdT                                                                        (3.8)

3.3 Table of Thermodynamic Properties

Thermodynamic properties may be found also in form of tables. Unfortunately, only
water has that of completed ones at various ranges of temperature and pressure. Other
fluids have to be calculated. Let’s review the following examples.

Example 1

Steam undergoes a change from initial state of 4500C and 3,000 kPa to a final state of
1400C and 235 kPa. Determine its enthalpy, H and entropy, S .

Initial condition P1 = 3,000kPa and T1 = 4500C
Final condition P2 =235kPa and T2 = 1400C

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From Table F2 (page 740), at P1 = 3,000kPa and T1 = 4500C, you will get

H 1  3344.6 J       and S1  7.0857 J         .
g                        gK

From Table F2, at P2 =235kPa and T2 = 1400C (413.15K), interpolate between 225kPa &
250 kPa and 400K & 450K, you will get

H 2  2744.5 J       and S 2  7.2003 J
g                        gK

So,

H  H 2  H 2  2744.5 J          3344.6 J        600.1 J       and
g                g                g

S  S 2  S1  7.2003 J         7.0857 J           0.0115 J
gK                  gK                gK

Exercise 1
Steam at 300psia and 5000F is throttled to 20psia.
a) What is the temperature of steam in its final state?
b) What is the entropy change?
c) What would be the final temperature and entropy change for an ideal gas?

3.4 Residual Thermodynamic Properties

Residual thermodynamic properties, M R also known as departure properties are the
difference between real thermodynamic properties and ideal thermodynamic properties.
For instance, at a certain pressure and temperature, a thermodynamic property is
supposed to have an ideal value, but in reality (through measurements or calculation that
has included noise factors) it does not. So, the difference is known as residual. In another
words, residual property is how much the real property deviate from the ideal property.

M R  M  M ig                                                                                    (3.9)

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Where ; M R residual property
M      real property
M ig   ideal-gas property

In this course, the two (2) main thermodynamics that we will focus on is enthalpy, H and
entropy, S. However, please note that in thermodynamics, actual H and S values is
relatively unimportant. What we want to find is actually their difference from initial to
final state for application in industrial (real) processes. In this case H and S .

Let say we have a fluid coming in a compressor and exiting at a certain pressure and
temperature as follows;

P1, T1                              P2, T2

Initial State                         Final State

Figure 1 Compressor

How are you going to calculate the changes in its enthalpy, H and entropy, S ?

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P1, T1
H , S
P2, T2

H 1R , S1R                                       H 2R , S 2R

H ig , S ig

Figure 2 Calculation Path

Now that we know WHAT we are suppose to calculate, lets find the correlation
(equations) to solve for enthalpy, H and entropy, S . The following table simplify on
the derivation equation that will be used to calculate both H and S.

Table 1 Derivation of Enthalpy Correlation

M R  M  M ig                                                                              (3.9)
H  H ig  H R                                                                              (3.10)

Where
T
H  H   C ig dT
ig       ig
0
p                                                                               (3.11)
T0

H 0 is ideal-gas enthalpy at reference conditions ( T0  298.15 K , P0  1 bar )
ig

Substitute (3.11) in equation (3.10) gives

T
H  H 0   C ig dT  H R
ig
p                                                                             (3.12)
T0

Note that

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T

C
ig
p    dT  R * ICPH  R * MCPH * T  T0                                                            (3.13)
T0

So,

                                    D 
               
T
C ig dT  R * T  T0 *  A  T0   1  T02  2    1  2 
B            C

T0
p
    2            3                 T0 
(3.14)

Equation (3.12) may be inserted in (3.13) to solve for H. The final equation is as follows;

                                                     D 
ig

B            C

H  H 0   R * T  T0  *  A  T0   1  T02  2    1  2    H R
T0  
                               (3.15)
                     2            3                       

How H can become H ? Remember that H is the difference between initial state and
final state. So,

H  H 2  H1                                                                                             (3.16)

From equation (3.15), if

                                                    D 
ig

B            C

H 1  H 0   R * T  T0 *  A  T0   1  T02  2    1  2    H 1R
T0  
                               (3.17)
                    2            3                       

And

                                                     D 
ig

B            C

H 2  H 0   R * T  T0  *  A  T0   1  T02  2    1  2    H 2R
T0  
                           (3.18)
                     2            3                       

Equation (3.16) becomes;

                                                    D 
                   
H   R * T  T0 *  A  T0   1  T02  2    1  2    H 2R  H 1R

B            C
T0  
(3.19)
                    2            3                       

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ig
H 0 is cancelled out because it is ideal-gas enthalpy at reference conditions

( T0  298.15 K , P0  1 bar ).

Table 2 Derivation of Entropy Correlation

M R  M  M ig                                                                                       (3.9)
S  S ig  S R                                                                                       (3.20)

Where
T
dT        P
S ig  S 0   C ig
ig
p                R ln                                                              (3.21)
T0
T         P0
S    ig
0    is ideal-gas entropy at reference conditions ( T0  298.15 K , P0  1 bar )

Substitute (3.21) in equation (3.20) gives

T
dT       P
S  S 0   C ig
ig
p                  R ln  S R                                                         (3.22)
T0
T        P0

Note that

T
dT
C                 R * ICPS  R * MCPS * ln 
ig
p                                                                                          (3.23)
T0
T

So,

T
dT                                  D    1    1  
Cp
ig
 R * ln  *  A   BT0   CT02  2 
                                                  (3.24)
T0
T               
     
              T0  2  ln   
                  

Equation (3.24) may be inserted in (3.22) to solve for S. The final equation is as follows;

                    D    1    1  
S  S 0  R * ln  *  A   BT0   CT02  2 
ig


 2  ln     S
R
(3.25)
                   T0               
                                 

Now that we know the overall equation that we need to solve, a new question arises. How
to solve for the residual thermodynamic properties?

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3.4 Solving Residual Thermodynamic Properties

Residual thermodynamic properties may be determined through the following methods;

a) Virial EOS
b) Cubic EOS
c) Lee-Kesler Generalized Correlation
d) Generalized Second Virial Coefficient Correlation

3.4.1 Virial EOS

A common form of the virial EOS is when it is truncated to two (2) or three (3) terms.
BP
Z 1           B  C 2                                                            (3.26)
RT

Based on the above three-term virial equation, residual Gibbs energy, residual enthalpy
and residual entropy may be found from the following equations;

GR         3
i)        2 B  C 2  ln Z                                                         (3.27)
RT         2
HR      B dB      C 1 dC  2 
ii)        T                                                              (3.28)
RT      T dT      T 2 dT  

SR             B dB    1  C dC  2 
 ln Z  T              
2  T dT  
iii)                                                                                  (3.29)
RT             T dT                 

3.4.2 Cubic EOS

The second applicable method to calculate residual thermodynamic properties is via the
cubic EOS. However, there are two (2) cases to consider;

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Case 1:   
1     1  b      1       Z   
I           1  b      ln  Z   
ln                                                                     (3.30)
                                 

Or

Case 2:   
b     
I                                                                                 (3.31)
1  b Z  

Thus, for residual Gibbs energy, residual enthalpy and residual entropy may be found
from the following equations;

GR
 Z  1  ln Z     qI                                                       (3.32)
RT

HR           d ln  Tr  
 Z 1                1 qI                                                   (3.33)
RT           d ln Tr        

SR                 d ln  Tr 
 ln z                  qI                                                  (3.34)
RT                   d ln Tr

But before you solve the equations, you need to find Z first (from Chapter 1) for either
vapor phase or liquid phase.

3.4.3 Lee-Kesler Generalized Correlation

Using Lee-Kesler generalized correlation, the following equations must be solve to get
residual enthalpy and residual entropy’

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HR

HR

 
HR
0
      1

(3.35)
RTc   RTc    RTc

SR

SR

 
0
SR              1

(3.36)
RTc   RTc    RTc

3.4.4 Generalized Second Virial Coefficient Correlation

The fourth and final method is a personal favorite of mine.

HR        0      dB 0      1      dB1 
 Pr  B  Tr         B  Tr
                                                        (3.37)
RTc              dTr              dTr 


SR          dB 0    dB1 
  Pr 
 dT                                                                    (3.38)
RTc         r       dTr 


Where;

0.422
B 0  0.083                                                                          (3.39)
Tr1.6

0.172
B1  0.139                                                                           (3.40)
Tr4.2

dB 0 0.675
 2.6                                                                             (3.41)
dTr   Tr

dB1 0.722
 5.2                                                                             (3.42)
dTr  Tr

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3.5 Two-Phase Systems

Preceding topics has concentrated only single (vapor) phase thermodynamic properties.
That is when you have inlet and outlet of the same phase (vapor in, vapor out). So, how
to solve for conditions where you have a liquid inlet and vapor outlet?

3.5.1 Liquid Vapor Pressure

An important property to be determined before hand is the vapor pressure of liquids. To
do this, there are four (4) methods available;

a) Rough estimation, a nearly linear relation between vapor pressure and
temperature.
B
ln P sat  A                                                          (3.43)
T

b) Antoine equation

B
ln P sat  A                                                          (3.44)
T C

Antoine equation is the most commonly method used to get vapor pressures. A, B
and C are constant values for fluids.

c) Wagner equation

A  B 1.5  C 3  D 6
ln P sat
                                                           (3.45)
1 
r

Where  1  Tr

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Wagner equation is less common, however it present a more accurate vapor
pressure values. Its constant A, B, C and D for fluids also not available in your
text book!

d) Lee-Kesler correlation

ln Prsat Tr   ln Pr0 Tr    ln Pr1 Tr                                      (3.46)

Where;

ln Pr0 Tr   5.92714 
6.09648
 1.28862 ln Tr  0.169347Tr6            (3.47)
Tr

ln Pr1 Tr   15.2518 
15.6875
 13.4721ln Tr  0.43577Tr6               (3.48)
Tr


 
ln Prnsat  ln Pr0 Trn
and
 
ln Pr1 Trn
(3.49)

Trn = reduced normal boiling point

Prnsat = reduced vapor pressure

3.5.2 Phase Transition

In case of two phase system, where you have an initial liquid inlet and vapor outlet
(which is common in industrial processes) at constant pressure and constant pressure
yields latent heat of phase transition (vaporization);

H lv  TS lv (Changes from liquid, l to vapor, v phase)                                          (3.50)

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An estimation of latent heats of vaporization for pure fluids (either liquid or vapor) at
their respective normal boiling point is given by equation by Riedel;

H n 1.092ln Pc  1.013
lv
                                                                               (3.51)
RTn      0.930  Trn

To estimate the latent heat of vaporization of a pure fluid at any temperature from a
known value (normal boiling point) is given by equation by Watson;

0.38
H lv  1  Tr      
                                                                             (3.52)
H n  1  Trn
lv




Both of the latter equation may be use concurrently.

3.5.3 Vapor and Liquid In Equilibrium

For vapor and liquid in equilibrium, i.e. saturated vapor and saturated liquid in
equilibrium, the total value of any extensive property is equal to total property of the
phase.

M  V ,U , H , S etc
         
M  1 xv M l  xv M v
M  M  x v M lv
l

Where x v is vapor quality.

3.5.4 Calculational Path For Two Phase System

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P1, T1
H , S
H lv , S lv
P2, T2

H 2R , S 2R
H 1R , S1R

H ig , S ig

Figure 3 Calculation Path for Two Phase System

Example 2

A steam power plant employs an adiabatic turbine. Steam enters the turbine at 6500C and
7,000 kPa and discharges at 20 kPa and 5000C. If you are asked to calculate the enthalpy
and entropy change of the process sketch the relevant calculational path.

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P1=7,000 kPa,
T1=6500C
H , S
P2=20 kPa,
T2=5000C

H 1R , S1R                                  H 2R , S 2R

H ig , S ig

Exercise 2

Propane gas at 6500C and 5 bar is discharged to the atmosphere at 500C. If you are asked
to calculate the enthalpy and entropy change of the process sketch the relevant
calculational path.

Example 3

Calculate compressibility factor, Z, residual enthalpy and residual entropy for ethanol at
450K and 5 bar using Peng-Robinson EOS.

a) Find Z
To find Z, use equation from Chapter 3;
Z 
Z  1    q
Z   Z   
bP    P          aT   Tr 
Where           r and q                which is solved iteratively, starting with
RT    Tr         brT    Tr
Z=1.

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From the table of pure properties for ethanol;
  0.645
Tc  513.9 K
Pc  61.48 bar
P    5                 T   450
Pr             0.0813 Tr           0.8757
Pc 61.48               Tc 513.9

Pr          0.0813
So,          0.0778         0.0072
Tr          0.8757

 Tr     0.37464  1.54226  0.26992 2 1  Tr0.5   1.1680
2
1
 Tr        0.457241.1680
q                                   7.8388
Tr           0.07780.8757 
Z  0.0072
Z  1  0.0072  7.83880.0072
Z  1  2 0.0072Z  1  2 0.0072

Z  0.0772
Z  1.0072  0.0564
Z  0.003Z  0.0174

Refer to Table 3.1.

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For PREOS, it is Case 2, where    .

b     
I                                                                   (3.31)
1  b Z  

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