ENGSC 2333 – Thermodynamics
Chapter 3
3-1
Objectives

To apply the energy balance to a system of interest
requires knowledge of the properties of the system
and how the properties are related. The objective of
Chapter 3 is to introduce property relations relevant
to engineering thermodynamics. We will focus on
the use of the closed system energy balance
introduce in Chapter 2, together with the property
relations considered in this chapter.
3-2
Some Concepts and Definitions








State principal
Simple compressible
system
p-υ-T surface
2-phase region
Triple line
Triple point
Saturation state
Vapor dome







Critical point
p-υ diagram
T-υ diagram
Subcooled
(compressed) liquid
Superheated vapor
2-phase liquid-vapor
mixture
Quality
3-3
3.1 Fixing the State
For simple, compressible systems, the state
principle indicates that the number of
independent intensive properties is two.
Intensive properties such as velocity and
elevation that are assigned values relative to
datum outside the system are excluded from
present considerations.
3-4
3.2 p-υ-T Relation
Figure 3.1
3-5
Examples
Virtual pvT diagram
3-6
3.2 Phase Diagram
Figure 3.1
3-7
3.2 T-υ Diagram
3-8
3.2 p-υ Diagram
Figure 3.1
3-9
3.2.3 Quality
For a two-phase liquid-vapor mixture, the ratio of the
mass of vapor to the total mass of the mixture is call
quality, represented as x.
x
mvapor
mtotal

mvapor
mvapor  mliquid
3-10
ENGSC 2333 – Thermodynamics
Chapter 3
3-11
3.3.1 Specific Volume
The specific volume of a two-phase liquid-vapor mixture
can be determined by using the saturation tables and
the definition of quality.
V  Vliq  Vvap
V Vliq  Vvap Vliq Vvap
 


m
m
m
m

mliqliq
m

mvapvap
m
 mliq 
 mvap 
liq  
vap
 
 m 
 m 
3-12
3.3.1 Specific Volume
The specific volume of a two-phase liquid-vapor mixture
can be determined by using the saturation tables and
the definition of quality.

mliqliq
m

mvapvap
m
 mliq 
 mvap 
liq  
vap
 
 m 
 m 
  (1  x) f  x g   f  x( g   f )
   f  x fg
3-13
3.3.2 Specific Internal Energy
The specific internal energy of a two-phase liquid-vapor
mixture can be determined by using the saturation
tables and the definition of quality.
u
mliquliq
m

mvapuvap
m
 mliq 
 mvap 
uliq  
uvap
 
 m 
 m 
u  (1  x)u f  xug  u f  x(u g  u f )
u  u f  xu fg
3-14
3.3.2 Specific Enthalpy
In many thermodynamic
analyses the sum of the
internal energy U and the
product of pressure p and
volume V appears.
Because the sum U + pV
appears so frequently, we
give this combination a
name, enthalpy, and a
distinct symbol, H.
H  U  pV
h  u  p
h  u  p
3-15
3.3.2 Specific Enthalpy
The specific internal energy of a two-phase liquid-vapor
mixture can be determined by using the saturation
tables and the definition of quality.
h
mliqhliq
m

mvap hvap
m
 mliq 
 mvap 
hliq  
hvap
 
 m 
 m 
h  (1  x)h f  xhg  h f  x(hg  h f )
h  h f  xhfg
3-16
Examples
For water at the following conditions, determine
the phase or phases present:
1.
2.
3.
4.
5.
6.
T = 40ºC, P = 0.09593 bar
T = 250ºC, P = 39.73 bar, υ = 0.04 m3/kg
T = 250ºC, P = 39.73 bar, υ = 0.0012512 m3/kg
T = 90ºF, υ = 500 ft3/lbm
P = 50 psi, υ = 0.01727 ft3/lbm
P = 50 psi, m = 10 kg
3-17
3.3.5 Evaluating specific heats
The intensive properties cv and cp are
defined for pure, simple
compressible substances as partial
derivatives of the functions u(T,v)
and h(T,p) respectively.
u 
cv 

T v
h 
cp 

T  p
We also use the specific heat ratio, k.
k
cp
cv
3-18
3.3.5 Evaluating specific heats
Figure 3.9 cp of water vapor
3-19
3.3.6 Incompressible substance model
Approximations for liquids using saturated liquid data:
v(T , p)  v f (T )
u (T , p )  u f (T )
h(T , p )  h f (T )  v f (T ) p  psat (T )
3-20
3.3.6 Incompressible substance model
For a substance modeled as incompressible, the specific
heats cv and cp are equal.
c p  cv
T2
u2  u1   c(T )dT
T1
h2  h1  u2  u1  v( p2  p1 )
T2
  c(T )dt  v( p2  p1 )
T1
3-21
3.3.6 Incompressible substance model
Assuming the specific heats are constant (not a function
of temperature)…
u2  u1  c(T2  T1 )
h2  h1  c(T2  T1 )  v( p2  p1 )
3-22
3.4 Generalized compressibility chart
The ideal gas law:
molar basis
pv  R T
pV  nR T
Where:
R
R
M
mass basis
pv  RT
pV  mRT
3-23
3.4 Generalized compressibility chart
For ideal gases:
molar basis
pv  RM T
pV  nRM T
Where:
R
R
M
mass basis
R
pv   T
M 
R
pV  m T
M 
Always use absolute pressures and
temperatures!!!
3-24
3.4 Generalized compressibility chart
Universal gas constant:
pv
lim
R
p 0 T
R  8.314 kJ/kmol  K
 1.986 Btu/lbmol  R
 1545 ft  lbf/lbmol  R
Compressibility factor:
Figure 3.10
pv
Z 
RT
3-25
3.4 Generalized compressibility chart
For compressible gases:
molar basis
pv  ZR T
pV  ZnR T
Where:
R
R
M
mass basis
pv  ZRT
pV  ZmRT
Always use absolute pressures and
temperatures!!!
3-26
3.4 Generalized compressibility chart
For compressible gases:
molar basis
pv  Z RM T
pV  ZnRM T
Where:
R
R
M
mass basis
R 
pv  Z  T
M 
R 
pV  Zm T
M 
Always use absolute pressures and
temperatures!!!
3-27
3.4 Generalized compressibility chart
Z values from Figures A-1,
A-2, and A-3 in appendices.
Figure 3.11
3-28
3.4 Generalized compressibility chart
The principle of corresponding states:
Reduced pressure:
p
pR 
pc
Reduced temperature:
T
TR 
Tc
Figure 3.2
3-29
3.4 Generalized compressibility chart
The principle of corresponding states:
Figure 3.12
3-30
3.5 Ideal gas model properties
pv  RT
u  u (T )
h  h(T )  RT
3-31
Examples
For water at 374.15 ºC and 219.9 bar (gage),
determine:
1.
2.
3.
PR
TR
Z
Assume Patm=1 bar
3-32
ENGSC 2333 – Thermodynamics
Chapter 3
3-33
3.4 Generalized compressibility chart
Compressibility factor:
pv
Z 
RT
Reduced pressured:
p
pR 
pc
Reduced temperature:
T
TR 
Tc
3-34
3.4 Generalized compressibility chart
The principle of corresponding states:
Figure 3.12
3-35
3.4 Generalized compressibility chart
Figure 3.3
3-36
3.5 Ideal gas model properties
pv  RT
u  u (T )
h  u (T )  RT
3-37
3.5 Ideal gas model properties
h  u  RT
dh du

R
dT dT
c p (T )  cv (T )  R
3-38
3.5 Ideal gas model properties
c p (T )  cv (T )  R
cp in Table A-19
Note: cv not given
For monatomic
gases, cp=(5/2)R
k
c p (T )
cv (T )
kR
c p (T ) 
k 1
R
cv (T ) 
k 1
3-39
3.6 U and H of ideal gases
T2
u (T2 )  u (T1 )   cv (T )dT
T1
T2
h(T2 )  h(T1 )   c p (T )dT
T1
3-40
3.6 Specific heat functions
cp
R
   T  T  T  T
2
3
4
Values of the constants are listed in Table A-21.
3-41
3.7 Specific heats… simplified
Using ideal gas tables (A-22 and A-23)…
Evaluate the change in specific enthalpy for air from a state where
T1=400 K to a state where T2=900 K.
3-42
3.7 Specific heats… simplified
Assuming constant specific heats (A-20)…
Evaluate the change in specific enthalpy for air from a state where
T1=400 K to a state where T2=900 K.
3-43
3.8 Polytropic processes of an Ideal Gas
For a polytropic process of a closed system…
pV n  constant
n 1
n0
n 1 0
3-44
3.8 Polytropic processes of an Ideal Gas
For a polytropic process of a closed system…
pV n  constant
Isobaric
n0
Isothermal
n 1
n 1 0
Also, when specific heats are constant, the value of the exponent n
corresponding to an adiabatic polytropic process of an ideal gas is
the specific heat ratio, k.
3-45
3.8 Polytropic processes of an Ideal Gas
Remember from Chapter 2…
2

1
p2  V1 
  
p1  V2 
n
p2V2  p1V1
pdV 
1 n
2

1
V2
pdV  p1V1 ln
V1
for (n  1)
for (n  1)
3-46
Valid for any gas.
3.8 Polytropic processes of an Ideal Gas
Using pV=mRT…
T2  p2 
  
T1  p1 
2

1
2
n
 V1 
  
 V2 
mR(T2  T1 )
pdV 
1 n

1
 n 1
V2
pdV  mRT ln
V1
n 1
for (n  1)
for (n  1)
3-47
Valid for ideal gases.