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CLB 20703
Chemical Engineering
Thermodynamics
Chapter 2:
Properties of Pure Substances

Objective of Chapter 2
The objective of this chapter is to
introduce the concept of a pure
substance and apply the ideal-gas
equations to solve the problems in
engineering thermodynamics

Outline
Thermodynamic Properties Of Pure
Substances
 PVT Diagram
 Ideal Gas Law
 Compressibility Factor
 Other Equation Of States


Pure Substance
A substance that has a fixed chemical composition
throughout is called a pure substance, e
...
: H2O,
N2, CO2
 A pure substance does not have to be of a single
chemical element or compound
 A mixture of chemical elements/compounds also
qualifies as a pure substance as long as the mixture
is homogeneous, e
...
Air
 However, a mixture of oil and water is
oil
not a pure substance - oil is not
homogeneous/soluble in water, forming
water
two chemically dissimilar regions


Pure Substance (cont’d)


A mixture of two or more phases of a pure
substance is still a pure substance as long as
the chemical composition of all phases is the
same

Example: a mixture of ice and
liquid water is a pure substance
because both phases have the
same chemical composition


Phases Of Pure Substances
A

phase is identified as having a distinct molecular
arrangement that is homogeneous throughout and
separated from the others by easily identifiable
boundary surfaces
...
For eg:
 Helium has two liquid phases;
 Iron has three solid phases
...


PROCESS 1-2

PROCESS 2-3

v2=vf

PROCESS 3-4

Tsat
Thermodynamic properties
of saturated liquid and
saturated
vapour
are
listed in Saturated Tables

v4=vg

PROCESS 4-5

Tsat
Properties of water
in the superheated
region
listed
in
Superheated Tables

v2=vf

v4=vg

SUMMARY OF PROCESS 1-2-3-4-5

Tsat

v2=vf

v4=vg

If the heating process at constant P is
repeated at other constant P, the phase
change lines look like as in Figure 1
...

Similarly for saturated vapour points 
saturated vapour line formed
...

 At
critical point – saturated liquid
conditions are similar to those of saturated
vapour
...

 The region between saturated liquid and
saturated vapour line  saturated liquidvapour or wet region or saturated region
...

 The region to the right of saturated vapour
line and above the critical point 
superheated vapour region
...

• The temperature trend of constant P lines on T-v diagram
increases as the volume increases
...


Rajah 5

• Above critical temperature and pressure  no significant
process of phase change occurs
...


P-v diagram that combines all the
three phases of solid, liquid and vapour

(a) For a substance that contracts on
freezing

(b) For a substance that expands on
freezing

P-T diagram of a pure substance
•P-T
diagram
of
a
pure
substance also known as phase
diagram – all the three phases
are separated by three lines
...

•The three lines meet at a triple
point – all the three phases are
in equilibrium
...
01oC and
0
...


Liquid
Solid
Vapour

Vaporization
Condensation
Melting/Fusion
Freezing
Sublimation

Vapour
Liquid
Solid

PROPERTY TABLES














Relationship between thermodynamic properties of most substances are
complex – most data simplified in tables  Property Tables
...

Property Tables list data in both SI and ES unit systems
...
Eg: Tables 1-6 and 1-6E,
respectively list properties of superheated vapour in SI and ES units
...

Enthalpy (H) = combination of internal energy (U), pressure (P) and
volume (V), correlated as:
H = U + PV
kJ
Enthalpy per unit mass  specific enthalpy (u):
h = u + Pv
kJ/kg
Enthalpy – important thermodynamic property in determining the energy
in and out of a control volume and in energy balance of a piston-cylinder
system that undergoes an isobaric process
...


TYPES OF PROPERTY TABLES
PROPERTY TABLES

SUPERHEATED
TABLE
(Table 1-6)

SATURATED ICEWATER VAPOUR
TABLE (Table 1-8)
COMPRESSED
LIQUID TABLE
(Table 1-7)

SATURATED
TABLE

Temperaturebased
(Table 1-4)

Pressurebased
( Table 1-5)

SATURATED TABLES
 Properties

of saturated liquid and saturated vapour for water are listed in Table
1-4 and 1-5:
 Both give similar data but arranged differently
...


 Subscripts

used in saturated tables:

 f = property for sat
...
vapour, and
 fg =the difference of similar property

=difference between ug and uf =ug-uf
...
liquid and sat
...
For example : ufg

= enthalpy or latent heat of vaporization – represents the total energy
required to vaporize a unit mass of saturated liquid at a given T or P
...


 hfg

Example of saturated tables – temperature based

Sat
...

temperature
(Tsat) –
temperature at
which liquid and
vapour phases
are in
equilibrium at a
given P

Quality and saturated region
 During

vaporization, a substance exists in two phases (liquid
and vapour) that consists of a mixture of saturated liquid and
saturated vapour
...


x
 x=0

mvapour
mtotal



mvapour
mliquid  mvapour



mg
m f  mg

(or 0%) for saturated liquid and x=1 (or 100%) for
saturated vapour, so the range for quality is 0  x  1
...
It has no
meaning in the compressed liquid or superheated vapor
regions
...
During the
vaporization process, only the amount of saturated liquid
changes, not its properties
...


 Consider

a tank that contains a
saturated liquid–vapor mixture
...


• Given,

V  V f  Vg
m  m f  mg
V  mv , V f  m f v f , Vg  mg v g

• Hence,
mv  m f v f  mg v g
v

mf v f



m

mg v g
m

• From quality definition:
x

mg
m



mg
m f  mg

• Hence,
mf
m

• 
Or,



m  mg
m

 1 x

v  (1  x)v f  xvg

v  v f  x (v g  v f )

The values of the average
properties of the mixtures are
always between the values of
the saturated liquid and the
saturated vapor properties:
yf y  yg



Any extensive property per unit mass within saturated region can be
determined in similar equation like specific volume:
v  v f  x (v g  v f )



If Y = any extensive property and y = any intensive property, Y/m,
hence:
y  y f  x( yg  y f )
 y f  x y fg



with yfg = difference of property y between saturated vapour and
saturated liquid
...

The above equation always rearranged to determine the quality, x, of
saturated mixture:
x

y  yf
y fg

SUPERHEATED VAPOUR TABLE
 Table

1-6 lists down the properties for superheated vapour
for water
...

 Superheated vapor is characterized by:
a) Lower pressure (P < Psat at given T)
b) Higher temperature (T > Tsat at given P)
c) Higher values of other properties such as specific
values of volume, internal energy, enthalpy and
entropy (y>yg at given P or T, with y=v, h, u, s)

Example of superheated vapour table

Data in first
row represents
the saturated
vapour
properties at
given P

Tsat

COMPRESSED LIQUID TABLE OF WATER
A

substance is considered as a compressed liquid when its pressure is
higher than Psat at a given temperature
...

 Data in Table 1-7 starts at 5 MPa
...

Hence, at high pressure, enthalpy can be estimated as:

h  h f @T  v f @T P  Psat@T 

 So,

compressed liquid is characterized by:
Higher pressure (P > Psat at given T)
b) Lower temperature (T < Tsat at given P)
c) Lower values of other properties such
as specific values of volume, internal
energy, enthalpy and entropy (ygiven P or T, with y=v, h, u, s)
a)

Example of compressed liquid table

The format of
Table 1–7 is
similar to the
superheated
vapor tables,
except the
saturated data
represent the
saturated liquid
properties

Tsat

PHASE CONDITIONS
PROPERTY
TABLES

COMPRESSED
LIQUID TABLE
IF,
a) TTsat, or
b) PPsat, or
c) yyf

SATURATED
TABLE

IF,
a) T=Tsat, or
b) P=Psat, or
c) y=yf, or
d) y=yg, or
e) yfyyg

SUPERHEATED
TABLE
IF,
a) TTsat, or
b) PPsat, oru
c) yyg

GUIDELINES FOR
DATA ACQUISATION
FROM PROPERTY
TABLES

Data interpolation
(a)Single

interpolation:
P or T

a

b

c

y

d

e

f

If b to be calculated and values of a, c, d, e and f found from PT:

 e  d  
ba
 c  a 
  f  d 
If e to be calculated and values of a, b, c, d and f found from PT:

 b  a 
ed 
 f  d 
 c  a  

Data interpolation
(b) Triple interpolation

P

P1

P

P2

T1

a

e

b

T

f

y=??

g

T2

c

h

d

T

If y at given T and P to be calculated and values of a, b, c and d that represent
properties of y’s are found in PT:
(i) Horizontal interpolation – f and g are firstly determined before calculating y
...


 P  P  
1
ea
b  a ;
1
 P2  P 

 P  P  
1
h c
d  c ;
1
 P2  P 

Both ways should end up with the same value for y
...
1
Using Property Tables, complete this table for H2O:
P (MPa)
(b)

0
...
725

(a)

v (m3/kg)

100

15

(d)
(e)

400

0
...
75

(f)
(g)

0
...
221
0
...
25

180

0
...
2
A rigid tank with a volume of 2
...
Now the water is slowly heated
...
Also show the process on a T-v diagram with respect to
saturation lines
...
3
A piston-cylinder device contains 0
...
The piston that is free to move has a mass of 12 kg and a
diameter of 25 cm
...
Now, heat
is transferred to R-134a until final temperature is 15oC
...


The Ideal Gas
Any equation that relates the properties of
Pressure ( P ),Temperature ( T ) and
Specific Volume ( v ) of a substance is
called an Equation Of State
...

 The Equation of State predicts P, T and 
behaviour of a GAS quite accurately but within
some properly selected region, i
...
it has
limitation
...

a
...

b
...
e
...


The Ideal Gas


Equation of State for ideal gas:
T 
P  R  atau Pv  RT
v

Where:
R = gas constant(different value for different gas)
P = absolute pressure in MPa, atau kPa
v = specific volume in m3/kg
T = absolute temperature in K



Pv  RT
V
P  RT
m
PV  mRT

PV  NRuT
P V  RuT
N
Pv  RuT

If a gas obeys this relation, it is called an ideal gas
...
314 kJ/(kmol K) = 8
...

Substance

R, kJ/kg
...
2870

Helium

2
...
2081

Nitrogen

0
...
) At very low pressures (PR << 1), gases
behave as an Ideal Gas regardless of
Temperature
...
) At high temperatures (TR > 2), Ideal Gas
behaviour can be assumed with good accuracy
regardless of Pressure except when PR>>1
...
) The deviation of a gas from Ideal Gas
behaviour is greatest in the vicinity of the Critical
Point
...
4
Determine the specific volume of refrigerant134a at 1 Mpa and 50OC, using
a) The ideal gas equation of state
b) The generalized compressibility chart
c) Determine the error involved

Other Equation Of State


Other Equation Of State are :
1
...

2
...

3
...

4
...


Other Equation Of State


Van der Waals Equation Of State
...

Accurate for
ρ≤0
...
5ρcr

SPECIFIC HEATS (CP, CV),INTERNAL ENERGY
(U) AND ENTHALPY (H)

 Before

the 1st Law of Thermodynamics can be applied to a system, the
change of internal energy (U) for a substance in the system has to be firstly
determined
...

 For ideal gases, the internal energy can be determined from the specific heats
...

 Two types of specific heats:
(a) specific heat at constant volume (Cv) – energy required to raise
the
temperature of the unit mass of a substance by
one degree as the volume
is maintained constant
...


• Other definition of specific heats:
a) CV = the change in the internal energy of a
substance per unit change in temperature
at constant volume
...


 h 
CP   
 T  P
for specific heats : kJ/(kg
...
oC) (???)
 Molar basis unit for specific heats (ĈV and ĈP): kJ/(kmoloC) = kJ/(kmol
...

 Units



Thus, the differential changes in the internal energy and enthalpy of an ideal gas
can be expressed as:

du  CV (T ) dT integratio n  u  u2  u1   CV (T ) dT  CV ,ave T2  T1 



kJ

kg 

dh  C P (T ) dT    h  h2  h1   C P (T ) dT  C P ,ave T2  T1 


kJ

kg 

2

1

2

integratio n

1

Change in internal energy or enthalpy for an ideal
gas during a process from state 1 to state 2
• An ideal gas undergoes 3 different
processes in the same temperature range:
Process 1-2a: Constant volume
Process 1-2b: P=a+bV, a linear relationship
Process 1-2c: Constant pressure

• All the 3 processes have same value of
internal energy and enthalpy:

ua  ub  uc  1 CV (T )dT
2

ha  hb  hc  1 C P (T )dT

P

2
a
1

2
b

2c

T2

T1

2

V

Specific Heat Relations of Ideal Gases
Definition of
enthalpy

h  u  Pv

Replaced by
ideal gas
equation

DIFFERENTIATION
Replaced by
definition of
h and u

dh  du  d RT 
CP dT  CV dT  RdT

CP  CV  R or CP  CV  Ru
Important relationship for ideal
gases - enables to determine
CV from CP and the gas constant R

R, CP and CV
have units of
kJ/kg
...
K
...

 The constant-volume and constant-pressure specific heats
are identical for incompressible substances:

CV  CP  C

kJ

kg
...

 The specific heats of incompressible substances depend on
T only

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Title: Pure Substance
Description: In this topic, student will learn the pure substance in thermodynamics. This level of study is for those in 1st year degree student.

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