UNIT-2_Part1_NOTES-PKNAG-SPATI.pdf

Properties of Pure
Substances
7.1 INTRODUCTION
A pure substance is defined as one that is homogeneous and invariable in chemical composition throughout
its mass. The relative proportion of the chemical elements, constituting the substance, is also constant.
Atmospheric air, steam–water mixture and combustion products of a fuel are regarded as pure
substance. A phase is a physically distinct, chemically homogeneous and mechanically separable portion
of a substance.
7.2 PROPERTY DIAGRAMS
The property diagrams for a pure substance are of immense importance from thermodynamic point of
view. In this Chapter, we will discuss the pressure-volume (P–V), pressure-temperature (P–T), tem-
perature-entropy (T–S) and enthalpy-entropy (h–s) diagram for water.
7.2.1 Pressure–Volume Diagram
Water is one of the commonly used pure substances that can exist in three different phases—solid, liquid
and gas. Addition of heat at constant pressure causes its phase to change that result to a change of its
volume. Water is a rare category of pure substance which unlike other substances manifests decrease in
volume when it is converted from solid to liquid, i.e., when ice is converted to water. To study the
complete behaviour of the pure substance, it is imperative to plot P–V diagram for the same. This is
portrayed in Fig. 7.1.
Let us consider a mass of ice at say –20ºC at atmospheric pressure. Addition of heat to ice will cause
its temperature to increase up to 0ºC accompanied by increase in volume. The initial state is denoted by
1 and the final state is denoted by 2. Further heating will cause change in phase from ice to water without
increase in temperature. The substance will absorb latent heat of fusion for change in phase. The final
CHAPTER
7

7.2 Engineering Thermodynamics and Fluid Mechanics
volume is represented by the point 3. It is interesting to note that this time volume decreases. Addition of
further heat is responsible for rise in temperature of water from 0ºC to 100ºC. Consequently volume
increases from 3 to 4. At the point 4, the entire liquid attains its boiling point temperature. Continuation
of heating at this stage will convert water (liquid) to steam or vapour (gas) by taking latent heat of
vapourization. The final state is represented by the point 5 and is called saturated vapour. The increase
in volume during this part is significantly high. Further heating will convert the saturated steam to what
is called superheated steam. The temperature of superheated steam is higher than that of saturated steam.
The states 2, 3, 4 and 5 are called saturation states implying that at this point change in phase occurs.
Water (liquid) being at the middle of three different phases (solid to liquid and liquid to vapour) there
are two saturated liquid states. In the state 3, the liquid is saturated in regard to solidification whereas
at 4, the liquid is saturated with respect to vaporization. The temperature at which water is converted to
complete steam is called saturation temperature corresponding to that particular pressure. The
difference in temperature between the superheated steam and the corresponding saturation temperature
is called degree of superheat.
A liquid existing at a temperature lower than the saturation temperature corresponding to its pressure
is called compressed liquid or subcooled liquid. Vapour existing at a temperature higher than the
saturation temperature corresponding to its pressure is called superheated vapour.
If this heating is carried out at different pressures, similar states would be obtained. The Locus of all the
saturated solid states is called saturated solid line. Likewise, locus of all the points corresponds to point 3
and 4 are called saturated liquid lines. The line joining all the points 3 is called saturated liquids line with
respect to solidification, whereas the line joining all the points, 4 is called saturated liquid line with
respect to vaporisation. Similarly, the line joining all the points, 5 gives rise to saturated vapour line.
The zone that lies on the left side of the saturated solid line is the solid zone. Similarly, other zones
are established. The zones are established by the following series.
Solid Æ Solid + Liquid Æ Liquid Æ Liquid + Vapour Æ Vapour
4
1 5 6
4
Triple point line
Saturated vapour line
Saturated liquid line w.r.t.
vaporization Critical state
Saturated solid
line
Saturated
liquid line
w.r.t.
solidif ication
Liquid+vapour
solid+vapour
V
P 6
6
1
1
Figure 7.1 P–V diagram of pure substance (e.g. water) that expands on freezing

Properties of Pure Substances 7.3
Triple Point It is a line on the P–V diagram where all the three phases can exist in equilibrium.
Although it is a line but it is called triple point. At this line the pressure and temperature of three
different phases of the substance remains same but specific volumes are different. For water, the triple
point temperature and pressure are 0.01ºC and 0.6117 kPa respectively. At pressure below the triple
point, no substance can exist in stable equilibrium in liquid phase. Addition of heat to solid below this
pressure, directly converts it to vapour by absorbing latent heat of sublimation. Hence, the zone below
the triple point line denotes the solid and vapour zone.
Critical Point It is interesting to note that as the pressure increases, the transition zone from liquid to
vapour becomes narrow. This implies that saturated liquid line (locus of points 4) and saturated vapour
line are inclined towards each other. At a specific pressure, the transition zone is converted to a point.
The point, at which the saturated liquid state and saturated vapour state are identical, is called
critical point. At pressures above this point, the liquid, if heated, immediately flashes into vapour
without manifesting a clear transition zone. Conversely, cooling of vapour converts it to water all of a
sudden. The pressure, specific volume and temperature at critical point are called critical pressure,
critical specific volume and critical temperature respectively. The critical point data for water are
given below.
Pcr = 221.2 bar vcr = 0.00317 m3
/kg tcr = 374.15ºC
It must be remembered that phase change takes place only at constant pressure and temperature.
For constant pressure heating, liquid is converted to vapour only at a particular temperature called
saturation temperature. Similarly, if the temperature remains constant, this phase change takes place at
a definite pressure called saturation pressure.
7.2.2 Pressure–Temperature Diagram
The change in volume of pure substance at constant pressure heating is analyzed by the P–V diagram.
The pressure-temperature (P-T) portrays the variations in temperature at constant pressure heating.
This is plotted in Fig. 7.2.
T
P
Fusion curve
Sublimation curve
Vapourization curve
Triple point
Solid
phase
region
Liquid phase region
Vapour phase region
2,3 4,5 6
2,3 4,5 6
1
1
Figure 7.2 P–T diagram of a pure substanace (e.g. water) that expands on freezing

The variation of temperature at any particular pressure is represented by a horizontal line. Several
such curves may be obtained which are parallel. However, heating at higher pressure is accompanied by
increase in saturation temperature. Note that on any particular curve, points 2 and 3 and points 4 and 5
coincides. This is due to the fact that between 2 and 3 there is change in phase (from solid to liquid)
and between 4 and 5, liquid is converted to vapour. Since during phase change the substance absorbs
or rejects only latent heat, the temperature during phase change remains unaltered. The curve passing
through the series of 2, 3 points is called the fusion curve while the curve when drawn through points
4 and 5 is called the vaporization curve. The sublimation curve is the locus of vapour pressure at
different temperatures. Three aforementioned curves meet at the triple point.
7.2.3 Temperature–Entropy Diagram
Temperature-entropy (T-S) diagram for pure substance is of immense importance from the
thermodynamic point of view. Water being one of the popular working substance for power plants, it
is imperative to investigate the nature of T–S plot when the substance is heated gradually.
Two different situations are encountered while heating ice (at –20ºC) so as to convert it to a
superheated steam at 200ºC.
(i) Temperature is increased continuously so long there is no change in phase.
(ii) Temperature remains constant during change in phase.
In the first case, change in entropy between any two finite temperature range Tf and Ti can be
calculated as ln
f f
i i
T T
p f
f i p
i
T T
mC dT T
dQ
S S S mC
T T T
D = - = = =
Ú Ú
And if temperature remains constant, DS becomes
mL
T
.
The T–S plot is shown in Fig. 7.3. Several such curves are plotted when heating is carried out at
different pressures. The dome–shaped region formed by the saturated liquid line and saturated
vapour line is called vapour dome.
4
1
2
5
5
2
2
6
4
3
3
Critical state
Triple point line
Liquid + vapour
solid + vapour
6
Liquid
Vapour
1
T
S
Figure 7.3 T–S diagram of a pure substance

7.2.4 Enthalpy–Entropy Diagram
In this context, the enthalpy vs entropy plot or h–s diagram is found to be valuable in the analysis of
steady-flow devices like turbines, compressors, nozzles etc.The h–s diagram is popularly known as
Mollier diadram.
From the thermodynamic property relationship (see Eq. (6.17)), we have
Tds = dh vdp
-
For constant pressure, the above equation becomes
Tds = dh
or,
P
h
s
∂
Ê ˆ
Á ˜
Ë ¯
∂
= T (7.1)
Equation (7.1) implies that the slope of the constant pressure lines is equal to the corresponding
saturation temperatures. As the pressure increases, the saturation temperature also increases. This is
the reason for which constant pressure lines (isobar) are divergent on the h-s diagram. The h-s dia-
gram (Mollier diagram) for water is shown in Fig. 7.4. The saturated liquid line and saturated vapour
line meets at critical point.
Triple point line
Solid + vapour
Critical point
Vapour
Liquid
L V
+
S
h
1 2
3 4
4
3
2
1
5
6 6
5
Figure 7.4 h–s (Mollier diagram) of a pure substance (water)
Constant temperature lines are distinctly visible in the superheat region. However, in the wet region
these lines coincide with constant pressure lines. It is noteworthy that in the superheated region,
enthalpy is more dependent on temperature alone. As the degree of superheat increases, the constant
temperature lines tend to become horizontal. Constant dryness fraction lines are also added in the
curve to calculate different properties of wet steam. From the curve, it is evident that reduction in
pressure at constant enthalpy results in the drying and superheating of wet vapour.

7.3 P-V-T SURFACE
According to state postulate (see Section 1.8), the state of a substance is described by two
independent, intensive properties. Once these are completely specified, all the other properties become
dependent properties. Any equation with two independent variables in the form f = f (x, y) represents
a surface in space; it is possible to represent the P-V-T behaviour of a substance as a surface in space
as shown in Fig. 7.5. Here, T and V are considered as independent variables and P as dependent. All
the points in the surface represent equilibrium states. Hence the quasi-static processes must lie on the
surface.
Essentially, this is a three-dimensional view where three different dimensions are P, V and T. It can be
concluded therefore, that the popularly used P–V diagrams and T–V diagram are merely the orthographic
projections of the original-three dimensional objects. Although such a three-dimensional view is very
comprehensive, in thermodynamic analysis use of two-dimensional diagrams are more popular.
V
T
Solid + vapour
Solid Liquid
Critical state
P
Vapour
Liquid + vapour
Triple state
Figure 7.5 A P-V-T surface for a substance which expands on freezing
7.4 QUALITY OR DRYNESS FRACTION
Quality or dryness fraction of a liquid-vapour mixture is defined as the ratio of mass of saturated
vapour to the total mass of mixture. That is,

x =
mass of saturated vapour
total mass of mixture
or, x =
g g
f g
m m
m m m
=
+
(7.2)
where f
m is the mass of saturated liquid, g
m is the mass of saturated vapour and m is the total
mass of liquid-vapour mixture.
Let V be the total volume of the mixture, f
V be the volume of the saturated liquid and g
V be the
volume of the saturated vapour.
Total volume of liquid and vapour can be expressed in terms of their specific volume as
Vf = f f
m v
Vg = g g
m v
where vf and vg are the specific volume of saturated liquid and saturated vapour respectively.
Specific volume of the mixture is then
v =
f g
V V
V
m m
+
=
=
f f g g
f g
m v m v
m m
+
+
=
f f g g
f g f g
m v m v
m m m m
+
+ +
= (1 ) f g
x v xv
- +
= ( )
f g f
v x v v
+ -
= f fg
v xv
+
Similarly, one can write
h = f fg
h xh
+
s = f fg
s xs
+
and
u = f fg
u xu
+
7.5 STEAM TABLES
From the foregoing discussion, it is possible to know the various properties of steam when it is either
saturated liquid or saturated vapor provided its pressure or temperature is known. The subscripts f

and g refer to saturated liquid and saturated vapour respectively. Steam table is nothing but a complete
data book that contains various properties of water in different phases for a given pressure or tem-
perature. It is given in the Appendix I. The properties of steam are given in three different tables:
saturated steam table-temperature base (Appendix I, Table A.1), saturated steam table-pressure base
(Appendix I, Table A.2), and superheated steam table (Appendix I, Table A.3),. In the saturated steam
tables, the properties of saturated liquid and saturated vapour are presented. It is known that the
pressure and temperature both remain constant during the phase transition. That means both are not
independent variables. One is independent and the other is dependent. When the temperature is chosen
as independent variable and the properties of steam are tabulated, the steam table is referred to as the
saturated steam table-temperature base. For saturated steam table-pressure base, pressure is the
independent variable. There is a separate table for superheated steam for different amount of degree of
superheat.
Example 7.1 Using the steam tables, estimate the saturation temperature and specific volume,
specific enthalpy and specific entropy of saturated liquid and vapour at 3 MPa.
Solution From the saturated steam table based on pressure (Appendix I, Table A.2), it is found
that saturation temperature of water corresponds to 3 MPa is sat 233.9 C
t =
Specific volume of saturated liquid is 3
0.001216 m /kg
f
v =
Specific volume of saturated vapour is 3
0.06668 m /kg
g
v =
Specific enthalpy of saturated liquid is 1008.4 kJ/kg
f
h =
Specific enthalpy of saturated vapour 2804.1 kJ/kg
g
h =
Specific entropy of saturated liquid is 2.6462 kJ/kg-K
f
s =
Specific entropy of saturated vapour is 6.1878 kJ/kg-K
g
s =
Example 7.2 Using the steam tables, estimate saturation pressure and the specific volume, specific
enthalpy and specific entropy of saturated liquid and vapour at 200°C.
Solution From the saturated steam table based on temperature (Appendix I, Table A.1), it is
found that saturation pressure of water corresponds to 200°C is sat 1.554 MPa
P = .
0.001156 m /kg
f
v =
0.1274 m /kg
g
v =
f
h =
g
h =
f
s =
g
s =

Example 7.3 Calculate the specific volume, specific enthalpy and specific entropy of wet steam of
dryness fraction or quality of 0.9 at 2 MPa pressure.
Solution From the saturated steam table (Appendix I, Table A.2), at 2 MPa
0.001177 m /kg
f
v =
0.09963 m /kg
g
v =
f
h =
g
h =
f
s =
g
s =
Dryness fraction of the mixture is given as 0.9
x =
The specific volume of the wet steam is
v = ( )
f g f
v x v v
+ -
= 3
0.001177 0.9(0.09963 0.001177) 0.08978 m /kg
+ - =
The specific enthalpy of the wet steam is
h = ( )
f g f
h x h h
+ -
= 908.8 0.9(2799.5 908.8) 2610.43 kJ/kg
+ - =
The specific entropy of the wet steam is
s = ( )
f g f
s x s s
+ -
= 2.4478 0.9(6.3417 2.4478) 5.9523 kJ/kg-K
+ - =
Example 7.4 If the specific entropy of steam at 1 MPa is 5.82 kJ/kg-K , determine its state.
Solution From the saturated steam table based on pressure (Appendix I, Table A.2), it is found
that at 1 MPa the specific entropy of saturated liquid and saturated vapour are
2.1391 kJ/kg-K
f
s = and 6.5873 kJ/kg-K
g
s = respectively
Since the given specific entropy lies between sf and sg, the state will be within the
vapour dome. Thus, temperature of the steam is same as the saturation temperature
corresponds to 1 MPa pressure. From Appendix I, Table A.2, we get sat 179.9 C
t =
Let x be the quality of the steam.
s = ( )
f g f
s x s s
+ -
5.82 = 2.1391 (6.5873 2.1391)
x
+ -
or, x = 0.8275

Example 7.5 Find the specific volume, specific enthalpy and specific entropy of steam at 5 bar and
400°C.
Solution From the saturated steam table (Appendix I, Table A.2), it is found that saturation
temperature of water corresponds to 5 bar is sat 151.86 C
t =
Since the temperature is greater than that of saturation temperature, the state would
be in the superheated region.
From the superheated steam table (Appendix I, Table A.3),
Specific volume of steam is 3
0.6173 m /kg
v =
Specific enthalpy of steam is 3271.9 kJ/kg
h =
Specific entropy of steam is 7.7938 kJ/kg-K
s =
Example 7.6 A rigid vessel of volume 0.2 m3
contains 1 kg of steam at a pressure of 0.8 MPa.
Evaluate the specific volume, temperature, dryness fraction, enthalpy and entropy of
steam.
Solution Specific volume of the steam 3
0.2
0.2 m /kg
1
V
v
m
= = =
From the saturated steam table (Appendix I, Table A.2), it is found that at 0.8 MPa
the specific volume of saturated liquid and saturated vapour are
3
0.001115 m /kg
f
v = and 3
0.2404 m /kg
g
v = respectively.
Since the given specific volume lies between vf and vg, the state will be within the
vapour dome. From the saturated steam table, it is found that saturation temperature
of water corresponds to 0.8 MPa is sat 170.4 C
t =
Let x be the quality of the steam. Thus,
v = ( )
f g f
v x v v
+ -
0.2 = 0.001115 (0.2404 0.001115)
x
+ -
or, x = 0.8312
h = ( )
f g f
h x h h
+ -
= 721.1 0.8312(2769.1 721.1) 2423.4 kJ/kg
+ - =
The total enthalpy of the wet steam is then
H = ( )( )
1kg 2423.4 kJ/kg 2423.4 kJ
mh = =

The specific entropy of the wet steam is
s = ( )
f g f
s x s s
+ -
= 2.0466 0.8312(6.6636 2.0466) 5.8843 kJ/kg-K
+ - =
The total entropy of the wet steam is
S = ( )( )
1kg 5.8843 kJ/kg-K 5.8843 kJ/K
ms = =
Example 7.7 10 kg of wet steam at a pressure of 0.2 MPa is contained in a rigid tank of volume
6.058 m3
. The tank is heated until the steam becomes dry saturated. Determine the
final pressure and the heat transfer to the tank.
Solution Specific volume of the steam 3
6.058
0.6058m /kg
10
V
v
m
= = =
From the saturated steam table (Appendix I, Table A.2), it is found that at 0.2 MPa
the specific volume of saturated liquid and saturated vapour are 3
0.001061 m /kg
f
v =
and 3
0.8857 m /kg
g
v = respectively.
Since the given specific volume lies between vf and vg, the state will be within the
vapour dome. From the saturated steam table, it is found that saturation temperature
t =
v = ( )
f g f
v x v v
+ -
0.6058 = 0.001061 (0.8857 0.001061)
x
+ -
or, x = 0.6836
h = ( )
f g f
h x h h
+ -
= 504.7 0.6836(2706.6 504.7) 2009.92 kJ/kg
+ - =
From the saturated steam table, it is found that for
3
0.6058m /kg
g
v = , the corre-
sponding pressure is 0.3 MPa
At 0.3 MPa, the specific enthalpy of saturated vapour 2725.3 kJ/kg
g
h =
Specific heat transfer to the tank is 2725.3 2009.92 715.38 kJ/kg
g
q h h
= - = - =
Total heat transfer is then ( )( )
10kg 715.38 kJ/kg 7153.8 kJ
Q mq
= = =
Example 7.8 One kg of water at 75ºC is heated at a constant pressure of 0.8 MPa until it becomes
superheated vapor at 200ºC. Find the change in volume, enthalpy, entropy and inter-
nal energy.

Solution From the steam table (Appendix I, Table A.2), it is found that saturation temperature
t =
The conversion of water at 75ºC to superheated steam at 200ºC can be divided into
the following distinct steps
(i) Conversion of water at 75ºC to saturated water at 170.4ºC
The increase in enthalpy for this change is 1 sat
( )
pw i
H m C T T
D = ¥ ¥ -
= 1 4.18 (170.4 75) 398.72 kJ
¥ ¥ - =
(ii) Conversion of saturated water to saturated steam without change in temperature.
The increase in enthalpy for this process is 2 ( )
g f
H m h h
D = -
From the saturated steam table (Appendix I, Table A.2), at 0.8 MPa
721.1 kJ/kg
f
h = and 2769.1 kJ/kg
g
h =
Hence, DH2 = ( )
1kg (2769.1 721.1)kJ/kg = 2048 kJ
-
(iii) Conversion of saturated steam into superheated steam of 200ºC
From superheated steam table (Appendix I, Table A.3), it is found that at 0.8 MPa
and 200ºC, 2839.3 kJ/kg
h =
The increase in enthalpy associated with this change is
DH3 = ( )
g
m h h
-
= ( )
1kg (2839.3 2769.1) kJ/kg = 70.2 kJ
-
Therefore total change in enthalpy becomes
DH = 1 2 3
H H H
D + D + D
= 398.72 2048 70.2 2516.92 kJ
+ + =
From the superheated steam table it is found that v = 0.2608 m3
/kg and vf = 0.001115 m3
/kg
Therefore change in volume becomes
DV = ( )( ) 3 3
( ) 1kg 0.2608 0.001115 m /kg 0.2597m
f
m v v
- = - =
From the relationship h u Pv
= + we have 1 1 1
h u Pv
= + and 2 2 2
h u Pv
= +
Hence, change in internal energy is
2 1
U U
- = 2 1
( )
H H P V
- - ¥ D
= ( )( )
3
2516.92 800 kPa 0.2597 m 2309.16 kJ
- =

Following the logic followed for enthalpy computations,
Change of entropy of water during conversion from 75ºC to saturated water at
170.4ºC is
DS1 =
170.4 273
ln 1 4.18ln 1.0127 kJ/K
75 273
sat
p
i
T
mC
T
+
= ¥ =
+
Change of entropy of water Conversion of saturated water to saturated steam is
DS2 = ( ) 1 (6.6636 2.0466) 4.617 kJ/K
g f
m s s
¥ - = ¥ - =
Change of entropy of water during conversion of saturated steam into superheated
steam of 200ºC is
DS3 = ( )
g
m s s
¥ - = 1 (6.8158 6.6636) 0.1522 kJ/K
¥ - =
Hence total change in entropy becomes
DS = 1 2 3 1.0127 4.167 0.1522 5.7819 kJ/K
S S S
D + D + D = + + =
Example 7.9 A vessel of volume 0.08 m3
contains a mixture of saturated water and saturated
steam at a temperature of 200°C. The mass of the liquid present is 10 kg. Find the
pressure, the mass, the specific volume, the enthalpy, the entropy and the internal
energy.
Solution From temperature based saturated steam table (Appendix I, Table A.1), at 200°C,
saturation pressure is sat 1.554 MPa
P =
0.001156 m /kg
f
v =
0.1274 m /kg
g
v =
f
h =
Specific enthalpy of saturated vapour is 2793.2 kJ/kg
g
h =
f
s =
Specific enthalpy of saturated vapour is 6.4331 kJ/kg-K
g
s =
Specific internal energy of saturated liquid is 850.6 kJ/kg
f
u =
Specific internal energy of saturated vapour is 2595.3 kJ/kg
g
u =
Total Volume of liquid is f f f
V m v
=
= ( )( )
3 3
10kg 0.001156m /kg 0.01156 m
=
Total Volume of vapour is 3
0.08 0.01156 0.06844 m
g f
V V V
= - = - =

Mass of vapour is found to be
mg =
0.06844
0.5372 kg
0.1274
g
g
V
v
= =
Total mass of liquid-vapour mixture is 10 0.5372 10.5372 kg
f g
m m m
= + = + =
Quality (or dryness fraction) of the mixture is
0.5372
0.05
10.5372
g
m
x
m
= = =
Specific volume of the mixture is
v = ( )
f g f
v x v v
+ -
= 3
0.001156 0.05(0.1274 0.001156) 0.00747 m /kg
+ - =
Specific enthalpy of the mixture is
h = ( )
f g f
h x h h
+ -
= 852.4 0.05(2793.2 852.4) 949.44 kJ/kg
+ - =
Total enthalpy of the mixture is then
H = ( )( )
10.5372kg 949.44 kJ/kg 10004.44 kJ
mh = =
Specific entropy of the mixture is
s = ( )
f g f
s x s s
+ -
= 2.3313 0.05(6.4331 2.3313) 2.5364 kJ/kg-K
+ - =
Total entropy of the mixture is then found to be
S = ( )( )
10.5372kg 2.5364 kJ/kg-K 26.7266 kJ/K
ms = =
Specific internal energy of the mixture is
u = ( )
f g f
u x u u
+ -
= 850.6 0.05(2595.3 850.6) 937.835 kJ/kg
+ - =
Total internal energy of the mixture is
U = ( )( )
10.5372kg 937.835 kJ/kg 9882.15 kJ
mu = =
SUMMARY
A pure substance is defined as one that is homogeneous and invariable in a chemical
composition throughout its mass.
A phase is a physically distinct, chemically homogeneous and mechanically separable
portion of a substance.

The pressure and temperature under which two phases can exist in equilibrium are
called saturation pressure and saturation temperature.
A liquid existing at a temperature lower than the saturation temperature corresponding
to its pressure is called compressed liquid or subcooled liquid. Vapour existing at a
temperature higher than the saturation temperature corresponding to its pressure is
called superheated vapour.
The condition of pressure and temperature under which all the three phases of a pure
substance can exist in equilibrium is called the triple point of the substance.
The point, at which the saturated liquid state and saturated vapour state are identical,
is called the critical point. At pressures above this point, a liquid, if heated, immediately
flashes into vapour without manifesting a clear transition zone. Conversely, cooling of
vapour converts it to water all of a sudden. The pressure, specific volume and
temperature at the critical point are called critical pressure, critical specific volume and
critical temperature respectively.
Quality or dryness fraction is defined as the ratio of mass of saturated vapour to the
total mass of mixture. That is,
x =
m m
g g
m m + mg
f
=
where mf is the mass of saturated liquid, mx is the mass of saturated vapour and m
is the total mass of liquid-vapour mixture.
Properties of a liquid–vapour mixture may be found by relations such as
v = f fg
v + xv
h = f fg
h + xh
s =
f fg
s + xs
u = f fg
u + xu
REVIEW QUESTIONS
7.1 What is a pure substance?
7.2 What is a phase?
7.3 What is a saturated state?
7.4 What is saturation temperature and saturation pressure?
7.5 What is a compressed liquid?
7.6 What is superheated vapour?
7.7 Define the quality of dryness fraction of a liquid–vapour mixture.
7.8 What is the critical point? State the values of critical pressure and critical temperature of water.
7.9 Sketch the P–V diagram for a pure substance and show the isotherms and constant quality lines on it.
7.10 Is it possible to convert a liquid into vapour phase without ever observing the phase transition? If so,
sketch the process on a P–T diagram.

NUMERICAL PROBLEMS
7.1 Using the steam tables, estimate the specific volume, specific enthalpy and specific entropy of
saturated liquid and vapour at 30 bar.
7.2 Using the steam tables, estimate the specific volume, specific enthalpy and specific entropy of
saturated liquid and vapour at 250ºC.
7.3 Calculate the specific volume, specific enthalpy and specific entropy of wet steam of dryness fraction
of quality 0.9 at 30 bar pressure.
7.4 If the specific entropy of steam at 30 bar is 5.22 kJ/kg-K, determine its state.
7.5 Find the saturation temperature specific volume, specific enthalpy and entropy of saturated vapour at
5 bar. Also find the latent heat of vaporisation of steam at that pressure.
7.6 A rigid vessel of volume 0.3 m3
contains 10 kg of steam at a pressure of 5 bar. Evaluate the specific
volume, temperature, dryness fraction, enthalpy and entropy of steam.
7.7 A vessel of volume 0.03 m3
contains a mixture of saturated water and saturated steam at a pressure of
30 bar. The mass of the liquid present is 6 kg. Findthe pressure, the mass, the specific volume, the
enthalpy, the entropy and the internal energy.
7.8 Suppose a closed and rigid vessel is initially filled with saturated water and saturated vapour at
100 kPa. On transferring energy as heat, the water is found to pass through the critical point.
Determine the volume of saturated vapour to the volume of saturated liquid with which the vessel is
initially filled.
7.9 A vessel of volume 0.03 m3
contains a mixture of saturated water and saturated steam at a temperature
of 250°C. The mass of the liquid present is 8 kg. Find the pressure, the mass, the specific volume, the
enthalpy, the entropy and the internal energy.
MULTIPLE-CHOICE QUESTIONS
7.1 In a P-V-T surface, the zone below the triple point is known as
(a) liquid zone (b) vapour zone (c) sublimation zone (d) none of these
7.2 Triple point of a pure substance is a point at which
(a) liquid and vapour exist together (b) solid and vapour exist together
(c) solid and liquid exist together (d) solid, liquid and vapour exist together
7.3 The latent heat of vapourization at the critical point is
(a) equal to zero (b) less than zero (c) greater than zero (d) none of these
7.4 The phase change from liquid to vapour is referred to as
(a) melting (b) vapourization (c) sublimation (d) solidification
7.5 The point that connects the saturated liquid line to the saturated vapour line is called the
(a) triple point (b) superheated point (c) critical point (d) compressed liquid point

UNIT-2_Part1_NOTES-PKNAG-SPATI.pdf

Recommended

Recommended

More Related Content

Similar to UNIT-2_Part1_NOTES-PKNAG-SPATI.pdf

Similar to UNIT-2_Part1_NOTES-PKNAG-SPATI.pdf (20)

More from YOGESH AHIRE

More from YOGESH AHIRE (11)

Recently uploaded

Recently uploaded (20)

UNIT-2_Part1_NOTES-PKNAG-SPATI.pdf