J2006 termodinamik 1 unit8

PROPERTIES OF STEAM
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PROPERTIES OF STEAM
OBJECTIVES
General Objective : To define the properties of wet steam, dry saturated steam and
superheated steam using information from the steam tables.
Specific Objectives : At the end of the unit you will be able to:
 define the word phase and distinguish the solid, liquid and
steam phases
 understand and use the fact that the vaporization process is
carried out at constant pressure
 define and explain the following terms: saturation
temperature, saturated liquid, wet steam, saturated steam, dry
saturated steam, , dryness fraction and superheated steam
 determine the properties of steam using the P-v diagram
 understand and use the nomenclature as in the Steam Tables
 apply single and double interpolation using the steam tables
 locate the correct steam tables for interpolation, including
interpolation between saturation tables and superheated tables
UNIT 8

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where necessary
8.0 Introduction
In thermodynamic systems, the working fluid can be in the liquid, steam or gaseous
phase. In this unit, the properties of liquid and steam are investigated in some details
as the state of a system can be described in terms of its properties. A substance that
has a fixed composition throughout is called a pure substance. Pure chemicals
(H2O, N2, O2, Ar, Ne, Xe) are always pure substances. We all know from experience
that substances exist in different phases. A phase of substance can be defined as that
part of a pure substance that consists of a single, homogenous aggregate of matter.
The three common phases for H2O that are usually used are solid, liquid and steam.
When studying phases or phase changes in thermodynamics, one does not need to be
concerned with the molecular structure and behavior of the different phases.
However, it is very helpful to have some understanding of the molecular phenomena
involved in each phase.
Molecular bonds are strongest in solids and weakest in steams. One reason is that
molecules in solids are closely packed together, whereas in steams they are separated
by great distances.
INPUTINPUT

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The three phases of pure substances are: -
Solid Phase
In the solid phase, the molecules are;
(a) closely bound, therefore relatively dense; and
(b) arranged in a rigid three-dimensional pattern so that they do not easily
deform. An example of a pure solid state is ice.
Liquid Phase
In the liquid phase, the molecules are;
(a) closely bound, therefore also relatively dense and unable to expand to fill a
space; but
(b) they are no longer rigidly structured so much so that they are free to move
within a fixed volume. An example is a pure liquid state.
Steam Phase
In the steam phase, the molecules;
(a) virtually do not attract each other. The distance between the molecules are
not as close as those in the solid and liquid phases;
(b) are not arranged in a fixed pattern. There is neither a fixed volume nor a
fixed shape for steam.
The three phases described above are illustrated in Fig. 8.0 below. The following are
discovered:
(a) the positions of the molecules are relatively fixed in a solid phase;
(b) chunks of molecules float about each other in the liquid phase; and
(c) the molecules move about at random in the steam phase.
(a)
(b)
(c)

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Source: Thermodynamics: An Engineering Approach, 3rd
Ed by Cengel and Boles
Figure 8.0 The arrangement of atoms in different phases

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8.1 Phase-Change Process
The distinction between steam and liquid is usually made (in an elementary manner)
by stating that both will take up the shape of their containers. However liquid will
present a free surface if it does not completely fill its container. Steam on the other
hand will always fill its container.
With these information, let us consider the following system:
A container is filled with water, and a moveable, frictionless piston is placed on the
container at State 1, as shown in Fig. 8.1. As heat is added to the system, the
temperature of the system will increase. Note that the pressure on the system is
being kept constant by the weight of the piston. The continued addition of heat will
cause the temperature of the system to increase until the pressure of the steam
generated exactly balances the pressure of the atmosphere plus the pressure due to
the weight of the piston.
Figure 8.1 Heating water and steam at constant pressure
At this point, the steam and liquid are said to be saturated. As more heat is added,
the liquid that was at saturation will start to vaporize until State 2. The two-phase
mixture of steam and liquid at State 2 has only one degree of freedom, and as long as
liquid is present, vaporization will continue at constant temperature. As long as
liquid is present, the mixture is said to be wet steam, and both the liquid and steam
are saturated. After all the liquid is vaporized, only steam is present at State 3, and
W
W
W
W
Liquid Steam Superheated
Steam
STATE 1 STATE 2 STATE 3 STATE 4

PROPERTIES OF STEAM
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the further addition of heat will cause the temperature of steam to increase at
constant system pressure. This state is called the superheated state, and the steam is
said to be superheated steam as shown in State 4.
8.2 Saturated and Superheated Steam
While tables provide a convenient way of presenting precise numerical presentations
of data, figures provide us with a clearer understanding of trends and patterns.
Consider the following diagram in which the specific volume of H2O is presented as
a function of temperature and pressure1
:
Figure 8.2-1 T-v diagram for the heating process of water at constant pressure
Imagine that we are to run an experiment. In this experiment, we start with a mass
of water at 1 atm pressure and room temperature. At this temperature and pressure
we may measure the specific volume (1/ρ = 1/1000 kg/m3
). We plot this state at
point 1 on the diagram.
If we proceed to heat the water, the temperature will rise. In addition, water expands
slightly as it is heated which makes the specific volume increase slightly. We may
plot the locus of such points along the line from State 1 to State 2. We speak of
1
Figures from Thermodynamics: An Engineering Approach, 3rd
20
100
300
1
2 3
4
T, o
C
v, m3
/kg
Compressed
liquid
Saturated
mixture
Superheated
steam

PROPERTIES OF STEAM
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liquid in one of these conditions as being compressed or subcooled liquid.
State 2 is selected to correspond to the boiling point (100 o
C). We speak of State 2
as being the saturated liquid state, which means that all of the water is in still liquid
form, but ready to boil. As we continue to heat past the boiling point 2, a
fundamental change occurs in the process. The temperature of the water no longer
continues to rise. Instead, as we continue to add energy, liquid progressively
changes to steam phase at a constant temperature but with an increasing specific
volume. In this part of the process, we speak of the water as being a saturated mixture
(liquid + steam). This is also known as the quality region.
At State 3, all liquid will have been vaporised. This is the saturated steam state.
As we continue to heat the steam beyond State 3, the temperature of the steam again
rises as we add energy. States to the right of State 3 are said to be superheated steam.
Summary of nomenclature:
Compressed or subcooled liquid (Between States 1 & 2)
A liquid state in which the fluid remains entirely within the liquid state,
and below the saturation state.
Saturated liquid (State 2)
All fluid is in the liquid state. However, even the slightest addition of
energy would result in the formation of some vapour.
Saturated Liquid-Steam or Wet Steam Region (Between States 2 & 3) Liquid
and steam exist together in a mixture.
Saturated steam (State 3)
All fluid is in the steam state, but even the slightest loss of energy from the
system would result in the formation of some liquid.
Superheated steam (The right of State 3)
All fluid is in the steam state and above the saturation state. The
superheated steam temperature is greater than the saturation temperature
corresponding to the pressure.

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The same experiment can be conducted at several different pressures. We see that as
pressure increases, the temperature at which boiling occurs also increases.2
Figure 8.2-2 T-v diagram of constant pressure phase change processes
of a pure substance at various pressures for water.
It can be seen that as pressure increases, the specific volume increase in the liquid to
steam transition will decrease.
At a pressure of 221.2 bar, the specific volume change which is associated to a phase
increase will disappear. Both liquid and steam will have the same specific volume,
0.00317 m3
/kg. This occurs at a temperature of 374.15 o
C. This state represents an
important transition in fluids and is termed the critical point.
2
P = 1.01325 bar
P = 5 bar
P = 10 bar
P = 80 bar
P = 150 bar
P = 221.2 bar
Critical point
374.15
T, o
C
v, m3
/kg
Saturated
liquid
Saturated
steam
0.00317

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If we connect the locus of points corresponding to the saturation condition, we will
obtain a diagram which allows easy identification of the distinct regions3
:
Figure 8.2-3 T-v diagram of a pure substance
The general shape of the P-v diagram of a pure substance is very much like the T-v
diagram, but the T = constant lines on this diagram have a downward trend, as shown
in Fig. 8.2-4.
3
P
v
Critical
point
Saturated liquid line
Dry saturated steam line
T2
= const.
T1
= const.
COMPRESS
LIQUID
REGION
WET STEAM
REGION
SUPERHEATED
STEAM
REGION
T2
> T1
T
v
Critical
point
Saturated liquid line
Dry saturated steam line
P2
= const.
P1
= const.COMPRESS
LIQUID
REGION
WET STEAM
REGION
SUPERHEATED
STEAM
REGION
P2
> P1

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Figure 8.2-4 P-v diagram of a pure substance
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…!
8.1 Each line in the table below gives information about phases of pure substances. Fill
in the phase column in the table with the correct answer.
Statement Phase
The molecules are closely bound, they are also relatively
dense and unable to expand to fill a space. However they are
no longer rigidly structured so that they are free to move
within a fixed volume.
i._____________
The molecules are closely bound, they are relatively dense
and arranged in a rigid three-dimensional patterns so that they
do not easily deform.
ii.____________
The molecules virtually do not attract each other. The
distance between the molecules are not as close as those in the
solid and liquid phases. They are not arranged in a fixed
pattern. There is neither a fixed volume nor a fixed shape for
steam.
iii.____________
8.2 Write the suitable names of the phases for the H2O in the P-v diagram below.
Activity 8A
P
v
( vi )
( ii )
( iv )
T2
= const.
T1
= const.
( i )
( iii)
( v )
T2
> T1

PROPERTIES OF STEAM
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Feedback To Activity 8A
8.1 i) Liquid Phase
ii) Solid Phase
iii) Steam Phase
8.2 i) Compress liquid region
ii) Saturated liquid line
iii) Wet steam region
iv) Dry saturated steam line
v) Superheated steam region
vi) Critical point

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8.3 Properties of a Wet Mixture
Between the saturated liquid and the saturated steam, there exist a mixture of steam
plus liquid (wet steam region). To denote the state of a liquid-steam mixture, it is
necessary to introduce a term describing the relative quantities of liquid and steam in
the mixture. This is called the dryness fraction (symbol x). Thus, in 1 kg of wet
mixture, there must be x kg of saturated steam plus (1 – x) kg of saturated liquid.
Figure 8.3-1 Liquid-steam mixture
The dryness fraction is defined as follows;
where mtotal = mliquid + msteam
(8.1)
(1 - x ) kg of liquid
x kg of steam
total mass = 1 kg
masstotal
steamsaturateddryofmass
fractiondryness =
total
steam
m
m
x =
INPUTINPUT

PROPERTIES OF STEAM
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8.3.1 Specific volume
For a wet steam, the total volume of the mixture is given by the volume of
liquid present plus the volume of dry steam present.
Therefore, the specific volume is given by,
Now for 1 kg of wet steam, there are (1 – x) kg of liquid and x kg of dry
steam, where x is the dryness fraction as defined earlier. Hence,
v = vf(1 – x) + vgx
The volume of the liquid is usually negligibly small as compared to the
volume of dry saturated steam. Hence, for most practical problems,
v = xvg (8.2)
Where,
vf = specific volume of saturated liquid (m3
/kg)
Figure 8.3-2 P-v diagram showing the location point of the dryness fraction
At point A, x = 0
At point B, x = 1
Between point A and B, 0 < x < 1.0
Note that for a saturated liquid, x = 0;
and that for dry saturated steam, x = 1.
Sat. liquid
Sat. steam
Sat. liquid
P
v
ts
A B
x = 0.2 x = 0.8
vf
vg
Sat. steam
steamwetofmasstotal
steamdryofvolumeliquidaofvolume +
=v

PROPERTIES OF STEAM
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vg = specific volume of saturated steam (m3
/kg)
x = dryness fraction
8.3.2 Specific enthalpy
In the analysis of certain types of processes, particularly in power generation
and refrigeration, we frequently encounter the combination of properties
U + PV. For the sake of simplicity and convenience, this combination is
defined as a new property, enthalpy, and given the symbol H.
H = U + PV (kJ)
or, per unit mass
h = u + Pv (kJ/kg) (8.3)
The enthalpy of wet steam is given by the sum of the enthalpy of the liquid
plus the enthalpy of the dry steam,
i.e. h = hf(1 – x) + xhg
h = hf + x(hg – hf )
∴ h = hf + xhfg (8.4)
Where,
hf = specific enthalpy of saturated liquid (kJ/kg)
hg = specific enthalpy of saturated steam (kJ/kg)
hfg = difference between hg and hf (that is, hfg = hg - hf )
8.3.3 Specific Internal Energy
Similarly, the specific internal energy of a wet steam is given by the internal
energy of the liquid plus the internal energy of the dry steam,
i.e. u = uf(1 – x) + xug
∴ u = uf + x(ug – uf ) (8.5)
Where,
uf = specific enthalpy of saturated liquid (kJ/kg)
ug = specific enthalpy of saturated steam (kJ/kg)
ug –uf = difference between ug and uf

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Equation 8.5 can be expressed in a form similar to equation 8.4. However,
equation 8.5 is more convenient since ug and uf are tabulated. The difference
is that, ufg is not tabulated.

PROPERTIES OF STEAM
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8.3.4 Specific Entropy
A person looking at the steam tables carefully will notice two new properties
i.e. enthalpy h and entropy s. Entropy is a property associated with the
Second Law of Thermodynamics, and actually, we will properly define it in
Unit 9 . However, it is appropriate to introduce entropy at this point.
The entropy of wet steam is given by the sum of the entropy of the liquid
plus the entropy of the dry steam,
i.e. s = sf(1 – x) + xsg
s = sf + x(sg – sf )
∴ s = sf + xsfg (8.4)
Where,
sf = specific enthalpy of saturated liquid (kJ/kg K)
sg = specific enthalpy of saturated steam (kJ/kg K)
sfg = difference between sg and sf (that is, sfg = sg - sf )
REMEMBER!
These equations are used very often and
are, therefore, important to remember!
v = xvg
h = hf
+ xhfg
u = uf
+ x(ug –
uf
)
s = sf
+ xsfg

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8.4 The Use of Steam Tables
The steam tables are available for a wide variety of substances which normally exist
in the vapour phase (e.g. steam, ammonia, freon, etc.). The steam tables which will
be used in this unit are those arranged by Mayhew and Rogers, which are suitable for
student use. The steam tables of Mayhew and Rogers are mainly concerned with
steam, but some properties of ammonia and freon-12 are also given.
Below is a list of the properties normally tabulated, with the symbols used being
those recommended by British Standard Specifications.
Table 8.4 The property of steam tables
Symbols Units Description
p bar Absolute pressure of the fluid
ts
o
C Saturation temperature corresponding to the pressure p bar
vf m3
/kg Specific volume of saturated liquid
vg m3
/kg Specific volume of saturated steam
uf kJ/kg Specific internal energy of saturated liquid
ug kJ/kg Specific internal energy of saturated steam
hf kJ/kg Specific enthalpy of saturated liquid
hg kJ/kg Specific enthalpy of saturated steam
hfg kJ/kg Change of specific enthalpy during evaporation
sf kJ/kg K Specific entropy of saturated liquid
sg kJ/kg K Specific entropy of saturated steam
sfg kJ/kg K Change of specific entropy during evaporation
These steam tables are divided into two types:
Type 1: Saturated Water and Steam (Page 2 to 5 of steam tables)
Type 2: Superheated Steam (Page 6 to 8 of steam tables)
8.4.1 Saturated Water and Steam Tables

Complete the following table for Saturated Water and Steam:
t Ps vg hf hfg hg sf sfg sg
o
C bar m3
/kg kJ/kg kJ/kg K
0.01 206.1
0.02337 8.666
100 1.01325
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The table of the saturation condition is divided into two parts.
Part 1
Part 1 refers to the values of temperature from 0.01o
C to 100o
C, followed by
values that are suitable for the temperatures stated in the table. Table 8.4.1-1
is an example showing an extract from the temperature of 10o
C.
Table 8.4.1-1 Saturated water and steam at a temperature of 10 o
C
t ps vg hf hfg hg sf sfg sg
0
C bar m3
/kg kJ/kg kJ/kg K
10 0.01227 106.4 42.0 2477.2 2519.2 0.151 8.749 8.900
Example 8.1
Solution to Example 8.1
From page 2 of the steam tables, we can directly read:
t Ps vg hf hfg hg sf sfg sg
o
C bar m3
/kg kJ/kg kJ/kg K
1 0.006566 192.6 4.2 2498.3 2502.5 0.015 9.113 9.128
20 0.02337 57.84 83.9 2453.7 2537.6 0.296 8.370 8.666
100 1.01325 1.673 419.1 2256.7 2675.8 1.307 6.048 7.355

Complete the missing properties in the following table for Saturated Water
and Steam:
p ts vg uf ug hf hfg hg sf sfg sg
bar o
C m3
/kg kJ/kg kJ/kg kJ/kg K
0.045 31.0 2558
10 0.1944
311.0 5.615
bar o
C m3
0.045 31.0 31.14 130 2418 130 2428 2558 0.451 7.980 8.431
10 179.9 0.1944 762 2584 763 2015 2778 2.138 4.448 6.586
100 311.0 0.0180
2
1393 2545 1408 1317 2725 3.360 2.255 5.615
PROPERTIES OF STEAM
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Part 2
Part 2 (Page 3 to 5 of steam tables) is values of pressure from 0.006112 bar to
221.2 bar followed by values that are suitable for the pressures stated in the
table. Table 8.4.1-2 is an example showing an extract from the pressure of
1.0 bar.
Table 8.4.1-2 Saturated water and steam at a pressure of 1.0 bar
bar o
C m3
1.0 99.6 1.694 417 2506 417 2258 2675 1.303 6.056 7.359
Note the following subscripts:
f = property of the saturated liquid
g = property of the saturated steam
fg = change of the properties during evaporations
Example 8.2
From page 3 to page 5 of the steam tables, we can directly read:

For a steam at 20 bar with a dryness fraction of 0.9, calculate the
a) specific volume
b) specific enthalpy
c) specific internal energy
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Example 8.3
An extract from the steam tables
20 212.4 0.09957 907 2600 909 1890 2799 2.447 3.893 6.340
a) Specific volume (v),
v = xvg
= 0.9(0.09957)
= 0.0896 m3
/kg
b) Specific enthalpy (h),
h = hf + xhfg
= 909 + 0.9(1890)
= 2610 kJ/kg
c) Specific internal energy (u),
u = uf + x( ug -uf )
= 907 + 0.9(2600 - 907)
= 2430.7 kJ/kg
P
bar
v m3
/kg
ts
= 212.4 o
C
v
u
h
s
vg
ug
hg
sg
x = 0.9
20
uf
hf
sf

Find the dryness fraction, specific volume and specific enthalpy of
steam at 8 bar and specific internal energy 2450 kJ/kg.
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Example 8.4
An extract from the steam tables,
8 170.4 0.2403 720 2577 721 2048 2769 2.046 4.617 6.663
At 8 bar, ug = 2577 kJ/kg, since the actual specific internal energy is given as
2450 kJ/kg, the steam must be in the wet steam state ( u < ug).
From equation 8.5,
u = uf + x(ug -uf)
2450 = 720 + x(2577 - 720)
∴ x = 0.932
From equation 8.2,
v = xvg
= 0.932 (0.2403)
= 0.2240 m3
/kg
From equation 8.4,
h = hf + xhfg
= 721 + 0.932 (2048)
= 2629.7 kJ/kg
P
bar
v m3
/kg
ts
= 170.4 o
C
v vg
x = 0.932
8

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8.3 The internal energy of wet steam is 2000 kJ/kg. If the pressure is 42 bar, what is the
value of dryness fraction?
8.4 Determine the specific volume, specific enthalpy and specific internal energy of wet
steam at 32 bar if the dryness fraction is 0.92.
8.5 Find the dryness fraction, specific volume and specific internal energy of steam at
105 bar and specific enthalpy 2100 kJ/kg.
Activity 8B

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Feedback To Activity 8B
8.3 Dryness fraction (x),
u = uf + x(ug -uf)
2000 = 1097 + x(2601 - 1097)
x = 0.6
8.4 Specific volume (v),
v = xvg
= 0.92(0.06246)
= 0.05746 m3
/kg
Specific enthalpy (h),
h = hf + xhfg
= 1025 + 0.92(1778)
= 2661 kJ/kg
Specific internal energy (u),
= 1021 + 0.92(2603 - 1021)
= 2476 kJ/kg
8.5 Dryness fraction (x),
h = hf + x hfg
2100 = 1429 + x(1286)
x = 0.52
Specific volume (v),
v = xvg
= 0.52(0.01696)
= 0.00882 m3
/kg
Specific internal energy (u),
= 1414 + 0.52(2537 – 1414)
= 1998 kJ/kg

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8.4.2 Superheated Steam Tables
The second part of the table is the superheated steam tables. The values of
the specific properties of a superheated steam are normally listed in separate
tables for the selected values of pressure and temperature.
A steam is called superheated when its temperature is greater than the
saturation temperature corresponding to the pressure. When the pressure and
temperature are given for the superheated steam then the state is defined and
all the other properties can be found. For example, steam at 10 bar and 200
o
C is superheated since the saturation temperature at 10 bar is 179.9 o
C. The
steam at this state has a degree of superheat of 200 o
C – 179.9 o
C = 20.1 o
C.
The equation of degree of superheat is:
Degree of superheat = tsuperheat – tsaturation (8.5)
The tables of properties of superheated steam range in pressure from
0.006112 bar to the critical pressure of 221.2 bar. At each pressure, there is a
range of temperature up to high degrees of superheat, and the values of
specific volume, internal energy, enthalpy and entropy are tabulated.
For the pressure above 70 bar, the specific internal energy is not tabulated.
The specific internal energy is calculated using the equation:
u = h – pv (8.6)
For reference, the saturation temperature is inserted in brackets under each
pressure in the superheat tables and values of vg, ug, hg andsg are also given.
A specimen row of values is shown in Table 8.5.2. For example, from the
superheated table at 10 bar and 200 o
C, the specific volume is 0.2061 m3
/kg
INPUTINPUT

Complete the missing properties in the following table for Superheated
Steam:
p
(ts)
t 300 350 400 450
40
(250.3)
vg 0.0498 v 0.0800
ug 2602 u 2921
hg 2801 h 3094
sg 6.070 s 6.364
PROPERTIES OF STEAM
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and the specific enthalpy is 2829 kJ/kg.
Table 8.4.2 Superheated steam at a pressure of 10 bar
p
(ts)
t 200 250 300 350 400 450 500 600
10
(179.9)
vg 0.1944 v 0.2061 0.2328 0.2580 0.2825 0.3065 0.3303 0.3540 0.4010
ug 2584 u 2623 2711 2794 2875 2957 3040 3124 3297
hg 2778 h 2829 2944 3052 3158 3264 3370 3478 3698
sg 6.586 s 6.695 6.926 7.124 7.301 7.464 7.617 7.761 8.028
Example 8.5
From page 7 of the steam tables, we can directly read
p
(ts)
t 300 350 400 450
40
(250.3)
vg 0.0498 v 0.0588 0.0664 0.0733 0.0800
ug 2602 u 2728 2828 2921 3010
hg 2801 h 2963 3094 3214 3330
sg 6.070 s 6.364 6.584 6.769 6.935

Steam at 100 bar has a specific volume of 0.02812 m3
/kg. Find the
temperature, degree of superheat, specific enthalpy and specific
internal energy.
PROPERTIES OF STEAM
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Example 8.6
First, it is necessary to decide whether the steam is wet, dry saturated or
superheated.
At 100 bar, vg = 0.01802 m3
/kg. This is less than the actual specific volume
of 0.02812 m3
/kg. Hence, the steam is superheated. The state of the steam is
at point A in the diagram below.
An extract from the superheated table,
p
(ts)
t 425
100
(311.0)
vg 0.01802 v x 10-2
2.812
hg 2725 h 3172
sg 5.615 s 6.321
From the superheated table at 100 bar, the specific volume is 0.02812 m3
/kg
at a temperature of 425 o
C. Hence, this is the isothermal line, which passes
through point A as shown in the P-v diagram above.
P
bar
v m3
/kg
ts
=311.0o
C
100
425o
C
vg
= 0.01802
v = 0.02812
A

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Degree of superheat = 425 o
C – 311 o
C
= 114 o
C
So, at 100 bar and 425 o
C, we have
v = 2.812 x 10-2
m3
/kg
h = 3172 kJ/kg
∴ From equation 8.6,
u = h – Pv
= 3172 kJ/kg – (100 x 102
kN/m2
)(2.812 x 10-2
m3
/kg)
= 2890.8 kJ/kg
Note that equation 8.6 must be used to find the specific internal energy for
pressure above 70 bar as the specific internal energy is not tabulated.

PROPERTIES OF STEAM
J2006/8/28
8.6 Steam at 120 bar is at 500 o
C. Find the degree of superheat, specific volume,
specific enthalpy and specific internal energy.
8.7 Steam at 160 bar has a specific enthalpy of 3139 kJ/kg. Find the temperature, degree
of superheat, specific enthalpy and specific internal energy.
Activity 8C

PROPERTIES OF STEAM
J2006/8/29
Feedback to Activity 8C
8.6 From the superheated table at 120 bar, the saturation temperature is 324.6 o
C.
Therefore, the steam is superheated.
Degree of superheat = 500 o
C – 324.6 o
C
= 175.4 o
C
So, at 120 bar and 500 o
C, we have
v = 2.677 x 10-2
m3
/kg
h = 3348 kJ/kg
u = h – Pv
= 3348 kJ/kg – (120 x 102
kN/m2
)(2.677 x 10-2
m3
/kg)
= 3026.76 kJ/kg
8.7 At 160 bar, hg = 2582 kJ/kg. This is less than the actual specific enthalpy of
3139 kJ/kg. Hence, the steam is superheated.
From the superheated table at 160 bar, the specific enthalpy of 3139 kJ/kg is located
at a temperature of 450 o
C.
The degree of superheat = 450 o
C – 347.3 o
C
= 102.7 o
C
At 160 bar and 450 o
C, we have v = 1.702 x 10-2
m3
/kg
u = h – Pv
= 3139 kJ/kg – (160 x 102
kN/m2
)(1.702 x 10-2
m3
/kg)
= 2866.68 kJ/kg

PROPERTIES OF STEAM
J2006/8/30
8.5 Interpolation
The first interpolation problem that an engineer usually meets is that of “reading
between the lines” of a published table, like the Steam Tables. For properties which
are not tabulated exactly in the tables, it is necessary to interpolate between the
values tabulated as shown in Fig. 8.5-1 below. In this process it is customary to use a
straight line that passes through two adjacent table points, denoted by α and β. If we
use the straight line then it is called “interpolation”.
Figure 8.5-1 Interpolation
The values in the tables are given in regular increments of temperature and pressure.
Often we wish to know the value of thermodynamic properties at intermediate
values. It is common to use linear interpolation as shown in Fig. 8.5-2.
From Fig. 8.5.2, the value of x
can be determined by:
INPUTINPUT
f(x)
x
α β
Interpolation
y
x
y2
y
y1
x1
x x2
(x2
, y2
)
(x , y)
(x1
, y1
)

Determine the saturation temperature at 77 bar.
PROPERTIES OF STEAM
J2006/8/31
There are two methods of interpolation:
i. single interpolation
ii. double interpolation
8.5.1 Single interpolation
Single interpolation is used to find the values in the table when one of the
values is not tabulated. For example, to find the saturation temperature,
specific volume, internal energy and enthalpy of dry saturated steam at 77
bar, it is necessary to interpolate between the values given in the table.
Example 8.6
The values of saturation temperature at a pressure of 77 bars are not tabulated
in the Steam Tables. So, we need to interpolate between the two nearest
values that are tabulated in the Steam Tables.
ts = 292.3 o
C
( )( )
( ) 1
12
121
12
12
1
1
x
yy
xxyy
x
yy
xx
yy
xx
+
−
−−
=
−
−
=
−
−
7580
5.290295
7577
5.290
−
−
=
−
−st
5
5.290295
2
5.290 −
=
−st
( ) 5.290
5
5.42
+=st
P
ts
80
77
75
290.5 ts 295
Figure 8.5-2 Linear interpolation

Determine the specific enthalpy of dry saturated steam at 103 bar.
Determine the specific volume of steam at 8 bar and 220o
C.
PROPERTIES OF STEAM
J2006/8/32
Example 8.7
kJ/kg
Example 8.8
From the Steam Tables at 8 bar, the saturated temperature (ts) is 170.4 o
C.
∴The steam is at superheated condition as the temperature of the steam is
220o
C > ts.
An extract from the Steam Tables,
p / (bar)
(ts / o
C)
t 200 220 250
(o
C)
8
(170.4)
v 0.2610 v 0.2933
hg −
−
=
−
−
2725
103 100
2715 2725
105 100
( )
hg =
−
+
3 10
5
2725
2719=gh
P
hg
105
103
100
2725 hg
2715
P
v
250
220
200
0.2610 v 0.2933

PROPERTIES OF STEAM
J2006/8/33
m3
/kg
v −
−
=
−
−
0 2610
220 200
0 2933 0 2610
250 200
. . .
v = 027392.

Determine the specific enthalpy of superheated steam at 25 bar and
320o
C.
PROPERTIES OF STEAM
J2006/8/34
8.5.2 Double Interpolation
In some cases a double interpolation is necessary, and it’s usually used in the
Superheated Steam Table. Double interpolation must be used when two of
the properties (eg. temperature and pressure) are not tabulated in the Steam
Tables. For example, to find the enthalpy of superheated steam at 25 bar and
320o
C, an interpolation between 20 bar and 30 bar is necessary (as shown in
example 8.9). An interpolation between 300o
C and 350o
C is also necessary.
Example 8.8
An extract from the Superheated Steam Tables:
t(o
C)
p(bar)
300 320 350
20 3025 h1 3138
25 h
30 2995 h2 3117
Firstly, find the specific enthalpy (h1) at 20 bar and 320 o
C;
At 20 bar,
kJ/kg
300350
30253138
300320
30251
−
−
=
−
−h
2.30701 =h
T
h
350
320
300
3025 h1 3138

0.9 m3
of dry saturated steam at 225 kN/m2
is contained in a rigid
cylinder. If it is cooled at constant volume process until the pressure
drops to180 kN/m2
, determine the following:
a) mass of steam in the cylinder
b) dryness fraction at the final state
Sketch the process in the form of a P-v diagram.
PROPERTIES OF STEAM
J2006/8/35
Secondly, find the specific enthalpy (h2) at 30 bar and 320 o
C;
At 30 bar,
kJ/kg
Now interpolate between h1 at 20 bar, 320o
C, and h2 at 30 bar, 320o
C in order
to find h at 25 bar and 320o
C.
At 320o
C,
kJ/kg.
Example 8.9
300350
29953117
300320
29952
−
−
=
−
−h
8.30432 =h
20302025
121
−
−
=
−
− hhhh
h −
−
=
−
−
3070 2
25 20
30438 3070 2
30 20
. . .
h = 3057
T
h
350
320
300
2995 h2 3117
P
h
30
25
20
h1
h h2

PROPERTIES OF STEAM
J2006/8/36

PROPERTIES OF STEAM
J2006/8/37
Data: V1 = 0.9 m3
, P1 = 225 kN/m2
= 2.25 bar, P2 = 180 kN/m2
= 1.80 bar
a) Firstly, find the specific volume of dry saturated steam at 2.25 bar.
Note that the pressure 2.25 bar is not tabulated in the steam tables and
it is necessary to use the interpolation method.
From the Steam Tables,
vg at 2.2 bar = 0.8100 m3
/kg
vg at 2.3 bar = 0.7770 m3
/kg
∴vg1 at 2.25 bar,
vg1 = 0.7935 m3
/kg
∴ Mass of steam
in cylinder, (m3
x kg/m3
)
= 1.134 kg
b) At constant volume process,
Initial specific volume = final specific volume
v1 = v2
x1vg1 at 2.25 bar = x2vg2 at 1.8 bar
1(0.7935) = x2 (0.9774)
= 0.81
20.230.2
8100.07770.0
20.225.2
8100.01
−
−
=
−
−gv
1
1
gv
V
m =
=
0 9
0 7935
.
.
9774.0
)7935.0(1
2 =x
P
bar
1.80
2.25
v m3
/kg
1
2
0.7935 0.9774
v1
= v2

PROPERTIES OF STEAM
J2006/8/38
8.8 Determine the specific enthalpy
of steam at 15 bar and 275o
C.
8.9 Determine the degree of superheat and entropy of steam at 10 bar and 380o
C.
8.10 A superheated steam at 12.5 MN/m2
is at 650o
C. Determine its specific volume.
8.11 A superheated steam at 24 bar and 500o
C expands at constant volume until the
pressure becomes 6 bar and the dryness fraction is 0.9. Calculate the changes in the
internal energy of steam. Sketch the process in the form of a P-v diagram.
Activity 8D

PROPERTIES OF STEAM
J2006/8/39
Feedback to Activity 8D
8.8
kJ/kg
8.9 Degree of superheat = 380o
C – 179.9o
C
= 200.1o
C
kJ/kg K
8.10 An extract from the superheated steam table:
t(o
C)
p(bar)
600 650 700
120 3.159 x 10-2
v1 3.605 x 10-2
125 v
250300
29253039
250275
2925
−
−
=
−
−h
2982=h
350400
301.7464.7
350380
301.7
−
−
=
−
−s
3988.7=s
T
h
300
275
250
2925 h 3039
T
s
400
380
350
7.301 s 7.464

PROPERTIES OF STEAM
J2006/8/40
130 2.901 x 10-2
v2 3.318 x 10-2
Firstly, find the specific volume (v1) at 120 bar and 650 o
C;
At 120 bar,
m3
/kg
Secondly, find the specific volume (v2) at 130 bar and 650 o
C;
At 130 bar,
v2 = 3.1095 x 10-2
m3
/kg
Now interpolate between v1 at 120 bar, 650o
C, and v2 at 130 bar, 650o
C in
order to find v at 125 bar and 650o
C.
At 650o
C,
600700
10x915.310x605.3
600650
10x915.3 222
1
−
−
=
−
− −−−
v
2
1 10x3.382 −
=v
600700
10x901.210x.3183
600650
10x901.2 222
2
−
−
=
−
− −−−
v
120130120125
121
−
−
=
−
− vvvv
T
v
700
650
600
3.159 x 10-2
v1
3.605 x 10-2
T
v
700
650
600
2.901 x 10-2
v2
3.318 x 10-2
P
v
130
125
120
v v2
v1

PROPERTIES OF STEAM
J2006/8/41
v =
3.246 x 10-2
m3
/kg
120130
10x382.310x.10953
120125
10x382.3 222
−
−
=
−
− −−−
v

PROPERTIES OF STEAM
J2006/8/42
8.11 Data: P1 = 24 bar
T1 = 500o
C
P2 = 6 bar
x2 = 0.9
Firstly, find the initial internal energy at 24 bar, 500o
C. Note that the
pressure 24 bar is not tabulated in the Superheated Steam Tables and it is
necessary to use the interpolation method to find the changes in the internal
energy of steam.
At 500o
C,
kJ/kg
Secondly, find the final internal energy at 6 bar where x = 0.9,
u2 = uf2 + x2( ug2 -uf2 )
= 669 + 0.9(2568 - 669)
= 2378.1 kJ/kg
∴ The changes in the internal energy of steam is,
(u2 – u1) = 2378.1 – 3112.8
= - 734.7 kJ/kg
2030
31163108
2024
31161
−
−
=
−
−u
8.31121 =u
P
u
30
24
20
3116 u1
3108
P
bar
6
24
v m3
/kg
1
2
221.8o
C
v1
= v2
500o
C
158.8o
C 500o
C
221.8o
C
158.8o
C
v1
= v2

PROPERTIES OF STEAM
J2006/8/43
You are approaching
success. Try all the questions
in this self-assessment section and check your answers with those given in the Feedback to Self-
Assessment on the next page. If you face any problem, discuss it with your lecturer. Good luck.
1. With reference to the Steam Tables,
i. determine the specific volume, specific enthalpy and specific internal
energy of wet steam at 15 bar with a dryness fraction of 0.9.
ii. determine the degree of superheat, specific volume and specific
internal energy of steam at 80 bar and enthalpy 2990 kJ/kg.
iii. complete the missing properties and a phase description in the
following table for water;
P
bar
t
o
C
x v
m3
/kg
u
kJ/kg
h
kJ/kg
s
kJ/kg K
Phase
description
2.0 120.2 6.4
12.0 1 2784
175 354.6 0.9
200 425
2. With reference to the Steam Tables,
i. find the dryness fraction and specific entropy of steam at 2.9 bar and
specific enthalpy 2020 kJ/kg.
ii. determine the degree of superheat and internal energy of superheated
steam at 33 bar and 313o
C.
iii. determine the enthalpy change for a process involving a dry saturated
steam at 3.0 MN/m2
which is superheated to 600 o
C and carried out at
constant pressure.
SELF-ASSESSMENT

PROPERTIES OF STEAM
J2006/8/44
Have you tried the questions????? If “YES”, check your answers now.
1. i. v = 0.11853 m3
/kg
h = 2600 kJ/kg
u = 2419.8 kJ/kg
ii. degree of superheat = 55 o
C
v = 2.994 x 10-2
m3
/kg
u = 2750.48 kJ/kg
iii.
P
bar
t
o
C
x v
m3
/kg
u
kJ/kg
h
kJ/kg
s
kJ/kg K
Phase
description
2.0 120.2 0.87 0.7705 2267 2421 6.4 Wet steam
12.0 188 1 0.1632 2588 2784 6.523 Dry sat.
steam
175 354.6 0.9 0.007146 2319.8 2448.1 5.0135 Wet steam
200 425 - 0.001147 2725.6 2955 5.753 Superheated
steam
2. i. x = 0.68
s = 5.2939 kJ/kg K
ii. Degree of superheat = 73.85o
C
u = 2769 kJ/kg
iii. h2 – h1 = 879 kJ/kg
Feedback to Self-Assessment

J2006 termodinamik 1 unit8

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