JJ207 Thermodynamics I Chapter 2

JJ 207 THERMODYNAMICS I

Chapter 2

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.

PHASES

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;

1


(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 Figure 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)

The arrangement of atoms in different phases

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.
A container is filled with water, and a moveable, frictionless piston is placed on the
container at State 1, as shown in Figure below. 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.

STATE 1 STATE 2 STATE 3 STATE 4
W
Heating water
W
W
and steam at
W constant pressure

Steam Superheated
Liqui Steam
d
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

2


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 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.

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 pressure:

T, oC

300 4

Superheated
steam
Saturated
2 mixture 3
100
Compressed
liquid
20 1

v, m3/kg

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/ = 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 liquid
in one of these conditions as being compressed or subcooled liquid.

State 2 is selected to correspond to the boiling point (100 oC). 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

3


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.

The same experiment can be conducted at several different pressures. We see that as
pressure increases, the temperature at which boiling occurs also increases.
T, oC
P = 221.2 bar
Critical point P = 150 bar

P = 80 bar
374.15
P = 10 bar
T-v diagram of
constant pressure
P = 5 bar phase change
P = 1.01325 bar processes of a pure
substance at various
pressures for water.

Saturated Saturated
liquid steam

It can be seen that as pressure increases, the specific volume increase in the liquid to
v,
0.0031
steam transition will7decrease. m3/kg

4


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 oC. This state represents an
important transition in fluids and is termed the critical point.
If we connect the locus of points corresponding to the saturation condition, we will
obtain a diagram which allows easy identification of the distinct regions:
Saturated liquid line
T
Dry saturated steam line
Critical
point P2 = const. T-v diagram of a
P2 > P1
pure substance
COMPRESS P1 = const.
LIQUID
REGION
SUPERHEATED
STEAM
REGION

WET STEAM
REGION

v

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.
Saturated liquid line
P
Dry saturated steam line
Critical
point

SUPERHEATED
STEAM
REGION

COMPRESS
LIQUID
REGION T2 = const.
WET STEAM
T2 > T1
REGION T1 = const.

v
P-v diagram of a pure substance

5


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.

Symbols Units Description

p bar Absolute pressure of the fluid
o
ts 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
The property of steam tables

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)

6


Saturated Water and Steam Tables

The table of the saturation condition is divided into two parts.

Part 1
Part 1 refers to the values of temperature from 0.01oC to 100oC, 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 10oC.

t ps vg hf hfg hg sf sfg sg
0
C bar kJ/kg kJ/kg K
m3/kg
10 0.01227 42.0 2477.2 0.151 8.749
106.4 2519.2 8.900
Saturated water and steam at a temperature of 10 oC

Example 1

Complete the following table for Saturated Water and Steam:

t Ps vg hf hfg hg sf sfg sg
o 3
C bar m /kg kJ/kg kJ/kg K
0.01 206.1
0.02337 8.666
100 1.01325

Solution
From page 2 of the steam tables, we can directly read:

t Ps vg hf hfg hg sf sfg sg
o 3
C bar m /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

7


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.

p ts vg uf ug hf hfg hg sf sfg sg
bar oC kJ/kg kJ/kg kJ/kg K
m3/kg
1.0 99.6 417 417 2258 2675 1.303 6.056
1.694 2506 7.359
Saturated water and steam at a pressure of 1.0 bar

f = property of the saturated liquid
g = property of the saturated steam
fg = change of the properties during evaporations

Example 2

Complete the missing properties in the following table for Saturated Water
and Steam:

o
bar C m3/kg kJ/kg kJ/kg kJ/kg K
0.045 31.0 2558
10 0.1944
311.0 5.615

Solution

From page 3 to page 5 of the steam tables, we can directly read:

o 3
bar C m /kg kJ/kg kJ/kg kJ/kg K
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 0.01802 1393 2545 1408 1317 2725 3.360 2.255 5.615

8


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.

x kg of steam
total mass = 1 kg

(1 - x ) kg of liquid

Liquid-steam mixture

The dryness fraction is defined as follows;

mass of dry saturated steam
dryness fraction 
total mass
msteam
x
mtotal

where mtotal = mliquid + msteam

Sat. steam
Sat. steam
Sat. liquid At point A, x = 0
Sat. liquid At point B, x = 1
P x = 0.2 x = 0.8 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.

A B

ts

vg v
vf

P-v diagram showing the location point of the dryness fraction

Specific volume

9


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,
volume of a liquid  volume of dry steam
v
total mass of wet steam

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)
vg = specific volume of saturated steam (m3/kg)
x = dryness fraction

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)

The enthalpy of wet steam is given by the sum of the enthalpy of the liquid plus the
enthalpy of the dry steam,

h = hf + xhfg

Where,
hf = specific enthalpy of saturated liquid (kJ/kg)

10


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,
u = uf + x(ug – uf )

Specific Entropy
The entropy of wet steam is given by the sum of the entropy of the liquid plus the
entropy of the dry steam,
s = sf + xsfg

Summary:

v = xvg
h = hf + xhfg
u = uf + x(ug – uf )
s = sf + xsfg

Example 3

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

Solution
An extract from the steam tables

20 212.4 0.09957 907 2600 909 1890 2799 2.447 3.893 6.340

a) v = xvg c) u = uf + x( ug -uf )
= 0.9(0.09957) = 907 + 0.9(2600 - 907)
= 0.0896 m3/kg = 2430.7 kJ/kg
b) h = hf + xhfg
= 909 + 0.9(1890)
= 2610 kJ/kg

11


P
bar x = 0.9

20

ts = 212.4 oC

v m3/kg
v vg
uf u ug
hf h hg
sf s sg

Example 4

Find the dryness fraction, specific volume and specific enthalpy of steam at 8 bar and
specific internal energy 2450 kJ/kg.

Solution
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).

u = uf + x(ug -uf) h = hf + xhfg
2450 = 720 + x(2577 - 720) = 721 + 0.932 (2048)
x = 0.932 = 2629.7 kJ/kg

v = xvg
= 0.932 (0.2403) P
= 0.2240 m3/kg bar x = 0.932

8

ts = 170.4 oC

v m3/kg
v vg
12


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 oC is superheated since the
saturation temperature at 10 bar is 179.9 oC. The steam at this state has a degree of
superheat of 200 oC – 179.9 oC = 20.1 oC. The equation of degree of superheat is:

Degree of superheat = tsuperheat – tsaturation

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

For reference, the saturation temperature is inserted in brackets under each pressure in
the superheat tables and values of vg, ug, hg and sg 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 oC, the specific volume is 0.2061 m3/kg and the
specific enthalpy is 2829 kJ/kg.

p
t 200 250 300 350 400 450 500 600
(ts)
vg v 0.206 0.232 0.258 0.282 0.306 0.330 0.354 0.401
10 0.1944 1 8 0 5 5 3 0 0
(179.9)
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
Superheated steam at a pressure of 10 bar

13


Example 5

Complete the missing properties in the following table for Superheated Steam:
p
t 300 350 400 450
(ts)
vg 0.0498 v 0.0800
40 ug 2602 u 2921
(250.3) hg 2801 h 3094
sg 6.070 s 6.364

Solution
From the steam tables, we can directly read
p
t 300 350 400 450
(ts)
vg 0.0498 v 0.0588 0.0664 0.0733 0.0800
40 ug 2602 u 2728 2828 2921 3010
(250.3) hg 2801 h 2963 3094 3214 3330
sg 6.070 s 6.364 6.584 6.769 6.935

Example 6
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.

Solution
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.

P
bar

A
100
425 oC

ts = 311.0 oC

v m3/kg
vg= 0.01802
v = 0.02812

14


An extract from the superheated table,
p
t 425
(ts)
vg 0.01802 v x 10-2 2.812
100 hg 2725 h 3172
(311.0) 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 oC. Hence, this is the isothermal line, which passes through point
A as shown in the P-v diagram above.

Degree of superheat = 425 oC – 311 oC
= 114 oC

So, at 100 bar and 425 oC, 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

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  and . If we use the
straight line then it is called “interpolation”.

f(x)

Interpolation

x
 
Interpolation

15


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.
y

y2 (x2 , y2)

y (x , y)

y1 (x1 , y1)

x
x1 x x2

Linear interpolation

From Figure above the value of x can be determined by:
x  x1 x 2  x1

y  y1 y 2  y1

x
 y  y1 x2  x1   x
 y 2  y1  1

There are two methods of interpolation:
i. Single interpolation
ii. Double interpolation

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 7
Determine the saturation temperature at 77 bar.

Solution
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.

16


P
t s  290.5 295  290.5
80 
77  75 80  75
77
t s  290.5 295  290.5
75 
2 5
ts
290.5 ts 295 24.5
ts   290.5
5

ts = 292.3 oC

Example 8
Determine the specific enthalpy of dry saturated steam at 103 bar.

Solution
P hg  2725 2715  2725

103  100 105  100
105

103 3 10
hg   2725
100
5

hg g2719 kJ/kg
h
2725 hg 2715

Example 9
Determine the specific volume of steam at 8 bar and 220oC.

Solution
From the Steam Tables at 8 bar, the saturated temperature (ts) is 170.4 oC.
The steam is at superheated condition as the temperature of the steam is 220oC > ts.

An extract from the Steam Tables,
p / (bar) t 200 220 250
(ts / oC) (oC)
8 v 0.2610 v 0.2933
(170.4)
P

250
v  0.2610 0.2933  0.2610
220 
220  200 250  200
200
v  0.27392 m3/kg
v
0.2610 v 0.2933 17


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 320oC, an
interpolation between 20 bar and 30 bar is necessary (as shown in example 8.9). An
interpolation between 300oC and 350oC is also necessary.

Example 10
Determine the specific enthalpy of superheated steam at 25 bar and 320oC.

Solution
An extract from the Superheated Steam Tables:

t(oC) 300 320 350
p(bar)
20 3025 h1 3138
25 h
30 2995 h2 3117

Firstly, find the specific enthalpy (h1) at 20 bar and 320 oC;

At 20 bar,
T
h1  3025 3138  3025
350 
320  300 350  300
320
300 h1  3070.2 kJ/kg

h
3025 h1 3138

Secondly, find the specific enthalpy (h2) at 30 bar and 320 oC;
T
h2  2995 3117  2995
350 
320  300 350  300
320
300 h2  3043.8 kJ/kg

h
2995 h2 3117
18


Now interpolate between h1 at 20 bar, 320oC, and h2 at 30 bar, 320oC in order to find
h at 25 bar and 320oC.

P h  h1 h  h1
 2
25  20 30  20
30
h  3070.2 30438  3070.2
.
 25
25  20 30  20
20
h  3057 kJ/kg.
h
h1 h h2

Example 11
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.

Solution
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,
v g1  0.8100 0.7770  0.8100

2.25  2.20 2.30  2.20

vg1  0.7935 m3/kg

V1
 Mass of steam in cylinder, m  (m3 x kg/m3)
vg1

= 1.134 kg

19


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)
1(0.7935)
x2 
0.9774
P = 0.81
bar

1 v1 = v2
2.25

1.80
2

v m3/kg
0.7935 0.9774

20


TUTORIAL
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 i._____________
no longer rigidly structured so that they are free to move
within a fixed volume.
The molecules are closely bound, they are relatively dense
and arranged in a rigid three-dimensional patterns so that they ii.____________
do not easily deform.
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 iii.____________
pattern. There is neither a fixed volume nor a fixed shape for
steam.

2. Write the suitable names of the phases for the H2O in the P-v diagram below.
( ii )
P
( iv )

( vi )

(v)

(i)
T2 = const.
T2 > T1
( iii)
T1 = const.

v
3. Answer question below:
a. The internal energy of wet steam is 2000 kJ/kg. If the pressure is 42 bar, what is
the value of dryness fraction?

b. Determine the specific volume, specific enthalpy and specific internal energy of
wet steam at 32 bar if the dryness fraction is 0.92.

4. Find the dryness fraction, specific volume and specific internal energy of
steam at 105 bar and specific enthalpy 2100 kJ/kg.

21


5. Steam at 120 bar is at 500 oC. Find the degree of superheat, specific volume,
specific enthalpy and specific internal energy.

6. Steam at 160 bar has a specific enthalpy of 3139 kJ/kg. Find the temperature,
degree of superheat, specific enthalpy and specific internal energy.

7 Determine the specific enthalpy of steam at 15 bar and 275oC.

8. Determine the degree of superheat and entropy of steam at 10 bar and 380oC.

9. A superheated steam at 12.5 MN/m2 is at 650oC. Determine its specific
volume.

10. A superheated steam at 24 bar and 500oC 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.

22


DEFINITION OF PERFECT GASES

Did you know, one important type of fluid that has many applications in
thermodynamics is the type in which the working temperature of the fluid remains
well above the critical temperature of the fluid? In this case, the fluid cannot be
liquefied by an isothermal compression, i.e. if it is required to condense the fluid, then
cooling of the fluid must first be carried out. In the simple treatment of such fluids,
their behavior is likened to that a perfect gas. Although, strictly speaking, a perfect
gas is an ideal which can never be realized in practice. The behavior of many
„permanent‟ gases, e.g. hydrogen, oxygen, air etc is very similar to the behavior of a
perfect gas to a first approximation.

A perfect gas is a collection of particles that:
 are in constant, random motion,
 have no intermolecular attractions (which leads to elastic collisions in which
no energy is exchanged or lost),
 are considered to be volume-less points.

You are more familiar with the term „ideal‟ gas. There is actually a distinction
between these two terms but for our purposes, you may consider them
interchangeable. The principle properties used to define the state of a gaseous system
are pressure (P), volume (V) and temperature (T). SI units (Systems International) for
these properties are Pascal (Pa) for pressure, m3 for volume (although liters and cm3
are often substituted), and the absolute scale of temperature or Kelvin (K).

Two of the laws describing the behavior of a perfect gas are Boyle‟s Law and
Charles‟ Law.

BOYLE’S LAW
The Boyle‟s Law may be stated as follows:
Provided the temperature T of a perfect gas remains constant, then volume, V of a
given mass of gas is inversely proportional to the pressure P of the gas, i.e. P  1/V
(as shown in Fig. 3.1-1), or P x V = constant if temperature remains constant.

P

P  1/V

1/V
Graph P  1/V

23


If a gas changes from state 1 to state 2 during an isothermal process, then

P1 V1 = P2 V2 = constant

If the process is represented on a graph having axes of pressure P and volume V, the
results will be as shown in Fig. below. The curve is known as a rectangular
hyperbola, having the mathematical equation xy = constant.

P

P1 1
PV = constant
P2 2
3
P3

V1 V2 V3 V

P-V graph for constant temperature

Example 12
A quantity of a certain perfect gas is heated at a constant temperature from an initial
state of 0.22 m3 and 325 kN/m2 to a final state of 170 kN/m2. Calculate the final
pressure of the gas.

Solution

From equation P1V1 = P2V2

  325 kN/m 2
P1
V2  V1 x  0.22 m 3 
   0.421 m 3
 2
P2  170 kN/m 

CHARLES’ LAW

The Charles‟s Law may be stated as follows:
Provided the pressure P of a given mass of gas remains constant, then the volume V of
the gas will be directly proportional to the absolute temperature T of the gas, i.e.
V  T, or V = constant x T. Therefore V/T = constant, for constant pressure P.

If gas changes from state 1 to state 2 during a constant pressure process, then

24


V1 V2
  constant
T1 T2

If the process is represented on a P – V diagram as before, the result will be as shown
in Fig. 3.2.

P

1 2

0 V
V1 V2
P-V graph for constant pressure process

Example 13
A quantity of gas at 0.54 m3 and 345 oC undergoes a constant pressure process that
causes the volume of the gas to decreases to 0.32 m3. Calculate the temperature of the
gas at the end of the process.

Solution

From the question
V1 = 0.54 m3
T1 = 345 + 273 K = 618 K
V2 = 0.32 m3

V1 V2

T1 T2

V2
 T2  T1 x
V1

 0.32 m 3 
 618 K  
 0.54 m 3 

 
 366 K

25


UNIVERSAL GASES LAW

Charles‟ Law gives us the change in volume of a gas with temperature when the
pressure remains constant. Boyle‟s Law gives us the change in volume of a gas with
pressure if the temperature remains constant.

The relation which gives the volume of a gas when both temperature and the pressure
are changed is stated as equation 3.3 below.

PV
 constant  R
T

i.e. P1V1 P2V2

T1 T2
No gases in practice obey this law rigidly, but many gases tend towards it. An
PV
imaginary ideal that obeys the law is called a perfect gas, and the equation R
T
is called the characteristic equation of state of a perfect gas.

The constant, R, is called the gas constant. The unit of R is Nm/kg K or J/kg K. Each
perfect gas has a different gas constant.

The characteristic equation is usually written
PV = RT

or for m kg, occupying V m3,
PV = mRT

Another form of the characteristic equation can be derived using the kilogram-mole as
a unit. The kilogram-mole is defined as a quantity of a gas equivalent to m kg of the
gas, where M is the molecular weight of the gas (e.g. since the molecular weight of
oxygen is 32, then 1 kg mole of oxygen is equivalent to 32 kg of oxygen).

From the definition of the kilogram-mole, for m kg of a gas we have,
m = nM

(where n is the number of moles).
Note: Since the standard of mass is the kg, kilogram-mole will be written simply as
mole.

Substituting for m from equations above
PV = nMRT or PV
MR 
nT

26


Now Avogadro’s hypothesis states that the volume of 1 mole of any gas is the same as
the volume of 1 mole of any other gas, when the gases are at the same temperature
and pressure. Therefore V/n is the same for all gases at the same value of P and T.
That is the quantity PV/nT is constant for all gases. This constant is called the
universal gas constant, and is given the symbol Ro.

i.e. PV
MR  Ro  or PV  nRoT
nT

or since MR = Ro then,
R
R o
M

Experiment has shown that the volume of 1 mole of any perfect gas at 1 bar and 1 oC
is approximately 22.71 m3. Therefore from equation 3.8

PV 1 x 10 5 x 22.71
R0    8314.4 J/mole K
nT 1 x 273.15

From equation 3.10 the gas constant for any gas can be found when the molecular
weight is known, e.g. for oxygen of molecular weight 32, the gas constant is

Ro 8314.4
R   259.8 J/kg K
M 32

Example 14
0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m2 and a
temperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m2
and the temperature is 83oC. If the gas is assumed to be a perfect gas, determine:

a) the mass of gas (kg)
b) the final volume of gas (m3)

Given:
R = 0.29 kJ/kg K

Solution

From the question
V1 = 0.046 m3
P1 = 300 kN/m2 P2 = 1.27 MN/m2 = 1.27 x 103 kN/m2
T1 = 45 + 273 K = 318 K T2 = 83 + 273 K = 356 K
R = 0.29 kJ/kg K

27


From equation PV = mRT

P1V1 300 x 0.046
m   0.1496 kg
RT1 0.29 x 318

the constant volume process i.e. V1 = V2

P1 P2

T1 T2

 P   3

T2  T1  2   318
1.27 x 10
     1346 K

 P1   300 

SPECIFIC HEAT CAPACITY AT CONSTANT VOLUME (CV)

The specific heat capacities of any substance is defined as the amount of heat energy
required to raise the unit mass through one degree temperature raise. In
thermodynamics, two specified conditions are used, those of constant volume and
constant pressure. The two specific heat capacities do not have the same value and it
is essential to distinguish them.

If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the
temperature of the gas by 1 degree whilst the volume of the gas remains constant, then
the amount of heat energy supplied is known as the specific heat capacity at constant
volume, and is denoted by Cv. The unit of Cv is J/kg K or kJ/kg K.

For a reversible non-flow process at constant volume, we have
dQ = mCvdT

For a perfect gas the values of Cv are constant for any one gas at all pressures and
temperatures. Equation above can then be expanded as follows :
Heat flow in a constant volume process, Q12 = mCv(T2 – T1)

Also, from the non-flow energy equation
Q – W = (U2 – U1)
mcv(T2 – T1) – 0 = (U2 – U1)
 (U2 – U1) = mCv(T2 – T1)

i.e. dU = Q

Note:
In a reversible constant volume process, no work energy transfer can take place since
the piston will be unable to move i.e. W = 0.

28


The reversible constant volume process is shown on a P-V diagram in Figure below:

P

P2 2

P1 1

V
V1 = V2
P-V diagram for reversible constant volume process

Example 15
3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC until
the temperature rose to 147 oC. If the gas is assumed to be a perfect gas, determine:

a) the heat flow during the process
b) the beginning pressure of gas
c) the final pressure of gas

Given
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K

Solution

From the question
m = 3.4 kg
V1 = V2 = 0.92 m3
T1 = 17 + 273 K = 290 K
T2 = 147 + 273 K = 420 K
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K

a) From equation, Q12 = mCv(T2 – T1)
= 3.4 x 0.72(420 – 290)
= 318.24 kJ

29


b) From equation, PV = mRT

Hence for state 1,
P1V1 = mRT1
mRT1 3.4 kg x 0.287 kJ/kgK x 290 K
P1    307.6 kN/m 2
V1 0.92 m 3

c) For state 2,
P2V2 = mRT2
mRT2 3.4 kg x 0.287 kJ/kgK x 420 K
P2    445.5 kN/m 2
V2 0.92 m 3

SPECIFIC HEAT CAPACITY AT CONSTANT PRESSURE (CP)

If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the
temperature of the gas by 1 degree whilst the pressure of the gas remains constant,
then the amount of heat energy supplied is known as the specific heat capacity at
constant pressure, and is denoted by Cp. The unit of Cp is J/kg K or kJ/kg K.

For a reversible non-flow process at constant pressure, we have
dQ = mCpdT

For a perfect gas the values of Cp are constant for any one gas at all pressures and
temperatures. Equation above can then be expanded as follows:

Heat flow in a reversible constant pressure process Q = mCp(T2 – T1)

RELATIONSHIP BETWEEN THE SPECIFIC HEATS

Let a perfect gas be heated at constant pressure from T1 to T2. With reference to the
non-flow equation Q = U2 – U1 + W, and the equation for a perfect gas
U2 – U1 = mCv(T2 – T1), hence,
Q = mCv(T2 – T1) + W

In a constant pressure process, the work done by the fluid is given by the pressure
times the change in volume, i.e. W = P(V2 – V1). Then using equation PV = mRT, we
have
W = mR(T2 – T1)

Therefore substituting,
Q = mCv(T2 – T1) + mR(T2 – T1) = m(Cv + R)(T2 – T1)

30


But for a constant pressure process from equation before,
Q = mCp(T2 – T1)

Hence, by equating the two expressions for the heat flow Q, we have
mCp(T2 – T1) = m(Cv + R)(T2 – T1)
Cp = Cv + R
Alternatively, it is usually written as

R = Cp - C v

SPECIFIC HEAT RATIO ()

The ratio of the specific heat at constant pressure to the specific heat at constant
volume is given the symbol  (gamma),

Cp
i.e. =
Cv

Note that since Cp - Cv= R, from equation above, it is clear that Cp must be greater
than Cv for any perfect gas. It follows therefore that the ratio Cp/Cv =  , is always
greater than unity. In general,  is about 1.4 for diatomic gases such as carbon
monoxide (CO), hydrogen (H2), nitrogen (N2), and oxygen (O2). For monatomic gases
such as argon (A), and helium (He),  is about 1.6, and for triatomic gases such as
carbon dioxide (CO2), and sulphur dioxide (SO2),  is about 1.3. For some hydro-
carbons the value of  is quite low (e.g. for ethane (C2H6),  = 1.22, and for iso-
butane (C4H10),  = 1.11.

Some useful relationships between Cp , Cv , R, and  can be derived.
From equation above

Cp - Cv= R

Dividing through by Cv
Cp R
1 
Cv Cv
Cp
Therefore  = , then,
Cv

31


R
 1 
Cv

R
Cv 
(  1)
Also, Cp = Cv hence substituting in equation above,

R
Cp = Cv =
(  1)
R
Cp =
(  1)

Example 16

A certain perfect gas has specific heat as follows

Cp = 0.846 kJ/kg K and Cv = 0.657 kJ/kg K

Find the gas constant and the molecular weight of the gas.

Solution

From equation R = Cp - Cv

i.e. R = 0.846 – 0.657 = 0.189 kJ/kg K
or R = 189 Nm/kg K

R0
From equation M =
R

8314
i.e. M=  44
189

32


NON-FLOW PROCESSES

A process occurs when a system‟s state (as measured by its properties) changes for
any reason. Processes may be reversible or actual (irreversible). In this context the
word „reversible‟ has a special meaning. A reversible process is one that is wholly
theoretical, but can be imagined as one which occurs without incurring friction,
turbulence, leakage or anything which causes unrecoverable energy losses. All of the
processes considered below are reversible and the actual processes will be dealt with
later.

Processes may be constrained to occur at constant temperature (isothermal), constant
pressure, constant volume, polytropic and adiabatic (with no heat transfer to the
surroundings).

Constant temperature (Isothermal) process (pV = C)

If the change in temperature during a process is very small then that process
may be approximated as an isothermal process. For example, the slow expansion or
compression of fluid in a cylinder, which is perfectly cooled by water may be
analysed, assuming that the temperature remains constant.

P
1
W

2

W

v
v1
Q
v2

Constant temperature (Isothermal) process

The general relation properties between the initial and final states of a perfect
gas are applied as follows:
p1V1 p 2V2

T1 T2
If the temperature remains constant during the process, T1 = T2 and the above relation
becomes
p1V1  p2V2

33


From the equation we can know that an increase in the volume results in a
decrease in the pressure. In other words, in an isothermal process, the pressure is
inversely proportional to the volume.

Work transfer:
Referring to the process represented on the p – V diagram in Figure above it is
noted that the volume increases during the process. In other words the fluid is
expanding. The expansion work is given by

2
W   pdV
1
2
c
= dV (since pV = C, a constant)
1
V
2
dV
= c
1
V
2
dV
= p1V1 
1
V
V  larger vol ume 
= p1V1 ln 2  smaller volume 
V1  

V2
= mRT1 ln (since p1V1 = mRT1)
V1
p V2 p
= mRT1 ln 1 (since  1)
p2 V1 p2

Note that during expansion, the volume increases and the pressure decreases.
On the p – V diagram, the shaded area under the process line represents the amount of
work transfer.

Since this is an expansion process (i.e. increasing volume), the work is done
by the system. In other words the system produces work output and this is shown by
the direction of the arrow representing W.

Heat transfer:
Energy balance to this case is applied:

U1 + Q = U2 + W

For a perfect gas

U1 = mcvT1 and U2 = mcvT2

As the temperature is constant

U1 = U2

34


Substituting in the energy balance equation,

Q=W

Thus, for a perfect gas, all the heat added during a constant temperature
process is converted into work and the internal energy of the system remains constant.

Adiabatic process (Q = 0)

If a system is thermally well insulated then there will be negligible heat transfer into
or out of the system. Such a system is thermally isolated and a process within that
system may be idealised as an adiabatic process. For example, the outer casing of
steam engine, steam turbines and gas turbines are well insulated to minimise heat loss.
The fluid expansion process in such machines may be assumed to be adiabatic.

P
1
W

2

W

v
v1
Thermal insulation
v2

Adiabatic (zero heat transfer) process

For a perfect gas the equation for an adiabatic process is

pV = C

Cp
where  = ratio of specific heat =
Cv

The above equation is applied to states 1 and 2 as:

p1V1  p2V2

35



p 2  V1 
 
p1 V2 

Also, for a perfect gas, the general property relation between the two states is given
by the equation below

p1V1 p 2V2

T1 T2

By manipulating 2 equations above the following relationship can be
determined:

 1
 1
T2  p 2   V 
    1
T1  p1   V2 

By examining the equations the following conclusion for an adiabatic process on a
perfect gas can be drawn:
An increase in volume results in a decrease in pressure.
An increase in volume results in a decrease in temperature.
An increase in pressure results in an increase in temperature.

Work transfer:
Referring to the process represented on the p-V diagram it is noted that the
volume increases during the process.
In other words, the fluid expanding and the expansion work is given by the
formula:

2
W   pdV
1
2
c
= 
dV (since pV = C, a constant)
1V
2
dV
= c 
1 V

p V  p 2V2
= 1 1 [larger pV- small pV]
 1

Note that after expansion, p2 is smaller than p1. In the p – V diagram, the
shaded area under the process represents the amount of work transfer.
As this is an expansion process (i.e. increase in volume) the work is done by
the system. In other words, the system produces work output and this is shown by the
direction of the arrow representing W.

36


Heat transfer:
In an adiabatic process, Q = 0.
Applying an energy balance to this case

U1 - W = U2

W = U1 – U2

Thus, in an adiabatic expansion the work output is equal to the decrease in
internal energy. In other words, because of the work output the internal energy of the
system decreases by a corresponding amount.

For a perfect gas, U1 = mcvT1 and U1 = mcvT1
On substitution
W = mcv(T1-T2) [larger T- smaller T]

We know
cp- cv = R
or
R
cv =
 1

mR(T1  T2)
W
 1

But, mRT2 = p2V2 and mRT1 = p1V1

Then the expression for the expansion becomes

p1V1  p 2V2
W
 1

Example 17
In an industrial process, 0.4 kg of oxygen is compressed isothermally from 1.01 bar
and 22o C to 5.5 bar. Determine the work done and the heat transfer during the
process. Assume that oxygen is a perfect gas and take the molecular weight of oxygen
to be M = 32 kg/kmole.

Solution

Data: m = 0.4 kg; p1= 1.01 bar; t1= 22oC
p2 = 5.5 bar; W=? Q=?

37


From the equation
R
R= 0
M
8314
=
32
= 260 J/kgK
= 0.260 kJ/kgK

For an isothermal process
Work input,
p
W = mRTln 2
p1
5.5
= 0.4 x 0.260 x (22  273) ln
1.01
= 52 kJ

In an isothermal process all the work input is rejected as heat.
Therefore, heat rejected, Q = W = 52 kJ

Example 18

In a thermally insulated reciprocating air compressor, air at 0.98 bar and 20oC is
compressed into one sixth of its original volume. Determine the pressure and
temperature of the air after compression. If the compressor cylinder contains 0.05 kg
of air, calculate the required work input. For air, take  = 1.4 and cv = 0.718
kJ/kgK.

Solution
Data : p1 = 0.98 bar; T1= 20 + 273 = 293 K
V2 1
 ; m = 0.05 kg; W = ?
V1 6

As the cylinder is well insulated the heat transfer is negligible and the process
may be treated as adiabatic.
Considering air as a perfect gas

p 2  V1 
From equation,  
p1 V2 
p2 = 0.98 x 61.4
= 12 bar

38


 1
T V 
From equation 2   1 
T1  V2 
T2 = 293 x 60.4
= 600 K
= 327oC

for an adiabatic compression process

W = mcv(T2-T1) [larger T- smaller T]
= 0.05 x 0.718 (600-293)
= 11 kJ

Polytropic process (pVn = C)

This is the most general type of process, in which both heat energy and work
energy cross the boundary of the system. It is represented by an equation in the form

pVn = constant

If a compression or expansion is performed slowly, and if the piston cylinder
assembly is cooled perfectly, then the process will be isothermal. In this case the
index n = 1.

If a compression or expansion is performed rapidly, and if the piston cylinder
assembly is perfectly insulated, then the process will be adiabatic. In this case the
index n = .

If a compression or expansion is performed at moderate speed, and if the
piston cylinder assembly is cooled to some degree, then the process is somewhere
between those discussed above. Generally, this is the situation in many engineering
applications. In this case the index n should take some value, which is between 1 and
 depending on the degree of cooling.

Some practical examples include:
compression in a stationary air compressor (n = 1.3)
compression in an air compressor cooled by a fan (n = 1.2)
compression in a water cooled air compressor (n = 1.1)

39


P
1 pVn=C
W P1

2
P2

W

v1 v2 v
Qloss

Polytropic process

At states 1 and 2:

p1V1n  p2V2n
or
n
p 2  V1 
 
p1 V2 

Also, for a perfect gas, the general property relation between the two states is given
by

p1V1 p 2V2

T1 T2

By the manipulation of 2 equations above the following relationship can be
determined:

n 1
n 1
T2  p 2  n V 
    1
T1  p1   V2 

By examining the equations the following conclusions for a polytropic process on a
perfect gas can be drawn as:

An increase in volume results in a decrease in pressure.
An increase in volume results in a decrease in temperature.
An increase in pressure results in an increase in temperature.

Work transfer:
Referring to the process represented on the p-V diagram it is noted that the
volume increases during the process.
In other words the fluid is expands and the expansion work is given by

40


2
W   pdV
1
2
c
= n
dV (since pVn = C, a constant)
1V
2
dV
= c n
1 V

p V  p 2V2
= 1 1 [larger pV- small pV]
n 1

Note that after expansion p2 is smaller than p1. In the p – V diagram, the
shaded area under the process represents the amount of work transfer.
Since this is an expansion process (i.e. increase in volume), the work is done
by the system. In other words, the system produces work output and this is shown by
the direction of the arrow representing W.

Heat transfer:

Energy balance is applied to this case as:

U1 – Qloss - W = U2
Qloss = (U1 – U2) – W
or
W = (U1 – U2) - Qloss

Thus, in a polytropic expansion the work output is reduced because of the heat loses.

Example 19
The combustion gases in a petrol engine cylinder are at 30 bar and 800oC before
V 8 .5
expansion. The gases expand through a volume ratio ( 2 ) of ( ) and occupy 510
V1 1
cm3 after expansion. When the engine is air cooled the polytropic expansion index n =
1.15. What is the temperature and pressure of the gas after expansion, and what is the
work output?

Solution V2 = 510 cm3
p2 = ?
P1= 30 bar t2 = ?
t1 = 800oC Qloss

W

State 1 State 2

41


Data: p1 = 30 bar; T1 = 800 + 273 = 1073 K; n = 1.15

V2
= 8.5; V2 = 510 cm3;
V1

t2 = ? p2 = ? W=?

Considering air as a perfect gas, for the polytropic process, the property relation is
given as:
n 1
V 
T2  T1  1 
 V2 

1.151
 1 
= 1073x  
 8.5 

= 778.4 K
= 505.4oC

n
V 
From equation p 2  p1  1 
V 2 
1.15
 1 
= 30 x  
 8.5 

= 2.56 bar

Now,
V2 = 510 cm3 = 510 x 10-6 m3
and,
V2
= 8.5
V1
Then,
510 x10 6
V1 
8.5
= 60 x 10-6 m3

Work output during polytropic expansion is given as:

p1V1  p 2V2
W = [larger pV- small pV]
n 1
(30 x10 5 )(60 x10 6 )  (2.56 x10 5 )  (510 x10 6 )
=
1.15  1

= 330 J
= 0.33 kJ

42


Constant volume process

If the change in volume during a process is very small then that process may be
approximated as a constant volume process. For example, heating or cooling a fluid in
a rigid walled vessel can be analysed by assuming that the volume remains constant.
Q
p 2 p 1

1 2
v v
Q

a) Heating b) Cooling

Constant volume process (V2=V1)

The general property relation between the initial and final states of a perfect gas is
applied as:
p1V1 p 2V2

T1 T2
If the volume remain constant during the process, V2 = V1 and then the above relation
becomes
p1 p 2

T1 T2
or
T2 p 2

T1 p1

From this equation it can be seen that an increase in pressure results from an increase
in temperature. In other words, in constant volume process, the temperature is
proportional to the pressure.

Work transfer:
Work transfer (pdV) must be zero because the change in volume, dV, during the
process is zero. However, work in the form of paddle-wheel work may be transferred.

Heat transfer:
Applying the non flow energy equation

Q – W = U2 – U1
gives Q – 0 = U2 – U1
i.e. Q = U2 – U1

43


This result, which is important and should be remembered, shows that the nett amount
of heat energy supplied to or taken from a fluid during a constant volume process is
equal to the change in the internal energy of the fluid.

5.3 Constant pressure process

If the change in pressure during a process is very small then that process may be
approximated as a constant pressure process. For example, heating or cooling a liquid
at atmospheric pressure may be analysed by assuming that the pressure remains
constant.

P
W

p 1 2

W

v
v1 v2 – v1
Q
v2

Constant pressure process

Consider the fluid in the piston cylinder as shown in Figure above. If the load on the
piston is kept constant the pressure will also remain constant.

The general property relation between the initial and final states of a perfect gas is
applied as:

p1V1 p 2V2

T1 T2
If the pressure remain constant during the process, p2 = p1 and then the above relation
becomes
V1 V2

T1 T2
or
T2 V2

T1 V1

From this equation it can be seen that an increase in volume results from an increase
in temperature. In other words, in constant pressure process, the temperature is
proportional to the volume.

44


Work transfer:
Referring to the process representation on the p-V diagram it is noted that the volume
increases during the process. In other words, the fluid expands. This expansion work
is given by
2
W   pdV
1

2
 p  dV (since p is constant)
1

= p (V2 – V1) (larger volume – smaller volume)

Note that on a p-V diagram, the area under the process line represents the amount of
work transfer.

W = area of the shaded rectangle
= height x width
= p (V2 – V1) (larger volume – smaller volume)

Heat transfer:
Applying the non flow energy equation

Q – W = U2 – U1

or Q = (U2 – U1) + W

Thus part of the heat supplied is converted into work and the remainder is utilized in
increasing the internal energy of the system.

Substituting for W in equation

Q = (U2 – U1) + p(V2 – V1)
= U2 – U1 + p2 V2 – p1 V1 (since p2 = p1 )
= (U2 + p2 V2) – (U1 + p1 V1)

Now, we know that h = u + pv or H = U + pV
Hence
Q = H2 – H1 (larger H – smaller H)

45


Example 20
The specific internal energy of a fluid is increased from 120 kJ/kg to 180 kJ/kg during
a constant volume process. Determine the amount of heat energy required to bring
about this increase for 2 kg of fluid.

Solution

The non flow energy equation is
Q – W = U2 – U1

For a constant volume process
W=0

and the equation becomes
Q = U2 – U1
 Q = 180 – 120
= 60 kJ/kg

Therefore for 2 kg of fluid
Q = 60 x 2 = 120 kJ

i.e. 120 kJ of heat energy would be required.

Example 21
2.25 kg of fluid having a volume of 0.1 m3 is in a cylinder at a constant pressure of 7
bar. Heat energy is supplied to the fluid until the volume becomes 0.2 m3. If the initial
and final specific enthalpies of the fluid are 210 kJ/kg and 280 kJ/kg respectively,
determine
a) the quantity of heat energy supplied to the fluid
b) the change in internal energy of the fluid

Solution
Data: p = 7.0 bar; V1 = 0.1 m3 ; V2 = 0.2 m3

a) Heat energy supplied = change in enthalpy of fluid
Q = H2 – H1
= m( h2 - h1 )
= 2.25( 280 – 210 )
= 157.5 kJ

46


b) For a constant pressure process
W = P(V2 – V1)
= 7 x 105 x ( 0.2 – 0.1)
= 7 x 104 J
= 70 kJ

Applying the non-flow energy equation
Q – W = U2 – U1
gives
U2 – U1 = 157.5 – 70
= 87.5 kJ

47


TUTORIAL

1. Study the statements in the table below. Mark the answers as TRUE or
FALSE.

STATEMENT TRUE or FALSE

i. Charles‟ Law gives us the change in volume
of a gas with temperature when the
temperature remains constant.
ii. Boyle‟s Law gives us the change in volume of
a gas with pressure if the pressure remains
constant.
iii. The characteristic equation of state of a
perfect gas is PV  R .
T
iv. Ro is the symbol for universal gas constant.
v. The constant R is called the gas constant.
vi. The unit of R is Nm/kg or J/kg.

2. 0.04 kg of a certain perfect gas occupies a volume of 0.0072 m3 at a
pressure 6.76 bar and a temperature of 127 oC. Calculate the molecular weight
of the gas (M). When the gas is allowed to expand until the pressure is 2.12
bar the final volume is 0.065 m3. Calculate the final temperature.

3. Two kilograms of a gas receive 200 kJ as heat at constant volume
process. If the temperature of the gas increases by 100 oC, determine the Cv of
the process.

4. A perfect gas is contained in a rigid vessel at 3 bar and 315 oC. The
gas is then cooled until the pressure falls to 1.5 bar. Calculate the heat
rejected per kg of gas.
Given: M = 26 kg/kmol and  = 1.26.

5. A mass of 0.18 kg gas is at a temperature of 15 oC and pressure 130
kN/m2. If the gas has a value of Cv = 720 J/kg K, calculate the:

i.gas constant
ii.molecular weight
iii.specific heat at constant pressure

48


iv.specific heat ratio

6. 1 m3 of air at 8 bar and 120 oC is cooled at constant pressure process
until the temperature drops to 27 oC.
Given R = 0.287 kJ/kg K and Cp = 1.005 kJ/kg K, calculate the:

i. mass of air
ii. heat rejected in the process
iii.
iv. volume of the air after cooling.

7. A system undergoes a process in which 42 kJ of heat is rejected. If the
pressure is kept constant at 125 kN/m2 while the volume changes from 0.20 m3
to 0.006 m3, determine the work done and the change in internal energy.

8. Heat is supplied to a gas in a rigid container.The mass of the container
is 1 kg and the volume of gas is 0.6 m3. 100 kJ is added as heat. If gas has
Cv = 0.7186 kJ/kg K during a process, determine the:

9. In the cylinder of a large engine, 1.0 kg of carbon dioxide at 527 o C
and 20 bar expands isothermally to a pressure of 1.4 bar. What is the final
volume of the gas?
Take R = 189 Nm/kgK for carbon dioxide.

10. 1 kg of nitrogen (molecular weight 28) is compressed reversibly and
isothermally from 1.01 bar, 20oC to 4.2 bar. Calculate the work done and the
heat flow during the process. Assume nitrogen to be a perfect gas.

11. Air at 1.02 bar, 22oC, initially occupying a cylinder volume of 0.015
m3, is compressed reversibly and adiabatically by a piston to a pressure of 6.8
bar. Calculate the final temperature, the final volume, and the work done on
the mass of air in the cylinder.

12. 0.112 m3 of gas has a pressure of 138 kN/m2. It is compressed to
690 kN/m2 according to the law pV1.4 = C. Determine the new volume of the
gas.

13. 0.014 m3 of gas at a pressure of 2070 kN/m2 expands to a pressure of
207 kN/m2 according to the law pV1.35 = C. Determine the work done by the
gas during expansion.

14. A cylinder containing 0.07 kg of fluid has a pressure of 1 bar, a volume
of 0.06 m3 and a specific internal energy of 200 kJ/kg. After polytropic

49


compression, the pressure and volume of the fluid are 9 bar and 0.011 m3
respectively, and the specific internal energy is 370 kJ/kg.
Determine
a) the amount of work energy required for the compression
b) the quantity and direction of the heat energy that flows during the
compression.

15. The pressure of the gas inside an aerosol can is 1.2 bar at a temperature
o
of 25 C. Will the aerosol explode if it is thrown into a fire and heated to a
temperature of 600o C? Assume that the aerosol can is unable to withstand
pressure in excess of 3 bar.

a. 0.05 kg of air, initially at 130o C is heated at a constant pressure of 2
bar until the volume occupied is 0.0658 m3. Calculate the heat supplied and
the work done.

b. A spherical research balloon is filled with 420 m3 of atmospheric air at
a temperature of 10o C. If the air inside the balloon is heated to 80oC at
constant pressure, what will be the final diameter of the balloon?

50

JJ207 Thermodynamics I Chapter 2

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