THERMODYNAMICS
Principles & Applications
Prof. Dr. Nuri KAYANSAYAN
DOKUZ EYLUL UNIVERSITY
Me c ha nic a l Engineering Department
www.nobelyayin.com
YAYIN NU
Teknik Nu
ISBN
: 591
: 50
: 978-605-133-493-6
© 1. Baskı, Nisan 2013
TERMODYNAMICS Principles & Applications
Prof. Dr. Nuri KAYANSAYAN
Copyright 2013, NOBEL AKADEMİK YAYINCILIK EĞİTİM DANIŞMANLIK TİC. LTD. ŞTİ. SERTİFİKA NU 20779
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KÜTÜPHANE BİLGİ KARTI
Kayansayan, Nuri
TERMODYNAMICS Principles & Applications / Prof. Dr. Nuri Kayansayan
1. Baskı, X + 526 s., 195x275 mm
Kaynakça ve dizin yok
ISBN 978-605-133-493-6
1. Heat 2. Energy 3. Exergy 4. Entropy
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“Thermodynamics is a funny subject. The first time you
go through it, you don’t understand it at all. The second
time you go through it, you think you understand it,
except for one or two small points. The third time you go
through it, you know you don’t understand it, but by that
time you are so use to it, it doesn’t bother you anymore.”
-Arnold Sommerfield-
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P1
BASIC CONCEPTS & DEFINITIONS
P4
ENERGY ANALYSIS OF SYSTEMS
1.1
Introduction ..................................................... 1
4.1
Introduction ................................................... 91
1.2
Dimensions and Units ...................................... 2
4.2
Energy of a system ........................................ 92
1.3.
The System Concept ........................................ 3
4.3
Forms Of Energy Transfer ............................ 96
1.4
The Property Concept ...................................... 6
4.4
Heat Transfer ................................................. 97
1.5
The Pressure and Temperature of a System .... 8
4.5
Work Transfer ............................................. 101
1.6
The State Concept ......................................... 18
4.6
Mechanical Work Transfer .......................... 102
1.7
The Equilibrium Concept .............................. 19
4.7
1.8
The Process Concept ..................................... 20
Other Forms Of Quasistatic Work
Transfer ....................................................... 107
References ..................................................... 21
4.8
Convective Transfer Of Energy ................... 108
Problems ....................................................... 21
4.9
The Energy Equation ................................... 110
4.10
Steps In Problem-Solving ............................ 112
4.11
Closed Systems ........................................... 112
4.12
Constant pressure process of closed
systems ........................................................ 118
4.13
An Introduction To Thermodynamic
Cycles.......................................................... 119
4.14
Steady State Flow Systems .......................... 124
P2
THERMODYNAMIC PROPERTIES OF
SYSTEMS
2.1
Introduction ................................................... 27
2.2
The State Principle ........................................ 28
2.3
The p-v-T Behavior Of Systems .................... 29
2.4
The Use Of Thermodynamic Tables .............. 35
2.5
The Specific Heats of a Pure Substance ......... 39
2.6
Gaseous Behavior of a Pure Substance .......... 40
2.7
The Ideal Gas Model ..................................... 45
2.8
The Specific Heats of Ideal Gases .................. 51
4.15
Problems ..................................................... 142
True and False ............................................. 158
Check Test 4 ................................................ 159
References ..................................................... 54
Problems ....................................................... 55
Transient flow systems ................................ 135
References ................................................... 142
P5
SECOND LAW EXERGY ANALYSIS OF
SYSTEMS
5.1
Introduction ................................................. 161
5.2
Equilibrium of Systems ............................... 164
5.3
Exergy of a System ...................................... 165
5.4
Exergy Loss of a System ............................. 167
5.5
Equation of Exergy ...................................... 172
P3
MASS ANALYSIS OF SYSTEMS
3.1
Introduction ................................................... 63
3.2
The Equation Of Continuity .......................... 63
3.3
The Mass Change Of a System ...................... 67
3.4
Integral Formulation Of The Continuity
Equation ........................................................ 69
3.5
Velocity Measurements ................................. 77
5.6
The exergy transfer by work ........................ 173
Flow Rate Measurements............................... 80
5.7
The Exergy Transfer by Heat ....................... 175
References ..................................................... 83
5.8
Entropy ........................................................ 182
Problems ....................................................... 84
5.9
The Entropy Change of a System................. 184
True and False ............................................... 88
5.10
The General Equation of Exergy ................. 185
Check Test 3 .................................................. 88
5.11
Exergy Analysis of Closed Systems............. 186
3.6
v
vi
CONTENTS
5.12
Exergy Analysis of Steady State Flow
Systems ....................................................... 189
5.13
Exergy Efficiency of Energy Conversion
Systems ....................................................... 194
References ................................................... 206
Problems ..................................................... 206
True and False ............................................. 216
Check Test 5 ................................................ 217
P8
POWER PRODUCING SYSTEMS
8.1
General considerations for power cycles...... 327
8.2
Four-stroke SI engine cycle ......................... 332
8.3
Four-stroke CI engine cycle ......................... 336
8.4
Gas Turbine Engine ..................................... 342
8.5
Improving the Thermal
Efficiency of Gas Turbine Engines .............. 349
8.6
The Jet Engine ............................................. 357
8.7
Stirling Engine ............................................ 359
8.8
A Simple Rankine-Cycle power plant .......... 362
8.9
Improving the Thermal
Efficiency of Rankine Cycle ........................ 367
P6
ENTROPY: A SYSTEM DISORDER
6.1
Introduction ................................................. 219
6.2
Entropy Balance for Closed Systems ........... 222
6.3
Entropy balance for open systems ............... 226
8.10
Cogeneration ............................................... 375
6.4
Temperature-Entropy (T-s) diagram ............ 228
8.11
Organic Rankine Cycle ................................ 380
6.5
Enthalpy-Entropy (h-s) Diagram ................. 232
References ................................................... 382
6.6
Some Relations for Flow Processing
Devices ........................................................ 232
Problems ..................................................... 383
6.7
Adiabatic Efficiencies of Steady Flow
Devices ........................................................ 243
Check Test 8 ................................................ 396
6.8
Thermodynamic relations ............................ 247
6.9
Relations on Specific Heats ......................... 267
6.10
Clausius-Clapeyron Equation ...................... 255
6.11
Use of entropy in design .............................. 257
References ................................................... 264
Problems ..................................................... 265
Steady flow systems .................................... 267
Isentropic flow ............................................. 272
True and False ............................................. 276
Check Test 6 ................................................ 277
P7
GAS MIXTURES & PSYCHROMETRY
7.1
Basic Definitions for Mixtures ..................... 279
7.2
p-v-T behavior of gas mixtures .................... 283
7.3
Moist Air and its Psychrometric
Properties .................................................... 291
7.4
Air conditioning processes .......................... 299
7.5
Cooling Tower Basics ................................. 304
7.6
Homogenous and Ideal Binary Solutions ..... 310
References ................................................... 315
Problems ..................................................... 316
Check Test 7 ................................................ 324
True and False ............................................. 395
P9
REFRIGERATION SYSTEMS
9.1
General considerations ................................ 399
9.2
Vapor compression refrigeration cycle ........ 401
9.3
Multi-Pressure Refrigeration ....................... 404
9.4
Refrigeration System Components............... 409
9.5
Heat Pump ................................................... 437
9.6
Vapor absorption refrigeration (VAR) ......... 438
9.7
Miscellaneous Refrigeration Methods ......... 446
References ................................................... 449
Problems ..................................................... 450
True and False ............................................. 462
Check Test 9 ................................................ 463
P
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E
This book is written for an introductory and intermediate level course in the subject of thermodynamics for engineering curricula. The approach followed in the text is to emphasize the physical concepts of
thermodynamics and the method of analysis that starts with identifying the underlying principles and definitions. The primary objective of the text is to help students develop an orderly approach in understanding
the obscure concepts such as the energy and the exergy of a system. In doing so, a total of two hundred
illustrative sample problems are provided, and on the average eighty unsolved problems with engineering
emphasis are contained at the end of each chapter. Example problems are set apart in a format different
from the text so that they are easy to identify and follow.
In solving the end chapter problems,
 Students should be able to incorporate with the meaning of physical principles associated with the
subject.
 Students should be able to use the control volume approach to identify the system.
 Students should be able to state the related assumptions.
 Students should be able to relate the mathematical results to the corresponding physical behavior,
and draw conclusions concerning the process or the system design from attendant analysis.
In addition, each chapter contains True-False and a Multiple Choice Test sections. Both of these tests
are aimed for self-check of student’s weakness or strength on the highlights of the related chapter.
Special care has been given to illustrations of both the main text and of the problems, and if necessary,
figures in color are used for making the subject more understandable. Further facilitate comprehension of
the subject, especially for students taking an elementary thermodynamics course in engineering curricula
for the first time, systems like flow machines or heat exchangers have been illustrated with their essential
components without going into unnecessary complexity.
The material has been selected carefully to include a broad range of topics suitable for two-semester
engineering thermodynamics course at the junior level. The material in the manuscript has been organized
around the following topics:
 Introductory concepts and definitions for system property, state, process, and equilibrium (Chapter 1)
 Methods of measuring pressure and temperature (Chapter 1)
 State principle, p-V-T behavior of a pure substance, the ideal gas behavior, specific heats (Chapter 2)
 Development and application of control volume approach to mass analysis (Chapter 3)
 Methods and instruments for measuring velocity and flow rate (Chapter 3)
 Development and application of control volume approach to energy analysis, steady and transient
flow systems (Chapter 4)
 Heat transfer analysis and overall heat transfer coefficient of systems (Chapter 4)
 Work transfer analysis (Chapter 4)
 Energy analysis of thermodynamic cycles (Chapter 4)
 Exergy of a system, the exergy change and the exergy loss of a system (Chapter 5)
 Application of control volume approach to exergy analysis, the exergy equation, exergetic efficiency of flow systems and cyclic devices (Chapter 5)
 Application of control volume approach to entropy analysis, fundamental equation of thermodynamics, isentropic flow, isentropic efficiencies of steady flow devices (Chapter 6)
 Applications of Maxwell’s relations, the Clausius-Clapeyron equation, the phase equilibria (Chapter 6)
vii
viii
CONTENTS

The use of entropy in design of thermal systems, specific applications for channel flow and for
wind turbine design (Chapter 6)
 Partial properties of a gas mixture component, ideal gas mixtures, rules for estimating the p-V-T
behavior of gas mixtures, the Orsat apparatus (Chapter 7)
 Moist air properties, the psychrometric chart, air conditioning processes, cooling tower basics (Chapter 7)
 Classification of power producing systems, cyclic properties of internal and external combustion
systems (Chapter 8)
 Efficiency analysis of Otto, Diesel, Brayton, Stirling, and Rankine cycles, low temperature applications of Rankine cycle, cogeneration (Chapter 8)
 Vapor compression refrigeration systems, and analysis of multi-pressure systems (Chapter 9)
 Refrigeration compressors, expansion devices, design of refrigerant condensers and evaporators,
properties of refrigerants, heat pump systems, basics of absorption refrigeration systems (Chapter 9)
The first part of the book (Chapters 1-5) contains material suitable for a Basic Course in Thermodynamics that can be taken by engineering students of all majors. The second part of the book (Chapters
6-9) is designed for an Applied Thermodynamics Course or for Thermodynamics II course in mechanical
engineering programs.
Due to industrialization and growth of world population, the increase in per capita energy consumption
is one of the prime causes of the need for efficient use of available energy resources in today’s world.
The book considers this fact in the selection and sequential presentation of the subject material
throughout the book. The conservation of mass, energy, and the non-conservation of exergy are covered
in sequence in Chapters 3, 4, and 5. A student, taking an introductory course in Thermodynamics, should
be able to calculate the amount of energy of a system as well as the maximum portion of that energy that
is available for use. Moreover, the efficiency of energy conversion systems is defined in two different
ways as the exergy-based efficiency and the energy-based efficiency. First, the exergy-based efficiency is
introduced. The energy-based efficiency of flow machines needs a substantial background in entropy and
is covered in Chapter 6. Moreover, subjects like reversible shaft work, multi-stage compression, incompressible and adiabatic flow processes and isentropic flows are applications of entropy and are studied in
Chapter 6. Similarly, the use of Maxwell’s relations in entropy-based design is exemplified by case studies.
The p-V-T behavior of gas mixtures, ideal and real, and properties of moist air are provided in Chapter 7.
As an introduction to air conditioning engineering, processes related to moist air are exemplified and the
design methods for cooling towers are illustrated. Chapter 8 deals with work producing cyclic systems as
predominantly used in today’s industrial applications, and explains system modifications for increasing the
cyclic efficiency. Special attention is given to low temperature applications of Rankine cycle for which the
temperature of heat source is in the range from 160°C to 200°C. Chapter 9 is about refrigeration systems,
and covers vapor compression refrigeration, heat pumps and the essentials of absorption refrigeration.
In this chapter, the thermodynamic analysis of compressors and expansion devices is explained, and the
design methodology for condensers and evaporators are provided. Special attention is given to thermal
design of cooling towers.
Where appropriate and especially in Chapters 7, 8, and 9 open-ended type design problems are included. Students could be assigned to work in teams to solve these problems. Design problems encourage
the students to spend more time exploring applications of thermodynamic principles to devices and flow
systems.
The book is well suited for independent study by students or practicing engineers. Its readability and
clear examples help to build confidence. When students finish the text, I expect them to be able to apply
the related principles and derived equations to a variety of systems including those they have not encountered previously.
Last but not least, my sincere appreciation goes to Dr. Mehmer Akif Ezan for his endless effort in
drawing the figures and reviewing the manuscript.
Prof. Dr. Nuri Kayansayan
"Izmir, April 2013"
L
I
S
T
O
F
S
Y
A
B
COP
CR
c
cp
cv
E
e
Area
Magnetic induction
Coefficient of performance
Compression ratio
Specific heat of liquid or solid
Constant pressure specific heat
Constant volume specific heat
Energy and Electric field strength
Specific energy
PR
p
ps
pi
pr
Q
q
F
G
g
H
h

S
s
Sg
S
I
I
i
Force
Gibbs function
Acceleration of gravity
Total enthalpy and magnetic field strength
Specific enthalpy and convective heat
transfer coefficient
Irreversibility and electric current
Irreversibility rate
Specific irreversibility
KE
ke
k
L
M
m
m
n
P
PE
pe
Kinetic energy
Specific kinetic energy
Thermal conductivity and specific heat ratio
Length
Molecular mass
Mass
Mass flow rate
Number of moles
Polarization
Potential energy
Specific potential energy
u
V
v
W
Q
R
g
T
Ts
Tr
U
W
X
x
Z
M
B
O
L
S
Pressure ratio
Pressure
Saturation pressure
Partial pressure of species i
Reduced pressure
Heat transfer
Heat transfer per unit area
Heat transfer rate
Individual gas constant and electrical
resistantance
Universal gas constant
Total entropy
Specific entropy
Entropy generation
Entropy generation rate
Temperature and torque
Saturation temperature
Reduced temperature
total internal energy and overall heat
transfer coefficient
Specific internal energy
Volume and velocity
Specific volume
Work transfer
Work transfer rate
Exergy
Vapor quality
Compressibility factor
G
R
E
E
K
L
E




b
s
Isentropic compressibility
Volume expansivity
Emissivity and exchanger effectiveness
Efficiency
Boiler efficiency
Isentropic efficiency


J




Exergetic efficiency
Isothermal compressibility

x

T
T
E
R
S
Angular displacement
Viscosity
Joule-Thomson coefficient
Kinematic viscosity
Density
Surface tension and Stefan-Boltzmann
constant
Exergy rate
Specific exergy
C
H
1
A
P
T
E
R
Basic Concepts & Definitions
1.1 Introduction
Why thermodynamics ? Thermodynamics is an engineering tool for many branches of
engineering and is used to describe processes that involve energy interaction. Thermodynamics can be stated as a generalization of an enormous body of empirical evidence with
no hypotheses concerning the type and the structure of systems. In short, thermodynamics
provides unique answers to such questions as following:
1. What is the maximum amount of work that may be obtained per liter of gasoline or
per kilogram of coal?
2. What is the ultimate efficiency or “the maximum ever possible efficiency” of an automobile engine or a power plant operating between two given temperature levels?
3. Under what conditions and how the natural way of heat flow can be reversed so
that heat can be transferred from a low temperature level to a higher temperature
level?
4. What general relations exist between the equilibrium properties of materials? including those for which there may be no experimental data or theoretical models?
A wide variety of other questions can be answered by thermodynamics, as evidenced
by the fact that a course on thermodynamics is found in the curricula of almost all branches
of engineering.
In performing engineering analysis on a real phenomenon, it is necessary that engineer
has the capability to describe the phenomenon he seeks to control. A complete description
generally needs geometric as well as dynamic similarity between the phenomenon and
its model. Due to minor effect of certain parameters, however, those parameters might
be ignored in the analysis. For instance, in analyzing the trajectory of a soccer ball by the
laws of classical mechanics, we completely ignore the molecular structure of the ball even
though there are events occurring at the molecular level. Therefore, a basic principle may
be expressed as follows,
1
2
THERMODYNAMICS
Principle 1: In engineering analysis of a real phenomenon, a model which
facilitates studying the desired features of the actual event should be
engendered.
Thus, the conclusions drawn by means of a particular model for the real phenomenon largely
depends upon the appropriateness of the model. The question, “how appropriate the model is?” is in
the context of the art of engineering and such a question is often answered by appealing to experience.
Thus, there are two fundamental models for the matter of the universe: a. The macroscopic model,
and b. The microscopic model. Even though each model is important and provides its own characteristics, the macroscopic model of matter will be discussed and employed in this text. In addition,
the thermodynamic analysis of mechanical systems may successfully be completed by the systematic
application of the following fundamental principles: a. The conservation of mass, b. The conservation
of energy, c. The non-conservation of exergy. The objective of this book is to develop and employ
these principles to the problems encountered in mechanical engineering applications.
1.2
Dimensions and Units
Dimensions are names that characterize physical quantities. Common dimensions include length
L, time t, mass m, and temperature T. In engineering analysis, any equation relating physical quantities must be dimensionally homogeneous. Dimensional homogeneity requires that the dimensions of
the terms on both sides of an equation must be the same.
Units are those arbitrary magnitudes and names assigned to dimensions that are adapted as standard
for measurements. The fundamental system of units chosen for scientific work all over the world is
the System’e Internationale, which is abbreviated as SI. The SI employs seven primary dimensions.
Those are: mass, length, time, temperature, electric current, luminous intensity, and the amount of
substance. The basic units for measuring these quantities are given Table 1.1.
Although the description of these basic units can be found in a text of any college physics, the
definition of a mole is important for engineering calculations. A mole is the amount of substance containing 6.023x1023 number of particles. A kilo-mole is 1000 times as large as a mole. For instance, 1
kmol of pure carbon contains 12 kg of carbon. The number of moles N of a substance is defined as,
N
m
M
(1.1)
Table 1.1 SI base units
Physical quantity
SI unit and symbol
Mass
kilogram (kg)
Length
meter
(m)
Time
second
(s)
Temperature
Kelvin
(K)
Electric current
ampere (A)
Luminous intensity
candela (cd)
Amount of substance
mole
(mol)
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 3
In Eq. (1.1), M is the molar mass. The molar mass of a substance is equal to its molecular mass.
For example, the molecular mass of oxygen gas is 32 kg/kmol.
All other SI units are secondary ones and are derivable in terms of these seven primary units. The
SI units of force is the Newton (N) and it is derived from Newton’s second law, F  ma . Thus, a net
force of 1N accelerates 1 kg of mass at one meter per second. In conjunction with the definition of
force, weight always refers to a force of attraction between the body and the Earth,
W  mg
(1.2)
where g is the acceleration of gravity and varies with the location of the body on the Earth.
Thus, the weight of a substance may vary but the mass is always constant. Force interactions have
two principal effects: They tend to alter the motion of the objects, and to deform the shape of objects.
In Figure 1.1a, the applications of a force F to a transitional spring tends to stretch it. Similarly, in
Fig. 1.1b, the attraction of the Earth has a tendency to alter the motion of the airplane from a level
flight to a vertical dive. An ideal transitional spring is a one-dimensional spring of zero mass that
can experience only transitional displacements along its axis. As shown in Figure 1.1a, for an ideal
spring, the relation between the applied force, F and the spring displacement, x is a linear one, and
expressed as,
F  Kx
(1.3)
where K is called the spring constant and has the units of N/m.
1.3
The System Concept
For a successful application of the fundamental principles to a particular phenomenon under
consideration, it is necessary to first identify the system.
Definition: (a) A system is a three-dimensional region of universe, not necessarily of constant volume
or mass, is set for purposes of analysis. (b) Everything that is apart from the system is referred to as
the surroundings. (c) The actual or the imaginary envelope separating the system and the surroundings is the boundary.
4
THERMODYNAMICS
Example 1.1: Consider a fluid flowing through a pipe of length L. Taking the pipe as a system, define the boundary of the
system.
Solution:
The boundary of the system is defined by dash line in Figure 1.2.
It should be noted that the thermodynamic system is merely an analytical model. However, the
specification of a system comprises the first step in the analysis. As shown in Fig. 1.3, the boundary
of a system may be rigid or moveable. A system with rigid boundaries is said to have a constant
volume.
The analysis of thermodynamic processes includes the study of the transfer of mass and energy
across the boundaries of a system. Thus, selection of an appropriate boundary makes the analysis
less difficult. The system described in Fig. 1.3 belongs to an important class of systems called open
system.
Definition: An open system is a system for which mass as well as energy may cross the boundaries
of the system.
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 5
The boundary used to define an open system is a surface called a control surface. The region of
space enclosed by this surface is called a control volume and actually is the open system itself. Whenever there is a mass transfer to or from the system, an energy transfer also simultaneously takes place.
However, energy transfer to or from the system may also be accomplished without mass transfer. As
shown in Fig. 1.4, another important class of systems in engineering consists of closed systems.
Definition: A closed system is a system for which no mass crosses the boundary. Although the quantity
of matter is fixed in a closed system, energy is allowed to cross the boundaries.
The closed system may be regarded as special case of the open system in the sense that mass and
energy cross the boundary of the open system, while energy but no mass crosses the boundary of the
closed system. A special form of the closed system is called an isolated system.
Definition: A system having fixed mass and energy is called isolated system. Neither mass nor energy
is allowed to cross the boundaries of an isolated system.
Example 1.2: Consider 1 kg of water being heated in a container open at the top. Define the boundaries of the system, and
classify the system as open or closed.
Solution:
It is obvious that as the liquid water heated, some portion of it will be evaporated. The vapor particles will cross the imaginary
boundary. Thus, assuming the system to be open or close mainly depends what percentage of the original mass evaporates
during the process. Therefore, for some instances, the system may be regarded as closed, in others, it may be taken as an open
system. To decide which model is more appropriate for a particular problem is part of the art of engineering analysis.
6
THERMODYNAMICS
Example 1.3: Consider a reciprocating compressor and classify the
system as open or close.
Solution:
To represent the system as open or close depends upon the portion of the
mechanical cycle that the compressor undergoes. For instance, during
a compression stroke, both valves are closed and the system may be
regarded as closed. However, if the overall cycle is considered, since
air enters and leaves through the valves, the system must be regarded
as open.
1.4
The Property Concept
Once a system has been selected for analysis, it must be
described in precise numerical terms. A system is described
in terms of its physical properties.
Figure 1.6 System schematic of a
reciprocating compressor
Definition: A property is any characteristic of a system that can be assigned a numerical value at a
particular instant of time without reference to the history of the system.
Examples of properties include pressure, temperature, mass, volume, density, electrical conductivity, acoustic velocity, thermal coefficient of expansion. The distinction between properties and
non-properties is of outmost importance. Mass is a property, but the amount of mass entering to the
system through a flow port, say in one hour, is not a property, because it depends on the history of the
system. Similarly, the population of Chicago at a particular time t is a property. However, the number of babies born in the last 24 hours is not a property, because this number may not be established
without a historical record. Regardless of the method of measurement, the value of the property is
unique and fixed by the condition of the system at the time of measurement. Thus, one may state that
a system characteristic is a property if it is a function of other properties.
In performing thermodynamics analysis, properties may be grouped into two different categories.
Those are called; 1.The extensive properties, 2.The intensive properties.
Definition: Thermodynamic properties whose values depend on the size of the system are called
extensive properties.
For instance, volume, mass, energy, exergy, and entropy of a system are extensive properties. If
a system is subdivided into a group of smaller sub-systems, the value of any extensive property for
each sub-system will depend on the size of the sub-system. The value of the same extensive property
for the composite system is simply the sum of the values of the extensive property for the constituent
sub-systems.
It is important to note that the independent variable for an extensive property is time. At a particular
time t, there is only one value for the system volume, mass, and energy, etc. Thus symbolically,
The amount of extensive property P 
P (t )  

of the system at time t

(1.4)
In engineering analysis of systems, one is usually interested in time rate of change of a particular
extensive property. The extensive property rate equation may be written as follows,
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 7
Time rate of change  The rate at which P  The net rate at which P is


 
 

of P contained within   is produced in the  +  transported into the system through 
a system
 system at time t
  the boundaries at time t


 
 

  P (t ) 

P (t )  
 cv  PT t 
 t 
(1.5)
Thus, at a particular time t, if P (t ) <0, it indicates that the amount of P in the system is decreasing
 (t )  0 , the amount of P contained in the system is being transported
at that instant. Similarly, if PT
outside of the system at that instant.
The net change in the amount of property P for a specified time interval (t1, t2) may be calculated
by the integration of Eq. (1.5).
t2
P t2   P t1   P (t )dt 

t1
t2
 P

 dt 

t


t1

cv
t2
 (t )dt
 PT
(1.6)
t1
The net change in the amount  The net amount of P produced within 


+
 of property P of the system   the system at the specified time interval 
The net amount of P



 transported into the system at the specified 
 time interval



The transfer of an extensive property into or from the system may be accomplished at some regions
of the boundary where there is a mass flow. If the transfer of an extensive property is due to flow of
mass, then it is called convective transfer.
There are certain other properties which do not change with the size of the system.
Definition: The thermodynamic properties whose values are fixed at an instant of time t and at each
point within the system are called intensive properties.
Pressure, temperature, density are
intensive properties of a system. In regard
to the definition, the independent parameters of an intensive property are position
and time. For instance, the temperature
of a particular point positioned at (x, y,
z) on a Cartesian-coordinate system, and
at a particular time is T(x, y, z, t). Similarly, the pressure of a particular point
at time t is p(x, y, z, t).
Example 1.4: As shown in Fig 1.7, an elastic
balloon is to be filled with helium gas flowing through a pressurized pipe. It is desired
to portray the behavior of the balloon and its
content. a) Define the system, its boundary,
8
THERMODYNAMICS
and its surroundings. b) At an instant of time t, determine if (i) mass of the system (ii) balloon’s age (iii) velocity of helium
flowing into the balloon (iv) the amount of helium entering to the balloon at a particular time interval are properties or not?
If so, specify the type of property.
Solution:
a.
The balloon is taken to be the system, and its boundaries and the surroundings are shown in the figure.
b.
(i) extensive property, (ii) not a property, (iii) intensive property, (iv) not a property.
A change in an intensive property  is determined by the end states and is independent of the
details of the change. In other words, the amount of change in an intensive property can be computed
without any knowledge about the process causing the change or the details of the path between the
end states. Hence,

2
1
d   2  1
(1.7)
As we know from Calculus, the integration of an exact differential is independent of path of the
integration and the integration result is simply the difference between the end values of the function.
Thus an infinitesimal change of a property can be represented by an exact differential. However,
every infinitesimal change may not be an exact differential. In accord with Calculus, if the relation
d   Mdx  Ndy is an exact differential, then the following condition must hold,
M N

y
x
(1.8)
This is a key relation in questioning whether a function is a property or not. The intensive properties are further classified into two groups:
a. Specific properties are the locally estimated limit values of extensive properties for a unit
mass or a volume. For instance, density of a substance is defined as,
  lim V  V 
m
V
(1.9)
Although mass and volume are both extensive properties, the ratio yields a specific property called
density. The specific volume v has been defined as the volume per unit mass of the matter. Thus, it
is the reciprocal of the density.
v
V 1

m 
(1.10)
b. Pure intensive properties are those intensive properties which are not specific. Temperature,
pressure, and velocity are examples of pure intensive properties.
1.5
The Pressure and Temperature of a System
1.5.1 Pressure and its measurement. The pressure and the temperature are two important properties and are frequently used in engineering analysis of systems. The pressure, p, in a fluid is defined
as the normal component of force per unit area acting on the boundary of the system. Denoting  A
as the smallest area so that fluid continuum is not violated, and  Fn is the force component normal
to surface  A , then the pressure, p, is defined as,
p  lim A A
 Fn
A
(1.11)
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 9
The SI unit for pressure is the Pascal Pa. Considering the definition of pressure, 1 Pa=1 N/m2,
since Pascal is relatively a small unit of pressure, multiples of Pascal, 1kPa (kilopascal)= 103 Pa, and
1MPa (mega-pascal)=106 Pa, are also used. Although not being within the Internationale System e
, two other units are widely used in industrial applications. These are the bar, 1 bar=105 Pa, and the
standard atmosphere, 1 atm  101325 Pa .
The thermodynamic pressure at a particular point in a system is called absolute pressure, and
measured relative to absolute zero pressure. However, most pressure measuring instruments indicate
the difference between the absolute pressure and the atmospheric pressure existing at the gage. This
reading is referred to as gage pressure. Instruments for measuring pressures above and below atmospheric level are respectively called manometer and vacuum meter.
Referring to Fig 1.8, the relation between the absolute pressure, p, and the pressure value indicated
by a manometer, pm, or a vacuum meter, pv, is as follows,
p  po  pm
(1.12)
p  po  pv
(1.13)
Pressure is usually measured by transferring its effect to a deflection through the use of a pressurized area and either a gravitational or elastic restraining element. Therefore the pressure measuring
instruments may be classified as following: 1.Direct acting elastic types, 2.Gravitational types, and
3.Electrical pressure transducers.
1. Direct acting gauges.The most common direct acting elastic type gauges are the aneroid barometer and the Bourdon gage.
10
THERMODYNAMICS
As shown in Figure 1.9, an aneroid barometer has a vacuum chamber with an elastic surface.
When pressure imposed on its surface, it deflects inward, and the needle rotates accordingly. Hence
it measures the absolute pressure. A Bourdon gage manometer, as presented in Figure 1.10, is a thin
walled metal tube bent into a form of C, and one end of C is fixed, the other end is closed but it is free
to move. When pressure is applied at the fixed end, the tube deflects like the deflection of the snake
like paper whistle and the pointer at the free end indicates the gage pressure.
The Bourdon gauge is a highly accurate but rather delicate instrument. It may be easily damaged.
In addition, it malfunctions if the pressure varies very rapidly.
2. The Gravity Type Manometers. As shown in Figure 1.11, the gravitational type manometers
are basically three types:
a. The simple type, b. The differential type, and c. The inclined type. One end of these manometers
is connected to a point where the pressure has to be determined. The other end is either open to atmosphere or has a connection to a point where the pressure difference to be measured. Depending upon
the pressure difference, liquid column rises in one direction. When equilibrium is reached, the force
balance on the manometer yields the following relations,
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 11
Case a:
p A  po   gh
Case b:
p A  pB  3 gh3   2 gh2  1 gh1
Case c:
p A  pB   gh sin 
(1.14)
Equation (1.14) also provides a method for measuring the pressure in terms of a liquid column
height. For instance, it might be shown that 1 bar of pressure difference corresponds to 9.8 meters of
water column.
3. The pressure Transducers. The electrical pressure transducer, Figure 1.12a, is a device that
converts displacement of a diaphragm to an electrical signal from which a reading can ultimately be
derived. For industrial applications, strain gage is more popular technology applied to this type of
transducers. The strain gage is constructed from either a metal foil or a semiconductor and bonded
to a pressure gathering diaphragm by using high strength epoxies. Generally four strain gages are
configured into a Wheatstone bridge, Figure 1.12b, on the diaphragm. When the voltage is applied
across points A, and C, the resistance change in the strain gages due to pressure causes a change
in the voltage output between B, and D. This voltage output is linearly proportional to the applied
pressure. As shown in Figure 1.12c, the pressure gathering diaphragm can be steel or ceramic and
is circular in shape. Since the response time of these instruments is very small, they are especially
suitable for applications at which pressure varies rapidly. The outputs of these transducers can be
recorded electronically.
Example 1.5: The vacuum tank in Figure 1.13 is fitted with a mercury manometer. When the tank is pumped down, the
manometer reads 745 mm. Inside the vacuum tank, there is a chamber which is divided into two compartments, and the
installed pressure gages read pA=4 atm, pB=1.5 atm. The atmospheric pressure is 1 atm.
a.
Determine the absolute pressures in two compartments
b.
Find the reading of gage C in atm.
c.
Determine the minimum force required to lift the lid up.
12
THERMODYNAMICS
Solution:
a.
Applying Eq.(1.12) to the manometer, p0-p3=745, and since 1 atm=760 mmHg, then p3=0.019 atm. Similarly, for gages
A and B, pA=p1-p3, pB=p1-p2, and the absolute pressures of the compartments are: p1=4.019atm, p2=2.519atm.
b.
Gage C indicates the pressure difference between 2 and 3. Thus, pC=p2-p3, pC=2.5 atm.
c.
The force to be applied to the upper lid is
d2 
F= 
 ( p  p3 ) ,
 4  0


and substitution of numerical values yields
F=1560.2 kgf
Example 1.6: To keep the gate in its vertical equilibrium position of Figure 1.14, what must be the density of the liquid
on the left tank for the given dimensions? Take the width of the plate W=1m.
Solution:
To keep the gate in its vertical equilibrium position, the torque created by hydrostatic forces on both sides of the plate must
be balanced. Hence,
5.5
5.5
1.5  g x  1.5dxW x  2.5 w g x  2.5dxW x
After simplification of the above expression, and completing the integrals for the indicated limits, one may obtain the
following,
33.332   20.240  w
or
  607.49 kg/m3
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 13
1.5.2
Temperature and its measurement.
Temperature is the property of a system that indicates the potential for heat transfer with other
systems. Therefore, two systems are said to be equal in temperature when there is no heat transfer
among them.
Definition: Thermal equilibrium of systems is characterized by the equality and uniformity of temperature.
Principle 2: If two systems are each equal in temperature to a third system,
then the temperatures of these two systems are equal. This principle is also
called zeroth law of thermodynamics.
This principle is utilized in measuring the temperature of systems by thermometers, thermocouple
wires, and other instruments explained below.
The SI unit for temperature is Kelvin (K), and the absolute temperature scale is called Kelvin
scale. The triple state of water (a state in which water vapor, liquid, and solid phases all coexist in
equilibrium) is internationally accepted to be 273.16 K on the Kelvin scale. Thus, at atmospheric
pressure, water freezes at 273.15 K. The relationship between the Kelvin and the Celsius temperature
scale is:
K= °C + 273.15
(1.15)
The following instruments are used for industrial applications of temperature measurements:
1.Thermistors, 2.Thermocouples, 3.RTD (Resistance Temperature Detectors), 4.IC Sensors,
5.Bimetalik Indicators, 6.Optical Sensors: a. Poyrometers, b. Infrared detectors, c. Liquid crystals,
7.Liquid Bulb Thermometers, 8.Gas Bulb Thermometers. Among these, the most versatile ones are
the thermistors, the thermocouples, the RTD’s, and the IC sensors.
Example 1.7: The density of mercury changes approximately linearly with temperature as,  m  14277.5  2.5T ( K ) . Due
to influence of temperature, the same pressure difference will be measured by different manometer heights. Suppose in New
York City, on a hot summer day the temperature is 40C and the pressure is the same as the pressure measured on a cold
Winter day of -10C. What will be the percent deviation in the manometer reading?
Solution:
Since we measure the same pressure for both cases, in accord with Eq. (1.14), 1gh1   2 gh2 , and
h2 1
. On the other

h1  2

h1  h2
 
 100  1  1   100 . Since T1=313K, and
h1

2

T2=263K, by stated formula, 1/2=0.9908. Then the percent error becomes, err%=0.91% which is less than 1-percent.
hand the percent deviation in height may be expressed as, Err% 
1.Thermistors.A thermistor, Figure 1.15a, is a semiconductor material with a well defined variation
of electrical resistance with temperature. The relationship between the temperature and the resistance
change is given as,
T  K R
(1.16)
As shown in Figure 1.15c, for most of thermistors, the temperature coefficient ( K  0 ) is negative
(NTC). That is the resistance decreases as the temperature increases.
14
THERMODYNAMICS
Since the thermistor output is directly related to the absolute temperature, there is no need for a
reference junction, and a calibration curve as given in Figure 1.15c is sufficient to convert the resistance
measured by the circuitry in Figure 1.15b to the temperature. The main disadvantage of a thermistor
is the possible self heating error in case of repeated measurements. Due to slow response, they are
more difficult to apply to transient processes.
2.Thermocouples.As shown in Figure 1.16, any two unlike conducting materials could be used
to form a thermocouple.
Principle 3: If the ends of a junction formed by two different conducting materials are at different temperatures then a potential
difference develops across the junction.
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 15
The net electromotive force generated on the circuit is due to the temperature gradient along the
wire, and is called the Seebeck effect. The Seebeck coefficient,  for a thermocouple wire is defined
as,
d


  dT



T  T   
i
0
 

(1.17)
where  indicates the electromotive force difference measured by a voltmeter, and To is the
reference junction temperature. In Eq.1.17, the nominal values of Seebeck coefficients () for certain
combinations of materials and the corresponding temperature ranges are provided in Table 1.2.
Table 1.2 The most common thermocouples and their temperature limits
Thermocouple
type
Metal
Seebeck
Coefficient(μV/oC)
Temperature
Range(oC)
Constantan
50
-210 to +760
Nickel
39
-270 to +1372
Constantan
38
-270 to +400
+
−
J
Iron
K
Nickel Chromium
T
Copper
Since the  values of thermocouples are small, the voltage outputs are also small. The output values
are typically in the milli-volt range. The size of the thermocouple wire is of some importance. Usually
the higher the temperature to be measured, the heavier should be the wire. As the size is increased,
however, the time response of the wire to temperature change increases and the couple becomes bulky.
Hence, some compromise between the response and the thermocouple life is required.
The thermocouple calibration data is used for determining the temperature corresponding to a
particular measured potential difference. As shown in Table 1.3, the experimental data for a particular
couple is tabulated and, in Figure 1.17a, the curve fit is used for temperature determinations.
Table 1.3 Calibration data for T-Type thermocouple
ξ (mV)
T(oC)
ξ (mV)
T (oC)
ξ (mV)
T (oC)
0.0
0
6.704
150
14.861
300
2.035
50
9.288
200
17.818
350
4.278
100
12.013
250
20.872
400
Thermocouples are only capable of measuring the temperature difference. To measure the temperature of an object, we need a known reference temperature. As shown in Figure 1.17b, the reference temperature is taken to be the temperature of ice and water mixture at sea level and is called
“ice point reference junction”. The use of large number of thermocouples in a particular application
is shown in Figure 1.18. This figure also describes the use of a single recording system by the zone
box application to the circuitry.
16
THERMODYNAMICS
Example 1.8: The temperatures at four points of an air-conditioning unit are measured by using copper-constantan thermocouples. The reference junction temperature is recorded as 20oC. If emf outputs in mV of these thermocouples are -1.620,
-1.053, +0.181, +2.215, determine the corresponding temperatures.
Solution:
Temperatures may be determined either by using Eq. (1.17) or by calibration curve values in Table 1.3. Since for large
temperature differences thermocouple behavior is nonlinear, the results will be more dependable, if the values in Table 1.3
are used.
From Table 1.3, (1)=2.035mV corresponds to a temperature difference of 50oC, and employing the linear interpolation method for
1=-1.620mV, one may find, T1  To  39.803 , where To, is the reference temperature of 20oC, and the measured
temperature becomes, T1=-19.803 oC.
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 17
Since the following two measurements are in same range of temperature calibration, the temperature values can be
determined by applying the same method and the corresponding temperatures become, T2=-5.87 oC, T3=24.44 oC.
For the last temperature, 4=2.215mV > 2.035mV, the difference has to be evaluated by the second calibration value
in Table 1.3. 4-(1)=2.215-2.035=0.18mV, on the other hand, (2)- (1)=4.278-2.035=2.243mV. Hence, 0.18mV correspond to 4.012oC. T4  To  50  4.012 , T4=74.012oC.
3. Resistance Temperature Detectors (RTD). Similar to thermistors, a resistance temperature detector
(RTD) is a thermally sensitive resistor composed of semi-conductor material. Because of being chemically stable, easy fabrication, and reproducible electrical properties, the platinum resistance sensor is
the most acceptable sensor. As shown in Figure 1.19, to eliminate the negative effect of connection
wires usually four-wire circuitry is used. Hence the measurement depends neither on the line resistance
nor on their variations due to temperature. In addition, no line balancing is required.
The operational principle of RTD is as follows: A digital multi-meter (DMM) uses a known current source to create a potential difference. The voltage drop across the RTD is independent of the
properties of the connecting wires. The voltage drop across RTD varies as the resistance changes in
accord with the temperature measured.
RTD’s are positive temperature coefficient (PTC) sensors whose resistance increases with temperature. The platinum resistance thermometers can cover a temperature range -200oC to +800oC, and
they are the most accurate sensors for industrial applications.
As explained above there are several different temperature-sensing technologies available for the
applicant to select the appropriate one. To find out the right technology, however, depends on the
characteristics of the target temperature (for instance; the number of measurement points, steady or
unsteady measurement, etc.), and on the system requirements such as cost, circuit size and design
time. In Figure 1.20, a comparison of the advantages and the disadvantages of these three temperature
measurement systems is presented and discussed as following:
Advantages
1. Thermocouple:
2. Thermistor:
a. Self powered
High output
b. Simple
Fast
c. Inexpensive
Two-wire ohm measurement
d. Variety of physical forms
3. RTD:
Most stable
Most accurate
More linear than thermocouple
18
THERMODYNAMICS
Disadvantages
1. Thermocouple:
2. Thermistor:
3. RTD:
a. Nonlinear for a wide range
Nonlinear
Expensive
b. Low voltage
Limited temperature range
Slow
c.
Fragile
Current source required
d. Least stable
Current source required
Small resistance change
e.
Self heating
1.6
Reference required
Least sensitive
The State Concept
Depending upon the system properties at an instant time t, the state signifies the condition of the
system at that instant. Therefore, specifying the thermodynamic state of the system is identical to define each extensive property at every location within the system at time t. This description is general
in defining the state of a system. However, it is sometimes convenient to have a local description in
terms of intensive properties.
Definition: The intensive state is the state of a point (x, y ,z) at time t, and is specified by all intensive
properties at that instant.
Homogeneous system. The system is said to be homogeneous at time t, if its intensive state is
the same throughout the system. Thus, for a homogeneous system, the intensive properties are independent of location. Hence, the pressure, the temperature, the density, etc. are all uniform throughout
the system. For instance, in transient temperature analysis of a copper block, a homogeneous system
model may appropriately be employed.
Steady-state system. If all properties of a system are independent of time, then the system is
called a steady-state system. In this case, the intensive properties of the system are time invariant,
but they may vary with position.
Example 1.9: Consider the heat exchanger of Fig.1.21, the cold water flows through the tubes of the exchanger and steam
condenses on the shell side. Both fluids are at constant flow rate. Define the system and the state whether the system is
steady or unsteady.
Solution:
Defining the exchanger as a system, it is a steady-state system. For instance, the water pressure assumes the value of p1 at
the inlet, and p2 at the outlet. These pressures are invariant with time. Because of fluid friction, however, the pressure at a
certain point along the flow is always less than the inlet value. Thus, the pressure exhibits a local variation.
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 19
1.7
The Equilibrium Concept
Definition: A thermodynamic equilibrium of a system is a state that cannot be changed without
interactions with its environment.
Principle 4: If two systems are in thermodynamic equilibrium with each other then they
are said to be in mechanical equilibrium (equality of pressures), in thermal equilibrium
(equality of temperatures), and in chemical equilibrium (equality of gibbs function) etc.
A system might be in mechanical equilibrium but not in thermodynamic equilibrium. Consider
a system consisting of two identical copper blocks one at the top of the other, isolated from environment, and initially at different temperatures. Such a system is obviously in mechanical equilibrium (all
forces are balanced) and cannot change its position by itself. However, this system is not in thermodynamic equilibrium. Due the temperature difference energy interaction will take place between the
blocks, and the system will change its state without interacting with the environment. The hot block
will cool down and the cold will get hot. When the temperatures of both blocks become uniform, the
thermodynamic equilibrium will be attained.
Like a ball in gravitational field, as described in Figure 1.22, a system might possess three different equilibrium states. The metastable equilibrium is a state that a finite change of state of the
20
THERMODYNAMICS
system may be produced by an infinitesimal change of state of the environment. There is always a
high possibility that the system might not return to its initial state. Thermodynamics is restricted to a
large degree to systems in stable states.
Definition: A system is said to be in stable equilibrium state if and only if a change of state of the
system is attained by a corresponding finite change in its environment.
In Figure 1.22, the state (3) where the ball is at the bottom of the curved surface is the stable
equilibrium state. The position of the ball can only be changed if it interacts with environment and
finite amount of energy is consumed. The system and its environment may always be return to their
initial states
1.8
The Process Concept
A process occurs when a system undergoes a change of state with or without interactions with its
surroundings. During the change of state, the system passes through a succession of states that form
the path of the process. Thus, the complete description of a process requires a specification of the
initial and final states, the path, and the type of interaction between the system and its surroundings
during the change of state.
Since the properties of a system define the state of the system only if equilibrium exists, how can
one describe the intermediate states of the process path if the actual process occurs only when equilibrium does not exist? This difficulty is overcome by the definition of a quasi-equilibrium process. For
a quasi-equilibrium process, the deviation of an intermediate state from equilibrium is infinitesimal.
Example 1.10: Consider a gaseous system in a piston-cylinder device. The gas is compressed by replacing small weights
one by one on the piston. Discuss whether the change of state is a process or not.
Solution:
In Figure 1.23, the initial and the final states of the gas is represented by (1) and (2) respectively. Considering the occurrence
of the process from the beginning to the end, all intermediate states are definable, and so is its path. Hence, the gaseous
system undergoes a quasi-equilibrium process.
Processes during which one property remains constant are designated by the prefix iso- before
the property. For example, a process for which the temperature is constant is called iso-thermal,
similarly the constant pressure process is called isobaric, and the constant volume process is called
iso-volumic process.
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 21
At the compressed state, state 2 in Fig 1.23b, if all the weights are removed at once, a rapid rising
of the piston will result with a spontaneous expansion of the gas. This type of a process is called
a non-equilibrium process. For a non-equilibrium process, the process path is not mathematically
definable, and only the end states before and after the process can be described.
References
1.
Y. A. Cengel and M. A. Boles, Thermodynamics An Engineering Approach, 5th edition, McGraw Hill Publications,
ISBN 978-0-07310-7684, 2005.
2.
I. Müller, A History of Thermodynamics, The Doctrine of Energy and Entropy, Springer-Verlag, ISBN 978-3-54046226-2, 2007.
3.
S. J. Blundell and K. M. Blundell, Concepts in Thermal Physics, Oxford University Press, ISBN 978-0-19-856770-7,
2006.
4.
P. R. N. Childs, Practical Temperature Measurement, Butterworth-Heinemann, ISBN 0-7506-5080-X, 2001.
5.
L. A. Gritzo and N. Alvares, Thermal Measurement: The Foundation of Fire Standards, American Society for Testing
and Materials International (ASTM), ISBN 0-8031-3451-7, 2003.
6.
D. J. C. Vazquez, and M. C. Sancho, Thermodynamics of Fluids under Flow, 2nd Edition, Springer Science, ISBN
-978-94-007-0198-4, 2011.
7.
“Instrumentation Reference Book, 3rd Edition, Edited by W. Boyes, Butterworth-Heinemann, ISBN -0-7506-7123-8,
2003.
Problems
Concepts
1.1
1.4
Which of the following represent a system in the
thermodynamic sense? For each that is a system,
describe the system boundaries.
a. System: The filament of an incandescent
lamp.
i. The mass, ii. The diameter, iii. The number of
hours of operation, iv. The electrical resistance,
v. The total watt-hours consumed.
a. An explosion, b. A bicycle pump, c. Two kilograms of air, d. A wave on the surface of a lake, e.
A force, f. An automobile, g. The volume inside an
evacuated tank, h. Five meters of copper wire, i. A
flow through a tube.
1.2
Which of the following are properties of the specified
system and which are non-properties?
b. System: A dry cell battery.
i. The mass, ii. The volume, iii. The voltage,
iv. The mass of each element in the battery.
Draw a schematic of the following systems and label
the boundaries. Also label each system as open, or
closed.
c. System: A clock spring.
i. The torque on the output shaft, ii. The total
energy transferred to the spring by the input
shaft, iii. The volume.
a. Rotating propeller of an air plane,
b. water pump in operation,
c. pressure cooker,
1.5
A system is left alone for a long time. During this
time, no mass, and no energy transfer have crossed
its boundary. May we state that this system is at
equilibrium? Explain.
1.6
A water tank used in a residential area initially
contains 120 L of water (ρw=1000kg/m3). The tank
outlet valve opens for watering the lawn at a rate of
10 liters per minute and meantime water is supplied
into the tank at a rate of 0.5 liters per second. Considering the mass of water in the tank as an extensive
property, evaluate the amount of water left after 10
minutes of operation.
d. electric light bulb in operation,
e. steam boiler for building heating including all
piping and radiators.
1.3
Three cubic meters of air at 25°C, and 1bar have a
mass of 3.51 kg.
a. List the values of three intensive and two extensive properties for this system.
b. If the local gravity g is 9.8m/s2, evaluate the
specific weight of the system as a property.
22
THERMODYNAMICS
Mass, volume, density
1.7
The density of air at atmospheric conditions of 1 bar,
and 20°C is 1.2 kg/m3. Calculate the amount of air
in kg in a conference room which has dimensions
20 m  15 m  3 m .
1.8
On the surface of the moon where the local gravity
g is 1.67 m/s2, 3.7 kg of a gas occupies a volume of
1.25 m3. Determine,
layer is uniform and is 3 mm. If 70-percent of the
tank volume is filled with water (ρw=1000kg/m3),
determine the total weight of the tank.
a. the specific volume of the gas in m3/kg,
b. the density in g/cm3,
c. the specific weight in N/m3.
1.9
The acceleration of gravity as a function of elevation
above sea level is given by g  9.807  3.32  106 z ,
where g is in m/s2 and z is in meters. Find the height,
in kilometers, above sea level where the weight of
a person will have decreased by a. 3 percent, b. 10
percent.
1.13
A pressurized tank of ammonia contains 12 kg of
liquid and 1.01 kg of vapor ammonia. Liquid ammonia
occupies a volume of 19.65L, and the remainder of the
tank volume is filled with vapor. For vapor specific
volume of 0.1492 m3/kg, define and determine,
a. the system and its boundaries,
1.10
b. the total volume of the tank,
A gas at 0.12 MPa is contained within a vertical
cylinder by a weighted piston of mass m, and 350
mm2 cross-sectional area. The outside atmospheric
pressure is 1 atm. Determine the value of mass m
in kilograms, if the local acceleration of gravity is
9.78 m/s2.
c. the specific volume of the liquid,
d. the density of the vapor,
e. the specific volume of the system consisting of
liquid and vapor mixture
Pressure
1.11
1.12
Two columns are connected to the same vacuum pump
as shown in Fig 1.24. One column contains water
and stands at 20 cm. The liquid containing column
stands at 32 cm. If the specific volume of water is
0.001m3/kg, then find the density of the liquid.
A polyethylene plastic water storage tank in Figure
1.25 (ρp=1600kg/m3) is cylindrical in shape with
an outside diameter of 120 cm, and a height of 150
cm. At all cross sections, the thickness of the plastic
1.14
A steel (ρs=7860 kg/m3) tank in Figure 1.26 has a
cross-sectional area of 3m2, 16m height and weighs
98100N and is open at the top. We want to float it in
the ocean (ρsw=1150kg/m3) so it sticks 10m straight
down by pouring concrete (ρc=1860 kg/m3) into the
bottom of it. How much concrete should we put in?
1.15
If the local atmospheric pressure is 920 millibar,
convert, a. an absolute pressure of 2.5 bar to a gage
reading in bar, b. a vacuum reading of 600 millibar
to an absolute value in bar, c. 0.75 bar absolute to
millibar vacuum, d. an absolute reading of 1.45 bar
to a gage reading in kilo Pascal.
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 23
1.16
A steam turbine is supplied with steam at a gauge pressure of 1.35MPa. After expansion through the turbine,
the steam flows into a condenser that is maintained
at a vacuum of 700 mmHg. The barometric reading
of the outside pressure is 750 mmHg. For mercury
density of 13600 kg/m3, express the inlet and the outlet
pressures of steam in kilo-Pascals absolute.
1.17
A submarine is cruising at a depth of 200 m in sea
water with a density of 1035 kg/m3. If the inside of
the submarine is pressurized to atmospheric pressure,
determine the pressure difference across the hull in
kilopascals for the local gravity of 9.75 m/s2.
1.18
The pressure rise due to wind striking a window
of a building p is approximated by the formula
p  V 2 / 2 , where ρ represents the air density,
and V is the wind speed. For air density of 1.2kg/
m3, calculate the force applied to a window of 3 mx2
m if the wind blows with a speed of 80 km/h.
1.19
A water manometer shown in Fig.1.27 is used to
measure the low pressure in a natural gas main. The
water level is 7 mm higher in the right-hang tube.
Determine the absolute pressure of the natural gas
in Pa, if a closed-tube barometer measuring local
atmospheric pressure has a reading of 748 mm of
mercury.
1.20
Two vacuum tanks are connected as shown in Fig.1.28.
Each tank is also connected to a separate vacuum
pump. If the atmospheric pressure is 755 mmHg, a.
Determine the readings on the pressure gages 1, 2,
and 3. b. Evaluate the absolute pressure of each tank.
c. The tanks are sealed to the base plate by rubber
gaskets which are modeled as ideal springs. For the
gasket detail in Fig.1.28b, if the gasket spring constant
is K=3x106N/m, evaluate the displacement of the
gasket after the tank is pumped down. Readings on the
manometers are: L1  25 cm and L2  15 cm .
1.21
As shown in Figure 1.29, the pressure at the bottom
of a pressurized water tank is measured by a multi
fluid manometer containing water (ρw=1000 kg/
m3),oil (ρo=800 kg/m3), and mercury (ρm=13600
kg/m3). Determine the pressure caused by the air
on water surface for h1=0.25 m, h2=0.32m, and
h3=0.51 m.
1.22
The mercury manometer of Fig.1.30 measures the
pressure difference between points 1 and 2 in a flexible
pipe through which water flows. Let the densities of
water and mercury respectively be 1000kg/m3, and
13600kg/m3, and calculate,
a. the pressure difference between points 1 and 2,
24
THERMODYNAMICS
b. the absolute pressure of point 2 for a pressure
reading of 2 atm. on manometer A.
1.25
Assume the outside pressure to be at 760 mmHg.
An inclined manometer in Figure 1.11c is always
much more sensitive than a simple manometer.
Determine the angle of inclination for making an
inclined manometer that is ten times as sensitive as
a simple manometer.
Temperature
1.26
1.23
A bicycle rider in Figure 1.33 has several reasons
to be interested in the effects of temperature on air
density. First of all, the aerodynamic drag force decreases linearly as the density decreases. Secondly,
the tire pressure will be affected by the change in
air temperature.
Figure 1.31 shows a schematic of a hydraulic testing
machine. The machine is designed to produce 1800N
at point B and 300MPa on the specimen. What is the
area ratio of sections A to B?
a. The variation of air density at atmospheric pressure (p0=100kPa) with respect to temperature is
approximated as,   348.432 / T (kg/m3). Write
a computer program to estimate the air density
for a temperature range between -15°C and 45°C
with 5°C increments at atmospheric pressure.
b. Considering the fact that the volume of the tire
does not change with temperature, the density of
air is approximately constant and is 5.946 kg/m3
at 500kPa, 20°C of tire respectively pressure and
temperature. Hence the pressure and temperature of
tire air may be related as p kPa   1.706T . Write
a computer program to estimate the tire pressure
for the same temperature range (-15°C to 45°C).
c. Graph your results for both cases and discuss
what engineering insight you gain from these
calculations.
1.27
1.24
The U-tube manometer in Figure 1.32 has a 1 cm
inside diameter and contains mercury. If 20 cm3 of
water is poured into the right-hand leg, what will the
free-surface height in each leg be after the sloshing
has died down?
An ice-bath reference junction is employed in conjunction with a copper-constant thermocouple. Using
the data of Table 1.3,
a. Draw a calibration curve for type-T thermocouple.
b. The following millivolt outputs are read for
four different conditions:-4.334 mV, 0.00 mV,
+8.133 mV, and +11.13mV, determine the corresponding measured junction temperatures.
CHAPTER 1 BASIC CONCEPTS & DEFINITIONS 25
1.28
1.29
Type-T thermocouples are employed for measuring
the temperatures at various points in air conditioning
system of a building. The reference junction temperature is taken to be 22°C. The following emf output
are supplied by various thermo-couples: -1.623 mV,
-1.088 mV, -0.169 mV, and +3.250 mV. Determine
the corresponding junction temperatures by linearizing
the calibration curve given in Figure 1.17a.
i.
It is always possible to define the path
of a quasi-equilibrium process.
j.
To define the end state properties of a
system, the path of a process has to be known.
k.
A thermal equilibrium within systems
is established by the equality of temperatures.
The temperature difference between the inlet and
outlet of a heat exchanger has to be measured. The
measuring and the reference junctions of type T
thermo-couple are embedded within the inlet and
outlet sections of the exchanger and an emf of
+0.395mV is read.
l.
A mechanical equilibrium of two systems
requires the equality of pressures.
m.
Two end states are sufficient to identify
a process
n.
In measuring unsteady pressures usually electromechanical transducer methods are
preferred.
o.
A pressure pick up should be insensitive
to temperature change and acceleration.
p.
The emf created by a thermocouple
with junctions at T1 and at T2 is not affected
by a temperature elsewhere in the circuit.
q.
If a thermocouple produces emf E1 when
its junctions are at T1 and T2, and E2 when at
T2 and T3, it will produce emf of “ E1  E2 ”,
when the junctions are at T1 and T3
r.
Resistance thermometers are mainly
stem sensing devices with a finite sensing length
and as such are best suited to immersion use.
s.
A thermistor is composed of ceramic like
semi-conducting material which has thermally
sensitive resistance.
t.
A thermistor may also be used as a
compensating electrical circuitry for fluctuating
ambient temperatures.
u.
A system is in thermodynamic equilibrium if the temperature, and pressure at all
points are the same and there is no any velocity
gradients.
v.
A property of a system is a characteristic
of the system which depend on how the system
reached the end state.
a. Is this a correct way of measuring the temperature
difference? Explain.
b. What additional information is essential and what
procedure may be followed to get an answer for
the temperature difference? Explain.
True and False
1.30
Answer the following questions with T for true and
F for false.
a.
No mass can flow across a system
boundary.
b.
If the absolute pressure in a tank
is 850mmHg and the atmospheric pressure
is 760mmHg, a pressure gage would read
0.25atm.
c.
A system boundary is defined as part
of the system that has rigid walls.
d.
In a closed system, convective transfer
of property is done at the moving boundary.
e.
A system is said to be at steady-state if
its mass does not change with time.
f.
Temperature is an extensive property,
while the system mass is intensive.
g.
Spontaneous processes are treated as
quasi-equilibrium processes.
h.
In a homogeneous system, the system
properties may vary with time and location.
26
THERMODYNAMICS
Check Test 1
Choose the correct answer:
1.
2.
An isolated system is a region where
7.
a. the transfer of energy and mass can take
place,
b. only energy may cross the boundaries,
c. no transfer of energy nor mass is allowed across
the boundaries,
d. the mass within the system is not constant.
An open system allows
a. both the energy and mass cross the boundary of
the system,
b. only energy cross the boundary but no mass,
c. only mass cross the boundary but no energy,
d. neither energy nor mass cross the boundary.
8.
If p, v, and T are properties of a system and R is a
constant. With respect to following relations, which
one describes a change in property, and what is that
property?
1
a. d  dv  Rdp ,
T
b. d   pdv  vdv ,
c. d  pdv  vdp ,
d. d  RdT  pdv .
9.
The approximate number of stories of a skyscraper
may be determined by a barometer. The readings at
the bottom and the top of the building respectively
are 98.2 kPa, and 96.505 kPa, and if the height of
one floor is 3 meters, then the number of floors on
the building are:
a. 47,
b. 48,
c. 49,
d. 50.
10.
The deepest point of an Olympic swimming pool is
2.2 meters below the water surface. The maximum
pressure difference between the top and the bottom
of the pool is
a. 0.125 bar,
b. 0.205 bar,
c. 0.215 bar
d. 0.255 bar
11.
State which one of the following is correct
a. pabs  pgauge  patm ,
Which one of the following is an intensive property
of a system?
a. Temperature,
c. Volume,
3.
Which one of the following is an extensive property
of a system?
a. Density,
c. Temperature,
4.
6.
b. Pressure,
d. Mass.
When two systems are in thermal equilibrium with
a third system, these two systems are said to be in
thermal equilibrium with each other. This statement
is called:
a. Kelvin Planck’s law,
c. Seebeck’s law,
5.
b. Mass,
d. Energy.
b. Euler’s law,
d. Zeroth law.
The temperature at which the volume of a gas vanishes
is called:
a. Absolute zero temperature,
b. Absolute temperature,
c. Absolute scale of temperature,
d. Gas volume cannot vanish.
b.
pgauge  pabs  patm ,
The absolute zero pressure will be obtained
c.
patm  pabs  pgauge ,
a.
b.
c.
d.
d.
pabs  pgauge  patm .
at sea level,
at the center of earth,
under vacuum conditions,
when the system molecular momentum becomes
zero.
12.
The absolute zero temperature is
a. 273°C,
b. 237°C,
c. -373°C,
d. -273°C.
C
H
2
A
P
T
E
R
Thermodynamic Properties of Systems
2.1
Introduction
Without any reference to a system of being either closed or not, the energy interaction
with other systems causes a change in the state of the system. As shown in Figure 2.1, during these energy interactions, one or several properties of the system might be altered and
then the system may reach to a new equilibrium state. The reverse is also true. That is, if a
system changes its state of equilibrium then the system must be in energy interaction with
some other system(s).
Hence there is a strong relation between the state of a system and the amount of energy
interaction. To determine the new state of a system due to energy interaction, or to find
out the amount of energy transferred due to change of its state, we have to have detailed
information about the properties of the system and cast a methodology for determining the
system properties. The purpose of this chapter is to provide information on the equilibrium
thermodynamic properties of homogeneous systems based on experimental observations.
27
28
THERMODYNAMICS
Among all the thermodynamic properties, there are some which are related by definitions. For
example, enthalpy, H is related to the internal energy U, the volume V, and the pressure p of a system as,
H = U+pV
(2.1)
Since the specific enthalpy is h = H/m, it may be described by the specific internal energy and
the volume as,
h = u+pv
(2.2)
The specific Gibbs free energy and the Helmholtz free energy are accordingly defined as,
g = h-Ts
(2.3)
a = u-Ts
(2.4)
where “s” is the specific entropy of a simple system. The physical meaning and the use of entropy
is explained later in detail in Chapter 5.
Certain other properties are related as a result of either experimental observations or of the first
and second laws of thermodynamics. These relations are called equations of state. One particular
Example of an equation of state is the relation among the pressure, the temperature, and the specific
volume of a fluid (liquid or gas) and described in alternative forms as following,
 p  p( v, T ) 


 v  v ( p, T ) 
T  T ( p, v ) 


(2.5)
It has to be reemphasized that these relations are deduced by empirical means and interrelate only
the intensive properties of a system.
2.2
The State Principle
As shown in Fig. 2.2, the thermodynamic condition of a point M(x, y, z) in a system at time t is described
by all the intensive properties. Some of these intensive properties are internal (T,p,v,ρ,u,h,...), and some are
external (V, ke, pe, ...). To define the state of a homogenous system, however, it is not necessary to know all
the intensive properties. Some are interrelated by definitions, and some are described by the state equations.
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 29
The minimum number of independent properties required to define the state of a system is determined by
the state principle. Although this principle is based on experimental observations of so many systems, it is
regarded as a fundamental law of thermodynamics. The state principle stated rather simply as,
Principle 5: Any two independent intensive properties are sufficient to establish the
thermodynamic state of a simple system. For each identifiable departure from the
requirements of a simple system, one additional independent property is necessary.
This principle requires the definition of a simple system.
Definition: A simple system is one for which in the absence of magnetic, electrical, shear strain, and
gravitational effects has a homogeneous and invariable chemical composition. It may exist in more
than one phase.
A mixture of liquid water and water vapor, or a mixture of ice and liquid water are all simple systems. Sometimes, a mixture of gases such as air may be considered as simple system as long as there
is no change of phase. For these systems, the minimum number of intensive properties required to fix
the state of the system is two. However, let us consider a system which is a mixture of air and water
vapor in equilibrium with liquid water, to define the state of such a system at least three independent
intensive properties like, pressure, temperature, and humidity ratio, are to be specified. As the complexity of the system increases, the number of properties for defining the system also increases.
Another important point about the state principle is that it describes which intensive properties
have to be considered as the independent properties. It is known from experimental observations that
during a phase change, the pressure and the temperature of a system are dependent. Considering their
measurement simplicity, the independent properties are depicted form among pressure, temperature,
and volume. Thus, for a simple system, if temperature T, and volume v are selected as independent
properties, the other intensive properties, according to the state principle, can be evaluated and the
relationship may be stated as following; p=p(T,v), h=h(T,v), u=u(T,v).
2.3
The p-v-T Behavior of Systems
Considering the state principle for a simple system, an intensive property is a function of two
other independent properties. That is, P1 = f(P2,P3), where P, in general, is any intensive property.
From a mathematical point of view, any equation involving two independent variables (such as P2,
30
THERMODYNAMICS
and P3) can be represented in a Cartesian space as a surface. As a consequence, the equilibrium states
of a simple system can be represented as a surface in space, where geometric coordinates are the
intensive properties of that state. Taking the pressure p, and the entropy s as independent properties, the enthalpy h of a simple system becomes the function of the pressure and entropy; h = h(p,s).
However, because of exhibiting the basic structure of matter, p-v-T surface of a simple substance will
be studied in this section.
In Fig. 2.3, two unique p-v-T behavior of matter is shown. As in the case of metals, some matter
contract on freezing. In Fig. 2.3a, it may be noted that a step decrease in specific volume takes place
when going from liquid to solid state. Water expands on freezing and Fig 2.3b schematically represents
the p-v-T surface of water. It may be noted that a step increase in specific volume takes place when
going from liquid to solid state.
A detail understanding of the following terms is important in studying the p-v-T surface of matter:
a. Phase (solid, liquid, gas), b. Sublimation, melting, vaporization, c. Triple point, d. Saturated
liquid, saturated vapor, liquid-vapor mixture, e. Critical point, f. Saturation temperature at a given
pressure, h. Superheated vapor, i. Compressed liquid, or sub-cooled liquid, j. Interpolations.
In addition, it is equally important to visualize and locate the states on the three coordinate projections of the p-v-T surface. Figure 2.4 schematically illustrates p-T, T-v, and p-v plane projections for
a simple system. In these figures, the discrepancy between the contracting and the expanding matter
upon freezing is also indicated. The dashed line in Figure 2.4a represents the melting curve for a
substance which contracts on freezing.
To understand the thermodynamic behavior of substances at various pressures and temperatures,
first, regions of low, medium, and high pressures are to be defined with respect to the critical pressure
of that particular matter under study. After selecting the pressure region to be studied, the state of
matter at different temperatures may be discussed, and also compared. Because of its abundance in
our environment and its importance in many technological and biological processes, water is one of
the most thoroughly studied substances. Hence we will study the thermodynamic behavior of water
and also draw some general rules for certain properties.
Let us start analyzing the low pressure behavior of water by assuming the pressure to be at 0.26
kPa. Consider 1 kg of ice initially at (-20°C, 0.26 kPa) in a piston-cylinder device, the change of state,
as a result of heating the ice at constant pressure, is displayed in Figure 2.4a by line AB . First, the
ice temperature increases to -10°C, and then the ice changes phase directly from solid to vapor. This
process is known as sublimation.
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 31
If the initial state was at (-20°C, 0.61 kPa),
the constant pressure heating process would
be indicated by line CD in Figure 2.4a. At this
pressure, when the temperature reaches 0.01°C,
further heating of ice would result in some ice
becoming vapor, and some becoming liquid.
This state is represented schematically in Figure 2.5, and is called triple point at which three
phases of matter are present in equilibrium.
The thermodynamic properties of matter are
fixed at the triple point and are characteristics
of matter. Table 2.1 provides the triple point
pressure and temperature for several substances.
Respect to the values given in Table 2.1, if a
Figure 2.4 Plane projections of p-v-T surface of a
solid carbon dioxide (dry ice) at a condition of
substance
(270K, 100kPa) is brought into an environment
at (298K, 100kPa), the dry ice starts to sublimate. It changes phase directly from solid to vapor.
TABLE 2.1
Triple-state of several substances
Substance
Oxygen (O2)
Nitrogen (N2)
Ammonia (NH3)
Carbon dioxide (CO2)
Water (H2O)
T(K)
54.36
63.18
195.4
216.55
273.16
p(kPa)
0.152
12.56
6.18
516.63
0.61
It should be pointed out that a simple system may exist in a number of different triple points.
A substance may have two solid phases and liquid phase, or two solid phases and a vapor phase in
equilibrium. Furthermore, three solid phases, like in iron-carbon equilibrium diagram, may coexist.
However, only one triple point involves the equilibrium of solid, liquid, and vapor phases.
32
THERMODYNAMICS
To study the thermodynamic behavior of water at medium pressure range, let us consider 1 kg of
water contained within a piston-cylinder arrangement as shown in Figure 2.6a. The pressure on water
is maintained constant at 100 kPa, and the initial temperature is at 25°C. As heat added to water, the
volume will be measured as a function of water temperature, with water allowed to reach equilibrium
prior to each measurement. The starting value for specific volume is 0.001003 m3/kg at a temperature
of 25°C. As the water temperature increases, the liquid water expands slightly increasing its specific
volume. For instance, at 80°C, the specific volume is 0.00103 m3/kg.
Eventually, the water reaches 99.63°C, and with more heat input the water starts to boil. During
the boiling process, two phases (liquid and vapor) coexist in equilibrium with the temperature constant
at 99.63°C (see Figure 2.6b). At a particular pressure, the temperature of a substance at which boiling
takes place is called the saturation temperature. The saturation temperature for water at 100kPa is
99.63°C. Similarly, at a particular temperature, the pressure at which boiling takes place is called the
saturation pressure. The saturation pressure for water at 99.63°C is 100kPa or 1 bar. On T-v diagram
of Figure 2.6, the point B represents the saturated liquid state. The addition of more heat to water at
saturated condition, the vaporization process will be completed, and saturated vapor at 99.63C will
be obtained (point C on the T-v diagram). The specific volume of water will increase from 0.001043
m3/kg to 1.6940 m3/kg. This is a factor of 1624 times increase in volume.
Under the saturation line BC , liquid and vapor coexist in equilibrium. As displayed in Figure 2.6b,
the two phases may exist in a container at separate locations, or as an intermingling of vapor and small
liquids droplets which is called fog. Table A1, in the appendices, presents the thermodynamic properties
of saturated water. However, to provide a value for v at a state in the region between the saturated liquid
and vapor states, the amount of liquid and vapor present in the mixture has to be known. Thus,
v
m f v f  mg v g
V V f  Vg


m m f  mg
m f  mg
(2.6)
In this expression, the subscript “f” refers to saturated liquid, and “g” to saturated vapor. Defining the quality of the mixture as x which is the ratio of the mass of the vapor to the total mass of the
mixture (x=mg/(mf+mg), equation (2.6) may be rearranged as,
v  1  x v f  xvg  v f  xv fg
(2.7)
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 33
It is obvious that the value of x may vary from zero (saturated liquid state, mg = 0) to unity (saturated
vapor state, mf = 0). In the liquid-vapor region, since the pressure and the temperature are dependent
properties, the state of a simple system is specified by (pressure, a property except temperature), or
by (temperature, a property except pressure). For instance, the value of v is completely specified
by the quality x, and the saturation pressure p, or by the quality x, and the saturation temperature T.
The other thermodynamic properties of a liquid-vapor mixture may also be described in terms of the
properties of saturated liquid and vapor states as following.
enthalpy:
h  h f  xh fg
(2.8)
internal energy:
u  u f  xu fg
(2.9)
entropy:
s  s f  xs fg
(2.10)
gibbs free energy:
g  g f  xg fg
(2.11)
Further addition of heat to saturated vapor at 99.63°C causes the temperature to increase above the
saturation temperature. The vapor at a temperature above the saturation temperature is called superheated vapor. The curve CD in Figure 2.6 represents this region. In a superheated vapor region, as the
data for water shown in Table A3, pressure and temperature are independent properties. For a fixed
pressure, the temperature, being greater than the saturation temperature, may be varied as desired.
As illustrated in Figure 2.6, performing similar experiments for pressures other than 100kPa,
similar T-v plots will be obtained. However, we know that for pressures below 100kPa, the water will
boil below 99.63°C, and for pressures above 100kPa, as in a pressure cooker, the boiling temperature
will be above 99.63°C. In Figure 2.6, the points B1, B, and B2 etc. correspond to the saturated liquid
states, and C1, C, and C2 etc. correspond to saturated vapor states. The locus of all such saturation
states is called the saturation curve.
In Figure 2.7, the right of saturation line is the superheated region. To the left of saturation line
is the liquid region which is called either compressed liquid region, since the pressure of the liquid
is greater than the saturation pressure at the given temperature, or sub-cooled liquid region, since the
temperature of the liquid is below the saturation temperature at the specified pressure.
Figure 2.7 The saturation curve separates liquid, vapor, and liquid-vapor mixture regions.
34
THERMODYNAMICS
The compressed liquid data for water is shown in Table A4 and the locations are schematically
presented in Figure 2.8. As seen from this figure, the functional dependence of compressed liquid
properties on pressure is rather weak. For instance, by using the saturated liquid data at 100°C as an
approximation for the compressed liquid state of T*=100°C and p*=100 bars, errors of 0.479%, 0.67%,
1.7% and 0.589 percent are made in values of v, u, h, and s respectively. When compressed liquid
data based on experimental information are available, they should be used. In the absence of such
data, however, the above comparison indicates that the compressed liquid state can be approximated
by using the property values of the saturated liquid state at the same temperature.
A careful observation of saturation curve in Figure 2.7 displays that the difference in specific
volume between the saturated liquid and vapor get smaller as the pressure on water increases. At a
pressure of 22.12 MPa liquid and vapor exist indistinguishably at 374.15°C. This point labeled as
“c.p.” in the figure is called the critical point, with the pressure and temperature at this point called
critical pressure and critical temperature. Depending upon the kind of a substance, the pressure and
the temperature assume specific values at the critical point. Hence the critical point is considered as
a property of matter. In Table 2.2, pc, vc,, and Tc values of the critical point for several substances are
given and the critical constants of some more common substances are supplied in the appendices.
The thermodynamic behavior of matter at high pressure region may be analyzed by considering
pressures above the critical pressure. If water at 30 MPa, and 25°C is heated by a constant pressure
process, a curve such as EFG in Figure 2.7 will be obtained. The typical feature of v vs T distribution
in this pressure region is that the curve is continues and no indication of phase change is present.
Therefore, at supercritical pressures, there is no liquid or vapor phase of a simple substance existing
in equilibrium.
TABLE 2.2
Critical point data for several substances
Substance
Water
Carbon dioxide
Refrigerant R11
Refrigerant R22
Dry air
Tc (°C)
374.15
31.04
198.07
96.01
-140.6
pc (kPa)
22120
7383
4380
4977
3769
vc (m3/kg)
0.00317
0.002137
0.0018
0.0019
0.00312
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 35
2.4
The Use of Thermodynamic Tables
In the preceding section, a brief introduction to the p-v-T behavior of simple systems is offered.
In presenting the state of matter accurately, however, tabulation of data is needed. Because of distinct
behavior of matter at different phases, three different tables are to be supplied for each matter. These
include saturation tables (liquid-vapor, and solid-vapor), superheated tables, and compressed liquid
tables. Tabular data are listed at convenient increments of the independent properties and include the
following intensive properties: the pressure p, the specific volume v, the temperature T, the specific
internal energy u, the specific enthalpy h, and the specific entropy s. A compilation of such data for
water is found in Tables A1, A2, A3, A4, A5 in the appendices. Similarly the thermodynamic data
for refrigerant 22 (R22) and carbon dioxide are presented in Tables A6 through A11.
Due to dependence of pressure on temperature or vice-versa at saturation, two tables (Tables A1,
and A2) are given for the presentation of data in this region. Both tables have the same information,
for convenience, Table A1 is used when temperature is given, and Table A2 when pressure is given
in a problem. In accord with the preceding section, the lower and the upper limits of liquid-to-vapor
saturation curve are respectively the triple and the critical points of matter. Thus, the saturation table
is arranged in a manner to cover this range. In the superheated vapor region, the first line of Table A3
is the saturated vapor data at that particular pressure.
For the compressed liquids, there is no great deal of data in the literature (see Table A4). In the
absence of such data, the approximation rule, explained in the previous section, might be applied.
Many scientific and engineering problems involve states of matter which do not fall on the grids
of data available for that substance. The interpolation of data becomes necessary.
As shown in Figure 2.9, choosing the interval for data to be the same as given in tables, reasonably accurate results might be obtained by linear interpolation. For instance, let us approximate the
temperature of superheated steam at (5 bars, 0.4249 m3/kg). In Table A3, the data for this state is
supplied and T=200°C, however, to evaluate the approximation error, the data at 0.4249 m3/kg are
assumed to be missing. The linear interpolation at pi=5 bars requires that,
v  vi
T  Ti

vi 1  vi Ti 1  Ti
(2.12)
where the subscript i represents the data evaluated at 0.4045 m3/kg in Table A3, and i+1 at 0.4646
m /kg. Substitution of table values into equation (2.12) yields the temperature at (5 bars, 0.4249 m3/
kg) as 200.36°C which is off by less than 0.1 percent of the tabulated value and is quite accurate.
3
36
THERMODYNAMICS
Double interpolation has to be applied for cases such as neither the pressure nor specific volume
of the given state correspond to the available data. First, an interpolation carried out for the pressure,
and next the specific volume is interpolated to determine the desired data. As a result, the following
interpolation formula can be derived for determining the temperature at a state (p, v),
T  Tij  a1 (Ti , j 1  Tij )  a2 (Ti 1, j  Tij )  a1a2 (Ti 1, j 1  Ti , j 1 )  a1a2 (Ti 1, j  Tij )
(2.13)
where,
a1 
p  pij
pi , j 1  pij
, a2 
v  vij
vi 1, j  vij
(2.14)
It should be noted that the double interpolation formula is general in the sense that any other intensive property can be evaluated by replacing T by the desired property in Eq (2.13) The following
examples illustrate the use of tables in solving certain problems of engineering interest.
Example 2.1: Determine the dryness quality (if saturated) or the temperature (if superheated) of the following substances
at given states. a. R22, p=500kPa, v=0.031 m3/kg b. water, p=5 bars, v=0.6 m3/kg
Solution:
a.
In determining whether the given state is in saturated or in superheated region, the specific volumes of the given
state and the saturated vapor state may be compared. The given state would be in superheated region for v  vg ,
and in saturated region if v  vg .
From Table A6, at p  500 kPa , vg  0.0469 m3 /kg , and v  vg , a saturated state. By Eq. (2.7), the quality is
v  v f / vg  v f  and due to negligible effect of the pressure on liquid volume, the value of v f
at p  497.567 kPa
may be taken. Thus,
x  0.031  0.0007783 / 0.0469  0.0007783 ,
b.
x = 0.655.
From Table A2, specific volume of saturated vapor at 5 bars, vg = 0.3749 m3/kg, v>vg, superheated vapor.
Using Table A3, and applying the linear interpolation method, Ti = 360°C, vi = 0.5796 m3/kg, Ti+1 = 400°C, and
vi+1 = 0.6173 m3/kg.
T  360
0.6  0.5796
,

400  360 0.6173  0.5796
T=381.6°C.
Example 2.2: The closed tank of Figure 2.10 contains saturated water at
150°C. Determine the pressure at the base if the liquid water level from the
base is 12m.
Solution:
From Table A1, at a saturation temperature of 150°C, the pressure at surface 1 is: p=475.8kPa, and vf = 0.00109 m3/kg. Since the liquid density,
 f  1 / v f , then  f  917 kg/m3 . The pressure at the base, by equation
(1.12), p2  p1   f gL , or
p2  475.8 
9.8  917  12
,
1000
p2 = 583.64 kPa.
Example 2.3: Having a volume of 2 L, a rigid tank contains 1 kg of liquid
and vapor water at 50°C. The mixture is heated until a single phase obtained.
a. Determine whether the final state is a saturated liquid or a saturated vapor
state. b. Solve the same problem for a tank volume of 200 L.
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 37
Figure 2.11 Heating of vapor at a constant volume increases the pressure
Solution:
a.
Due to the rigidity of the tank, the heating is a constant volume process, and v2  v1  V / m . For V = 0.002 m3,
m=1kg, the specific volume becomes, v1  0.002 m3 /kg . Since v1  vc , the liquid and vapor mixture will be at a
saturated liquid state at the end of the heating process. Such a result might seem peculiar at a first glance. However,
one should recall that the tank pressure and temperature increase as heat added.
b.
In this case, the initial specific volume, v1  0.2 m3 /kg , and v1  vc . Thus, heating at a constant volume will result
with a saturated vapor state.
Example 2.4: A rigid and insulated tank in Figure 2.12 has a volume of 0.0081 m3 and contains 0.05 kg of water at 15 bar.
By a cooling process, the tank loses heat to surroundings at the base and the steam temperature decreases. Determine a. the
initial temperature, b. The temperature at which vapor becomes saturated, c. the quality of water if the final pressure in the
tank is 10bars.
Solution:
a.
The initial specific volume of water, v1 
V
0.0081 m3
, v1  0.162 m3 /kg

m
0.05 kg
For v1  0.162 m3 /kg , p1  15 bar , from Table A3,
b.
T1 = 280°C.
If the saturated vapor state represented by 2, v2  v1  0.162 m3 /kg , then Table A1 would yield, vi  0.1565 m3 /kg ,
T i=190°C, and vi 1  0.1941 m3 /kg , Ti 1  180o C . After a linear interpolation, the temperature at 2 is,
o
Ti
i 1  188.53 C .
c.
From Table A2, at p3 = 10 bar, v3 = 0.162 m3/kg, vg3 = 0.1944 m3/kg, vf3 = 0.00112 m3/kg, and by Eq. (2.7),
x
0.162 0.00112
,
0.1944 0.00112
0.8323 .
Figure 2.12 Cooling of vapor at a constant volume reduces the pressure
38
THERMODYNAMICS
Example 2.5: The piston-cylinder apparatus of Figure 2.13 contains 5 kg R22 at 10°C, 80% liquid and 20% vapor by
mass. As the refrigerant heated, the piston rises, and reaches the stops when the cylinder volume becomes 200 L. Estimate
the refrigerant temperature as the piston just touches the stops, and plot the process on p-v diagram.
Solution:
Since the stops don’t exert any force at the final state, the heating is a constant pressure process. From Table A6, at T1  10o C ,
the pressure is p1  354.284 kPa , and p2  p1 .
At state 2, V2  0.2 m3 , the total mass, m = 5 kg, and the specific volume, v2 
V2
 0.04 m3 /kg which is less than
m
the saturated vapor specific volume of 0.065342m3/kg at 354.284 kPa. Thus the final state is still a saturated state, and the
final temperature remains the same, T2 = 10°C. Including the saturated line, the p-v plot of the process is illustrated in
Figure 2.13.
Example 2.6: Consider the piston-cylinder arrangement in Figure 2.14a, which contains 0.1 kg of saturated vapor at 140°C,
and the piston has a cross sectional area of 0.05 m2. The linear spring, having a spring constant of 100 kN/m, initially
touches to the piston surface and exerts no force. Through a heating process, the pressure in the cylinder becomes 500 kPa.
Determine the final temperature of steam and plot the process on p-v and T-v diagrams.
Solution:
A careful study of the geometry of the systems reveals that due to presence of spring the vapor pressure and volume are
interrelated as,
p  p1 
K V  V1 
A2p
(2.15)
This relationship is derived by considering the equilibrium of forces applied on the piston surface for a volume V  V1 .
From Table A1, at T1 = 140°C, p1 = 361.3kPa, v1 = 0.508m3/kg, and V1  mv1 , or V1 = 0.0508 m3. For p2 = 500kPa, and
K=100kN/m, Eq. (2.15) yields the final volume, V2  0.05426 m3 , and the specific volume, v2  0.542 m3 /kg . Since
v2  v2g , the final state is in superheated vapor region, and through Table A3, the temperature at (500kPa, 0.5426m3/kg)
may be estimated by linear interpolation as following,
T2  320 0.5426  0.5416

360  320 0.5796  0.5416
or
T2=321°C.
The p-v and T-v plots of the process are shown in Figures 2.14b and c respectively. In contrast to the linear relation
between p and v, v vs T representation is non-linear.
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 39
2.5
The Specific Heats of a Pure Substance
In addition to the tabulated thermodynamic properties, the specific heats of a pure substance
under no phase change condition play equally important role in numerous engineering applications.
A method of describing the specific heat of a substance is to state the slope of the line formed by
the intersection of a given plane with an equilibrium surface. For instance, the intersection of the
constant pressure plane with a “h-p-T” surface yields a curve whose slope defines the specific heat
at constant pressure as,
 h 
cp  

 T  p
(2.16)
Similarly, the slope of a curve obtained by the intersection of a constant volume plane with a
“u-v-T” surface defines the specific heat at constant volume as following,
 u 
cv  

 T v
(2.17)
Both of these expressions contain only thermodynamic properties from which one may conclude
that the constant-pressure and the constant-volume specific heats are also thermodynamic properties
of a substance. Moreover, since the enthalpy of a substance has a higher value than internal energy,
cp, has to be numerically greater than cv.
Figure 2.15 displays the distribution of cp for water at 1atm pressure. Notice that although cp varies strongly with T for solid water (ice), it is nearly independent of T for the liquid and vapor phases.
The difference between cp and cv for any substance may be determined by making use of Maxwell
relations and expressed as following,
 p   v 
c p  cv  T 
 

 T v  T  p
(2.18)
If a substance is in solid or in liquid phase, both the thermal expansion coefficient, ( v / T )p,
and the pressure coefficient, ( p / T )v, are very small in value and the difference between cp and cv
is generally insignificant, cpcv. Therefore, for liquids and for solids, a single value for each phase
is assigned to the specific heat and simply indicated by ' c ' without any subscript. Moreover, cp of a
solid or a liquid substance depends only weakly on pressure except states near the critical point. Thus,
40
THERMODYNAMICS
the data shown in Figure 2.15 for solid and liquid water can be safely used for pressures other than 1
atm. On the other hand, cp of water vapor varies significantly with pressure especially for states near
the saturation line and near the critical point.
Example 2.7: Using the data in Table A3, evaluate the specific heats c p , and cv of superheated vapor at 10 bars pressure,
and 200°C temperature.
Solution:
Equation (2.16) may be approximated as,
h h 
c p   i 1 i 
 Ti 1  Ti  p  pi
and from Table A3, at pi=10 bars,
Substitution of these data into the defining relation results with
u u 
Similarly, cv   i 1 i 
 Ti 1  Ti v  vi
Ti = 200°C,
hi = 2827.9 kJ/kg
Ti+1 = 240°C
hi+1 = 2920.4 kJ/kg
c p  2.312 kJ/kgK .
and from Table A3, at vi = 0.206 m3/kg, Ti = 200°C,
Ti+1 = 400°C
Note that vi+1 = vi, but the pressure at i+1 is 15 bar. Thus, the definition of cv yields,
2.6
ui = 2621.9 kJ/kg
ui+1 = 2951.3 kJ/kg
cv  1.647 kJ/kgK .
Gaseous Behavior of a Pure Substance
Gas and vapor are often used as synonymous words. Vapor usually implies a gas state of a pure
substance which is liquid or solid at atmospheric conditions that corresponds to 1 atm pressure and
25°C temperature. Similarly, when a substance is vapor at atmospheric conditions, it is customarily
called a gas. The property tables for the vapor region provide accurate information, but they are bulky
and are presented in discrete intervals. As illustrated by the previous examples, interpolation is needed
for states not corresponding to the tabulated ones. A more practical and desirable approach would be,
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 41
if possible, to provide relations among the properties that are sufficiently general and accurate. Any
equation that relates the pressure, the temperature, and the specific volume of a substance is called
an equation of state.
As shown in Figure 2.16, experiments performed on numerous gases show that as the absolute
pressure approaches zero (p0), “ pv 1” assumes the same value regardless of the nature of the gas.
The limiting value of pv at zero pressure is the same for all gases at the same temperature. Similar plots would result for data at other temperatures except that the value of zero pressure intercept
differs for every new temperature.
Consequently, the pv product at zero pressure is proportional to absolute temperature as following,
T
1
(lim pv)
 p 0
(2.19)
where  is the constant of proportionality, and is called the universal gas constant. Numerous
experiments performed on various gases at a condition identical with the triple-point of water have
showed that  assumes a numerical value of 8.3144 kJ/kmol K. Sometimes, instead of , it is preferable to use the specific gas constant which represented symbolically as R. These two gas constants,
however, are related as following,
  MR
(2.20)
In Eq. 2.20, M is the molar mass of a gas, and R varies with respect to the molar mass of the gas
concerned. In the appendices, Table A17 provides R values for a number of gases.
Figure 2.16 Experimental data for the variation of p v with pressure at a given
temperature for several gases
1
a bar over a property stands for a molar property.
42
THERMODYNAMICS
Besides this peculiar behavior of gases at low pressure, it is desirable to relate p, v, and T at high
pressures and at regions near the saturation and near the critical point. The p-v-T representation of a
gas may be provided for a wide region of pressures and temperatures by the introduction of a correction factor called the compressibility factor Z. It is defined as,
pv
T
Z=
(2.21)
where, p and T are the absolute pressure and temperature respectively at a given state. As shown
in Fig. 2.17, the plot of experimental data for various gases at different temperatures indicate that Z
assumes a limiting value of unity as the pressure is lowered to zero.
lim Z 
p 0
1
(lim pv )  1
T p  0
(2.22)
For all gases, independent of temperature, Z = 1 at very low pressures. However, as indicated in
Fig. 2.17 at high pressures, Z is a function of both pressure and temperature, Z = Z(p,T).
The question of what exactly constitutes the low pressure and the high temperature essentially
depends on the nature of the gas under consideration. For a particular gas, the pressure and the
temperature at a given state is low or high relative to the critical pressure and temperature of that
gas. Referring to Table 2.2, a temperature of “-50°C” is high enough for air which has a critical
temperature of “-140.6°C”, but this temperature is low for carbon dioxide. Therefore, gases behave
differently at a given temperature and pressure, but behave very much the same at temperatures
and pressures normalized with respect to their critical temperatures and pressures.
pr 
p
pc
,
Tr 
T
Tc
(2.23)
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 43
Where pr and Tr are the ratios of absolute pressures and temperatures, and are respectively called
the reduced pressure and the reduced temperature. In Figure 2.18, experimentally determined Z
values of ten gases for reduced isotherms Tr are plotted on a Z vs pr chart.
Through a curve fitting to available data, the p-v-T behavior of gases is estimated by an average
deviation of less than 5-percent. Therefore the following principle may be stated.
Principle 6: The compressibility factor for all gases is the same at
the same reduced pressure and temperature.
Figure 2.18 Experimentally determined Z values for various gases
This principle is also called the rule of corresponding states, and curve fitting to data for all
gases, the generalized compressibility chart as shown in Figure 2.18 is obtained. Recently, Lee and
Kesler provided an analytical expression which satisfies the generalized compressibility values fairly
accurately for reduced pressures in the range 0  pr  10 , and expressed as following,
Z  pr vr  / Tr  1 

  
A1 A2 A3
A
 
 2  5  3 4 2     2  exp   2 
vr  vr  vr  Tr vr  
vr  
 vr  
where, A1  a11  a21 / Tr  a31 / Tr2  a41 / Tr3 A 2  a12  a22 / Tr A 3  a13  a23 / Tr
A 4 =0.042724   0.65392   0.060167
(2.24)
44
THERMODYNAMICS
The aij type constants of this relation for a simple substance are provided in appendices. In Figure
A1, the pseudo reduced volume which is defined as vr   pc v / RTc , represents iso-volumetric lines on
“Z vs p” chart. It should be noted that the generalized chart is only an approximation. In the absence
of experimental data, however, the generalized chart yields results that are accurate to within a few
percent, and such an error is considered to be reasonable for many engineering design problems.
Example 2.7: 2 kg of Butane (C4H10) in a piston cylinder assembly undergoes a process from (5 MPa, 500K), to (3 MPa,
450K). Determine the volume change of Butane.
Solution:
From Table A.17, pc=38 bar, Tc=425K, M=58.12 kg/kmol, reduced values at state1 are, pr1  p1 / pc  50 / 38  1.315 ,
Tr1  500 / 425  1.176 , the corresponding, Z1, by Figure 2.18, Z1 = 0.68. Similarly, pr 2  30 / 38  0.789 , Tr 2  450 / 425  1.058 ,
and by Figure 2.18, Z2 = 0.74. The gas constant is, R=8.314/58.12=0.143 kJ/kgK.
RT
0.143  500
, v1=0.00972 m3/kg
The specific volume at state 1 is, v1  Z1 1  0.68 
p1
5000
and at state 2 becomes, v2  Z 2
or V  2 (0.0158 0.00972)
0.143  450
RT2
, v2=0.0158m3/kg. The total volume change is: V  m(v2  v1 )
 0.74 
3000
p2
0.012 m3 .
2.6.1 The equations of state. Even though the generalized compressibility chart estimates the
p-V-T behavior of real gases fairly accurately, it is always preferable to have an equation of state that is
applicable to all kinds of gases. Unfortunately there is no such a relation that represents the gas phase
region of all substances accurately. The most of the equations are accurate only for densities less than
the critical density, and a few are reasonably accurate to densities 2.5 times the critical density.
There are two experimental facts that may not be neglected in constructing an equation of state
for real gases. 1. Molucules attract each other with a force which is inversely proportional to the
square of the average distance between them, F  1 / d 2 . 2. Molecules have physical dimensions and
occupy certain volume and this volume usually is not negligible with respect to the volume that the
gas spreads.
Taking into account these two factors, an early
attempt was made by van der Waals in 1873, and
suggested the following equation of state,
a

p 2
v


 ( v  b )  T

(2.25)
where, a / v 2 , represents additional pressure caused
by intermolecular forces and b is the volume occupied
by molecules due to their physical dimensions.
As shown in Figure 2.19, the critical isotherm
fulfills three conditions at the critical point (cp): 1. It
Figure 2.19 Van der Waals isotherms
passes through the critical point, 2. Its slope is zero,
3. It is a deflection point. Mathematically these thre
 2 p 
 p 
e conditions corresponds to;    0  2   0 . Evaluating these derivatives at p  pc , T  Tc
 v T
 v T
and by using Eq. (2.24) yield
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 45
a
272Tc2
64 pc
b=
Tc
8 pc
(2.26)
Eq. (2.25) together with the critical point data of a substance determines a, and b constants of van
der Waals equation. Table 2.3 provides the constants of van der Waals equation of state for several
gases.
Table 2.3 Constants of van der Waals equation of state
a
(m kPa/kmol2)
365.4
557.1
797.31
232.4
136.8
552.6
Substance
Carbon dioxide
Ethane
R22
Methane
Nitrogen
Water vapor
6
b
(m /kmol)
0.0428
0.065
0.077
0.0427
0.03864
0.03042
3
The van der Waals equation of state presents deviations in a region especially close to the critical
point. There are other equations of state with better approximation to experimental data but they are
more complex. One of them is Beattie-Bridgeman equation.
p
RT
B A

(1  C ) 1    2
v
v v

(2.27)
c
 a
 b
where, A  Ao 1   B=Bo 1   C= 3
vT
 v
 v
The five adjustable parameters, a, b, c, Ao, and Bo are determined by curve fitting to experimental
data. This equation of state is very accurate if it is used within the range of the data that the adjustable
parameters were derived.
2.7
The Ideal Gas Model
In some design problems, because of the need for iteration through the chart, the process of relating ‘p’, ‘v’, and ‘T’ of real gases might be time consuming and tedious. A direct relationship among
the pressure, the specific volume, and the temperature is always preferred. At certain regions of p,
and T, the following model approximates the real gas behavior.
Principle 7: Kinetic theory of gases: a. Gas molecules are small compared with average distance between them. The volume occupied by the gas molecules themselves is
neglected, b. Gas molecules collide without loss of kinetic energy, c. Gas molecules
exert practically no forces on one another except when they collide.
The gas behavior which closely approximates the above stated conditions is called ideal gas
behavior. In accord with principle 5, the coefficients of van der Waals equation of state (Eq. (2.26))
are both zero; a=b=0, and ideal gas is defined as following.
46
THERMODYNAMICS
Definition: The ideal gas model of a real gas satisfies the following criteria: a. pv  T , and b.
u  u (T ) .
The first of the defining criteria requires that the compressibility of the gas has to be unity. Referring to the compressibility chart in Figure 2.18, a. At very low pressures, pr  1 , gases obey the
ideal gas model regardless of temperature, b. At high temperatures, Tr  2 , gases display ideal gas
behavior for reduced pressures in the vicinity of 4, c. The deviation from ideal gas model is greatest
in a region near the critical point.
Let us consider water vapor at pressures lower than 20 kPa, since, pr  0.0009 , the vapor can
be treated as an ideal gas for such pressures. In air-conditioning applications, the pressure of water
vapor in air is very low, and can be treated as ideal gas. In steam power plant applications, however,
due to high pressure, the ideal gas model for vapor yields unacceptable errors.
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 47
When the equation pv  T plotted on a p-v-T coordinate system, the generated surface would
appear as shown in Figure 2.20 where the constant T-plane represents hyperbolas because the product
pv is constant. The p-T projection of the surface, however, represents straight lines and indicates that
absolute pressure linearly varies with absolute temperature at a constant volume.
In applying the relationship, pv  T , difficulties usually arise due to failure of properly considering the units involved. Table 2.4 may be used for the application of ideal gas equation in various
forms and the units involved.
Referring to the second of the ideal gas defining criteria, the internal energy of ideal gas is only a
function of temperature. This result seems contradicting the state principle but it is not. At very low
pressures, the effect of pressure on the internal energy of real gases becomes negligible. One should
also recall that ideal gas model represents the real gas behavior at very low pressures. Furthermore,
the substitution of pv  T into enthalpy definition, h  u (T )  pv , yields,
h  u (T )  T
(2.28)
Which shows that the enthalpy of ideal gas is also a function of temperature only, h  h(T ) .
Example 2.8: Evaluate the volume occupied by 2 kg of Refrigerant 22 (R22) at 700 kPa, and 50°C by a. the experimental
values in superheated table, b. the principle of corresponding states, c. the van der Waals equation of state, and d. ideal gas
equation of state.
Solution:
a.
From Table A7, the specific volume of R22 at (700 kPa, 50°C) is 0.0408 m3/kg, and the total volume occupied,
V  mv , V  0.0816 m3.
b.
For R22, Table 2.2 yields pc  4977 kPa, Tc  96.01o C , and corresponding reduced pressure and temperature
respectively are pr  0.1406 , Tr  0.875 . The generalized compressibility chart, Figure 2.18, yields Z = 0.92.
From Table A21, R=0.0961 kJ./kgK, and substituting these values into pv  ZRT ,
v
c.
0.92  0.0961  323
700
or
The van der Waals equation of state, v 
v = 0.04081 m3/kg
which is 0.02% off from the true value.
T
8.314 x323
 b , assume v=2 m3/kmol, v (1) 
 0.077 , v(1)=3.06
a
797.31
p 2
700 
v
22
8.314 x323
 0.077 , v(2)=3.497 m3/kmol. Hence v=0.04043m3/kg which is 0.08% off from the
797.31
700 
3.062
experimental value.
m3/kmol, v (2) 
d.
The equation of state for ideal gas, pv  RT , and v 
0.0961  323
or v = 0.0445 m3/kg which is in 9% error
700
for estimating the specific volume.
Table 2.4 Various forms of equation of state for ideal gas
Equation
Units
p= RT
= V/n
pV = nRT
p:kPa
V:m3
n:number of kilomoles
R = 8.3144 kJ/kmolK
48
THERMODYNAMICS
Equation
n = m/M
pV = m(R/M)T
R = R/M
pV = mRT
v = V/m
pv = RT
ρ = 1/v
p = ρRT
Units
T:K
M:kg/kmol
p:kPa
V:m3
m:kg
m:kg/kmol
R:kJ/kgK
T:K
Example 2.9: A tank having a volume of 5 m3 is filled with methane (CH4) at 8.5 MPa and -23°C. Due to poor insulation,
the gas warms up to 17°C after a period of time. Estimate on the basis of compressibility chart, a. the mass within the tank,
and b. the final pressure in the tank.
Solution:
a.
From Table A21, pc  46.4 bar, Tc  190.7 K , R = 0.5183 kJ/kgK, M = 16.04, and pr1  8500 / 4640  1.83 ,
Tr1  250 / 190.7  1.31 . By Figure 2.18, Z1 = 0.73, and the specific volume at state 1 is:
v1 
Z1RT1
0.73  0.5183  250
, v1 
 0.0111 m3/kg. The amount of methane in the tank,
p1
8500
m V/
b.
450.45 kg .
The pseudo reduced volume is constant for the heating process, and
Tc
8.3144  190.7
vc 

 0.3417 m3/kmol.
pc
4640
The molar specific volume, v1  Mv1 , v1  0.178 m3 /kmol and the pseudo reduced volume at state 2 is
Thus, for Tr 2  1.52 , and vr 2  0.52 , Figure 2.18 yields pr 2 , and the final pressure becomes, p2  10.904 MPa.
Example 2.10: As shown in Figure 2.21, a laterally insulated vertical cylinder with 0.001 m2 cross sectional area initially
contains air at 100 kPa, 20°C, and a frictionless piston of 10 kg mass rests on the first stop. As a result of a heating process,
determine, a. the air temperature at which the piston just starts raising, b. the position of the piston for air temperature of
700 K, and c. the temperature at which the piston touches the second stop.
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 49
Solution:
a.
The required pressure for raising the piston, p2  p1  m p g / Ap which yields p2  200 kPa . At state 2, where
the piston just starts raising, V2  V1 , and the equation of state for ideal gas yields,
p2 T2
p
200

or T2  2 T1 , T2 
 293 , T2 = 586 K.
p1 T1
p1
100
b.
At state 3, where the air temperature is 700 K, if the piston is not at the upper stoppers, then V3 < 0.0004 m3, and
p3  p2 . This condition can be checked by the ideal gas equation, since p3 = p2, , and
V3  0.000238 m3 which is less than 0.0004 m3. Thus the height of the piston from the base is: 0.238 m.
c.
V V
V
When the piston just touches the upper stopper, p4  p3  p2 , and 4  2 ,T4  T2 4 , T4 = 1172 K.
T4
T2
V2
Example 2.11: A U-tube manometer with a cross sectional area of 5 cm2 contains mercury (ρm=13600 kg/m3). One end of
the manometer is open to the atmosphere while the other end is sealed off and contains air. The manometer initially is in
equilibrium with the environment of 100 kPa, 300 K, and as shown in Figure 2.22, the mercury column is 15 cm high. After
pouring 50 cm3 of mercury into the manometer, determine a. the final volume occupied by air, b. the difference in height of
two mercury columns, c. the final pressure of the trapped air in the sealed column.
Solution:
a.
Note that before and after pouring the mercury, the temperature of air is the same, T2 = T1 = 300 K. Thus, for air
the ideal gas equation reduces to p1V1  p2V2 . Air pressures at state 1 and 2are calculated as following:
p1  po   gL1 , L1 = 15 cm, and p2  po   gL2 ,
where according to the figure, L2 = 25-2y cm. Air volumes at state 1 and 2: V1 = 150 cm3, V2 = 5(30-y) cm3.
Thus,
100  1.36  15 150  100  1.36  (25  2 y ) 5  (30  y ),
Solving for y yields, y = 1.94 cm. Therefore, the final volume becomes, V2 = 140.3 cm3.
b.
c.
L1 = 15 cm, L2 = 25-2y, L2-L1 = 10-2y, or L2-L1 = 6.12 cm.
p2  po   gL2 , or p2  100  1.36  25  2  1.94  , p2 = 128.72 kPa.
50
THERMODYNAMICS
Example 2.12: A vertical and weightless tank with an open bottom has a height of 1 m and submerged into liquid water.
The open end initially just touches the water surface and the tank containment is air at 1bar, 25°C. a. Drive a relation between
the pressure in the cylinder and the height of the cylinder above water surface, b. Determine the height L for air pressure of
1.05 bar, c. Evaluate the air pressure for L=0.
Note: Assume that the temperature of air in the cylinder is always at 25°C.
Solution:
a.
Since temperature is constant for the submerging process, the relationship between the pressure and volume of air is:
poVo  pV ,
or
po Lo  p L  y 
On the other hand, the air pressure in the tank is p  po   gy . Solving for y, and substituting into above relation
yields L in terms of pressure p as,
p
p  p0
L
 0
L0
p
 gL0
b.
From Table A1, at 25°C,  = 997.1 kg/m3, p = 1.05 bars, p0 = 1 bar, L0 = 1 m, and g = 10 m/s2, L/L0 = 0.451, or
L = 0.451 m.
c.
Let  = p/p0, and for L/L0 = 0, above relation yields the following equation,  2    0.0997  0 , by which
α = 1.089, or p=1.089 bar.
Example 2.13: As shown in Figure 2.24, an insulated vessel of 0.1 m2 cross sectional area is divided into two identical
compartments by a weightless, frictionless, and dia-thermal piston which is initially at rest on stops. The lower compartment
contains 1 kg of liquid water and steam mixture at 10 bar, and the upper one holds air. As a result of a heating process, the
piston moves upward by 10 cm. Determine a. the final pressure and temperature in both compartments, b. the quality of
vapor at the final state.
Solution:
a.
Since the piston is weightless and frictionless, both compartments have the same pressure at the initial as well as
the final states, pa1 = pb1 = p1, p1 = 10 bars, and pa2 = pb2 = p2 . Moreover, dia-thermal piston necessitates that Ta1 =
Tb1=T1, T1=179.9°C, and Ta2 = Tb2 = T2. For the ideal gas behavior of air,
 V  T 
p2   a1  2  p1
 Va 2   T1 
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 51
Because of saturated conditions at the final state, the pressure p2, and the temperature T2 are dependent variables, and
trial and error method has to be applied. Let p2 = 13 bars, interpolation through Table A2 yields T2 = 464.6 K, and Va1 =
0.05 m3, Va2 = 0.04 m3. Substitution of these values into the relation, as given above, result as, 1312.8, which closely approximates the final state.
b.
The final volume of compartment a, Va2 = 0.06 m3, ma  1 kg , and va2 = 0.06 m3/kg. From Table A2, for p2 = 13
bars, v f 2  0.00114 m3 /kg , vg 2  0.161 m3 /kg , and Eq. (2.7) yields,
x2
2.8
0.06 0.00114
or
0.15486
2
0.368 .
The Specific Heats of Ideal Gases
Previously it has been determined for an ideal gas that both the internal energy and the enthalpy
are only functions of temperature. Thus, with respect to equations (2.16) and (2.17), the partial derivatives become ordinary ones, and the specific heats of ideal gases can be stated as follows,
The constant pressure specific heat, c po 
The constant volume specific heat, cvo 
dh
dT
du
dT
(2.29)
(2.30)
We know that as the pressure is lowered, the behavior of a real gas approaches ideal gas behavior.
Hence, the symbols c po , and cvo signify values of real gas specific heats at zero pressure.
As shown in Figure 2.25, a typical characteristic of monatomic gases, such as Argon, Helium,
Neon, etc., is that c po of these gases is constant over a wide range of temperature. The kinetic theory
of gases predicts the value of c po for monatomic gases as  or 20.8 kJ/kmol K. As one may notice
from Figure 2.25, an appreciable change in c po values for gases having molecules with two or more
atoms takes place when the temperature interval is rather large. For instance, in a temperature range
from 300 K to 1300 K, of carbon dioxide changes by 65 percent.
52
THERMODYNAMICS
The derivative of equation (2.28) with respect to temperature is,
dh du


dT dT
(2.31)
Substituting equations (2.29) and (2.30) into (2.31) would result the following relationship between
the specific heats of ideal gases.
c po  cvo  
(2.32)
This simple relationship between c po and cvo is important, because knowing one of them allows
the other one to be calculated. Equation (2.32) also indicates that both of the specific heats the same
functional dependence on temperature, and the c po distribution is just displaced by the amount of 
respect to cvo line. Thus, the value of cvo for monatomic gases is or is 12.471 kJ/kmol K.
Sometimes it is preferred to describe the specific heats in terms of their ratio which is defined as,
k T   c po / cvo . Then, together with equation (2.32), the constant pressure and the constant volume
specific heats become,
c po (T ) 
k
1
, cvo (T ) 

k 1
k 1
(2.33)
The integration of equations (2.29) and (2.30) for a specified temperature interval results as,
T2

h(T2 )  h(T1 )  c po (T )dT
T1
(2.34)
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 53
T2

u (T2 )  u (T1 )  cvo (T )dT
(2.35)
T1
These equations respectively provide the change of enthalpy and internal energy of ideal gases, and
they are valid for all processes regardless of its path. Besides, through the use of enthalpy definition,
Eq. (2.2), the enthalpy and the internal energy changes for ideal gases may be interrelated as,
 h   u  T
(2.36)
Evaluation of h or u for monatomic gases is a straight forward procedure, because c po and cvo
are constants, and can be set outside of integration sign. Thus for monatomic gases,
T2

 h  c po dT  c po T
(2.37)
T1
T2

u  cvo dT  cvo T
(2.38)
T1
For other types of ideal gases, we need equations for describing in terms of temperature. These
equations are generally in the form of polynomials with undetermined coefficients.
c p /   ao  a1T  a2T 2  a3T 3  a4T 4
(2.39)
Depending upon the type of a gas, the coefficients may be evaluated by curve fitting to the experimental data. For a number of gases used in industrial processes, the numerical values of these
coefficients are provided in Table A22. Substituting the coefficients of Table A22 into equation (2.39)
and evaluating it for a particular temperature yield the variation of with respect to temperature. Such
a variation for some common gases is tabulated in Table A23.
In determining the enthalpy change of an ideal gas, after substituting equation (2.39) into (2.34),
the integration has to be carried out. The calculations, especially for repetitive cases, become lengthy
and tiresome. However, a close observation of Figure 2.25 indicates that the specific heat of gases
with single atomic structure is not a function of temperature at all. For di-atomic atomic gases, due
to low slope of c po distribution, it is convenient to use the arithmetic average, c po ,av , evaluated for
the given temperature interval as,
c po ,av 
c po (T1 )  c po (T2 )
2
(2.40)
Thus, the change in enthalpy and in internal energy becomes,
 h  c po ,av T
(2.41)
u  cvo ,av T
(2.42)
54
THERMODYNAMICS
Such an approximation especially yields negligible errors when the temperature interval is relatively small, say a few hundreds of Kelvin. If the molecular structure contains more than two atoms
and if the range of integration is greater than a few hundreds of Kelvin, then the integration given in
Eq. (2.34) has to be carried out with temperature dependent specific heat.
A further rough approximation method is that the specific heat at the initial state might be used
as the average value. This method is especially appropriate when the value of the final temperature
is yet unknown.
As a consequence, for gases, the evaluation of h and u may be carried out by one of the following four methods. These methods are in the order of decreasing accuracy.
1. Use tables based on experimental data. This method may require interpolation of data.
2. Use of predetermined equation for c po , and carry out the integration of enthalpy expression
for the specified temperature interval. Specifying the gas temperature, such integration results
are given for air and for carbon dioxide in Tables A24 and A25 respectively.
3. Use an arithmetically averaged specific heat for the specified temperature interval.
4. Use the specific heat at the initial state and assume it to be constant.
Example 2.14: Evaluate the specific enthalpy change of carbon dioxide that is heated from 50°C to 250°C at a pressure
of 1 bar by use of a. tables, b. the specific heat equation, c. the averaged specific heat, d. the specific heat at the initial
temperature.
Solution:
a.
From Table A11, at T1 = 50°C, h1 = 417.9 kJ/kg, and T2 = 250°C, h2 = 609.4 kJ/kg, ∆h = h2-h1, ∆h = 191.5 kJ/kg.
b.
From Table A22, the coefficients of specific heat equation (Eq. (2.39)) for carbon dioxide, a0 = 2.401, a1 = 8.735x10-3,
a2 = -6.607x10-6, a3 = 2.002x10-9, a4 = 0, and
c p /   2.401  8.735  103T  6.607  106 T 2  2.002  109 T 3
where and T are respectively in kJ/kmol K, and in Kelvin. Through Eq. (2.34),
h = 8.314,
h = 8398.902 kJ/kmol, and for M = 44 kg/kmol, h = 190.884 kJ/kg, which underestimates the enthalpy change by
0.32 percent error.
c.
From Table A23, at T1 = 323 K, cpo = 0.869 kJ/kg K at T2 = 523 K, cp0 = 1.0287 kJ/kg K, and by Eq. (2.40),
cp0,av=0.9489 kJ/kg K, h  0.9489  200 , h  189.77 kJ/kg which underestimates the enthalpy change by 0.903
percent error.
d.
From Table A23, at T1=323 K, cp0 = 0.869 kJ/kg K, and h  0.869  200 , h  173.8 kJ/kg, an underestimation
of 9.2 percent.
As can be deduced from these results, the largest error has done by the last method. However, this method still yields
reasonably accurate results if the temperature interval is kept less than a few hundreds of Kelvin.
References
1.
R.E. Sonntag, C. Borgnakke, and G.J. Van Wylen, Fundamentals of Thermodynamics, 6th edition, Wiley Publications,
ISBN 0-471-15232-3, 2003.
2.
J. H. Gross, Mass Spectrometry, 2nd edition, Springer-Verlag, ISBN 978-3-642-10709-2, 2011.
3.
T. Al-Shemmeri, Engineering Thermodynamics, Ventus Publishing ApS, ISBN 978-87-7681-670-4, 2010.
4.
M.J. Moran, H.N. Shapiro, D.D. Boettner, and M.B. Bailey, Fundamentals of Engineering Thermodynamics, 7th edition,
John Wiley and Sons, ISBN 13-978-0470-49590-2, 2011.
5.
Y.A. Chang and W.A. Oates, Materials Thermodynamics, John Wiley and Sons, ISBN 978-0-470-48414-2, 2010.
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 55
Problems
Table readings
2.1
Complete the following table for water substance:
Pressure
Temperature
p (bar)
T (°C)
a.
125
b.
15
110
d.
270
e.
45
f.
70
140
g.
6
400
h.
10
i.
5
Quality %x
h(kJ/kg)
u(kJ/kg)
(if applicable)
2400
2520
0.02
2920.4
55
270
2520
2300
l.
45.1
0.75
2538
m.
0.11
3100
Complete the following table of properties of
refrigerant-22.
Temperature
T (°C)
a.
10
b.
200
c.
50
d.
-20
Pressure
p (kPa)
Specific volume,
v (m3/kg)
Enthalpy
h (kJ/kg)
Internal energy,
u (kJ/kg)
Quality %x
(if applicable)
229.79
0.0895
250
180
e.
200
f.
450
g.
90
h.
5
i.
2.3
Internal energy,
0.32
k.
2.2
Enthalpy
300
c.
j.
Specific volume
v(m3/kg)
80
0.0633
326.23
220
0.06
Determine the required data for water for the following specified conditions:
a. the pressure and the specific enthalpy of saturated
liquid at 25°C,
b. the temperature and the specific volume of
saturated vapor at 6 bar,
c. the specific volume and internal energy at 0.5
bar and 200°C,
d. the specific volume and the enthalpy at 10 bars
and quality of 70 percent,
380.8
e. the temperature and the internal energy at 1 MPa
and an enthalpy of 3565.6 kJ/kg,
f. the quality and the specific volume at 0.7 bar,
and an internal energy of 1850 kJ/kg,
g. the internal energy and the specific volume at
300°C and an enthalpy of 2500 kJ/kg,
h. the pressure and enthalpy at 440°C and an internal
energy of 2930 kJ/kg,
i. the temperature and the specific volume at 6
MPa, and an enthalpy of 292.98 kJ/kg.
56
THERMODYNAMICS
Applications of table reading
2.4
2.5
as a saturated vapor. Show the path of the process
on a p-v diagram, and determine the change of the
following properties between the initial and the
final states: a. the specific enthalpy, b. the specific
volume.
Consider 2 kg of water at its triple point. The volume
of liquid phase is equal to that of solid phase, and the
volume of the vapor phase is 104 times that of the
liquid phase. Determine the mass of water in each
phase.
A storage vessel has a volume of 50 Liters and
contains a mixture of liquid and vapor nitrogen at
1 bar pressure. Due to poor insulation around the
tank, heat is transferred from the surroundings at
20 °C, and the content of the vessel passes through
its critical state. Determine,
a. the amount of nitrogen in the vessel,
2.9
A piston-cylinder device initially contains carbon
dioxide at 1 MPa, and 25°C, and occupies a volume
of 50 L. The fluid is compressed to 10 MPa, and
1 L. Determine the final temperature and enthalpy
change of carbon dioxide.
2.10
A rigid vessel is charged with carbon dioxide at 20°C.
If the initial charge contains the correct proportions
of liquid and vapor, the carbon dioxide will pass
through the critical state when heated with the filling line closed. Determine the proper proportions by
volume of liquid and vapor carbon dioxide so that
the desired change of state will be produced when
the mixture is heated.
2.11
A rigid tank containing saturated water vapor at
200°C has a volume of 0.6 m3. Due to heat transfer
to surroundings, the temperature in the tank drops
to 80°C in one hour. For the final state, determine
a. the pressure in the tank,
b. the ratio of liquid mass to the mass of vapor,
c. the quality,
d. show the process on T-v diagram including the
saturation line.
b. the initial proportions by volume of liquid and
vapor nitrogen in the vessel.
2.6
Water contained in a piston-cylinder apparatus initially is at 5 bar, 320°C, and occupies a volume of
0.02 m3. It is compressed at constant pressure until
it becomes a saturated vapor. Determine,
a. the mass of water,
b. the final temperature,
c. the amount of enthalpy change of the fluid in
kilojoules.
2.7
Complete the following table of properties of Carbon
Dioxide.
Temperature
T (°C)
a.
-10
b.
Specific volume,
v (m3/kg)
10
d.
75
f.
250
20
350
i.
25
j.
-80
Quality %x
(if applicable)
300
100
500
h.
Entropy
s (kJ/kg K)
0.61
e.
g.
Enthalpy
h (kJ/kg)
0.01
1969.1
c.
k.
2.8
Pressure
p (kPa)
650
2.180
10000
3.00
15000
0.3
3481.7
Refrigerant-22 at 200 kPa, with a specific volume
of 0.1519 m3/kg expands at a constant temperature
until the pressure falls to 50 kPa. Then the refrigerant is cooled at a constant pressure until it exists
42
2.12
A rigid tank containing saturated water vapor at 120°C
has volume of 0.1 m3. Due to heat transfer, the tank
is cooled to -20°C. Determine the mass percentage
of solid water at this state.
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 57
2.13
Consider a system consisting of 1 kg of water vapor,
1 kg of liquid, and 1 kg of solid water all in thermodynamic equilibrium. The system is to experience
heat transfer until all of the mass exists as a vapor.
Two different methods of heat transfer are to be
considered, one at constant volume, and the other
at constant temperature. Determine,
a. the temperature, the pressure, and the volume
of the complete system at the initial state,
b. the proportions by mass of liquid and vapor
when all of the solid has melted for each of the
heat transfer processes,
c. the final temperature, the pressure, and the volume
for each of the two heat transfer processes.
2.14
A closed and rigid container having an internal volume
of 0.2 m3 contains 0.5 kg of steam at a pressure of
0.5 bars. The system is heated quasi-statically until
the final absolute pressure is 3 bars.
a. Including the saturation line of water, show the
process on a T-v diagram,
b. Determine the initial and the final temperatures
of the container,
c. Estimate the quality of steam in the initial and
the final state.
2.15
As shown in Figure 2.26, 5 kg of water at 20°C contained in a vertical cylinder by a frictionless piston with
a mass such that the pressure on water is 15 bars. The
water experiences a quasi-static heat transfer causing
the piston to rise until it reaches the stops at which
point the volume inside the cylinder is 0.4 m3. The
heat transfer process continued until the water exists
as a saturated vapor. Including the saturation line of
water, show the process on a T-v diagram, and find
the final pressure of vapor in the cylinder.
2.16
A boiling water nuclear reactor contains m kg of saturated liquid water at 250°C. In case of reactor pressure
vessel fail, a secondary containment structure has to
be provided to avoid the spread of radioactive water.
Assuming that the space between the reactor and the
containment is initially at zero pressure, determine
how many times larger the secondary containment
chamber be if the maximum design pressure is 2 bar
and the specific internal energy of water after expansion is 1080 kJ/kg.
2.17
The piston-cylinder apparatus shown in Figure 2.27
is fitted with a leak proof, frictionless piston upon
which are mounted weights of sufficient magnitude
that a pressure of 60 bar is required to support the
piston and the weights.
Initially the piston rests on stops such that the volume
trapped between the piston and the end of cylinder is
0.03m3, and contains 2 kg of water at a pressure of
2 bar. Water experiences a quasi-static heat transfer
until its temperature reaches 500°C.
a. Show the path of the process on p-v diagram,
b. Determine the temperature at the initial state,
c. What fraction of the total mass is liquid in the
initial state,
d. What fraction of the initial volume is occupied
by the liquid water,
e. Determine the temperature of water as the piston
just begins to lift off of the stops.
58
THERMODYNAMICS
Compressibility chart
2.18
2.19
2.21
Water at 50 bars and 40°C changes state to 150 bars
and 100°C. Determine,
a. the change in specific volume, and in specific
internal energy on the basis of compressed
liquid data,
b. the same quantities by using saturated liquid
data,
c. the percentage of error involved when the approximation in b is applied.
2.22
A piston-cylinder device initially contains refrigerant-22
at 500 kPa, 100°C, which occupies a volume of 10
L. The fluid is compressed to 3 MPa pressure and 2
L volume. Determine,
a. the final temperature and the enthalpy change
based on the tabulated data,
b. the final temperature and the enthalpy change
based on the ideal gas model with a constant
specific heat of cp,
c. the final temperature based on the generalized
compressibility chart,
d. the percentage error in estimating the final
temperature by the generalized compressibility
chart,
e. the percentage of error in ∆H due to ideal gas
assumption.
2.23
Steam initially at 16.0 MPa and 400°C, expands
isothermally until its volume is doubled. Determine
the final pressure, if
a. the ideal gas equation is applied,
b. the generalized compressibility chart is used,
c. the steam table is used.
2.24
A tank with a volume of 10 m3 contains propane
(C3H8) initially at 100°C and 35 bar. Estimate the
mass of the gas by using
a. an ideal gas model,
b. the compressibility chart.
It is claimed that below a pressure of 2 bars, steam
may be assumed to behave as an ideal gas. Consider
a saturated vapor which is at a pressure of 10, 50, and
100 kPa respectively. Calculate the specific volume
by using the ideal gas equation of state and compare
with the tabulated values. Evaluate the percentage
of error for each pressure.
Show that the molar density of ideal gases is the same
at a particular p and T. Calculate the molar density
of acetylene (C2H2) gas at 320 K and 30 bars
a. by assuming ideal gas behavior,
b. by using the generalized compressibility chart,
c. percent of error in (a) due to ideal gas assumption.
2.20
A vertical cylinder of Figure 2.28 is fitted with a
piston restrained by a spring and held by a pin. The
cylinder cross-sectional area is 0.1 m2, and contains
carbon dioxide at 1.5 MPa, 0°C, with an initial volume of 0.05 m3. Together with an ambient pressure
on the piston, the piston weight causes a pressure
of 200 kPa. The spring force would be zero for a
cylinder volume of 0.03 m3. Considering a spring
constant of 400 kN/m, determine the final pressure
in the cylinder when the pin is pulled.
The gas is withdrawn until the mass in the tank is
one-third of the original mass, and the temperature
remains the same. Estimate the final pressure in the
tank,
c. by using the compressibility chart,
d. by the ideal gas model.
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 59
2.25
A system having an initial volume of 1 m3 is filled
with steam at 25 bar and 360°C. The system is cooled
at constant volume to 200°C. Then, by a constant
temperature process, steam is compressed by heat
rejection until ending with saturated liquid water.
a. Sketch the processes on a p-v diagram relative
to the saturation line.
b. Determine the total internal energy change of
steam.
c. Estimate the constant pressure specific heat at
the initial state by using steam tables.
d. Evaluate the change in cp of steam between the
initial and the final state.
2.26
A diver operating at a depth of 40 m releases a
spherical bubble with a diameter of 3 cm. The surrounding water has a density of 1030 kg/m3 at a
uniform temperature of 18°C. Because of surface
tensions on the bubble, the pressures of the air and
the surrounding water are related as,
pa  pw 
a. the final pressure, and volume of air in the
reservoir,
b. the mercury height, ht, at the tube
2.28
As shown in Figure 2.30, a piston-cylinder arrangement
which contains carbon dioxide is surrounded with
a well insulated steam jacket. The saturated steam
at 1.5 bar enters to the jacket and exits as saturated
liquid at the same pressure. Carbon dioxide in the
cylinder is initially at 5 bar pressure and occupies a
volume of 0.1 m3. The gas expands and in 3 minutes
the volume increases by 0.05 m3. Determine,
a. the change in specific enthalpy of steam,
b. the type of process the gas undergoes,
c. the internal energy change of Carbon dioxide
by using tables and by assuming ideal gas behavior,
d. sketch the process on T-v diagram.
2.29
It is desired to store 5 kg of hydrogen gas in an
auxiliary storage tank for the space shuttle project.
If the tank pressure is 15bars and the gas has to be
maintained at -200°C, determine the volume of the
tank by using the compressibility chart.
2
r
where,   0.4 N/m . Assume that the water surrounding the bubble is inviscid and that the air can
be modeled as an ideal gas with constant specific
heat. Determine,
a. the mass of air in the bubble,
b. the diameter of bubble at the water surface where
the pressure is 100 kPa,
c. Express the bubble volume V in terms of the water
depth z, and Sketch of the graph of V vs z,
d. Repeat the problem for =4 N/m.
2.27
The manometer shown in Figure 2.29 consists of
a reservoir partially filled with mercury. The cross
sections of the tube and the reservoir are respectively
At and 4At. Initially the manometer is in equilibrium
with the outside environment (T0 = -25°C, p0 = 760
mmHg). The valve at the top of the reservoir is closed
trapping an air column 30 cm high with a volume
of 3000 cm3. The manometer is brought indoors
(Ti =20°C, pi =760 mmHg) and allowed to come to
equilibrium with indoors’ environment. Assuming
an ideal gas behavior for air, determine
60
2.30
2.31
THERMODYNAMICS
One kilogram of water at 320°C and 60 bar is contained in an insulated cylinder and piston apparatus.
Consider a process in which the water system expands
to atmospheric pressure (1 bar) and a final volume
of 1 m3.
a. Considering the given information, in what
domain of the property surface (superheated
vapor, two-phase, sub-cooled liquid) is the final
state located.
b. Determine the temperature and the entropy of
the system at the final state.
The piston-cylinder apparatus shown in Figure
2.31 is separated into two compartments by a rigid
partition. The pressure of the surroundings is 1 bar,
and the piston has a mass of 1000 kg with a cross
sectional area of 0.1 m2. Both compartments contain
the same ideal gas, R=0.15 kJ/kgK, cv=0.83 kJ/kg K,
with identical masses of 0.1 kg, and are initially in
thermal equilibrium with its environment at 20°C.
The initial pressure of the bottom compartment is
4bars. The partition ruptures and the pressure in two
compartments equalize. The apparatus finally comes
into equilibrium with its environment. Determine,
a. whether the piston moves upward or downward,
and how many meters,
b. the change in internal energy and in the enthalpy
of the gas.
b.
If the pressure of a substance is lower
than the saturation pressure for a particular temperature of the substance, it is in a compressed
liquid state.
c.
If a fluid is at a saturated vapor state,
the quality is unity.
d.
e.
The universal gas constant expressed
as =8.314 kJ/kmol K.
f.
Specific volume and temperature are
independent properties in two-phase region.
g.
For an ideal gas specific volume is a
function of temperature only.
h.
At the critical point, solid, liquid, and
vapor exist in equilibrium.
i.
The quality of a two-phase liquid vapor
mixture is defined as the ratio of the mass of
vapor to the mass of the liquid.
j.
If the temperature of a liquid is higher
than the saturation temperature corresponding
to the pressure of the fluid, it is in a superheated
vapor state.
k.
The change in property value between
two states is independent of the process linking
the two states.
l.
At pressures above the triple point, no
indication of phase-change can be traced.
m.
The compressibility factor of a gas may
be taken to be unity at the critical point.
n.
As the pressure of a gas is reduced, the
pV term becomes proportional to the temperature.
o.
According to the ideal gas model, as
the gas temperature is reduced to absolute zero
at a particular pressure, the volume occupied by
the gas vanishes.
True and False
2.32
Answer the following questions with T for true and
F for false.
a.
Temperature and pressure are invariably
sufficient to completely specify intensive state
of matter.
For an ideal gas, cvo  c po  R .
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 61
p.
At the critical state, the saturated liquid
and the saturated vapor specific volumes intercept.
q.
3.
As the pressure of water increases,
a. the boiling temperature of water increases and
the enthalpy of evaporation increases,
b. the boiling temperature of water increases and
the enthalpy of evaporation decreases,
The constant volume specific heat for
h
a substance is defined as 
 .
  T v
c. the boiling temperature of water decreases and
the enthalpy of evaporation increases,
d. the boiling temperature of water decreases and
the enthalpy of evaporation decreases.
r.
s.
The compressibility factor assumes
approximately the same value for all gases at
the same reduced pressure and temperature.
4.
As the pressure of steam increases,
a. the enthalpy of saturated vapor increases,
The specific heats of diatomic gases
are independent of temperature.
b. the enthalpy of saturated vapor decreases,
c. the enthalpy of saturated vapor remains the
same,
t.
The freezing temperature of water
decreases as pressure increases.
u.
The pressure and temperature are independent during the phase change.
v.
d. the enthalpy of saturated vapor first increases
then decreases.
5.
Liquid water initially at 0C is heated at constant
pressure, the specific volume
a. first increases and then decreases,
The critical point of a substance is the
equilibrium of solid, liquid, and vapor phases.
b. first decreases and then increases,
c. increases steadily
w.
The equations of state usually suffer
representing the p-V-T behavior at the triple
point.
d. decreases steadily.
6.
Check Test 2
Choose the correct answer:
1.
The latent heat of vaporization of a substance at the
critical point is
7.
A 100 cm3 of can contains refrigerant 22 at saturated
liquid state at room temperature of 20°C. Due to
small crack on the can, leak develops and the final
pressure in the can becomes to 2 bar. The amount
of refrigerant escaped is
a. 90.3 g,
b.
100.3 g,
c. 110.3 g,
d.
120.3 g.
The volume occupied by vapor portion of 1kg of
water vapor mixture is
a. is a negative number,
a. xvg,
b.
xvf,
b. is a positive number,
c. xv
d.
x2vg.
c. equal to zero,
d. is a very large number.
2.
8.
A 2.5 m3 of rigid vessel contains steam at 10 bar,
240°C. The mass of steam is
The safety valve of a rigid tank containing 10 kg of
ideal gas at 250 kPa, and 38C suddenly opens, and
after evacuating 25% of the total mass and dropping
the pressure to 200 kPa, the valve closes. The final
temperature in the tank is
a. 5.3 kg, b.
4.38 kg,
a. 58.8C,
b.
58.5C,
c. 5.38 kg, d.
3.38 kg.
c. 48.8°C,
d.
-58.8°C.
62
9.
10.
11.
THERMODYNAMICS
The internal energy of saturated vapor may be evaluated as,
a.
h fg  pvg ,
b.
hg  pvg ,
c.
h fg  pvg ,
d.
hg  pv f .
12.
5 kg of water at 1 bar, x1=0.5 is heated at constant
volume to a final state of x2=0.85. The final pressure
becomes
a. 1.695 bar,
b. 1.895 bar,
c. 1.795 bar,
d. 1.995 bar,
13.
The specific heat at constant pressure of propane gas
is required at 6 bar, 70°C. The enthalpy data at the
same pressure but at 60°C and 80°C are respectively
given as 576.00kJ/kg, and 615.4kJ/kg. Then the
specific heat value is
The specific latent heat of vaporization, h fg , of
water is measured for three different temperatures
as, T1=75°C, T2=100°C, T3=125°C. The relation
between the corresponding h fg values is
a.
h fg1  h fg 2  h fg 3 ,
b.
h fg 3  h fg1  h fg 2 ,
c.
h fg 2  h fg 3  h fg1 ,
d.
h fg 3  h fg 2  h fg1 .
Sublimation takes place when the pressure of a
substance is
a. above the triple point,
b. below the triple point,
c. above the critical point,
d. below the critical point.
a. 1.97 kJ/kgC,
b.
2.97 kJ/kgC,
c. 3.97 kJ/kg°C,
d.
14.
0.97 kJ/kg°C.
10 kg of steam at 600C is contained in a 182 L of
tank. The tabulated pressure value is represented
by pt , the pressure evaluated by the van der Waals
equation of state is pvw and the pressure computed by
using the compressibility chart is pcf . The relation
between these three pressures may be stated as
a.
pt  pvw  pcf ,
b.
c.
pvw  pcf  pt ,
pcf  pt  pvw ,
d.
pcf  pt  pvw .
C
H
3
A
P
T
E
R
Mass Analysis of Systems
3.1
Introduction
In this chapter an intensive property rate balance for mass is developed by the conservation of mass principle. The mass conservation principle, however, is not applicable to all
processes occurring in nature. In nuclear reactions such as fission or fusion related processes,
due to conversion of mass into energy or vice versa, this principle does not imply. For most
events of interest in mechanical engineering field, the mass conservation principle is strictly
appropriate, and provides means of relating time rate of change of system mass with the
mass rate entering and exiting the system.
Principle 8: Conservation of mass: a. Mass is an extensive property of a system, b. The mass of a system is neither destroyed nor
produced.
3.2
The Equation of Continuity
Mass being an extensive property of a system, as described in Section 1.4, the extensive
property rate equation together with Principle 8 may be applied to a system as follows,
Time rate of change of mass   The net rate of mass transported into the system 


 (3.1)
contained within a system   through the boundaries at an instant time t

63
64
THERMODYNAMICS
As shown in Figure 3.1, for a system enveloped by a control surface, the conservation of
mass principle becomes,
 dm 
 dt  

cv
 m (t )   m (t )
i
i
e
(3.2)
e
 dm 
where, 
 = Time rate of mass accumulation within the system at an instant time t.
 dt cv
 m (t ) = The sum of mass rates transported into the system at an instant time t.
i
i
 m (t ) = The sum of mass rates transported out of the system at an instant time t.
e
e
The analysis of flow processes begin with selecting a region or space called control volume.
Mass might enter or leave the system at several ports and the boundary or the control surface
around the system should be capable of identifying the stream lines at each port so that a precise
application of Eq. (3.2) can be achieved. For instance, the T-elbow of an ordinary shower in
Figure 3.2a serves as a mixing chamber for hot and cold water flows, and to identify clearly the
mixing phenomenon, three stream lines of the cold water (1), the hot water (2), and the mixed
outlet (3) have to be identified on a schematic for fully describing the process (see Figure 3.2b).
As in Figure 3.2c, some portions of the boundary are impermeable, but some sections permit
bulk mass flow.
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 65
The continuity relation given by equation (3.2) is general in the sense that it is appropriate for open as
well as for closed systems. For closed systems, the mass is invariant with time, that is m cv  0 .
The constant mass condition, however, does not necessarily guarantee that the system is closed.
The mass of an open system may still be time invariant when the sums of mass rates entering and
exiting the system are identical. If the mass of that particular open system is constant, then there
will be no accumulation of mass within the system. The flow processes with the system mass kept
constant are called steady flow processes, and the continuity relation for such processes is:
 m (t ) =  m (t )
i
i
e
(3.3)
e
To assure that the system is closed, the system boundary should be impermeable entirely. Thus,
substituting m i (t )  0 , and m e (t )  0 into equation (3.2) results with time invariant mass for closed
systems.
Example 3.1 Two identical tanks of 0.5 m3 in volume contain Hydrogen gas at states of (600 kPa, 20°C) in one and (150
kPa, 30°C) in the other tank. The valve on the line connecting the tanks is opened and tanks come into thermal equilibrium
with the environment at 15°C. Determine the final pressure in the tanks.
66
THERMODYNAMICS
Solution:
For the system in Figure 3.3, there is no flow of mass crossing the control surface. The system is closed, and undergoing process
may be considered as a constant mass process. Thus, m t1   m t2  . The system mass at t  t 1 ; p1 A  600 kPa , T1  293 K ,
V1 A  0.5 m3 , m1 A 
600  0.5
150  0.5
 0.246 kg , p1B  150 kPa , T1B  303 K , V1B  0.5 m3 , m1B 
 0.0595 kg ,
4.157  293
4.157  303
and the total mass at t1 is, m t1   0.246  0.0595  0.3055 kg . Hence, at t  t2 , m t2   0.3055 kg , T2 = 288 K, V2 =
3
0.5+0.5 =1 m , and p2 
0.3055  4.157  288
1
p2  365.75 kPa
or
Example 3.2: As in Figure 3.4, 0.2 m3 of cylindrical rigid tank is subdivided into two sections equal in volume by a partition. Section A contains refrigerant 22 at 30C, 80% vapor, 20% liquid by volume, while section B is evacuated. The valve
on the line connecting the two sections is opened and the flow of vapor refrigerant is allowed until the pressures at both
sections are the same. For a final temperature of 30C, determine the change in the quality of R22.
Solution:
The initial mass of the system, m 0   m A 0   mB 0  , where mB 0   0 . Considering that VA  0.1 m3 , and V fA  0.02 m3 ,
VgA  0.08 m3 , and the saturation tables of R22 yield v f  0.0008519 m3 /kg , vg  0.01974 m3 /kg . The mass of R22,
m A (0) 
0.02
0.08

 27.528 kg.
0.0008519 0.01974
Thus, the initial quality of the mixture is x1 
m Ag
m A ( 0)
or x1  0.147 . For a system in Figure 3.4, since the system is
closed,
m 0   m A t2   mB t2   m2 , and V2 = 0.2 m3, then v2 
x2 
v2  v2 f
v2 fg
V2
 0.00726 m3 /kg . Finally, the quality at state 2 is
m2
 0.339 .
Hence the change in quality becomes, x  x2  x1 , or x  0 192 .
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 67
3.3
The Mass Change of a System
In many engineering analyses rather than the mass rate of change, the amount of change for a
particular time interval is needed. To calculate the mass change of a system, Equation (3.1) has to be
integrated over a time interval from t1 to t2 as following,
t2
m(t2 )  m(t1 )cv   t
in
1
m i (t )dt 

exit
t2
t1
m e (t )dt
(3.4)
where, m   m t2   m t1  cv is the mass change of the system within a time interval ∆t. In
words, equation (3.4) states that the change in the amount of mass contained within a system over
a time interval ∆t equals to the difference between the total of masses flowing into the system and
leaving the system during the same time interval ∆t.
Example 3.3. The cylindrical tank in Figure 3.5 has 0.5 m2 cross sectional area and 2 m height, and contains water vapor and
liquid mixture each of which occupies 50% of the total volume at 250C. A valve located at the bottom of the tank is opened
and liquid water is drawn off. Due to heat transfer from the surroundings, the mixture temperature is kept constant during
the discharging process. If the liquid level of the tank decreases by 200 mm, determine the amount of water drawn off.
Solution:
For the control surface in Figure 3.5, there is no mass flowing into the system. Thus, mi t   0 , and Equation (3.4) be
comes,  m t2   m t1 
cv
  me
Let case 1 and 2 represent respectively the initial and the final states of the process. Then for case 1,
m(t1 ) 
Vf1
vf1

Vg 1
v g1
where V f 1  Vg1  0.5 m3, and with the specific volumes of saturated liquid and vapor at 250C, one may obtain
m t1   409.98 kg
For case 2, V f 2  0.5  0.2  0.5  0.4 m3, Vg 2  0.5  0.1  0.6 m3. Since the temperature is kept constant, the specific
volumes for both states are the same and, the final mass is:
68
THERMODYNAMICS
0.4
0.6

 331.97 kg . The mass of water flowing out of the tank is, me  409.98  331.97 , or
0.00125 0.0501
m( t 2 ) 
me  78 01 kg
Example 3.4: The container in Figure 3.6 has a volume of V (m3) and is initially empty. At time t = 0, the valve on the
connection line opens and the tank liquid of density ρ (kg/m3) flows into the container at a mass rate of m i  ct . Determine
the time needed for completely filling the container.
Solution:
Considering the control surface in Figure 3.6, there is no mass flow out of the system, and equation (3.4) reduces to
t2
m(t2 )  m(t1 )cv   m i (t )dt
t1
t*
*

m(t )  ctdt , m(t * ) 
0
1 *2
ct
2
Letting t1  0 and the time when the filling is completed be t2  t * , the container being empty at t1 = 0, m(0) = 0.
Hence,
t *  2V / c
Since m(t*) = V, then
Example 3.5: Modeling an automobile tire as a system with a fixed volume, air pressure in the
tire basically depends on the temperature. At a surrounding temperature of 25C, the manometric
pressure of the tire is 250 kPa. For an internal volume of 0.5 m3, determine,
a.
the pressure as the tire temperature increases to 50C,
b.
the amount of air to be discharged for reducing the pressure to its initial value.
Solution:
a.
For ideal gas behavior of air, the initial mass is
m1 
p1V1
, and
RT1
p1  p0  pm , then m1 
On the other hand,
b.
m3
p1V1
T1
p2V2
2
and
350  0.5
 2.046 kg
0.287  298
1
2
2
350 323
298
2
379 3 kPa
Let the state of air at (350kPa, 50C) represent the state 3 of air in the tire. Then the amount of air remained becomes,
350 0 5
0 287 323
m3 1 887 k
kg Finally
i ll the
h amount off di
discharged
scharged air is
is, m
m3
1 -m
,
0 159 kg.
kg
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 69
3.4
Integral Formulation of the Continuity Equation
As with the lumped analysis given by equation (3.2), the first step in the derivation of an approximate integral formula is to consider a control volume fixed in space through which mass flows.
In Figure 3.8, the system defined by its boundaries coincides with the control volume boundary at
t=t, and occupies another volume in space during a time interval t. Then the rate of change of mass
within the control volume becomes,
d
m(t  t )  m(t )  dm 
 dV



t
t 0
 dt cv dt cv

lim
(3.5)
The control volume remains its original
position and the regions A and C in
Figure 3.8b represent respectively the
mass entering and leaving the control
volume. The outgoing and incoming
flows of mass through the control surface
may be evaluated by the shaded cylinder
in Figure 3.9. For the control surface in
this figure, the height of the cylinder is
(V∆t)n, and the volume V t ndA .
Thus the mass flow rate becomes, ρ(V∆t)
ndA. In this formulation, V denotes the
velocity of the flow, n is the outward
normal vector, and dA represents the
area element on the control surface. Thus the flow rate of through dA is found to be ρVndA.
The integration of this quantity over a control surface yields the net rate of mass flow as,
 m1 m2 


t 
t 0  t
lim 

cs

V .ndA
(3.6)
70
THERMODYNAMICS
 
where   V  n represents the mass flux transported into the system at time t in a manner perpendicular to the system boundary.
Principle 9: The mass flux transferred
 across the system boundary
 at an
instant of time t is determined by V . The parameters ρ and V represent the density and the velocity of incoming and outgoing flows and are
evaluated at the boundary conditions.
Combining equations (3.5) and (3.6), the integral form of the conservation of mass principle
becomes,
d
 dt

d
 dt


V .ndA 
cv
cs



 dV 
V .ndA  0 

cv
cs




 dV  
(3.7)
This integral has to be carried out along the boundary surface where there is a flow of mass.
Equation 3.7 is actually the application of Reynolds Transport Theorem (RTT) to the principle of
“Conservation of Mass”. In words, the Reynolds Transport Theorem is stated as,
Rate of change of property B Rate of change of property B in CV
The net efflux of property B through the control surface
 dB 
 dB 
 dt    dt  

 sys 
cv
 
b V .dA
 
cs

(3.8)
Where, b=B/m, is the specific property. For the case of mass balance, B=m, and b=1 and substituting these values into equation (3.8), one exactly ends up with equation (3.7). In chapter 4, the
Reynolds Transport Theorem will be applied to the energy of a system. For such a case, B=E, and
then the general form of conservation of energy principle will be obtained.
Definition: “One dimensional flow” Flow fields that are characterized by a single component of
velocity are called one-dimensional flows.
For a one-dimensional flow, instead of actual velocity distribution on the system boundary, a mean
value of velocity is used for calculating the mass flow rate. As shown in Figure 3.10, at a particular
location of the system boundary, the mean value of velocity may be determined as,
Vm,i 
1
Ai

Ai
VdA
(3.9)
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 71
For instance, the actual velocity distribution at cross section A-A is illustrated in Figure 3.11a,
and the average value of the velocity corresponding to the same mass flow rate is presented in Figure
3.11b.
72
THERMODYNAMICS
In a manner similar to Equation (3.9), the average values of thermodynamic properties of the flow
such as the pressure, the temperature of a fluid crossing the system boundary may be evaluated.
Definition: For one-dimensional flows, all intensive internal properties such as p, T, ρ, u, etc.,of the
flowing fluid at instant of time t are constant in value across the boundary of the system.
Experience reveals that the use of mean values for the intensive properties at the flow boundaries
mathematically simplifies many engineering analyses. Therefore, the inlet mass flow rate at surface
1 in Figure 3.10 is:
m 1  1V1 A1
(3.10)
where ρ1 and V1 respectively represent the mean values of density and the velocity of the fluid
crossing the boundary. These values are evaluated at the boundary conditions. Similarly, the mass
flow rate at the outlet in Figure 3.10 is:
m 2   2V2 A2
(3.11)
For systems with single inlet and outlet and having one-dimensional flow, substitution of Equation
(3.10) and (3.11) into equation (3.7) results as,
 dm 
 dt   1V1 A1   2V2 A2

cv
(3.12)
 dm 
At steady-state flow conditions, the term 
 is zero, and ρVA evaluated at any cross-section
 dt cv
will be independent of time and location. Thus, for one-dimensional and steady state flows, the conservation of mass principle is simplified to
VA  Constant
(3.13)
In differential form, this relationship might be described as follows,
d  dV dA


0

V
A
(3.14)
The incompressible fluid model is a model of such fluids for which the density does not vary with
D
 0 and hence, for incompressible, one-dimensional,
time and location. For incompressible fluids,
Dt
and steady flows the continuity equation reduces to
V  VA  Constant
(3.15)
Where, V is called the volumetric flow rate.
Example 3.6: Air flowing through insulated channels 1 and 2 enters a mixing chamber as shown in Figure 3.12. Determine
the velocity of air at the exit of the chamber. State 1: p  1.5 bar , T  25o C , V  2.5 m3 /min , state 2: p  1.5 bar ,
1
1
1
T2  60 C , V2  5 m /min , and state 3: p3  1.3 bar , T3  47.4 C , A3  0.5 m .
o
3
o
2
2
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 73
Solution:
Due to steady flow process, there is no flow accumulation in the system of Figure 3.12. Thus,
1V1   2V2  3V3
p
. SubstituRT
tion of the determined densities and volume flow rates into the above relation yields the exit flow
For ideal gas behavior of air, the densities at each port may be calculated by  
rate as V3  8.653 m3 /min . Considering the definition of volumetric flow rate, the velocity of air at
the exchanger exit is:
V3 
V3
8.653

,
A3 60 x 0.5
V3  0.288 m / s
Example 3.7: As shown in Figure 3.13, a rigid bottle 0.2 m2 in cross sectional area contains saturated liquid and
vapor mixture of R22 at +5C. As the valve is opened, the saturated vapor flows through a heat exchanger and exits the
exchanger at 500kPa in pressure, 50C in temperature. Meanwhile the temperature in the bottle is kept constant by heat
transfer from the surroundings. It is observed that the liquid level in the bottle drops by 100mm in 30 minutes time duration of the discharging process. Determine the discharge velocity of the refrigerant for a cross sectional area of 0.01m2
at the exchanger exit.
Solution:
For a control surface in Figure 3.13, the continuity equation requires that
m(t2 )  m(0)cv   2V2 A2t2
At a constant temperature of +5C, the mass change in the bottle,
m(t2 )  m(0)cv  V (
1
1
 )
vg v f
where V  Ab L , and vg  0.0395 m3 /kg , v f  0.0007889 m3 /kg , and V  0.02 m3. Thus,
m(t2 )  m(0)cv  24.846 kg
74
THERMODYNAMICS
From property tables, v2  0.0586 m3 /kg at the exchanger exit where p2  500 kPa , T2  50o C . In addition, sub-
stitution of t2  1800 seconds, and A2  0.01 m 2 into the continuity equation yields,
V2  0.01  1800
 24.846 and the
0.0586
discharge velocity becomes, V2  0 08 m/s .
Example 3.8: As shown in Figure 3.14, refrigerant 22 enters a compressor at a state of 200kPa, 40C with a
mass flow rate of 0.1kg/s. For refrigerant entrance velocity of 6 m/s, determine the inlet pipe diameter of the
compressor.
Solution:
Applying the continuity relationship of one-dimensional flow to the
AV
inlet of the compressor results as, m 
v
Where A is the tube cross-sectional area and equals to  d 2 / 4 .
Rearranging the continuity equation for the diameter yields,
d

4 mv
V
From property tables, v  0.1467 m3 /kg ,
d=
4  0.1  0.1467
,
6
or
d  55.8 mm .
Example 3.9: As shown in Figure 3.15, air is uniformly distributed into a system by a pipe of 1m in diameter and 5 m in
length. The pipe contains 30 holes of each 3 cm in diameter. The gage pressure in the pipe is 80 kPa, and the outside pressure is atmospheric. The volume flow rate through the holes is estimated by V  0.67 Ah (2p /  )0.5 , where Ah(m2), is the
hole cross sectional area, ρ (kg/m3), is the air density, and ∆p (Pa), represents the pressure drop through the hole. If air is at
a temperature of 27oC, determine the velocity of air at the entrance to the distribution pipe.
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 75
Solution:
For ideal gas behavior of air, the density of air in the pipe is:  
Ah  30 
180
p
 2.09 kg/m3
, 
0.287  300
RT
3.14  0.032
 0.0211 m 2 , V  0.67  0.0211  (2  80000 / 2.09)0.5  3.911 m3/s
4
3.911
On the other hand, V  ApV , and Ap  0.785m 2 , V 
or V  4.98 m/s.
0.785
Example 3.10: A horizontal water storage tank of 4m long and 2m in diameter, as shown in Figure 3.16, initially contains
water at a depth 0.5m. The tank has to be emptied by a 5cm diameter discharge hole located at the bottom of the tank.
Since the velocity through the hole is determined as V  2 gh where h (m) is the water depth, and g (m/s2) represents the
gravitational acceleration. Estimate the time required to empty the tank.
Solution:
With respect to the geometry in Figure 3.16, the cross sectional area of water is, A 
cupied is V 
2
1
LD 2 

  sin 2  , the change in volume for change in time t is: dV  LD 1  Cos 2 d  VAp dt . Since

4 
2

4
V  2 gh and h  R (1  Cos ), Ap 
dt 
2 LD 2
d
2
D2 
1

  sin 2  and the volume oc4 
2

gD
d2
substituting these values into the differential volume and simplifying results as,
4
Sin 2 d
. Integration and substituting the numerical values yield, t  608.19 s or t=10.13 minutes.
(1  Cos )
76
THERMODYNAMICS
Example 3.11: Saturated water vapor at p=100 kPa flows steadily through a porous plate as shown in Figure 3.17. The
vapor is sucked constantly at the plate surface with velocity of 0.1 mm/s. The velocity profile at section CD is given as
3
u
 y
 y 2
 3    2   . For a plate width of 1m, evaluate the mass flow rate through section AD.
uo

 
 
Solution:
For the specified problem, U o  2.5 m/s , V  0.0001 m/s , b  1 m , L  4 m ,   0.002 m , and by Table A1, =0.598kg/m3,
The mass flow across CD: m CD   b


0
udy   b U o
1
 (3  2
0
1.5
)d where = y/, or m CD  0.00209 kg/s
The mass flow across BC: m BC  VA  0.598  0.0001  4  1  0.000239 kg/s
The mass flow across AB: m AB  0.598  2.5  0.002  1  0.00299 kg/s
The mass flow across AD: m AD  m AB  (m CD  m BC )  0.00299  (0.00209  0.000239) or m AD  0.000661 kg/s
Example 3.12: An air conditioning unit receives outside air-water vapor mixture at 35°C, 100kPa with vapor to air mass
ratio of 0.029. This mixture is cooled to 20°C and the vapor-air mass ratio becomes 0.015 at the same pressure. The volume
flow rate of the mixture and the specific volume at the exit conditions respectively are 0.1 m3/s and 0.85 m3per kilogram of
dry air. For one-hour of process time, determine the mass of condensate.
Solution:
 a1  m
 v1  m
 a2  m
 v2  m
 l2
For a unit in Figure 3.18 operating at steady state conditions, the mass balance is satisfied by m
 a represents dry air mass flow rate and is constant in the process, m
 a1  m
 a 2 . Then, m
 l2  m
 v1  m
 v 2 . Let the mass
where m
ratio of vapor to air be represented by   m v / m a , then the mass rate of condensate becomes
 l2  m
 a ( 1   2 ) .
m
The mass flow rate of dry air, m a 
V2
0.1
, m a 
 0.1176 kg/s and m liq 2  0.1176  (0.029  0.015)  1.64  103
v2
0.85
kg/s. The amount of mass condensed in one-hour is,  mliq 2  m liq 2 t or,  mliq 2  5.92 kg .
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 77
3.5
Velocity Measurements
Flow velocity and flow rate measurements are usually needed for industrial process control and
thermodynamic analysis of systems. The velocity of a flow field is usually measured indirectly by the
following instruments: 1. The Pitot Tube, 2. The Vane Anemometer, 3. The Hot Wire Anemometer,
4. Laser Doppler Anemometer.
3.5.1 The Pitot Tube. As shown schematically in Figure 3.19a, the pitot tube measures the difference between stagnation and static pressures of the measurement point at which the velocity to be
determined.
In Figure 3.19, structurally two different tubes, Prandtl and Brabbee, are given and the accuracy
of the measurement is found to depend on the shape of the tip. In accord with the measured height,
h, the velocity of the fluid near the tip is,
V  C 2 gh(  m /   1)
(3.16)
Where “C” is the velocity coefficient and has to be determined by calibrating the Pitot tube. For
gas flows, assuming ideal gas behavior, the velocity at a particular point is calculated as,
78
THERMODYNAMICS
 2kRT   p   
1
V1  C 
  o   1 
k

1
  p1   

(3.17)
Where, po, is the stagnation pressure, and p1, T1 are the pressure and the temperature at a particular
point for which the velocity to be determined.
Example 3.13: The velocity of air in a duct is measured to be 40.1 m/s. The same velocity is measured by a pitot tube and the
recorded pressure is 0.12 m of water column. Take the density of air as 1.2 kg/m3, and calculate the velocity coefficient, C.
Solution:
Applying Eq. (3.16), C 
V
, or C 
2 gh(  m /   1)
40.1
,
2  9.81  0.12  (1000 / 1.2  1)
C  0.905
3.5.2 The Vane Anemometer. As shown in Figure 3.20, the drag force that is caused by the moving
stream on the vanes tends to rotate the vane. As the intensity of the drag force increases due to increase
in air velocity, the rotational speed of the vane anemometer also increases. The vane anemometers
are used to measure air velocity in large flow fields.
The calibration of these anemometers is done by measuring the vane rotational speed at pre-determined wind velocities. Hence for a particular rotational speed there is a corresponding air velocity.
3.5.3 The hot wire anemometer. Electrically heated thin platinum wire (1mm long and 0.005mm in
diameter) is placed perpendicular into a flow field as in Figure 3.21. The heat transfer rate from the
hot wire anemometry (HWA) to the fluid stream is mainly by convection (see section 4.4).
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 79
If the wire current is kept constant, the wire reaches an equilibrium temperature, Tw, and the following relation holds between the power input to the wire and the temperature difference between
the wire and the fluid, Tw  To , as,
I 2R
 A  B V
Tw  To
(3.18)
where, R is the resistance of the wire, I, represents the instantaneous current through the wire, ρ
and V are respectively the density and the velocity of the free stream. The constants A and B in Eq.
(3.18) are to be determined by the calibration curve of the probe. The hot wire anemometry is well
suited for measuring velocity fluctuations in turbulent flow. HWA may record velocity fluctuations
with a frequency up to 105Hz, and velocities as small as 0.02m/s. HWA being sensitive to low velocity
flows, it is also used in boundary layer analysis.
Example 3.14: The velocity data of air (ρ = 1.2kg/m3) flowing through a 24 cm pipe are recorded by a HWA and tabulated by
the following table. Determine the mean velocity, the ratio of maximum to mean velocity, and the mass flow rate of air.
Solution:
r(cm)
Apply Eq. (3.9) to get the mean velocity,
1 
Vm  2 Vo r12 
R 
Vm 

(Vi  Vi 1 )ri ri 
i 1


V(m/s)
r(cm)
V(m/s)
0
9.7
7
6.8
1
9.6
8
5.9
2
9.4
9
4.8
3
9.2
10
3.5
4
8.7
10.5
2.9
5
8.2
11
2.4
6
7.5
11.5
1.0
1 9.7  12  9.5  2  1  1  9.3  2  2  1  8.95  2  3  1  8.45  2  4  1  7.85  2  5  1  7.15  2  6  1  


122 6.35  2  7  1  5.35  2  8  1  4.15  2  9  1  3.2  2  10  0.5  2.65  2  10.5  0.5  1.7  2  11  0.5
The velocity at each strip is the average of bottom and the top velocities. Hence the mean velocity is Vm=4.71m/s,
Vmax=9.7m/s,
Vm
max  2.05
Vm
and m   AVm  1.2  0.0452  4.71 ,
m =0.2554 kg/s.
3.5.4 Laser Doppler Anemometer. In hot wire anemometer applications, dirt in the flow might be
deposited on the wire resulting with an insulation effect, or due to high temperature of the wire, fluid
might be decomposed in the measurement region. Moreover, due to finite size of the probe, the flow
will be disturbed. If these drawbacks become dominant at a particular application then Laser Doppler
Anemometer (LDA) is used and is actually the ideal instrument for non-intrusive measurement of
velocity and turbulence in gas and liquid flows.
As shown in Figure 3.22, a laser of fixed wave length serves as a source of light and optical
components split the laser beam into a reference beam and a secondary beam that intersect at the
measurement particle.
The frequency of light scattered by a moving particle shifts by an amount that is proportional to
the speed of the particle. LDA uses this principle in measuring the velocity of a particle. Both the
80
THERMODYNAMICS
frequency shifted and the unshifted beams are collected at the photo-detector. At the output, a frequency tracking filter locks onto the modulation frequency to get the Doppler frequency. The Doppler
frequency is linearly related to the velocity through the optical system geometry.
This method needs particles for measurements. The particles must be small enough to follow the
flow. In liquid flows such particles are naturally present, but in gas flows, particles 10-6 m in size are
seeded. To transfer the light beam, the fluid medium must be transparent. LDA applications include
pipe flows, flow inside engine cylinders, flow between pump impeller blades, and combustion processes.
3.6
Flow Rate Measurements
The volume flow rate in ducts is mainly measured by two types of instruments; 1.Obstruction
Meter, 2.Rotameter or Flow Meter.
3.6.1 Obstruction Meters. Venturi, Nozzle, and Orifice meters are the three kinds of obstruction
meters commonly used for measuring volume flow rate through pipes and ducts.
a. Venturi Meters. The Venturi meter was first used by J.B. Venturi in 1797 in Italy. As shown in
Figure 3.23, decreasing the cross sectional area of the flow, some pressure head is converted to velocity. The head differential can be measured between the upstream and the throat section to estimate
the flow rate as following:
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 81
V  CA2
2p

 1  4
(3.19)

where, ρ, is the fluid density, and β = D2/D1, for Venturi meters; 0.25    0.50 . “C” represents
the discharge coefficient and varies in the range between 0.935 and 0.988.
The upstream section converges with an angle of 21o from the pipe axis, and the diverging section
has an angle of 5o to 7o. In connecting the Venturi meter to a pipeline, a distance of 10D1 straight pipe
section upstream of the Venturi is required. The overall loss in a Venturi is in between 10% to 20%
of the total pressure drop.
Example 3.15: A venturi meter has to be designed for measuring the water flow rate through a 300 cm diameter horizontal
pipe. The estimated discharge through the pipe is 15 m3/s, and the pressure drop at the venturi is limited to 250 kPa. Calculate
the throat diameter of the venturi. Take the discharge coefficient as 0.95.
Solution:
Referring to Eq. (3.19), the relation between β and A2 is,
1  4

A2
0.95
2  25000
1000
 1.416
15
Substituting,  D 2 / 4 , for A2 and then rearranging the above relation yields, D2  0.948(1   4 )0.25 . This equation is solved
for D2 by trial and error method. Assume =0.35, find D2, D2=0.944 m, and β=0.314, then finally venture throat diameter
is D2=0.945 m.
b. Flow Nozzles. Flow nozzles operate on the same principle as Venturi meters. As shown in Figure
3.24, there is upstream converging section, but there is no downstream diverging section to reduce the
energy loss of the flow. Therefore, the head loss tends to be much higher than Venturi meter.
The upstream pressure is measured at a distance between 0.5D1 and D1. The downstream pressure measured at the outlet of the nozzle. Equation (3.19) is used to estimate the flow rate through
the duct. The discharge coefficient, C, in this equation, however, varies in the range of 0.7 and 0.9
for nozzles.
c.Orifice Meters. As shown in Figure 3.25, an orifice meter is a thin plate with an opening at the
center. The orifice opening is usually circular. The sudden area of contraction in these instruments
leads to a higher pressure loss compared to other two. The upstream and downstream pressures are
measured respectively at a distance of D1 and 0.5D1 from the plate.
82
THERMODYNAMICS
With respect to flow pressure drop, the measured flow rate through the orifice plate is expressed as,
V  KA2 2p / 
(3.20)
where, K, is the orifice plate coefficient. As described in Figure 3.26, K, depends on the ratio of
plate diameters and on the Reynolds number of the flow which is defined as, Re  VD 1 /  .
Example 3.16: A 30cm diameter pipe carries oil, (ρ = 880kg/m3, μ=0.799kg/ms), and to measure the flow rate an orifice of 15cm
diameter is fitted to the pipe. A mercury manometer reads the pressure drop across the orifice as 0.95m. Find the oil flow rate.
Solution:
D2/D1=0.5, and assume high Reynolds number flow. Then, by Fig. (3.26), K=0.62, ∆p = ρmgh = 13.6x9.81x0.95=126.745 kPa.
2  126745
A2=3.14x0.152/4, A2=0.0176m2. Substitute these values into Eq. (3.20), V  0.62  0.0176 
 0.185 m3/s
880
The Reynolds number condition should be checked. A1  0.07 m 2 , V  0.185 / 0.07  2.645 m/s , and
Re  880  2.645  0.3 / 0.799  874 . Respect to Fig. (3.26), K  0.72 , and the corrected flow rate becomes, V  0.214 m3/s.
Figure 3.26 Variation of K coefficient with respect to orifice plate size and
the flow Reynolds number
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 83
3.6.2 Rotameters. As shown in Figure 3.27, this instrument has a tapered glass tube, and a stainless
steel float moving freely inside the tube. As the fluid flows through the instrument, the forces acting
on the float establishes an equilibrium state for which the float assumes a position inside the tube.
The instrument must be installed vertically, and essentially the float motion is linear with the flow
rate. Rotameters are not suitable for very high pressures and for liquids with large number of particles
in it. Having an uncertainty of 5% of the full scale, the accuracy of a rotameter is less than that Venturi
or Orifice meters. The advantage of rotameters over the other obstruction meters is that the flow rate
capacity may be easily changed by changing the float shape or the glass tube.
Example 3.17: A rotameter is calibrated for water, ρ=998 kg/m3, at standard conditions (p=1atm, T=20oC). However, the
rotameter is used to measure the flow rate of oil (ρ=880 kg/m3) and the scale on the rotameter indicates 12 L/s. Determine
the actual flow rate of oil.
Solution:
The flow rate through the rotameter is V  AV , where A is the flow area between the tapered tube and the float. Hence
the flow rate ratio becomes, Vo / Vw  Vo / Vw . The aerodynamic suspension of float is obtained by the same pressure drop
across the float, pw  po . Considering that the pressure drop is proportional with the square of velocity,  wVw2  oVo2
 
, Vo / Vw   w / o ,
Hence,
Vo  12 x1.066 ,
or
Vo  12.79 L/s.
References
1.
M. E. Gurtin, E. Fried, and L. Anand, The Mechanics and Thermodynamics of Continua, Cambridge University Press,
ISBN 978-0-521-40598-0, 2009.
2.
R. B. Bird, W. E. Steward, and E. N. Lightfoot, Transport Phenomena, 2nd Edition, John Wiley & Sons, ISBN 978-0471-41077-5, 2001.
3.
Roger C. Baker, Flow Measurement Handbook, Cambridge University Press, ISBN 978-0-521-48010-9, 2000.
4.
L. Theodore, F. Ricci, and T. Van Vliet, Thermodynamics for Practicing Engineer, John Wiley & Sons, ISBN 978-0470-44468-9, 2009.
5.
R. P. King, Introduction to Practical Fluid Flow, Butterworth-Heinemann, ISBN 07506-4885-6, 2002.
6.
R.S. Brodkey, and H. C. Hershey, Transport Phenomena – A unified Approach, McGrawHill Inc., ISBN 0-07-007963-3,
1988.
84
THERMODYNAMICS
Problems
Mass change of a system
3.1
A 1 m3 tank containing superheated steam at 500 kPa,
200°C is connected to another tank containing 5 kg of
steam at 150°C, 200 kPa. Opening the valve on the
connection line, the entire system is allowed to reach
an equilibrium temperature of 160°C. Determine,
a. the volume of the second tank,
b. the final equilibrium pressure.
3.2
A rigid tank 0.2 m3 in volume contains equal volumes of liquid and vapor Refrigerant-22 at 40°C.
Additional R-22 is charged into the tank until the
final mass is 200 kg. For a final temperature of 40°C,
determine
a. the amount of mass entering the tank,
b. the final volume of the liquid.
3.3
A rigid tank contains 5 kg of air at 200 kPa in pressure and 20°C in temperature. Air is added to the
tank through an opening, and the pressure and the
temperature respectively rise to 300 kPa, 50°C.
Determine the amount of air added to the tank.
3.4
Consider a steady flow system with a single inlet
i =
and outlet. The mass flow rate at the inlet is m
 e = 25t (kg/h) where
50 (kg/h), and at the exit is m
t denotes time in hours. Determine,
a. the rate of change of the system mass at t = 1,
2, 3 hours,
b. the system mass change from t1 = 0 to t2 = 2
hours.
3.5
A refrigerant charging bottle 0.1 m3 in volume
contains liquid-vapor mixture of Refrigerant-22 at
a temperature of 30°C, and the vapor occupies 40%
of the total volume. In a charging process, the valve
on the connection line opens, and the vapor refrigerant at 30°C, and 1 bar, flows through at a velocity
of 10 m/s. For a flow cross sectional area of 0.002
m2 determine the decrease in the amount of liquid
refrigerant in the bottle when the charging process
lasts 5 minutes.
Steady flows
3.6
A vacuum pump is used to pump a vacuum over
a bath of liquid nitrogen. The volume flow rate of
vapor flow into the vacuum pump is 1.8 m3/s. The
pressure and the temperature at the vacuum inlet
respectively are 30 Pa, -40°C. Through the use of
compressibility chart, determine the mass flow rate
of vapor nitrogen entering the pump.
3.7
A boiler feed pump delivers 20 kg/s of water at
250°C, and 50 bars.Determine the volume flow rate
of flow in m3/s.
3.8
A cylindrical tank of diameter d0 closed at the bottom
is partially filled with an incompressible fluid. As
shown in Figure 3.28, a cylindrical rod of diameter
di is lowered into the liquid at a velocity of V0. Express the velocity V of the liquid escaping through
the clearance between the tank and the rod in terms
of the given parameters.
3.9
As shown in Figure 3.29, an incompressible fluid
flows in a pipe of radius R. At section1, the velocity
is uniform over the cross-section with a value of V1.
At section 2, the velocity varies with radius accord-

ing to the relation V  Vc 1  r / R 
2
 where Vc is
the velocity at the tube centerline. Demonstrate that
V1 / Vc  0.5 .
Figure 3.29 Velocity distribution of a fluid
flowing through a round tube
3.10
Turbulent pipe flow velocity distribution is given
as V  Vc 1  r / R 
1/ n
, where Vc is the center line
velocity and R is the pipe radius. Determine the ratio
of the average velocity, V , to the centerline velocity
( V / Vc ) for n=4, 6, 8, 10.
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 85
3.11
As shown in Fig. 3.30, the parallel plates of length 2L
are separated by a distance of b0 at t = 0. The upper
plate moves with a constant velocity V downward
direction and the liquid filling the gap between the
plates is squeezed out. Assuming b<<L, show that
at a particular (x, t), the average velocity in the gap
is u 
Vx
.
b
 o  Vt 
3.12
A double-pipe heat exchanger in Fig. 3.31 contains
three identical and circular inner pipes with a diameter
of d, an outer pipe of diameter, D. If the velocity
of the fluid in the pipes and in the annular region
has to be the same for the same flow rates at both
sides, then determine the ratio of diameters of the
inner to the outer.
3.13
A steady flow of air through a circular pipe is shown
in Figure 3.32. At cross-section 1, the volume flow
rate is 0.15m3/s, and the pressure and temperature
respectively are 1.2 bar, 100°C. Determine the air
velocities at each section for the case that d1 =
40 cm, d2 = 18 cm, d3 = 60 cm, ρ2 = 0.6 ρ1, and
ρ3 = 1.2 ρ1.
3.14
Air enters steadily to a converging round tube at 5
bar, and 400°C with a mass flow rate of 600 kg/h,
and exits at 1bar, 50 m/s. For an area ratio of inlet
to exit A1/A2 = 3, and an inlet cross sectional area
of 500 cm2. Determine,
a. the inlet velocity,
b. the temperature and the diameter at the exit
conditions.
3.15
An air compressor draws air at a flow rate of 0.2 kg/s
through an inlet section where the pressure and temperature respectively are 0.9 bar, 20 °C. Determine the
inlet section diameter for an air velocity of 10 m/s.
3.16
As shown in Figure 3.33, in a combustion chamber
of a jet engine, 8000 m3/min of air having a density
of 1.15 kg/m3 enters through a cross-sectional area
of 0.1 m2. Fuel in the ratio of 0.015 kg fuel/kg air
is injected into the combustor. The combustion
products having a density of 1.35 kg/m3 exit the
combustor through an area of 0.08 m2. Determine
the exit velocity of combustion gases.
3.17
In Figure 3.34, a primary fluid stream of high velocity entrains a secondary stream of the same fluid at
low velocity at section 2. As a result of mixing in
a constant diameter mixing tube, the streams are
thoroughly mixed and uniform in velocity at section
3. Consider the case for which saturated steam as
secondary fluid is at
p2 = 40 kPa, V2 = 3 m/s, and the primary fluid being
vapor at 100°C is at the same pressure. For V1 = 30
m/s,
A1 = 0.01 m2, and A3 = 0.1 m2, determine the mixture
velocity at the ejector exit.
Figure 3.34 A schematic of a jet ejector
86
3.18
THERMODYNAMICS
In Figure 3.35, air water vapor mixture with vapor to
air mass ratio of 0.02 enters the cooling coil of an air
conditioning unit at 40°C, 1 bar, and V  400 m3 /h .
At the coil exit, the mixture temperature drops
to 20°C, the mass ratio becomes 0.015, and the
flow cross-sectional area decreases to 1/2 of the
inlet section. For v1  0.92 m 3/kg dry air, and
A 1 = 1m2, determine,
Figure 3.37
a. the mass flow rate of dry air,
b. the rate of water condensation,
c. the exit velocity if the mixture assumes a specific
volume of v2 = 0.85 m3/kg dry air at the exit.
3.21
A circular cylinder of diameter d=50 cm and
length L=1.5m was immersed in a steady twodimensional incompressible flow. Measurements
of velocity were made at the boundaries of the
control surface shown in Fig. 3.38 and Vo=10
m/s. The x-component of the velocity at the exit
of control surface boundary was approximated as
indicated by the sketch. Determine the total mass
flow rate through surfaces AD and BC.
Figure 3.35
3.19
For the globe valve shown in Fig. 3.36, the distance between the disc and the seat is 5 mm and
the diameter of the opening is 4 cm. If water is
flowing through the valve at a rate of 40 L/min,
find the velocity of water at the opening section
of the valve.
Figure 3.38
Unsteady flows
3.22
The pistons in Fig. 3.39 have diameters of Da  9
cm, Db  3 cm but the piston B moves three times
faster than piston A ( Vb  3Va ). Determine if the
water level in the tank is rising, falling, or not moving at all.
Figure 3.36
3.20
Air flows past an object in a pipe of 2 m diameter
and exits as a free jet as shown in Fig. 3.37. The
velocity distribution at the upstream is uniform at
10 m/s. As indicated in the figure, at the pipe exit,
the velocity distribution is non-uniform. Determine
the velocity V3.
Figure 3.39
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 87
3.25
The tank shown in Figure 3.42 initially contains 500
kg of brine (salty water) containing 10-percent of
salt by mass. An inlet stream of brine, that contains
20-percent of salt by mass, flows into the tank at a
rate of 10 kg/min. The mixture in the tank is kept
uniform by stirring action of a paddle-wheel. The
brine is also removed from the tank by an outlet pipe
at a rate of 5 kg/min. Determine,
a. the amount of salt in the tank as a function of
time,
b. the time elapsed when the amount of salt in the
tank is 150 kg.
Figure 3.40
3.23
A cylindrical tank with an open top as in Figure
3.40 initially contains water that occupies 1/200 of
the tank volume. As the rain starts falling at a rate
of 10t (kg/s), the valve at the bottom of the tank is
opened, and the water outlets at a constant flow rate
of 20 kg/s. For a tank volume of 4 m3, express the
amount of water in the tank as a function of time,
and determine the filling time of the tank.
3.24
As piston moves upward in Fig. 3.41 for the exhaust
stroke of the cycle, the mass flow rate through the
exhaust port is given as m  0.65 pAv / RT where
p and T are the cylinder pressure and temperature
respectively, and Av, is the gas escape area of the
valve. Consider the case for which the diameter of the
cylinder is 5 cm, and the piston is moving upward at
a speed of 25 m/s. At an instant for which the clearance between the piston and the head is 12 cm, and
the valve opening is 1.5cm2, the gas pressure and
the temperature are measured as 300 kPa, 627oC
respectively. For a gas constant of 0.32 kJ/kgK,
determine how the rate of gas density is changing
in the cylinder.
Figure 3.42
Figure 3.43
3.26
As shown in Figure 3.43, a large tank of volume V
filled with air at an initial pressure p1 and temperature
T1. The tank is situated within a large test chamber
whose pressure is maintained constant at p0. When the
valve is opened, drive a differential equation by which
the time rate of change of tank pressure is related to
 e , at the exit.
T1, V and the mass flow rate, m
Velocity and flow measurements
3.27
Figure 3.41
A venturimeter with throat diameter of 10 cm and
a coefficient of discharge 0.95 measures the water
flow rate of a pipeline and the pressure difference
across the venturimeter is determined to be 15 Pa.
The same flow rate of the pipeline is also measured
by an orifice meter with 10cm diameter and the
88
THERMODYNAMICS
pressure difference across the orifice is found to be
30Pa. Determine the discharge coefficient of the
orifice meter.
3.28
3.29
3.30
3.31
A pitot tube is used to measure the velocity of water
in a pipeline. If the mercury manometer attached to
it indicates a reading of 8 cm, evaluate the water
velocity in the pipe for a velocity coefficient of
0.92.
Oil (ρ=800kg/m3) at a rate of 0.15 m3/s flows through
a pipe of 30 cm in diameter. In calibrating a venturimeter with 0.15 m throat diameter, the same flow
rate is measured and the mercury manometer attached
to the meter indicates 0.28 m of reading. Evaluate
the discharge coefficient of the venturimeter.
A sharp-edge concentric orifice is employed to
measure the flow rate of water in a pipeline with 10
cm of diameter. If D2/D1=0.4, and if the flow rate
is 0.045m3/s, evaluate the differential pressure for
water temperature of 10 oC.
Oil (ρ=910 kg/m3) flow rate is to be metered by
a 5  2.5 (5 cm diameter pipe with 2.5 cm throat)
venture. The maximum flow rate is limited by 120
liters per minute, and the ambient temperature will
be in the range of +5oC and +35oC. At these conditions, what maximum differential pressure must be
accommodated on the manometer to be used?
True and False
3.32
e.
One-dimensional flow fields can be
described by a single component of velocity.
f.
Mean thermodynamic values at a particular cross-section are the area averages over
that cross-section.
g.
For one-dimensional steady flow of an
incompressible fluid, if the flow cross-section is
constant, then the velocity remains constant.
h.
The density of an incompressible fluid
is constant with respect to time but might change
along the flow path.
i.
An ideal fluid is a good approximation
of real fluid if the viscosity is small.
j.
Compressible flow is flow of gases.
k.
Turbulent flow is an unsteady flow.
l.
Incompressible flow is defined as flow
when the density varies with time and location.
m.
In a pitot tube the opening perpendicular
to the flow direction measures the static pressure.
n.
The coefficient of discharge of a venturimeter is the ratio of the actual flow rate to
the ideal flow rate.
o.
A hotwire anemometer is an electrically
heated wire placed in a flow field to measure
velocity.
p.
Pitot tube is used to measure the dynamic
pressure.
Answer the following questions with T for true and
F for false.
a.
If the mass of a system is time invariant
then the system under consideration must be a
closed system.
b.
The boundaries of a closed system do
not allow any mass interaction.
c.
For one-dimensional flow, all properties
are constant with time at the flow boundary.
d.
A steady-state flow process requires
that the rate of change of the system mass has
to be constant.
Check Test 3
Choose the correct answer:
1.
The integrated circuits of a laptop computer are to
be cooled by a fan whose flow rate is 0.25 m3/min.
For air density of 0.75 kg/m3 at the location of the
computer, the mass flow rate of air is
a. 1.12 g/s
c. 3.12 g/s,
b.
d.
2.12 g/s,
4.12 g/s.
CHAPTER 3 MASS ANALYSIS OF SYSTEMS 89
2.
Water enters a flat channel as shown in the figure
with a uniform velocity of 3 m/s. For a channel
height of h the exit velocity distribution is given as
u  u0 1  2 y / h  . Then the velocity u0 at the exit
centerline is
a. 4 m/s,
c. 6 m/s,
3.
4.
An incompressible and inviscid fluid with density of
800 kg/m3 is flowing through a diverging channel
as in the figure. The velocity at section 1 is
a. 10.5 m/s
b. 11.5 m/s,
c. 12.5 m/s,
d. 13.5 m/s.
7.
Ammonia enters a 100 m3 storage tank at 10 bar, 100°C
through a 10 cm diameter pipe with a velocity of 30
m/s. At the outlet pipe, 20 cm in diameter, ammonia
leaves the tank at 5 bar, 60°C with a velocity of 30
m/s. The rate of change of the density of ammonia
in the tank is
5 m/s,
7 m/s.
The following two-dimensional flow satisfies the
continuity relation,
a.
u  A sin xy  , v   A sin xy 
b.
u  x y ,v  x y
c.
u  2 x 2  cy , v  3 y 2
d.
u  x  2 y , v  2x  y
The following two-dimensional flow represents an
incompressible steady flow,
a.
5.
b.
d.
6.
u  4 xy  2 y 2 , v  3 x 2  6 xy
2
a. 0.016 kg/m3s decreasing
2
b.
u  x  y , v  7  2 xy
b. 0.026 kg/m3s increasing
c.
u  x/ y ,v  y/ x
c. 0.016 kg/m3s increasing
d.
u  2x  y , v  x  4 y
Flow of air through a duct of constant cross sectional
area is cooled by cold water coil as in the figure.
The air density at the inlet and outlet respectively
are ρ1=1.01 kg/m3 and ρ2=1.15 kg/m3. The percent
increase/decrease in the velocity of air is
a. 13.8 decrease
b. 12.2 increase,
c. 13.8 increase,
d. 12.2 decrease.
d. 0.026 kg/m3s decreasing
8.
Steam at 3 bar, 280°C flows through a pipe of 40
cm in diameter at a speed of 35 m/s as in the figure.
At a junction point, steam splits into three branches
with equal diameters of 20 cm. If the pressure and
temperature in smaller pipes are 1.5 bar and 200°C,
the steam velocity in smaller pipes is
a. 59.6 m/s
b. 69.6 m/s
c. 79.6 m/s
d. 89.6 m/s
90
9.
THERMODYNAMICS
A tank with a square cross sectional area (a=1m)
contains water as in the figure. At side 1, water
flowing into the tank through a pipe 0.15 m in
diameter, and at side 2, water discharging through
a pipe of diameter 0.05 m. The rate of water level
is rising or falling in the tank as
11.
a. 0.011 m/s rising
12.
The coefficient of discharge of an orifice meter is
in the range of
a. 0.85 and 0.9
c. 0.92 and 0.98
b. 0.021 m/s falling,
c. 0.021 m/s rising,
13.
A storage tank has a cross sectional area of 0.12
m2 and contains mixture of liquid-vapor of refrigerant R134a at 30C. Due to small crack on the
top surface of the tank, saturated vapor flows out
and the liquid level drops by 8 cm. The change in
refrigerant mass is
a. 11.428 kg
b.
11.528 kg
c. 11.066 kg
d.
10.066 kg
15.
b. pressure
d. flow rate.
The velocity without causing any disturbance on the
flow may be measured by
a.
b.
c.
d.
10.
b. venturimeter
d. orifice meter.
An anemometer measures
a. viscosity
c. velocity
14.
0.7 and 0.85
0.65 and 0.7.
Which one is more accurate in measuring a flow
rate?
a. pitot tube
c. rotameter
d. 0.011 m/s falling.
b.
d.
hotwire anemometry
orifice mater
pitot tube
laser-doppler anemometry.
Pitot tube measures
a. the stagnation pressure
b. the dynamic pressure
c. the static pressure
d. the difference between dynamic and static pressure.
C
H
4
A
P
T
E
R
Energy Analysis of Systems
4.1
Introduction
Since the early ages of human history, man has been aware of the fact that a force was
required for causing a change in surroundings. It was also realized that the effort for affecting
such a change was proportional to the product of the applied force and the distance traveled.
The magnitude of the resulting change, however, depends upon the capacity of the system
exerting the force. The capacity of the system causing a physical change in its environment
is called energy. Thus energy is the capacity either latent or apparent to exert a force through
a distance. In this chapter, the principle of energy conservation and the application of this
principle to energy interactions of thermodynamic systems are studied.
Principle 10: Principle of energy conservation. a. Energy is an extensive property of
a system, b. Energy can never be destroyed nor created but can be transformed into
another form. This principle is also called the first law of thermodynamics.
Since energy is an extensive property of a system, with respect to Equation (1.5), the energy
conservation of a system may be stated as,
 Time rate of energy 
 accumulation within 




a system at an instant 


of time t

The net rate of energy 
 transported into the 




 system at that instant 


of time t
(4.1)
The term on the left hand side of the above expression represents the change in the
system energy, and the term on the right is the energy transferred across the boundaries of
the system.
91
92
THERMODYNAMICS
4.2
Energy of a System
The energy of a system is considered to be either external or internal in nature. The external forms
of energy are those associated with the motion and the position of a system’s mass. For instance, the
kinetic energy is the energy of mass due to its motion. At an instant of time t, a system of mass m
having a velocity V(t) with respect to earth’s surface possesses the following kinetic energy,
Ek(t) =
1
2
mV 2(t)
(4.2)
Where units of m, and V are kg and m/s respectively, and the kinetic energy Ek is in joules.
The potential energy is the energy stored in mass due to its position in the gravitational field with
respect to a reference position. At an instant of time t, a system of mass m elevated by a distance Z
with respect to a reference surface in a gravitational field has the following potential energy,
Ep(t) = mgZ(t)
(4.3)
Where g is the gravitational constant, taking the units of m and Z in kilograms and in meters
respectively, the associated potential energy is measured in Joules.
The internal form of energy is called internal energy, and related to the mass composition of the
system. The portion of the system energy which exceeds the sum of kinetic and potential energies is
named as the internal energy. Therefore, the energies of all sort which can not be classified as kinetic
or potential energies are determined to be the internal energy of a system. In mathematical terms, the
system internal energy at an instant of time t may be defined as,
U t   E t    Ek t   E p t 
(4.4)
The following examples illustrate the meaning of internal energy more rigorously.
1. An Automobile Battery
Consider a battery situated on a table and upon closing the circuit
switch, the stationary fan starts turning.
Thus the system causes a change in its surroundings. Since the
potential and kinetic energies of the battery are zero, this change is
provided by the chemical energy consumption. Considering the definition given by equation (4.4), the stored chemical energy is the internal
energy of the battery.
Figure 4.1 Stored energy in an
2. A compressed gas
accumulator is internal energy
Consider a gas compressed in a stationary piston-cylinder apparatus
for which the initial pressure is greater than the pressure of the
surroundings  p  po . Upon pulling the locking mechanism, the
piston rises up, and pushes the surrounding air to a certain extent.
Since the kinetic and potential energies of the system are zero, the
change in the surroundings is caused by the internal energy of the
gas. For this particular case, the internal energy is the sum of kinetic
and potential energies of all molecules comprising the gas, and for
vapor-liquid mixtures, can be calculated by equation (2.9), and for
Figure 4.2 Compressed gas posan ideal gas, by equation (2.35).
sesses internal energy
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 93
3. A compressed spring or an extended bar
A stationary and massless spring in Figure 4.3a is compressed by X meters from its free length.
Upon releasing the locking mechanism, the spring expands, and causes a change in its environment.
Because of zero kinetic and potential energy, this change is provided by the internal energy of the
spring and calculated as follows,
Figure 4.3 The stored energy of a compressed spring or an extended bar is internal energy.
U (t ) 
1
KX 2 (t )
2
(4.5)
Where K is the spring constant in kN/m, and X is the elongation from free length in meters.
Similarly, in extending the elastic bar of Figure 4.3b, the internal energy stored in the bar may
be determined as,
U t  
AE
( L )2
2 L0
(4.6)
Where E, A, and L0 are the modules of elasticity in (kN/m2), the cross-sectional area of the bar in
(m ), and the initial length in (m) respectively.
As a consequence, the energy of a system at an instant of time t may be determined by the sum
of three energy terms as follows,
2
E(t)= U(t)+Ek(t)+Ep(t)
(4.7)
Energy, like pressure, volume, and temperature, is a state property of the system, and its numerical
value at an instant of time t is path independent. In other words, the way in which the system arrives
at a particular state at time t is of no concern in determining the system energy. The derivative of
equation (4.7) with respect to time yields the rate of change of system energy at an instant of time t
as follows,
E (t )  U (t )  E k (t )  E p (t )
(4.8)
94
THERMODYNAMICS
As defined in chapter 1, a system which is closed to energy transfer across the boundaries during
a change of state is an isolated system. Thus, for an isolated system, the time rate of energy accumulation at an instant of time t must be zero.
U (t )  E k (t )  E p (t )  0
(4.9)
For a change of state of an isolated system within a time interval (t1, t2), the integration of equation (4.9) yields,
U (t2 )  U (t1 ) Ek (t2 )  Ek (t1 )  E p (t2 )  E p (t1 )  0
(4.10)
U  Ek  E p  0
(4.11)
or,
As indicated in Figure 4.4, the energy transforms from one form into another within the boundaries during a change of state of an isolated system. A pure mechanical system in which no frictional
interactions occur is a special form of an isolated system. The total energy of a pure mechanical system
consists of kinetic and potential energy and the sum remains constant during a change of state.
Figure 4.4 Energy transformations for an isolated system
For example, if a ball thrown vertically up into the air, the ball leaves thrower’s hand with a
certain velocity, hence, with a certain kinetic energy. When the ball reaches its maximum height, its
velocity vanishes and the ball possesses only potential energy. Thus the initial kinetic energy of the
ball is transformed into gravitational potential energy.
Example 4.1: As shown in Figure 4.5, a 2 kg mass is released from rest in the position 1 and slides along the smooth and
frictionless rod. For a linear spring of K = 400 N/m, having a free length x0 = 50 cm, determine the velocity when the mass
strikes the support at 2.
Solution:
The mass-spring system sliding on a frictionless rod can be considered as an isolated system, and the internal energy change
is due to the elongation of the spring.
U 
1
1
K ( x12  x22 )   400  0.36  0.715   71 Joules
2
2
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 95
The potential energy change is
Ep = mg (x2 – x1) = 2 × 10 × (0 – 0.846) = 16.92 Joules
and the change in kinetic energy becomes,
1
1
Ek  m(V22  V12 )   2  (V22  0)  V22
2
2
Substitution of these energy terms into equation (4.11) yields,
V22
 16.92  71  0
V2 = 9.37 m / s
Example 4.2: In stress release of molded machine parts, one method
is to submerge the part into a liquid nitrogen tank and retain it in until
a thermal equilibrium is established. A 5 kg piece with a specific heat
of 0.4kJ/kg K is initially at 300 K, and submerged into a 3 kg liquid
and vapor mixture of nitrogen at 1 bar. To get a final temperature
equal to the saturation temperature of nitrogen at 1 bar pressure, estimate the initial percentage of liquid nitrogen in the mixture. Take
ufg= 176.9 kJ/kg for nitrogen.
Solution:
Considering the tank and its constituents as an isolated system, because of no change in kinetic and in potential energies,
equation (4.11) reduces to,
U n  U m  0
where Um is the internal energy change of the machine part and calculated as,
Um = mcT = 5 × 0.4 × (77 – 300) = –445.4 kJ.
For nitrogen, U n  mn x2  x1 u fg, since ufg = 176.9 kJ/kg, then the difference in quality becomes,
x2  x1 
U m
445.4

 0.839
mnu fg 3  176.9
At the final state, nitrogen is in a state of saturated vapor, x2 = 1.0, and by using the above result, x1 = 1 – 0.839 = 0.161.
Thus, at the initial state, 83.9 % of nitrogen by mass must be at liquid state.
Example 4.3: A rigid and insulated tank 20 L in volume is initially evacuated.
As shown in Figure 4.6, a 1 L glass-flask placed in the tank is full of liquid
water at 7 bars, 150°C. Determine the final temperature of water when the
glass-flask bursts and the water and water vapor occupies the entire tank.
Solution:
Considering the system boundary in Figure 4.6, the system composed of
the rigid tank and the glass-flask is an isolated system, and E(t2) = E(t1). Because of zero change in kinetic and potential energy terms, equation (4.11)
reduces to U2 = U1. From saturated steam tables, the internal energy of water
at 150°C, u1 = 681.68 kJ/kg, the specific volume v1 = 0.00109 m3/kg, and the
V
mass of water becomes, m  1  0.917 kg. The specific volume at state 2,
v1
V
v2  2  0.0218 m3/kg, and because of constant internal energy u2 = u1 = 631.68 kJ/kg.
m
Due to expansion of water to a higher volume, the final pressure in the tank will be less than 7 bars. Assuming liquid-vapor
mixture at the final state, the quality of the mixture may be defined in terms of the specific volume difference and the internal
energy difference as well. Considering the fact that both of these definitions must yield the same result for the quality,
x2 
v2  v2 f
v2 fg

u2  u2 f
u2 fg
96
THERMODYNAMICS
Assuming a pressure, p2 < 7 bars, and through the use of saturated steam tables, the above relation has to be satisfied
by trial and error method. Thus the temperature at the final state becomes 135°C.
Example 4.4: As shown in Figure 4.7, an insulated cylindrical tank 0.05 m2 in cross-sectional area is subdivided into two
equal compartments by a frictionless piston. One section contains 1 kg of air at 10 bars, 200°C, and the other is evacuated.
A linear spring of 100 kN/m spring constant supports the piston at the
evacuated section and initially is at its free length. Upon releasing the
locking mechanism, the air expands and a final equilibrium state is
reached. Determine the final pressure and temperature of the air.
Solution:
The boundary in Figure 4.7 defines the system as an isolated one. The
change in kinetic and potential energy being zero, equation (4.11)
reduces to
U 2  U1 a  U 2  U1 b  0
In section b, at the initial state, the spring is at its free length,
and thus U1b  0 . Assuming air as an ideal gas, the change in internal
energy is, U 2  U1 a  mcv T2  T1 
Considering the geometric constraints of the cylinder, the final
pressure of the tank is
p2 
K V2  V1 
A2
Together with ideal gas equation of state, pV  mRT , defining the constant volume specific heat as cv = R/(k-1), the
internal energy change of air may be transformed into
mcv T2  T1  
K (V2  V1 ) 
1 
p1V1 
V2  .
( k  1) 
A2

Finally, the internal energy of spring at state 2 becomes, U 2b 
1 (V2  V1 )2
.
K
2
A2
K (V2  V1 ) 
1 V2  V1 
1 
K
V2  .

 p1V1 
2
k
(
)

2
1
A
A2


2
Thus, the energy conservation principle requires that
Substituting p1 = 1000 kPa, V1 = 0.135 m3, k = 1.4, K = 100 kN/m, A = 0.05 m2, into the above relation yields,
V22  0.1575V2  0.00022  0
Solving for V2 results as V2 = 0.1448 m3. Then the final pressure of air is,
p2 
K V2  V1  100  0.1448  0.135 

, or p2 = 392 kPa. The use of p2V2  mRT2 , gives the final temperature as,
A2
0.052
T2 = 197.7 K.
4.3
Forms of Energy Transfer
In addition to transformation of energy from one form into another within the system boundaries,
transfer of energy across the boundaries may also take place. Substituting equation (4.8) into (4.1),
the energy conservation principle becomes,
U (t )  E k (t )  E p (t ) 
The net rate of energy 


 transported into the 
 system at time t 


(4.12)
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 97
The transfer of energy into or out of a system can be performed in three different forms as follows:
a. convective transfer of energy, b. heat transfer of energy, c. work transfer of energy. The convective transfer of energy is associated with flow processes. Whenever there is a flow of mass across the
boundaries of a system, the energy contained within the mass is also transferred. A certain amount of
mass flowing in or out of a system carries its internal, kinetic, and potential energy in or out of the
system. This is called convective transfer of energy. Besides, the energy transfer may be achieved
either by heat or work transfer.
4.4
Heat Transfer
Definition: If the transfer of energy between two systems is solely due to temperature difference, then
the transferred energy is called heat.
Heat is transferred from a system at higher temperature to a system at lower temperature. It is
not necessary for systems to be in contact. The physical ways by which heat can be transferred are
threefold: a. Conduction, b. Convection, and c. Radiation.
4.4.1 Heat transfer by conduction. This mode of heat transfer occurs when there is a temperature
gradient across a body and is performed by purely internal motion of molecules of the media –Diffusion. Higher temperatures are associated with higher molecular energies and when they collide with
low energy molecules the transfer of heat energy takes place. No external motion of the media is
involved in the transfer process. The simplest heat conduction is one dimensional heat flow and can
be described by the contact of a metal bar to a hot and a cold copper blocks as indicated in Figure 4.8.
For such a case, the temperature distribution within the bar is only a function of location measured
from the face of the bar.
T
, Where,
n
T
(K/m), the
Q (W), is the heat transfer rate, k (W/mK), the thermal conductivity of the material,
n
temperature gradient within the body, and, A (m2), is the cross sectional area perpendicular to the
heat flow direction. Therefore, the heat transfer rate through the brick in Fig. 4.8 can be estimated as,
T
, T  T1  T2 . Hence decreasing the thickness, and increasing the cross sectional area,
Q  kA
L
or the thermal conductivity of the material will increases the heat transfer rate. The thermal conductivity being a function of temperature plays an important role in conduction heat transfer and is a
property of the material. In Table 4.1, experimentally determined thermal conductivities of various
substances are given.
Heat transfer rate in conduction mode is determined by Fourier’s law as Q   kA
98
THERMODYNAMICS
Table 4.1 Thermal Conductivity [k(W/mK)] of various materials at 0°C
Metals
Non-metalic solids
Copper (pure)
385
Marble
Aluminum (pure)
202
2.2
Liquids
Gases
Mercury
8.21
Hydrogen
0.175
Glass, window 0.78
Water
0.556
Helium
0.141
0.054
Air
0.024
Iron (pure)
73
Oak
0.17
Ammonia
Carbon steel 1%C
43
Sawdust
0.059
Lubricating oil SAE50 0.147
Water vapor(sat.) 0.0206
Chrom-nicel steel
(18%Cr), (%8Ni)
16.3
Glass wool
0.038
R134a
Carbon dioxide
0.073
0.0146
Example 4.5: As shown in Fig. 4.8b water heater is covered up with an insulation layer over a total surface area of 2.5 m2.
The inside and the outside surface temperatures of the layer are 82°C and 20°C respectively. For glass wool insulation,
determine the thickness if the heat loss is limited to 250 W.
Solution:
Assuming that the thermal conductivity of the insulation material has not changed appreciably with temperature, then we may
62
T
use Table 4.1. For glass wool, = 0.038W/mK, and the thickness of the insulation is L  kA
, or L  0.038 x 2.5
,
250
q
L = 23 mm
4.4.2 Convective heat transfer. The convection heat transfer
mode is comprised of two mechanisms: i. Random molecular motion (diffusion), and ii. Energy transferred by bulk or macroscopic
motion of the fluid. As shown in Fig. 4.9, the convection heat
transfer takes place when a cool fluid flows over a warm body or
vice versa. The fluid adjacent to the body forms a thin layer called
the boundary layer. Since the velocity of the fluid at the surface of
the body is reduced to zero, at this point, the heat is transferred by
conduction and the moving fluid carries the conducted heat away.
Hence the temperature gradient at the body surface depends on
the rate at which the fluid carries the heat away.
Figure 4.8b
The overall effect of convective heat transfer is expressed by Newton’s cooling law as, Q  hA(Ts  T ) ,
Where, Q , (W), is the heat transfer rate, A (m2), the heat transfer surface area, Ts, and T∞ are the surface
and fluid temperatures respectively, and, h (W/m2K), is the heat transfer coefficient.
Depending upon the nature of the fluid flow, two types of convection occur. If the motion of the
fluid is due to some external means such as a fan, a pump, or a blower, then this type of flow is called
forced convection. If the flow is induced by buoyancy effects caused by the density differences, then
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 99
the flow is called natural convection or free convection. Fig. 4.11a shows the forced convective cooling of a circuit board by a fan, and Fig.4.11b illustrates the cooling of hot circuit components in still
air. Since the hot air on the board surface is lighter than the surrounding air, it moves upward and the
space left behind is filled by cold air and hence free convection of air will be created. The approximate
values of heat transfer coefficients for different modes of convection are provided in Table 4.2.
Table 4.2 Values of convection heat transfer coefficients (h W/m2K) for various convection modes
Natural convection,
∆T = 30°C
Vertical plate in air
L = 300mm,
4.5
Air flow over a square plate, V
= 2 m/s, L = 200 mm
12
Horizontal cylinder in air,
D = 50mm
6.5
Air flow over a square plate,
V=35m/s, L=750mm
75
Horizontal cylinder in
water, D = 20mm
890
Air at p = 200kpa flowing in
a tube, V = 10m/s,
D = 25mm
65
Upper surface of hot
horizontal and square
plate in air L = 300mm
6.0
 = 0.5kg/s
Water at m
flowing in D = 25mm
diameter tube
3500
Forced convection
Air flowing across a tube,
V = 50m/s, D = 50 mm
180
Air flowing across a
tube bundle of staggered
arrangement with 10 rows, V
= 6m/s, D = 16 mm,
St = Sl St/D = 1.9
142
Boiling of water
Boiling in a pool or a
container
2500-35000
Lower values are used
for film boiling and
upper values are for
nucleate pool boiling
Boiling as flowing in
a tube
5000-100000
Lower values are used
for bubbly flow regime
and upper values are for
annular flow regime.
Condensation of water
vapor at 100kpa
Vertical (plates and tubes)
surfaces
4000-11500
Lower values are for
laminar and upper values
are for turbulent film
condensation
Outside of a horizontal tube
9500-25000
As the tube diameter
decreases upper values
are used. For a vertical
tier of N tubes, the heat
transfer coefficient has to be
multiplied by N

1
4
100
THERMODYNAMICS
Example 4.6: As shown in Fig. 4.11, annealed steel plates (L = 0.75m) suspended vertically and are at a temperature of
350°C. The plates have to be cooled by air flowing at a speed of 35m/s. If air is at a temperature, To = 25°C. Determine the
initial rate of heat transfer from a single plate.
Solution:
As indicated in Fig.4.11, the heat loss through the surfaces of plates is
due to solely by convection. If we assume that Vc/Va1 then the heat
transfer coefficient of the surface becomes h = 75W/m2K (Table 4.2). The
heat transfer surface is the both sides of the plate, A  2 x (0.75 x 0.75)  1.125
o
o
m 2 , and Ts  350 C , To  25 C , then Q  75 x1.125 x (350  25) or
Q  27421.875 W
4.4.3 Radiative heat
transfer. All substances emit energy by
means of electromagnetic
radiation propagated as
a result of temperature
difference is called
Figure 4.11
thermal radiation. Both
conduction and convection require the presence of a medium for
the transfer of energy. In radiation, however, energy is transferred
by electromagnetic waves, or photons and there is no need for a
material medium. An ideal thermal radiator is so called black body
and is a perfect emitter and also perfect absorber.
The heat transfer rate that a black body emits is proportional
to the forth power of its absolute temperature and calculated by, Q   A(T 4  T04 ) . This is called
Stefan-Boltzman law of thermal radiation, and  is the constant of proportionality ( = 5.668 × 10-8W/
m2K4). Since thermal radiation is proportional to the forth power of temperature, this model of heat
transfer becomes more important with rising temperature levels and may be totally dominant over
conduction and convection at very high temperatures. Therefore, thermal radiation is important in
fires, furnaces, rocket nozzles, solar energy collectors etc..
Real bodies emit thermal radiation less than the ideal black body, and the ratio of emissive powers
of the real to the ideal at the same temperature is called emissivity, ε, of the body. The emissivity of
surfaces varies in the range of 0 ≤  ≤ 1, and for a black body  = 1. In general, the radiative properties are usually difficult to measure and often display erratic behavior. In addition, due to complexity
inherent in the analysis of thermal radiation, we often result with approximate answers. For instance,
consider the radiation heat exchange between two bodies at temperatures T1 and T2 respectfully. It
is very clear that not all the radiation leaving a surface will reach on the other surface. Hence the
radiation exchange between two surfaces can be expressed as, Q  F Fv A(T14  T24 ) , where, F , is
the emissivity function and contains emissivity of both surfaces and, Fv , represents the complex view
factor function. Due to complexity of this relation, we will examine rather simple problem of a body
at a temperature, T1, is completely enclosed by much larger surface, A1 / A2  0 ,at a temperature,
T2. For this case, the net radiation heat transfer from surface 1 to surface 2 is, Q12  1 A1 (T14  T24 ) ,
where, subscript 1 indicates the parameters of the smaller surface.
Example 4.6: A horizontal hot water pipe 8m long and 10cm in diameter has surface emissivity,  = 0.7, and is maintained at a temperature of 77°C in a large room where the air is at 27°C, and 100 kPa. The walls of the room are at 37°C.
Determine the heat loss of the pipe
Solution:
Since the pipe is surrounded by air, due to temperature difference between the pipe surface and air, heat transfer by natural
convection will take place around the horizontal cylinder. Respect to the values of heat transfer coefficients given in Table
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 101
4.2, the value of h may be taken as h  7.0 W/m2K. The reason that the present heat transfer coefficient will be larger than
the listed values in Table 4.2 is due to larger temperature difference and larger pipe diameter. The heat loss by convection,
Q c  hA(Ts  To ) or Q c  7  2.512  (77  27 ) , Q c  879.2 Watts.
 350
310

The heat loss by radiation on the other hand is, Q r   1 A1 (T1  T24 ) or Q r  0.7 x 2.512 x55.668(
)4  (
)4  ,
100
100


.
Q r  575.177 Watts. The total heat loss from the pipe becomes Q = 1454.377 Watts.
4
Even though the surface temperature of the pipe is not very high, the heat loss due to radiation is not negligible and
almost 40 % of the total loss comes from radiation.
As explained above, in all modes, the essential property which causes transfer of heat is the temperature difference, and the transfer process terminates when the equality of temperatures is attained.
Unlike system energy, heat is not a property. For instance, one may not state that a system possesses
10 kiloJoules of heat. The amount transferred depends upon the initial and the final states of interacting systems as well as the path followed. In other words, the way of interaction alters the amount of
heat transferred.
The differential quantity of heat during a process connecting two adjacent states is denoted as Q
and the integration between two end states is:
2
 Q  Q
12
(4.13)
1
The symbol  indicates that small quantity of heat is path dependent, and integration may not be
carried out without knowledge of the process. By way of sign convention heat added to a system is
taken to be positive, and the heat loss is indicated by a negative sign.
Definition: If the boundaries of a system prevents heat transfer during a change of state, then the
  0.
system is called adiabatic, and for the undergoing process, Q
A process taking place in a well insulated container can be considered adiabatic. Similarly, the thermodynamic process occurring in a rocket nozzle is adiabatic since the gas travels at very high velocity
and there is little time for heat transfer from the hot gas. In the opposite extreme where the time scale
of the process is considerably larger than the time to reach thermal equilibrium, the boundary can be
modeled as diathermal.
Definition: Diathermal boundary refers to a boundary across which the temperature gradient is
zero even in the presence of heat transfer. If n is the direction normal to the boundary, then for a
diathermal boundary T  0 .
n
4.5
Work Transfer
Definition: The transfer of energy between two systems due to difference in an intensive property
other than temperature is called work transfer.
For instance, in mechanical systems the intensive property which causes work transfer is the pressure
or stress, and in electrical systems is the voltage potential. To define the work transfer in mathematical
terms, the process must be a quasistatic one. This means that the system be very nearly in equilibrium
at all times. A process proceeding in such a way that only one property kept slightly different at an
102
THERMODYNAMICS
instant of time is called an idealized process. During such a process, the system internally will be
close to the state of equilibrium at all times.
Principle 11: Work transfer across the boundaries of a system for quasistatic processes is product of an intensive property other than temperature
and the difference in an extensive property.
Therefore, the general formula of work interaction for a quasistatic process may be stated as,
W  Fk dX k , or W12 
X2
 F dX
k
k
(4.14)
X1
Similar to heat transfer, the work transfer is also path dependent, and evaluation of equation (4.14)
requires the process as well as the end states to be known.
As shown in Figure 4.13a, the expansion of a gas by lifting the small masses one by one is a
quasistatic process, and the gas work can be calculated by equation (4.14). It is possible to represent
quasistatic processes on two-dimensional diagrams where the coordinates represent thermodynamic
properties.
As in Figure 4.13b, the area below the process curve represents the amount of work transfer, and
such diagrams are extremely useful in analyzing design problems. The sudden expansion process in
Figure 4.13c which occurs by pulling off the locking mechanism is non quasistatic. During the process, the pressure of the gas may not be uniform within the cylinder and the piston may have varying
acceleration. All intermediate states are of non-equilibrium character. Even though the initial and the
final states are the same, due to unknown path in Figure 4.9d, the work done by a sudden expansion
process may not be determined by equation (4.14).
The sign convention of work is that the work done by a system is positive, and the work done
to the system is considered to be negative. Depending upon the type of work transferred across the
boundaries of a system, in equation (4.14), Fk and Xk represent different properties. Several forms of
quasistatic work transfer are discussed in the following sections.
4.6
Mechanical Work Transfer
In numerous engineering applications, work transfer across a system boundary is achieved either
by pressure which creates normal forces, or by shear stress which is a result of torque. The following
cases are considered as forms of mechanical work transfer.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 103
4.6.1 Work transfer by moving boundary
As in Figure 4.14a, due to pressure difference between the system and its surroundings, the piston
as well as the system boundary moves. Assuming uniform pressure distribution at the piston and gas
interface, the force exerted on the piston at an instant of time t is p(t)A. With respect to equation (4.14),
the quasistatic work transferred by change of state form 1 to 2 is:
Figure 4.14 Mechanical work done by the moving boundary
 Mechanical work 


transferred by the  =
 moving boundary 


2
2
1
1
 p(t ) AdX   pdV
(4.15)
where the term “ Adx ” is replaced by dV . The work done by the boundary movement is also called
pdV work, and a method of calculating the work is to know the p-V relationship of the process.
Suppose a gas, in a piston-cylinder arrangement, is expanded or compressed by a polytropic process
defined as pV n = Constant and the mechanical work done by the gas has to be determined for n = 0,
1, , n, and k.
For n = 0, the relation, pV n = Constant, reduces to a constant pressure process, and by equation
(4.15) the work done is:
W12  p(V2  V1 )
(4.16)
For n = 1, the pressure and the volume relationship becomes pV = Constant. Thus, at an intermediate state, the gas pressure p is p1V1/V and substitution into equation (4.15) results in,
V 
W12  p1V1 ln  2 
 V1 
or
 p 
W12  p1V1 ln  1 
 p2 
(4.17)
Notice that pV = Constant process of an ideal gas requires the temperature of the gas to be constant.
Then the isothermal process of an ideal gas is represented by pV = Constant. It should be emphasized,
however, that isothermal process of a real gas or a vapor does not necessarily obey this relation.
For n = , the gas undergoes a constant volume process. Since dV = 0, by equation (4.15) the
work done becomes zero.
For n = n, the gas pressure at an intermediate state is p1(V1/V)n, and equation (4.15) yields,
104
THERMODYNAMICS
W12 
p1V1  p2V2
n 1
(4.18)
where n n. Together with ideal gas equation, the polytropic relation provides a relationship
between the temperatures and the pressures or volumes of the two states as follows,
T2
p
( 2)
T1
p1
n 1
n
(
V2 1 n
)
V1
(4.19)
Thus equation (4.18) may be modified and expressed in terms of pressure ratio as,
W12 
n 1
p1V1 
p2 n 

1
(
)


n  1 
p1

(4.20)
For n = k in which k represents the ratio of specific heats as c p / cv , then the process is called
reversible adiabatic and is discussed in detail in Chapter 5. Replacing n by k in equation (4.18) or in
(4.20), the work of an ideal gas done by such a process may be calculated.
In Figure 4.15, p-V diagram of the processes discussed above are supplied and the relative steepness of each process is indicated.
Figure 4.15 p-V illustration of several processes
Example 4.7: A reciprocating compressor initially contains 0.1 m3 of air at a state of 0.95 bar, and 67°C. The compression
process is quasistatic and represented by pV1.3 = Constant. The final volume is 0.02 m3. Determine the mass of air contained
in the compressor cylinder and the work of compression in kilojoules.
Solution:
Assuming ideal gas behavior for air, p1V1 = mRT1, and the amount of mass in the compressor cylinder becomes,
m=
95  0.1 , or m = 0.097 kg. With respect to relation, p V 1.3  p V 1.3 , the final pressure is
1 1
2 2
0. 287  340
1.3
 0.1 
p2  0.95  
  7.697 bars. Together with the polytropic exponent n = 1.3, substitution of the data into equa 0.02 
tion (4.18) provides the work of compression as,
W12 
95  0.1  769.7  0.02
1.3  1
or W12 = –19.646 kiloJoules.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 105
Example 4.8: An open diving bell in Figure 4.16 is lowered to a
depth of 10 meters. How much work is done on the air in the diving
bell? Suppose that the air has an initial volume of 1 m3 at 1 atmosphere pressure and also that the air is compressed isothermally. The
density of water is 1000 kg/m3.
Solution:
The pressure at state 1: p1 = 100kPa. The pressure at state 2:
p2  p1   gZ  100 
1000  9.8  10
 198 kPa
1000
For ideal gas behavior of air, the p-V relationship of an isothermal
process is pV = Constant, and by equation (4.17),
W12  p1V1 ln
p1
100
,
 100  1  ln
p2
198
or W12 = –68.3 kiloJoules.
The negative sign indicates that the work is done on the air.
4.6.2
Work Transfer by Mass Flow
Consider one-dimensional flow of a fluid crossing the system boundary at time t with a flow rate of.
For an inlet pressure of pi and a velocity of Vi , the force exerted by the fluid entering the system is
pi A i , and the rate of work done is pi AiVi. By the continuity relation, AiVi = mivi, the rate of mechanical
work transported into the system may be stated as,
The rate of mechanical work done by the in flow of mass 

   m i ( pv )i
 transported into the system at an instant of time t

(4.21)
Similarly, as shown in Figure 4.17, for the mass flowing out of the system, the rate of mechanical
work done by the system is:
106
THERMODYNAMICS
The rate of mechanical work done by the mass transported 

  m e ( pv)e
out of the system at an instant of time t


(4.22)
Thus the net rate of mechanical work of the system due to a flow process is the sum of equation
(4.21) and (4.22).
The net rate of mechanical work of the system 

  m e ( pv)e  m i ( pv)i
 due to mass flow at an instant of time t

(4.23)
Example 4.9: Steam enters steadily into a tank as shown in Figure 4.17 at the state of 10 bars, 200°C, and flowing through
a valve at the exit, leaves the tank at a pressure of 1 bar and a temperature of 120°C. For the same mass flow rate of 0.1 kg/s
at both parts, determine the work rate of the system.,
Solution:
i m
 e  0.1 kg / s. With respect to equation (4.23), the flow
From steam tables, vi = 0.206 m3/kg, ve = 0.1793 m3/kg, and m
work of the system is:
W flow  0.1  (100  0.1793  1000  0.206) ,
4.6.3
or

W
flow = –18.807 kW.
Work Transfer by Rotating Shaft
A rotating shaft crossing a system boundary indicates transfer of mechanical work. For the gear system
in Figure 4.18, an external force F acts at a distance r from the center of the shaft. The work required
to move the force through a differential distance dx becomes Fdx.
Considering the relation between the force and the torque, F = T / r, and between the displacement
and the angular rotation, dx  rd , the rotational work of shaft becomes,  W  Td. For a rotational
speed of n (rev/s), the angular displacement may be expressed as d  2 ndt . Thus the rate of work
transferred by a rotating shaft is:
Rate of mechanical work 


transferred by a

  2 T n


rotating shaft


Figure 4.18 A schematic of shaft work
(4.24)
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 107
Example 4.10: Due to shear stresses on the blades of a blender, the torque
for turning the blades is provided by an electric motor as shown in Figure 4.19.
Suppose that the shaft rotates at 1000rpm, and draws a power of 200 Watts from
the electric motor. Compute the torque developed by the motor.
Solution:
In equation (4.24), W shaft  200 , n = 16.67 rev/s, and the developed torque
is,
T
4.6.4
W shaft
2 n

200
2  3.14  16.67
or
T  1. 91 Joules
Work Transfer by Surface Area Change
In processes like stretching a plastic sheet, inflating a balloon, forming a vapor bubble or a liquid droplet, the surface tension  is the force per unit length for
changing the surface of a system. The work done for a differential change in surface area is  W   dA
, and in integral form, becomes
A
 Mechanical work transferred  2

   dA
 by changing surface area  A1
(4.25)
Example 4.11: A certain balloon is constructed of a material such that the surface tension is linearly proportional to the
surface area. Initially the surface tension of the balloon is 500 N/m, and contains 0.5 kg of air at 500 kPa, 27°C temperature.
Determine the work done by doubling the surface area of the balloon.
Solution:
mRT1 0.5  0. 287  300
=
= 0.0861 m3 , for a spherical balloon the diameter d1 is 0.547
500
p1
m, and the surface area, A1 = 0.939 m2. Because of the relation,  = CA, the constant of proportionality C is 532.48 N/m3.
The initial volume of the balloon, V1 =
An expression for  may be stated as,
  532.48 A
Substitution of this relation into equation (4.25) and then integrating we may get,
 A

Wsurface  266.24 A12 ( 2 )2  1 ,
tension
 A1

For A2 / A1  2 , Wsurface = 703.67 Joules.
tension
4.7
4.7.1
Other Forms Of Quasistatic Work Transfer
Work Of An Electric Charge
A galvanic cell is a device for converting chemical energy to electric energy by means of developed
potential difference across the poles of the cell. The work required for moving a differential amount
of charge dQc through a potential difference  is  W   dQc . If the charge is removed at a constant
rate I, then dQc  Idt , and the rate of electrical work transferred is:
108
THERMODYNAMICS
 The rate of electrical 


 work transferred by   I
a potential difference  


(4.26)
Thus, whenever there is a flow of electric charge crossing a system boundary, the energy transfer
mode is work, and is calculated by equation (4.26).
4.7.2
Work of magnetization
For a substance situated in a magnetic field with a magnetic strength H, the work required to
change the magnetization of the substance by a differential amount dM is,  W  o HdM , where 0
is the permeability of the free space. For a rate of change M in the magnetization of the substance,
the rate of magnetic work transfer may be described as,
The rate of magnetic 


 work transfer in a    0 H M
 magnetic field H 


4.8
(4.27)
Convective Transfer of Energy
Energy transfer by convection takes place in flow processes. As shown in Figure 4.20, as the mass
crosses the system boundary, the energy contained within the mass also becomes part of the system
energy. For one-dimensional flow, the amount of energy transferred to the system by a unit mass at
an instant of time t is as following:
e(t )  u (t )  ek (t )  e p (t )
(4.28)
and,
ek (t ) 
V 2 (t )
, e p (t )  gZ (t )
2
(4.29)
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 109
where V(t) indicates the mass velocity at the system boundary, and Z(t) is the altitude of inlet point
with respect to a reference plane. Thus the convective energy transferred to the system at a mass flow
rate of is:
The rate of energy transferred in

by convection at an instant of time

  m i (u  ek  e p )i
t
(4.30)
By similar arguments, the convective energy transferred from the system at a mass flow rate of is:
The rate of energy transferred out of the

system by convection at an instant of time

 e (u  ek  e p ) e
 = m
t
(4.31)
As a consequence, the net rate of energy accumulation in the system due to mass inlet and outlet
may be expressed as follows,
The net rate of energy



accumulation by convection   m i (u  ek  e p )i  m e (u  ek  e p )e
at instant of time t



(4.32)
Example 4.12: Consider the tank given in Example 4.9 where the steam steadily enters the tank at a velocity of 10 m/s,
and exits at a velocity of 1 m/s. At a height difference of 2 meters between the inlet and outlet ports, and for the same thermodynamic conditions determine the energy transferred to the system by convection.
Solution:
Having the same mass flow rate at both ports, the rate of energy stored in the system by convection becomes,
The net rate of energy 
1 2




2
 accumulated in the   m ui  ue   Vi  Ve  g Z i  Z e 
2


 system by convection 




By using steam tables, at the inlet conditions (10 bars, 200°C), ui = 2621.9 kJ/kg, similarly at the outlet (1 bar,120°C),
ue = 2537.3kJ/kg. ui – ue = 84.6 kJ/kg.
The differences in kinetic and in potential energy terms are as follows,
1 2
1
Vi  Ve2 
100  1  0.0495kJ / kg
2
2  1000
1
g Z i  Z e  
 9.8  2  0.0196 kJ / kg
1000


For m  0.1 kg/s, substituting these values into the above expression yields,
The net rate of energy 


 transferred into the  = 0.1 × (84.6 + 0.0495 + 0.0196) or (The convection energy) = 8.466 kW.
 system by convection 


110
THERMODYNAMICS
4.9
The Energy Equation
As explained in previous sections, energy transfer into or out of a system can be achieved by
three different modes. For a simultaneous existence of all these modes in a system, a general energy
balance may be stated as follows,
Time rate of
The net rate

 
  The net rate   The net rate 
energy accumulation  
 
  of work 
of energy
of heat

 
 
 

 within a system  =  transported into    transported    transferred 

  the system by  into the system   by the system 
at an instant

 
 
 


 convection at time t   at time t   at time t 
of time t
(4.33)
Because of the sign convention of work, a negative sign appears in front of the work transfer term.
As shown in Figure 4.21, for several inlet and outlet ports on the system, the first term on the right
hand side of equation (4.33) is given as,
 The net rate of 
 energy transported 



   m i (u  ek  e p )i   m e (u  ek  e p )e
e
 into the system by  i
convection at time t 
(4.34)
The net rate of heat transfer to the system is:
 The net rate of heat 


transferred to the system  Q


at time t


(4.35)
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 111
The work transfer, the last term on the right hand side of equation (4.33), can be categorized into two:
a. work transfer due to mass flow, b. work transfer at the non-flow boundary regions.
 The net rate of 
 work transferred 


at
the
flow
regions
   m e ( pv ) e   m i ( pv ) i

i
 of the boundary  e




at time t
(4.36)
Revising equation (4.23) for a system with several flow ports, the first term on the right hand side
of equation (4.36) becomes,
 The net rate of 
 work transferred 


at the flow regions   m e ( pv ) e   m i ( pv ) i
i
 of the boundary  e




at time t
(4.37)
The second term in equation (4.36), on the other hand, describes a non-flow type of work transfer. As in Figure 4.17, this can be either shaft work, electrical work, moving boundary work etc., or
several of them occurring simultaneously. Denoting such a work transfer symbolically as , the net
rate of work transferred by the system may be formulated as,
 The net rate of 
work transferred 


   m e ( pv ) e   m i ( pv ) i  W

by
the
system
i
 e



at time t
(4.38)
Substituting equations 4.8, 34, 35, 38 into 4.33 and recalling the definition of enthalpy (equation
(2.21)), some terms may be combined. As a result, a general energy equation for a system may be
stated a follows,



U  E k  E p  Q  W   m i h  ek  e p   m e h  ek  e p
i
i
e

e
(4.39)
This relation is appropriate for lumped analysis of systems with mass flowing at discrete points.
For continuous nature of mass flow along the system boundary, similar to the method in section 3.4,
an integral formulation of energy can be derived as follows,


d 
  (u  ek  e p )dV     (h  ek  e p )V ndA  Q  W
dt V
cs



(4.40)
Equation 4.40 is exactly the application of Reynolds Transport Theorem (RTT) to the principle of
“Conservation of Energy”. In equation (3.8), taking B  E , and b  e and performing the necessary
transformations for flow work, exactly the same result of Eq. (4.40) can be found.
112
THERMODYNAMICS
4.10
Steps In Problem-Solving
Thermodynamics as an engineering science suggests a definite method in problem solving which
is not only extremely useful in its context but also fairly common to most engineering analysis. In
thermodynamic analysis of systems, a systematic answering of technical questions requires the following steps to be followed.
1. Draw a figure describing the system. Define the system boundary and introduce suitable notation.
2. Decide which energy interactions are more important than the others, and indicate them with their
signs on the system boundary.
3. Write down the energy equation for the given system in appropriate form, and if necessary, write
down the mass conservation equation as applicable to the system defined.
4. Prepare a chart illustrating the physical data of the given system and consider the following questions: Could you model the system as an ideal gas? Could you use the tables? Do you think that
the given data is sufficient for the solution? Could you derive something useful from the data?
5. Define the type of the process. Is the process isothermal? Adiabatic? Is it a constant pressure
process? or a combination of several processes? Sketch the process in an appropriate diagram.
This might help understanding the problem.
6. State your assumptions. For instance, is the change in kinetic and/or potential energy negligible?
For a system not insulated, can the process be assumed as adiabatic? For a particular problem,
stating proper assumptions yield reasonable results.
7. Solve the governing equations for the desired parameters. In the solution, did you use all the data
given? Did you use the whole conditions? Have you taken into account all essential notions involved in the problem? Did you check all the related equations for the dimensional uniformity?
The first two steps of the above explained procedure are actually restatement of principle 1 and, in
deriving a solution for a problem, have a major importance. It is highly essential to draw a boundary
around the system schematic and to indicate the energy transfer forms crossing the boundary. Besides, synthesis of the given information helps in solving the problem. In step 5, the aim for drawing
a process diagram is to illustrate the change of state in thermodynamic coordinates like p-V plane.
Process diagrams are especially useful for understanding of complex problems.
To exercise the solution methods of energy equation with the guidance of the above explained
steps, in addition to isolated systems, the engineering systems have been classified as closed and open
systems, and the details of applications are presented in the following sections.
4.11
Closed Systems
Since the boundary of a closed system does not allow the transfer of mass into or out of the system,
the terms under the symbol  in equation (4.39) are all zero. Thus the energy balance for a closed
system becomes,
U  E K  E P  Q  W
(4.41)
Integrating this rate equation along the path of a process within a time interval (t1, t2) yields a
relationship between the energy change of a system and the amount of heat or work transfer.
U  E K  E P  Q12  W12
(4.42)
where the change in energy components of internal, kinetic, and potential energy of a system may be
calculated as explained in section 4.2.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 113
Example 4.13: As shown in Figure 4.22, 1 kg of air at 1 bar in pressure and 30°C in temperature is contained in a rigid
and adiabatic container. Inside the container is a propeller which connected to a shaft. Attached to the shaft is a rope which
is also attached to a mass of 50 kg that is initially held in equilibrium 2 m above the ground level by means of a supporting
plate. Pulling the supporting plate, the mass falls to the ground, turning the shaft, and air assumes a new equilibrium state.
Determine the final temperature and pressure of air.
Solution:
Consider the tank as a system, the kinetic and the potential energy change of the system are zero, and because of no transfer
of heat, Q12 = 0. Equation (4.42) reduces to
U2 – U1 = – W12 .
The work of lowering the weight is,
W12   mgL  500  2  1 kJ
The internal energy change of air is,
U 2  U1  mcv T  1  0.718  T2  30 
The
energy
balance
requires
that
1  0.718  T2  30    1 Solving for T2 results as,
T2 = 31.39°C.
The final pressure of the gas in the container is determined by the ideal gas equation.
p2  p1
T2
304.39
 1
T1
303
p2 = 1.004 bar
Example 4.14: In Figure 4.23a a laterally insulated piston-cylinder apparatus 0.5 m2 in cross sectional area contains an
ideal gas for which R = 0.5 kJ/kgK. The initial temperature of the gas is 250K and is heated by a resistor of 10 with a
potential difference across the wire being 35 V. Assume that the heater circuitry is left open for one hour, and during this
time interval, as presented in Figure 4.19b the gas pressure varies linearly with respect to piston movement. Determine a.
the heat transfer, b. the work transfer, c. the constant volume specific heat of the gas.
Solution:
a.
The gas being a system, by equation (4.26), the rate of heat transfer is,
  I  35  3.5  122.5 Watts
Q
114
THERMODYNAMICS
The amount of heat transferred in one hour becomes,
  122.5  3600 or Q = 441 kiloJoules.
Q12  Qt
12
b.
Because of linear relation in between p and x, the gas work is,
W12
400 200
1 5 0.5
1.5
05
2
12
1
225 kJ.
With respect to equation (4.42), the internal energy change of the gas is, U = Q12-W12 = 441-225 = 216 kJ.
c.
The mass of the gas in the cylinder, m 
p1V1 400  0. 25
 0.8 kg,

RT1
0.5  250
The gas temperature at the final state is,
T2 
p2 V2
200  1

 500 K
mR
0.8  0.5
By equation (2.35), the internal energy change of the gas may be described as, 216 = 0.8 × cv × (500 – 250), and solving
for cv yields cv = 1.08 kJ/kgK.
Example 4.15: An accordion like device which consists of metal bellows as shown in Figure 4.24a can be used to measure
the strength of a man. The bellows at its free length of 30 cm, the instrument is filled with air at 1 bar and 20°C and sealed
off. The diameter of the device is constant at 5 cm, and the spring constant of the bellows is 2 kN/m. Suppose that 1 kN of
force is exerted on this system and the compression process can be either isothermal or adiabatic.
a.
Determine by which process the device will be compressed more, b. Sketch p vs V of both processes on pV diagram,
c. Compute the heat transfer, the work transfer, and the change in internal energy of air. Assume that air can be
modeled as an ideal gas.
Solution:
a.
For a quasistatic change of state, the gas pressure, the spring force and the applied force can be interrelated as,
p
1.
KL0 X
F
( )
A L0
A
Isothermal process: Because of constant cross sectional area, the relation between the gas pressure and the amount of
compression,
po Lo  p Lo  X  , or, p  po
Let  = X/L0, and
Lo
Lo  x
F
KL0
= 306.1,
= 510.2 then the force balance results with the following equation,
A
A
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 115
100
 306.1  510.2
1 
Solving for  yields, = 0.675
2.
or Xi = 20.25 cm.
Adiabatic process: Due to constant cross section, air pressure and the amount of compression are interrelated as follows,
p  po
1
1   k
100
The force balance for this process results as,
 306.1  510. 2 by which = 0.577 or Xa = 17.32 cm. Thus,
(1   )1.4
Xa < Xi
b.
p-V illustration of both processes are given in Figure 4.24b.
c. 1. Isothermal process: Due to constant temperature, the internal energy of air is zero, U = 0. The work transfer
by equation (4.17) is:
W12i  p1V1 ln
V2
 0.065 kJ
V1
Equation (4.42) reduces to Q12 = W12. Thus the heat transfer is, Q12i = –0.065 kJ.
2. For adiabatic process: Q12 a = 0, and in equation (4.20) taking n = k = 1.4, the adiabatic work transfer, W12a = –0.0059 kJ.
The energy balance requires that U2 – U1 = –W12a = 0.0059 kJ.
Example 4.16: As shown in Figure 4.25, a cylinder with thermally insulated walls is initially divided into two equal compartments of volume V1 by a moveable, frictionless, and insulated piston. Each compartment contains m kilograms of gas which
can be modeled as an ideal gas with a constant specific heat cv, and a gas constant R. Initially the gases at both compartments
are at the same pressure p1 and temperature T1 and the value of k is 1.50. The compartment on the left contains an electrical
27
p1 . Determine a. the heat transfer
8
and the final temperatures of gases in each compartment in terms of m, cv, and T1, b. the electrical work transfer.
resistor and due to electrical work transfer, the final pressure of the gas becomes p2 
Solution:
a.
Because of thermal insulation, the heat transfer to both compartments is zero, .
Due to adiabatic compression of the gas in compartments B, and referring to Eq. (4.19), the final temperature may be
expressed as,
T2 B  T1 (
Similarly, the final volume of compartment B is,
p2
)
p1
k 1
k
3
 T1
2
116
THERMODYNAMICS
1
V2 B  V1 (
p1 k 4
)  V1
p2
9
Since the total volume of cylinder is constant at a value of 2V1, the final volume of compartment A becomes,
V2 A  2V1  V2 B 
14
V1
9
The final temperature of A, by the gas equation, is,
T2 A  T1 (
b.
p2 V2 A
21
)(
)  T1
p1 V1
4
For the entire cylinder as a system, the energy balance requires that U A  U B  Welectrical
where, U A  mcv (
3
1
21
17
T1  T1 )  mcvT1 , and U B  mcv ( T1  T1 )  mcvT1 ,
2
2
4
4
19
mcvT1 .
4
Example 4.17: A cylinder of fixed volume V in Figure 4.26a is divided into two compartments of volumes VA and VB by
means of rigid and diathermal wall fitted with a check-valve. The valve opens when the pressure in A exceeds the pressure
in B by 0.5 bars. A frictionless piston is fitted into compartment A and the entire system is in thermal communication with
the surroundings at 30°C. Both compartments are filled with air and initially p1A = p1B = 2 bars, V1A = V1B = 0.05 m3.
Then the energy equation yields, Welectrical  
a.
Determine p-V relation of air for both compartments when the piston moves inward and sketch the variation.
b.
Calculate the heat transfer, the work transfer and the change in internal energy of the system for V2A = 0.5 V1A.
Solution:
a.
Due to diathermal walls, the temperature of air in both compartments is the same as the surroundings. Because of
the discontinuity when the check valve opens, the problem should be analyzed in two steps.
1.
For a pressure in the range, 2  pA  2.5 bars, the check valve is closed and pV relationship for compartment A is pAVA
= 10, and the volume of compartment when valve opens is V2A = 0.04m3.
2.
For a pressure pA > 2.5 bars, one should consider the gas equation for the entire system as,
p AVA  pBVB  ma  mb RT
Substituting PB = PA – 50, VB = 0.05m3, R = 0.287 kj/kgK, mA + mB = 0.229 kg into above expression, the p-V relationship for compartment A becomes,
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 117
p AVA  0.05 p A  22.41
Thus, at V2A = 0.025 m3, solving this relationship for p2A, yields, p2A = 298.8 kPa. In Figures 4.22b and c, p-V distribution
of gases in compartments A and B are respectively displayed.
b.
1.
The work transferred by the system is W13 = (W13)A + (W13)B. Because of constant volume of compartment B,
(W13)B = 0. Similar to analysis in a, the work transfer of compartment A is calculated in two steps.
For p A  2.5 bar, and for isothermal compression process, equation (4.17) yields,
3
(W23 ) A 

0.025
2
2.

pdV  22.41
0.04
dV
V  0.05
For p A  2.5 bar, the relation between pressure and volume is, p A 
3
(W23 ) A 

0.025
pdV  22.41
2

0.04
W112 A  2.23 kJ
or
dV
V  0.05
22.41
and pdV type of work becomes,
V  0.05
(
2233 ) A
4.086 kJ.
Thus, the work transfer of the system is: W13 = –2.231+(–4.086) = –6.317 kJ.
Since the change in internal energy of the system is zero, the energy equation reduces to Q13 = W13 and the heat transfer
becomes,
Q13 = –6.317 kJ.
Example 4.18: Figure 4.27 represents schematically an air rifle. Air charged into the reservoir of volume V accelerates the
bullet down the barrel when the valve opens quickly. The initial pressure and temperature of air in the reservoir are 40 bars
and 77°C. Assume that the atmospheric pressure is 1 bar and the spherical bullet has a mass of 0.01 kg, and that the barrel
diameter is
1 cm. To provide an exit velocity of 100 m/s to the bullet,
determine the minimum volume of the reservoir and the
length of the barrel for an adiabatic expansion of air.
Solution:
Considering the bullet and the compressed gas as a system, the kinetic energy change of the system:
Ek 
1
1
mbVb2 
 0.001  1002  0.05 kj.
2
2000
Since the final pressure is 1 bar, the final temperature of air due to adiabatic expansion is
p 
T2  T1  2 
 p1 
k 1
k
By equation of state, the initial density of air is r 
 100 
 350  

 4000 
0.285
 122.3 K
p
4000

 39.82 kg / m3 , then the mass of air in the rifle is,
RT 0.287  350
m  rV  39.82V
118
THERMODYNAMICS
The internal energy change of air becomes, U  mcv T2  T1   39.82V  0.718  122.3  350   6510.1V
1
 p1  k
The final volume occupied by air, V2  V   = 13.92V,
 p2 
The work done by pushing the surroundings by the volume change is, W12 = p0(V2-V) = 100 (13.92V – V) = 1292V
Substituting the above evaluated terms into equation (4.42), one may obtain, -6510.1V+0.05 = 0-1292V and solving
for V yields,
V = 9.58 10-6m3
The change in air volume is, V = V2 – V = 12.92V = 123.7 10-6 m3 which corresponds to the volume of the barrel,
AL = 123.7 10-6m3. Thus, for a cross sectional area of 0.785 10-4m2, the barrel length has to be
L = 1.575 m.
4.12
Constant Pressure Process of Closed Systems
Consider a piston-cylinder apparatus in which the fluid expands isobaricly. The objective is to
determine the amount of heat transfer by such a process. For a stationary system obviously no change
in kinetic and potential energy would take place and by equation (4.42), the heat transfer may be
evaluated as,
Q12 = (U2 – U1) + W12
(4.43)
Where, W12 represents the work transfer at constant pressure. Thus, substituting equation (4.16)
into (4.43) and rearranging yields,
Q12 = (U2 + pV2) – (U1 + pV1)
(4.44)
Notice that the pressure p represents the fluid pressure at any state of the process, and then p = p1 = p2.
As given by equation (2.1), using the definition of enthalpy, the above result may be expressed as,
Q12 = (H2 – H1)p
(4.45)
In closed systems, the enthalpy difference represents the amount of heat transferred by a quasistatic
process at a constant pressure. In equation (4.39), on the other hand, the enthalpy term has entirely
a different meaning. There, due to a flow process, the enthalpy, signifies the sum of internal energy
and flow work of the mass which inlets or outlets the system.
Example 4.19: A long and vertical pipe encapsulated by a frictionless 2 kg piston has a cross-sectional area of 2 cm2 and
initially contains 1L of liquid water and vapor mixture of which the volumetric ratios respectively are 20% and 80%. Due to heat
leak through the walls of the device, as shown in Figure 4.28, the vapor condenses and the piston moves downward at a rate of
0.003 m/s. Assume that the atmospheric pressure is 100kPa, and determine, a. the rate of condensation, b. the rate of heat loss,
c. the distance of the piston from the base when the mixture comes into thermal equilibrium with the surroundings at 30°C.v
Solution:
a.
The volume rate swept by the piston, V  AV p  2  104  3  103  6  107 m3/s. The mixture pressure may be
calculated as,
p  po 
mg
2  9.8
 100 
 198
A
2  104  103
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 119
From saturated steam tables, the specific volume of saturated steam at this
pressure, vg = 0.885 m3/kg, then the rate of condensation becomes,
g 
m
b.

V
6  107

vg
0.885
or
 g  6. 78  107 kg / s
m
Since the condensation takes place at a constant pressure, the rate of heat
transfer, by equation (4.45), is:
 m
 g h fg  6. 78  107  2201. 9
Q
or
  1. 492 Watts
Q
c. The total mass of water contained in the device is,
m
Vf 1 Vg1 0. 001  0. 2 0. 001  0.8



 0.1895 kg .
v f 1 vg1
0. 00106
0.885
The volume occupied at 30°C, V2 = mv2 = 0.1895 0.001004 = 0.00019 m3.
Thus the height of the piston at the final state becomes,
L2 
4.13
V2 0.00019

or L2  0 95 m
A 2  104
An Introduction to Thermodynamic Cycles
Definition: A thermodynamic cycle is a sequence of processes that eventually returns the working
fluid to its original state.
As shown in Figure 4.29, since the original state of the working fluid has to be regained, a thermodynamic cycle is considered as a closed system. A periodic repeat of certain processes is not sufficient to
constitute a thermodynamic cycle. For instance, an aircraft jet engine continually takes in fresh ambient
air and fresh ambient air and discharges hot combustion gases to the atmosphere.
Eventually these gases cool down in the atmosphere to the same temperature as the ambient temperature. Such a sequence of processes can be considered as a cycle only if air at the original state is
supplied to the system. To illustrate the concept of thermodynamic cycle, as shown in Figure 4.30a,
consider a fluid contained in a piston-cylinder apparatus that undergoes heat interaction with two heat
120
THERMODYNAMICS
reservoirs at different temperatures. Suppose that the fluid exercises the following four processes and
regains its original state.
(1-2): Adiabatic compression to the temperature of high temperature reservoir, (2-3): The fluid at
the temperature of high temperature reservoir, isothermally heated, (3-4): Adiabatic expansion to the
temperature of low temperature reservoir, (4-1): Isothermal cooling of the gas to the initial pressure
and temperature. The cycle produced by these four processes is called the Carnot Cycle and will be
further studied in Chpt.5.
Equation (4.41) being valid for a particular state of the system in Figure 4.30b, it may be integrated
along the cycle as follows,
 U  E
K
  Wdt
 E P dt   Qdt
 
The initial and the final states of a cycle being identical,
balance for a cycle results as,
 Q   W , or Q
net
= Wnet
 U  E
(4.46)
K
 E P dt  0 , and the energy
(4.47)
This equation states that since the energy of a system remains unchanged as a result of cyclic processes, an amount of energy equal to the net work output must have flowed into the system during the
cycle as the net heat input. As indicated by the shaded region in Figure 4.30b, the net work transfer,
Wnet, of a cycle equals to the area enclosed by the cycle on p-V representation and this is proved to be
true for any cycle in Example 4.20.
Definition: Thermodynamically a power cycle is represented by  W  0 , and the reverse cycle
through which power is consumed in completing the cycle is represented by  W  0 .
4.13.1 Power cycle. An Example for power cycles is the Organic Rankine Cycle (ORC) technology.
ORC is used for electricity production from biomass, geothermal heat, and from waste heat. Due to
required low temperature of evaporation, instead of water as the working fluid, organic substances
(Fluorocarbons, Silicone oils etc.) are used. As shown in Figure 4.31, the working fluid (Pentane) is
heated by a low temperature heat source in the evaporator, and allowed to pass through a turbine to
produce the required power, and then is condensed in the condenser.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 121
The performance of power cycles is measured by a parameter called thermal efficiency. The
thermal efficiency of a power cycle is the ratio of the net work output to the heat energy input and
represented by  as follows,

Wnet
 100%
QH
(4.48)
For a power cycle, as in Figure 4.31b, interacting with two heat reservoirs, the net work output,
by equation (4.47), may be expressed as, Wnet = QH – QL. Accordingly, rearranging equation (4.48)
yields,
h  1
QL
QH
(4.49)
122
THERMODYNAMICS
The condition of QH  QL always exists for power cycles. Thus, with respect to above relation, the
thermal efficiency of power cycles is always less than unity. As explained in Chapter 5, the efficiency
of a power cycle strongly depends on the temperature of the heat source. As the temperature of the heat
source is higher, the thermal efficiency is the better. Since ORC cycles are used for low temperature
heat utilization, the thermal efficiency of these cycles is rather low and is around 20-percent.
4.13.2
Refrigeration Cycle.
For the reverse cycle, since the Wnet < 0, the energy in the form of work is consumed. For instance, in vapor refrigeration cycles, the heat is transferred from a medium at low temperature to a
medium at high temperature at the expense of work energy consumption. As used for an automobile
air conditioning system, the elements of a mechanical refrigeration cycle are shown in Figure 4.32.
The performance of such systems is measured by a parameter called the coefficient of performance,
COP in short, and is defined as follows,
br 
QL
WC
(4.50)
Where QL represents the useful heat energy extracted from refrigerated space, and WC is the work
energy consumed by the compressor of the cycle. This reverse cycle is also used as a basic heat pump
cycle that extracts heat from the outside air in winter and transfers heat to the indoor air. For a heat
pump then the useful output is the heat supplied to the indoor air and represented by QH. Thus COP
of a heat pump is expressed as,
 hp 
QH
WC
(4.51)
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 123
Example 4.20: A 1 kg of air contained in a piston-cylinder apparatus occupies an initial specific volume of 0.47 m3/kg at
250°C and executes a cycle between two heat reservoirs which are respectively at 250°C and 150°C in temperatures. As shown
in Figure 4.33, the cycle proceeds as follows, (1-2): Expanding isothermally to specific volume of 1.19 m3/kg. (2-3): Cooling
to 150°C at a constant volume. (3-4): Isothermal compression to specific volume of 0.47 m3/kg. (4-1): Heating to 250°C at
a constant volume. Show that equation (4.47) holds for this cycle and determine the thermal efficiency of the cycle.
Solution:
The net heat transfer to the system may be evaluated as, Qnet = Q41 + Q12 + Q23 + Q34
From Figure 4.33b, one may deduce that T1 = T2, T3 = T4, V1 = V4, V2 = V3, then the heat transfer for each process of
air is as follows,
Q41 = U1 – U4 = mcv(T1 – T4) = 1 × 0.718(523 – 423) = 71.8 kJ, Q12 = W12 = mRT1 ln
139.44kJ
Q23 = U3 – U2 = mcv(T3 – T2) = 1 × 0.718(423 – 523) = –71.8 kJ, Q34 = mRT3 ln
-112.78 kJ
V2
1.19
=
 1x0. 287  523  ln
V1
0. 47
V4
0. 47
=
 1x0. 287  423  ln
V3
1.19
Thus, the net amount of heat transferred through the cycle is Qnet = 71.8 + 139.44 + (-71.8) + (-112.78) = 26.66 kJ.
Similarly the net amount of work transferred by the cycle may be expressed as, Wnet = W14 + W12 + W23 + W34
Because of constant volume processes at (4-1) and (2-3), the corresponding work terms W41 and W23 are zero, in addition,
W12  mRT1 ln
V4
0. 47
V2
1.19
 1x0. 287  423  ln
= -112.78 kJ.
= 139.44 kJ, W34  mRT3 ln
 1x0. 287  523  ln
V
1.19
V1
0. 47
3
Then, Wnet = 0 + 139.44 + 0 + (-112.78) = 26.66 kJ, and the equality of Qnet and Wnet is obvious from the numerical
results.
In equation (4.48), QH represents the total amount of heat transferred by the high temperature reservoir. Therefore,
QH = Q41 + Q12 = 71.8 + 139.44 = 211.24 kJ. For Wnet = 26.66 kJ/kg, the thermal efficiency of the cycle becomes,

26. 66
 100 %
211. 44
or
  12. 62 %
Example 4.21: A heat pump supplies heat to the indoor air at 29°C by extracting 10kW of heat from the outside air at
-7°C. For a power input of 3.5kW, determine the heat supplied in one hour and COP of the heat pump.
Solution:
With respect to equation (4.47), Q H  Q L  WC  10  3.5  13.5 kW or 48600 kiloJoules per hour. Thus, the coefficient of
performance of the heat pump becomes,
124
THERMODYNAMICS
b hp 
4.14
Q H 13.5 , or b = 3.85
hp

W C
3.5
Steady State Flow Systems
As described in Chapter 1, for a system at steady-state conditions, the energy of the system is invariant
with time. Thus the terms on the left hand side of equation (4.39) disappear. In general the principle
of energy conservation for such systems becomes,

1
 m  h  2 V
e
2

 gz  
e

1
 m  h  2 V
i
2

gz   Q  W
i
(4.52)
Notice that the expression above involves mass flow rates of incoming and outgoing fluids. Therefore,
in solving equation (4.52) for a particular system, the application of the principle of mass conservation
is a necessity. The mass conservation in the form of equation (3.2) is valid for systems operating at
steady-state conditions,
 m i   m e . In words, the sum of flow rates entering the system equals to
the sum of flow rates leaving the system.
In this section, energy analysis of certain equipment which are extremely important in engineering applications will be discussed. Considering their specific functions and the way the flowing fluid
behaves in such devices, the various forms of equation (4.52) will be considered. For instance, many
engineering systems operating under steady-state conditions have two ports, one inlet and one outlet,
and for such systems the conservation of mass and energy simplifies to
m i  m e
(4.53)
1
1
(h  V 2  gz )e  (h  V 2 gz )i  q  w
2
2
(4.54)
where q and w are respectively the heat and work transfer of the system per unit mass of fluid flowing
through the system.
4.14.1
Nozzles and Diffusers
Nozzles and diffusers are devices used for changing the velocity of a flowing stream. In steam or
gas turbine applications, a nozzle increases the velocity of a fluid at the expense of a pressure drop
in the flow direction. A diffuser, however, increases the fluid pressure in the flow direction at the
expense of a decrease in velocity. In centrifugal compressors, the increase in pressure of an accelerated fluid is provided by such devices.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 125
Under subsonic or supersonic flow conditions, the general shapes of a nozzle or a diffuser are shown
in Figure 4.34. A nozzle for subsonic flow or a diffuser for supersonic flow must have a decreasing
cross sectional area in the flow direction as in Figure 4.34a. For supersonic flows, however, the opposite occurs. In Figure 4.34b, the spray atomizing nozzle, as used for humidification processes, is a
supersonic nozzle with increasing cross-sectional area in the flow direction. Besides, if a fluid has to
be accelerated from subsonic to supersonic velocity, a converging-diverging nozzle as in Figure 4.34c
must be used. In such applications, the fluid assumes the sonic flow conditions at the throat.
Since both of these devices are essentially ducts, no shaft work is involved, and for most conditions,
due to high velocity of the fluid, the heat transfer through the walls is negligible. The potential energy
change may also be neglected. Owing to single inlet and outlet, equations (4.53) and (4.54) are applicable
to such devices. With respect to these assumptions, equation (4.54) may be simplified and reduced to
hi 
Vi 2
V2
 he  e
2
2
(4.55)
By rearranging this relation, the velocity of a fluid at any cross-section of a nozzle or a diffuser
may be calculated as,
V e  Vi2  2hi  h e 
Example 4.22. Steam enters a nozzle with a mass flow rate of 9kg/min at a
stagnation enthalpy of 2780 kJ/kg. At the nozzle exit, the steam
has a velocity 1070 m/s and a specific volume 18.75 m3/kg. Determine, at the nozzle exit, a. the enthalpy of the steam, b. the
cross-sectional area of the nozzle.
(4.56)
126
THERMODYNAMICS
Solution:
a. A stagnation enthalpy of a fluid is defined as the sum of the enthalpy and the kinetic energy and represented by
h .Thus,
h1  h1 
and according to equation (4.55), h2  h1 
V22
10702
, or
 2780 
2
2000
V12
= 2780 kJ/kg
2
h2 = 2207.5 kJ/kg.
b. For one dimensional flow, considering the continuity relation as in the form of equation (3.10), the cross-sectional
area at the exit becomes, A2 
4.14.2
 2 0.15  18.75
mv
or A 2  0. 00262 m2 .

V2
1070
Throttling devices
A throttling device is an apparatus that reduces the pressure of the flow by an obstruction. An
orifice, a valve, a long capillary tube, or a porous plug are throttling devices.
A reduction in pressure by an orifice in Figure 4.36a is used as a measure of the velocity of a fluid
flowing through a pipe. In Figure 4.36b, reducing the cross-sectional area of the flow by a valve, a
greater pressure drop across the valve occurs and the flow rate decreases. The throttling devices are
widely used in refrigeration units, and are also utilized in reducing the power of an engine.
In a throttling process, no work interaction is involved, and usually the heat transfer of the process
is negligible. In addition, the changes in kinetic and potential energies can be neglected. Thus, for a
throttling process, equation (4.54) simplifies to
h i = he
(4.57)
by which it is stated that a throttling process can be considered as a constant enthalpy process.
Example 4.23: As shown in Figure 4.37, compressed water at 20 bars in pressure and 200°C in temperature is adiabatically
throttled to 15 bars by flowing through a valve. Determine, at the exit of the valve, a. the temperature and the quality of
water, b. the velocity for A2 = 2A1, and V1 = 10 m/s.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 127
Solution:
a.
The enthalpy of the compressed liquid may be assumed to be the
same as the enthalpy of saturated water at 200°C. Then h1 = 852.45
kJ/kg, and by equation (4.57), h2 = h1 = 852.45 kJ/kg. By saturated
steam data at p2 = 15 bars, h2f < h2 < h2g, which states that water
at the valve exit is a mixture of liquid and vapor. Thus the exit
temperature becomes,
T2 = 198.3°C, and the quality is,
x2 
b.
h2  h2 f 852. 45  844.84

, or x2 = 0.0039
1947. 3
h2 fg
The specific volume at the valve exit is, v2 = v2f + x2v2fg = 0.00115 + 0.0039  0.1306 = 0.00166 m3/kg. Considering the continuity equation of one-dimensional flow, the exit velocity becomes,
 A  v 
1 0.00166
V2   1  2  V1  
 10 , or V2 = 8.3 m/s
A
v
2 0.001
 2  1 
As can be deduced by this result, neglecting the change in kinetic energy is an appropriate assumption for throttling
processes.
4.14.3
Mixing chambers
In chemical industry, certain fluids at different temperatures are mixed at appropriate ratios of mass
so that a desired temperature is obtained. In a central air-conditioning system, outdoor air is mixed
with refrigerated air for providing the desired conditions at the indoor. The mixing of fluids in such
devices must be at the same pressure. There is no work transfer through such devices, and the chambers
are usually insulated. The change in kinetic and potential energies of fluids entering and leaving the
chamber may be neglected. Accordingly, the continuity and the energy equations become,
 m i   m e
(4.58)
 m i hi   m e he
(4.59)
As shown in Figure 4.38, steam-water mixers provide
relatively inexpensive and instant source of low pressure hot
water by utilizing existing steam and cold water supplies.
The temperature of water is easily controlled by using steam
and water valves fitted to the inlets, and a thermometer at
the outlet monitors the water temperature.
Example 4.24: To supply hot water for a process, steam at 10 bars and
0.2 in quality inlets the mixing chamber in Figure 4.38 at a flow rate of 1
kg/s. After throttling to 5 bars, the steam is mixed with 2 kg/s of water at
5 bars, and 30°C. At the chamber exit, determine the mass flow rate and
the temperature of the mixture.
Solution:
With respect to the boundary around the chamber, equations (4.58), and
(4.59) become,
m 1  m 2  m 3 and m 1h1  m 2 h2  m 3h3
128
THERMODYNAMICS
 3 = 3 kg/s.
Thus, at the chamber exit, m
By using the steam tables, h1 = h1f+x1h1fg = 762.81+0.22015.3= 843.42 kJ/kg, h2 = 125.79kJ/kg, and considering the
enthalpy relation, the enthalpy at the chamber exit becomes, h3 = 365 kJ/kg. Since, h3 < h3f at p3 = 5 bars, the water, at the
chamber outlet is a compressed liquid, and the temperature is,
T3 = 87.3°C.
4.14.4
Turbines, Compressors, Pumps, and Fans
These machines are used in a large number of processes frequently encountered in industry varying from power production to refrigeration, and from liquid transporting in pipe lines to air handling
duct systems.
Turbines. In a turbine, the energy content of the working fluid that can be liquid or gas is converted
into mechanical shaft work. As the fluid does work on the rotating blades of this machine, the pressure
and the temperature of the fluid decrease in the flow direction. A single stage and axial flow steam
turbine is presented in Figure 4.39. As shown in the figure, the steam is first accelerated by nozzles,
then the blades that are attached to the turbine wheel transmit the work of steam to a rotating shaft
which in turn operates the generator. The change in potential energy between the inlet and the outlet
of a turbine is usually neglected. Unless stated, the heat losses through the turbine casing are small in
comparison to the enthalpy changes. Then, for a turbine, equation (4.54) reduces to

V2
w  h 

2

2
 
  h  V
 
2
i 



e
(4.60)
Compressors and Pumps. Compressors and pumps are utilized in compressing or raising the pressure of a fluid. A compressor that uses a gas or a vapor as the working fluid can be either rotating or
reciprocating type. In a rotary compressor, the fluid is accelerated and the kinetic energy is increased
by the shaft work done on the impeller. Then, in the diffuser section, deceleration of fluid causes an
increase in fluid pressure. Figure 4.40a displays a rotary compressor which might be used for raising
the pressure of the inlet air stream of a jet engine. In a house hold refrigerator, however, a reciprocat-
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 129
ing compressor is used to raise the pressure of the refrigerant vapor. A cutaway view of a pump is
shown in Figure 4.40b. Pumps are used to raise the pressure of liquids.
Both machines consume work energy. Again the change in potential energy between the inlet
and outlet of a compressor is negligible. For compressors with a large power capacity either fins are
attached to the compressor body or water is circulated through the jackets of the compressor cylinders
for cooling. Hence, in general, the energy equation for a compressor might be written as,

V2
w  q  h 

2

2
 
  h  V
 
2
i 



e
(4.61)
Fans and Blowers. Fans are almost universally used for the circulation of air or other gases through
low pressure systems. As shown in Figure 4.41, a centrifugal fan is widely used for moving large or
small quantities of air over an extended range of pressures at the expense of work consumption. The
heat transfer through the housing of a fan and the change in potential energy of the fluid are usually
neglected. Then. equation (4.54) simplifies to

V2
w  h 

2

2
 
  h  V
 
2
i 



e
(4.62)
130
THERMODYNAMICS
Example 4.25: A small steam turbine working at a partial load produces 100kW of power at a flow rate of 0.3kg/s. The
steam as in Figure 4.42 is throttled before entering the turbine from 1.5MPa, 300°C to a state of 1 MPa in pressure. For
an exhaust pressure of 10kPa, determine a. the state of steam at the turbine outlet, b. the diameter of the exhaust duct for a
stream velocity of 20m /s.
Solution:
a.
In equation (4.60), neglecting the
change in kinetic energy, the outlet enthalpy becomes, h 3 = h 1-w where
h1 = 3037.6kJ/kg, and the turbine specific
 100
W
power is: w 

 333. 3 kJ/kg.

m
0. 3
Thus, the enthalpy relation yields, h3= 2704.3
kj/kg at p3 = 10 kPa. Using the superheated
steam tables, the temperature and the specific
volume at the turbine outlet are:
T3 = 108.8°C, v3 = 17.6 m3/kg.
b.
,
 3 d 23 0. 3  17. 6
 3  A 3V3 or A 3  mv


Assuming one-dimensional flow in the exhaust duct, mv
= 0.264m2,
4
20
V3
and the duct diameter becomes,
d3 = 0.58m.
Example 4.26: An axial flow compressor in Figure 4.43 intakes air at 100kPa, 27°C, and compresses to 500kPa, and 227°C
with an outlet velocity of 100m/s. For a shaft power of 50 kW, determine the amount of air flowing through the compressor
in one hour.
Solution:
Assuming the compression process to be adiabatic, equation (4.61) reduces to

V2
w   h 
2

 
V2
  h 
 
2
1 



2
Since the atmospheric air is stagnant, V1 = 0, and h1 – h2 = cp(T1 – T2) = 1.005 (300 – 500) = –201 kJ/kg. Then,
1002
 206 kJ/kg
2000


W
W
50
 
, or
Considering the definition of specific shaft work w as, w 
, Then the mass flow rate becomes, m


m
w 206
w  201 
m  0.242 kg/s by which the amount of air flowing through the compressor in one hour is: 871.2 kg.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 131
Example 4.27: A reciprocating compressor in Figure 4.44 is cooled
by water circulated through the water jackets of the compressor.
The compressor intakes air at (1 bar, 25°C), and compresses to 7
bars, 80°C. The cooling water circulating at a flow rate of 0.3 kg/s,
inlets at 15°C and exists 35°C. The volumetric flow rate of air at
the inlet being 0.05m3/s, determine the shaft power.
Solution:
The mass flow rate of supplied air, m a  i  Vi
where r i 
pi
100

 1.169 kg/
RTi 0.287  298
m3, and m a  1.169 0.05 = 0.058kg/s. Neglecting the change
in kinetic energy of air, equation (4.61) simplifies to w  q  hi  he  , and the amount of heat removed by water is,
Q w  m w cw T2  T1   0.3  4.18  35  15  = 25.08 kW
  Q
 = –25.08kW,
The heat loss of air during the compression process equals to the heat removed by water. Then, Q
a
w
Q a 25.08
ferred per unit mass of air is: q 
= –432.41 kJ/kg,

m a
0.058
and the heat trans
hi – he = cpa(Ti – Te) = 1.005 (25 – 80) = –55.275 kJ/kg. Substituting these values into the energy equation yields,
w = –432.41 + (–55.275) = –487.685 kJ/kg.
Then the shaft power of the compressor is:
 = –28.285 kW.
  0.058  (487.685) or W
W  mw
4.14.5
Heat exchangers
A heat exchanger is a device that provides the transfer of heat energy between two or more fluids at different temperatures. Heat transfer between the fluids take place through a separating wall.
Since the fluids are separated by a heat transfer surface, they do not mix. Common examples of heat
exchangers are the shell and tube exchangers, automobile radiators, condensers, evaporators, air
preheaters, and dry cooling towers.
A heat exchanger as in Figure 4.45 consists of active heat exchanging elements such as a matrix
containing the heat transfer surface, and the passive fluid distributing elements such as headers, inlet
and outlet nozzles, baffles, and seals. Usually there are no moving parts in a heat exchanger.
The heat transfer to the environment through an heat exchanger is usually small in comparison to
the transfer of heat between two fluids. There is no work transfer in such devices and the changes in
kinetic and in potential energies of both fluids can be neglected.
132
THERMODYNAMICS
Besides, due to unmixed fluids, equation (4.52) is reduced to
m c he  hi c  m h hi  he h
(4.63)
Example 4.28: Figure 4.46a presents a layout of a room air conditioner in which refrigerant R22 is used as the working
fluid. The warm room air at 100kPa, 27°C enters the evaporator of such a device at a flow rate 6m3/min. As shown in Figure
4.46b, the refrigerant at 245 kPa in pressure and 0.3 in quality flows into the tubes of the evaporator at a rate of 1kg/min.
and exits as a saturated vapor at the same pressure. Determine the air temperature at the evaporator exit.
Solution:
In this example, the hot and the cold fluids are respectively the room air, and the refrigerant. The mass flow rate of air is:
100
 ,   pi 
 a  i V
m
 1.161 kg / m3 , and
i
i
RTi 0. 287  300
 a  1.161 
m
6
= 0.116kg/s
60
The enthalpy change of the refrigerant is: (he – hi)c = (1 – x1)hfg = 0.7 220.33 = 154.231 kJ/kg
Then, by equation (4.63), 0.0166 154.231 = 0.116 1.005 (27 – Te), and solving for Te, the temperature of air at
the evaporator exit becomes, Te = 5.03°C.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 133
4.14.6
Flow in ducts and pipes
There is no need to signify the engineering importance of transporting a liquid or a gas between
two stations. In a flow of a fluid in a restraining channel or a duct a drop of pressure occurs. This
pressure drop is called fluid friction. The magnitude of fluid friction depends on various factors: fluid
velocity, diameter or shape of duct section, the condition of its surface, density, viscosity, pressure
and temperature of the fluid, and the type of flow, viscous or turbulent.
A piping or a duct system has only one inlet and outlet. Hence, the energy equation in the form
of equation (4.54) is applicable to such systems. In certain applications, in addition to flow friction,
heat transfer takes place. There are numerous examples for such a case like; flow of water through
the tubes of a boiler, flow in pipes of a heat exchanger, or in flow of a refrigerant through the tubes of
an evaporator or a condenser. In these applications transfer of heat cannot be neglected in the energy
equation. For a flow of air in distribution ducts of an air-conditioning system, however, the walls of
the channel are insulated and the system is taken to be adiabatic. In cases of installing a fan, a pump, or
an electric heater to a pipe system, then the work transfer term in equation (4.54) has to be retained.
Let us consider a flow of an incompressible fluid in an insulated duct, the enthalpy change for an
incompressible fluid,

h = u + vp, and u = cT
(4.64)
Due to fluid friction, certain amount of heat will be produced at the conduit walls. Because of
insulation, this heat is regained as an increase in the internal energy of the fluid. However, if the fluid
is assumed to be non-viscous, then there is no friction and the flow process becomes isothermal. Thus,
for an incompressible, non-viscous flow of a fluid through an insulated duct together with Eq. (4.64),
Eq. (4.54) simplifies to
1


w    v  p2  p1   V22  V12  g z2  z1 
2




(4.65)
This is known as the Bernoulli equation in literature.
Example 4.29: A hair dryer is basically a duct in which a 100 W electric resistor and a 50 W fan are installed. The fan
steadily sucks in the ambient air at the 100kPa, 20°C, and forces it over a resistor such that the exit temperature of air is
50°C. For a cross-sectional area of 20 cm2 at the dryer exit, determine, a. the exit velocity of air, b. the volumetric flow rate
of air at the inlet.
Solution:
a.
With respect to the system defined in Figure 4.45, the mass flow rate, m  r 2 A2V2 , where,
r2 
p2
100

 1.078 kg / m3 , and m  0.00215V2
RT2 0.287  323
134
THERMODYNAMICS
The enthalpy change of air is: h2 – h1 = cp(T2 – T1) = 1.005 (50 – 20) = 30.15 kJ/kg
 W
 W
  –0.1 + (–0.05) = –0.15 kW
The total work transferred to the duct is: W
e
f
There is no change in the potential energy of this flow, and the surface of the duct is assumed to be insulated. Then, equation (4.54) becomes,
V22
30.15 
2

0.15
, where the velocity of ambient air is taken to be zero, V1 = 0. Rearranging this expression
0. 00215V2
results as, V23  60.297V2  139.534  0
which yields,
V2 = 2.147 m/s.
b. The mass flow rate, m  0.00215V2  0.00215  2.147  0.0046 kg/s. Respect to inlet conditions, m  1V1
m 0.0046
p1
100
= 1.189 kg/m 3, and the volumetric flow rate at the inlet is: V1 
, or


RT1 0. 287  293
1 1.189
1 
V1  0.00386 m3/s.
Example 4.30: A house with an internal volume of 500m3 is to be heated by the circulation of hot air as in Figure 4.48. Air
flows through an insulated duct in which a 20kW electrical heater is installed. The circulation of air is provided by a 200W
fan. The inside temperature and pressure initially are 10°C and 95 kPa. The total heat loss through the walls and the ceilings
of the house is 450 kJ/min. For a change of 5°C in the temperature of air stream between the duct inlet and outlet, determine,
a. the time duration required for a final temperature of 25°C in the building, b. the average flow rate of air in the duct.
Solution:
a.
Considering the house as a system, it should be treated as a closed system for which there is no change in kinetic
and potential energies, and equation (4.42) is reduced to
U = Q12-W12
The
mass
of
air
in
the
building
is:
p1V
95  500

 584.82 kg. Neglecting the change
RT1 0.287  283
in the internal energy of the walls, the change in the internal
energy of air becomes identical to the internal energy change of
the system. Thus,
U = mcv(T2 – T1) = 584.82 0.718 (25 – 10) = 6298.51
kJ. Due to constant rate of heat loss and power supply,
  7.5t , W  W  W t = –20.2t, and the energy
Q  Qt
m
12
12

e
f

balance results as, 6298.51 = –7.5t – (–20.2t)
Solving for t yields,
b.
t = 495.95s.
In this case, the channel has to be considered as a system
for which the change in kinetic and potential energies are
negligible, and due to insulation, equation (4.54) simplifies
to
he – hi = –w
where, he-hi = cp(Te-Ti) = 1.0055 = 5.025 kJ/kg,
and w 
W
20.2 . Substitution into the above relation yields,

m a
m a
5.025 =
 20.2 
 
 . The mass flow rate of air becomes, m a = 4.019 kg/s.
 m a 
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 135
4.15
Transient Flow Systems
Consider the process that occurs in charging of compressed gas tanks. During this process, the
mass flow rate of the charged fluid as well as the pressure and the temperature of the fluid in the tank
will vary with time and such systems are called transient systems. One of the numerous applications
of transient systems is to run a turbine by the pressurized air of a storage tank, and thus fulfilling the
energy need of a facility during a cut-off period.
In these systems, the shape and the boundary might vary with time, and the boundary can be in
motion with respect to a reference point. Besides, an accurate calculation of the terms involved in
equations of mass and energy is required. However, analytic methods can still be devised to get results reasonable for a system under study. Thus, to get a solution, the following simplifications can
be imposed on transient systems:
1. The thermodynamic state of every point within the system boundary is the same at an instant
of time. The system properties are single valued. However, the state of the system is time
variant.
2. The state of the fluid flowing into the system is time invariant. This assumption can be fulfilled by extending the system boundary to a region where the fluid properties do not vary
with time. The flow rate of incoming fluid, however, might be a function of time.
3. The fluid flowing out of the system is assumed to be at the same state as the system. This assumption requires that the fluid is essentially in equilibrium at all times. Hence, a discharge
of a fluid must be slow enough that the process is quasistatic.
In regard to these assumptions, the kinetic energy change is neglected. Unless indicated, the change
in potential energy is usually negligible.
4.15.1
Transient charging processes
Charging a storage tank empty or partially filled from a large supply source with a fluid different from
or the same as the fluid in the tank is a typical problem of charging processes. As shown in Figure
4.49, the fluid generally inlets the system at a single port and there is no exit. During the process, the
system might be in energy interaction with other systems. Hence, regarding the above stated assumptions, the energy equation may be reduced to
U  m i hi  Q  W
(4.66)
Integration over a time interval of (0-t) yields,
t
U (t )  U (0)  hi  m i dt  Q12  W12
(4.67)
0
In addition, the conservation of mass requires,
t
m(t )  m(0)   m i dt
(4.68)
0
Substitution of equation (4.68) into (4.67) results with the energy equation of the charging processes. The process takes place in a time interval of (0-t).
136
THERMODYNAMICS
m(t)u(t) – m(0)u(0) = m(t) – m(0)hi + Q12 – W12
(4.69)
where m(t) and u(t) respectively are the mass and the specific internal energy of the system at
time t, and m(0), u(0) are the same parameters evaluated at the initiation of the process. On the other
hand, the terms Q12 and W12 represent the heat and work transfer of the process.
Example 4.31: As shown in Figure 4.50, a piston-cylinder apparatus containing 0.1 kg of saturated steam at 10 bars is
connected through a valve to a supply line which carries steam at 20 bars in pressure and 500°C in temperature. The valve
is opened and the steam is allowed slowly to flow into the apparatus until temperature reach 300°C. During the charging
process, 90 kJ of heat is transferred to the surroundings. Determine the amount of steam charged to the apparatus.
Solution:
With respect to the system boundary, the energy balance requires,
m2u2  m1u1  m2  m1 hi  Q12  W12
where the subscripts 1 and 2 respectively refer to the initial and the final states in the
apparatus. From steam tables, at state 1 (p1 = 10 bars, x1 = 1.0), u1 =2583.6 kJ/kg, v1 =
0.1944 m3/kg, and at state 2 (p2 = 10 bars, T2 = 300°C), u2 = 2793.2 kJ/kg, v2 = 0.2579
m3/kg. The enthalpy of the charged steam is: hi = 3467.6 kJ/kg. Let m represent the
amount of steam charged into the cylinder. Then, m2 = m1 + m. Since the work transfer
takes place at a constant pressure,
W12  p1 V2  V1 
where, V1 = m1v1 = 0.1 0.194 = 0.0194 m3, and V2 = m2v2 = (0.1+m) 0.2579 =
0.02579 + 0.2579m. Then the work transfer becomes, W12 = 1000 (0.02579 +
0.2579m-0.0194) = 6.39 + 257.9m. In addition, Q12 = –90 kJ, substituting these
evaluated terms into the energy equation yields,
(0.1+m) 2793.2 – 0.1 2583.6 = 3467.6m + (–90) – 6.39 – 0.2579m, and solving for m results as m = 0.174 kg.
Example 4.32: A thermodynamicist is attempting to model the process of balloon inflation by assuming an evacuated and
elastic casing to behave as a spring opposing the expansion. As air at 10 bars, and 20°C is admitted through the valve, the
spring is compressed and the pressure in the gas space becomes p = p1+K(L-L1), where the spring constant K is 500kPa/m.
The initial condition in the gas space is: p1 = 1 bar, T1 = 20°C, L1= 0.2 m. For a piston area of 0.02 m2, determine the temperature of air in the gas space when L2 = 0.8 m.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 137
Solution:
For the system boundary shown in Figure 4.51, Q12 = 0, and equation (4.68) simplifies to
m2u2  m1u1  m2  m1 hi  W12
The system does a pdV type of work against the spring.
L2

W12  pdV 
 p1  K L  L1 AdL  p1A L2  L1   KA
L2  L1 2
L1
2
or
W12 = 1000.02
( 0.8  0. 2) 2
= 3 kJ.
2
For idealgas behavior of air,
m1u1 
p1V1
100  0.004
cvT1 
 0.718  1 = 1 kJ, = 16.01 kJ,
RT1
0.287
m2u2 
p2V2
400  0.016
cvT2 
 0.718  16.01
RT2
0.287
The enthalpy of charged air, hi = cpTi = 1.005293 = 294.46 kJ. Substituting these terms into the energy equation results
as, 16.01-1 = 294.46m-3, and solving for m yields, m = 0.061 kg.
The final mass of air in the casing, m2 = m1+m = 0.00475+0.061 = 0.0657 kg, and through the equation of state,
T2 
4.15.2
p2 V2
400  0. 016
, or T2 = 339.4 K.

m2 R 0. 0657  0. 287
Transient discharging processes
The process of discharging a fluid through a pressurized vessel is physically similar to the charging process. However, the conservation of energy principle yields totally different results for these
two phenomena. For a discharging process, the change in kinetic and in potential energies of both the
system and the outgoing fluid is assumed to be negligible. In addition to the temperature difference
between the system and the surroundings, the rate of the process that is how fast the phenomenon
taking place governs the heat transfer. It is always assumed that the outgoing fluid possesses the same
thermodynamic properties as the system has. This assumption requires a discharging process to be
slow enough that the process is quasi-static. Hence, for time varying properties of the system, the
discharged fluid exhibits the same time variation for those properties. Besides, there is no flow of
mass into the system. As a consequence, the energy equation might be rearranged as follows,
U   m e he  Q  W
(4.70)
138
THERMODYNAMICS
The change in the internal energy of the system which discharges fluid for a time interval of
(0 - t ) is:
t
U (t )  U (0)    m e he dt  Q12  W12
(4.71)
0
Moreover, the conservation of mass principle applied to a such system yields,
t
m(t )  m(0)    m e dt
(4.72)
0
As a specific Example for such a process, consider a pressurized tank equipped with a gas releasing
valve contains an ideal gas. As shown in Figure 4.52, at a certain pressure p, the valve opens and the
gas is adiabatically released. It is desired to determine a relationship between the mass remained in the
tank and the gas temperature. Considering the differential form of equations (4.71) and (4.72) as,
dU  me he  Q  W
(4.73)
dm  me
(4.74)
Since the process is adiabatic, Q = 0, and the absence of a moving boundary, or a shaft crossing
the system boundary, etc., the transfer of work is also zero, W = 0. Due to change of system mass,
the differential form of the internal energy of the system is: dU  mdu  udm . At an instant of time,
the enthalpy of discharging fluid is the same as the system enthalpy, he = h. As a result, equation
(4.73) may be rearranged to yield,
mdu  (h  u)dm
(4.75)
Referring to the definition of enthalpy and internal energy of an ideal gas, equation (4.75) may
be reduced to
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 139
dm cv dT
 
m
R T
(4.76)
Integration between the initial and the final states yields,
 m2 


 m1 
k 1

T2
T1
(4.77)
 1 k 


Recall that for a reversible and adiabatic process of an ideal gas, Tp k   Constant . Then the
mass ratio in equation (4.76) may be expressed in terms of the pressure ratio as,
k
 m2 
p

  2
p1
 m1 
(4.78)
This result is unique in the sense that it provides a direct relationship between the mass and the
pressure of the gas remaining in the tank for a process of no work and no heat transfer interactions.
Example 4.33: The insulated tank in Figure 4.52 has a volume of 1.5 m3 and contains air at 10 bars, 227 C. The valve
automatically opens at this pressure, and air is released quasi-statically. The process stops when the tank pressure is reduced
to 1 bar. Determine a. the final temperature, b. the ratio of the discharged mass to the initial mass of the tank.
Solution:
a.
The initial mass of air in the tank, m1 
1
p1V 1000x1.5

 10. 452 kg, and equation (4.78) gives the final mass as,
RT1 . 287 x500
1
m2  p2  k  1 1.4
  
 0.193, m2  0.193 x10.452  2.017 kg. Thus the air temperature at the final state becomes,
m1  p1   10 
m 8.435

 80.7%
m1 10.452
Example 4.34: A 2m3 tank contains air at 2Mpa, and 500C. The tank is connected to a turbine which exhausts into the
atmosphere as in Figure 4.53. The tank, the connection pipes, and the turbine are insulated, and their volumes are negligible.
After opening the valve, the air expands quasi-statically. Determine, a. the total mass flowing through the turbine until the
flow stops, b. the work delivered to the turbine shaft.
b.
The amount of discharged air is: m  m1  m2  10.452  2.017  8.435 kg
Note: Air is always assumed to exhaust to atmospheric conditions at 1 bar, and 27C.
140
THERMODYNAMICS
Solution:
a. The flow of air through the turbine will stop when the final pressure in the tank is 1bar. Due to quasistatic and
adiabatic expansion of air, by equation (4.78), the mass remained in the tank at p2 = 1 bar is,
1
1
m2  p2  k  1 1.4
  
 0.117,
m1  p1   20 
where
m1 
p1V
2000 x 2

 18.03 kg, and m2 = 0.117 × 18.03 = 2.1 kg
RT1 0.287 x 773
Thus the amount of mass flowing through the turbine becomes, m  m1  m2  18.03  2.1 or m = 15.93 kg
b. In applying equation (4.71) to the system in Figure 4.49, the enthalpy of air being constant at the exit,

t2
t1
t2
m e he dt  he  m e dt  he (m1  m2 )
t1
and Q12 = 0 . Thus rearranging equation (4.70) for work transfer yields, W12  (m1u1  m2u2 )  he (m1  m2 )
p V
100x2
The final temperature of air, T2  2 
 331.8 K , considering that u = cvT, and h = cpT, the work transfer
m2 R 2.1x0. 287
may be evaluated as,
W12  0. 718x(18. 03x773  2.1x331.8)  1. 005x300x15. 93 or W12 = 4703.72 kJ
There are many other types of transient flow problems of theoretical as well as practical interest. The
solution of each type depends on the restrictions and idealizations placed on the process. In every case,
it is better to begin with the basic or fundamental equations and to proceed logically from that point.
A discussion of some other type of transient flow systems is presented in the following examples.
Example 4.35: The pressure cooker in Figure 4.54 has an internal volume of 0.5 m3, and operates in the following manner:
A small amount of water placed in the cooker experiences a positive heat transfer increasing the water pressure to 2 bars.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 141
At this pressure the vent valve opens automatically and vents the contents of the cooker to the atmosphere in a manner such
that the pressure in the cooker is held constant at 2 bars. After the cooking time is elapsed, the vent valve is closed and the
heat transfer becomes negative. When the pressure returns to 1 bar, the door is opened and the food is removed. In the event
that the vent valve fails to operate when the heat transfer is positive, the pressure will rise until the cooker explodes. To
prevent such a catastrophe, the cooker is equipped with a burst disk that ruptures at a pressure of 3.5 bars which is less than
the explosion pressure. Consider the case in which the cooker contains nothing but 3 kg of water at a pressure of 1 bar when
the cooker is sealed. The cooker now experiences a positive heat transfer until the pressure reaches 3.5 bars whereupon the
burst disk ruptures. The content of the cooker vents rapidly to the atmosphere. a. Determine the temperature and the quality
of water in the cooker just before the burst disk ruptures. b. If the venting process is modeled as adiabatic, estimate the mass
of water that escapes from the cooker. c. Calculate the work done against the atmosphere.
Solution:
a.
The heating process until the burst disk ruptures is a constant volume process. Therefore, v2 = v1 = 0.166 m3/kg,
and from steam tables, at p2 = 3.5 bars, v2g = 0.524 m3/kg. Since, v2 < v2g , the state of water before bursting is a
saturated state. Thus the temperature at state 2 is, T2 = 138.9 C
The quality of water just before the disk ruptures, x2 
b.
v2  v2 f
v2 g  v2 f

0.166  0.001078
, or x2 = 0.315
0.524  0.001078
This is a non-quasi-stastic discharging process, it occurs suddenly. To get a result, the system has to be treated
as closed. Letting the final state after the disk ruptures represented by 3, for the closed system in Figure 4.54, the
energy equation is:
m(u3  u2 )  W23
where, W23 = mp0(v3 - v2) and substitution yields, u3  p0v3  u2  p0v2  h3 . Recall that after bursting a mechanical
equilibrium establishes and p3 = p0 = 100 kPa. Evaluating the internal energy at state 2 as,
u2  u2 f  x2 (u2 g  u2 f )  583.95  0.315 x(2546.9  583.95)  1203.8 kJ/kg . Thus the enthalpy at state 3 is:
h3  1203.8  100 x0.166  1220.4 kJ/kg .Then the final quality of water, x3 
h3  h3 f
h3 g  h3 f

1220.4  417.46
 0.355 , and the
2258
specific volume becomes, v3  v3 f  x3 (v3 g  v3 f )  0.602 m3 /kg . Thus the change in volume due to rupture is:
V  m(v3  v2 )  3 x(0.602  0.166)  1.306 m3 . Accordingly, the amount of discharged mass is:
m 
c.
V 1.306 or m = 2.169 kg.

v3 0.602
The work done against the atmosphere may be determined as follows,
W23  mp0 (v3  v2 )  3 x100 x(0.602  0.166) , W23 = 130.8 kJ.
Example 4.36: The device shown in Figure 4.55 is designed to provide hot air for defrosting the windshield of an automobile. Atmospheric air flows into a rigid, insulated chamber where it is heated by an electrical heating element. Assume
that steady-state conditions exist initially with the outlet temperature of air being T0. Shutting off the power to the heating
element at t = 0, T2 will decrease with time. Drive an expression for T2 as a function of time.
142
THERMODYNAMICS
Figure 4.55
Solution:
For t > 0, there is no heat and work transfer to the system in Figure 4.55. There is also no change in the system kinetic and
potential energies. Because of single inlet and outlet, equation (4.39) simplifies to
  mu  m i hi  m e he , where considering the mass conservation, m e  m i  m
mu
Substituting m e into the energy equation and rearranging results as, m (u  he )  mu  m i (hi  he )
Since the outgoing fluid has the same properties as the system, u  he   RT . The mass of air in the system at an instant
pV 1  dT 
 

, where the system pressure
of time t is, m  pV , and the time rate of change may be expressed as m
R T 2  dt 
RT
dT
and hi - he  c p (T1 - T ). After substitution of these
and the volume are assumed to be constant. Additionally, u  cv
dt
dT
dT
m
pV
, the following differential equation results,
relations into the energy equation and letting m0 

 1 dt and
T T1  T m0
RT1
T(0) = T0 and T(0) = T0
Together with the initial condition, the integration yields an expression for the outlet temperature as follows,
 m 1
 T1  T  T0 



 T  T  T   exp   m
0 
 0

References
1.
W.P Graebel, Advanced Fluid Mechanics, Elsevier
Academic Press, ISBN 978-0-12-370885-4, 2007.
2.
J.F. Gülich, Centrifugal pumps, 2nd Edition, SpringerVerlag, ISBN 978-3-12823-3, 2010.
3.
J.M. Gordon, and K.C. Ng, Cool Thermodynamics,
Cambridge International Science Publishing, ISBN
1-8983-2690-8, 2001.
4.
M. Kaufman, Principles of Thermodynamics, Marcel
Dekker Inc., ISBN 0-8247-0692-7, 2002.
5.
A.N. Berris, and B.J. Edwards, Thermodynamics of
flowing systems with internal microstructure, Oxford
University Press, ISBN 0-19-507694-X, 1994.

t 

6.
R.P. King, Introduction to Practical Fluid Flow,
Elsevier Butterworth-Heinemann, ISBN 075064885-6, 2002.
7.
S.L. Dixon, Fluid Mechanics and Thermodynamics
of Turbomachinery, 5th edition, Elsevier ButterworthHeinemann, ISBN 07506-7870-4, 2004.
Problems
Isolated systems
4.1
Consider the mechanical system shown in Figure 4.56
in which a mass M suspended from a string in a gravity
field g. Below the mass is a rigid platform of mass m
(m<< M) which is mounted on two identical springs
with spring constant K. The string supporting the mass
M is cut, and the mass falls from a height L striking
the platform midway between the two springs.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 143
c. Suppose that section A contains steam at 2.5 MPa,
400C, and has a volume of 0.1 m3. Determine
the final equilibrium state of steam when the
diaphragm is ruptured.
4.7
An insulated and rigid tank is divided into two compartments. The compartment A contains 0.1 kg of
saturated liquid water at 3 bars, and the other contains
0.4 kg of steam at 3 bars, 200C. The membrane is
ruptured, and the fluids of the two compartments
are mixed.
a. Determine the final temperature of the mixture.
b. Calculate the amount of water to be contained
in compartment B for obtaining saturated vapor
at 3 bars after the mixing process completed.
4.8
A well-insulated cylinder 0.1 m2 in cross-sectional
area contains 5 kg of steam at 300C, and 10 bars.
As in Figure 4.58, the section behind the piston is
evacuated, and a linear spring of K = 80 kN/m is
inserted. Initially, the spring is unstressed, and the
piston is held by a stop at its position. The stop is
removed and the steam expands to a new equilibrium
state at which the pressure is 6 bars. Determine,
Figure 4.56
a. Defining the system, determine the equilibrium
height of the platform by stating your assumptions about the system.
b. Suppose now that the mass M is replaced by a
smaller mass at a value of M/10, while the height
is 10L. What effect does this have on the answer
to (a)?
4.2
A 20 kg mass of copper at 60C and a15 kg mass
of aluminum at 70C are dropped into an insulated,
steel tank which has a mass of 8 kg and contains 250L
of water at 20C. Determine the final equilibrium
temperature.
4.3
A 10 cm3 cube of ice at 0C is dropped into an
insulated glass which contains 250 cm3 of water at
20C. Determine the final equilibrium temperature
after the ice has melted. Assume the latent heat of
fusion for water to be 333 kJ/kg.
4.4
An unknown mass of copper at 80C is dropped
into a well insulated tank which contains 0.15 m3
of water at 25C. If the final temperature is 28C,
find the mass of copper.
4.5
A casting of 50 kg is taken from an annealing
furnace at a temperature of 450C and is plunged
into an insulated tank containing 400 kg of oil at a
temperature of 27C. The casting and the oil reach
a thermal equilibrium, and the final temperature is
37C. The specific heat of oil is 2kJ/kgK. Determine
the specific heat of the casting.
4.6
A shock tube is a device that is used to produce
high velocity gas flows. As shown in Figure 4.57,
the section A contains air initially at 14 bars, 27C,
and VA = 0.1VB, and section B is evacuated. In the
operation, the diaphragm is suddenly ruptured and
the gas adiabatically expands to fill the entire tube.
For an ideal gas behavior of air, determine,
Figure 4.58
a. the volume ratio of the final state to the initial,
b. the final state of steam.
4.9
A 0.08 m3 container is divided by a partition into
two parts as in Figure 4.59, and both parts contain
nitrogen (ideal gas, M = 28 kg, k = 1.4). Part a has a
volume of 0.03 m3, at 300 kPa, 25C. While in part
B, nitrogen is at 600 kPa, 225C. After rupturing
the partition, the gas mixes and a new equilibrium
state is attained. Determine the final temperature
and pressure for a well insulated container.
Figure 4.57
a. the final temperature and pressure,
b. the work required to restore the system to its
initial state.
Figure 4.59
144
4.10
THERMODYNAMICS
As shown in Figure 4.60, a vertical and insulated tank
has a volume of 40L and separated by a piston into
two equal parts. The air is initially at 1 bar, 250C,
and water is at 250C with a quality of 0.6. the 100
kg piston has a cross-sectional area of 80 cm2, and is
held in place by a stop. Releasing the stop, the piston
moves upward and an equilibrium state is reached.
For a diathermal piston,
a. determine the final temperature, the pressure
and the volume of air.
b. Suppose now that the steam is replaced by oxygen gas at 35 bars, 250C, determine the same
quantities stated in (a) for this case.
reached a temperature of 650C and a pressure of 30MPa
and exploded. The steam then expanded into the boiler
room having a volume of 250 m3. The engineer feels
that the final pressure of steam may have been high
enough to cause the walls of the boiler room to burst.
Thus he wants to calculate this pressure by assuming
that the boiler room is sealed and the initial contents
of the room are negligible. Determine,
a. the final pressure in the room after the explosion,
b. the final quality of steam.
Closed systems
4.13
An insulated copper container with a mass of 3 kg
contains 20 kg of water. The water is stirred by a very
light paddle-wheel connected to an electric motor as
shown in Figure 4.62. Due to stirring action, the temperature of the system consisting of water and copper
rises at a rate of 1C per minute. Make a table showing
the rates of heat transfer and work transfer and the rate
of internal energy change of the system consisting of
Figure 4.60
4.11
A well insulated container 0.16 m3 in volume is
divided internally into two equal parts by a rigid,
adiabatic partition. As shown in Figure 4.61, a small
connecting tube through this partition equipped with
a valve accessible to the outside. Side A contains
oxygen gas at 37C and 1.5 bars, and side B also
contains oxygen but at 97C, 7 bars. Both gases are
ideal with a specific heat of cp = 0.918kJ/kgK.
a
The valve is quickly opened and the pressures
on both sides are rapidly equalized. Assume no
conduction of heat takes place, and then determine
the temperatures and pressures on both sides.
b. The valve is left open and eventually the temperatures become equal on both sides. Determine
the final pressure and the temperature.
Figure 4.62
a.
b.
c.
d.
4.14
the water,
the container,
the water and the container,
the water, the container, and the motor.
A 0.5 kg of air is enclosed in a stepped cylinder. As
in Figure 4.63, the cross-sectional area of the large
section is related to the small by A1 = 2A2 , and the
initial state of air is 3 Mpa and 327C. As air cools
down, the piston descends and reaches the step. The
cooling process continues until the final temperature
is 7C. Determine,
Figure 4.61
4.12
A marine engineer is investigating the possible cause of
a recent submarine disaster, and comes upon the idea that
it may have been due to the explosion of a faulty steam
boiler. He hypothesizes that the 4 m3 boiler accidentally
Figure 4.63
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 145
a. the temperature when the piston reaches the
step,
b. the final pressure of air,
c. the amount of heat and work transfers for the
complete process.
d. Show the process on p-V diagram.
4.15
Ammonia at a state of 200kPa and 20C is contained
in a piston-cylinder apparatus and occupies a volume
of 1.5 m3. A 750 kJ of work is transferred to ammonia
by compression and its temperature increases to
40C. During the compression process, 1500 kJ of
heat is transferred to surroundings at 20C. Through
an energy analysis, determine whether the given
process is feasible or not.
4.16
A piston-cylinder machine contains nitrogen initially
at 2 bars, 117C, and 0.25 m3. The piston moves
with negligible friction until the pressure rises to 5
bars. The process is described by, V = 0.4 - 0.05p,
where V and p are in m3 and in bars respectively.
Determine,
a. the final temperature,
b. the work done, and the amount of heat transfer.
4.17
Oxygen is contained within a piston-cylinder assembly initially at 500 kPa, 200C, and occupies a
volume of 0.04 m3. The gas expands according to
the process described by pV1.15 = Constant , until the
temperature reaches 97C. Determine the value of
the work done and the heat transferred.
4.18
A heat exchanger containing steam at 1.5 bars, 160C,
has a volume of 0.15 m3 on the steam side. Because
of exchanger maintenance, the inlet and the outlet
valves are closed and the exchanger is allowed to
cool to a final temperature of 25C. Determine the
heat transfer and sketch the process on a p-V diagram
relative to saturation line.
4.19
A rigid tank contains 5 kg of refrigerant 22 at 5 bars,
75C. A paddlewheel within the tank adds energy at
a constant torque of 150 Nm for 1000 revolutions.
At the same time, the system is cooled to a final
temperature of 15C. Determine the heat transfer
and also sketch the process on p-V diagram relative
to saturation line.
4.20
A 2kg of water contained in a piston-cylinder device
is initially at 300C. The substance undergoes a
constant-temperature process with a volume changing
from 0.02 m3 to 0.15 m3. The measured work output
is 890 kJ. Determine,
a. the final pressure,
b. the heat transfer. Sketch the process on p-V
diagram, relative to saturation line.
4.21
A 2.5kW resistance heater is placed in a 0 08 m3
container filled with water at 27C and 1 bar. The
heater is allowed to operate for 30 minutes. The
mass of heater is 2 kg, and its specific heat is 0.6kJ/
kgK. For an insulated container, determine the final
temperature of water and the heating element.
4.22
A 1 kW of electric heater is placed at the base of a
laterally insulated cylinder which contains 2 kg of
water at 15C. As shown in Figure 4.64, a 25 kg of
frictionless piston covers water at the top, and has a
cross-sectional area of 0.05 m2. For an atmospheric
pressure of 100 kPa, determine the time required to
boil off 0.8 kg of water.
Figure 4.64
4.23
Consider two separate systems. One is a piston-cylinder
arrangement and the other a rigid tank both of which
contain 10 kg of the same ideal gas initially at the same
pressure, temperature, and the volume. As a result of
heating, the temperature of the gas in both systems
increases by 10C. Determine the amount more heat
to be supplied to the piston-cylinder assembly.
Note: The molar mass of the gas is 25 kg/kmol.
4.24
A balloon made of thin elastic material contains air at
200 kPa, 20C, and occupies a volume of 5 m3. The
inside pressure of the balloon is proportional to its
volume and is represented by p = bV - 100 , where p
is in kPa, and V in m3. The air is heated by a burner
to a final temperature of 150C. The surroundings
outside of the balloon is at 1 bar in pressure and
20C in temperature. Determine,
a. the final pressure in the balloon,
b. the work and the heat interactions during the
process.
Figure 4.65
Note: Assume that the heat transfer through the
balloon material is negligible.
146
THERMODYNAMICS
4.25
As shown in Figure 4.65, a laterally insulated vertical
cylinder contains saturated steam. Due to atmospheric
pressure and the piston weight, steam is initially at
10 bars in pressure and occupies a volume of 0.15
m3. The steam is heated until the piston reaches the
stops with a resulting volume of 0.2 m3. Then an additional of 30 kJ of heat is supplied. Determine the
final temperature of steam.
4.26
A 200L of steam at 10 bars, 200C is contained in a
piston-cylinder arrangement. As the steam is cooled,
the piston in Figure 4.66 descends and reaches the
locking mechanism. At this instant, the position of
the piston is fixed by the locking mechanism. Then
the content of the cylinder is heated so that the final
state of water is at the critical state.
a. Evaluate the spring constant for an atmospheric
pressure of 100kPa.
b. At what position of the piston, the refrigerant
becomes saturated vapor?
c. Calculate the heat transfer for the complete
process.
d. Sketch the overall process on a p-V diagram,
relative to saturation line.
4.28
As shown in Figure 4.68, a vertical column of mercury
2 m in height rests on a frictionless, massless, and
non-conducting piston which has 0.1 m2 of crosssectional area. A pocket of air trapped behind the
piston has a height of 20 cm and provides sufficient
pressure to support the mass of mercury at an initial
temperature of 27C. It is desired to raise the piston
by heating the trapped air. As the piston rises, the
displaced mercury spills over the tube wall. To
displace all the mercury, determine,
Figure 4.66
a. Determine the heat lost and gained respectively
by the cooling and the heating processes.
b. Sketch the overall process on a p-V diagram,
relative to saturation line.
4.27
The laterally insulated piston-cylinder arrangement in
Figure 4.67 contains 4 kg of R22 at a temperature of
-5C, and a quality of 1%. Saturated vapor has to be
obtained by heating the refrigerant. When the piston
impacts the mechanical stop, the cylinder volume
becomes 10L. The piston has a cross-sectional area
of 0.1 m2 and considered to be massless. The spring
is linear and assumes its free length when the piston
is at the base.
Figure 4.66
Figure 4.68
a. the amount of work and the heat transferred by
air.
b. Illustrate the process on p-V and T-V diagrams.
Note: Assume that the density of mercury is 13600
kg/m3.
4.29
A room with dimensions of 3 m x 6 m x 8 m is to
be heated by a heater which supplies 30,000 kJ of
heat per hour. As shown in Figure 4.69, a fan of 200
W in power provides circulation currents for air.
Heat leaks through the walls of the room at a rate
of 10,000kJ per hour. For an initial temperature of
0C, determine the time required for bringing the
room temperature to 20C.
Figure 4.69
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 147
4.30
volumes of 0.4 m3. The refrigerant 22 having 4 kg
of mass is initially at 50C. A heater is installed on
the Freon side of the cylinder. Due to heat transfer,
the final pressure in the cylinder becomes 500kPa.
Determine, Assuming ideal gas behavior for R22, with
c p  0.74 kJ/kgK , cv  0.57 kJ/kgK determine,
The piston-cylinder apparatus shown in Figure 4.70 is
divided into two compartments by a frictionless piston.
Initially both of the compartments are filled with air
at 1 bar, 27C. The piston is 5 cm in diameter, and
the cross-sectional area of the piston rod is negligible.
An actuator capable of exerting a maximum of 1 kN
of force on the piston rod is activated. Determine,
a. the final temperatures in both sections,
b. the amount of heat transferred by the heater.
Figure 4.70
a. the distance traveled by the piston if the process
is quasi-static and adiabatic,
b. the distance traveled if the process is carried out
quasi-statically and isothermally.
c. In each case, calculate the transfer of work and
heat for all the gas.
d. Suppose that the rod is pulled rather than pushed.
Would the answers to questions a, b, and c have
any change?
4.31
As shown in Figure 4.71, a cylinder with thermally
insulated walls is initially divided into two equal
compartments of 0.02 m3 in volume by a moveable,
frictionless, and insulated piston. Each comparment
contains 1 kg of carbon monoxide modeled as ideal
gas with a constant volume specific heat of 0.744
kJ/kgK, and a gas constant of 0.296 kJ/kgK. The
gases at both compartments are initially at the same
temperature of 300 K, and at the same pressure. Due
to electrical work transfer, the final pressure becomes,
p2 = 2p1. Determine,
Figure 4.72
4.33
A well insulated piston-cylinder assembly contains
ideal gas at 10 bars, 27C. As shown in Figure 4.73,
a mechanical stop prevents the piston from moving against the surroundings at 1 bar. The cylinder
height containing the gas is 40 cm and the piston
cross-sectional area is 0.02 m2. The 250 kg piston
is tightly fitted into the cylinder and is frictionless.
When the stop is released, the piston moves until it
impacts to another mechanical stop where the gas
volume has just double.
a. the final temperatures of gases in each compartment,
b. the electrical work transfer.
Figure 4.73
a. As an engineer, you are asked to estimate the temperature and the pressure after the expansion.
b. Repeat the problem for a case that the cylinder
is rotated by 90 before tripping the mechanical
stop.
Figure 4.71
4.32
As shown in Figure 4.72, R22 and saturated steam
are contained in an insulated cylinder separated by
a non-conducting, frictionless piston into two equal
c. Solve the same problem for a case that the cylinder
is rotated 180 before the tripping action.
Note: Assume the constant volume specific heat and
the gas constant to be 0.8kJ/kgK, and 0.4kJ/kgK
respectively.
148
4.34
THERMODYNAMICS
A cylinder containing 0.5 kg of steam at 3Mpa is
maintained at 400C by a constant temperature bath.
As shown in Figure 4.74, a 200 kg of insulated piston
whose area is 200 cm2 is held in place by means of
a stop. The stop is released, and the piston moves
upward, oscillates for a while and assumes a new
equilibrium state. For an atmospheric pressure of 1
bar, determine the work done by the steam.
500C while cylinder B is empty. Opening the valve
on the connecting line, the steam is allowed to flow
from A to B through the turbine until piston in A
comes to the base.
Figure 4.76
Figure 4.74
4.35
As shown in Figure 4.75, a piston (A) and a piston rod
(B) are fitted inside a cylinder of length 500 mm, and
area of 60 cm2. The piston weighs 100 N, the piston
rod is 12 cm2 and weighs 50 N. On the top of the rod,
but outside of the cylinder, a 200 N of weight (C) is
placed. Initially air in (D) is at atmospheric pressure
while the piston is positioned in the middle of the
cylinder. The initial temperature is 40C everywhere.
Assume that the cylinder is insulated and the piston
rod is non-conductive. Determine the final pressure
and the temperature in (E),
For a final temperature of 200C in cylinder B,
determine the work delivered by the turbine.
4.37
The device shown in Figure 4.77 is equipped with
an insulated horizontal cylinder, two frictionless
pistons and an orifice plate. The plate divides the
volume of the cylinder into two compartments. The
pistons are clamped together in such a fashion that
when they move, the distance between their faces
remains constant. The cylinder initially contains 1 kg
of air at a pressure of 10 bars, and a temperature of
100C. An actuator which is external to the system
moves the pistons so that 30 kJ of work is done to
the system. Determine the final temperature and the
pressure of air.
Figure 4.77
Figure 4.75
a. for a diathermal piston,
b. for an adiabatic piston.
4.36
Two vertical and insulated piston-cylinder assemblies
are connected through an adiabatic turbine as shown
in Figure 4.76. The piston weights and the cylinder
cross-sections are such that pressures of 1.5Mpa
and 0.15Mpa are necessary for moving the pistons
respectively in cylinders A and B. The cylinder A
initially contains 3 kg of steam at a temperature of
4.38
As shown in Figure 4.78, a vertical and insulated cylinder
is fitted with a 30 kg of frictionless and adiabatic piston
whose area is 15 cm2. The volume trapped between the
piston face and the bottom of the cylinder is partitioned
off into two sections by a rigid and adiabatic membrane.
The upper section contains 10 kg of saturated liquid
water while the lower section has 1 kg of water at a
pressure of 7 bars and a temperature of 200C. At the
moment when the membrane is ruptured, the system
comes into a new equilibrium state.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 149
Figure 4.78
Assuming that the surroundings is at a pressure of
100kPa, determine, a. the quality and the volume
of the final equilibrium state, b. the work transfer
experienced by the system.
4.39
Figure 4.80
Thermodynamic cycles
4.41
Consider a piston-cylinder assembly with an unstreched spring just touching on the piston surface as
shown in Figure 4.79. The cylinder contains 4 kg
of saturated liquid water at 100kPa, and the piston
is assumed to be massless. It is determined that as
the heat supplied, the water passes through saturated
vapor state at 250kPa. For a final pressure of 300kPa,
calculate the amount of heat to be supplied to the
system.
A closed system, containing 2 kg of steam, undergoes
a cycle as shown in Figure 4.81. The cycle consists
of three quasistatic processes which are straight lines.
Determine, a. the heat transfer of each process, b.
the net work transfer of the complete cycle, c. the
efficiency of the cycle.
Figure 4.81
4.42
Figure 4.79
4.40
As in Figure 4.82, carbon dioxide undergoes a cycle by
the following successive processes. (1-2) isothermal expansion and the work done is W12 = 10 kJ, (2-3) isochoric
heating and the amount of heat transferred is Q23 = 18
kJ, (3-1) isobaric compression to the initial state. The
shaded area in the figure is 12 kJ, and the internal energy
of carbon dioxide at state 1 is 75 kJ. Determine,
A vertical and insulated piston-cylinder arrangement
shown in Figure 4.80 is divided into three compartments by means of two diathermal diaphragms, and
one floating piston. Each compartment contains the
same mass of air at a value of 1 kg. However, the
pressures of the compartments are p1A = 1 bar, p1B
= 2 bars, and p1C = 3 bars.
The three compartments are initially at the same
temperature of 27C. The two diaphragms are punctured at the same time, and the air assumes a new
equilibrium state. Determine the final equilibrium
temperature of air.
Figure 4.82
a. the internal energies at states 2, and 3,
b. the heat transfer of each process,
c. the enthalpy change of the gas for the complete
cycle.
150
4.43
4.44
THERMODYNAMICS
d. Is the cycle a power or a refrigeration cycle?
Explain.
rejects 700 kJ of heat to another. The heat engine
performs 1000 kJ of mechanical work. Determine,
A 2 kg of ideal gas with a molar mass of 44 kg is
contained in a piston-cylinder device. The gas, initially
occupying a specific volume of 0.48 m3/kg at 20C,
undergoes a cycle as following: (1-2) isothermal compression to a volume of V1/2, (2-3) isobaric expansion
to a volume of 3V1/2, (3-1) returning to its initial state
by a process along a straight line on pV coordinates.
a. Sketch the cycle on a pV diagram.
b. Determine the net work output and the efficiency
of the cycle.
Note: Assume that k = 1.5 for the gas.
a. the amount of heat exchange with the third
reservoir,
Figure 4.83 presents a cycle undergone by an ideal
gas with constant specific heat cv and k = 1.6. The
temperature and the pressure at state 1 respectively are
300C, and 14 bars. In addition, the volumes at state
1 and 3 are given as 0.1 m3, and 0.4 m3 respectively.
One of the processes of (1-2) or (3-1) should be
isothermal and the other is adiabatic. Determine,
a. which one is isothermal, and which one is adiabatic?
b. the pressures and the temperatures at state of 2
and 3,
c. the volume at state 2,
d. the net work output and the efficiency of this
cycle.
b. the efficiency of the cycle.
4.48
A ton of refrigeration is defined as the heat absorption
rate by which 1 ton of water at 1 bar, and 0C freezes
in 24 hours and becomes ice at the same pressure and
temperature. A refrigeration cycle produces 5 tons
of refrigeration by rejecting heat to surroundings at
a rate of 25 kW. Determine,
a. the power input,
b. the coefficient of performance,
c. the cost of operation for one day, if the electricity costs 10 cents per kilowatt-hour, and the
refrigeration system is on one-third the time.
4.49
The piston-cylinder arrangement shown in Figure
4.84 contains 1.5 kg of water at 50C and is used for
raising toys to a certain distance by cyclic processes
in a toy manufacturing facility. The piston weight
causes a pressure of 200 kPa, and initially rests on the
lower stop. The toy, together with the piston weight,
causes a pressure of 500kPa when it is mounted on
the piston. The content of the cylinder is heated until
its temperature reaches 200C. At this instant, the
piston reaches the upper stop, and the mounted toy
is removed. Then the water is cooled down to its
initial temperature of 50C.
a. Draw the cycle on a p-v diagram relative to
saturation line.
b. Determine the heat and work interactions of the
cycle.
c. Compute the efficiency of the cycle
4.45
Argon, k = 1.67, M = 40, and initially at 30 bars
undergoes a cycle in a piston-cylinder apparatus
consisting of the following three quasistatic processes:
(1-2) isobaric expansion to a volume of nine times
the initial volume, (2-3) isochoric cooling, (3-1)
adiabatic compression to the initial state. The piston
has a cross-sectional area of 200 cm2 and a stroke
of 18 cm. The crank-shaft that moves the piston by
a connecting rod rotates at a speed of 1500 rpm.
a. Sketch the cycle on a p-V diagram.
b. Calculate the power produced and the efficiency
of the engine.
4.46
4.47
A refrigeration unit having a coefficient of performance of 0.85 absorbs heat from a refrigerated space
at a rate of 10 kW. Determine the rate of heat that is
rejected to surroundings in kilojoules per hour.
Exchanging heat with three heat reservoirs, a heat
engine completes integral number of cycles and
absorbs 1500 kJ of heat from one reservoir, and
Figure 4.84
4.50
An air compressor intakes air at 1 bar pressure and
delivers at 4 bars by a sequence of processes as shown
in Figure 4.85. The operation of the compressor is
as follows: (1-2) air is withdrawn into the cylinder
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 151
at 1 bar pressure when the cylinder volume expands
from V1 to V2. (2-3) the air in the cylinder is compressed quasistatically and adiabatically until the
cylinder volume is V3. (3-4) some of air is delivered
at a constant pressure of 4 bars by decreasing the
volume to V4. (4-1) after the delivery stroke, the air
remaining in the cylinder expands adiabatically until
the cylinder volume is V1.
a. the mass flow rate of the refrigerant,
b. the exit enthalpy,
c. the exit temperature, and the exit area.
4.56
Saturated liquid water at 150C flows steadily through
a throttle valve to a pressure of 2 bars.
a. Determine the temperature of water at a crosssection downstream of the throttle valve where
the kinetic energy is negligible.
b. Suppose that the mass flow rate of water is 0.5
kg/s, and the valve cross-sectional areas at the
inlet and outlet are each 20 cm2. Then, justify the
negligible kinetic energy change for throttling
processes.
4.57
A well insulated throttling calorimeter shown in
Figure 4.86 is used for determining the quality of
liquid-vapor mixture. The pressure of steam in the
pipe line is 30 bars, and the pressure and the temperature in the calorimeter chamber are measured
to be 1 bar, and 150C respectively. Determine the
quality of steam in the pipe line.
Figure 4.85
a. Under what circumstances the sequence of
processes constitutes a cycle? Explain.
b. Modeling air as an ideal gas with cv = 0.718 kJ/
kgK, and R = 0.287 kJ/kgK, compute the work
transferred to the compressor, and the net work
of air.
Steady flow systems
4.51
Air is admitted to an adiabatic nozzle at 4 bars,
250C, and at a velocity of 30 m/s. At the nozzle
outlet where the pressure is 2 bars, the air flows out
with a velocity of 300 m/s. Determine the ratio of
the exit area to the entrance.
4.52
Air at 150 kPa, 35C and with a velocity of 150 m/s
enters a nozzle at an inlet volume flow rate of 1 m3/s
and leaves the nozzle at 100 kPa with a velocity of
200 m/s. The volume flow rate at the exit is 1.45 m3/s.
For a steady flow of air, calculate the heat transfer
rate from the surface of the nozzle.
4.53
Steam at 2 MPa and 400C enters a turbine nozzle
at a velocity of 120 m/s. The cross-sectional area at
the entrance of the nozzle is 1 cm2. At the nozzle
exit, the pressure and the temperature of steam are
respectively 150 kPa, and 150C. For an adiabatic
process, determine,
a. the velocity and the area at the nozzle exit.
b. Suppose that heat is removed from the nozzle
surface at a rate of 40 kW. Determine for this
case, the exit velocity and the cross-sectional
area.
4.54
Air at a pressure of 0.5 bars and a temperature of
-30C enters a diffuser with a velocity of 500 m/s,
and leaves with a velocity of 50 m/s. For an adiabatic
diffuser, determine the air temperature at the exit
4.55
Refrigerant 22 enters an adiabatic diffuser at 2 bars,
25C, with a velocity of 150 m/s. The inlet area is
8 cm2. At the diffuser outlet, the pressure is 3 bars
and the refrigerant velocity is 65 m/s. Determine,
Figure 4.86
4.58
Saturated liquid carbon dioxide at a pressure of
5.7MPa undergoes a throttling process to a pressure of 0.198Mpa. Determine the fraction of carbon dioxide which is converted into dry ice at
0.198Mpa of pressure. Take the enthalpy of sublimation h fs  571 kJ/kgK .
4.59
As shown in Figure 4.87, steam flows at a rate of 0.2
kg/s through an insulated porous plug with identical
inlet outlet cross sections. The condition at the inlet
is 2bar and 300C, and at the outlet is 1.2 bars.
a. Determine the velocity ratio of the outlet to the
inlet.
b. Compute the error involved in neglecting the
kinetic energy change for a cross-sectional area
of 25 cm2.
Figure 4.87
152
THERMODYNAMICS
4.60
In a certain facility, 1000 kg per hour of saturated
steam at 1.5MPa pressure is needed. The existing
boiler supplies steam at 2MPa in pressure and
300C in temperature. In order to obtain steam at
desired conditions, the steam of the boiler is mixed
with water at 2MPa and 40C in a mixing chamber.
The steam, thus obtained by the mixing process, is
throttled down to 1.5MPa pressure at the chamber
exit.
a. Sketch the process.
b. Determine the mass flow rates of the cold water
and the steam from the boiler.
4.61
In an air-conditioning unit, two streams of air are
mixed in a mixing chamber to form a third stream.
The conditions of three streams and the volumetric
flow rates are given in the table below. Determine
the temperature of the third stream exiting from the
chamber.
1
2
3
Pressure (bar)
1.5
1.5
1.5
Volumetric flow rate(m3/min)
2.5
5
-
Temperature(C)
25
60
-
Stream no.
4.62
The small high-speed turbine in a dentist’s drill is
driven by compressed air. Air inlets the turbine with
a velocity of 2 m/s at a condition of 500kPa, 27C,
and exits at 100kPa and 0C. For an exit velocity
of 30 m/s, the turbine produces a power of 50 W.
Determine the mass flow rate of air through the
adiabatic turbine.
4.63
A small steam turbine produces 75 kW output with
a steam flow rate of 0.2 kg/s. Steam at 1.4 MPa,
250C is throttled to 1.15MPa before entering the
turbine and expands to an exit pressure of 10kPa
in the turbine. Heat transfer from the turbine to the
surroundings amounts to 40kJ per kg of flowing
water. Determine the quality (if two phase) or the
temperature (if superheated) of water at the turbine
exit.
4.64
Figure 4.88
4.66
The centrifugal air compressor of a gas turbine receives air from the ambient where the pressure is 1
bar and the temperature is 27C. At the compressor
discharge, the pressure is 4 bars and the temperature
is 200C. For a mass flow rate of 100 kg/min, the
exit velocity is determined to be 100 m/s. Compute
the power required to drive the compressor.
4.67
A compressor steadily intakes 30 kg/min of refrigerant
22 at 80kPa and 0C through a pipe with an inside
diameter of 10 cm. The compressor discharges the
refrigerant at 600kPa and 150C through a pipe with
4 cm inside diameter. During the process, 45000 kJ/h
of heat is lost to the surroundings. Determine,
a. the inlet and the outlet velocities,
b. the power input to the compressor.
4.68
A water pump mounted on the ground floor of a
skyscraper is used to deliver water to the observation
platform that is 300 meters above the ground level.
Determine the power input to the pump for delivering 3 kg/min of water at a discharge pressure of 2.5
bars. Assume that the temperature of water does not
change during the process.
4.69
Crude oil is in a cavern at 35C and 5 bars, and is
located 500 meters below the ground level. The oil
is pumped to the surface through a pipe with 10
cm inside diameter. The volume flow rate of oil is
20m3/h, and the local gravity is 9.75 m/s2. For oil
density of 850 kg/m3, determine,
In a power plant, steam at the outlet of the boiler
is at 15 bars, 350C and is supplied at a rate of 1.5
kg/s to an adiabatic turbine that is located 20 meters
below the boiler exit. The steam velocities at the
inlet and outlet of the turrbine are 5 m/s and 40 m/s
respectively. Determine,
a. the power output of the turbine for an exit condition of 0.05 bars and a quality of 0.95,
b. the error in power due to neglecting the effect
of elevation and the change in velocity.
4.65
1 bar, the steam enters a food dryer and exhausts as
a saturated liquid at the same pressure. The heating
capacity of the dryer is 7 MW. If the heat transfer
from the surface of the turbine is 104kJ/h, determine
the power supplied by the turbine.
In a food processing unit, steam at 20 bars, and
500C enters a turbine at a rate of 3 kg/s as shown
in Figure 4.88. Leaving the turbine at a pressure of
a. the exit velocity,
b. the required pump power.
4.70
A fan described in a manufacturer’s table is rated
to deliver 500 m3 of air per minute with a pressure
increase of 30 mm of water column when running at
300 rpm and requiring 4 kW of power. It is known
that the delivered volumetric flow rate, the increase
in pressure, and the power required are directly proportional to the first, second, and the third power of
the fan speed respectively. For a fan speed of 400
rpm, determine,
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 153
a. the volumetric flow rate,
b. the increase in pressure, and the power required.
4.71
A steady-flow steam generator operating at atmospheric
pressure uses 15 kg of water per minute at 25C and
transforms it into steam at 100C. Determine the heat
capacity rate of the steam generator.
4.72
Water at 65 bars and 30C enters a steam generator
at a rate of 200 000 kg/h. The super heated steam at
60 bars and 550C leaves the generator.
a. Determine the heating capacity rate of the steam
generator.
d. the inlet water velocity if the pipe diameter at
the inlet is 10 cm.
4.76
Steam is discharged from the turbine of a power plant
at a rate of 2500 kg/h and at conditions 0.1 bars and
a quality of 0.9. The discharge of the turbine enters a
condenser where the steam is converted into saturated
liquid at the same pressure by the flow of cooling water
through the tubes of the condenser. Determine,
a. the heat removal rate in the condenser,
b. the mass flow rate of cooling water for a temperature rise of 10C.
4.77
The ducts of an air heating system have the crosssectional dimensions of 30 cmX40 cm. The heated air,
flowing with a velocity of 5 m/s at 85C and 1 bar, is
transported through an uninsulated section. As a result
of heat losses, the temperature of air drops by 5C.
Determine the heat loss from air to the surroundings.
4.78
Steam enters a long and inclined pipe with an inlet
diameter of 5 cm at 5 bars, 300C, and at a velocity
of 10 m/s. At the pipe exit that is elevated by 30
meters, the conditions are 3 bars, and 250C, and the
diameter of the pipe is 10 cm. Determine,
a. the exit velocity of steam,
b. the rate of heat loss of the pipe.
4.79
A 3kW pump is used to pump well water that is 5
meters below the ground level to a storage tank 15
meters above the ground. The transporting pipe is
insulated. Due to frictional effects, however, the
water temperature rises by 0.03C during the process.
Determine the mass flow rate of water.
b. Compute the fraction of this capacity that is used
for the phase change in the generator.
4.73
Refrigerant 22 at 10 bars, 50C, flows at 8 kg/s
through a heat exchanger, leaving at 9 bars and
100C. Refrigerant exchanges heat with a stream
of steam entering at 2 bars, 200C.
a. If the steam leaves the exchanger as saturated
vapor at 2 bars, determine the mass flow rate of
steam.
b. For the same inlet and outlet conditions, the
refrigerant flow rate is reduced to 4 kg/s, and the
flow rate of steam remains the same. Determine
the outlet temperature of the steam.
4.74
As shown in Figure 4.89, carbon dioxide at a state
of 7 bars, 525C inlets to a shell-and-tube heat exchanger with a velocity of 150 m/s. The gas has to
be cooled by refrigerant 22 flowing through tubes.
At the outlet of the exchanger, conditions of carbon
dioxide are 3.5 bars, 125C. The inlet and the outlet
states of the refrigerant and other related parameters
are given in the figure. Assume ideal gas behavior
for R22, with c p  0.74 kJ/kgK , and Determine,
Unsteady flow systems
4.80
The weightless piston of
the insulated cylinder in
Figure 4.90 is initially
at the base, and the uncompressed spring is
just touching the piston
surface. Steam flowing
through a pipe at 15 bars,
300C slowly enters the
cylinder as the valve is
opened. The charging
process terminates when
the pressure inside the
cylinder reaches 10 bars
Figure 4.90
Determine the final temperature and the amount of steam in the cylinder if
the final volume is 0.1 m3.
4.81
A storage tank initially contains 2 kg of air at 1
bar, 27C. Atmospheric air at 1 bar, 27C is to be
compressed and supplied into a tank. Assuming
that both the tank and the compressor are adiabatic,
determine the shaft work of the compressor if the
final pressure in the tank is 14 bars.
Figure 4.89
a. the exit velocity of carbon dioxide,
b. the mass flow rate of refrigerant 22.
4.75
Water enters a heat exchanger at 0.5MPa and 140C
at a rate of 24 kg /min and it leaves the exchanger at
0.5MPa and 60C. The water is cooled by passing air
through the heat exchanger at an inlet volume flow
rate of 100 m3/min. The air at the inlet is 110kPa,
25C and the exit pressure is 110kPa. Calculate,
a. the exit temperature of air,
b. the heat transfer rate,
c. the inlet cross sectional area for air flow if the
inlet velocity of air is 25 m/s,
154
THERMODYNAMICS
4.82
A 1.5 m3 of refrigerant storage vessel contains 700
kg of liquid R22 in equilibrium with its vapor that
fills the rest of the vessel. The temperature in the
vessel is 0C. An additional amount of 200 kg of
saturated liquid refrigerant at -5C is pumped into
the vessel. Neglecting the pump work, determine the
heat transfer necessary for the content of the vessel
to remain at its original temperature and pressure.
4.83
Steam is used to move a frictionless lift cabin upwards in a certain application. As shown in Figure
4.91, the steam below the cabin is initially saturated
vapor at 500kPa with a volume of 0.002 m3 and the
steam source is at 1MPa and 300C. After opening
the valve, 0.15 kg of steam from the source slowly
flows in until the volume below the cabin becomes
0.04m3. Determine,
temperature bath at 300C. While the walls of tank
B are insulated, both tanks are identical and each
one has an internal volume of 0.2 m3. The valves of
the tanks are left open until the pressure equilibrium
with the main is obtained.
Figure 4.92
a. Determine the heat transfer of tank A.
b. Compute the mass of steam supplied to each
tank and the mass ratio of tank A to B at the
final state.
4.87
As shown in Figure 4.93, an evacuated tank of 100
L in volume is charged by two mains. First, 1.5 kg
of water at 15MPa, 300C is charged by main A,
and then steam at 5MPa, 600C is admitted into the
tank through main B. The desired final state in the
tank is saturated vapor at 5MPa. Determine,
a. the amount of steam supplied by main B,
b. the heat transfer to the tank.
Figure 4.91
a. the final state of steam,
b. the heat transferred from the cylinder.
4.84
A space capsule with an internal volume of 1.5 m3
is initially evacuated, and is on the surface of Mars.
Owing to small cracks on the capsule, the atmosphere
of Mars slowly penetrates into the capsule. In a few
hours of time duration, temperature and pressure
equilibrium are established between the inside of the
capsule and the Mar’s atmosphere whose pressure is
70kPa. Assuming that the Mar’s atmosphere behaves
as ideal gas, compute the amount and the direction
of heat transfer during the process.
4.85
A rigid and insulated tank is initially evacuated.
Atmospheric air at 1bar, 20C is allowed to leak
into the tank until the pressure reaches 1 bar. a.
Determine the final temperature of air in the tank, b.
Suppose that the tank initially contains air at 0.1bar,
and 20C. Determine the final temperature of air for
this case.
4.86
Steam at 15 bars, 300C flows with a velocity of
100 m/s through a main as shown in Figure 4.92.
Two tanks are connected to this main by pipes. Both
tanks contain steam at 5 bars and 300C. Tank A has
diathermal walls and is surrounded by a constant
Figure 4.93
4.88
A pipeline contains steam flowing at 40 bars with an
unknown temperature T. In determining the value of
T, the following experiment is proceeded. A tank of 1
m3 in volume, initially containing saturated vapor at 1
bar, is attached to the pipe. Opening the valve located
between the tank and the pipe, the steam is allowed to
flow until the tank pressure reaches 20 bars. At this
instant, a thermometer, measuring the inside temperature
of the tank, registers 350C. The heat loss during the
filling process is determined to be 198 kJ. Estimate the
temperature of steam, T, in the line.
4.89
It is desired to pump well water at 10C from 5 meters
below the ground level to a storage tank that is 65
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 155
dangerously brittle at low temperatures, and the drop
of gas temperature to such a low level might result
with a hazardous operation. Therefore, it is desired
to determine the time variation of temperature in the
bottle. A 1 m3 of bottle initially contains nitrogen gas
at 140 bars and 0C. During the blow down period,
it is known tthat the gas pressure in the bottle varies

as, p  2 100 p0 , where t is in seconds. For a negligible heat transfer, estimate the gas temperature after
2 minutes of operation.
meters above the ground and occupies a volume of
50 m3. The tank is leak proof, and initially contains
air at 1 bar, 10C. It is desired to fill half of the tank
with water. For a final temperature of 10C in the
tank, determine the shaft power of the pump.
4.90
A vessel of 0.2 m3 in volume contains nitrogen gas
at 1 bar, 17C. The vessel has a safety valve that
prevents the pressure rising above 1.5 bars. When
the pressure reaches 1.5 bars, the valve opens, and
the excess nitrogen is released. The vessel is heated
until the final temperature of nitrogen becomes 300C.
Determine,
a. the temperature at which the valve opens,
b. the amount of nitrogen released,
c. the heat transfer of the process.
4.91
A 5L pressure cooker has an operating pressure of
180kPa and initially 1/4 of volume is filled with vapor,
and the rest is liquid water. Heat is supplied at a rate
of 500 W to the cooker. If the pressure cooker is not
allowed to run out of liquid, estimate the maximum
time duration of the heating process.
4.92
A 40L bottle of refrigerant 22 at 20 bars, 35C is in an
environment at 1 bar, 35C. A small crack develops
on the bottle, and the refrigerant escapes slowly to
the atmosphere so that the temperature in the bottle is
always at 35C. For a final state of 1 bar, 35C in the
bottle, determine the heat transfer of the process.
4.93
4.94
A tank with a volume of 2 m3 is half filled with liquid R22 and the remainder is filled with vapor at a
pressure of 909kPa. The content of the tank is heated
until one-third of the liquid by mass evaporates. An
automatic valve allows saturated vapor to escape at
such a rate that the pressure remains constant within
the tank. Determine the heat transfer required.
As shown in Figure 4.94, a 1 m3 insulated rigid
vessel has a connection valve and a paddle wheel.
The vessel initially contains air at 27C and 400kPa.
When the valve is opened to allow air to escape, the
paddle wheel begins to turn in such a way that air
temperature remains constant during the process.
As the air pressure drops to 250 kPa, the valve is
closed and the paddle wheel stops. Determine the
work done on air during the process
4.96
An insulated piston-cylinder
apparatus equipped with a
linear spring as shown in
Figure 4.95 initially contains 0.2 m3 of steam at 10
bars, 400C. As the valve at
the bottom of the cylinder
is opened, the steam flows
out, the piston moves down,
and the spring unwinds. At
the final state, the cylinder
contains 0.1 m3 of saturated
vapor at 6 bars. Determine,
Figure 4.95
a. the amount of steam discharged during the
process,
b. the average enthalpy of the steam at the outlet.
4.97
A tank of 0.8 m3 in volume contains 5% liquid water
by volume and the rest is water vapor at 2bars. The
container is heated until the pressure reaches 100
bars. Then, as shown in Figure 4.96, the vapor is
allowed to flow through an adiabatic turbine. During
the process, the heating continues, and the pressure
is maintained at 100 bars. The vapor at the outlet
of the turbine is always at 1 bar and a quality of 80
percent. During the process, it is also known that the
enthalpy of steam at the outlet of the tank linearly
varies with its temperature. Determine,
Figure 4.96
Figure 4.94
4.95
In many industrial installations, occasions arise when
compressed gas bottles are rapidly blown down. These
bottles are constructed of carbon steel that becomes
a. the temperature of water when pressure reaches
100 bars in the tank,
b. the mass of discharged steam for a final temperature of 700C in the tank,
c. the total work output of the turbine,
d. the amount of heat transfer.
156
4.98
THERMODYNAMICS
A storage tank 100 m3 in volume contains steam at 5
bars, 300C. As shown in Figure 4.97, the stored steam
is use to drive a turbine at times of high demand for
electric power. Assume that both the tank and the turbine
are insulated. Due to discharge process, the pressure
of steam drops, and at the final state, it becomes saturated vapor at 2 bars in the tank. Determine the power
delivered by the turbine, if the conditions at the turbine
outlet are 1 bar and 0.98 in quality.
Figure 4.98
4.100 Consider the following process which simulates the
operation of the Newcomen steam engine which
was used during the 18th century for pumping
water from mines in England. As shown in Figure
4.99, the cylinder 50 cm in diameter is fitted with a
frictionless piston and the volume of the cylinder is
0.3 m3 when the piston is against the stop.
Figure 4.97
Miscellaneous systems
4.99
The working fluid of the refrigeration system shown
in Figure 4.98 is R22. The mass flow rate of the
refrigerant through the cycle is 0.015 kg/s, and the
power consumed by the compressor is 1 kW. The
conditions of the refrigerant at several points of the
cycle are given in the table below. Determine,
a. the heat transferred through the compressor,
b. the cooling capacity of the evaporator,
Initially the piston rests against the stop and the
cylinder contains saturated water vapor at 1 bar in
pressure. Liquid water at 1.4 bars, 20C is sprayed
into the cylinder. The resulting condensation creates a
pressure drop. When the pressure inside the cylinder
decreases to 0.7 bars, the atmospheric pressure is
sufficient to overcome the load on the piston, and
the piston begins a downward travel. Assume that
the process is adiabatic, determine the mass of water
that must be sprayed into the cylinder to cause the
piston to travel 1 meter.
c. the mass flow rate of the cooling water,
d. the heat transfer rates of pipes at sections (8-1)
and (2-3),
e. the pipe diameters for sections (1-8) and (2-3)
for a maximum of 5 m/s refrigerant velocity.
1
2
3
4
0.2
1.4
1.4
1.3
0
100
80
30
5
6
7
8
Pressure (MPa)
1.3
0.2
0.2
0.2
Temperature (C)
32
-25
-25
-1
State
9
10
Pressure (MPa)
-
-
10
20
State
Pressure (MPa)
Temperature(C)
State
Temperature (C)
Figure 4.99
4.101 Water 10MPa and 150C enters the boiler of a steam
power plant with low velocity. The volumetric flow
rate entering the boiler is 0.02 m3/s. Exiting the
boiler at 10MPa with low velocity, water enters an
adiabatic turbine and leaves the turbine at 10kPa and
90 % quality with an average velocity of 200 m/s.
The power output of the steam turbine is 20.8MW.
Assuming steady flow process, calculate,
a. the mass flow rate of steam,
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 157
b. the temperature of steam at the turbine inlet,
c. the heat transfer rate in the boiler,
d. the cross-sectional area of the flow at the turbine
exit.
4.102 You have undoubtedly observed that a jet of water
issuing from a faucet necks down. As shown in Figure
4.100, suppose that the flow issuing from the faucet
has a uniform velocity V0, and the faucet diameter at
the exit is d0, determine the diameter of the jet as a
function of the distance downstream of the nozzle.
Figure 4.100
4.103 The system shown in Figure 4.101 is used to produce
liquid refrigerant 22. Refrigerant, entering to a heat
exchanger at a state of 20 bars and 100C, flows
through a throttling valve and inlets to an insulated
tank where liquid and vapor mixture is stored at -5C.
The saturated vapor part is let flowing through the
exchanger and is heated to a temperature of 80C.
Determine the ratio of liquid mass obtained at port
4 to the refrigerant mass supplied at port 1.
Figure 4.102
4.105 To reduce the gas storage costs, two companies A
and B have built a common storage tank as in Figure 4.103. The tank is 3 meters in diameter and 30
meters long. To decide how much gas each company
uses between refills, a thin piston is placed in the
tank. The piston can move freely and the pressure
is always the same on both sides. As company A
uses gas, the piston moves left, and visa-versa for
company B. When the gas company makes a refill,
how much of gas is used by each company has to
be decided. They can measure the position of the
piston, and if necessary can install instruments such
as thermometers, pressure gauges, in either or both
ends of the tank. List the minimum instruments that
can be recommended, and describe by this list how
the amount of gas consumed by both companies can
be determine at the time of refilling.
Note: The gas is assumed to behave as ideal gas.
The piston is adiabatic. The walls of the tank are
well insulated and have very low heat capacity. For
refilling process, the gas company positions the piston
at the center of the tank, equalizes the temperature
at both compartments and meters the total amount
of gas added.
Figure 4.101
4.104 As shown in Figure 4.102, an insulated and perfectly
mixed tank contains a heating coil that adds 6000
kJ/min of heat to water in the tank. Water flows into
and out of the tank at a constant volumetric flow
rate of 0.35m3/min, and is stirred by an agitator that
consumes 4 kW of power. Initially the liquid water
in the tank is at 20C and occupies a volume of 3.5
m3. The temperature of water flowing into the tank
is 65C. Determine,
a. the temperature of water in the tank as a function
of time,
b. the tank temperature as time assumes very large
values.
Figure 4.103
4.106 An insulated test chamber, as shown in Figure 4.104,
has a volume of 20 m3 and initially contains air at a
state of the surroundings (p0 = 1 bar, T0 = 20C). An
electric heater located in the supply channel consumes
power at a rate of 20 kW and is turned on when the
outside air is circulated through the chamber at a
rate of 1 kg/s. A 500 W paddle mixes the content of
the chamber so that air is at the same temperature as
the chamber at the outlet. Determine an expression
for the chamber temperature as a function of time.
158
THERMODYNAMICS
Figure 4.104
4.107 As shown in Figure 4.105, an automobile refrigeration unit absorbs 1kw of power from the shaft
of the engine car, and works with a performance
coefficient of 1.2. The car might be taken to be a
rectangular prism with overall dimensions of 1.2m ×
1.5m × 2.0m (the car cross section is 1.2m × 1.5m),
and cruises with an average speed of 100km/h that
creates a heat transfer coefficient of 80W/m2K
on the outside surface. Initially, the inside of the
car is at thermal equilibrium with surroundings at
T(0) = To = 37°C.
a. Estimate the time needed to drop the inside
temperature to 18°C,
b. What power the refrigeration unit consumes to
keep the inside temperature at 18°C?
e.
The enthalpy change of an ideal gas in
isothermal process is zero.
f.
In a throttling process of an ideal gas,
the pressure is invariable.
g.
The heat absorbed or rejected by a polytropic process is k  n  / k  1 x(work done)
h.
The temperature of an ideal gas decreases
as it flows through an adiabatic nozzle.
i.
Water and Helium having the same
masses are at the same pressure and temperature. Water requires more energy to raise its
temperature.
j.
The internal energy of an ideal gas
depends only on temperature.
k.
The work done by a polytropic process
is  p2v1  p1v1  / n  1 .
l.
The change in energy as well as in work
by a system can be represented by an exact differential.
m.
A water pump operating at a constant
rotational speed is a steady flow machine.
n.
No work is associated with the rotation
of a shaft in a well lubricated journal bearing.
o.
Since the air density changes, due to flow
through a compressor, the flow is considered to
be unsteady.
p.
Insulating a compressor is always a
good idea.
q.
The term m h in open system applications is often neglected.
r.
The specific flow energy is represented
by h  V 2 / 2  gz .
Figure 4.105 Refrigeration unit of an automobile
s.
The nozzle decelerates the flow and the
diffuser accelerates it.
t.
Nitrogen with a mass of 2kg at 1000K
is cooled by withdrawing heat at a rate of 500W
at constant volume. The temperature of Nitrogen
changes at a rate of -335K/s.
u.
Water inlets a boiler pipe with a speed
of 1m/s at a pressure of 1bar. Water is saturated
vapor at the pipe exit. If the pipe diameter is
kept constant, then the water velocity will be
the same.
v.
An initially empty cylinder is filled
with air at 300K. If the cylinder is insulated,
the temperature of air will not change after the
filling process.
w.
It is not possible to compress an ideal
gas isothermally in an adiabatic cylinder.
True and False
4.108 Answer the following questions with T for true and
F for false.
a.
As a result of 1st law, the heat transfer
is equal to internal energy change if the process
is adiabatic.
b.
The internal energy includes the kinetic
and the potential energy.
c.
The work done by a free expansion
process is minimum work.
d.
A process that does not involve heat
transfer is called adiabatic process.
CHAPTER 4 ENERGY ANALYSIS OF SYSTEMS 159
x.
A turbine is an energy conversion machine which first converts the flow energy into
kinetic energy by stationary blades and then to
shaft work by the rotary blades.
y.
A diffuser is an adiabatic device which
decreases kinetic energy but increases the flow
enthalpy.
z.
On a hot summer day, a student turns his
fan on when he leaves his room in the morning.
When he returns in the evening, the room must
be warmer than the neighboring room.
5.
6.
Check Test 4
Choose the correct answer:
1.
2.
3.
4.
Liquid-vapor mixture of H2O is initially at 10bar
with a quality of 90% is contained in a rigid and
well insulated tank. The mass of water is 2kg. The
electric resistance heater in the tank transfers energy
to water at a constant rate of 70W for 1.8h. The final
temperature of water in the tank in 0C is,
a. 215
b. 235,
c. 255,
d. 275.
A 2kg block of ice with an initial temperature of
-250C is dropped into an insulated tank containing
10kg of water initially at 200C. Assuming that the
specific heats of solid ice and water respectively are
2.0kJ7kgK, and 4.18kJ/kgK, and the heat of melting
of ice is 335kJ/kg, the final temperature of tank water
in 0C is,
a. -30C
b. 30C,
c. 00C,
d. -50C.
On a cold winter day, outside temperature is -100C, and
yet, the outside surface temperature of the mechanical
engineering building is -90C. The thickness of the
building wall is 25cm and the thermal conductivity is
0.6W/mK. The convective heat transfer coefficients
on the outer and the inner wall surfaces are 28W/m2K,
and 5W/m2K respectively. The interior temperature
of the building is,
a. 11.2
b.
15.2,
c. 20.2,
d.
25.2.
7.
8.
Refrigerant R134a with a mass of 0.5kg is at 4bar,
500C is contained in a piston-cylinder arrangement.
The refrigerant is cooled under constant volume until
its temperature is -40C, and then is compressed to
be saturated liquid at -40C. The total amount of heat
removed in kJ might be,
a. 114
b.
119,
c. 124,
d.
129.
Air enters a compressor operating at steady state at
1.05bar, 300K with a volumetric flow rate 20m3/min,
and exits at 12bar, 400K. Heat transfer occurs at a rate of
3kW to surroundings. Neglecting kinetic and potential
energy effects, and assuming ideal gas behavior, the
shaft power of the compressor in kW is,
a. 43.8
b.
40.8,
c. 35.8,
d.
45.8.
Air enters a household electric furnace at 250C, 1bar
with a volumetric flow rate of 20m3/min. The furnace
delivers air at 500C, 1bar to a duct system with three
branches consisting of two 15cm diameter ducts and
30cm duct. The air velocity in smaller ducts is 3m/s,
and hot air in large duct is used for laundry drying
and leaves the duct at 270C. Hence the amount of
heat absorbed by the laundry in kW is,
a. 6.4
b. 7.4,
c. 8.4,
d. 9.4.
The exhaust gas from a gas turbine is used to generate steam to drive a steam turbine. The exhaust gas
is approximated as air, and the air stream as shown
in Figure 4.106 exchanges heat with water in the
heat exchanger. The air stream with 5.2kg/s of mass
flow rate enters at a temperature 727°C, and 1bar
pressure, and leaves at 227°C, 1bar. Water at 30bar
of operating pressure enters the exchanger at 20°C
and exits at 440°C. The steam produced expands in
an adiabatic turbine to a pressure of 0.5bar, and at
turbine exit water is saturated vapor. The turbine
power output in kW is,
a. 490
b. 590,
c. 540,
d. 440.
A system consisting of 1kg of water initially saturated
vapor at 2000C undergoes a power cycle composed
of the following processes: (1-2): Constant pressure
heating, (2-3): Constant volume cooling to 5bar, 2000C.
(3-1): Isothermal compression with loosing 57.18kJ
of heat. The efficiency of this cycle in percent is:
a. 25.2
b.
30.2,
c. 15.2,
d.
19.2.
Figure 4.106 Use of exhaust gases for power
generation
160
9.
THERMODYNAMICS
A scuba divers’ tank of 0.6m3 is to be filled with
atmospheric air at 1bar, 27°C by an adiabatic compressor as shown in Figure 4.107. Assume air to be
an ideal gas with c p  1.005 kJ/kgK, and k  1.4 ,
and the tank is initially empty. For an adiabatic tank,
if the final pressure in the tank is 10bar, the amount
of work done by the compressor in kJ is,
a. 310
b. 360,
c. 410,
d. 460.
10.
As shown in Figure 94.108, 6kg of two-phase propane
is contained in a rigid tank initially at 10 bar, and the
quality is 0.5. Heat transfer from a higher temperature
external source (possibly a fire nearby the tank) to
the propane in the tank occurs until temperature has
increased to 120°C. During the process a pressurerelief valve allows propane to escape maintaining
constant pressure in the tank. The total amount of
energy transferred by heat in kJ is,
a. 1446
b. 949,
c. 496,
d. 1246.
Figure 4.107 Filling of a scuba divers tank
Figure 4.108 A typical mass discharge process
C
H
5
A
P
T
E
R
Second Law: Exergy Analysis of Systems
5.1
Introduction
In today’s world of raising the use of technologically complex products, the assessment of
a potential for work production is especially vital for those countries starving for inexpensive
energy in available form. Almost all of our daily activities depend upon the availability of
work energy. The activation of machinery and equipment, pumping liquids, compressing
gas, conveying goods, transforming people are made possible by the transformation of work
energy. In this chapter, we start our discussion with comparing the quality of heat energy
with work energy. Then, we will define the exergy of a system which is the maximum useful
work that could be obtained from the system at a given state and at a specified environment,
and apply the exergy principles to a system undergoing a specific process. We will continue
with determining the loss of exergy of a system through a process, and evaluate the system
performance on the basis of exergetic efficiency of the system.
Work Energy Versus Heat Energy
Work energy is essentially produced in the form of electrical energy. Although some
portion of electrical energy is produced by hydro-electric power plants, and by wind generators, the most of it comes from fossil-fuel consumption or from nuclear reactors.
The method of producing work by consuming fuel is accomplished through a process
called combustion. In combustion, a heat reservoir is generated, and this heat reservoir is
a gaseous medium at high temperature. To design and built equipment, or to improve the
existing equipments that convert the heat energy of such a high temperature reservoir into a
useful work with the least amount of energy loss is one of the prime subjects of mechanical engineering.
The quality of the heat energy produced is important as much as its quantity. In winter,
we usually plug in an electrical heater to warm up or raise the temperature of a room. As
verified by this simple energy conversion device that the complete conversion of the work
energy into heat is possible. However, the reverse process of converting heat into work, as
illustrated in Figure 5.1, some portion of the heat energy has to be lost. A complete con-
161
162
THERMODYNAMICS
version is unachievable and impossible. For this reason, work is classified as the energy of higher
quality respect to heat and such a classification cannot be made solely by the use of first law of thermodynamics.
The maximum portion of heat energy that can be converted into useful work is called exergy. As
explained in the following sections, the higher the temperature of the reservoir that supplies the heat
energy, the higher the possibility of converting more heat energy into work. Hence the thermal energy
stored at high temperatures has higher quality of heat energy than the thermal energy supplied by
reservoirs at low temperatures. For instance, the amount of thermal energy stored in sea water is immense. However, because of its low temperature, its quality is very low and it is frequently useless.
Another important observation about heat energy is that as heat is transferred from a system
to another or transformed into other form, its quality always lessens and changes in descending
order. Some of the factors that might exist in a process that reduce the quality of heat energy are:
1.Mechanical friction, 2.Electrical resistance, 3.Turbulence in a fluid flow, and 4.Transfer of heat
through a temperature difference.
Why the second law?
Second law of thermodynamics provides a basis for measuring the quality of energy and for determining the loss in quality of energy at a particular process. A large number of events exist in nature
that can not be explained solely as a consequence of the first law of thermodynamics. For instance,
we all know from observations that if two objects at different temperatures are brought into contact,
a thermal equilibrium will be attained and temperatures will be equal after a certain time elapsed.
Since an energy interaction between these two systems takes place, in accord with the principle of
energy conservation, heat energy received by one system has to be identical with the energy rejected
by the other. However, this principle does not indicate anything about how the equilibrium will be
established. As long as the energy is conserved, one may easily state that cooling down the cold object
by heating up the hot one is perfectly an acceptable process. Obviously, observations reveal that the
process does not proceed in that direction. Even this very simple event of heat exchange between two
objects indicates the need for principles other than the conservation principle that play an important
role in understanding and analyzing the factors affecting the physical event. The following events
take place only in one direction:
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 163
1. Cooling of a coffee mug. As illustrated in Figure 5.2, two coffee mugs at different temperatures
brought into thermal contact. After a long period of time, both mugs assume an equilibrium temperature (T1<Te<T2). The reverse event, however, cannot be done. It is impossible to bring the two mugs
to their initial temperatures without consuming any work.
2. Two tanks at different pressures. As shown in Figure 5.3, two tanks having the same gas at different pressures are brought to mechanical equilibrium by turning on the connecting valve. Both tanks
assume an equilibrium pressure  p1  pe  p2  . The reverse event cannot be accomplished. It is
impossible to bring the two tanks at their initial pressure without consuming any work.
Each and every process in nature occurs in a particular direction, and second law of thermodynamics secures that there is no any arbitrary path for a process to take place.
Second law of thermodynamics particularly assists engineers in the following subjects:
1. Makes possible to measure the quality of energy.
2. Provides means for defining a criteria for ideal performance of energy conversion systems.
3. Predicts the natural direction of processes.
4. Determines the final equilibrium state of instant processes
Considering its role on the above indicated subjects, second law is not a law of conservation. In
mathematical terms, therefore, it can not be expressed as equality, and there is no single statement
in expressing the law. Differing from conservation principles, several statements that are identical in
essence can be used in describing the law. Presenting these statements without getting certain maturity
164
THERMODYNAMICS
on the subject is like describing an elephant by a person who has never seen and has no idea at all
about an elephant. If that person touches the leg of the elephant, he may think that it is the trunk of a
tree, if he grabs its tail, he might think that it is a sneak, or contacting its body, he may assume that he
is against a wall. Each one of these imaginations is part of the reality and helps identifying the beast.
However, none of them is capable of expressing her totally. This is exactly analogous to providing
different statements for describing the second law as whole. Each statement views the law from a
different point of view but none of them is sufficient to describe the law completely.
5.2
Equilibrium of Systems
Definition: If there is no any possible way to draw work energy out of a system that is composed
of two systems, then these two systems are said to be in mutual equilibrium.
Mutual equilibrium of two systems may take place at several different modes. For instance, if the
temperatures of two systems A and B are equal, then thermal equilibrium is said to be established.
Similarly, if the pressures of these two systems are equal, then the mechanical equilibrium, and if
there is no further chemical reaction within the systems then a chemical equilibrium exists. The
thermodynamic equilibrium of two systems, on the other hand, requires establishment of equilibrium
at all modes like thermal, mechanical, chemical etc. If two systems are not in thermodynamic equilibrium, then a method can be devised to extract work out of the composite system until both systems
are brought into mutual equilibrium.
Principle 12: If two systems are not in mutual equilibrium with each other,
these two systems can be brought into equilibrium by extracting work energy out of the composite system. If the two systems are in mutual equilibrium, then the work that can be taken out of the composite system is zero.
This principle is basically a result of experimental observations and is considered a part of the
second law of thermodynamics. In bringing two systems into mutual equilibrium, there is certainly
more than one method in each of which the resulting work energy output will not be the same.
Example 5.1: Discuss two methods in bringing the pressurized air of a piston-cylinder device as shown in Figure 5.4 to
mutual equilibrium with surroundings.
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 165
Solution:
Method 1: Remove the locking mechanism on the piston. As the enclosed air suddenly expands the piston eventually assumes a new position which will be the equilibrium state of air with the surroundings.
In this method, as described in Figure 5.4, the useful work output of the composite system is zero, W12=0.
Method 2: Instead of sudden pull of the pin, small weights located on the piston are slide to the corresponding shelves
one by one as the piston moves upward. This process continues until the equilibrium with surroundings is established. As
shown in Figure 5.4, since the weights are lifted up to certain distances, certain amount of work has been drawn out of the
composite system, W12>0. Finally, the two subsystems are brought into thermodynamic equilibrium. The amount of useful
work taken out of the composite system is as following:
W12  Mgha  ( M  m) g (hb  ha )  ( M  2m) g (hc  hb )  ( M  3m) g (hd  hc )
As can be concluded by this example, there are several possible processes in bringing two systems
into mutual equilibrium. As far as the maximum amount of the work energy produced, however, there
is only one particular process that is more advantageous with respect to others in bringing the system
into mutual equilibrium with its environment.
5.3
Exergy of a System
In accord with the experimental studies carefully done in energy conversion systems, energy contained in a system can not be converted totally into useful work. Some portion of the system energy
will be discarded as waste energy. Hence it is desirable to have a property that enables us to determine
the useful work potential of a given amount of energy at a specified state of the system.
Example 5.2: Discuss the relation between the useful and the actual work of the expanding gas in a piston-cylinder apparatus as shown in Figure 5.5 and express how useful work can be determined.
Solution:
Due to “pushing” action of the surroundings at a pressure of
p0 by a volume change of V, a portion of the actual work
of the system is unrecoverably lost. Thus,
Wuseful  Wactual  p0 V
On the other hand, the actual work of the system is the
result of the 1st law application as,
Wactual  W12  U 2  U1   Q12
Definition: Exergy of a system at an instant of
time is the maximum possible useful work that
can be extracted from the system as the system
comes into equilibrium with its surroundings. Like
energy, exergy is a property of the system.
Similar to defining the reference state for
energy, the thermodynamic conditions of the
surroundings is taken to be the reference state for
system exergy. In other words, if a system is in mutual equilibrium with the surroundings then useful
work potential of the system is zero, so the exergy of the system is zero. The reverse is also true. If
166
THERMODYNAMICS
the exergy of the composite system “the system and the surroundings” is zero, then the system must
be in equilibrium with its surroundings. The properties of exergy can be stated as following,
1. Exergy defines the maximum amount of useful work that can be transferred by the composite
system at an instant time, t . At a particular state of a system, there is only a single value for the
maximum work. In evaluating this maximum work, it is not essential to know how the system has
assumed its present state at time t . Exergy is a property of the system and is not path depended. It is
exactly like p, T , v, h, u of the system. Exergy depends on the state of the system and on the state
of its environment.
2. Exergy of a system assumes the value of zero when the system is in mutual equilibrium with the
surroundings. Other than this, the exergy of a system is always positive. Negative exergy is meaningless.
3. As shown in Figure 5.6, as the system changes its state from an initial to an intermediate state, its
exergy also changes. However, only some portion of this change can be converted into work. Regardless of how carefully the process proceeds and how professional methods are applied in transforming
energy into work, the useful work will always be less than the change of exergy of the system.
4. The exergy of a system increases the more it deviates from its environment. Whenever energy loses
its quality, exergy is destroyed.
5. The surroundings is assumed to be unaffected by the changes of the system. The surroundings is at
a constant state of pressure po and temperature To . Hence, in estimating the exergy of a system, the
state of the surroundings is taken to be the reference state.
These are the properties of system exergy and care must be exercised as the change for a particular
system is evaluated. The following Example is helpful on this basis.
Example 5.3: As shown in Figure 5.7, air with a velocity of 10m/s flows through a wind generator having a propeller with
a diameter of 12 meters.
a.
Discuss under what conditions the work provided by the wind generator becomes the maximum.
b.
Determine the maximum work.
c.
Evaluate the change in exergy of air.
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 167
Solution:
a.
Assuming that the air stream has the same pressure and temperature as the surroundings, decelerating it to zero velocity
without any loss of energy will yield the maximum work. In
doing this, turbine must be frictionless and free of wake behind
the propeller blades.
b.
To get the maximum work, the air stream is stationary at the
turbine exit, V2 = 0, and the kinetic energy change becomes,
V12  V22 102

 0.05kJ / kg .The mass flow rate of
2
2000
AV1 113.04 x10

 1334 kg / s
air through the turbine is, m 
v1
0.847
ek 
. and,
c.
 ek , W max  66 7kW .
Wmax  m
Air stream, as a system, at an initial state of (p0, T0, V1) is
brought into equilibrium with the surroundings, and the
maximum work obtainable due to such change is equivalent
to exergy change of the system. Then, the exergy change of
the system becomes 66.7kW.
5.4
Exergy Loss of a System
As illustrated by Example 5.3, it is not possible to reduce the air velocity to zero at the turbine exit,
and also not possible to produce a friction free turbine. Therefore, for all real processes undergoing
between a system and its environment, a portion of the system exergy will be lost, and will not be
converted into useful work. In other words, the ability of a system to produce work will be reduced.
In all engineering processes, the following factors may cause a loss of exergy of the system:
a. Electrical resistance
b. Inelastic deformations
c. Viscous fluid flow
d. Friction: solid-to-solid, solid-to-fluid, fluid-to-fluid
e. Shock waves
f.
Damping of vibrating systems
g. Fluid behavior at sudden expansion
h. Fluid flow through a valve or throttling process
i.
Heat transfer by temperature difference, “thermal friction”
j.
Sudden occurrence of chemical reactions
k. Mixing of liquids or gases having different chemical compositions
l.
Process of osmosis
m. Phase existing in another phase
n. Mixing of fluids at different pressures and temperatures
Definition: If one or some of the above indicated factors does not exist in a process occurring between
a system and its environment, then the process is called a reversible process. The exergy loss of a
system undergoing a reversible process is zero.
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THERMODYNAMICS
Since there is no exergy loss for a system undergoing a reversible process, the exergy of a system at
an instant of time “ t ” also signifies the maximum possible work that can be obtained until the system
comes into mutual equilibrium with its environment. The reverse is also true. For a reversible process,
without the need of any additional energy, the amount of work just extracted from the system can be
supplied back, and hence the conditions at the initial state of both the system and the environment can
be restored. Consequently, there will be no traceable change in the system nor in its environment.
Principle 13: For systems undergoing a reversible process, exergy is conserved. Otherwise the total exergy of both the system and its environment will
be lessened. The exergy loss is always positive, and becomes zero at the limit
for a reversible process.
Example 5.4: Analyze the following systems in Figure 5.8 and the related reversible processes.
Solution:
In Figure 5.8a, assume that the piston-cylinder device is adiabatic and the gas in the cylinder expands quasi-steadily by the
motion of a frictionless piston. Hence, the amount of work done by the gas is  pdV . This energy is stored as rotational
kinetic energy on the fly wheel. In the reverse process, due to rotation of the fly wheel, the piston is moved slowly forward.
The gas is brought to its initial state. Eventually, the fly wheel stops. Without any need of additional energy or the need
of any external interference, both the gas and its environment (the fly-wheel) are brought to their original states. Thus, the
process is reversible.
In Figure 5.8b, the flow in a nozzle is adiabatic and frictionless. As the enthalpy of the flowing fluid decreases, the fluid
particles are accelerated, and the kinetic energy of the fluid increases. At the diffuser section, however, the frictionless fluid
is slowly decelerated, due to increase in pressure, the fluid enthalpy increases. At the exit of the diffuser, the state of the
fluid is identical with the state at the nozzle inlet. Since the nozzle-diffuser system is perfectly insulated, the flowing fluid
does not undergo any energy interaction with its environment, and the environment is always at its initial state. Hence, the
process of frictionless flow of fluid through a converging-diverging nozzle is an Example for reversible process.
As shown in Figure 5.8c, the motion of a free pendulum in a frictionless medium is an Example for reversible process.
Since there is no frictional loss, the energy of the pendulum transforms between kinetic and potential without any reduction
in the amount.
As presented in Figure 5.8d, the slow extension or contraction of an ideal elastic spring is again a reversible processes.
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 169
Example 5.5: One of the factors that leads to the exergy loss of a system is heat transfer at a finite temperature difference.
As shown in Figure 5.9, which is initially at a state of (p1, T1) where p1>p0 and T1>T0. Explain two different methods to
bring the system into mutual equilibrium with its environment at (p0, T0) and show that heat transfer at a finite temperature
difference causes exergy loss.
Solution:
Method 1: First expand the gas by reversible and adiabatic process until the final temperature is at surroundings, T2=T0.
Then, compress the gas by reversible and isothermal process so that the final pressure becomes the pressure of surroundings. In compression, heat is transferred to surroundings to keep the temperature constant. Thus the system is brought from
state (p1, T1) to state (p0, T0) completely by a reversible manner. The shaded region in Figure 5.9 represents the maximum
amount of work that can be extracted from the system by changing from the initial state (p1, T1) to the state of the surroundings, and hence it is the exergy of the system at state 1. As explained by method 2, by a simple heat transfer, this amount
of work is totally lost.
Method 2: The gas in the piston-cylinder device can be brought into the state of surroundings by cooling it down to the surroundings temperature at constant volume. The amount of work output of the system by this process, however, will be zero.
The exergy level of the system by constant-volume cooling process is decreased, but no transfer of work is accomplished.
The total exergy of both the system and its environment is lessened.
Considering the results of the above explained methods of Example 5.5, one may conclude that
the transfer of heat at a finite temperature difference causes a decrease in the system exergy.
Principle 14: Heat transfer at a finite temperature difference between
two systems causes a loss of exergy of the combined system. The exergy
loss proportionally lessens with the decrease of the temperature difference between two systems. The decrease in temperature difference
lessens the thermal friction.
Example 5.6: Another factor that causes a loss in exergy of a system is “sudden expansion”. As illustrated in Figure 5.10,
the gas contained in a piston-cylinder device has to be brought into equilibrium with its environment. The gas is assumed
to expand in a frictionless medium. Explain two methods in bringing the gas from its initial state (p1, T0), where p1 > p0, to
the state of the surroundings at (p0, T0) and prove that sudden expansion causes a loss of exergy.
170
THERMODYNAMICS
Solution:
Method 1: As shown in Figure 5.10a, after releasing the locking mechanism of the piston, the gas expands without any
frictional loss, the piston oscillates around the state (0) and eventually comes to a stop at (p0, T0). Thus the amount of work
that can be extracted from this system is, p0 (V0  V1 )
Method 2: Instead of using a locking mechanism, suppose that small weights are located on the top of the piston, and the
frictionless piston moves slowly as the weights are laterally slide one by one. Since the piston-cylinder device is not insulated, the gas expands reversibly and isothermally. The amount of work extracted from the system is, p1V1ln( p1 / p0 ) , and is
exactly identical to the exergy change of the system. The difference in the amount of work of these two methods is indicated
by the shaded region in Figure 5.10b (the area (1-0-a-1)) and corresponds to exergy loss of sudden expansion.
Comparing the two processes of Figure 5.10, a sudden expansion process causes a loss of exergy
of the system, and the amount of exergy loss is indicated by the shaded region in p-v plot of Figure
5.10b.
Principle 15: All processes involving a sudden pressure drop or a sudden expansion result in an exergy loss. Decrease in exergy loss is directly proportional
with the decrease of pressure difference between the two states.
Example 5.7: As shown in Figure 5.11, mass A, on an inclined plane, has to be brought from state (1) to state (2). Show
that as an outcome of the frictional motion the system losses exergy.
Solution:
As shown in Figure 5.11a, on a frictionless surface, mass A can be brought to state (2) by the required minimum mass of
mB1+mB2. For a surface with friction, however, an extra mass of mB3 is required. Since the frictional effects are opposite to
motion, in restoring the mass A to its initial state, it is not enough to release mass B3. Both masses B2, and B3 have to be left
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 171
on the reference plane. As shown in Figure 5.11b, when the mass A comes back to its initial state, the work potential of the
system composed of masses A and B is reduced.
As a consequence of the above experiment, due to friction, the work potential of systems are always reduced. Initially the work potential of the system in Figure 5.11 is mB  mB  mB gL , however,
after a frictional process restoring the system into its initial state, the work potential of the composite
system becomes . Thus, due to friction, the system has undergone an exergy loss process.
1
2
3
Principle 16: Friction causes a loss of exergy. The more the
friction at a process is lessened, the less the exergy loss due
to friction occurs.
Definition: The amount of exergy loss by an irreversible process measures the irreversibility of that
process. The irreversibility of a reversible process is zero. No exergy loss takes place by a reversible
process.
The irreversibility that may take place at a particular process may be gathered in the following
two groups,
a. The external irreversibility
Such irreversibility occurs when a system interacts with its environment at a pressure, or temperature difference, or friction exists during that interaction. Thus, such processes of the system are
called externally irreversible.
b. The internal irreversibility
Such irreversibility takes place when there is heterogeneous distribution of properties within the
system. Occurrence of local temperature, velocity, and pressure gradients within the system results
in sudden processes. Such sudden processes, on the other hand, yield a loss of exergy. For instance,
consider the flow of a viscous fluid through a channel, due to fluid viscosity, particles flow at different
velocities, and the rubbing action between the fluid particles results in heating effect. As explained
previously, heat transfer indicates a loss of exergy of the system. Hence viscous flow is an internally
irreversible process.
Example 5.8: As shown in Figure 5.12, consider the use of an oil filled column heater for heating up a room. The device
consists of electric heating elements immersed in oil, which heat their outer casing. Determine if the heating process is, a.
externally irreversible, b. internally irreversible, or c. totally irreversible process.
172
THERMODYNAMICS
Solution:
a.
The surface of the heater is hotter than
the surroundings, Ts  To , and considering principle 14, the process is externally
irreversible.
b.
Due to resistivity of the wires, the flow
of current through the heating element
is done by a voltage difference. In fact,
the rubbing action between the moving
electrons causes heating. Hence the process is also internally irreversible.
c.
A process which is both internally, and
externally irreversible is called totally
irreversible, or simply irreversible.
5.5
Equation of Exergy
If a system transfers energy across
its boundaries, and if some portion
of that energy can be converted into
useful work, then the system is said to be capable of transferring exergy. Thus, the energy and the
exergy are two non-separable concepts of a system, and without the transfer of energy, transfer of
exergy is unthinkable and cannot be accomplished.
Principle 17: As a result of energy transfer, a system transfers a certain
amount of exergy. The reverse is also true. The transfer of exergy indicates that an energy transfer takes place.
Since the system exergy is single valued at a particular state of a system and is the property of the
system, like the other properties, the rate of exergy change of a system may be stated as following,
Time rate of


exergy accumulation 


 within a system  =


at an instant


of time t


 The net rate   The rate

 of exergy
 of irreversibility 

 

 transported into    production 
 the system  at an instant of 

 

at time t  
time t


(5.1)
As explained in section (4.3), transfer of energy from one system to another can be done in three
different modes. Similarly, the transfer of exergy through a system can be carried out in three different modes.
a. The exergy transfer by work
Such an exergy transfer takes place when a property difference other than temperature exists
between two systems.
b. The exergy transfer by heat
The exergy transfer due to transfer of heat energy occurs when there is a temperature difference
between two systems.
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 173
c. The exergy transfer by convection
Such an exergy transfer is provided by the flow of a fluid into and out of the system. The net rate
of exergy accumulation at the system due to transfer of energies by work, heat and by convection are
as following,
The net rate of exergy 


 transfer by work
  W
at instant of time t



(5.2)
The net rate of exergy 


 transfer by heat
   Q
at instant of time t



(5.3)
The net rate of exergy



accumulation by convection  
at instant of time t



 m    m 
i
i
i
e
(5.4)
e
e
Consequently, the net rate of exergy transported into the system becomes,
The net rate of exergy



accumulation in the system  
at instant of time t



 m    m 
i
i
i
e
e
  Q  W
(5.5)
e
Let us represent the rate of irreversibility produced by a real process at an instant of time as İ. As
all we know the value of İ is zero for reversible processes and is always positive for irreversible ones,
I  0 . Substituting Eq. (5.5) into (5.1) and rearranging yields,
 

 m    m 
i
i
i
e
e
  Q  W  I
(5.6)
e
 , represents the rate of exergy accumulation in the system at an instant of time t. As you
where, 
may also note that the rate of exergy destroyed at a particular process,  destroyed , is identical to the rate
of irreversibility produced by that process,
 destroyed  I
(5.7)
In order to make use of equation (5.6) in analyzing engineering systems, one has to explore how
the terms like, W ,  Q and  i depend on the system properties. Figure (5.13) illustrates schematically exergy balance of a system interacting with other systems at all modes of energy transfer.
5.6
The Exergy Transfer by Work
The transfer of exergy by a difference in a property other than the temperature is called exergy
transfer by work. If the work transfer is due to motion of system boundaries, the useful portion of
such a work transfer is obtained by subtracting the work done against the environment. In general,
the rate of exergy transfer by work is formulated as following,
174
THERMODYNAMICS
dV
W  W  p0
dt
(5.8)
Example 5.9: A rigid and insulated tank as in Figure 5.14 contains 1kg
of air at a pressure of 1bar and temperature of 27C. Work is done against
the air by running the fan and the final temperature becomes 527C.
The increase in exergy of air is determined to be 100kJ. Calculate the
exergy loss by this process, and give reasons for such losses.
Solution:
Since the tank is adiabatic, the energy equation yields, U 2  U 1  W12 ,
for ideal gas behavior, U 2  U 1  mcv ( T2  T1 ) , and, W12  358 kJ .
Since the tank is rigid, no transfer of work takes place by volume
change. Respect to Eq. (5.8), for a reversible process, the increase of
the exergy of the system would be 358kJ.
For this particular problem, however, the exergy change of the
system is given as   100 kJ , and the exergy transfer by work is W  358 kJ. Substituting into Eq. (5.6) yields,
100  358  I12
Thus the irreversibility of the process becomes, I12  258 kJ.
The loss of exergy is basically due to the frictional effects between the following surfaces:
a.
The fan blades and the fluid
b.
The tank surface and the fluid
c.
Fluid-to-fluid
d.
The shaft surface and its housing
e.
The type of bearing used in the housing
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 175
As can be deduced from this example, contrary to energy conservation, exergy is not conserved. Due
to friction, certain amount of exergy is lost during the transfer of work from one system to another.
5.7
The Exergy Transfer by Heat
Heat Engine - In converting heat energy into useful work (exergy), a heat engine running between
the heat source and the environment is required. To determine the maximum useful work (exergy) the
heat engine must run in reversible manner.
Definition: A heat engine is a cyclic device which operates between two thermal reservoirs, and
produces positive work after completion of a cycle.
Let Q be the amount of heat energy received from high temperature reservoir. The maximum
portion of this heat energy that can be converted into work identifies the exergy of Q. Since we use a
cyclic machine in this conversion, there would not be any exergy loss at all processes of the cycle, if
and only if the heat engine undergoes a reversible cycle. Thus, each and every process of a reversible
cycle must be reversible. As explained in section (4.13), Carnot cycle is an Example for a reversible
cycle.
Definition: A cycle is called a reversible cycle if each and every process of the cycle is reversible. If one of
the processes is irreversible that cycle is called irreversible cycle. Carnot cycle is a reversible cycle.
Carnot cycle. Named after a French engineer and army officer, Nicolas Sadi Carnot, he described
his cycle in 1820 as explained in the following example.
Example 5.10: Assuming an ideal gas behavior for the working fluid, determine the cyclic efficiency of a heat engine
undergoing a reversible Carnot cycle.
Solution:
As explained previously, Carnot cycle contains the following processes:
(1-2) Isothermal process at temperature T1, heat is transferred reversibly from a heat reservoir at temperature T1 to working
fluid at the same temperature. The amount of heat transferred is:
Q1  mRTln
V2
V1
(5.9)
(3-4) Isothermal process at temperature T0, heat is transferred
reversibly from the working fluid at T0 to an environment at the same
temperature. The amount of heat transferred is:
Q0  mRT0ln
V3
V4
(5.10)
The two other processes of Carnot cycle are reversible and adiabatic
processes, and the 1st law states that
dU   pdV , or cv
For the process of (2-3)
dT
dV
 R
V
T
cv ln
(5.11)
T0
V
  Rln 3
T1
V2
(5.12)
176
THERMODYNAMICS
cv ln
For the process of (4-1)
Combining Eqs. (5.12) and (5.13) yields,
T1
V
  Rln 1
T0
V4
(5.13)
V3 V2

V4 V1
(5.14)
Dividing Eq. (5.9) by (5.10) side by side and considering the result of Eq. (5.14), one may get the following relation
between the amount of heat transfer and the temperatures of the reservoirs as,
Q0 T0

Q1 T1
or
Q1 Q0

T1 T0
(5.15)
Thus, for Carnot cycle, the ratio of the heat transfers is equal to the ratio of absolute temperatures of the respective reservoirs. Considering the definition of the cyclic efficiency (Eq. (4.49)), then the efficiency of the Carnot cycle becomes,
Valid only for a reversible cycle
 1
Q0
T
1 0
Q1
T1
(5.16)
In accord with this result, one may state that the efficiency of any reversible cycle is independent
from the properties of the working fluid. All the reversible engines working between the same two
heat reservoirs yield the same efficiency. The efficiency of a reversible heat engine is unique. The
thermal efficiency of an actual heat engine, however, is much less than the value to be determined by
Eq. (5.16) and varies in the range between 10 percent and 40 percent.
Principle 18: (i) The thermal efficiency of an irreversible heat engine is always less than the efficiency of a reversible engine operating between the same heat reservoirs. (ii) All reversible heat
engines, irrespective of their cyclic differences, provide the same efficiency, and its value is the
maximum under giving conditions. (iii) The efficiency of a reversible heat engine is unique.
Respect to Eq. (5.16), the net work output of the Carnot cycle becomes,
Wrev  Q(1 
T0
)
T
(5.17)
A refrigeration system. This cycle is the reverse of heat engine cycle. The overall effect of the cycle
is to transfer heat from a low temperature medium to a high temperature by consuming work and finds
a wide application in our daily activities.
Definition: A cycle that transfers heat from low temperature medium to high temperature one by
consuming work is called refrigeration cycle.
Principle 19: (i) A refrigeration cycle is the reverse cycle of a heat engine. (ii) When
operated on the reverse order, a reversible heat engine becomes a refrigeration
cycle.(iii) The coefficient of performance (βr) of a refrigeration system is always less
than the reversible one operating between the same heat reservoirs. Hence a reversible refrigeration cycle consumes the minimum amount of work for the same heat
transfer rate.
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 177
In terms of heat interactions, this principle states that no difference exists between a reversible heat
engine and a reversible refrigeration cycle. Thus, Eq. (5.15) is still valid for the reversible refrigeration. Substituting Eq. (5.15) into (4.50) and (4.51) yields the performance coefficients for a reversible
refrigeration and a reversible heat pump respectively as,
r 
1
T
1
T0
 hp
T 
 
T
  0
T

  1
 T0

 hp   r  1
(5.18)
Similar to reversible heat engine, the coefficient of performance COP  of a reversible refrigeration cycle is independent from the properties of the working fluid. Equation (5.18) provides the
maximum value of the coefficient of performance of a refrigeration system operating between specified
low and high temperature reservoirs. In Figure 5.16, again Carnot cycle is exemplified as a reversible
refrigeration cycle and the cycle components are illustrated.
Example 5.11: As shown in Figure 5.17, a reversible heat engine running between the thermal reservoirs at 227C and
27C supplies work to a reversible refrigeration system that cools down a space at -50C and rejects heat to an environment
 / Q .
at 27C. Evaluate Q
1
2
Solution:
 T 
In regard to Eq. (5.16), the work output of the reversible heat engine is, Whe  Q1  1  0  , and the amount of work consumed
T1 

T0
1

 T 
Q
T
by the reversible refrigeration is, Whe  Q1  1  0  . Since, W he  W r , The heat ratio becomes, 1  2
T
T1 
Q 2

1 0
T1
Substituting the temperature values as, T1 = 500K, T0 = 300K, T2 = 223K, into above relation yields,
Q 1
 0.862 .
Q 2
178
THERMODYNAMICS
Example 5.12: As shown in Figure 5.18, a reversible heat pump is to be used for heating a farm house. To keep the inside
temperature at 22C, the required heat supply for an outside temperature of 2C is determined to be 12x104 kJ/h. The compressor of the heat pump is rated as 10 kW.
a.
Calculate the number of hours that the heat pump works on one day period.
b.
Instead, the heating of the farm house could be done by electric heaters. Determine, on daily basis, the useful energy
conserved by using the heat pump
Solution:
a.
Q 
For this particular case, the high and the low temperature
reservoirs are the farm house (T = 295K), and the surroundings (T0 = 275K) respectively. The amount of heat
supplied by the reversible heat pump is,
W hp
, Q  10  147.5 kW . The heat loss of the
275
T0
1
1
295
T
120000
house is, Q h 
 33.3 kW
3600
The number of hours that the pump will be on is,
Q
33.3
t hp  h x 24 , t hp 
x 24 ,
147.5
Q
b.
t hp  5.418 hours
The electrical energy consumed by the heat pump on daily
 ) , E  10 x5.418  54.18 kWh
basis is, E hp  (Wt
hp
hp
The electrical energy daily consumed by the heater is, Eh  33.3  24  799.2 kWh
Instead of heating by electrical heaters, the use of heat pump saves up to (799.2 - 54.18) = 745.02 kWh of electrical energy.
Example 5.13: Heating of a convention center in winter has to be done by warm air heated by a heat pump. The total heat
loss of the center to outside air is 200kW. As shown in Figure 5.19, the indoors air is circulated through a heat exchanger.
The atmospheric air is at 100kPa and -15C of temperature.
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 179
Determine,
a.
The amount of heat supplied by the heat pump,
b.
The mass flow of air flowing through the heat exchanger,
c.
The power consumption of the heat pump,
d.
The coefficient of performance of the heat pump.
Solution:
a.
The amount of heat supplied by the heat pump to warm
air equals to the amount lost by the center, Q = 200kW.
b.
In terms of the inlet and the outlet enthalpies, the heat
 a ( hi  he )  mc
 p ( Ti  Te ) ,
supplied to warm air is, Q  m
200
 19.9 kg / s
1.005( 30  20)
The amount of heat extracted from the environment is,
a 
m
c.
 Q 0  Q
dT 
or  Q 0  m a c pT0

Q0  m a c pT0lnT
T0
T
T
Q 0  17316
. kW . Then the power consumption of the heat pump is, W hp  Q  Q 0 , W hp  200  17316
.  26.84 kW
This result also represents the minimum work required to accomplish the heating task.
d.
The coefficient of performance of the pump is,
200
Q
hp  
hp 
 7.45 . This value of the coefficient of performance is the highest attainable under given condi26.84
Whp
tions.
Example 5.14: A heat pump is used for heating a greenhouse in winter and cooling in summer. Inside temperature of the
greenhouse has to be kept constant at 20C throughout the year. Heat leaks through the outside walls and ceiling of the
house. The heat leaking rate is 2000 kilo Joules per hour per 1C temperature difference (2000kJ/hC).
For a period of one day, in winter, the temperature of the environment varies as, T ( K )  273  10 Sin   t  , calculate daily
0
 
energy consumption of the heat pump.
 12 
a.
Calculate the average coefficient of performance on daily basis.
b.
By switching to cooling mode, the same device with the same amount of work input at the compressor is used for cooling purposes in summer. Determine the maximum value of the outside temperature.
c.
Calculate the average coefficient of performance of refrigeration on daily basis.
Solution:
The temperature difference between inside and outside is T  Ti  T0 , or T  20  10Sint / 12 and the rate of heat
leaking from the green house is Q  2000 x  20  10Sin  t / 12  . Then, for a reversible heat pump, the rate of heat taken from
T
2000
the surroundings ( Q 0 ) is Q 0  0 Q , Q 0 
x 20  10 Sin( t / 12)x 20  10 Sin( t / 12) . For a period of 24 hours,
Ti
293
the total amount of heat supplied to the greenhouse and taken from the surroundings respectively are,
24
Q

24
  960000 kJ
Qdt
Q0 
0
 Q dt  886279 kJ
0
0
Daily energy consumption of the heat pump becomes Whp  Q  Q0 ,
b.
Daily average value of the performance coefficient is, hp 
Q
,
Whp
Whp  960000  886279  73721 kJ
hp 
960000
 13
73721
180
c.
THERMODYNAMICS
For the case of running the same system in refrigeration mode, the amount of heat leaking per day is,
Q  48000(T0  293) kJ/day and energy consumption of the compressor is still the same Wr = 73721 kJ/day. Then the
amount of heat rejected to surroundings becomes, Q0 = 48000(T0 – 293) + 73721 kJ/day. Since the system performs reversible cycles,
d.
48000( T0  293 48000( T0  293)  73721

, the solution of this equation yields, T0 = 314K.
293
T0
For surroundings at T0 = 314K, and the greenhouse at T = 293K, the amount of heat to be extracted from the house per
day is Q  48000(314  293)  1008000 kj/day kj/day, and Wr =73721 kJ/day. Then, in refrigeration mode, the daily
1008000
average coefficient of performance is,   Q
r 
r  13.6
Wr
73721
The Example problems given above are illustrative in the sense that explains how much of heat energy at the upper most limit can be converted into useful work or how a certain amount of work can
be utilized in the most advantageous way so that the maximum heat energy can be transferred from
a low temperature medium to a higher one.
An actual heat engine is less efficient than a reversible one operating between the same two thermal
energy reservoirs. Similarly an actual refrigerator or a heat pump has a lower coefficient of performance
than a reversible one operated at the same conditions. As stated by Eq. (5.15), for a reversible heat
engine, Qo , is the minimum amount of heat that is rejected to the low temperature reservoir. Since
T1 and To are the same for an actual cycle, and (Q0 ) actual  (Q0 ) rev , then Eq. (5.15) becomes,
 Q1 Q0 
 

 T1 T0  actual
(5.19)
This relation is further studied in Chapter 6 for cyclic processes.
Providing the maximum of work that can be extracted from heat energy of Q, equation (5.17)
specifies the exergy transfer by heat energy of Q from a reservoir at a temperature of T . Hence, in
determining the exergy content of heat energy, a reversible heat engine has to be run between the
reservoir supplying the heat and the environment. Then the rate of exergy becomes,
T

Q  Q  1  0 

T
(5.20)
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 181
As indicated by the diagram in Figure 5.20, in order to convert the maximum possible portion of
heat energy into work, it must be processed through a reversible system.
For the case of simultaneous transfer of heat from isothermal surfaces at different temperatures,
the total exergy transferred to the system becomes,
n 
T
 Q    1  0  Q j
Tj 
j 1 
(5.21)
Where T j is the temperature of j th surface transferring heat at a rate of .
Example 5.15: As shown in Figure (5.21), a laterally insulated and cylindrical concrete rod has one end at 227C and the other
at 77C. The rate of heat conduction through the rod is 500 kJ/h. For surroundings at a temperature of 27C, determine,
a.
the exergy transfer by heat conduction,
b.
the irreversibility of the process.
Figure 5.21 Heat conduction through a concrete rod
Solution:
a.
Taking the rod as a system, the transfer of heat occurs at surface of 1 and 2 respectively. With respect to Eq. (5.19)
then,
 T 
 T 
 300 
 300  ,
Q  Q1  1  0   Q 2  1  0  . Where Q 2   Q 1  500kj / h . Substitution yields, Q  500 1 

  500 1 


 350 
T1 
500

 T2 
 Q  128.58kj / h
b.
Since the concrete rod is at steady state, there will be no accumulation of exergy in the system. There is no exergy transfer
to or from the system other than the transfer of exergy by heat. The exergy equation (Eq. (5.6)) for this particular case
becomes. Then the irreversibility rate of the process is,
I    128.58 kj/h
Q
Even though the rate of exergy supplied to the concrete rod is 200 kJ/h, the exergy available at the
exit of the rod is 71.42 kJ/h. Due to the temperature difference between the two ends of the rod, an
exergy loss takes place. If there was no thermal friction, both ends would be at the same temperature
for transferring the same amount of heat. Hence, there wouldn’t be any loss of exergy.
Example 5.16: A blender is on when the blades are in a fluid at 97C and transfers work to the fluid at a rate of 360kj/h.
As shown in Figure 5.22, the bowl containing the fluid is laterally insulated and transfers heat to surroundings at 27C from
the bottom surface at a rate of 360 kJ/h. Calculate,
a.
the rate of net energy supplied to the system,
b.
the rate of exergy accumulation of the system,
c.
the rate of irreversibility of this process.
182
THERMODYNAMICS
Solution:
a.
The rate of work and heat transfers respectively are
W  360 kj/h , Q  360 kj/h . Considering the energy bal
ance of the system, E  Q  W , E  360  ( 360) ,
The system is at steady state, its energy is not altered.
A system with constant energy cannot have any exergy accumulation.
 = 0.
Hence, 
W  360 kj/h
The exergy transfer by work is, W  W
The exergy transfer by heat is,
 T 
 300 
Q  Q  1  0   360 1 
. kj / h . Respect to the
  68108
 370 

T
.
 360  I and the rate
equation of exergy balance (Eq. (5.6)), 0  68108
of irreversibility of the process becomes, I  291.89 kj/h
The system receives exergy at a rate of 360 kJ/h and rejects exergy by
heat transfer at a rate of 68.108 kJ/h. Since the difference is not accumulated
within the system, it must be an exergy loss. Such a loss occurs due to heat transfer at a finite temperature difference.
5.8
Entropy
In previous sections, we have studied the exergy balance of a system for a particular process, and
discussed the factors that might cause an exergy loss. In this section, a system property, which is an
important parameter in evaluating the exergy of a system at an instant of time t , is introduced.
As shown in Fig. (5.23), a piston-cylinder device placed in an environment at  po , To  contains
gas at an initial state of  p1 , T1  and reaches to a final state of  p2 , T2  by a reversible process. In
accord with Eq. (5.6), the rate of exergy change of the system becomes,
    

Q
W
or
  1  T0  Q  W  p V 

0

T 

(5.22)
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 183
The energy balance of the system requires that Q  E  W
and
 = pV
 . Substitution of
W
these relations into Eq. (5.22) and rearranging results as,
T
d  E  p0V    dE
dV
p


dt 
T0
dt
 dt
(5.23)
Since the right hand side of Eq. (5.23) is completely related to the system properties (energy,
pressure, and volume), the left hand side must also be a property of the system. After all, E, V, and
 are all properties of the system and (p0, T0) are constants of the environment. Hence the combina( E  p0V   )
tion of these properties in the form of
must also be a property of the system. Such a
T0
definition of property is called the entropy of the system. Thus the entropy of a system at an instant
of time t is expressed as,
S
E  p0V  
T0
(5.24)
For a stationary system, the energy of the system, E, equals to the internal energy . Modifying
Eq. (5.24) accordingly and substituting into Eq. (5.23) yields,
dS dU
dV

p
dt
dt
dt
T
(5.25a)
or
TdS  dU  pdV
(5.25b)
This is called fundamental equation of thermodynamics or the first TdS equation, and evaluates the rate of change of the system entropy with respect to other properties like u, p, v, and T of
the system. Since the enthalpy of the system is defined as H = U + pV, then the derivative of internal
dU dH
dV
dp
. Thus, Eq. (5.25a) can be modified
energy with respect to time becomes,

p
V
dt
dt
dt
dt
accordingly to yield,
T
dS dH
dp

V
dt
dt
dt
(5.26a)
or
TdS  dH  Vdp
(5.26b)
This is the second TdS equation and relates the entropy to H , V , p, and T of the system. Even,
though both equations (5.25) and (5.26) are derived for reversible processes, they are equally applicable to irreversible processes with the same end states. A close examination of both expressions
reveals that they both contain terms which are properties, and computing the change of a property
is not related to the type of the process, depends only on the end states. Thus, both equations can be
used in evaluating the entropy change at a particular process.
184
THERMODYNAMICS
5.9
The Entropy Change of a System
After giving the definition of new property “entropy” of a system, evaluation of its change respect
to other properties has to be described. If there is no change of phase within the system, the integration
of equations. (5.25b) and (5.26b) between initial and final states of a process results as,
dU

1 T
2 dH
S 2  S1 

1 T
S 2  S1 


2
p

dV 

T

2V
dp 

1 T


2
1
(5.27)
Entropy change of solids and liquids. The following assumptions are usually made in evaluating
the entropy change of solids and liquids:
a. The internal energy and the enthalpy of solids and liquids are usually strong function of temperature. The effect of the pressure or the volume change can be neglected.
b. For liquids and especially for solids, specific volume at a particular state assumes such a
small value that the “pv” term in enthalpy formulation can be neglected; h~u. In addition, the
change in volume due to change in temperature is so small that such a change can be neglected;
dv~0.
In accord with the above stated assumptions, Eq. (5.27) can be simplified as,
s2  s1 

2
1
c
dT
T
(5.28)
Instead of temperature variation of the specific heat , defining an average value for a particular
liquid or solid transforms Eq. (5.28) into,
s2  s1  c0 ln
T2
T1
(5.29)
Entropy change of ideal gases. Recalling that the internal energy and the enthalpy change of an
ideal gas respectively are du = cvdT, dh = cpdT. Together with the equation of state, substituting these
relations into Eq. (5.27) and rearranging yields,
T2
v 
 Rln 2 
T1
v1 

T
p
s2  s1  c p 0ln 2  Rln 2 
T1
p1 
s2  s1  cv 0ln
(5.30)
The above set of equations reveals that s = f1(T, v) or s = f2(p, v). Entropy can also be described
in the following form of s = f3(p, v) by substituting (p2/p1)(v2/v1) for T2/T1 in Eq. (5.30).
s2  s1  c p 0ln
v2
p
 cv 0ln 2
v1
p1
(5.31)
Entropy change of a system with two-phase in equilibrium. For saturated states, the system pressure and the temperature are not independent properties. If the temperature stays constant then the
pressure is constant and Eq. (5.27) reduces to
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 185
hg  h f
sg  s f 
s  sf 
or
T
h  hf
(5.32)
T
In addition, substituting the enthalpy of a saturated state as h = hf + xhfg into Eq. (5.32) results as,
s  s f  x( s g  s f )
(5.33)
This relation can be used in evaluating the entropy of liquid-vapor or solid-liquid mixtures.
5.10
The General Equation of Exergy
Considering the definition of entropy, the time rate of exergy change of a system at a particular
time t may be expressed as,
  E  p V  T S

0
0
(5.34)
This result can also be used in evaluating the exergy transfer by pure convection.
The rate of exergy



 transfer by convection   m i ui  u0   p0 vi  v0   T0 si  s0   kei  pei 
at time t



(5.35)
 i i , however, the
In evaluating the rate of exergy transferred into the system by a fluid stream, m
exergy transfer by work has to be taken into account. Pressing the matter at the inlet by a pressure
difference of  pi  p0  requires the following amount of work to be done on the system,
The rate of exergy 


 transfer by work   m i  pi vi  p0 vi 
at time t



(5.36)
The total rate of convective exergy transfer into the system is the sum of the exergy transferred by
pure convection, Eq. (5.35), and the work of compression, Eq. (5.36). Hence, the convective exergy
transfer of a fluid stream can be formulated as,
 m   m h  h   T s  s   ke  pe 
i
i
i
i
0
0
i
0
i
(5.37)
i
Similarly, the rate of exergy transferred by the system in convective mode to a fluid stream is,
 m   m h
e
e
e
e
 h0   T0 se  s0   kee  pee 
(5.38)
Together with the exergy transfer by heat and by work Eq. (5.6) can be expressed in the following general form,
 E  p0V  T0 S  
cv
 m h  h   T s  s   ke  pe    m h
i
n
+
i

0
i
0
i
T0  
Q j  W  p0V  I
j 
 1  T
j 1 
0
i
e
e
 h0   T0 se  s0   kee  pee  
(5.39)
186
THERMODYNAMICS
In addition to the measures described in section (4.10), the following steps have to be considered
in exergy analysis of systems:
1. For closed systems, mass analysis is not a factor. However, since the exergy equation contains
terms like, Q j , and W , then the exergy and the energy equations have to be solved simultaneously.
Besides, depending upon the number of unknowns of the problem, additional relations may be devised
by utilizing the property relations of the system.
2. For control volume applications, together with property relations and specifics of that particular
process, three equations, mass, energy and exergy have to be solved simultaneously. Particularly, for
steady flow systems, due to irreversibilities of the system, I , the relations are not system free.
Example 5.17: Consider universe as an isolated system, and analyze the variation of energy and of the exergy content with
respect to processes occurring in it.
Solution:
As shown in Figure 5.24, for the universe the energy


  Wdt
 . For an isolated
change is, E2  E1  Qdt
t
t
system, Q  W  0 . Hence, the energy content of
the universe becomes constant. In addition to the
absence of heat and work transfer, there is no mass
.
flow in and out, m
  m  0
i
e

 .
Then, Eq. (5.39) reduces to  2  1   Idt
t
As shown in Figure 5.24 Due to irreversibilities
involved in all processes, the exergy content of
universe is always diminished.
5.11
Exergy Analysis of
Closed Systems
For closed systems, no mass transfer
is allowed through the boundaries of the
i  m
 e  0 . For a particular prosystem; m
cess between the initial and the end states
of 1 and 2, Eq. (5.38) reduces to
n
 2  1 

T0 
 Q12 j  W  p0V 12  I12

j 
 1  T
j 1
(5.40)
The lost work due to heat
transfer irreversibilities
The actual work and heat
transfer by the system
The lost or unrecoverable
work on atmosphere
The exergy change or the maximum useful
work obtainable from the system
Consider a stationary system, i.e. ep=ek=0, with respect to Eq. (5.34), the exergy change becomes,
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 187
 2  1  (U 2  U1 )  p0 (V2  V1 )  T0 ( S2  S1 )
(5.41)
Hence the exergy of a stationary system at a given state is,
1  m u1  u0   p0 v1  v0   T0 ( s1  s0 ) 
(5.42)
Example 5.18: Air is contained in a tank of 0.4m3 at a pressure and temperature of 400kpa, 300K respectively. Determine,
a.
the maximum useful work that can be obtained.
b.
the exergy change of air upon doubling its volume by a reversible and adiabatic process.
c.
the exergy change of air upon doubling its volume by a reversible and isothermal process.
Assume p0=101kpa, T0=300K.
Solution:
a.
The mass of air in the container, m 
pV
400 x0.4

, m = 1.858 kg
RT 0.287 x300
v1 = 0.4/1.858 = 0.215m3/kg, v0 = 0.287 × 300/101=0.852 m3/kg, s1-s0 = -0.287ln(400/101) = -0.395 kJ/kgK. Since the initial
and the final temperatures are the same and for an ideal gas, u1-u0=0. Hence by Eq. (5.42),
1  1.858 x 0   101x 0.215  0.852   300 x 0.395 
1  100.634 kJ. This is the maximum amount of energy that
can be extracted until the system attains mutual equilibrium with its environment.
k
b.
Since the process is reversible and adiabatic, I12 = 0,
kPa, and the work done due to expansion, W12 
V 
Q12 = 0, V2 = 0.8m , and p2  p1  1   151.57
 V2 
3
400 x0.4  151.57 x0.8
, W12=96.86kj. The exergy change of the system by
0.4
Eq. (5.40) is:
 2  1  98.86  101x0.4 =-56.46 kJ. This is the amount of exergy decrease in air due to adiabatic expansion.
Exergy change due to expansion into
Exergy change due to due to work
188
THERMODYNAMICS
c. For a reversible and isothermal process, Q12  0, but Tj=T0, hence there is no exergy change due to heat transfer.
For a reversible process, I12=0, and W12=p1V1ln(V2/V1), W12=110.885 kJ. Substituting these results into Eq. (5.40) yields,
  0  110.885  101x0.4   0 ,   70.485 kJ. Decrease in exergy for isothermal process is bigger. However, for
none of the processes, exergy decrease can be greater than the exergy value of the system at the initial state.
Example 5.19: Referring to the Example problem (4.14) and Figure 4.23, for the same change of state of the gas, determine:
a.
The exergy change of the gas
b.
The total exergy transfer by work
c.
The irreversibility of the process
The surroundings is at p0 = 100kPa, T0 = 25C.
Solution:
a.
Considering the results obtained by the solution of Example (4.14), m  0.8 kg, U  216 kJ, V=0.75 m3 ,
cv  1.08 kJ/kgK, c p  1.58 kJ/kgK .
2
2 

By Eq. (5.30), the entropy change of the gas, S  0.8  1.08ln  1.58ln
  1.153 kJ/K and the exergy change of the
4
0.5 

or   -52.594 kJ.
system is,   U  p0 V  T0 S  216  100 x 0.75  298 x1153
.
b.
The exergy transfer by the current flow through the circuit is  elec  441 kJ and the exergy transfer by
the expansion of the gas,  exp  W12  p0 V  225  100x 0.75  150 kJ. The total exergy transfer by work becomes,
W  441  150  291 kJ.
c.
Evaluating Eq. (5.4) for
  -52.594kJ, Q  0 , W  291 kJ yields,
52.594  0  291  I12 , or
I12 = 343.594 kJ.
The process of heating the gas by an electrical resistor causes an energy waste of 343.594 kJ.
Example 5.20: An insulated room of a dormitory is at 2.5mx3mx4m of dimensions and contains air at p1=100kPa, T1=25C.
The student, being in hurry for his thermo exam, leaves the 50W-fan on and the fan runs for 8 hours. Determine the loss of
exergy for such a case.
Solution:
The mass of air in the room, m 
p1V1
100 x30

 35.07 kg
RT1 0.287 x 298
The room final temperature, 1 st law of thermodynamics applied to the room, U  W12 , W 12 =-50x8x3600=-1440kJ. Thus, 35.07 x 0.781x (T2  25)  ( 1440),
p2 
100
x 350.5 , p2=117.6kPa and the entropy change of the room
298
becomes, S  35.07 x 0.781x ln  117.6    4.45 kJ/K . Respect
 100  

to changes in internal energy, entropy, and the volume of the room;
U  1440 kJ,
S  4.45 kJ/K,
V  0 , the exergy change
of air is   1440  0  298  4.45  113.9 kJ Considering that,
Q  0,  W  1440 kJ and =1139kJ in Eq. (5.38) yields,
113.9  0  ( 1440)  I12 , and the amount of wasted energy is
I12 = 1326.1 kJ.
T2  77.5 C The final pressure, p 2/T 2=p 1/T 1,
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 189
Example 5.21: An elastic and spherical tank contains 2 kg of refrigerant R134a at 0C with 30% vapor quality. It has been
stated that the inside pressure varies linearly with the diameter of the tank. The refrigerant reaches a thermal equilibrium
with the surroundings at 25C, determine,
a. The exergy loss of the system
b. The reveresible work that could be obtained by the heating process
Solution:
a.
Since the diameter of the sphere is proportional to 1/3-power of its volume, pV relationship of the heating process becomes,
pV 1/3  C . Using the R134a tables for T1 = 0C, and x1 = 0.3, p1 = 0.294 MPa, v1 = 0.0213 m3/kg, and the constant C
of the proposed relation is 1.06. In addition to this relationship taking into consideration the super heated vapor data, the
pressure and the volume at the final state becomes p2=0.4MPa, v2=0.0543 m3/kg. The work done by the elastic tank is
p V  p1V1
W12  2 2
 23.186 kJ and the internal energies at both states are u1=253.24 kJ7kg, u2=392.25 kJ/kg. The total
1 n
changes in the internal energy, the entropy and the volume are, U  278.02 kJ , S=1.082 kJ/K, V  0.066 m3 .
Thus the overall exergy change of the system becomes,   278.02  6.6  293 x1.082  32.406 kJ . Since at the
boundary of the system, Tj=To, then Q  0, and  W  W12  po V  16.586 kJ .
Substitution of the above evaluated parameters into exergy relation provides the wasted energy of the process.
32.406  0  (16.586)  I12 , or I12=15.82 kJ.
Figure 5.27 Schematic of the problem
b.
To determine the amount of reversible work that could be obtained by this process, let I12=0 and substitute
Q  0,  W  Wrev  6.6, and   -32.406 into exergy relation (Eq. 5.40) yields, 32.406  0  (Wrev  6.6) or
Wrev=39.006kJ.
In fact, the amount of reversible work just figured out is identical with the sum of the actual work W12 and irreversibility I12 of the process.
5.12
Exergy Analysis of Steady State Flow Systems
Since the exergy of a system is the part of the system energy that can be used, and since the mass
and the energy of steady state flow systems are invariable with time, then the exergy of such systems
is fixed in time. Hence, for steady flow systems, the following can be stated.
 dm 

  0,
 dt  cv
 dE 

  0,
 dt  cv
 d 

 0
 dt  cv
(5.43)
For systems at steady state, the system exergy is invariable with respect to time, and in Eq. (5.39),
let
then rearranging the exergy equation for steady flow systems yields,
 E  p0V  T0 S   0 ,
cv
190
THERMODYNAMICS
X Q  X W 
 m    m 
 I
(5.44)
1
  (h  h0 )  T0 ( s  s0 )  V 2  gz
2
(5.45)
e
e
e
i
i
i
where, the flow exergy at a particular state is,
If the steady flow system under study is a single inlet and outlet system with negligible kinetic
and potential exergy changes, then Eq. (5.44) reduces to
xQ  xW  he  hi   T0 se  si   i
(5.46)
The lost work due to
irreversibilities of the process
The actual work and heat
transfer to the flow
The lost or unrecoverable
energy of the flow
The net exergy or the maximum useful
work transferred to the flow
Since the system has a single inlet and outlet, the mass flow rate at the inlet and the exit are the same.
In Eq. (5.46) xQ  X Q / m and xW  X W / m represent respectively the exergy transfer by heat and
work per kg of flowing fluid.
The following sample problems illustrate the use of these results for various engineering applications.
Example 5.22: Determine the maximum power that can be obtained from a hydroelectric power station located at the foot
of a 450m dam as shown in Figure 5.28. Water with a mass flow rate of 1500 kg/s enters the discharge pipe at 100 kPa, and
20C and flows downward. Assume that the pipe diameter is constant throughout and the water pressure and temperature
at the outlet is the same as at the inlet. The surroundings is again at 100 kPa, and 20C.
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 191
Solution:
a.
The flow exergy of water at state1 has to be calculated by Eq. (5.45). Since flowing water is at the temperature and pres-
s1  s0  0 , and being stationary, V=0. Then,  1  gz1  9.81x 450 / 1000
 =4.414 kJ/kg. To get the maximum work, the process must be reversible, in Eq. (5.44), I  0 . Since there is no
sure of surroundings, h1  h0  0,
1
heat transfer, Q  0,
X Q  0 . Finally, Eq. (5.44) reduces to X W  m  1 , or X W  1500 x 4.414  6621 kW,
and Wrev  X W . Hence the maximum power is Wrrev = 6621 kW
Example 5.23: As shown in Fig. 5.29, for a regenerative thermal power plant, steam is extracted and enters the feed water
heater at (1MPa, 200C) with a flow rate of 0.5 kg/s, and leaves the heater as saturated liquid at the same pressure. Likewise,
the feed water enters the heater at (2.5MPa, 50C), and is heated to a temperature 10C below the exit temperature of the
steam. Neglect the heat loss at the outer surface of the exchanger and determine,
a. The flow rate of feed water
b. The exergy loss of the process.
Solution:
a.
The energy balance on the heater yields, m s (h1  h2 )  m w (h4  h3 ) For T 2=180C, T 3=50C, T 4=170C, and
h1-h2=2065.09 kJ/kg, h4-h3=504 kJ/kg, the above stated energy equation yields the mass flow rate of steam,
ms=2.048 kg/s.
b.
Together with the energy balance, assuming that the change in kinetic and potential energies of both fluids are
  0 in Eq. (5.44) results as, I  T m ( s  s )  m ( s  s ) ,
negligible and substituting X Q  0, X
W
0
s 2
1
w 4
3
s2-s1=-4.556 kJ/kgK, s4-s3=cln(T4/T3)=1.326kJ/kgK, and the irreversibility rate for surroundings at 25C becomes,
I  298 x 0.5(4.556)  2.048(1.326)  130.4kW , the exergy loss is then,  destroyed  130.4 kW.
Example 5.24: Liquid water (200 kPa, 20C) at a flow rate of 2.5 kg/s is mixed with steam (200 kPa, 300C) in a mixing
chamber as shown in Fig. 5.30, and the chamber loses heat at a rate of 600 kJ per minute to surroundings at 25C. Water
leaves the chamber at (200 kPa, 60C). Determine,
a.
The mass flow rate of steam,
192
THERMODYNAMICS
b.
The work lost due to mixing process,
c.
The percent exergy reduction and suggest methods to reduce the exergy loss.
Solution:
a.
Together with the continuity relation, the energy balance for the chamber yields, Q  m 1 (h3  h1 )  m 2 (h3  h2 ) . For
h3-h1=168 kJ/kg, h3-h2=-2819.8 kJ/kg, Q =-10 kW, and m 1 =2.5 kg/s, then m 2 =0.152 kg/s.
b.
Considering that X Q  0,
X W  0 and T j  T0 , the exergy equation (Eq. 5.44) for this particular application
turn out to be, I  m 1 (h1  h3 )  m 2 (h2  h3 ) T0 m 1 ( s1  s3 )  m 2 ( s2  s3 ) , where s2-s3=7.0615 kJ/kgK, s1-s3=cLn
(T /T )=-0.537 kJ/kgK. Then I  90.207 kW.
1
3
Due to mixing of two fluids at different temperature (highly irreversible process), a useful energy at maximum value
of 90.207 kW becomes unavailable. In other words, this amount of exergy is lost.
c.
For two streams entering the mixer, the exergy supply rate is X supplied  1   2 , and
For stream 1, 1  m 1 h1  h0   T0 s1  s0   2.5 x 83.96  104.89   298 x 0.2966  0.3674  , 1  0.345kW.
For stream 2,  2  m 2 h2  h0   T0 s2  s0   0.152 x 3071.8  104.89   298 x 7.892  0.3674  ,  2  110.136
kW.
I
x100%
The exergy supplied, X supplied  110.481 kW. Finally, percent of exergy destruction becomes, X dest %  
X supplied
or X dest  81.69 %.
The process is highly irreversible. Essentially two factors contribute to the irreversibility of the process.
1.
Heat transfer through the mixing chamber. The chamber should be insulated and heat leaks should be reduced as much
as possible.
2.
Work can be extracted from two streams separately for best performance and then the fluids can be mixed for the lowest
exergy destruction.
Example 5.25: The combustion gas products flow at a speed of 80 m/s through the nozzle of a turbo-jet engine at (260 kPa,
747C), and exits at (70 kPa, 500C). For a surroundings temperature of 17C, determine,
a.
The gas speed at the nozzle exit
b.
The rate of exergy decrease of the gas for a flow rate of 0.1 kg/s.
Assume cp=1.15 kJ/kgK, and k=1.3 for the expanding gas.
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 193
Solution:
a.
With no heat and work transfer at the boundaries of the nozzle, the steady state energy equation for this particular case
V2
V2
becomes, h1  1  h2  2 , and for the given data the gas velocity at the nozzle exit is, V2=757.9m/s.
2
2
b.
The exergy equation (Eq. 5.44) takes the following form for this particular application,
1


I  m (h1  h2 )  (V12  V22 )  T0 m ( s1  s2 ) and I   m T0 ( s1  s2 ) . Assuming ideal gas behavior, the entropy change
2


can be calculated by Eq. 5.28, and the gas constant, R 
k 1
c p , can be used. The rate of exergy decrease due to friction
k
and mixing of fluid particles becomes, I  0.841 kW.
Example 5.26: Steam enters to an adiabatic turbine at (6 MPa, 600C) with a velocity of 80m/s and exits at (50 kPa, 100C)
and the speed at the exit is 140 m/s. For ambient conditions of 100 kPa, 25C, the turbine shaft power is measured to be 5
MW. Determine the rate of exergy loss in the turbine.
Solution:
The following simplifications can be done on exergy equation (Eq. 5.41) for this particular case,  Q  0,  W  5MW
1


and also recall that for steady flow systems the shaft work is expressed as, W  m (h1  h2 )  (V12  V22 )    W . Thus the
2


exergy relation, Eq.5.44, is reduced to I  m T ( s  s ) , where the mass flow rate is calculated by the above expression as,
0 1
2
m  5.15 kg/s. Then for s = 0.527 kJ/kgK, T0 = 298K, the exergy loss is
 destroyed  810.04 kW.
194
THERMODYNAMICS
5.13
Exergy Efficiency of Energy Conversion Systems
For energy management the most important figure is the efficiency of the system. In other words,
the following question is very frequently asked: How much of the useful energy is left from a given
energy source? In chapter 4, we specified the efficiency of a power cycle on the basis of energy ratio
as the work output of the cycle divided by the heat energy input.
Similarly, for energy conversion machines, the energy based efficiency would be,
e 
energy output
energy supplied
(5.47)
Unfortunately, the energy based efficiency is not a useful parameter in energy management. Due
to absence of comparison with the ideal process, this definition of efficiency makes no reference to
the performance of the system. An engineer would have no idea how well or how poor the device
performs for that particular process.
Example 5.27: Feed water with 3kg/s mass flow rate enters to a power plant boiler, as shown in Figure 5.33, at 10 bar,
50°C, and exits as saturated vapor at the same pressure. The boiler consumes 0.22 kg/s natural gas with lower heating value,
Hu=42000 kJ/kg. Determine the energy based efficiency of the boiler.
Solution:

Referring to Eq. (4.63), the energy out may be calculated as, Q out  m s (h2  h1 )  3 x(2778.1  209.33) , Q
out = 7706.31kW.
The energy supplied by combustion of fuel, Q  m H  0.22 x 42000 , Q = 9240kW. Hence by Eq. (5.47), the energy
in
f
u
efficiency of the boiler is: e  Q out / Q in  7706.31 / 9240 , e = 83.4 %
in
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 195
This Example illustrates only that the present design of the boiler allows 83.4% of the energy supplied
by the burner is transferred to water. Is this a good performance? What if heat leaks through the boiler
walls were reduced by better insulation and if the exit temperature of stack gases were reduced to the
temperature of surroundings, then the boiler efficiency will approach to unity, i.e. e  1 . This simply
indicates the proper supply of heat energy to the product. On this basis, however, do you think that
the performance of the boiler becomes the best?
The use of energy based efficiency in assessing and improving the energy conversion systems is misleading and mostly confusing. The reduction of energy losses of a process can clearly be identified by exergy
analysis. Exergy plays an important role in increasing the efficiencies of energy systems. Because the
location, the type, and the magnitude of wastes and loses can be quantified by exergy analysis. Accordingly, more meaningful efficiencies are evaluated with exergy rather than energy analysis.
Definition: The exergetic efficiency or “second law efficiency“ of a system is the ratio of exergy
recovered through a process or a cycle to the exergy supplied.
x 
exergy recovered
exergy destroyed
1
exergy supplied
exergy supplied
(5.48)
In Eq. (5.48), the exergy supplied indicates the amount of maximum useful energy supplied to the
system, and the exergy recovered is the amount of maximum useful energy left after the process
through the device. Since exergetic efficiencies are always a measure of the approach to the ideal
process or the ideal cycle, the exergy analysis also identifies the margin available to design more
efficient energy systems by reducing inefficiencies.
Example 5.28: Determine the exergetic efficiency,  x , of the power plant boiler in Figure 5.34, operating at the same
conditions as given in Example 5.27.
Solution:
Referring to Figure 5.34, the energy balance for the boiler requires that Q in  Q e  Q out  0 . Then the heat energy removed
 = -1533.69 kW.
by stack gases is Q e   9240  (7706.31)  or Q
e
 T 
 T 
Since the boiler operates at steady state, the exergy balance, Eq. (5.44), requires that I  1  0  Qin  1  0  Q e
T
i 

 T0 
0
 m s h1  h2   T0 s1  s2  or I  4464.528 kW .
196
THERMODYNAMICS

The exergy supplied =  1 

T0  
4464.528
 0.35  x = 35 %
 Qin  6876.828 kW . Hence by Eq. (5.48),  x  1 
Ti 
6876.828
As provided by the above example, if one compares the energy based, and the exergy based efficiencies of the same boiler the lower exegetic efficiency for steam production reflects the fact that
fuel is consumed by combustion and only 35-percent of the useful energy is transferred into water.
In addition to the irreversibilities involved in combustion phenomena, the required temperature
difference to transfer the heat against thermal friction caused a reduction in exergy efficiency of
the boiler.
Similar approach can be used to determine the exergy efficiency of industrial furnaces and heaters. The methodology implemented here can be extended for the development of exergy efficiencies
for other components of energy conversion devices. In the following sections, the exergy efficiency
of most common devices is formulated.
5.13.1
Nozzles and Diffusers
These devices are used as accelerators (nozzle) or decelerators (diffuser) for fluid streams. At
steady state conditions, and for adiabatic system, the energy balance equation, Eq. (4.55), reveals that
the energy efficiency of these devices is 100-percent. Due to irreversibilities of the flow; friction, expansion to lower pressure etc., the useful energy of the flow stream will decrease, and the description
of the efficiency with respect to energy becomes meaningless. Referring to Eq. (5.44), the destructed
exergy by these devices is,
I  m ( 1   2 )
(5.49)
 1 . Substituting these results into Eq.
Besides, the exergy supplied to a nozzle or a diffuser is m
(5.48), the exergy efficiency of nozzle or a diffuser may be described as,
x  1 
I / m  2 useful energy at the exit


1
 1 useful energy at the inlet
(5.50)
Since the flow exergy always decreases, this result of efficiency is applicable to both nozzles and
diffusers.
Example 5.29: Determine the exergetic efficiency of the nozzle of Example (5.25). Surroundings at T0=17C, p0=100
kPa.
Solution:
Referring to Figure 5.31, the flow exergy of the gas stream at the inlet, Eq. (5.45),  1  839.5  290  1.15  ln(1020 / 290)
 0.2653  ln(260 / 100)  802 / 2000 ,  1  497.02 k/kg. Similarly, at the exit,
 2  555.45  290 x 1.15 xLn(773 / 290)  .2653 xLn(70 / 100)  757.92 / 2000 ,  2  488.27 kJ/kg. By Eq.
(5.50), the exergy efficiency of the nozzle is
nx  98.2%.
The same result could be obtained by using the numerical
value of the irreversibility term, I , of Example (5.25).
5.13.2
Turbines, Compressors, Pumps, and Fans
These machines convert fluid energy to electrical energy or vice-versa. Operating at steady state
and no heat transfer to surroundings, the exergy recovered by a turbine can be calculated by Eq.
(5.44) as,
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 197
(5.51)
Wt  m  1   2   I
Exergy destruction
Decrease in the exergy of the
Shaft power
Since the exergy supplied to the turbine is m  1  2  , then the exergy efficiency becomes,
tx 
Wt / m
w

 1  2 wrev
(5.52)
Considering the definition of exergy,  1  2  represents the maximum useful work, Wrev, that
could be extracted from the fluid stream at the same inlet and outlet conditions. Hence, as in Eq. (5.52),
two different expressions for the exergetic efficiency of a turbine may be provided.
For a compressor, a pump, or a fan operating at steady state with no heat transfer to surroundings,
and stating,  X w  Wc in Eq. (5.44), the exergy equation reduces to the following form,
m  2   1   Wc  I
Shaft power
Increase in the exergy of the
(5.53)
Exergy destruction
Since the exergy efficiency is the ratio of the increase in flow exergy to the exergy supplied, then for
adiabatic compressor or a fan the efficiency becomes,
cx 
 2   1 wrev

wc
Wc / m
(5.54)
Example 5.30: Determine the exergy efficiency of the adiabatic compressor explained in Example (4.26). Surroundings at T0=27C, p0=100kpa.
Solution:
Referring to Figure 4.43, the compressor intakes the air at atmospheric conditions, and compresses it to (500 kPa, 227C).
The air exit velocity is 100 m/s and the mass flow rate is 0.242 kg/s. The shaft power input is given as, Wc  50 kW .
In accord with Eq. (5.45), the exergy change of air stream is,


 2  1  1.005 500  300   300 x 1.005 xln 500 / 300   0.287 xln 500 / 100   1002  02 / 2000 , or
 2  1  190.559 kJ/kg. The exergy supplied to the compressor is, wc  206.6 kJ/kg, and by Eq. (5.54),
cx 
190.559
206.6
5.13.3
cx  92.2%
Heat Exchangers
Common to all heat exchangers, heat transfer between the fluids take place through a separating
wall. Since the fluids are separated by a heat transfer surface, they do not mix. Then, for steady state
and adiabatic heat exchanger, exergy equation, Eq. (5.44), becomes,
m h  1   2 h  m c  4   3 c  I
(5.53)
Exergy destruction
Decrease in the exergy of the hot
Increase in the exergy of the cold
198
THERMODYNAMICS
Decrease in the
exergy of the hot
Increase in the
exergy of the cold
Exergy destruction
Considering the fundamental definition, Eq. (5.48), the exergy efficiency of a heat exchanger
becomes,
m (   3 )c
 hx  c 4
(5.54)
m h  1   2 h
Example 5.31: Water enters the tubes (d=10cm) of an air heater at 0.5 MPa and 140C at a rate of 24 kg /min and leaves
the exchanger at 0.5 MPa and 60C. The cold fluid is air and enters the heat exchanger with a volume flow rate of 100 m3/
min. The velocity of air at the inlet is 25m/s, and inlet conditions are 110 kPa, 25C. The air pressure at the exit is 110 kPa.
If surroundings is at T0=25C, p0=100 kPa, calculate,
a.
The exit temperature of air,
b.
The efficiency of the exchanger.
Solution:
a. Referring to Figure 5.35, the energy equation for steady state conditions becomes, m w h1  h2   m a c pa T4  T3  .
Assuming ideal gas behavior for air 3  p3 / RT3  110 / 0.287 x 298  1.286 kg/m3 and m a  33  2.147 kg/s.
use Table A3 for water thermodynamic properties, then
T4  T3 
0.4 589.13  251.13
2.147 x1.005
 62.65 K, T4 = 87.65C
b. Assuming that the change in water velocity is negligible, then the exergy change of water stream becomes,
m w  1  2   0.4 x 589.13  251.13  298 x 1.7391  0.8312   26.978 kW.
m/s
Assuming that the cross section area of air at the exit is the same as the inlet, 3V3   4V4 or V4 
T4
V3  1.21x 25  30.25
T3
The exergy change of air stream is,


1


m a  4  3   2.147 x 1.005 x 87.65  25   298 x1.005 x ln(360.65 / 298) 
30.252  252   12.843 kW.
2000


CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 199
To get the efficiency of the exchanger, apply Eq. (5.54),  hx 
12.843
 0.476 , hx  47.6%
26.978
This result means that the 52.4% of the useful energy at the inlet of the exchanger is lost, whereas referring to the
energy efficiency, there is no loss, and the energy efficiency of the device is 100-percent which is “NOT TRUE!!”. The
useful energy of hot fluid is not completely transferred to the cold.
5.13.4
Mixing Chambers
Mixing chambers are found a wide range of applications in industry. In addition to the processes
explained in Section 4.14, they are commonly used as direct contact heat exchangers for regeneration of the feed water of a power plant, or as a thermocompressor, or ejector to reduce the venting of
low pressure steam. As shown in Figure 5.36, if the objective is to recover the latent heat content of
the low pressure suction vapor for process use, the device is called a thermocompressor. When high
pressure steam is available in pressures between 15-20bars, thermocompressors can economically
be used for compression ratios up to 6:1. If the objective is to pull a vacuum on a process vessel, the
mixing chamber is called ejector.
At steady state conditions with no heat and work transfer with its surroundings, the exergy balance yields,
m 1  1   3   m 2  3   2   I
Exergy
destruction
Decrease in the
exergy of high
pressure stream
Increase in the
exergy of low
pressure stream
Together with the relation in Eq. (5.55), the exergy efficiency of a mixing chamber is,
(5.55)
200
THERMODYNAMICS
 mx 
m 2  3   2 
m 1  1   3 
(5.56)
Example 5.32: An industrial facility vents 5000 kg/h of saturated steam at atmospheric pressure. The wasted steam
can be converted into useful low pressure process steam by boosting its pressure to 2 bar, saturated steam. The available saturated motive steam is at 15 bar. Assume surroundings is at 27C, and determine,
a.
The required mass flow rate of motive steam (kg/s)
b.
The efficiency of the thermocompressor under given operating conditions.
Solution:
a.
Referring to Figure 5.37, the energy equation at steady state conditions yields the mass flow rate of high pressure
steam as,
m 1  m 2
b.
h3  h2 
h1  h3 
or
m 1  1.389 x
2706.3  2675.5
 0.50 kg/s
2792.2  2706.7 
Together with the tabulated values of steam, the exergy increase of low pressure steam is,
 3  2  2706.7  2675.5   300 x 7.127  7.36 
or
 3  2  101.1 kJ/kg
Similarly, the exergy decrease of high pressure steam is,
 1  3  2792.2  2706.7   300 x 6.44  7.127   291.6 kJ/kg
Hence, the exergy efficiency of the thermocompressor is,
mx 
1.389 x101.1
 0.963
0.50 x 291.6
or
mx  96.3%
The thermo-compressor requires 0.5 kg/s saturated steam at a pressure of 15 bar, to produce a discharge of 1.889
kg/s of 2bar pressure steam, and doing this the efficiency of the thermocompressor is 96.3-percent.
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 201
5.13.5
Cyclic devices
As described in Section 4.13, the cyclic devices are categorized in two groups, 1. Heat engines,
and 2. Refrigeration machinery.
As shown in Figure 5.38a, after a combustion process, a reservoir of high temperature, T1, is produced. The exergy of the heat energy, Q1, released by the combustion process is Q1 1  T0 / T1  , and is
identical to the maximum useful work “reversible work” obtainable at these given conditions. Hence
the exergetic efficiency of the heat engine is,
 Hx 
Wnet
W

 net 

 T  Wrev  rev
Q1 1  0 
 T1 
(5.57)
Similarly the exergy supplied to the refrigeration system in Figure 5.38b is Wc , and the exergy
recovered may be expressed as,
Q 02 1  T2 / T0   Q 02 T2 / T0 T0 / T2  1  Q 2 T0 / T2  1
(5.58)
Reversible
which represents the reversible work obtainable by the heat extracted from the low temperature
environment. Hence the exergy efficiency of the refrigeration system is,
 Rx
T

1
Q 2  0  1

T
  Wrev  COPrev  COP
  2

1
COPrev
Wc
Wc
COP
(5.59)
This result is also applicable to the heat pump applications of refrigeration systems.
The exergy based efficiency is especially useful for devices carrying out a single process. For
instance, a device performing a heat transfer process, there is no way of having any idea about its
performance by simply looking into its energy based efficiency. Because, for all operating conditions,
202
THERMODYNAMICS
the energy based efficiency would be 100-percent. Whereas, the exergy based efficiency signifies
the performance of the device, and depending upon the operating conditions, the device performance
and the exergetic efficiency will alter. In case of cyclic devices, however, why do we need exergy
based efficiency? Is energy based efficiency not sufficient for measuring the performance of the
cycle? The performance of a cyclic device can only be measured by comparing the actual useful work
recovered with the ideal one at the given conditions of the cycle. Two cycles at the same operating
conditions might have the same energy based efficiency but perform differently. An improvement in
performance might not be implemented if an increase in total cost would result. In industry, decisions
are usually made on the basis of total cost.
Example 5.33: Consider two heat engines both of which having the same energy based efficiency of 45-percent. Engine1 operates between heat reservoirs of 1000K and 300K, and Engine2 between 700K and 300K. Which engine performs better?
Solution:
At a first glance, one might state that since both engines convert the same fraction of heat energy into work, both perform
equally well. However, the performance is measured by comparing the actual useful work output with the ideal one.
Engine 1: The reversible engine efficiency at the given conditions is,  rev1  1  300 / 1000  0.7 ,   0.45 . Hence
the exergy efficiency becomes,  Hx1 
1
0.45

 0.642
rev1 0.70
Engine 2: Similarly the reversible and the energy based efficiencies are,  rev 2  1  300 / 700  0.571, and   0.45 .
The engine exergy efficiency is  Hx 2 
2
0.45

 0.788
rev 2 0.571
Comparison shows that Engine 2 performs better than Engine 1 which means that Engine 2 converts more of available
energy into work than Engine 1.
5.13.6
Space and Hot Water Heaters
Water heating accounts for about 18-percent of the total energy used in a typical house. Most of four
person households use 800L to 1000L of hot water per day with 16000kwh-20000kwh energy consumption per year. If we consider 15-18 years of a typical life cycle for a residential water heater, the energy
efficient features of these devices will provide long term benefits. Especially in hot climates, air heat
pumps are preferred and used as water heaters. Since the heat pump takes the heat from the room and
transfers it to water in the storage tank, not only the water is heated but also the room is cooled.
Assuming that the heater is perfectly insulated and then the exergy equation, Eq. (5.44),
reduces to
m w  2   1   We  I
(5.60)
Exergy destruction
Electrical energy input
Increase in the exergy of water
Then, the exergy efficiency of the water heater is expressed as following,
Wx 
m w  2   1 
W
(5.61)
e
It is possible to relate the exergy and the energy based efficiencies of liquid heaters by considering
the energy based efficiency definition. Due to heat losses through the outer surface of the heater, the
energy based efficiency may be expressed as,
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 203
e 
Q in
W
(5.62)
e
where, Qin represents the rate of heat transferred to liquid and is,
 T2  T1   mc
 T
Q in  mc
(5.63)




  T  T0 ln 1 
By Eq. (5.45), the exergy increase of liquid is, m  2  1   mc
T
T1

  and together
 
with eqs. (5.63) and (5.62), substituting this relation into Eq. (5.61) and rearranging results as,
 x / e  1 
T0  T 
ln 1 

T 
T1 
or
 x / e  1 
1
ln 1  ay 
y
(5.64)
where, T1 and T0 are the temperatures of liquid at the inlet and surroundings respectively, and a  T0 / T1
and y  T / T0 .
The performance behavior of an electric heater for a liquid having surroundings temperature at the inlet (a=1) is represented in Figure 5.39. In this graph, the energy based efficiency
is taken to be constant and is assumed to be unaffected by the temperature rise range ( T ) of
204
THERMODYNAMICS
liquid which is taken to be between 5°C and 60°C. The typical behavior of these heaters is that
even though the energy based efficiency is close to unity, the exergy based efficiency is only a
small fraction of ηe.
Space heaters deliver heat by radiation or by convection. As shown in Figure 5.41a, radiant heaters
emit heat from a glowing red bar, and directly transfer the heat to people and the objects. They are
appropriate for rooms with high ceilings where it is difficult to retain warm air.
Convective heaters are effective in closed insulated rooms with average ceiling heights. As shown
in Figure 5.41b, the radiant convective heaters combine both radiant and convection effects in space
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 205
heating. For both the electrical resistance and the natural gas fired space heaters, the energy efficiency
is nearly 100-percent and there are almost no energy losses. Yet the exergy efficiency of such devices
is typically less than 10-percent. Indicating that by system improvements it is possible to heat the
same space by consuming only one-tenth of the electricity used at the present.
Example 5.34: A water heater supplies water at 62C. At steady state conditions, the rate of heat input to the heater is
1 kW. Due to heat leaks through the insulation on the outer surface, 2-percent of the energy input is lost to surroundings.
Assuming surroundings temperature at 300K, determine the energy and the exergy based efficiencies of the heater if the
energy is supplied by,
a.
an electrical resistance heater
b.
an air heat pump with overall COP=4.0
Solution:
a.
Electrical resistance heater:
Energy analysis, Qin  Wel  Ql  0.98 kW, and energy efficiency of the heater,
We 
energy out 0.98

x100%  98% Exergy analysis, exergy supplie d=1 kW, Exergy recovere
energy in
1
 T 
0.102
 300 
d = Qin 1  0   0.98 x 1 
  0.1023 kW, Wx  1 x100% Wx  10.2%
T
335


1 

b. Air heat pump:
Energy analysis, energy in, Wel  1 kW, and energy out, Qin  3.92 kW at a temperature of T1=335K. The energy
based COP becomes, COP  3.92  3.92
1
Wx
Exergy analysis, exergy supplied, 1kW, exergy recovered, 3.92 x 1  300   0.4095 kW. The exergy efficiency,
 335 
0.4095

x100%  40.95% Wx  40.95%
1
or
By Eq. (5.18), COP rev
335
300

 9.568 and by Eq. (5.59),
COP  3.92
335
Wx 

 40.95%
1
COP rev 9.568
300
.
The same result is obtained. The air heat pump is much better than the resistance heater. If daily consumption of hot water
is high, then the use of heat pump for water heating will be a better choice.
206
THERMODYNAMICS
The exergy efficiencies and loses due to exergy destruction describe the system better and provide
more meaningful information. It is actually a measure of how the operation of the system approaches
the ideal or theoretical upper limit. Together with information in system performance, a better understanding of factors affecting the efficiency is attained, and efforts to improve the performance can be
better directed. However, efficiency improvements require creativity and engineering, and involve
trade off with other factors such as economics and environmental impact.
References
1.
B. R. Bakshi, T. G. Gutowski, and D. P. Sekulic, Thermodynamics and the Destruction of Resources, Cambridge
University Press, ISBN 978-0-521-88455-6, 2011.
2.
“Modeling and Computation in Engineering” Editor, Jinrong Zhu, CRC Press, ISBN 978-0-415-61516-7, 2011.
3.
I. Dincer, and M. A. Rosen, Exergy, Energy, Environment and Sustaniable Development, Elsevier Publications,
2007.
4.
C. P. Kothandaraman, and R. Rudramoorthy, Fluid Mechanics and Machinery, 2nd edition, New age international
publisher, ISBN 978-81-224-2558-1, 2007.
5.
Y. A. Cengel, and R. H. Turner, Fundamentals of Thermo-fluid Sciences, McGrawHill Inc., ISBN 978-0-072-97675-6,
2004.
Problems
Exergy balance and closed systems
5.1
A 0.8 m of tank contains air. a. Plot how the total
stored energy and total stored exergy of air in the
tank change as the pressure is increased from 10 kPa
to 10 MPa with temperature held constant at 27°C.
b. Repeat the plot for the specific stored energy and
specific stored exergy. c. Plot how the stored energy
and stored exergy of air in the tank change as the
temperature is increased from -100°C to +100°C
with pressure held constant at 100 kPa. d. Repeat
the plot for the specific stored energy and specific
stored exergy. Assume atmospheric conditions to
be 100 kPa, 27°C
5.2
Water of 2.5 kg undergoes a process from an initial
state where the water is saturated vapor at 150°C,
the velocity is 55 m/s, and the elevation is 12 m to
a final state where the water is saturated liquid at
25°C, the velocity is 15 m/s, and the elevation is 2
m. Determine the exergy of water at the initial state,
the final state and the change in exergy between these
two states. Assume T0= 25°C, p0=1 atm.
5.3
5.4
b. the change in exergy of air upon doubling its
volume in reversible adiabatic process,
c. the change in the exergy of air upon doubling
its volume in reversible isothermal process
d. the change in exergy of air upon doubling its
volume in an adiabatic process with no work.
3
Determine the exergy change for the following
processes, and assume surroundings at 300K and
101 kPa,
a. heat interaction of 550 kJ between a reservoir
at (+55C) and the surroundings,
b. heat interaction of 550 kJ between a reservoir
at (-55C) and the surroundings,
c. heat interaction of 550 kJ between the two
reservoirs (+55C) and (-55C).
Consider surroundings being at 300K and 101 kPa,
and assuming air as an ideal gas, for 0.5 kg of air at
0.4 MPa, 0.4m3, determine the following,
a. the initial exergy content of air,
5.5
As shown in Figure 5.43, the inner and outer surfaces
of a 1.0 cm thick 3.5 mx8 m glass aperture in winter
are 22°C and 3°C respectively. If the surroundings
is at To=0°C, then determine,
a. the amount of heat loss through the glass over
a time period of 5 hours
b. the exergy destruction associated with this
process.
5.6
A mass of 1.2 kg of oxygen in a cylinder and piston
assembly as in Figure 5.44 expands by an internally
reversible isothermal process at 440K from 3.2 MPa
to 0.18 MPa. Assume that the surroundings is air
at 300K and 1.02bar. Discuss the requirement for
additional reservoirs to accomplish the process and
determine,
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 207
a. the work and heat interactions of the oxygen
b. the maximum work of the process
5.7
One kg of an ideal gas having constant heat capacity
cp=1.2 kJ/kgK, and k=1.3 is compressed adiabatically
from 100 kPa and 10C to 0.5 MPa. The process
is irreversible and requires twice the work than a
reversible adiabatic compression from the initial to
the final state. Determine,
a. the work required,
b. the entropy change of the gas?
c. the exergy destruction of the process?
5.8
Water in a piston-cylinder device as in Figure 5.45
is at 100 kPa, 27°C. The piston has stops mounted
so Vmin=0.02 m3 and Vmax=0.4 m3. The weight of the
piston is such that inside pressure of 2.5 MPa will
float it. If 10 MJ of heat is supplied from a heat
source at 327°C, find,
a. the total change in exergy of water
b. the total exergy destruction
5.9
An adiabatic vessel as in Figure 5.46 is divided by a
partition into two parts. One part (A) contains 20 kg
of water at 20C at 100 kPa, and the other (B) 1 kg
of water at 500C and 20 MPa. The surroundings is
air at 100 kPa, and 20C. The partition is ruptured
and the contents of the vessel mix. Determine the
exergy destruction of this mixing process.
5.10
A thin elastic balloon contains 30 g of nitrogen (ideal
gas) at 200K and 0.4 MPa. The balloon exerts on its
contents a pressure difference which is proportional
to its volume. The surroundings is at 1bar, 300K.
The nitrogen undergoes a heat interaction with a
reservoir at 400K until mutual equilibrium is attained.
Determine,
a. the final pressure of the nitrogen
b. the work and the heat interactions during the
process,
c. the change in exergy of nitrogen and of the balloon.
5.11
5 kg of air at 550K and 4 bar is enclosed in a closed
system. a. Determine the exergy of the system if the
surrounding pressure and temperature are 1 bar and
290K respectively. b. If the air is cooled at constant
pressure to the atmospheric temperature, determine
the change in exergy of the system.
5.12
Employing the ideal gas model determine the change
in specific entropy between the indicated states, in
kJ/kgK. Solve in two ways: Use the appropriate
ideal gas table, and a constant specific heat value
from appendices.
a. air, p 1 =100kPa, T 1 =20C, p 2 =100kPa,
T2=100C
b. air, p1=1bar, T1=27C, p2=3bar, T2=377C
c. carbon dioxide, p1=150kPa, T1=30C, p2=300kPa,
T2=300C
d. carbon dioxide, v1=1.1m3/kg, T1=300K, v2=0.75m3/
kg, T2=500K
e. nitrogen, p1=2000kPa, T1=800K, p2=1000kPa,
T2=300K
5.13
Adiabatic cylinder in Figure 5.47 contains 0.15m3
of air at 40C under a floating piston which exerts
a pressure of 1.2Mpa. The volume above the piston
is 0.05m3 and its completely evacuated. At a certain
moment the piston breaks up and falls to the bottom
of the cylinder. Assume air to be an ideal gas, and
determine,
208
THERMODYNAMICS
4. The volume of the system is doubled in an
isothermal process by heating with the help of
a reservoir at the lowest temperature allowable
to complete the process.
5. The volume of the system is doubled by expansion
into an empty vessel connected to the system by
a pipe.
For each of the above processes, determine,
a. the final state of the system,
b. the work and heat interactions of the system,
c. the useful work of the system,
d. the maximum work for the same extreme states
of the system and of the additional reservoir,
e. the change in exergy of the system and of the
reservoir (if one is used)
f. the exergy loss of the process.
Figure 5.47
a. the entropy change of air,
b. the change in exergy of the entire system if the
surroundings is at 102kpa, 22C.
5.14
Air (considered ideal gas) is contained in an insulated
rigid tank at 200C and 200 kPa. A paddle wheel
inserted in the volume does 750 kJ of work on the
air. If the volume is 2 m3 compute the change of
entropy of the system ΔS in kJ/K.
5.15
The pressure-volume diagram of a Carnot power
cycle executed by an ideal gas with constant specific
heat ratio of k is shown in Figure 5.48. Repeating
calculations done in the text show that
5.17
A light adiabatic envelope contains 50 g of nitrogen
(ideal gas, M=28, k=1.4) at 0.4 MPa, 240K. The
envelope exerts an additional pressure on its content
proportional to its enclosed volume. The nitrogen is
heated by a reservoir at 400C till the temperatures
equalized. The surroundings is at 0.1 MPa, 300K.
Determine,
a. the final state of the nitrogen
b. the heat and work interactions of the nitrogen,
c. the change of exergy of the nitrogen,
d. the exergy destruction of the process
5.18
Liquid initially at the temperatures T1 and T2 (see
Figure 5.49 below). The membrane ruptures and
eventually the system attains an equilibrium state
with temperature Tf = (T1 + T2 )/2. Each mass is incompressible with constant specific heat c. Compute
the amount of exergy lost due to mixing process in
terms only of m, c, T1, T2 and To.
Figure 5.48
Figure 5.49
5.16
Consider the following descriptions of a number of
processes that take place in a closed system which
contains 1 kg of water at 180C and 0.7 MPa. The
surroundings is at 300K and 1.02bar and there are
no other reservoirs unless otherwise stated.
1. The volume of the system is doubled at a constant
pressure by heating with the help of a reservoir
at the lowest temperature allowable to complete
the process.
2. The volume of the system is doubled at a constant
pressure by rapid stirring of the system.
3. The volume of the system is doubled in an
adiabatic process.
5.19
15 kg of water is heated in an insulated tank by a
churning process from 300K to 340K. If the surrounding temperature is 300K, find the loss in exergy for
this process.
5.20
Calculate the unuseable energy in 55 kg of water at
55°C with respect to the surroundings at 5°C. Take
the pressure of water as 1 atmosphere.
5.21
2.1 kg of air at 6 bar, 90°C expands adiabatically is
a closed system until its volume is doubled and its
temperature becomes equal to that of the surroundings which is at 1 bar, 5°C. Determine,
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 209
a. the maximum work obtainable,
b. the change in exergy of air,
c. the exergy loss or the irreversibility of the process.
Assume for air, cv = 0.718 kJ/kg K, R= 0.287
kJ/kg K.
5.22
A bicyclist in Figure 5.50 rides the bicycle at a speed
of 5 m/s. To keep the speed constant, a trust force of
20N has to be created on the pedals. Determine the
loss in useful energy and discuss where that energy
is gone.
5.25
Data: The specific heat of water, cp=4.18kJ/kgK, the
specific heat of ice, cp=2.1kJ/kg K and the enthalpy
of fusion of ice (latent heat), hfs=333.5kJ/kg.
5.26
A power cycle operating between two reservoirs
receives energy QHby heat transfer from a hot reservoir at TH=1100K and rejects energy QC by heat
C
transfer to a cold reservoir at TC =300K. For
each
of the following cases determine whether the cycle
operates reversibly, irreversibly, or is impossible:
a. QH = 800 kJ, Wcycle = 480 kJ,
b. QH = 800 kJ, QC=200 kJ,
c. Wcycle = 800 kJ, QC =200 kJ,
d. η = 50%.
5.27
At steady state, a refrigeration cycle removes
18,000 kJ/h of energy by heat transfer from a space
maintained at –40°C and discharges energy by heat
transfer to surroundings at 20°C. If the coefficient
of performance of the cycle is 25 percent of that of
a reversible refrigeration cycle operating between
thermal reservoirs at these two temperatures, determine the power input to the cycle, in kW.
5.28
One kilogram of air as an ideal gas executes a Carnot
power cycle having a thermal efficiency of 60%.
The heat transfer to the air during the isothermal
expansion is 40 kJ. At the end of the isothermal
expansion, the pressure is 5.6 bar and the volume is
0.3 m3.Determine
a. the maximum and minimum temperatures for
the cycle, in K.
b. the pressure and volume at the beginning of the
isothermal expansion in bar and in m3 respectively.
c. the work and heat transfer for each of the four
processes in KJ.
d. sketch the cycle on p-v coordinates.
5.29
A 25 kg of thermal system with 0.7 kJ/kgK specific
heat is initially at 600C temperature and undergoes
a thermal interaction with a cyclic heat engine which
produces 10 kJ of work per cycle and rejects 15 kJ
of heat to a reservoir at 27C. Determine,
a. the amount of heat interaction with thermal
system per cycle.
b. the number of cycles the system will operate.
c. the amount of work the heat engine will produce.
d. the amount of work to be produced by a reversible
heat engine working at the same conditions.
5.30
A refrigeration cycle operating between two reservoirs receives energy QC from a cold reservoir at
TC =250K and rejects energy QH to a hot reservoir
Figure 5.50
5.23
On a cold winter day in Erzurum, it is -17°C outside.
Yet, as shown in Figure 5.51, the temperature of the
inside of the mechanical engineering building is
+17°C. The thickness and the height of the building
wall respectively are L=40cm, H=5.5m. The wall
thermal conductivity is 0.69 W/mK, and the convective heat transfer coefficient between the inner wall
surface and interior is 4.5 W/m2K. Determine,
Figure 5.51 Exergy loss due to thermal friction
a. the heat loss throgh the wall for an inner surface
temperature of 12°C,
b. the outer surface temperature of the wall,
c. the rate of exergy lost due to heat loss through
the wall
5.24
3 kg of gas (cv=0.81 kJ/kg K) contained in a rigid
tank is initially at 2.5 bar and 400K receives 600
kJ of heat from an infinite source at 1200 K. If the
surrounding temperature is 290 K, find the loss in
exergy due to heat transfer from the source.
1.2kg of ice at 0°C is mixed with 12 kg of water at
27°C. Find the loss in available energy when the
system reaches an equilibrium temperature. Assume
the temperature of the surroundings as 15°C.
210
THERMODYNAMICS
at TH =300K. If QC=1400kJ and Wcycle=140 kJ, the
cycle operates,
a. Reversibly, b.Irreversibly, c.It is impossible
5.35
Calculate the decrease in exergy when 20 kg of water
at 90°C mixes with 30 kg of water at 30°C, the pressure being taken as constant and the temperature of
the surroundings is 10°C. Take cp of water as 4.18
kJ/kg K.
5.36
As shown in Figure 5.53, a flywheel whose moment
of inertia is 0.75 kg m2 rotates at a speed 3200 r.p.m.
in a large heat insulated system, the temperature of
which is 20°C. If the kinetic energy of the flywheel
is dissipated as frictional heat at the shaft bearings
which have heat capacity of (mc)b =10 kJ/K. Calculate,
a. the rise in the temperature of the bearings when
the flywheel has come to rest,
b. the greatest possible amount of the above heat
which may be returned to the flywheel as highgrade energy,
c. the amount of kinetic energy lost,
d. the final r.p.m. of the flywheel, if it is set in
motion with this available energy
Figure 5.52 Cross section of a steam actuator
5.31
A cylinder maintained at 260C by a bath is divided
by a stopped piston into two parts. Part A contains
0.01 m3 water at 30 MPa, and part B contains 2.49 m3
of steam at 0.2 kPa (see Figure 5.52). The surroundings is at 102 kPa, and 20C. The stopped is removed
and the two parts reach equilibrium. Determine,
a. the final state in the cylinder,
b. the work and heat interaction,
c. the maximum work associated with the process.
5.32
A cylindrical rod of length L insulated on its lateral
surface is initially in contact at one end with a wall
at temperature TH and at the other end with a wall
at a lower temperature Tc. The temperature within
the rod initially varies linearly with position x according to T ( x)  T   TH  TC  x . The rod is
H



L
(a) Flywheel of a car engine

then insulated on its ends and eventually comes to a
final equilibrium state where the temperature is Tf.
Evaluate Tf in terms of TH and TC, and determine the
exergy destruction in term of m, cp, TH, TC, Tf, and
To, where m is the mass and cp is the specific heat
of the rod..
Hint: Since the rod is adiabatic, Eq. 5.39 reduces
to
I12  m u1  u2   T0 s1  s2 
5.33
2.2 kg of air is compressed polytropically from 1
bar pressure and temperature of 290K to a pressure
of 6.8 bar and temperature of 420 K. Determine the
exergy loss of the process if the sink temperature is
290K. Assume R=0.287 kJ/kgK, cp =1.004 kJ/kg K
and cv= 0.716 kJ/kg K.
5.34
Air at 500K, occupies a volume of 0.5 m , receives
7200 kJ/min from a source at 1000K, and its volume
doubles. Assuming that the temperatures of system and
the source remain constant during the heat transfer,
determine the decrease in exergy after the process.
Take the temperature of atmosphere as 300 K.
(b) A mechanical flywheel on a compressor
3
Figure 5.53 Typical uses of flywheels
5.37
A small block of copper is to be cooled from room
temperature of 300K to liquid helium temperature
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 211
of 4.2K by means of a container of liquid helium.
Since liquid helium is very expensive, $30.0 per liter,
the most economical scheme for cooling should be
used. You are to make an engineering evaluation of
two proposed schemes:
1. Assume that the copper block is dropped very
quickly into the liquid helium.
2. Assume that the copper block is lowered slowly
into the container so that helium gas leaving
the container is always in thermal equilibrium
with the copper block. You have to answer the
following questions:
a. How much liquid helium is required by scheme
1?
b. How much liquid helium is required by scheme
2?
c. Is one of the processes reversible? Give some
justification.
d. Determine the exergy loss, if any, in both
schemes.
To aid the solution of the problem the following
data is given.
Helium; p=1bar, T=4.2K, uf=9.2kJ/kg, ug=24.8kJ/
kg, hf=10kJ/kg, hg=30.9kJ/kg, sf=3.47kJ/kgK,
sg=8.43kJ/kg, vf=8.01L/kg, vg=59.83 L/kg. Additionally assume that at 1bar pressure, gaseous
helium is ideal at all temperatures above the
normal boiling point.
Copper; (mc)=0.01 kJ/K
Steady flow systems
5.38
At an industrial facility, 3 cm thick and 60 cmx60
cm in dimensions bronze plates (=7000 kg/m3,
cp=0.45 kJ/kgK) are heated from 24°C to 500°C by
passing the plates at a rate of 250 plates/min through
an oven at 700°C. Determine the rate of exergy loss
of this heating process.
5.40
A geothermal source provides 4.5 kg/s of hot water
at 800 kPa, 150°C flowing into an adiabatic flash
evaporator that separates vapor and liquid at 200 kPa.
Sketch the problem schematically and determine,
a. the exergy of three fluxes (one inlet and two
outlets)
b. the rate of exergy loss for this process
A 2.6 kg/s flow of steam at 15 bar and 640°C should
be brought to 400°C by spraying liquid water at 15
bar and 27°C. Sketch the problem and determine,
a. the cold water flow rate
b. the exergy destruction of the process if the surroundings is at 27°C.
An inventor claims to have developed a device
requiring no work input or heat transfer, yet able
to produce at steady state hot and cold air streams
as shown in Figure 5.54. Employing the ideal gas
model for air ignoring kinetic and potential energy
effects, evaluate this claim by exergy analysis.
Figure 5.54
5.43
Liquid water enters an adiabatic piping system at
17°C at a rate of 7.5 kg/s. It is observed that the
water temperature rises by 0.6°C in the pipe due to
friction. If the environment temperature is also 17°C,
evaluate the rate of exergy destruction due to flow
in the pipe.
5.44
Air enters a nozzle steadily at 350 kPa, 87°C with a
velocity of 25 m/s and exits at 95 kPa, and 375 m/s.
The heat loss from the nozzle to the surroundings
at 22°C is estimated to be 3.2 kJ/kg. Determine,
a. the temperature of air stream at the nozzle exit
Steam at 0.5 MPa, 360oC flows through a 10-cmdiameter pipe with a velocity of 25 m/s. Determine
the rate of exergy transported by the flow, what would
the rate of exergy transported if air instead of steam
were flowing through the pipe? Take atmospheric
conditions as 100 kPa and 25°C.
5.39
5.41
5.42
b. the amount of exergy destroyed per kg of air
Figure 5.55 Convergent-divergent nozzle
5.45
As shown in Figure 5.55, carbon dioxide (CO2) enters
a nozzle at 2.5bar, 760°C, 75m/s and exits at 0.8bar,
640°C. Assuming the nozzle to be adiabatic and the
surroundings to be at 1bar and 17°C. Determine,
a. the velocity at the nozzle exit,
b. the drop in useful energy between the inlet and
the exit.
c. What would the exit velocity be if carbon dioxide entered the nozzle at 150m/s and the same
pressure and temperature?
212
THERMODYNAMICS
5.46
Steam expands in an adiabatic turbine from 8 MPa
and 440°C to a pressure of 50 kPa as saturated vapor.
For a steam flow rate of 1.8 kg/s, determine the
maximum possible power output of the turbine.
5.47
In a turbine as shown in Figure 5.56, the air expands
from 7 bar, 600°C to 1 bar, 250°C. During expansion
9 kJ/kg of heat is lost to the surroundings which is at
1bar, 15°C. Neglecting kinetic energy and potential
energy changes, and assume air behaves like an ideal
gas with cp=1.005 kJ/kg K, determine per kg of air,
a. the decrease in exergy,
b. the maximum work obtainable,
c. the irreversibility of the process.
5.49
Air (ideal gas) enters a compressor operating at steady
state at 17°C, 1bar and exits at a pressure of 5 bar.
KE and PE changes can be ignored. If there are no
internal irreversibilities, evaluate the work and heat
transfer in kJ/kg for an isothermal compression.
Figure 5.58
5.50
A centrifugal air compressor as in Figure 5.58
compresses air at the rate of 20 kg/min from 1 bar
to 2 bar. The temperature increases from 20°C to
120°C during the compression. Determine actual
and minimum power required to run the compressor. The surroundings is at 20°C. Neglect the heat
interaction between the compressor and surroundings
and changes in potential and kinetic energy.
5.51
Figure 5.59 provides steady state data for a throttling
valve in series with a heat exchanger. Saturated liquid
Refrigerant 134a enters the valve at 40°C with a mass
flow rate of 0.25 kg/s and is throttled to -10°C. The
refrigerant then enters the heat exchanger, exiting
as a saturated vapor with no significant decrease in
pressure. In a separate stream, liquid water enters the
heat exchanger at 25°C and exits as a liquid at 5°C.
Take the heat capacity of water to be 4.18 kJ/kgK.
Stray heat transfer to the surroundings and kinetic and
potential energy effects can be ignored. Determine,
a. the mass flow rate of liquid water,
b. decrease in exergy of refrigerant due to throttling
process,
c. overall decrease in exergy of R134a,
d. increase in exergy of water
e. the exergy loss of overall process.
Figure 5.56 Exergy loss of air byflowing
through a turbine
5.48
Figure 5.57 shows a two-stage steam turbine with
reheat in between. Both stages may be considered
adiabatic. Steam at 14 MPa, 400C is supplied at
a rate of 105 kg/h to the high pressure turbine and
comes out as saturated vapor at 3.0MPa. Then it is
reheated at constant pressure by means of a reservoir
at 550C until its temperature reaches 400C, and
finally after passing through the low pressure turbine
it is exhausted as saturated steam at 150 kPa. The
surroundings is at 100 kPa, and 5C. Determine,
a. the total power supplied by the turbine,
b. the heat transfer rate of the heat reservoir,
c. the maximum power that can be obtained from
the steam and the reservoir combination.
d. the irreversibility of the process
Figure 5.57
Figure 5.59
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 213
5.52
A vapor stream, condensing at 450°C in a certain
process, transfers heat to saturated liquid water at
250°C. The resulting saturated steam at 250°C is
used in a power cycle which rejects heat at 32°C.
What is the fraction of the available energy in the
heat transferred from the process vapour at 450°C
that is lost due to the irreversible heat transfer at
250°C.
5.53
Each person in a family of five takes a 5-minute
shower everyday. The average flow rate through
the shower is 8L/min. City water at 20°C is heated
to 60 °C in an electric water heater and tempered to
45°C by cold water at the T-elbow of the shower.
Determine the amount of useful energy destroyed
by this family per year as a result of taking daily
shower.
5.54
In a double-pipe paralel flow type heat exchanger
as in Figure 5.60, waters enter at 40°C and leaves at
60°C while oil (density=820 kg/m3, specific heat=2.6
kJ/kg K) enters at 200°C and leaves at 90°C. If the
surrounding temperature is 27°C determine the
exergy loss of the process for oil mass flow rate of
1.5 kg/s. The surroundings is at 300K.
Figure 5.61
5.57
Figure 5.60
5.55
In Figure 5.61, hot gases of a steam boiler transfer heat
to water which vaporizes at constant temperature. In
a certain case, the gases are cooled from 1100°C to
500°C while the water in tubes evaporates at 180°C.
Take the specific heat of gases as 1.005 kJ/kg K, and
the latent heat of water at 180°C as 2015 kJ/kg, and
assume all the heat transferred from the gases goes
to the water. For 104 kg/h water vapor generation,
determine the increase in unavailable energy due to
this irreversible heat transfer process.
5.56
Water (cp=4.18 kJ/kgK) at a flow rate of 0.35 kg/s
is heated from 20C to 60C by flowing through a
heat exchanger. The hot fluid of the exchanger being
geothermal water (cp=4.31 kJ/kgK) is at 165C and
the flow rate is 0.28 kg/s at the inlet. The heated water
passes through the tubes of 10 mm diameter with a
velocity of 0.5 m/s. Neglect the tube thickness, and
determine,
a. the number of tubes,
b. the exergy loss rate due to heat exchange process
in the exchanger.
As in Figure 5.62, a 15 cm diameter pipe of length
12m containing hot water at 75C is losing heat to
the surroundings at 5C by natural convection. The
rate of heat loss is calculated by Q=UAT Where U
is the overall heat transfer coefficient and U=9.5 W/
m2K for this case, A is the heat transfer surface area,
m2, and T is the temperature difference between the
water and the surroundings. Determine,
a. the heat loss rate through the pipe
b. the rate of exergy wasted during this process,
hint: assume the wall temperature is the same
as the fluid temperature.
c. suppose that the hot water inlets the pipe at 90C
and exits at 60C by losing the same amount of
heat as calculated in case a. Evaluate the exergy
wasted in this case.
Figure 5.62
Efficiency of systems
5.58
As shown in Figure 5.63, 0.08 kg of steam initially at
1MPa, 240C in a piston-cylinder device expands to
300 kPa, 160C by doing work and releases 3 kJ of
heat to surroundings at 100 kPa, 27C. Determine,
a. the exergy values at the initial and the final
states
214
THERMODYNAMICS
c. the average overall heat transfer coefficient in
W/m2K.
b. the reversible work obtainable from steam
between the end states
c. the irreversibility or the exergy loss of the process
d. the exergy efficiency of the process
Figure 5.63
5.59
The gear train of car gear transmission system as
shown in Figure 5.64 operates at steady state. The
gear system runs in lubrication oil and the power
input, and output of the box respectively are 85 kW,
and 81.1 kW. The surface temperature of the box
is 37°C. Assume surroundings at a temperature of
27°C and determine,
Figure 5.65
5.61
In a steady flow machine, air having a mass flow
rate of 0.5 kg/s enters the system at a pressure of 10
bar and 200°C with a velocity of 100 m/s and leaves
at 1.5 bar and 27°C with a velocity of 50 m/s. The
temperature of the surroundings is 27°C and pressure
is 1 bar. Determine,
a. the type of the flow machine
b. the reversible work and the actual work for an
adiabatic process,
c. the irreversibility of the system,
d. the exergy based efficiency of the system.
Take for air : cp =1.005 kJ/kg K, R =0.287 kJ/kg K.
Figure 5.64 Gear transmission system of a car
a. the heat transfer rate through the box,
b. the energy based efficiency of the gear train,
c. the rate of exergy destruction due to power
transmission,
d. the exergy based efficiency of the system.
5.60
As shown in Figure 5.65, a house is maintained at an
average temperature of 22°C in winter by circulating the interior air through a duct at which 10 kW
of electric power is supplied by electric resistance
heaters. The house has 400m2 of heat transfer surface
area and the outdoors temperature is at 0°C. Evaluate,
a. the exergy based efficiency of the heater,
Hint: Consider a reversible heat pump working
between the given heat reservoirs.
b. the rate of useful work lost in kW,
Figure 5.66 Cross sectonal view of a water
pump
5.62
The pump of a water distribution system as shown
in Figure 5.66 is powered by a 3 kw electric motor
whose energy efficiency is 85-percent. The water flow
rate through the pump is 10 L/s. If the pressures just
at the inlet and outlet of the pump are measured to
be 100 kPa and 280 kPa respectively, determine,
a. the energy efficiency of the pump,
b. the rate of exergy destruction of the pumping
process,
c. the exergy efficiency of the pump.
5.63
Combustion gases enter a gas turbine at 950°C, 800 kPa
and 100 m/s and leave at 650°C, 400 kPa and 220 m/s.
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 215
Taking cp=1.15 kJ/kgK and k=1.3 for the combustion
gases, and surroundings at 100 kPa, 27°C, find,
a. the exergy of the combustion gases at the turbine
inlet,
b. the work output of the turbine under reversible
conditions,
c. the exergy based efficiency of the turbine.
5.64
Air at 20°C as shown in Figure 5.67 is to be heated
to 55°C by mixing it in steady flow with a quantity
of air at 100°C. Assume that the mixing process
takes place at constant pressure and the chamber
is adiabatic. Neglecting the changes in kinetic and
potential energy, determine,
a. the ratio of mass flow of air initially at 100°C
to that initially at 20°C.
b. the exergy efficiency of the heating process, if
the atmospheric temperature is at 20°C.
Figure 5.67 Mixing chamber
5.65
A liquid is heated at approximately constant pressure
from 15°C to 85°C by passing it through tubes which
are immersed in a furnace. The furnace temperature
is constant at 1450°C. Calculate the exergy efficiency
of the heating process when the atmospheric temperature is 15°C. Take the specific heat of liquid as
6.35 kJ/kgK.
5.66
The condenser in a refrigerator receives R134a at
8 bar and 60°C and it exits as sub-cooled liquid at
30°C. The flow rate is 0.05 kg/s and the condenser
has air flowing in at an ambient temperature of 17°C
and leaving at 27°C.
a. the exergy change of each stream
b. the exergy based efficiency of the exchanger
5.67
In a power station, the saturated steam is generated
by transferring heat from hot gases in a steam boiler
as shown in Figure 5.68. Find the increase in unavailable energy due to irreversible heat transfer process.
The gases are cooled from 950°C to 500°C and all
the heat from gases transferred to water. Assume
water enters the boiler as saturated liquid and leaves
as saturated vapor at 200ºC. Take the cp for gas as,
cpg= 1.0 kJ/kgK, and the temperature of surroundings as, 20°C. Obtain the results on the basis of 2.2
kg/s of vapor flow rate. Determine also the exergy
efficiency of the boiler.
Figure 5.68 A three drum steam boiler
5.68
The exergy based efficiency of a refrigeration plant
is 42-percent. The refrigerated space is maintained
at +1°C by removing heat at a rate of 382 kJ/min,
while the surroundings is at 27°C. Determine,
a. the COP of the refrigeration plant,
b. the electric power input, if the energy based
efficiency of the electric motor driving the
compressor is 82%.
c. the amount of exergy lose in kW-h for eight
hours of operation.
5.69
As shown in Figure 5.69, the high temperature heat
source for a cyclic heat engine is a SSSF heat exchanger where R134a enters at 100°C, saturated vapor
and exits at 100°C, saturated liquid with a flow rate
of 3.2 kg/s. Heat is rejected from the heat engine to
an exchanger where air enters at 125 kPa, 25°C and
exits at 110 kPa, 75°C. The rate of exergy destruction
of the overall process is 178 kW. Determine,
Figure 5.69
a. the mass flow rate of air,
b. the power produced by the heat engine,
c. the exergy efficiency of the heat engine.
Miscellaneous systems
5.70
A 30L can of compressed air is suspended in a large
atmosphere the pressure of which is 0.5Mpa and the
temperature is 40C. Initially the pressure of air in
216
THERMODYNAMICS
the can is 1.5Mpa. A small hole is made in the can
allowing air to escape slowly to the atmosphere,
and the pressure reduced to atmospheric pressure.
The process is carried out slowly enough so that the
temperature of the air in the can is always the same
as its environment. Determine,
a. the amount of mass escaping the can,
b. the heat interaction of the can during the process,
c. the exergy destruction of the process.
5.71
A granite rock of Ephesus (density 2700 kg/m3,
specific heat 1.017 kJ/kg.K) with a mass of 3200
kg heats up to T = 53°C during daytime due to solar
heating. Assuming the surroundings to be at 27°C,
determine,
a. the maximum amount of useful work that could
be extracted from the rock,
b. what would the answer be in “a” if the rock
temperature was less by 5°C,
c. what would the answer be in “a” if the ambient
temperature decreased by 5°C?
5.72
The 0.2 m3 tank of Figure 5.70 initially contains
saturated vapor R134a at 28°C. The tank is charged
to 1 MPa pressure, and the refrigerant is in thermal
equilibrium with surroundings at the final state of
28°C. The supply line carries R134a at 1.5 MPa and
30°C. Determine,
a. the heat transfer of the process
b. the wasted work potential.
cooled down to +20°C by passing through plate type
of heat exchangers and cold water at +15°C is used
for this purpose. Finally, the milk is refrigerated
back +5°C in a cold storage room. To save energy,
the plant installs a regenerator and preheats the hot
water entering the boiler. The combustion gases
are at 820°C, and 1bar pressure (cp = 1.10 kJ/kgK)
and the temperature drops to T10 = 300°C before
entering the regenerator. The stack gases leave the
system at T11 = 140 °C. Assume the temperature of
surroundings be at 15°C and evaluate,
a. the fuel consumption rate, take the heating value
of natural gas as Hu = 37800 kJ/kg,
b. the combustion gas mass flow rate,
c. the mass flow rate of hot water through the
circulation pump,
d. the temperature, T4, at the exit of regenerator,
e. the exergy efficiency of the regenerator
f. overall exergy loss due to pastuerization process
Figure 5.71 Milk pasteurization plant with
regeneration
True and False
5.74
Figure 5.70
5.73
In a diary industry in Chicago, milk at +5°C is pasteurized continously at +70°C at a rate of 5.5 L/s for 24
h/day and 365 day/year (cm= 3.93 kJ/kgK, m=1.023
kg/L). First, the milk is heated to the pasteurizing
temperature by hot water which is at T1=85°C at the
exit of a natural gas fired boiler having an energy
efficiency of 78%. The temperature of hot water
drops to T2 = 37°C at the exit of the exchanger. As
shown in Figure 5.71, the pastuerized milk is then
Answer the following questions with T for true and
F for false.
a.
To increase the exergy transfer by
heat transfer from high temperature to low
temperature the temperature difference should
be decreased.
b.
The exergy change of a system is expressed as du  p0 dv  T0 ds .
c.
A system becomes dead when the
enthalpy of the system assumes a value at environmental conditions.
d.
A sudden acceleration of a car causes
more fuel exergy loss than a slowly accelarating
one.
e.
Any reversible cycle converts exactly
the same amount of exergy as it receives.
CHAPTER 5 SECOND LAW EXERGY ANALYSIS OF SYSTEMS 217
f.
Energy is degraded each time as it flows
through a finite temperature difference.
u.
There are processes for which the exergy
destruction becomes negative.
g.
The exergy content of universe is not
affected by the irreversibilities involved in daily
activities.
v.
When an automobile breaks to rest
its exergy due to kinetic energy is completely
destructed.
h.
Direct mixing of warm stream with
a cold one would result with higher mixing
temperature than mixing after extracting some
work from warm stream.
w.
i.
j.
Exergetic efficiency of a heat engine
cannot be greater than its thermal efficiency.
k.
In domestic hot water systems, the
exergy destruction may be reduced by using
the discarded water for heating the fresh cold
water.
l.
Check Test 5
The exergy change of a flowing stream
is determined by dh  T0 ds .
The maximum work obtainable from a
Choose the correct answer:
1.
2.
Exergy based efficiency is for comparing
two systems. However, energy based efficiency
measures the performance of a system.
n.
The actual work which a system does
is always less than the system exergy change,
and the difference represents the irreversibility
of the process.
o.
A nozzle with small pressure gradient
is better than a nozzle having large pressure
gradient. Both operate at the same inlet and
outlet conditions.
p.
q.
s.
The exergy of a system is totally independent of the energy and the entropy of that
system.
t.
3.
The exergy of a block of ice decreases
as it melts.
b.
145
c. 150
d.
158
All four engines, operating between temperatures
957°C and 27°C, reject heat at a rate of 400 kW to
low temperature reservoir. The heat rate received
and the work produced by each engine is indicated
below. Which engine is reversible?
Engine
Q H (kW )
Wnet (kW )
A
B
C
D
1500
1400
1800
1600
1100
1058
1400
1200
A
b.
B
c. C
d.
D.
A 10 g of bullet travelling horizontally with a speed
of 800 m/s at a height of 30m above the ground is at
200°C and has a specific heat of 0.16 kJ/kgK. For
p0  100 kPa, T0  270 C , the exergy of the bullet
in kJ is,
In comparing two heat exchangers
operating with the same fluids at the same inlet
conditions, the heat exchanger with larger heat
transfer surface area performs better.
When you heat a glass of water you
increase its exergy.
132
a.
A water heater with two gradual heating
tanks consumes less energy than a heater with
single tank.
r.
0.1kg of water at 100 kPa, 2000C can be isothermally
compressed in a piston-cylinder arrangement to a
final volume of 6.1% of the initial volume. Which
one of the following might represent the work done
on the system in kJ?
a.
non-flow system is h1  h0   T0 s1  s0 
m.
An evacuated space has a negative
exergy
a.
4.
3.261
b.
2.261
c. 4.261
d.
3.521
100 g of ice at 00C is mixed with 1kg of lemonade at
400C. Assume lemonade has the same properties as
water, c p  4.18 kJ/kgK, and the latent heat of ice
is 333.5 kJ/kg. For T0  270 C , the final temperature
and the destructed exergy respectively are,
a.
29.10 C 3.9kJ
b.
27.10 C 3.9kJ
c.
29.10 C  4.9kJ
d.
27.10 C  4.9kJ
218
5.
THERMODYNAMICS
For an environment of p0  100kPa, T0  270 C , the
unavailable energy of 50 kg of water at 90°C in kJ is
a.
8,951.9
c. 10,951.9
6.
11,951.9.
b.
11.42
c. 12.42
d.
13.42.
Air expands through a turbine from (5 bar, 500°C)
to (1 bar, 200°C). During the expansion process, 50
kJ/kg of heat is lost from turbine surface at 350°C
to surroundings at ( p0  100kPa , T0  270 C ). The
work (kJ/kg), the exergy destruction (kJ/kg), and the
exergetic efficiency respectively are,
241.5, 14.56, 80%
b.
c. 251.5, 14.56, 95%
9.
d.
10.42
a.
8.
9,951.9
One end of a fire hose 5 cm in diameter is held 30
m above the ground to extinguish a building fire.
Water at 27°C exits the hose at a speed of 15 m/s.
For an environment of p0  100kPa, T0  270 C ,
if water is sucked from a well 5 meters below the
ground level, the minimum amount of power (kW)
needed for sprayed water is
a.
7.
b.
34%
b.
35%
c.
36%
d.
37%.
Same fluids having the same mass flow rate and
temperature drop flow through an heat exchanger.
The exit temperatures of both hot and cold fluids and
the temperature of the environment respectively are
The , Tce , T0 . If the temperature change is T , the
exergy efficiency of the exchanger becomes
T0  T
ln 1 
T  Tce
 T
T
1  0 ln 1 
T  The
1
a.






c.
T0  T
ln 1 
T  Tce

T
T
1  0 ln 1 
T  The






d.
T0  T
ln 1 
T  Tce

T
T
1  0 ln 1 
T  The






1
10.
d. 251.5, 14.56, 90%.
a.






1
251.5, 14.56, 85%
The energy based efficiency of an automobile engine
is 23% and the efficiency of a reversible engine operating at identical conditions is 62 %. The exergetic
efficiency of this engine is,
b.
T0  T
ln 1 
T  Tce

T
T
1  0 ln 1 
T  The
1
11.
12.
A flywheel having 1.5 kgm2 of moment of inertia
rotates at a speed of 3000r.p.m. in an insulated space.
The shaft bearings for which m=3.2 kg, and c=2.1
kJ/kgK heat up due to friction and the flywheel
comes to rest. The final temperature of the bearings
in K and the exergy destroyed (kJ) by this process
respectively are,
a. 302, 70.6
b. 303, 72.6
c. 304, 71.6
d. 304, 72.6
The temperatures of cold and hot water streams at
the inlet of a heat exchanger are 1000 C and 200 C
respectively. Both fluids having the same mass flow
rate and the same temperature change ( T  400 C ),
the exergy based efficiency of the exchanger is,
a. 33.1%
b. 37.1%
c. 35.1%
d. 39.1%
A water heater, having 10% energy loss through
the insulation, raises the water temperature from
0
T1  270 C to T2  47 C . The exergetic efficiency
of water heater is,
a. 3.0%
b. 2.9%
c. 2.8%
d. 2.7%.
C
H
6
A
P
T
E
R
Entropy: A System Disorder
6.1
Introduction
It is clear from the previous chapter that entropy is a useful property and serves as a
valuable tool in exergy analysis of engineering devices. However, we do not exactly know
what the entropy means. Our understanding of entropy will deepen as we continue making
use of it.
When viewed microscopically, entropy is a measure of molecular disorder, and the
entropy of a system increases whenever the molecular randomness or uncertainty of the
system increases. As shown in Figure 6.1a, we know from experiments that a paddle-wheel
inserted into a tank containing a gas at high pressure and temperature will not be rotated.
This is because the gas molecules are disorganized and we cannot extract useful energy
from disorganized molecules. Similarly, in Figure 6.1b, the magnetic field around the unmagnetized iron is randomly oriented. This randomness is what causes the magnetic field of
each domain to be cancelled out by the magnetic field of another domain. As a result there
is no single northpole or southpole. A bunch of north and south poles cancels each other
strength and no work can be produced.
As shown in Figure 6.2, however, consider the case of organizing the gas particles to
a certain extent by letting it flow through a pipe as in Figure 6.2a, this time, since the molecules are partially organized to flow in the same direction, it will be possible to extract
some useful energy. After the flow process, however, the energy of the gas is degraded,
and the ability to do work is reduced. Due to increase in molecular disorder an increase in
entropy will be noted. Similarly, as in Figure 6.2b, when the molecules of an iron bar are
realigned, that piece becomes a powerful magnet with a single north and south poles and
creates a magnetic field.
219
220
THERMODYNAMICS
From statistical point of view, entropy indicates the uncertainty about the positions of molecules
at any instant. At high temperatures, even for solids, the molecules oscillate with high frequency and
create relatively large uncertainty about their position. As the temperature decreases, however, these
oscillations fade away, and molecules become motionless at absolute zero. Thus, at absolute zero,
the entropy of a pure substance is zero. This is because there is no uncertainty about the position of
molecules.
Principle 20: The entropy of a pure substance at absolute zero temperature is
zero. This principle provides a reference for determining the absolute entropy
of pure substances and especially used in analyzing combustion processes.
A system can only generate, not destroy entropy. This means that the natural direction of a change
in state of a system is from order to disorder. However, energy conversions may proceed in such a
way that the entropy of a system may decrease. Charging a battery, freezing ice cubes, or formation
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 221
of sand dunes in desert are examples entropy reduction processes. In each of these examples, order
has been won from disorder and entropy has decreased. If the system together with its environment
is considered, the total effect is an increase in disorder. In Figure 6.3, you may clearly see that the
sand has ordered itself into ripples which are caused by the wind blowing over the desert. Hence, the
increase in the order of the sand was accompanied by a larger increase in the disorder of the wind and
the overall effect is an increase in the disorder.
As a result, the quantity of energy is always preserved during an actual process but the quality
is bound to decrease. After a process, the system energy becomes disorganized to a certain extend.
Therefore, the concept of entropy, as a measure of disorganized energy, indicates that the energy, the
exergy, and the entropy balances of a system are not independent from each other.
In fact, as shown in Figure 6.4, these three are interrelated. There are cases for which each of these
three properties is free of others. For instance, electrical energy is an entropy free energy. Similarly,
air at atmospheric conditions contains energy but no exergy (dead state). The most systems appear
222
THERMODYNAMICS
at the shaded region in Figure 6.4. Steam used in industrial applications possesses energy, exergy,
and entropy.
Referring to Figure 6.4, the following relation makes possible to evaluate the third one by knowing the other two.
(Entropy balance) = (Energy balance) – (Exergy balance)
(6.1)
Depending upon the system properties, this general statement can be cast into rigorous equations
as described in the following sections.
6.2
Entropy Balance for Closed Systems
6.2.1 Closed Systems
Referring to the energy and exergy equations for a closed system as given in chapters 4, and

 
  X  I and the side by side difference of these two equations yields,
5, E  Q  W and 
Q
W
 T 
E  
o
j
Q j
Tj
 poV  I .
Together with the definition of entropy (Eq. 5.24), one may state that the rate of entropy change
for close systems is,
Q j
I

S  
T0
j Tj
(6.2)
or expressing in words,
Since there is no work term in Eq. (6.2), the increase in entropy of a system, the increase in the
system disorder, is solely caused by heat transfer. However, if the process is an irreversible one then
the degree of irreversibility is another cause for the entropy production of the system. Thus, for an
adiabatic system, the entropy production of the system results from the irreversibility of the process
and identified as the entropy generation of the system. The entropy generation mathematically expressed as,
I
I
S gen 
or ( S 2  S1 ) gen  12
T0
T0
(6.4)
For an adiabatic system, the entropy generation of the system is proportional to the irreversibilty
of the process and the constant of proportionality is 1/T0. If a system is adiabatic and the process is
reversible, then Eq. (6.4) reduces to S gen  0 . In other words, a reversible and adiabatic process is
also called constant entropy process or simply “isentropic process”.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 223
Recall that an isolated system is defined as constant energy system. As a result of Eq. (6.4), one
may also define the isolated system in terms of the system entropy.
Definition: As an isolated system undergoes a process, the system entropy always increases. If the
process is a reversible one then the system entropy is kept constant. Hence, in general, for an isolated
system, one may state that S iso  0 .
The results of Eq. (6.2) should not be misunderstood. For instance, a constant entropy process does
Q j
I

not always mean a reversible and adiabatic process. For a process, if 
then respect to Eq.
T0
j Tj
(6.2) the system entropy change becomes zero. However, the system is not adiabatic and the process
Q j
I
is not a reversible one. Moreover, for a process, if 
then the system entropy change

T0
j Tj
could be negative. It means the system losing so much heat energy that the uncertainty in molecular
disorder is reduced.
Example 6.1: An ideal gas in a piston-cylinder device changes states through an isentropic process (constant entropy).
Determine the p-V relationship for that gas.
Solution:
Referring to the entropy equation for ideal gases (Eq. 5.28), for an isentropic process, s 2  s1  0 , and the equation transc
c
c / cv
forms into  p2 / p1  v  v1 / v2  p or  p2 / p1   v1 / v2  p
and considering the definition of the ratio of specific heats as,
k = cp/cv, the equation for the process results as, pv k  Constant .
Example 6.2: Drive the isentropic shaft work relation for an incompressible substance flowing through a pipe.
Solution:
The transport of liquids in pipes is of primary importance in numerous engineering designs. A pump may or may not be part
of the control volume, and the pipe may have different diameters at different sections of the flow, or the fluid may undergo
a considerable change in elevation. Considering all these possibilities, and referring to Eq. (4.54), the energy equation is
in the following form,
1
1
(h  V 2  gz )e  (h  V 2 gz )i  q  w
2
2
For incompressible flow, h  u  vp and the above equation for shaft work can be modified as,
 wshaft  u  vp  ke  pe  q
For a case of fluid having the same temperature as surroundings, or for flow through an insulated pipe, the heat transfer will
be zero ( q  0 ). In addition, due to internally reversible flow process, s  0 , and Eq. (5.27) states that T2  T1 and that
u  0 . Thus the energy equation for isentropic and incompressible process reduces to
 wshaft  vp  ke  pe
Hence the shaft work for an incompressible, frictionless fluid flowing through a pipe is due to change of pressure, velocity
and elevation in general. In case of excluding the pump work from the control volume, the steady flow energy equation
becomes,
vp  ke  pe  0
or
v(p2  p1 ) 
V22  V12
 g( z2  z1 )  0 Bernoulli’s equation
2
224
THERMODYNAMICS
This is known as the Bernoulli’s equation in fluid mechanics, and is used for energy analysis of isentropic flow of incompressible fluids through pipes and ducts.
Example 6.3: Refrigerant R12 is compressed isentropically from (1bar, x=0.957) to a pressure of 6 bars in a piston-cylinder
device. Determine,
a.
the final temperature of the refrigerant
b.
the specific work done to the refrigerant
Solution:
Together with Eq. (5.31), the table values of entropy at saturated state yield the initial entropy as s1 = 0.6877 kJ/kgK, and
for isentropic processes s1=s2, p2=6 bar. Hence the temperature at the final state becomes T2 = 22C, saturated vapor state.
Simplifying the energy equation for a closed system with no heat transfer, W12   u2  u1  where u1  u1 f  x1u1 fg  151.77
kJ/kgK, and at the final state (p2=6 bar, x2=1.0) u2=179.09kJ/kg. Thus the energy equation yields W112  27.32 kJ/kg.
Example 6.4: The gear train of a car is composed of ten
helical gears each of which is made out of pressed steel
(c=0.465 kj/kgK) and each of the first five has a mass of 1kg
and each of the other five is 2 kg. In a gear manufacturing
plant The gears are initially at 350C and submerged into
a tank containing 200L of water at 22C. For an ambient
temperature at 22C, determine the amount of entropy generated and irreversibility due to heat treatment of gears.
Solution:
For a system that consists of gears and water in the tank,
energy balance yields, mg u2  u1 g  mw u2  u1 w  0 ,
and substitution of numerical values gives the final temperature as, T2=24.713C.
297.713
S g  15  0.465  ln
 5.1504 kJ/K, similarly for water,
The entropy change of the gears is
623
297.713
S w  200  4.18  ln
 7.616 kJ/K, and the entropy generation of the process is the same as the entropy change
295
of the overall system. Thus, S gen  S w  S g  2.4390 kJ/K. Then by Eq. 6.4, the irreversibility of the process is
I12  T0 S gen  719.514 kJ.
6.2.2
Thermodynamic cycles
Previously a thermodynamic cycle is treated as a closed system, thus the time rate of entropy
production of the working fluid can be expressed by Eq. (6.2), and the cyclic integral respect to time
may be expressed as following,
 
 Sdt

Q

T
I
 T
(6.5)
0
Together with the definition of a thermodynamic cycle, since entropy is a system property, the
  0 . For all cyclic processes, the
cyclic integral of entropy of the working fluid must be zero; Sdt
irreversibility is never negative;  I  0 , then, respect to Eq. (6.5), one may conclude that for all
possible cycles the following statement must hold.

Q
 T
0
(6.6)
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 225
As stated for a cycle, this relation leads to Clasius inequality which was first described by German
Physicist R.J.E.Clasius in 1870. It is an important criterion in testing the 2nd law appropriateness of
a cycle and as can be deduced from this statement that the cyclic integral of Q/T is always negative
for real cycles. Specifically, for reversible cycles, it assumes the value of zero at the limit.
Example 6.5: A heat engine withdraws 325 kJ of heat energy from high temperature reservoir at 1000K and rejects 125 kJ
of heat to a low temperature reservoir at 400K. The net work output of the engine is claimed to be 200 kJ. Is it possible to
run such an engine at these conditions?
Figure 6.6 Schematic of the problem
Solution:
For a cycle first law states that Qnet  Wnet or 325-125=200 kJ and the engine satisfy the energy balance requirement.
However, Applying Eq. (6.5) to the proposed heat engine yields,
Q
 T

Q1 Q001 325 125



 0.0125 kJ/K
T1
T0 1000 400
0 which tells us that the engine is inappropriate in terms of the second
law and is not possible to run such an engine at engine by these conditions.
Example 6.6: The ice surface of an ice skating ring in İzmir experiences
heat transfer with air in the arena at a rate of Q r  2500T0  Tr  kJ/h,
and is to be withdrawn by a heat pump that is used to warm up a close
by facility. The heat loss of the facility is Q  5000T  T kJ/h.
h
h
0
a.
Determine the minimum driving power for the heat pump for
Tr=-12C, Th=21C, and T0=0C.
b.
Determine the ambient temperature for which the heat exchange
of the heat pump with the surroundings changes sign.
Solution:
a.
First law applied to the refrigeration system yields
Q  Q  Q  W . Assuming a reversible cycle, Clasius
0
r
h
p
Q
Q
Q
inequality becomes 0  r  h  0 and together with the
T0 Tr Th
energy equation, the compressor work can be expressed as,
 T 
 T 
W p  1  0 Q h  1  0 Q r . For a reversible cycle, the
T
h 
 Tt 

minimum amount of work is used at the compressor and equals
to 2.465kW.
226
THERMODYNAMICS
b.
At the temperature where the heat exchange between the pump and the surroundings changes sign, the heat transfer
2500(T0  Tr ) 5000(Th  T0 )
must be zero, Q 0  0 and together with Clasius inequality,
. Substituting numeri
Tr
Th
cal values for Tr and Th, the surroundings temperature may be determined as T0=282K.
6.3
Entropy Balance For Open Systems
Similar to the principles applied for formulating the energy equation, the mass crossing the
system boundary possesses entropy as well as energy. Therefore, the net entropy transfer due to
convection may be expressed as,
The net rate of entropy



accumulation by convection  
at instant of time t



In evaluating the term, 
j
Q j
Tj
 m s   m s
i i
(6.7)
e e
i
e
, in Eq. (6.2), a system has to have a finite number of isothermal surfaces.
For a case of continuous variation of temperature along the system boundary, however, this term has
 q 
to be transformed into an integral and expressed as,   dA . Hence, in a general format, the rate of
T
A

entropy change of a system may be calculated as following,
 dS 
 q 
 dt    T  dA 

cv A  

 m s   m s
i i
i
e e
 S gen
(6.8)
e
where q (kW/m2) presents the rate of heat transfer per unit surface area of the system. Unlike the exergy
equation (Eq. 5.37), in Eq. (6.8), there is no need for identifying the reference state of the process,
and entropy analysis can be performed without referring to any state of the system.
One may recall that the system energy and the exergy are invariant in time for Steady State Steady
Flow Systems (SSSFS). Since entropy is a system property, likewise the system entropy is invariant
 dS 
with time for SSSFS. Hence, substituting 
  0 into Eq. (6.8) and rearranging results as,
 dt cv
q
 m s    T  dA   m s  S
e e
e
i i
A
i
gen
(6.9)
If a SSSF system is adiabatic having single inlet and outlet, then Eq. (6.9) may be reduced to the
following form
se  si  sgen
(6.10)
Due to irreversibilities in the flow; friction between surface and particles, and within particles, sgen
is always a positive quantity and hence for SSSF and adiabatic systems with single inlet and outlet,
Eq. (6.10) reveals that se  si is true.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 227
As a consequence, one may state that exergy loss through a process results with entropy gain
through that process. For processes through which the exergy is conserved (reversible processes), the
entropy is also conserved, and does not increase. This is called the principle of entropy increase.
Principle 21: Entropy is a system property. Through all real processes, some
entropy is generated and this generation of entropy is due entirely to the non-conservation of exergy.
This principle assists us in identifying the real processes that might exist and the direction through
which these processes may possibly take place.
Example 6.7: As shown in Figure 6.8, the ducting of a ventilation system operating at steady state is well insulated, and
the pressure is very nearly 1atm throughout. Assuming ideal gas model for air with cp=1.005 kj/kgK, and ignoring kinetic
and potential energy effects, determine
a.
the temperature of air at the duct exit,
b.
the pipe diameter at the exit,
c.
the rate of entropy production within the duct in kW/K
Solution:
a.
Neglecting the changes in kinetic and potential energies, the energy equation for a control volume around Figure 6.8,
results as, m 1h1  m 2h2  m 3h3 and the conservation of mass yields, m 3  m 1  m 2 . The mass flow rates at the corresponding sections, m 1  1 ( AV )1 1 =
and m 2   2
100
 1  1.603 kg/s . Similarly,  2  1.149 kg/m3
=1.283 kg/m3 and m
0.287 x 283
 D2
V2 . After substituting the numerical values, m 2  4.464 kg/s and m 3  6.067 kg/s . For ideal
4
gases, h=cpT, the energy equation may be expressed in terms of temperatures, and the temperature of air at the
exit becomes, T3 
b.
1.603  282  4.464  303
 297.7 K
6.067
Since m 3  3 A3V3 and 3 
100
 1.17 kg/m3 , then the diameter of the circular cross section becomes
0.287  297.7
D3  1.73 m .
c.
Since the system is at steady state and insulated Eq. (6.9) reduces to S gen  m 1 ( s3  s1 )  m 2 ( s3  s2 ) , and using the
entropy relation for ideal gases, Eq. (5.30), the entropy generated through the mixing action is, S gen
2.412 x10 3 kW/K .
228
THERMODYNAMICS
As mentioned previously, the principle of entropy increase, Eq. (6.8), allows us to identify processes
that could never occur, irrespective of the details of the system. A process that violates equation (6.8)
also violates the second law and is obviously impossible. The following Example numerically illustrates how to use this principle.
Example 6.8: As shown in Figure 6.9, two entering streams of air mix to form a single exiting stream. A hard to read photocopy of the data sheet indicates that the pressure of the exiting stream is either 1.8MPa or 1.3MPa. Stray heat transfer and
kinetic and potential energy are negligible. Assuming the ideal gas behavior for air with cp=1.02 kJ/kgK, determine if either
or both these pressure values can be correct
Solution:
The energy equation, m 1h1  m 2 h2  m 3h3 , yields the exit temperature as, T3  (1.2  900  2.2  500) / 3.4  641.1K . For
steady state conditions, the entropy equation, Eq. (6.8), becomes S gen  m 1 ( s3  s1 )  m 2 ( s3  s2 ) . Since the entropy generation term is always positive, m 1 ( s3  s1 )  m 2 ( s3  s2 )  0 . Substituting the entropy difference for ideal gases, Eq.(5.30), the
relation for the exit pressure is ln( p30.975 / 1.269)  0.14261, or p3  1.476 Mpa. The correct exit pressure is then 1.3MPa.
6.4
Temperature-Entropy (T-s) Diagram
2

In chapter 4, the work due to moving boundary of the system was expressed as, W12  pdV , where
1
p presents the pressure at the moving boundary. For an ideal process, however, the system properties,
like pressure, temperature etc., must be uniform throughout the system. Gradients of any property
should not be allowed. Otherwise, due to internal irreversibility, an exergy loss will take place. As a
2
consequence, the work done due to change of system volume may be represented by
 pdV , where
1
p must be the system pressure not the pressure at the boundary at any instant of time t. For an ideal
process then the area under the curve in p-V diagram represents pdV type of work of the system. Due
to non-uniformities in thermodynamic properties for an irreversible process, however, the system
work is always be less than the shaded area in Fig. 6.10.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 229
Figure 6.10 The Ideal mechanical work of the moving boundary is represented by the shaded area.
Similarly, for a reversible change of state, the system should not experience any temperature gradients, and at an instant of time t, the system temperature must be fixed at a value of T. Hence, for a
Q
reversible process of a closed system, Eq. (6.8) reduces to S  . Integrating this relation between
T
the initial and the final states yields,
2

Q12  TdS
(6.11)
1
This relation provides us to calculate the amount of reversible heat transferred through a closed
system and is represented by the shaded area in Fig. 6.11a. Recall that for a non-ideal process, the work
2
transfer of the system is always less than;
 pdV . Likewise, for an irreversible heat transfer process,
1
2

one may state that the relation between the heat transfer and the entropy variation is Q12  TdS .
1
Up to this section, the pressure, p and the volume, V are used for representing the processes of
the system. Likewise, since T and s are two independent properties of the system, they can also be
utilized in identifying various reversible processes. On this respect, Fig. 6.11b shows isentropic, isochoric, isobaric, and isothermal processes. Referring to Eq. (5.28), isochoric and isobaric processes
are logarithmic, and since cp>cv, the slope of isochoric process is greater than isobaric one.
230
THERMODYNAMICS
Figure 6.11 The Ideal heat transfer to a system and T-s presentation of various ideal processes
Example 6.9: As shown in Fig. 6.12, steam enters to an adiabatic turbine at (16 bar, 350C) and outlets at 1bar pressure as
saturated vapor. Determine,
a.
if the turbine can run under given conditions,
b.
the mass flow rate for a shaft power of 150kW,
c.
the mass flow rate if the turbine operates isentropic for the same shaft power.
d.
Show both processes on T-s diagram.
Solution:
a.
Referring to the steam tables, the entropy at state 1 is s1=7.069kJ/kgK and at state 2, s2=7.359kJ/kgK, and the
results are consistent with Eq. (6.10); s2> s1. Hence, the turbine can run under these conditions.
b.
Considering the energy equation (Eq. (4.60)), and h1=3145.4kJ/kg, h2=2675.5kJ/kg, the mass flow rate can be
computed as 0.319kg/s.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 231
c.
For an isentropic turbine s1=s3=7.069kJ/kgK. At the
turbine exit, steam is at saturated state and the vapor
quality is calculated by the entropy equation for twophase systems (Eq. (5.31)) as x3=0.952. Then the enthalpy
at state 3 is evaluated by, h3=h3f + x3(h3g-h3f)=2567 kJ/
kg. Thus the mass flow rate of steam for isentropic
turbine can be determined by Eq. 4.60 as 0.259kg/s.
The isentropic turbine consumes 18.8% less steam for
producing the same shaft power.
d.
The graphical illustration is given in Figure 6.13. As
shown in the figure, the exit temperature of the steam
expanding through an isentropic turbine is less than
the one obtained by adiabatic turbine
Example 6.10: As shown in Fig 6.14, an adiabatic air compressor compresses the ambient air at rate of 0.05 kg/s, from (1 bar,
25C) to (5 bar, 210C). Determine,
a.
if it is possible to run the compressor,
b.
the shaft power consumed by the compressor,
c.
the shaft power if an isentropic compressor were used for the same conditions.
d.
Show both processes on T-s diagram.
Solution:
a.
Assuming that the air behaves like an ideal gas with constant specific heats, then the entropy change of air due to
483
5
compression may be evaluated by Eq. (5.30) as, s2  s1  1.005 ln
 0.287 ln , or s 2  s1  0.0243kJ/kgK  0
298
1
.
Thus it is possible that the compressor might work under these conditions.
b.
For an adiabatic compressor the shaft power is W shaft
c.
p 
T
If air is compressed through isentropic process, then the air exit temperature by 2 s   2 
T1  p1 
m h1 h2   9.296kW .
k 1
k
becomes,
 p T1  T2 s   8.713kW
T2s=471.4K, and the shaft power is, Wrev  mc
d.
The graphical illustration is given in Fig. 6.15. As shown in the figure, fluid temperature at the outlet of an adiabatic
compressor is larger than the one obtained by isentropic compression to the same pressure.
232
THERMODYNAMICS
6.5
Enthalpy-Entropy (h-s) Diagram
Another diagram that is very important in engineering for analyzing steady-state and steadyflow systems is the enthalpy-entropy diagram. As described by Eq. (4.52), enthalpy is an important
parameter of flow systems for energy analysis. In addition, entropy is a property used in defining
the exergy (available energy) of systems. Hence, for flow systems, and for processes involving both
energy and exergy interactions may clearly be illustrated on an h-s diagram. The h-s diagram is also
called Mollier diagram after the German scientist R. Mollier. The general features of this diagram
are shown in Fig. 6.16.
Recall that at a saturated state on T-s diagram, the constant pressure curve is horizontal. Unlike
the T-s diagram, the constant pressure lines continue declining even for saturated states. In this graph,
the saturation temperature at a particular pressure can be determined by the intersection of constant
pressure and constant temperature lines at the saturated vapor line.
The measured vertical distance between the predefined states on this diagram presents the enthalpy
change, h, and that means the shaft work of an adiabatic turbine or work consumed by an adiabatic
compressor. In addition, the enthalpy change, h, might mean a kinetic energy change for a flow
through a nozzle, or through a diffuser. The horizontal distance between two specified states on this
graph indicates the entropy generated, ∆s, through an adiabatic flow process.
6.6
6.6.1
Some Relations for Flow Processing Devices
The reversible mechanical work of steady flow machines
In accord with Eq. (5.37), for a reversible process of a pure, compressible substance flowing
through a steady-state device, the conservation of exergy in differential form may be stated as,
q
 w
0   dh  T0 ds  dke  dpe  q  T0
(6.12)
T
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 233
where, dh  q  vdp , and the entropy balance relation (Eq. 6.8) for a reversible process becomes
q
ds 
. Substituting these relations into Eq. (6.12) yield the reversible shaft work of a steady flow
T
machine as,
e

wrev   vdp 
i


1 2
Vi  Ve2  g( zi  ze )
2
(6.13)
For a negligible change in kinetic and potential energies, the above relation reduces to
e
wrev    vdp
(6.14)
i
These relations provide quantitative results if a functional relationship between v and p is known,
and lead to the maximum work output or minimum work input for unit mass flowing steadily and
reversibly through turbines, compressors, and pumps. For a polytropic and reversible compression or
expansion process, pv n  Constant , and the integral of Eq. (6.14) yields
wrev
n 1


n
n
  p2  n

p1v1     1

 p2v2  p1v1  
n 1
n 1
  p1 



(6.15)
Example 6.11: Ammonia at 2.0 bar, -15C enters a compressor with a volumetric flow rate of 0.02 m3/s. The refrigerant is
compressed to a pressure of 10 bar by a reversible process expressed as, pv1.15  Constant . Determine,
a.
the power required,
b.
the rate of heat transfer.
Solution:
Since the pressure-volume relation for the compression process is given as, pv n  p1v1n , you may use Eq. (6.15)
as w 
n
( p2v2  p1v1 ) , and V2  0.02 x(2 / 10)1/1.15  0.00493 m3 /s . Substituting into work expression yields,
1 n
1.15
W 
(1000 0.00493 200 0.02)
0.15
7.13 kW
Apply 1st law to control volume around the compressor, Q  W  m (h2  h1 ) , m  V1 / v1  0.02/0.605=0.033 kg/s , then
v2  0.605 x(0.0049 / 0.012)  0.148 m3/kg, and from Ammonia table h2=1549.4kj/kg. Substituting the results into energy
7.13 0.033 (1549.4 1428.51)
equation yields, Q  7.13
6.6.2
3.14 kW
Multi Stage Vapor Compression and İnter Cooling
In high pressure ratio compressors (p2/p1>10), the amount of work consumed by the compressor
becomes comparable with the work generated by the turbine in a cyclic process. Similarly, in refrigeration systems, the performance of a single stage vapor compression system is adequate as long as
the temperature difference (the temperature lift) between the evaporator and the condenser is small.
However there are cases where the temperature lift can be quite high. To keep the frozen food refrigerated, the evaporator temperature can be as low as -30C, and in chemical industries, the liquefaction
of certain gases might require -150C of evaporator temperature.
234
THERMODYNAMICS
Figure 6.17 Mechanical refrigeration cycle and (h-p) representation for various
evaporator temperatures
Figure 6.17 indicates that decrease in evaporator temperature reduces the amount of heat removed
by the evaporator; (h1  h4 )  (h1a  h4 a )  (h1b  h4b ) . However, the amount of work consumed by
the compressor increases; (h2  h1 )  (h2 a  h1a )  (h2b  h1b ) . In short, as the evaporator temperature
drops, the single stage compression becomes inefficient and impractical and two-stage compression
becomes a necessity.
In Equation (6.15), v1, is the specific volume of the gas at the inlet, and the specific work input
to the compressor is directly proportional to this value. Hence, at a specified pressure, reducing the
inlet volume by cooling also reduces the work input as indicated by the shaded area in Figure 6.18a.
It can be demonstrated that, in Figure 6.18b, (h2  h1 )  (h4  h3 )  (h2b  h1 ) .
Intercooling of a gas may be achieved by using a flash tank as in the case of refrigeration systems,
or water cooled heat exchanger which is commonly used in air compressors. In addition to reducing
the compressor work input, intercooling also reduces the exit temperature of the compressed gas which
results with a better lubrication and a longer compressor life.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 235
Determining the intermediate pressure for two-stage compression process so that the total work
consumed attains a minimum value is essential in engineering applications. Assuming that the compressors work reversibly, the total work input may be stated as,
wrev
n 1
n 1




n
n
  p2  n

  pi  n

p1v1     1 
p1v1     1

n 1
  p1 
 n 1
  pi 





(6.16)
In this expression, we assume that the gas is cooled down to the inlet temperature before entering
the second stage. Hence the only variable in Eq. (6.16) is the intermediate pressure, pi, and by letting
the derivative of work equal to zero, one may obtain the following,
pi 
p1  p2
(6.17)
On the basis of reversible conditions and for two-stage compression, the intermediate pressure
has to be the geometric average of the end pressures for minimum work consumption.
Example 6.12: The R22 refrigeration system shown in Figure 6.19 uses two-stage compression with inters cooling by a
flash chamber. The evaporator capacity is 200 kW and operates at -30C, and the refrigerant at state 1 is saturated vapor.
For the flash chamber and the condenser pressures of 600 kPa, and 1500 kPa respectively, determine the total power
required by the two isentropic compressors.
Solution:
The mass flow rate through compressor1 is determined by the energy balance on the evaporator, qe  m 1 h1  h8  , h8=h7=51.8kJ/
kg, and h1=237.7 kj/kg, then m 1  1.075kg/s . Since the compressors are isentropic, s2=s1=0.9787 kJ/kgK, p2=600kPa,
h2=268.2 kJ/kg, and the power consumed by the first compressor is Wc1  m 1 h2  h1   1.075 268.2  237.78   32.7 kW .
The rate of refrigerant flowing through the second compressor may be calculated by the energy balance around
the flash chamber as, m 1 h2  h7   m 3 h3  h5  or, m 3  1.075  268.2  51.8  / 252  93.5   1.466 kg/s .Hence,
Wc 2  m 3 h4  h3   1.466  273.9  252   32.095 kW . The total power consumed is Wt  Wc1  Wc 2  64.795 kW .
236
THERMODYNAMICS
Discussion: The geometric average of the evaporator and the condenser pressures (pi=495.6 kPa) does not match with the
flash chamber pressure. Simply, the gas is not an ideal gas, and it is not cooled down to the inlet temperature of the first
compressor at the intermediate state.
6.6.3 Adiabatic Flow of Incompressible Fluids
In case of neglecting the gravitational potential energy changes for adiabatic and single inlet and
outlet devices, Eq. (5.54) yields,
wshaft   h  ke 
(6.18)
The enthalpy change for an incompressible fluid may be determined as,
h  cT  vp
(6.19)
Referring to Eq. (6.8), for an adiabatic process, s  s e  s i  0 , and together with Eq. (5.27),
one may conclude that the fluid temperature at the exit of the device is greater than the inlet (Te > Ti).
However, in case of isentropic flow, Te = Ti. Thus the isentropic process of an inviscid fluid is also an
isothermal process and the change in fluid internal energy is zero. Depending upon the reversibility
of the flow, the shaft work of a flow machine involving a unit amount of incompressible fluid may
be stated in two-fold:
 vp  ke
wshaft 
- cT  vp  ke 
(isentropic flow)
(actual flow)
(6.20)
Equations 6.13, 14, and 6.19 form the basis for efficiency analysis of various flow devices.
6.6.4
Isentropic Flow of Compressible Fluids
In accord with Eq. (6.10), if the flow is adiabatic and reversible ( sgen  0 ), then the entropy of
the flow stays constant throughout the flow.
Definition: An isentropic flow is an adiabatic and frictionless flow (reversible) for which the entropy
of the fluid is constant throughout the flow field.
Although no real flow is entirely isentropic, the major portion of many flows of engineering
practice can adequately be modeled as steady, one-dimensional and isentropic. As shown in Figure
6.20a, in internal duct flows, the effects of viscosity and heat transfer are usually restricted to thin
layers adjacent to the wall and the rest of the flow field can be assumed to be isentropic. Similarly, in
external flows, the effect of viscosity and heat transfer can be assumed to be restricted to the boundary layers, wakes, and shock waves and the rest of the flow field, as indicated in Figure 6.20b, can be
treated with adequate accuracy as isentropic flow.
As Explained in Example 6.1, for an isentropic change of state of an ideal gas with constant specific heats, the following relations hold,
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 237
p
 constant
k
and
T2   2 
 
T1  1 
k 1
p 
 2 
 p1 
k 1
k
(6.21)
In addition, recalling that the speed of sound in an ideal gas is a  kRT , then for an isentropic
flow,
 
a2
T
 2  2 
a1
T1  1 
k 1
2
k 1
 p  2k
 2 
 p1 
(6.22)
For steady and adiabatic flow of an ideal gas through a stream tube in Figure 6.18, the energy
equation yields,
238
THERMODYNAMICS
1
1
c pT1  V12  c pT2  V22
2
2
k 1 2
1
M1
T2
2

T1 1  k  1 M 2
2
2
or
(6.23)
where, M 1 ( M 1  V1 / a1 ) and M 2 are the Mach numbers at specified states. Substituting Eq. 6.23
into Eq.6.21, one may obtain the following for isentropic flow conditions,

1
p2 

p1 1 

k
k  1 2  k 1
M1 
2
k 1 2 
M2 
2

and

1
2 

1 1 

1
k  1 2  k 1
M1 
2
k  1 2 
M2
2

(6.24)
Example 6.13: Consider compressible and frictionless flow through a stream tube. For such a flow, the pressure-velocity
1
relation, and the temperature-velocity relations are respectively given as, dp   VdV , and dT   VdV . Applying the
cp
ideal gas equation,
d
dV
 M 2
.

V
a.
Show that the density-velocity relation is
b.
Calculate percent variations in density and temperature for Mach numbers; M=0.1, 0.3, 0.4.
Solution:
After differentiating the gas equation one may obtain,
pressure-velocity relation the pressure ratio becomes,
differential form is
d  dp dT
. In addition to gas equation considering the above



p
T
dp
dV
V2
, where, M 2 
. Similarly, the temperature ratio in
  kM 2
p
V
kRT
d
dV
dV
dT
dV
  kM 2
 k  1M 2
. Substituting these two ratios into the gas equation,
  k  1M 2

V
V
T
V
and simplifying results as,
d
dV
 M 2
. Therefore, at M=0.1, the fractional change in density will be 1% of fractional

V
change of velocity. At M=0.3, the fractional change in density will be 9% of fractional change of velocity, and at M=0.4,
the fractional change in density will be 16% of fractional change of velocity.
Let us calculate the temperature variation of air (k=1.4) with Mach numbers at M=0.1, 0.3, 0.4. The fractional change
in temperature will respectively be 0.4%, 3.6%, and 6.4% of the fractional change in velocity.
Comments. The temperature variation becomes increasingly important as Mach number increases. Especially, for flows
at M>0.3, the compressibility effect and the temperature variation have to be considered and become significant.
Example 6.14: Consider compressible and frictionless flow through a stream tube and determine how the flow cross sectional
area changes with respect to Mach number.
Solution:
After differentiating the continuity equation one may obtain,
variation from Example 6.13, substitution yields,
These variations are indicated in Figure 6.22.

d  dA dV


 0 . Since we know the density-velocity

A
V

dA
dV
. It means that for M<1, dA<0, and for M>1, dA>0.
 M 2 1
A
V
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 239
Stagnation conditions. If the flow at any point in the fluid stream is brought to rest isentropically
then the stagnation conditions are obtained. As shown in Figure 6.23, if the velocity is zero at some
point in the flow then the measured values of pressure, temperature and density are stagnation values.
Hence, substituting M2=0 in Eqs. 6.23, and 24, the stagnation parameters become,
 k 1 2 
To  T1 1 
M1 
2


k
 k  1 2  k 1
po  p1 1 
M1 
2


1
 k  1 2  k 1
o  1 1 
M1 
2


(6.25)
The critical conditions are those conditions that will exist if the flow is isentropically accelerated or
decelerated until the Mach number is unity. Hence, for a supersonic nozzle, the critical conditions
take place at the minimum area called the throat. The critical conditions are denoted by the symbols
p* , T * ,  * , A* , V * and are formulated by letting M1 = 1 in Eq. (6.25).
240
THERMODYNAMICS
k
1
 k  1  k 1
po  p 

 2 
 k  1
To  T 

 2 
 k  1  k 1
o   

 2 
*
*
*
(6.26)
Under steady conditions, the mass flow rate of isentropic gas flow can be described in terms the
stagnation properties, the flow cross sectional area, and the flow Mach number as following,
 p
m   AV  
 RT
k

 A kRT M  p0 A
RT0

M
k 1
2  2( k 1)
 k 1
1  2 M 


(6.27)
For specified stagnation values and the flow area, the maximum flow rate is obtained at the condidm
 0 . This yields M=1 and the flow Mach number assumes unity at the minimum
tion for which
dM
cross-sectional area, A*. Thus the maximum flow rate is:
k 1
*
m max  p0 A
k  2  2( k 1)
RT0  k  1 
(6.28)
It is important to note that for constant stagnation properties, the maximum flow rate is a linear
function of the throat area. This fact is widely used in the design of flow meters, medical devices,
and mass flux control systems. Hence there are four distinct possibilities for the maximum flow rate
as following,
a. p0, 0, and A* are fixed
:
m max is also fixed.
b. p0 increases only
:
m max linearly increases.
c. A increases only
:
m max linearly increases.
d. T * increases only
:
m max decreases.
*
Example 6.15: Consider hot combustion gases (k=1.33) flow in an adiabatic and no work system. At one section p1=14
bar, T1=500K, V1=125m/s, and A1=500mm2. At a downstream section M2=0.8. Assume isentropic flow conditions, R=0.28
kj/kgK and calculate,
a.
p2, T2, V2 and A2,
b.
the stagnation pressure and temperature,
c.
the critical values at the throat and the throat cross sectional area.
Solution:
a.
The speed of sound at section 1 is, a1  1.33  280  500  431.5 m / s and M1=0.289. In regard to Eqs. (6.23)
and (6.24),
1.33
T2
p
 0.916 , T2  458 K , 2  0.916 0.33 , p2  9.83bar , V2  a2 M 2 , a2  412.98 m/s and
T1
p1
V2  330.39 m/s . Since m  1V1 A1   2V2 A2 and  2  7.66 kg/m3, 1  10 kg/m3, the mass flow rate becomes
m  0.625 kg/s, and A2  0.625 / 7.66  330.39   246 mm2.
b.
0.33


By Eq. (6.25), the stagnation pressure and temperature are T0  500  1 
 0.2892  ,
2


1.33
 0.33
 0.33 , T  506.8 K , p  1479.39 kPa
p0  1400  1 
 0.2892 
0
0
2


CHAPTER 6 ENTROPY: A SYSTEM DISORDER 241
c.
The critical values are determined by Eq. (6.26), p* 
1479.39
1.33
2.33  0.33

 2 


 799.4kpa , T * 
506.8
 435.02K , and
 2.33 
 2 


 *  6.56kg/m3 . Since M*=1 at the throat, V *  1.33  280  435.02  402.49 m/s , and the continuity at the
throat yields, A* 
0.625
 236.7 mm 2
6.56 x 402.49
Comments. To accelerate the combustion gases, the channel must converge in the flow direction and if the cross sectional
area assumes the critical values, then the gas flow becomes sonic.
Figure 6.24 Operation of converging nozzles at various back pressures
Converging and diverging nozzles. As shown in Figure 6.24, let us consider a pressurized tank with
a converging nozzle attached is situated in a controllable environment. Initially, the stagnation pressure and the pressure of the surroundings are identical. Then the surroundings pressure is lowered
gradually and the flow accelerates. The gas pressure at the nozzle exit is identical with the back
pressure until the flow Mach number assumes unity, M=1, or pb= pe= p*. However, when the back
pressure is lowered further, pb<p*, the gas pressure at the nozzle exit does not respond to this change
and stays at the critical value. As shown in Figure 6.24c, the flow is called chocked at this condition
and is not possible to accelerate the fluid further. The mass flow rate at this condition is determined
by Eq. (6.28). Hence the converging nozzles are useful only for flows for which pb  p * and flow
Mach number is M  1 . The objective of making a convergent-divergent nozzle (sometimes called a
Delaval nozzle) is to obtain supersonic flow.
Figure 6.25 A typical converging-diverging nozzle and the pressure variation for various
flow conditions
242
THERMODYNAMICS
Lowering the back pressure at the nozzle exit relative to the pressure at the nozzle inlet, one may note
several different operational modes as are presented in Figure 6.25b.
Case a: “Venturi mode” For pressures above the first critical; pb  p fc , the nozzle operates like a
venture. The velocity at the throat is less than the sonic value. The nozzle is not chocked and the flow
rate is less than the maximum.
Case b: “the first critical mode” For back pressures at the first critical value pb  p fc the fluid velocity reaches the sonic speed at the throat, and the diverging section operates like a diffuser. The fluid
decelerates, and the flow is subsonic between 2 and 3.
Case c: “the third critical mode” For back pressures at the third critical value pb  ptc the fluid velocity reaches the sonic speed at the throat, and at the diverging section the fluid continues acceleration.
The flow is subsonic from 1 to 2, sonic at 2, and supersonic between 2 and 3. For cases b and c, the
flow variations are identical from the inlet to the throat.
Case d: As shown in Figure 6.25b, in a region, p fc  pb  ptc , a normal shock develops at a section
between the throat and the exit plane which causes a sudden increase in pressure and the flow is nonisentropic.
One may notice by Figure 6.25 that the first and the third critical points represent the only operating conditions that satisfy the following:
a. The Mach number is unity (M=1) at the throat,
b. Isentropic flow throughout the nozzle,
c. The pressure at the nozzle exit is the same as the surroundings.
Example 6.16: As shown in Figure 6.26, a converging-diverging nozzle discharges air into surroundings at 100 kPa of
pressure. A duct having 0.1m2 cross sectional area feeds the nozzle with air at 6.80 bar, 175C, and a velocity such that
the Mach number M1=0.25. The area at the nozzle exit is such that the flow exit pressure exactly matches the surroundings
pressure. Assume perfect gas, and steady, one-dimensional and isentropic flow, calculate,
a.
the mass flow rate,
b.
the throat area,
c.
the nozzle exit area.
Solution:
a. The density and the velocity at state 1; 1  p1 / RT1  680 / (0.287 x 448)  5.28 kg/m3, a1  1.4 x 287 x 448  424.27 m / s
and V1  106.06m / s m  1V1 A1  5.28 x0.1x106.06  56.095 kg/s.
 0.4

 0.4

b. Flow stagnation values are, T0  448  1 
 0.252 
0.252   453.6 K, p0  680  1 
2
2




3.5
 710.2 kpa. At
the throat, M*=1, by Eq. (6.26), p*  375.19 kpa, T *  378 K, and  *  375.19 / (0.287  378)  3.458 kg/m3. The
mass flow rate will be the same at the throat, A*  56.095 / (3.458 x 424.27)  0.038 m2.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 243
c.
 (710.21 / 100)0.285  1 
Since p3 = 100 kPa, by Eq. (6.25), the Mach number at the nozzle exit is M 3  

0.2


0.5
 1.937
and T3  453.6 / (1  0.2 x1.937 2 )  259.2 K, a3  322.7 m/s, V3  625.1 m/s, 3  1.344 kg/m3. Then the continuity equation yields the exit area as, A3  56.095 / (1.344 x625.1)  0.066 m2.
If the surroundings pressure is below the
third critical value, the nozzle operates internally as though it were at the third critical
point but expansion waves or shock waves
occur outside the nozzle. The shock waves
that occur in a plane normal to the direction
of the flow are called normal shock waves.
As shown in Figure 6.27, the flow process
through the shock wave is highly irreversible, but relations for the flow properties
before and after the shock may be developed
by applying and combining the conservation
of mass and energy equations that yields so
called Fanno line in literature. Similarly,
combining the conservation of mass and
momentum equations yields the Rayleigh
line. On h-s diagram, these lines intersect at two points at which all three conservation relations are
satisfied. One of these points represents the state before the shock and the other after the shock. The
details of the flow behavior after the occurrence of a shock is explained in gas dynamics texts.
6.7
Adiabatic Efficiencies of Steady Flow Devices
The loss of exergy, occurring partly due to inherent properties of the flowing fluid like viscosity,
and partly due to unavoidable drawbacks on the system design, cause degradation of the performance
of flow machines. Therefore it is always useful to have parameters that compare the actual performance
of a machine to that working under ideal conditions.
It should be realized that the actual flow machines for many engineering practices are closely
modeled as adiabatic machines. The ideal process for an adiabatic device takes place when the flow
is reversible and thus isentropic. Therefore, an appropriate measure for the performance of an actual
device is to compare the actual process with the isentropic one, and such comparison is named as
adiabatic or isentropic efficiency of the flow machine.
Definition: The comparison of actual performance of single inlet and outlet adiabatic devices at
steady state with the ideal performance obtained by isentropic processing of the fluid to the same
outlet pressure is defined as the isentropic or adiabatic efficiency of that device.
Such a definition of efficiency is appropriately used in determining the performance of turbines,
nozzles, compressors, diffusers, and pumps.
6.7.1 Adiabatic Turbine Efficiency
Adiabatic or isentropic turbine efficiency is the ratio of the actual shaft power, w a , to the isentropic power w s , obtained by the isentropic expansion of the fluid to the same exit pressure and is
formulated as,
w
t  a
(6.29)
w s
244
THERMODYNAMICS
Referring to Eq. (4.60), for a negligible kinetic energy change across a turbine, the isentropic efficiency may then be determined in terms of the enthalpy change as,
t 
h1  h2
h1  h2 s
(6.30)
The h-s presentation in processing of a fluid flowing through an adiabatic turbine or a nozzle
is shown in Fig. 6.28. The isentropic change in enthalpy is always greater than the adiabatic one,
(h1  h2 s )  (h1  h2 ) and hence  t  1 . For pre-specified relationship between p and v, the isentropic
enthalpy variation in Eq. (6.30) may be calculated by Eq. (6.14), or may be evaluated from tables
by using the constant entropy constrain. In general, the isentropic efficiencies of turbines vary in the
range between 80% and 90%.
6.7.2 Adiabatic Nozzle Efficiency
The function of a nozzle is to accelerate the fluid to a predetermined exit velocity. Usually the
fluid velocity at the inlet is negligibly small respect to that at the exit. Thus the nozzle isentropic efficiency may be defined as the ratio of actual kinetic energy at the nozzle exit to that provided by an
isentropic expansion from the same inlet conditions to the same exit pressure.
n 
Va2 / 2
V s2 / 2
(6.31)
V2
Va2
 h1  h2 , and s  h1  h2 s . Hence the nozzle efficiency may be expressed in terms
2
2
of the enthalpy change of the fluid as,
Where,
n 
h1  h2
h1  h2 s
(6.32)
This expression is analogous to turbine efficiency and the same process line on h-s diagram as
used for a turbine also applies to the expansion through an adiabatic nozzle. Nozzles for subsonic
flow have greater efficiency than the supersonic ones. In general, nozzle efficiencies are rated above
90-percent.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 245
6.7.3 Adiabatic compressor efficiency
A compressor is a device which increases some of the thermodynamic properties like pressure
and temperature of a gas through compression. In doing this, a compressor consumes shaft work. The
isentropic compressor efficiency is defined as the ratio of the isentropic work, w s , to the actual work,
w a , required for the same inlet conditions and the same outlet pressure.
c 
w s
w a
(6.33)
Neglecting the changes in kinetic energy of the fluid across the adiabatic compressor and referring to Eq. (4.60), the isentropic efficiency of a compressor may be evaluated in terms of the enthalpy
difference of the fluid as,
c 
h2 s  h1
h2  h1
(6.34)
Figure 6.29 presents the adiabatic compression process of a gas through a compressor in h-s diagram. As can be seen by this figure, the entropy of the flowing gas increases as it approaches to the
exit. The adiabatic efficiencies of gas compressors range roughly between 75% and 85%.
If an effort is made to cool the gas during the compression by using a water jacket or fins, the
ideal process for such a case is considered to be a reversible isothermal process. Then the isothermal
efficiency of a cooled compressor is then the ratio of the isothermal work, w T , to the actual work, w a ,
required for the same inlet conditions and the same outlet pressure and is defined as,
isother .comp 
w T
w a
(6.35)
6.7.4 Adiabatic Diffuser Efficiency
Like a compressor, a diffuser accomplishes the reverse process of what a turbine does so as a
diffuser acts as the reverse of a nozzle. However, it is much more difficult to arrange for an efficient
246
THERMODYNAMICS
deceleration of flow than it is to obtain an efficient acceleration. There is a natural tendency in a diffusing process for a fluid to break away from the walls of the diverging passage. Experiments have
shown that the maximum permissible angle of divergence of a rectangular channel with one pair of
sides diverging is about 11-degrees. If the divergence is too rapid, this may result in the formation
of eddies with a consequence of lost kinetic energy. Hence the isentropic efficiency of a diffuser is
defined as the ratio of the isentropic enthalpy increase to the actual increase in enthalpy for the same
inlet conditions and the same outlet pressure.
d 
h2 s  h1
h2  h1
(6.36)
6.7.5 Adiabatic Pump Efficiency
The design difference between a compressor and a pump lies in the type of fluid compressed. The
device used for increasing the pressure of an incompressible fluid is called pump. Hence, the adiabatic
pump efficiency is defined similarly to that of a compressor. For the same inlet conditions and same
w
outlet pressure, the pump efficiency is,  p  s . Together with Eq. (6.20), for incompressible fluids
w a
with negligible kinetic energy change, this expression may be transformed into the following,
p 
w s
vp

w a cT  vp
(6.37)
The pump efficiencies may range from 50% to 85%.
Example 6.17: Air from the environment where the conditions are 101 kPa, 25C is compressed at a rate of 0.3kg/s to 600
kPa by an adiabatic compressor with an isentropic efficiency of 75% (see Figure 6.30). The compressed air is then cooled
at a constant pressure to 45C in a heat exchanger by water that enters at 101 kPa, 25C and exits at 100 kPa, 40C. The air
is treated as ideal gas with M=29, k=1.4. Determine,
a.
the power consumed by the compressor,
b.
the minimum power associated with compressing the air to the same exit pressure
c.
the irreversibility of the process.
Solution:
a.
b. For reversible and adiabatic compression of an ideal gas, the
compressor exit temperature becomes,
T2 s  298  (600 / 101)
(1.4 1)
1.4
 495.1K . H e n c e t h e m i n i m u m

power to be consumed, Wcs  0.3  1.005  (495.1  298) or. By
Eq. (6.33), the actual power consumed by the compressor is
Wc  Wcs / c  59.425 / 0.75  79.234 kW
.
c.
The actual exit temperature of air;
T2  298  79.234 / (0.3  1.005)  560.8K ,
and the energy balance around the heat exchanger yields water
0.3  1.005  (560.8  318)
 1.167
m a s s f l o w r a t e a s , m w 
4.18  (40  25)
kg/s. After applying Eq. (6.9) with q  0 to the control volume in Figure 6.30, the generated entropy is
where, s3  s1  1.005  ln(318 / 298)  0.287  ln(600 / 101)  0.446 kJ/kgK and
S gen  m a ( s3  s1 )  m w ( s5  s4 )
s  s  4.18  ln(313 / 298)  0.205kj/kgK . Then the entropy generation rate becomes, S  0.1054 kW/K and
5
4
the irreversibility rate, I  To S gen  298 x0.1054  31.419 kW
gen
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 247
6.7.6 Adiabatic heat exchanger effectiveness
The exchanger heat transfer effectiveness is indicated by  and compares the actual heat transfer
  T
 p
rate between the two fluids computed by q  mc
h
h , in
 p  Tc , out  Tc , in 
 Th , out or by q  mc
c
to the thermodynamically limited and maximum possible heat transfer rate qmax of the exchanger.
Hence, the effectiveness is formulated as,

q
(6.38)
qmax
In equation (6.38), qmax , is the reversible heat transfer and can only be realized if the temperature difference between the two fluids is infinitesimally small. This, in turn, requires an infinitely
large heat transfer surface area, and the fluid with minimum heat capacity experiences the maximum
 p  Tmax ,
temperature difference as, Tmax  Th , in  Tc , in . Thus, qmax is calculated as, qmax  mc
c
 p   mc
 p  , and qmax  mc
 p  Tmax , if mc
 p   mc
 p  . Referring to Figure 6.31, having
if mc
c
h
h
h
c
identical flow rates of the same fluid in both flow circuits of the exchanger, the heat exchanger effectiveness may be expressed as,

6.8
h2  h1
h3  h1
(6.39)
Thermodynamic Relations
As we all know, the only properties of a pure substance that are measurable by laboratory instruments are the mass, volume, pressure, temperature, specific heat, thermal conductivity and the viscosity of a fluid. Occasionally we may measure the heat flux at a surface, but we are certainly unable to
measure the properties like the internal energy, or the entropy of a fluid. Then the natural question
arises as how the values of thermodynamic properties that cannot be measured can be evaluated by
the experimental data of the measured properties. In chapter 2, we formulated the internal energy, the
248
THERMODYNAMICS
enthalpy, and later in chapter 5, the entropy of an ideal gas. Besides the use of tables, however, nothing
is mentioned about formulating and determining these properties of a fluid in terms of the measured
ones. Such questions can only be answered by studying the thermodynamic relations among fluid
properties. In studying the thermodynamic relations, the following topics have to be cooperated:
1. Definitions. A property definition plays an important role in expressing the differential form with
respect to other properties and frequently used property definitions are represented in Table 6.1.
Table 6.1 Definition of some selected properties
Property
Definition
Property
Definition
Enthalpy
h  u  pv
Isothermal bulk
modulus
 p 
B  v  
 v T
Helmholtz free
energy
f  u  Ts
Isentropic bulk
modulus
 p 
Bs  v  
 v  s
Gibbs free energy
g  h  Ts
Joule-Thomson
coefficient
 T 
J  

 p  h
Volume expansivity
factor
1  v 
 
v  T  p
Joule coefficient
 T 
j 

 v u
Isothermal
compressibility
1  v 
   
v  p T
Speed of sound
 p 
a  v 2  
 v  s
Isentropic
compressibility
1  v 
   
v  p  s
Constant temperature
coefficient
 h 
KT   
 p T
2. State principle. As studied in chapter 2, this principle can be used in forming an exact differential of a property respect to other two independent properties.
Example 6.18: The internal energy of an ideal gas with constant specific heat is u  u0  cv T  T0  and the entropy exT
v
pressed as, s  s0  cv ln  R ln . Determine a rigorous relationship in the form of u  u s, v  for the internal energy
T0
v0
of an ideal gas.
Solution:
Referring to entropy relation,
 T  v 
s  s0
 ln  
 
cv
 T0  v0 
1 k
R / cv
or
T  v 
 
T0  v0 
 s  s0 
exp 
 , and rearranging the internal energy
 cv 
 1 k
 s  s0  
v
equation yields, u  u0  cvT0   exp 
  1
 v0 
cv  



3. Fundamental equation of thermodynamics. As stated by equation (5.25), using the fundamental
equation of thermodynamics, the exact differentials of enthalpy, Helmholtz free energy, and Gibbs
free energy can be expressed as following,
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 249
du  Tds  pdv 
dh  Tds  vdp 




df   pdv  sdT 
dg  vdp  sdT 
(6.40)
4. Mathematical properties of exact differentials. If z  z x, y  and if the exact differential expressed as, dz  Mdx  Ndy then the following relations hold,

 z  
 z 
N   
M   
 x  y
 y  x 



 M   N 





 

 y  x  x  y




  z   y   x   1
  x  y  z  x  y  z



(6.41)
The last relation in equation (6.41) is called cyclic relation among the variables. With the exact
differential relations as stated in Eq. (6.41), utilizing Eq. (6.40) in order, one may obtain relations
between p, v, T, and s of a fluid as indicated in Table 6.2.
Table 6.2 Maxwell relations
Differential equation
 z 
M  
 x  y
 z 
N  
 y  x
 M   N 

 

 y  x  x  y
du  Tds  pdv
 u 
T  
 s v
 u 
p   
 v  s
 T 
 p 
 v     s 

s
 v
dh  Tds  vdp
 h 
T  
 s  p
 h 
v 
 p  s
 T   v 

  
 p  s  s  p
250
THERMODYNAMICS
df   pdv  sdT
dg  vdp  sdT
 f 
p   
 v T
 f 
s  

 T v
 p   s 
 T    v 

v  T
 g 
v 
 p T
 g 
s  

 T  p
 s 
 v 
 T     p 

p
 T
In Table 6.2, one may note that one property derivable by two different derivatives, and using
these relations, a change in an immeasurable quantity of a system can be evaluated by measuring the
change in p, T, and v.
Example 6.19: As shown in Figure 6.33, a fluid completes a differential reversible cycle which consists of two constant
 T   p 
volume and two constant entropy processes. Show that  
    for the cycle.
 v  s  s v
Solution:
In p-V diagram, the net work done by the cycle is the shaded area and is expressed as,
 p 
 W   s  dsdv . Similarly, in
v
T-s diagram of Figure 6.33, the net heat transferred to the cycle is
 T 
  Q    v  dvds , where dv has a negative value.
s
Referring to the first law of thermodynamics for cycles, the two shaded regions in Figure 6.30 have to be identical. Therefore,
 p 
 T 
 p 
 T 
 s  dsdv    v  dvds or  s     v  .

 v

s
 v
  s
Comments: Considering the above example, the physical importance of Maxwell relations can be demonstrated.
Example 6.20: Show that Joule-Thompson coefficient  of an ideal gas is zero.
Solution:
 h 
 h 
Since functional form of enthalpy of a fluid is h  h  p, T  and in differential form, dh  
 dT    dp . Refer
T

p
 p T
 T 
 h   p   T 
 h 
and   
ring to Table6.1, c p  
 . Employing the cyclic relation in Eq. (6.41),   

 
  1 then
 T  p
 p h
 p T  T h  h  p
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 251
 h 
     c p . Hence the differential form of enthalpy becomes dh  c p dT   c p dp . As studied in chapter 2, enthalpy of
 p T
an ideal gas is only a function of temperature and c p  0 then only choice in dh expression is   0 .
As an application of Maxwell’s relations, a general expression for evaluating the change in
enthalpy of a pure substance can be derived. Considering the differential form of enthalpy as,
 h 
dh
ds
 h 
dh  
dT    dp and by Eq. (6.40),
T
 v for constant temperature this expression

dp
dp
 T  p
 p T
 h 
 s 
 s 
 v 
transforms into    T    v . Utilizing Maxwell’s relation in Table 6.2 as,     

 T  p
 p T
 p T
 p T
 h 
 v 
and back substitution yields,    v  T 
 . Hence the change in enthalpy between two pre T  p
 p T
defined states is expressed as,

h2  h1  


T2
T1



c p dT 
 p  p1 

 v   
v T 
  dp 


T

 p  

T2
p2 
p1
(6.42)
The information needed to integrate the first term in equation (6.42) is the variation of cp along
an isobar line, and totally gives the enthalpy change of an ideal gas. For the integration of the second
term, however, an equation of state which relates p, v, and T of the substance or the compressibility
 v 
chart is needed. Then the derivative, 
 can be evaluated and the integral can be carried out along
 T  p
isotherm at T2. This term also presents the amount of deviation of enthalpy change from the enthalpy
change of ideal gas behavior.
Example 6.21: Considering the compressibility factor of a real gas, derive an equation for enthalpy deviation so that the
amount of deviation from ideal gas enthalpy change can be evaluated in terms of the reduced values.
Solution:
Since the state equation for real gas is pv  ZRT and in differential form, vdp  pdv  ZRdT  RTdZ . For constant pressure,
ZRT RT 2  Z 
 v 
T

 

 . Hence the second term of Eq. (6.42) can be rearranged as v 
p
p  T  p
 T  p
RT 2  Z 
 v 
p
 

 .
p  T  p
 T  p
Then the enthalpy deviation in terms of reduced values can be expressed as,
h2  h1   h2*  h1* 
RTc
 Tr22

pr 2 
pr 1
Z  dpr


 Tr  pr pr
(6.43)
In accord with this relation, generalized enthalpy correction chart has been prepared in literature
and enthalpy deviation can be estimated graphically.
Similar methodology may be applied to evaluate the change of entropy of a pure substance by
measurable quantities. Since the entropy of a pure substance can be expressed in functional form as,
 s 
 s 
s  s  p, T  or as, s  s v, T , the change in entropy will be, ds    dp  
 dT or by Eq.

p
 T  p
 T
252
(6.40),
THERMODYNAMICS
dh
ds
dp
T
v
and for constant pressure,
dT
dT
dT
 h 
 s 
 T   T  T  . Hence

p

p
cp
 s 
 T   T . More
p
 s 
 v 
over, utilizing Maxwell’s relations,     
 and substitution into ds expression yields,
 T  p
 p T

s2  s1  


T2
T1
cp

dT 


T  p  p1 

p2
p1

 v 
 T  dp 

 p T T
2
(6.44)
Example 6.22: Considering the compressibility factor of a real gas, derive an equation for entropy deviation from the ideal
gas entropy change and express the entropy deviation in terms of reduced values.
Solution:
Consider the differential form of equation of state of a real gas as, vdp  pdv  ZRdT  RTdZ , for constant pressure,
v RT  Z 
 v 
 Z 
 v 
p
  ZR  RT  T  or  T   T  p  T  . Together with the ideal gas entropy change expression, the
 T  p

p

p

p
entropy deviation may be expressed as,
s*  s

R

p2 
p1
v
T  Z 
1

 
 dp .

 RT p  T  p p 
Considering the result of Example 6.22 and rearranging, the entropy deviation of a real gas may
be expressed as,
s*  s
 Tr 2
R

pr 2 
pr 1
Z  dpr


 Tr  pr pr
(6.45)
Example 6.23: A piston-cylinder device in Figure 6.34 contains 1.8kg of water vapor at 280C, and 0.44m3 of volume.
After a reversible and isothermal expansion process, vapor assumes a final volume of 0.6m3. Assuming that water vapor
obeys Van der Waal’s equation of state, determine
a.
percent deviation in initial pressure,
b.
the amount of heat transferred.
Solution:
a.
Referring to Table (2.3), a  552.6 m6 kPa/kmol and b  0.03042 m3/kmol. The mole number is n  1.8 / 18  0.1 kmol
and the initial molar volume is v1  0.44 / 0.1  4.4 m3/kmol. Eq. (2.24) yields, p1 
8.314  553
552.6

 1023
4.4  0.03042 4.42
kPa. Tabulated value for p1 is p1  1000 kPa, and the percent deviation becomes, err%=2.3-percent.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 253
 s 
 s 
Consider s  s v, T and differential change in s becomes, ds    dv  
 dT , since the process is isothermal,
 v T
 T v
b.
R
 s 
 p 
dT = 0, and utilizing Maxwell relations,    
  v  b . For a reversible process, q12 

v

T
 T 
v

v2
v1
Tds  RT

v2
v1
dv
,
vb
 v b 
or Q12  nRT ln  2
 , v2  0.6 / 0.1  6.0 m3/kmol, and Q12  143.39 kJ.

v
b
 1

6.9
Relations on Specific Heats
In chapter 2, the relationship between cp and cv for an ideal gas is derived but no further explanation is provided for the case of a pure substance. The purpose of this section is twofold: 1. To derive
and express a general relationship between the two specific heats of a pure substance, 2. To analyze
the effect of pressure and volume on both specific heats and express the results in term of measurable
quantities.
6.9.1
General relation between cp and cv
dT  v 

Consider the two functional forms of entropy of a fluid as follows, ds  c p
T  T  p
dT  p 
dp  cv
dv , using this equality, one may express the temperature difference dT as,

T  T v
1
 v 
 p 
T
dp 
T

 dv . Moreover, referring to the state equation as,
c p  cv  T  p
c p  cv  T v
 T 
 T 
T=T(p,v), dT may also be expressed as, dT  
 dp  
 dv . Comparing the coefficients of
 v  p
 p v
these two equations and making use of cyclic relation, one may end up with a general expression for
evaluating the difference in specific heats as,
dT 
1




2
 p   v 
 p   v 
c p  cv  T 
     T      

T
T

v 
p
 v T  T  p
or
c p  cv  TvB 2
(6.46)
Referring to Table 6.1, B in Eq. (6.46) is the isothermal bulk modulus, and  is the volume expansivity of the substance. Since for any fluid the quantities, T, v, B,  are always positive, cp will
always be greater than the value of cv.
Example 6.24: Evaluate the difference in specific heats of water at 20C. Take the bulk modulus and volumetric expansion
coefficient for water at 20C respectively as, B=2.15x109Pa, and =0.207x10-31/K.
Solution:
Assuming, v  0.001002 m3/kg and respect to Eq. (6.46), c p  cv  293  1.002  103  2.15 x109  0.207 2 x106  27.05 J/
kgK, or c p  cv  0.027 kJ/kgK. As can be seen this difference is considerably small with respect to bulk value of cp (0.6percent). For liquids the difference is usually neglected and is not considered in the calculations.
6.9.2
Dependence of Specific Heats on Pressure and Volume
We already know that both specific heats are functions of temperature, and in fact, empirical equations in polynomial form are cast for many fluids but we certainly do not know much about how the
254
THERMODYNAMICS
pressure or volume variations affect the specific heats as the temperature is kept constant. Consider
the exact differential of entropy of a fluid at single phase as, ds  c p

  cp / T
of Maxwell’s relations in Table 6.2, one may state that, 
 p

dT  v 
dp and making use

T  T  p


  v / T 
p


T

T 


 p
or rear-
ranging yields the effect of pressure on constant pressure specific heat at isothermal conditions as,
  2v 
 c p 

  T  2 
 p T
 T  p
(6.47)
Here, we need the state equation for determining the pressure effect on cp. Similar analysis can be
carried out for determining the effect of volume variation on cv by considering the second equation
dT  p 
dv . Applying Maxwell’s relations to this relaof entropy in differential form as, ds  cv

T  T v


  p / T  
  cv / T 
v
 and after rearrangement the effect of volume on constant
tion yields, 
 
v
T





T 
v
volume specific heat at isothermal conditions may be expressed as,
 2 p 
 cv 

T
 2
 v 

T
 T v
(6.48)
Here again we need the information between p, T, and v for evaluating the effect of volume
change on cv.
Example 6.25: Neon gas obeys Clausius equation of state as, p v  b   RT at a state of 100 MPa, and 300K. Evaluate,
a.
the effect of pressure and volume respectively on cp and on cv of Neon gas,
b.
the amount of heat transfer to 1.5 kg of Neon gas if gas expands from 100 MPa to 50 MPa reversibly and isothermally,
c.
the Joule-Thomson coefficient  J of the gas.
Take, b  0.0022 m3/kmol, c p  6.4 kJ/kmolK for Neon gas.
Solution:
a.

Referring to the equation of state, v / T p  R / p and  2v / T 2

not vary respect to pressure, by Eq. (6.47), c p / p

or  2 p / T 2
  0 . Hence constant pressure specific heat does
p
T  0 . Similarly by the state equation, p / T v v  b   R
 v  b   0 . Hence, in equation (6.48), cv / v T  0 . For Neon gas, the specific heats are inv
dependent of pressure and volume variations.
b.
Since, v / T p  R / p , and the process is reversible and isothermal,
0
 dT R 
dp Q   nRT ln p2 , the molar mass of Neon, M=20.17kg, and
 Q  Tds  T  c p
 dp   c p dT  RT
12
p1
T
p 
p

CHAPTER 6 ENTROPY: A SYSTEM DISORDER 255
n  1.5 / 20.17  0.074 , Q12  0.074 x8.314 x300ln
c.
50
 127.93 kJ.
100
 h 
 v 
RT
T
 
v
 v

T

p


 T 
b
0.0022
p
 T
p



substituting the values yields,  J  
J  
 or  J  
cp
cp
cp
 h 
6.4
 p h
 T 

p
 J =-3.4x10-4 K/kPa. Negative sign indicates that the gas becomes hot as it expands to lower pressure.
6.10
Clausius-Clapeyron Equation
The Clausisus-Clapeyron equation provides a relationship between the pressure and the temperature
of a fluid undergoing a phase changing process and frequently used for phase equlibria. It is also a wellknown Example of how a property that cannot be measured directly (enthalpy of fusion or enthalpy
of evaporation) can be evaluated by the measurements of pressure, temperature, and volume.
Principle 22: If two phases of a substance are in thermodynamic equilibrium
at a pressure of p and temperature of T, then the Gibbs free energy of these
phases must be identical.
6.10.1
Liquid-Gas Equilibrium at Pressure p, and Temperature T
Consider a fluid at saturated state at (p, T) and both phases (liquid and vapor phases in this case)
are in mutual equilibrium in a piston-cylinder apparatus. The condition for unconstrained equilibrium
of both phases at the same temperature and pressure is that the Gibbs function of both phases has to
be equal.
g f ( p, T )  g g  p, T 
dg f  dg g
or
(6.49)
Due to change in pressure and temperature, the change in Gibbs values of both phases can be
evaluated by Eq. (6.40) as,
dg f  v f dp  s f dT
and
dg g  vg dp  sg dT
(6.50)
Combining these two relations at the equilibrium of these two phases and rearranging yields,
sg  s f
 dp 
 dT   v  v

 sat
g
f
(6.51)
This expression gives the rate of change of pressure with temperature on the line representing the
liquid-vapor equilibrium. As shown in Figure 6.35a, since for many substances sg  s f , and vg  v f ,
the equilibrium line has a positive slope. Furthermore the phase change takes place at constant pressure;
the entropy change may be expressed as, sg  s f  h fg / T and substituting into Eq. (6.51) yields,
h fg
 dp 
 dT   Tv

 sat
fg
(6.52)
256
THERMODYNAMICS
Special solution of the above relation (Eq. 6.52) may be obtained at pressures very low respect to
the critical pressure of a fluid ( p  pc ). For such a case, the specific volume of saturated liquid may
be neglected ( vg  v f ) and the gas phase may be assumed to behave like an ideal gas, vg  RT / p .
dp h fg dT
. For moderate changes of pressure and temperature,

p
R T2
the enthalpy of evaporation does not change appreciably, and may be assumed to be constant. Thus,
Thus Eq. (6.52) may be rearranged as,
 p  h fg  1 1 
ln  2  
  
 p1  R  T1 T2 
(6.53)
The above equation represents the relation between pressure and temperature along the liquidvapor phase equilibrium line, and the slope of the line is expressed as,
d ln p 
d 1 / T 

h fg
R
(6.54)
As illustrated in Figure 6.35b, for higher values of slope, an appreciable change in saturation pressure will not affect the saturation temperature, and the saturation temperature will almost be constant.
This is a desired property for some fluids to accommodate at working conditions; especially refrigerants are desired to have such a property.
6.10.2 Solid-liquid equilibrium at pressure p, and temperature T
Similar analysis can be carried out for solid-liquid phase equilibrium. If both phases are in mutual
equilibrium, then their Gibbs free energy values, so do the differential changes have to be equal.
dg f  dg s
v f dp  s f dT  vs dp  ss dT
(6.55)
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 257
Furthermore, for the phase change process, s f  ss  hsf / T and vsf  v f  vs , the slope of (p, T)
line representing the solid-liquid equilibrium in the phase diagram becomes,
hsf
 dp 
 dT   Tv

 sat
sf
(6.56)
In short, (C-C) equation provides the slope of phase equilibrium line at a particular pressure and
temperature.
Example 6.26: Estimate the sublimation pressure of water at -40C by using the following data of water at the triple point:
ptp  0.61 kPa , Ttp  0.01 C and hsg  2834.8 kJ/kg and compare the result with the tabulated value.
Solution:
a.
Considering that vg  vs Equation (6.53) may be applied to the present problem by some modifications as,
 p  2834.8  1
1 
R  8.314 / 18  0.461 kJ/kgK, and ln  2  


 or p2  0.0249  0.61  0.0151 kPa. The
p
0.461
273
233


 1
tabulated value of sublimation pressure at -40C is 0.013 kPa, and C-C equation estimated the pressure by 16%
error.
Example 6.27: The enthalpy of melting of benzene (C6H6) at p  100 kPa and T  5.5 C is 7.165 kJ/kg and its density in
liquid and solid phases respectively are 894 kg/m3, and 1014 kg/m3. Determine the dp / dT  slope of the phase equilibrium
line at this pressure temperature.
Solution:
a.
7.165
1
1
 dp 
4
Respect to Eq. (6.56), vsf 

 1.323 x104 m 3/kg, then 
  278.5  1.323  10  194.46
dT
894
1014

 sat
kPa/K
6.11
Use of Entropy in Design
In today’s technological world, the applications of entropy are widespread, from engineering
fluid mechanics and thermodynamics, to information and coding theory, economics, and biology.
Entropy serves as a key parameter in achieving the upper limits of performance and quality in many
engineering problems. As the future technologies move toward the theoretical limits, the entropy and
the second law will have a significant role in the design.
258
THERMODYNAMICS
The entropy-based design is an emerging design methodology that incorporates the second law with
computational fluid dynamics (CFD), and experimental techniques. The methodology may shed some
light on various flow processes, ranging from optimized flow configurations in an aircraft engine to
highly ordered crystal structures (low entropy) in a turbine blade. Hence, the entropy based design extends the methods like exergy analysis and entropy minimization to more complex configurations.
In the development of a workable or an acceptable design that satisfies the prescribed requirements and the constraints, first as indicated in Figure 6.36, the entropy generation in a fluid that has
temperature and velocity gradients along the flow direction will be discussed.
Together with the definition of compressible and incompressible Newtonian fluids, as schematically described in Figure 6.37, rewriting the entropy transport equation (Eq. (6.8)) for an infinitesimal
control volume, δV, one may determine a way of calculating the local entropy generation for such
fluids as,
2
sgen 
k  T   ij ui

 
T xi
T 2  xi 
(W/m3K)
(6.57)
In Equation (6.57), k is the thermal conductivity of the fluid, and τij is the viscous shear arising
from the velocity gradients in the flow and is expressed as,
 u u j
 ij    i 
 x j xi
 2 uk 
 

 3 xk ij 


(Pa)
(6.58)
where, μ, is the dynamic viscosity and δij is the kronecker delta. One may note that the last term
 u

in Eq. (6.58) will vanish for an incompressible flow  k  0  .
 xk

Eq. (6.57) is a positive definite expression for the entropy generation rate. The first term on the
right represents entropy generation due to heat transfer across the fluid stream for a finite temperature
difference. If the fluid is isothermal then there will be no temperature gradient across the fluid body
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 259
and this term will be zero. The second term is the local entropy generation due to viscous dissipation
(conversion of kinetic energy into internal energy through fluid friction). This term will be zero if the
fluid is ideal or the flow is inviscid (μ=0). Thus, for the flow of isothermal, ideal, and incompressible
fluids, sgen  0 .
In general, the vector form of Eq. (6.57) for entropy production of laminar flow is expressed as
following:
sgen 
k T  T 
T
2


T
(6.59)
where  is viscous dissipation function and is due to velocity gradients in the flow. For twodimensional flow (2D flow), this function is expressed as,
 u v  2
 u  2  v 2  

      2       
x   y   

 y x 


(6.60)
and the first term in Eq. (6.59) represents the entropy generation due to heat transfer across temperature gradients in the fluid body.
6.11.1 Case study 1: Channel design
As shown in Figure 6.35, consider an isothermal and incompressible viscous fluid flowing through
a horizontal channel which consists of two plates of length l and spaced 2h apart. The fully developed

velocity profile of the flow is u  uc 1   y / h 
2
and the average velocity is related to the centerline
2
velocity as, u  uc .
3
The head loss, hl , due to entropy generation in the fluid may be formulated by integrating the
irreversibilities over the entire channel as following,
hl 
1
Ts dV
 V gen
mg

(6.61)
260
THERMODYNAMICS
The objective is to show that the relationship of Eq. (6.61) also holds for calculating the frictional
u
0,
losses of isothermal fluid flowing through a 2D channel. Since the flow is fully developed,
x
2
 u 
and v=0. Then the viscous dissipation function reduces to     and by the velocity distribution
 y 
2u y
u
equation,
  c2 . In addition, for an isothermal fluid, T  0 . Hence, the entropy rate genery
h
ated in the fluid becomes,
2
sgen
4  uc2 y 2
  u 
   
T  y 
Th 4
or
sgen 
9u 2 y 2
Th 4
(6.62)
Substituting this result into Eq. (6.61) and integrating over the fluid volume ( dV  1  dy  1 ) with
m  2h  u yields,
2
l
p
 24   u
hl  


 Re  2 g 2h   g
(6.63)
where, Re represents the flow Reynolds number and is defined as, Re   u (2h) /  .
Example 6.28: Water at 290K flowing isothermally with an average velocity of 0.05m/s through a duct with a height of
0.02m. Determine variation of entropy production rate with the channel height and the energy loss for a channel length of
2 meters.
Solution:
The viscosity of water at T=290K is μ=1.14x10-3 kg/ms, h=0.01 m, and referring to Eq. (6.59), for isothermal flow, T  0 ,
and sgen  8.844 y 2 . Since the range of y is 0.01  y  0.01 , the variation of entropy generation rate is as shown in
Figure 6.39.
Due to viscous effects, the entropy production rate is at maximum near the walls and is zero at the center. Water density,
=998kg/m3, and the flow Reynolds number is Re 
998  0.05  0.02 2
1.14  103
 875.4 , and by Eq. (6.61) the head loss for the entire
 l or El  0.998 9.81 0.697
channel is hl  0.697 m . Since the rate of energy loss is calculated by El  mgh
6.823 W
Since the energy loss, El , is totally equivalent to useful work, it also represents the exergy loss for the channel flow.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 261
The entropy production rate in turbulent flows may be determined by subdividing the fluid entropy
into mean and fluctuating components and by applying the entropy transport equation (Eq. (6.8)) to
an infinitesimal control volume, δV. However, the resulting equations are too complex for the content
of this text and are not studied here.
6.11.2
Case Study 2: Fluid Machinery Design
As shown in Figure 6.40, the mechanical power generated by steam, gas, or wind turbine highly
depends on the shape of the blades, the velocity field, and the system construction. Since the maximum
turbine power developed is related to the change of internal energy, kinetic, and potential energies of
the fluid as given by Eq. (6.13), the actual power output is then expressed as following,
n

1
w    vdp  Vn2  V12  g ( zn  z1 )  
2
 1




j n
T s
j gen , j
(6.64)
j 1
In equation (6.64), beginning from the inlet, j=1, to the exit, j=n, the entropy generation term is
summed up over the n sections throughout the entire turbine. In accord with this result, the power
output is maximized when the total entropy production over the control volume of the turbine is
minimized. Hence, the turbine stages have to be designed in such a way that the energy loss due to
viscous mixing, flow separation and the losses due to inlet and exit to a stage are minimized. One way
to achieve this goal is to apply design modifications to blade shape, gap spacing, and the thickness by
using CFD for studying the flow field in a stage or by experimental techniques. In such experiments,
the Particle Image Velocimetry (PIV) system is usually used for recording the necessary data of local
entropy production rates.
Wind turbines. The wind turbines are the special type of flow machines that are used to extract
the exergy of the wind. Locations having annually averaged wind velocities of 5m/s or greater are
considered to be suitable for wind turbine installation. The wind turbines are separated into two basic
types by which the way the turbine spins. As shown in Figure 6.41, 1.(HAWT) The horizontal axis
wind turbine that has blades that spins around a horizontal axis, 2.(VAWT) The vertical axis wind
turbine such as Darrieus, and Savonius.
262
THERMODYNAMICS
The main components of a typical HAWT are indicated in Figure 6.41a. If the wind data for a
particular wind field is available then the average wind velocity over the given time period is calculated as,
n
V
i
V
i 1
(6.65)
n
In addition, the energy based mean velocity becomes,


VE  



1/3

Vi 
i 1

n 


n

3
(6.66)
With respect to V and VE values, the type of the wind turbine may be decided.
Example 6.29: As shown in Figure 6.42, wind pressures well away from an axial wind generator are identical (pi=pe) and
the wind velocity distribution around the turbine is shown in Figure 6.42c. Formulate,
a.
the power extracted from the turbine with respect to inlet and exit wind velocities,
b.
the condition for which the maximum power is obtained,
c.
the value of the maximum power,
d.
the exergy based ideal efficiency of the rotor.
Assumptions: constant density, no temperature gradient through the fluid, negligible potential energy change. In figures,
a and b respectively represent front and rear surfaces of the propeller.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 263
Figure 6.42 Shaft power formulation for an axial rotor
Solution:
a.
1 2
1
Vi  pa va  Va2 . Simi2
2
1 2
1 2
1
larly the energy equation for (b) to (e) is, pb vb  Vb  pe ve  Ve , where, Va  Vb  Vturbine  Vi  Ve  .
2
2
2
1
Then the pressure difference between the front and the rear of the turbine is pa  pb   Vi 2  Ve2 . The
2
1
2
2
force and the power generated on the rotor respectively then are, F  A  pa  pb    A Vi  Ve and
2
1
W  FVturbine   A Vi  Ve  Vi 2  Ve2
4
The energy equation between (i) and (a), and for isothermal fluid becomes, pi vi 

b.





The condition by which the maximum power results is:
dW
 0 . In other words, the optimum value of Ve has to
dVe
be determined. After differentiating the power expression with respect to Ve and setting to zero yields Ve 
c.
8
Substituting this result into the power expression yields the maximum power as, Wmax 
 AVi3
d.
The exergy based ideal efficiency of the rotor is then  max 
1
Vi .
3
27
max
the max. exergy recovered
or
the exergy of the wind
8
 AVi3
27

 0.59
1
 AVi3
2
As discussed in this example, the actual efficiency of the rotors is usually below this limiting
value and is around 30%-40%. The actual efficiency basically depends on the following factors: 1.
The blade tip to wind speed ratio (tsr), tsr  Vtip / V , 2. The spillage and other losses. In Figure 6.43,
264
THERMODYNAMICS
the efficiencies of various rotors are indicated. For high speed propellers, the efficiency attains a
maximum value at a tip-speed-ratio between 6 and 7.
References
1.
Y. Jaluria, Design and Optimization of Thermal Systems, 2nd edition, CRC press, ISBN 978-0-8493-3753-6, 2008.
2.
A. Ben-Naim, A Farewell to Entropy: Statistical Thermodynamics Based on Information, World Scientific Publishing,
ISBN 978-981-270-706-2, 2008.
3.
I. Dincer, and Y. A. Çengel, Energy, Entropy, and Exergy Concepts and Their Roles in Thermal Engineering, MDPI,
Entropy, Vol. 3, pp.116-149, 2001.
4.
P. O’Keefe, G. O’Brien, and N. Pearsall, The Future of Energy Use, Earthscan Publishing, ISBN 978-1-84407-504-1,
2010.
5.
R. Petela, Engineering Thermodynamics of Thermal Radiation-for solar power utilization, McGraw-Hill Inc., ISBN
978-0-07-163963-7, 2010.
6.
G.F. Naterer, and J.A. Gamberos, Entropy-Based Design and Analysis of fluids Engineering Systems, CRC Press, ISBN
13-978-0-8493-7262-9, 2008.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 265
Problems
6.7
tropy change, S  S2  S1 and entropy production,
,
S gen , may assume the following signs,
One kilogram of water initially at 160C, 1.5
bar undergoes an isothermal, internally reversible
compression process to the saturated liquid state.
Determine the work and heat transfer, in each in
kJ. Sketch the process on p-v and T-s coordinates.
Associate the work and heat transfer with areas on
these diagrams.
6.8
a.
b.
c.
d.
e.
One kilogram of air is initially at 1 bar and 450 K.
A process takes place that brings the air to a final
state of 2 bar and 350 K.
a. Calculate the entropy change during the process
b. Is this an adiabatic process? Explain.
6.9
Two kg of superheated steam at 400C, 600 kPa is
cooled at constant pressure by transferring heat from
a cylinder until the steam is completely condensed.
The surroundings are at 25C. Determine the net
entropy change of the universe.
6.10
Nitrogen initially at 102 kPa, 17C with a volume of
0.05m3 is in a piston-cylinder device and compressed
by a reversible and isothermal manner to 4.5bar of
pressure.
a. Determine the heat flow and the work transfer.
b. Sketch the process on T-s and h-s diagrams.
6.11
For each of the following systems, specify whether
the entropy change during the indicated process is
positive, negative, zero, or indeterminate.
a. One kilogram of water vapor undergoing an
adiabatic compression process
b. Two kilogram mass of nitrogen heated in an
internally reversible process
c. One kilogram of Refrigerant 134a undergoing
an adiabatic process during which it is stirred
by a paddle wheel.
d. Half kilogram mass of carbon dioxide cooled
isothermally.
e. Two kilogram mass of oxygen modeled as an
ideal gas undergoing a constant pressure process
to a higher temperature
f. Two kilograms of argon modeled as an ideal
gas undergoing an isothermal process to a lower
pressure.
6.12
In a reversible process the rate of heat transfer to
the system per unit temperature rise is given by
Entropy applications and closed systems
6.1
6.2
6.3
When a closed system with internal irreversibilities
undergoes a process from state 1 to state 2, the en-
negative and positive, respectively.
negative and zero, respectively.
positive and negative, respectively.
positive and zero, respectively.
all of the above.
A piston cylinder assembly contains 0.02kg of steam
at 3.0MPa, and a volume of 0.002m3. The state of the
system has changed by a process that may be described
as a straight line on a T-s diagram to 0.2MPa, 0.014m3.
The surroundings is at 0.1MPa, 25C.
a. Determine the heat and work interactions associated with the process,
b. If only one reservoir is used in what range could
its temperature be?,
c. Select a reasonable temperature for the reservoir
and find the maximum work associated with the
process.
An inventor claims that the electricity-generating
unit shown in the Figure 6.44 below receives a heat
transfer at the rate of 263kW at a temperature of
300K, a second heat transfer at the rate of 369kW
at 400K, and a third at the rate of 527kW at 555K.
For operation at steady state, evaluate this claim.
Figure 6.44 Schematic of the problem
6.4
6.5
6.6
Air is contained in an insulated rigid volume at 20C
and 200 kPa. A paddle wheel inserted in the volume
does 720 kJ of work on the air. If the volume is 2m3
compute the change of entropy of the system ΔS in
kJ/K.
A 0.5kg of steam is compressed isentropically from
1MPa and 280°C to 8.0MPa. Determine the final
temperature of steam.
dQ
 0.35kj/K . Find the change in entropy of
dT
the system if its temperature rises from 300K to
550K.
6.13
Two kg of superheated steam at 400C, 600kPa is
cooled at constant pressure by transferring heat from
a cylinder until the steam is completely condensed.
The surroundings are at 27C. Determine the entropy
generation of the process.
6.14
0.2kg of ice cube at -10C is exposed to surroundings at 27C. Take ice specific heat and latent
State the conditions required for the following expressions to be valid. a. dS  0 , b. dS 
dS 
Q
, d. du  Tds  pdv , e.
T
Q
, c.
T
 dS 0
266
THERMODYNAMICS
T0 There are no other heat transfer process during
heat of fusion respectively as c p  2.093kj/kgK ,
the cycle,
h fs  333.3kj/kgK , and determine,
a. the entropy generated by this process,
b. the amount of minimum work required to bring
water at 27C back into ice at -10C
6.15
6.16
a. Show that the net work developed per cycle

is given by Wcycle  Q1  1 
T0
T1

  T0 S gen ,



where S gen entropy generation rate due to ir-
A perfectly insulated piston-cylinder device contains
ammonia at 1400 kPa, 70C. The piston moves in
a reversible manner and the temperature drops to
-25C. Since the amount of work done by ammonia
is 350kJ, determine the initial volume occupied by
ammonia.
reversibilities within the cyclic system.
b. If the heat transfers Q1 and Q 0 are with hot
and cold reservoirs that are respectively at Th
and Tc , what is the relationship of T1 to the
Argon in a light bulb as in Figure 6.45 is at 95kPa
and heated from 22C to 65C by electric input. At
steady state conditions, the bulb emits 60W of power.
If the bulb volume is 50cm3, calculate,
temperature of the hot reservoir and Th and the
relationship of T0 to the temperature of the cold
reservoir Tc ?
c. Obtain an expression for Wcycle if there are (i)
no internal irreversibilities, (ii) no internal or
external irreversibilities.
6.19
attains an equilibrium state. Assume each mass is
incompressible with constant specific heat c.
a. the entropy generated until the bulb reaches
steady state conditions.
b. the entropy generation rate at steady state conditions.
a. Show that the amount of entropy generated is
 T T 
S gen  mc ln  1 20.5 
 2 T1T2  


Hint: Neglect the glass mass and the radiation effects
6.17
At steady state conditions, the curling iron as shown
in Figure 6.46 consumes 18watts of power, and the
surface attains 85C. Determine the rate of heat
transfer and the rate of entropy production for 27C
of surroundings temperature.
An isolated system of total mass m is formed by mixing two equal masses of the same liquid initially at
temperatures of T1 and T2 . Eventually, the system
b. Demonstrate that S gen must be a positive
value.
6.20
A cylindrical rod of length L=50cm insulated on its
lateral surface is initially in contact at one end with a
wall at temperature T1  100 C and at the other end
with a wall at a lower temperature, T2  20 C. The
temperature within the rod initially varies linearly with
 T1  T2
 L
position x according to T ( x)  T1  

x.

The rod is then insulated on its ends and eventually
comes to a final equilibrium state where the rod
temperature is T f .
a. Evaluate T f in terms of T1 and T2 and calculate
the numerical value of T f .
b. Show that the amount of entropy generated is
6.18
A system undergoing a thermodynamic cycle receives
Q1 at temperature T1 and discharges Q 0 at temperature


T2
T1
S gen  mc 1  ln T f 
ln T2 
ln T1 
T1  T2
T1  T2


CHAPTER 6 ENTROPY: A SYSTEM DISORDER 267
6.21
The pneumatic lift of Figure 6.47 uses atmospheric
air at 27C, 1bar to increase the pressure of the vessel
that initially contains air at 30 bar, 27C with a volume
of 0.5m3. An inventor claims that the pressure in the
vessel can be increased to 100 bar by compressing
the content adiabatically while consuming 6500kJ
of work. Is the inventor’s claim possible, explain?
6.24
6.22
A piston-cylinder device in Figure 6.50 contains
R134a at -15C, and 60% quality with a volume of
80L. The cylinder is brought into a room of 27C,
meantime, electric current with 8A intensity and
15V voltage drop passes through the resistor in the
cylinder. It is predicted that in 1 hour time period,
the temperature of the refrigerant will rise 55C. Is
this possible? Explain.
As shown in Figure 6.48, 0.5kg of air in a pistoncylinder assembly undergoes two internally reversible
processes in series. For the first process, the pressure
is constant and the second process, the entropy is
constant. For each process, determine the work and
the heat transfer.
Steady flow systems
6.25
6.23
A piston-cylinder device in Figure 6.49 contains
0.5kg of water at 600kpa, and 600C. The piston
cross sectional area is 0.08m2 and restrained by a
linear spring with a spring constant of 8kN/m. The
system cools down to room temperature at 27C.
Calculate the entropy generated by this process.
Air enters a 3600kW turbine operating at steady state
with a mass flow rate of 18kg/s at 800°C, 3bar and
a velocity of 100m/s. The air expands adiabatically
through the turbine and exits at a velocity of 150m/s.
The air then enters a diffuser where it is decelerated
isentropically to a velocity of 10m/s and a pressure
of 1bar. Employing the ideal gas model,
a. Show the two processes on a T-s diagram.
b. Determine the pressure and the temperature of
the air at the turbine exit.
c. Calculate the rate of entropy production in the
turbine, in kW/K.
268
6.26
6.27
6.28
THERMODYNAMICS
A salesperson claims to have a 100kW steam turbine.
The steam enters the turbine at 10 bar, 300C and
leaves the turbine at 3bar, 200C. The temperature
of the surroundings is 25C.
a. Explain if the processes in the turbine is adiabatic
or not?
b. As a further explanation, the salesperson states
that to get 100kW of power at turbine shaft,
the turbine consumes 0.2kg/s of steam. Is this
possible?
As shown in Figure 6.51, a steam turbine having an
isentropic turbine efficiency of 85-percent drives
an air compressor with an isentropic compressor
efficiency of 78-percent. The steady state operating
data are provided on the figure. Assume the ideal
gas model for air, and ignore stray heat transfer and
kinetic and potential energy effects. Determine the
mass flow rate of the steam entering the turbine, in
kg of steam per kg of air exiting the compressor, i.e.
m steam / m air .
For a steam compressor, a salesperson states that the
steam enters the compressor at a pressure of 1 bar,
100C, and leaves 15 bar, 400C. The temperature
of the surroundings is 25C.
a. Is this an adabatic compressor? Explain.
b. The salesperson further states that for a steam
flow rate of 0.35kg/s, the power consumed by the
copressor is 62kW. Is this possible? Explain.
6.29
A compressor is used to bring saturated water vapor
at 1bar to 10 bar where the actual exit temperature
is 420°C. Find the isentropic compressor efficiency
and the specific entropy generation.
6.30
Steam enters a turbine at 320°C and exhausts at 50
kPa. It is estimated that the isentropic efficiency of
the turbine is 75-percent. If the steam is not to be
in the two-phase region at the exhaust, what will be
the maximum turbine inlet pressure? Hint: Consider
h-s diagram of the process and apply trial and error
method.
6.31
An adiabatic air compressor is to be powered by a
direct-coupled adiabatic turbine that is also driving a
generator. Steam enters the turbine at 12.5MPa and
500°C at a rate of 25kg/s and exits at 10kPa and a
quality of 0.92. Air enters the compressor at 98kPa
and 295K at a rate of 10kg/s and exit at 1MPa and
620K. List your assumptions and determine,
a. the net power delivered to the generator by the
turbine,
b. the isentropic efficiencies of both turbine and
the compressor,
c. the total rate of entropy generation within the turbine and the compressor during this processes.
6.32
Exhaust gases at 1.9bar, 750C exits from a jet engine
powering an airplane and enter an adiabatic nozzle
at a rate of 180kg/s, and at a velocity of 85m/s. The
pressure of the surroundings at the airplane level is
0.6bar. Assume exhaust gas behaving like air,
a. Determine the maximum possible speed at the
nozzle outlet.
b. For an isentropic nozzle efficiency of 92%
evaluate the gas speed at the nozzle outlet.
c. Determine the rate of entropy change of exhaust
gases as they pass through the nozzle for case a
and case b.
6.33
A turboprop engine consists of a diffuser, compressor,
combustor, turbine, and nozzle. The turbine drives
a propeller as well as the compressor. Air enters
the
3
diffuser with a volumetric flow rate of 83.7m /s at 40
kPa, 240K, and a velocity of 180m/s, and decelerates
essentially to zero velocity. The compressor pressure
ratio is 10 and the compressor has an isentropic efficiency of 85%. The turbine inlet temperature is 1140
K, and its isentropic efficiency is 82%. The turbine
exit pressure is 50kPa. Flow through the diffuser and
nozzle is isentropic. Considering variable specific
heat for air and neglecting kinetic energy except at
the diffuser inlet and the nozzle exit, determine
a. the power delivered to the propeller, in MW.
b. the velocity at the nozzle exit, in m/s.
6.34
Air is compressed at steady state from 1 bar, 300 K,
to 6 bar with a mass flow rate of 4 kg/s. Each unit of
mass passing from inlet to exit undergoes an ideal
process described by pv1.27  Constant . The exergetic efficiency of the compressor is 0.92. If kinetic
and potential energy changes are negligible,
a. Calculate the compressor power, in kW
b. If heat transfer occurs at a rate of 4.695 kJ per kg
of air flowing to cooling water circulating in a
water jacket enclosing the compressor, determine
the exit temperature of the compressed air.
6.35
After graduating you get an interview with TEK for
an engineering job at one of their power plants. As
is the custom, the prospective employees are given a
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 269
mass flow rate of steam entering the first turbine
stage is 2.63x105kg/h.
a. Assuming that the turbines are isentropic, determine the power output of each turbine stage
and the total power of the system.
b. Sketch T-s diagram for overall process of steam
between states 1 and 4.
c. Suppose each turbine stage operates with an
isentropic efficiency of 88%, then repeat (a)
and (b).
tour of the control room. Having passed MECH3033,
you recognize a lot of the system diagrams showing
how the water flows through the various components.
You are feeling confident. As your potential future
boss is talking to you about all the great benefits they
offer, you accidentally lean on a lever. Everyone
in the room starts to scramble and you turn around
to look at what you did. The lever is marked THE
MAIN TURBINE VALVE and you have moved
it from the open position to somewhere between
the open and closed position. This valve is located
after the boiler and just before the turbine inlet. The
pressure and temperature of the water just before
the valve are 20bar and 400C, respectively. The
pressure between the valve and the turbine inlet is
15bar and the turbine outlet pressure is 0.7bar. For
a turbine isentropic efficiency of 85%,
a. show the processes through the valve and turbine
on a T-s diagram,
b. determine the turbine work output per unit
mass,
c. what is the effect of your action on the turbine
work output?
6.36
6.37
6.38
6.39
Air enters an insulated diffuser operating at steady
state at 1 bar, -3C, and 260m/s and exits with a
velocity of 130m/s. Employing the ideal gas model
and ignoring potential energy change, determine
a. the temperature of the air at the exit and the
diffuser isentropic efficiency,
b. the maximum attainable exit pressure.
Steam flowing at a rate 0.65kg/s enters at 10 bar,
400°C and 50m/s the nozzles of a Curtis Wheel. A
total of 10 identical nozzles are located on a row of
the wheel and the pressure at the exit of nozzles is
2 bar.
a. Calculate the steam velocity at the nozzle exit
for isentropic nozzles, and for nozzles having
isentropic efficiency of 83-percent.
b. Determine the inlet and the outlet cross sectional
areas of each nozzle for both cases of part (a).
Liquid water at a temperature of 25°C, and 1.5 bar
with a mass flow rate of 1.1kg/s is used as part of
non-adiabatic mixing chamber to cool steam with
a mass flow rate of 1.2 kg/s from a pressure of 1.5
bar and a temperature of 200°C to a pressure of 1.5
bars and a quality of 40%. The surface temperature
of the mixing chamber is 127°C. Determine,
a. the heat transfer from the mixing chamber,
b. the entropy generation in the mixing chamber
c. the maximum surface temperature that heat
transfer can take place.
Water is the working fluid of a two-stage turbine
system as shown in Figure 6.52. Superheated vapor
enters the first turbine stage at 8MPa, 480C and
expands to 0.7MPa. Then, it is reheated to 480C
before entering the second turbine stage, where it
expands to the condenser pressure of 0.6 bar. The
Figure 6.52 Two stage turbine with reheating
6.40
An adiabatic air compressor is to be powered by a
direct-coupled adiabatic turbine that is also driving a
generator. Steam enters the turbine at 12.5MPa and
500°C at a rate of 25kg/s and exits at 10kPa and a
quality of 0.92. Air enters the compressor at 98kPa
and 295K at a rate of 10 kg/s and exit at 1MPa and
620K. List your assumptions and determine :
a. the net power delivered to the generator by the
turbine,
b. the rate of entropy generation within the turbine
and the compressor during these processes.
6.41
Liquid water enters a 10kW pump at 100kPa pressure
at a rate of 1.5kg/s. Determine the highest pressure
the liquid water can have at the exit of the pump.
Neglect the kinetic and potential energy changes of
water, and take the specific volume of water to be
0.001m3/kg.
6.42
Fire fighters working in a shipyard desire to have a
pump with the following minimum requirements; 1.the
minimum height of the sea water jet at the exit of the
pump must be 25 meters, 2.the minimum flow rate of
sea water through each pump must be 0.052m3/s. The
pump suggested by a company consumes 6.5kW of
power, and has an efficiency of ηp=0.72. Determine
if this pump is suitable for the fire fighters.
6.43
The room air at 100 kPa, 30°C with a flow rate of
5m3/min enters the evaporator section of an air
conditioner unit. The refrigerant (R134a) of the
system at the inlet of the evaporator is saturated
liquid at -20°C, and leaves as saturated vapor at the
same pressure with a mass flow rate of 1.2kg/min.
If the heat transfer from the room of 30°C to the
evaporator is 0.3kW, determine,
270
THERMODYNAMICS
a. the exit temperature of air,
b. the rate of entropy generation due to heat transfer
process.
6.44
As shown in Figure 6.53, air (cp=1.005 kJ/kg°C) is
to be heated by hot gases in a cross-flow heat exchanger before entering the furnace for combustion.
At the exchanger inlet, air is at 95 kPa and 27°C
and volumetric flow rate is 2.1m3/s. The combustion gases (cp=1.10kJ/kg°C) enter the exchanger at
200°C with a mass flow rate of 3.7 kg/s and leave
at 135°C. Determine,
a. the rate of heat transfer to the air,
b. the outlet temperature of the air,
c. the rate of entropy generation.
a. determine the rate of entropy generation,
b. what would be the outlet temperature if the
exchanger were infinitely long?
c. define the exchanger effectiveness, and evaluate
for cases (a) and (b).
6.47
A family of four consumes averagely 120L of hot
water at 45°C every day and 365 days in a year. As
shown in Figure 6.55, an electric resistance placed
in water pipe leading to the shower head heats the
water from 18°C to 45°C. To conserve energy, the
drained water at 38°C passes through a heat exchanger
to preheat the incoming water. For exchanger effectiveness of 0.72, evaluate,
Figure 6.53 Structural view of a cross flow
heat exchanger
6.45
6.46
Brass plates 60cmx90cmx3cm in dimensions
(=8530kg/m3, cp=0.118kj/kgK) initially are at
25C and are to be heated to 500C by passing
through an oven at 700C. The plates pass through
the oven at a rate of 460plates/hour. Determine,
a. the rate of heat transfer in the furnace,
b. the rate of entropy generation due to heat transfer
process.
Figure 6.55 The shower system
a. the energy consumed by the shower per year in
kWh without an exchanger on the line,
b. the energy consumed by the shower per year in
kWh with the exchanger on the line.
c. If the cost of 1kWh of energy is 0.25TL, calculate the annual saving due to exchanger on the
line.
d. Determine the annual reduction in entropy
generation in kJ/K due to installation of the heat
exchanger. Take the surroundings temperature
at 18°C.
In a parallel flow type heat exchanger in Figure 6.54,
12kg/min of water enters at 35°C and leaves at 70°C
while oil (=820kj/kgK, cp=2.6 kJ/kg K) inlets the
same exchanger at 150°C and exits at 90°C. Assuming that the exchanger is perfectly insulated,
6.48
Figure 6.54 A parallel flow heat exchanger
Milk is pasteurized continuously at 70°C at a rate of
5.5L/s (cm=3.93kJ/kgK, =1.023kg/L) for 24 hours a
day and 365 days a year. The milk at 22°C is heated
to the pasteurizing temperature by hot water at 85°C
that is heated in a natural-gas-fired boiler with an efficiency of 78%. The pasteurized milk is then cooled
by cold water at 20°C before it is finally refrigerated
to the storage temperature of 5°C. As shown in Figure
6.56, to save energy, the plant installs a regenerator
that has an effectiveness of 72% and heats up the
water returned from pasteurization process at 37°C.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 271
Take the heating value of natural gas as Hu=37800kJ
per normal m3, and determine,
a. the amount of water to be circulated,
b. how much natural gas in percent the regenerator
will save per year?
c. the percentage of annual reduction in entropy
generation.
6.50
As shown in Figure 6.58, air with a mass flow rate
1.2kg/s at 95kPa, 10°C is compressed to 2.0MPa by
isentropic two-stage compressor. At the intermediate pressure, the air is cooled down to 10C by an
intercooler. Determine,
a. the intermediate pressure for the minimum power
input to the compressor,
b. the total power required,
c. the mass flow rate of water through the intercooler
for 10C temperature change of the cooling
water.
d. Show the overall compression process of air on
T-s diagram
e. Repeat (b), (c) and (d) for isentropic efficiency
of 0.8 of each stage of the compressor.
Figure 6.56 Continuous pasteurization of milk
6.49
A geothermal heat pump operating at steady state
with refrigerant22 as the working fluid is shown
schematically in Figure 6.57. The heat pump uses
15°C water from wells as the thermal source. The
operating data for a day in which outside air temperature is -7°C is shown in the figure. For the heat
pump, determine,
a. volumetric flow rate of heated air in m3/min,
b. the isentropic compressor efficiency and the
power consumed by the compressor,
c. the coefficient of performance of the heat pump
at given conditions,
d. the volumetric flow rate of water through the
geothermal wells in liters/min.
e. the entropy generation rate of the system.
Figure 6.57 Schematic of a geothermal
heat pump
Figure 6.58 Two-stage compression with
intercooling
6.51
In Figure 6.52, suppose the number of turbine stages
having the same pressure ratio and the same reheat
has increased to a large number. Sketch T-s diagram
of the process and show that the total work obtained
becomes identical with isothermal process as the
stage number goes to infinity.
6.52
In Figure 6.58, suppose the number of compressor
stages having the same compression ratio and the
same inter cooling has increased to a large number.
Sketch T-s diagram of the process and show that
the total work consumed becomes identical with the
isothermal compression process as the stage number
goes to infinity.
6.53
A company needs vapor at 180C with 10% of quality.
It is expensive to purchase high pressure boilers to
produce such a low quality but high pressure vapor.
The company has saturated vapor available at 1.5
bar. As shown in Figure 6.59, an engineer claims that
272
THERMODYNAMICS
raising the pressure of the available vapor to 10bar
and mixing it with high pressure water at 80bar, and
20C will do the job. The engineer also states that
the relation between the mass flow rates must be as,
Isentropic flow
6.56
For a perfect gas with k=1.2, determine the Mach
number that will yield a temperature ratio of
T/T0=0.91. Evaluate also the pressure ratio p/p0 at
this flow conditions.
6.57
Carbon dioxide at a temperature and pressure of 340K
and 1.5 bar respectively is flowing with a velocity
of 200m/s. Evaluate,
a. the sonic velocity and the Mach number at this
state,
b. the stagnation density
6.58
An airplane is flying at an altitude of 6000m where
the temperature and pressure respectively are -18C,
0.53 bar. If the speed of the plane is 980km/h, evaluate the temperature and the pressure on the nose of
the plane.
6.59
Methane gas flowing in an adiabatic and no work
system with negligible change in potential energy
assumes at section 1 the following values for pressure, temperature and velocity: p1=12bar, T1=500K,
and V1=125m/s. If M2=0.8 at a downstream section,
determine,
a. the temperature, the pressure and the velocity
at section 2,
b. the area ratio, A2/A1.
6.60
Oxygen gas with stagnation values, T 0 = 285
C, p0 = 7 bar, enters a device with a cross sectional area
A1 = 0.1m2, and Mach number M1 = 0.2 at the entrance.
There is no heat, work transfer nor losses due to flow
of the gas through the device, and leaves the device
at atmospheric pressure of p2 = 1bar. Evaluate,
a. 1, V1, and the mass flow rate,
b. M2, V2, T2, 2, and the crosssectional area at the
exit, A2.
6.61
A perfect gas flows through an adiabatic, no work
and no loss system. Show that the sonic velocity at
the stagnation conditions, a0, is related to the sonic
velocity at Mach number unity (M=1), a*, as fol-
m w  2.62m v .
a. Check if the relation m w  2.62m v is true.
b. Analyze the process if it is thermodynamically
possible.
Figure 6.59 Producing low quality but high
pressure vapor
6.54
Water as in Figure 6.60 flows isothermally at 300K
through a pipe of 5mm in diameter with an average velocity of 0.1m/s. Assuming that the flow is
fully developed ( x  Le ), determine the variation
of entropy generation rate with the radius, and the
energy loss for a pipe length of 255 meters.
Figure 6.60 Development of velocity profile as
fluid flows through a pipe
1/2
6.55
The pipe in Figure 6.60 is subject to constant heat
flux at its surface at a value of 5W/m. Water inlets the
pipe at a temperature of 300K. All other parameters
and dimensions are the same as in problem (6.54).
Assume both the velocity and the temperature profiles
are fully developed and determine,
a. the variation of average temperature with respect
to x,
b. the variation of entropy generation rate with
(r, x),
c. the energy loss for the same pipe length.
lowing,
6.62
a*  2 

a0  k  1 
.
As shown in Figure 6.61, two venturi meters are
installed on a 25cm diameter insulated duct. Air flows
through the venturi meters isentropically, because
of friction through the connecting pipe, however the
flow is nonisentropic. Besides the flow conditions for
air are such that M1=M4=1. The diameter at section
1 is 10cm, and for throat pressures of p1=275kPa,
p4=225kPa, determine,
a. the entropy change at the connecting duct,
b. the diameter at section 4.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 273
c. the throat area if all irreversibilities are at the
diverging section of the nozzle.
d. the mass flow rate.
6.68
Figure 6.61 Flow through a duct
6.63
 dx  V 2
holds,  dp  f  
 VdV
 d  2
Nitrogen stored in a large tank is at 400K and 1.5
bar. The gas leaves the tank through a convergent
nozzle whose outlet area is 25cm2. The pressure of
surroundings is 1 bar. For no loss flow, evaluate,
a. the velocity of nitrogen at the nozzle exit,
b. the mass flow rate,
c. the maximum value of the flow rate obtained
by lowering the ambient pressure.
6.64
Air flows through a convergent-divergent nozzle.
Air is at a state of 20 bar, 40C at the nozzle inlet,
and the pressure at the nozzle exit is 2 bar. Assuming
isentropic flow, for a throat cross sectional area of
0.002m2, evaluate,
a. the nozzle cross sectional area at the exit,
b. the mass flow rate of air.
6.65
Air at a flow rate of 10kg/s flows through an adiabatic
duct. At one section of the duct p1=2 bar, T1=650C,
and the cross sectional area is 0.005m2. At another
section 2 in downstream direction, the flow Mach
number is 1.2. Determine,
a. the area A2 if the flow is frictionless, and sketch
the general shape of the duct,
b. the area A2 if the entropy change between these
two sections is 0.042 kJ/kgK
Hint: Trial-and-error method should be used in
solving section b.
6.66
Carbon monoxide expands adiabatically from 7 bar,
285C through a converging-diverging nozzle to a
pressure of 1.4 bar. If the velocity of the gas at the
nozzle inlet is negligible, evaluate,
a. the ideal exit Mach number,
b. the nozzle efficiency for actual Mach number
of 1.6.
c. the entropy change of the flow.
d. Sketch the T-s diagram showing the ideal and
the actual processes with pertinent pressure and
temperature values.
6.67
Air at a state of T1=25C, p1=10 bar enters a converging-diverging nozzle with negligible velocity.
The Mach number at the exit is 2.0, and the cross
sectional area is 0.1m2. For a nozzle efficiency of
0.92, calculate,
a. the actual values of p, T, and p0 at the exit,
b. the Mach number at the exit for isentropic
flow,
As shown in Figure 6.62, consider steady, onedimensional, isothermal flow of an ideal gas with
no shaft work, flowing through a horizontal duct
having a constant area A, and a perimeter P. Show
that
a. as a result of momentum principle the following
b. as a result of continuity and the equation of state
d  dp
dV


the following holds,

p
V
c. combining the results of parts (a) and (b) yields,
 fdx
d   kM 2


  2(kM 2  1)  d
Figure 6.62 Relation between density and
friction for isothermal flow
Unsteady systems
6.69
As shown in Figure 6.63, a pneumatic lift in a mechanic shop is used to lift a 1230kg Mercedes C180
by means of an adiabatic compressor. Initially the car
is at the ground level and the air in the cylinder is at a
pressure that balances the piston weight, m p  200kg
and the temperature is atmospheric, T1=20C. The
cylinder cross sectional area is 0.07m2 and initially
the piston is 0.8m away from the bottom of the
cylinder. The car is raised by 1.7m by pumping air
into the cylinder by a compressor having isentropic
efficiency of 0.82. The entire system is insulated and
the atmospheric conditions are p0=100kpa, T0=20C.
Determine,
a. the temperature of air in the cylinder when the
piston just starts moving upward,
b. the final temperature in the cylinder,
c. the work required just to initiate the piston moving,
d. the total work done by the compressor,
e. the total entropy generated by this process.
274
THERMODYNAMICS
6.72
Instead of air in the tank of Figure 6.65, it contains
water with 60% by volume of saturated vapor in
equilibrium with its liquid at 5bar pressure. The tank
content is heated until the pressure reaches 20 bar
and then the valve opens, the steam flows through
the adiabatic turbine with isentropic efficiency of
84% and exits to atmospheric pressure of 1 bar.
The process continues until tank content becomes
saturated vapor at 20 bar. Determine,
Figure 6.63 Pneumatic lift application
6.70
As in Figure 6.64, a cylinder of 1m in length and
0.5m in diameter is divided by an adiabatic and
frictionless piston and the left side contains air at
1bar, 27C. Initially, the air occupies the entire cylinder. The valve at the top of the cylinder opens and
carbon dioxide gas from main at 10 bar, 27C flows
slowly into right side and the flow stops when the
cylinder pressure becomes 10 bar. Assuming ideal
gas behavior for both air and carbon-dioxide with
constant specific heats, determine,
a. the final temperature of carbon-dioxide in the
cylinder,
b. the final temperature of air,
c. the amount of CO2 entered the cylinder
d. the entropy change of air and CO2.
e. Explain if the overall process is reversible or
not?
Figure 6.65 The maximum work delivered by
pressurized air
a. the mass of steam flowing through the turbine,
b. the temperature of steam at the turbine exit,
c. the amount of work delivered at the turbine
shaft.
Thermodynamic relations
6.73
Using the Tables of ammonia estimate the following
properties of saturated vapor ammonia at -10C.
a. the volume expansivity, 
b. the isothermal compressibility, 
c. the isothermal and isentropic modulus of elasticity, B, and Bs.
6.74
Find a relation for the change of Gibbs free energy
with temperature at constant volume, G / T v .
Express the relation in terms of the volume expansiv-
Figure 6.64 Carbon-dioxide filling process
6.71
As in Figure 6.65, a tank of 5m3 of volume contains
pressurized air at 20 bar, 400C. The tank is connected to an air turbine which exhausts the air into
the atmosphere at (po=1bar, To=7C) by a reversible
and adiabatic process. Neglect the volume occupied
by the connecting pipes and determine,
a. the amount of air discharged through the turbine
until the flow stops.
b. the amount of work received at the turbine
shaft.
c. Calculate the exergy change of air and compare
this change with the shaft work.
ity  , the isothermal compressibility  .
6.75
Using mathematical and thermodynamic relationships, write algebraic expressions for the following
partial derivatives: a. S / U  , b. V / U T ,
v
c. F / V s
6.76
Using, du  Tds  pdv and appropriate Maxwell’s
relations find out a relation for u / p T that only
parameters p, v, T,  and  are involved. Determine
the value of this partial derivative, u / p T , for
an ideal gas.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 275
6.77
Water vapor initially at 280bar and 440C is heated
isothermally to a pressure of 320bar. Estimate the
ideal gas deviation in specific enthalpy and entropy
by considering the critical point data as, pc  22.12
Starting with the fundamental equation,
dh  Tds  vdp and using Maxwell’s relations,
express h / v T and h / T v in terms of p,
T, and v.
MPa, Tc  374.15 C for water. Take a  0.147 and
Hint: Use relations like cv  u / T v  T s / T v
b  0.087 .
wherever applicable.
6.78
6.83
Show that the dependence of entropy on Gibbs free
energy and enthalpy may be described as following:




cp
c p   Ts
ds   
 dg  
 dh
 T 1   T s  Tc p 
 T 1   T s  Tc p 
6.79
The triple point data of water is given as following:
A well insulated vessel is divided into two compartments by a partition. The volume of each compartment
is 1m3. One compartment initially contains 10kg of
Argon at 20C, and the other is evacuated. The partition ruptures and the gas is allowed to equilibrate.
Assume that the gas obeys Van der Waal’s equation
of state and determine the final temperature.
ptp  0.61kpa, Ttp  0.01C .
Phase
Enthalpy
(kj/kg)
Volume
(m3/kg)
Entropy
(kj/kgK)
Solid
-333.4
0.00109
-1.221
Liquid
0.01
0.001
0.000
Vapor
2501.4
206.1
9.156
a. Using the triple point data, estimate the melting
temperature of ice at 10MPa.
b. A 60kg man is ice-skating on blades as shown in
Figure (6.66). The area of contact of each blade
is 0.11cm2. Determine if the ice under the blades
will melt for ice temperature of -3C.
Hint: Consider, T  T V , U  , dU=0 for this particular application, and express T / V u by cyclic
relation. The atomic mass of Argon is 39.948kg.
6.80
a.
6.81
Which of the following curves identifies the correct
dependence of specific Gibbs free energy g on pressure p at a constant temperature?
b.
c.
d.
Copper 22kg of mass at 27C and 100kPa is first
heated reversibly to 227C and then compressed
isothermally to 100Mpa of pressure. Determine,
a. the volume at its final state,
b. the entropy change,
c. the work done,
d. the heat transfer,
e. the internal energy change of copper.
Figure 6.66 Ice skating blade
6.84
The saturation pressure and the enthalpy of vaporization of R22 at 20C are respectively tabulated as
9.1 bar, and 187.28kJ/kg. Estimate the saturation
pressure at 0C and determine the percent error by
comparing with the tabulated value of 4.9811 bar.
Hint: Consider that the final state is reached by two
processes as; isobaric heating and then reversible isothermal compression. Express V, S and U accordingly.
Assume, c p (kj / kgK )  0.355  T / 104  T 2 / 106 ,
=894kg/m3,   51x106 K 1 ,   6.6 x109 kpa 1 .
6.82
For a certain range of pressures and temperature;
1.5  pr  6 , and 1.1  Tr  1.5 , the compressibility factor may be expressed as, Z  

a  bpr Tr3/2
.
Figure 6.67 Transfer of liquid oxygen by
double walled pipe
6.85
As illustrated in Figure 6.67, to reduce the heat losses
in transferring liquid oxygen, the transfer line is
designed with a double walled pipe and the annular
276
THERMODYNAMICS
space is filled with CO2 gas at ambient pressure.
The liquid oxygen flows through the line at 100K,
and most of the carbon-dioxide gas in the annulus
will freeze, and accumulate on the inner wall. The
pressure in this space will be the sublimation pressure of CO2 at 100K. Since no available data at this
low temperature, carbon-dioxide pressure will be
approximated by the data at the triple point. Evaluate the pressure of CO2 by considering that at triple
b.
Entropy of a closed system performing
a reversible process can never decrease.
c.
The entropy of R134a increases as it
flows through an expansion valve.
d.
If an ideal gas compressed isentropically, its temperature always goes up.
e.
The outlet of an actual turbine is at
higher energy state than that of the same turbine
operating ideally with the same inlet state and
the outlet pressure.
f.
The internal energy of a system and
its surroundings is not conserved during an irreversible process.
g.
When a closed system with internal
irreversibilities undergoes a process from state
1 to state 2, the entropy change is negative and
the entropy production is positive.
h.
It is possible to compress air adiabatically from (1bar, 150C) to (2bar, 300C) inside a
closed device.
i.
The relation pv k  Constant corresponds
to an incompressible process.
An elastic rod has a length X and temperature T
when a tensile force of F is applied on its ends. The
internal energy u and the entropy s of the system
j.
Mollier diagram is especially useful for
analyzing flow systems in energy interactions.
are determined by the following relations; u  cT ,
k.
In reducing the work consumption of
high pressure ratio compressors, multi-stage
compression with intercooling is applied.
l.
The effectiveness of a heat exchanger
compares the actual heat transferred with the heat
that would be transferred by an infinitely large
exchanger operating at the same conditions.
m.
The stagnation conditions are those that
would be obtained when the flow at any point
in the fluid stream was isentropically brought
to rest.
n.
The critical conditions for a flow in a
nozzle are those that would exist if the flow is
accelerated until the Mach number is unity.
o.
When M  1 , velocity and area change
are in the same direction, and for M  1 , velocity
and area change are in the opposite direction.
p.
The figure on the right represents a
subsonic nozzle or a supersonic diffuser.
point, Ttp  56.6C , ptp  510 kPa , and hsg  574.5
kJ/kg.
6.86
In throttling of gases, the change in temperature with
respect to change in pressure at constant enthalpy
conditions is called Joule-Thomson coefficient J .
Depending upon the volume expansivity, the specific
volume, the constant pressure specific heat and the
temperature of the gas, show that J can be determined
as, J 
 v 
T
 v
 T  p
cp
. Referring to the generalized
equation of state for real gases, pv  ZRT , show
also that J 
6.87
RT 2  Z 
.
pc p  T  p
2
s  c ln T  KX where c and K are constants.
a. Show that an equation of state for this system
is F  2 KX  0 .
b. The elastic rod performs a Carnot cycle between
heat reservoirs at 87C and 27C. The minimum
length at higher temperature is 2 meters, and
the work extracted per cycle is 100 Joules. For
c  3.0 J/K, and K  5.5 Joules/m2K, evaluate
the maximum length of the rod at higher temperature for engine efficiency of 10-percent.
6.88
 dp 
Use the C-C equation, h fg  Tv fg 
 , to
 dT  sat
estimate the enthalpy of vaporization of R134a at
24C. Assume that for pressure difference of p  1
bar, the saturation temperature change is given as,
T  5.14 K.
True and False
6.89
Answer the following questions with T for true and
F for false.
a.
The entropy of a material can never
decrease.
CHAPTER 6 ENTROPY: A SYSTEM DISORDER 277
q.
If air inlets a channel at M  1 , to get
a test section for supersonic flow conditions, a
converging-diverging type channel configuration
has to be designed.
r.
The Clapeyron equation discussed in the
text is used to compute the relationship between
the specific heats c p and cv .
3.
Liquid water is to be compressed by a pump whose
isentropic efficiency is 70% from 2bar to 50bar at
a rate of 0.12m3/min. The required power input in
kW to the pump is,
a. 13.71
s.
4.
Considering the definitions,
c p  h / T p  T s / T p and
cv  u / T v  T s / T v , you may conclude that the slope of constant v curve is bigger than the slope of constant p curve in a T-s
diagram.
t.
u
It can be shown that
cv / T T  T 
2
p / T
2
.
b. 137.1
c. 6.71
d. 9.71.
A horizontal cylinder as shown in Figure 6.68 is
separated into two compartments by an adiabatic and
frictionless piston. One side of the cylinder contains
0.3m3 of N2 and the other side contains 0.1kg of air
both initially at 270C, and 1bar. The sides of the cylinder and the air end are well insulated. Heat is added
to the nitrogen side from a heat source at 2270C until
the pressure of air rises to 120kPa. The total entropy
generated in kJ/K during this process is,
a. 0.0134
b. 0.134
c. 0.114
d. 0.124.
v
The relation
p / v T v / T p T / p v  1 is valid
for all p-V-T state relations in solids, gases,
and liquids.
v.
A body in equilibrium with a reservoir at
constant pressure and temperature will have the
lowest possible value of Gibbs free energy.
w.
If the viscous dissipation function of
a fluid is zero then no entropy generated by an
isothermal flow.
Figure 6.68
5.
Check Test 6
Choose the correct answer:
1.
An adiabatic air compressor with 85-percent isentropic
efficiency is powered by direct coupled turbine that
is also driving a generator. Steam enters the turbine
at 10bar, 4400C at a rate of 10kg/s, and exits as
saturated vapor at 10kPa. Air enters the compressor
at 98kPa, 290K at a rate of 10kg/s, and exits at 5bar.
The power delivered in MW by the generator with
an efficiency of 95% becomes,
a. 0.711
c. 0.511
d. 1.111.
For surroundings condition at 98kPa, 290K, the
total rate of entropy generation in kW/K by the
entire system of problem 1 including the turbine,
the compressor, and the generator might be,
a. 3.29
a. 0.227
b. 2.33
b. 5.29
c. 4.33
c. 7.29
d. 5.33.
d. 9.29.
6.
b. 0.911
A jet transport aircraft with a pressurized cabin
cruises at 10km altitude where the outside pressure is 30kPa. The cabin temperature and pressure
initially are 230C and 76kPa respectively. The cabin
volume is 22m3. Air escapes through a small hole
approximated as a converging nozzle with an exit
area of 0.0015m2. After a certain time interval, the
cabin pressure decreases by 30%. Assuming that
the temperature in the cabin remains constant, an
approximate value of the average mass flow rate of
the escaped air in kg/s is,
a. 3.33
2.
Air from a large reservoir is isentropically flowing
through a converging-diverging nozzle that has
throat and exit cross sectional areas of 0.001m2
and 0.0015m2 respectively. At the reservoir air is
at 310kPa, and 310K, and discharges into a region
where the pressure is 290kPa. The mass flow rate
of air in kg/s through the nozzle is,
b. 0.127
c. 0.0127
d. 0.327.
278
7.
THERMODYNAMICS
Using the methodology of your choice, you may
compute V / U T of a substance in terms of the
measurable properties as the isothermal compress-
9.
ibility,  , the volume expansivity,  , pressure, p ,
and temperature, T as following,
T 

a.  p 
 

c. p 
T
,

1
,
 T

b. 
 p
 

d.
1
,
T
p.

a. Evaporator, 0.1051,
b. Condenser, 0.0138,
c. Condenser, 0.1051,
d. Compressor, 0.1051.
10.
8.
The boiling temperature of naphthalene at 1bar
pressure is 218°C. The latent heat of vaporization
at this temperature is given as, h fg  338.28kJ / kg .
Then the boiling pressure of naphthalene at room
temperature of 27°C in kPa might be,
a. 0.227,
b. 0.427,
c. 0.117,
d. 0.327.
Rapid freezing of fresh fruit requires air at -34°C.
A refrigeration system with R134a produces air at
this temperature by operating at -40°C of evaporating and +46°C of condensing temperatures. The
refrigeration system rejects heat to surroundings at
300C, and the isentropic efficiency of the system
compressor is 90-percent. The refrigerant leaves the
evaporator by +40C of superheat. The evaporator
cooling load at steady conditions is 20kW. On these
operating conditions, the device that generates the
most amounts of entropy and its entropy generation
rate in kW/k are as follows,
In a water cooled ammonia condenser, ammonia enters at 12bar, 80°C with a mass flow rate of 200kg/h,
and exits at 12bar, 300C. The cooling water inlets the
condenser at 150C and outlets at 230C. Ignoring the
heat transfer from the outer surface of the condenser,
then the entropy generation rate of the exchanger in
W/K becomes,
a. 11.65,
b. 6.65,
c. 16.65,
d. 22.65.
C
H
7
A
P
T
E
R
Gas Mixtures & Psychrometry
7.1
Basic Definitions for Mixtures
A large number of thermodynamic problems are related to the mixture of different pure
substances. As depicted in Figure 7.1, the design of an air pre-heater used for the heat recovery of waste heat of a fossil fuel fired power plant certainly requires knowledge in gas
mixture thermodynamics. Both fluids of Figure 7.1, air and the flue gas are gas mixtures.
Flue gas produced by a combustion process of coal is a mixture of gases like; N2, H2O, CO,
CO2, NO, and SO3. Then the question is how the mixture properties can be computed with
respect to constituents and the composition.
The mixture properties depend not only on the properties of the individual gases (called
components) but also on the amount of each component. Hence, rules have to be developed
for estimating the mixture properties like the pressure, energy, enthalpy, entropy, etc., by
the similar properties of each component and the mixture composition.
279
280
THERMODYNAMICS
As shown in Figure 7.2, let us study the properties of a binary mixture (a mixture of two components). Any extensive property of a mixture, H, may be formulated as, H  f  p, T , n A , nB  . Then,
the change in H at the temperature and pressure of the mixture is,
dH p,T
 H 
 H 

dn A  
dnB


 n A  nB , p ,T
 nB nA , p ,T
(7.1)
 H 
 H 
and H B  
respectively define the partial properIn this formulation, H A  


 n A  nB , p ,T
 nB  nA , p ,T
ties if both components A and B are at the same temperature and pressure of the mixture. Similarly, if
H is considered to be formulated as, H  f V , T , n A , nB  , then the partial properties may be defined
as each component is at the same temperature and volume of the mixture as,
dH V ,T
 H 
 H 

dn A  
dnB


 n A  nB ,V ,T
 nB  nA ,V ,T
(7.2)
Definition: The partial properties of a component ‘i’ in a mixture of n components may be defined in
two different methods: a. HA, represents the partial rate of change in that property caused by the only
*
change in the amount of A at the same temperature and pressure of the mixture. b. H A represents
the partial rate of change in that property caused by the only change in the amount of A at the same
temperature and volume of the mixture.
The above definitions are applicable to all extensive properties of a mixture as, V, U, H, S, F, and
G. Hence for a binary mixture, both definitions yield,
H p ,T  H A n A  H B nB 




*
*
H V ,T  H A n A  H B nB 
(7.3)
Two types of fractions are defined and used in mixture analysis. For molar or volume analysis,
the mole fraction ( yi ) is used. For gravimetric analysis, the mass fraction ( xi ) is determined and both
of these fractions are expressed as following,
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 281
yi  ni / n , n 
n
i
and xi  mi / m , m 
i
m
(7.4)
i
i
In calculating the mean molecular mass of the mixture, however, the total mass of the system has
to be known as, nM m  n1M 1  n2 M 2  ..  nn M n , then the mixture molecular mass becomes,
Mm 
yM
i
(7.5)
i
i
After determining the mean molecular mass, conversion between mole and mass fractions can
be done as following,
xi 
Mi
yi
Mm
(7.6)
The most common gas mixture of nature is the atmospheric air, and for all practical purposes, the
composition of atmospheric dry air is considered to be constant as provided in Table 7.1.
Table 7.1 Properties of Atmospheric dry air
Constituents
Molecular
weight
Symbol
Volumetric
(%)
Gravimetric
(%)
yi M i
Oxygen
31.99
O2
20.95
23.14
6.702
Nitrogen
28.01
N2
78.09
75.53
21.873
Argon
39.95
Ar
0.93
1.28
0.372
Carbon dioxide
44.01
CO2
0.03
0.05
0.013
yM
i
i
 28.96
Since the existence of Argon and Carbon dioxide is negligibly small, for approximate calculations,
the atmospheric dry air may be assumed to be as indicated in Table 7.2.
Table 7.2 Approximate composition of atmospheric dry air
Molecular Volumetric (%)
yi  ni / n
Constituents weight
Gravimetric (%)
xi  mi / m
Oxygen
31.99
21.00
23.14
Nitrogen
28.01
79.00
75.53
3.76:1
3.29:1
N2/O2 ratio
Example 7.1 A tank of 0.5m3 of volume contains 0.35kg of carbon monoxide (CO), 0.65kg of carbon dioxide (CO2), 1kg
of sulfur trioxide (SO3). Calculate,
a.
the mass fraction, the mole fraction of each constituents and the molar mass of the mixture.
b.
the amount of carbon monoxide to be pumped into the tank to increase the mass ratio to 20-percent.
282
THERMODYNAMICS
Solution:
a.
The following table may be prepared for mass and the mole fractions of the constituents.
Constituents
mi
Mi
ni  mi / M i
xi
yi
yi M i
CO
0.35
28
0.0125
0.175
0.315
8.82
CO2
0.65
44
0.0147
0.325
0.37
16.28
SO3
1.00
80
0.0125
0.500
0.315
25.20
 n  0.0397  x  1.0  y  1.0  y M
i
i
i
i
i
 50.3
As indicated on the table above, the molar mass of the mixture is 50.3kg.
b.
To increase the carbon monoxide mass fraction to 20-percent, the following relation should hold, 0.35  x   0.2  2  x 
and the mass of CO to be pumped x  0.0625 kg.
In chapter 2, first, p-v-T behavior of ideal gas model of a pure substance is studied. Then p-v-T
behavior of real gasses followed. The same sequence may be considered for studying the properties
and p-v-T behavior of non-reacting ideal gas mixtures and real gas mixtures. However, similar to the
state principle introduced for a single substance, one has to know the number of independent properties
(degrees of freedom) required to identify the state of a mixture of n components. In determining the
degrees of freedom of a gas mixture with n components, Gibbs phase rule may be applied.
Principle 23: Gibbs Phase Rule: For a system consisting of n components
and p phases in equilibrium, the number of independent properties is calculated by f  n  p  2 .
Hence, for a binary gas mixture, n = 2, since both substances are in gas phase, p=1, then the
number of variables required to fix the state of a system is, f = 2 – 1 + 2 = 3. Thus, three properties,
usually p, T, and x1 are specified to identify a particular state of the mixture.
Example 7.2 Determine the number of independent properties to be specified for water under phase equilibrium as in
Figure 7.3.
a.
Two phases are in equilibrium,
b.
Three phases are in equilibrium.
Solution:
a.
For two phases in equilibrium, the number of components of the system, n  1 , and the number of phases,
p  2 , then f  1  2  2  1 . Only the temperature
or the pressure can vary.
b.
If the three phases are in equilibrium, p  3 and
n  1 then f  1  3  2  0 . As described in chapter
2, there is no independent property at the triple point.
The triple point takes place at a particular value of
pressure, temperature, and volume.
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 283
7.2
7.2.1
p-v-T Behavior of Gas Mixtures
Ideal gas mixtures
Principle 24: When the effect of dissimilar molecules on each other in a mixture
is negligible, then each gas component in the mixture behaves like it exists alone
at the mixture temperature and volume.
If a gas mixture obeys the above principle then it is called ideal gas mixture. For an ideal gas
mixture, then the partial property of component ‘i’ becomes the property of the pure substance at the
same volume and temperature of the mixture. For instance, the internal energy of a mixture is ex-
U  U 0 V ,T
pressed
as, is
component





 n u  u  
i
0
i
i V ,T
. For an ideal gas mixture, the internal energy of each
independent of volume and expressed as, u  u0 i  cvi T  T0  . Thus, substitution
yields the internal
energy of the mixture as following,

u  u0  


 y c  T  T 
i vi
(7.7)
0
i
The same procedure may be followed for determining the enthalpy and the entropy of an ideal
mixture as,

h  h0  


 y c  T  T 
s  s0 
i pi
(7.8)
0
i
 y s  s 
i
(7.9)
0 i
i
If the ideal gas mixture is treated as a single gas, then the corresponding specific heats may be
expressed as,
c pm (kJ/kmolK) 
yc
cvm (kJ/kmolK) 
i pi
i
yc
i vi
(7.10)
i
or
c pm (kJ/kgK) 
x c
i pi
i
cvm (kJ/kgK) 
x c
i vi
(7.11)
i
The same principle, however, does not work for relating the intensive properties; properties like
pressure and specific volume of the components.
Example 7.3 Express the mole volume of a binary mixture in terms of partial mole volumes of constituents at the same
temperature and volume as the mixture. Show that intensive mole properties of a mixture cannot be represented by partial
properties.
284
THERMODYNAMICS
Solution:
Each component occupying the same volume V, the specific mole volumes expressed as, v A  V / n A , vB  V / nB . In accord with Eq. (7.2), the following relation should hold, v  v A  vB . However, n  n A  nB and if we divide both sides by
the mixture volume V,
1 1
1
n n A nB

or 
and this result is quite different from Eq. (9.2).


v v A vB
V
V
V
Hence for determining the intensive properties of a mixture Dalton’s rule of additive pressures
or Amagat’s rule of additive volumes have to be utilized instead.
Principle 25: (a) The pressure of a gas mixture is the sum of contributing pressures
of each component gas. The partial pressure of component ‘ i ’ is evaluated when
all other components are removed but the volume and the temperature remained the
same. This is known as Dalton’s rule of additive pressures. (b) The volume of a gas
mixture is the sum of the volumes of each component evaluated at the mixture pressure
and temperature. This is also known as the Amagat’s rule of additive volumes.
As shown in Figure 7.4, if one applies the Dalton’s rule to a binary mixture, the pressure of the
mixture at V and T becomes, p V , T   p A V , T   pB V , T  . Here, the partial pressure of each component is quite different from the definition given by Eq. (7.3). In regard to Amagat’s rule, however,
the volume occupied by the mixture at (p, T) is expressed as, V  p, T   VA  p, T   VB  p, T  . These
results can be generalized for a mixture with n component as,

 p V , T  


V p, T 

 


n
 p V ,T 
i
i 1

Vi  p, T  
i 1

n

(7.12)
Since both expressions in the set of Eq. (7.12) yield identical results for ideal gases, there is no difference between these two rules for ideal gas behavior. In fact, for an ideal gas component being at the
temperature and volume of the mixture, the partial pressure is piV  ni T , or being at the temperature
and pressure of the mixture, the partial volume is pVi  ni T and the ratio of these two expressions
to pV  nT yields,
yi 
ni pi Vi


n
p V
(7.13)
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 285
The mole fraction of component i indicates the volume ratio as well as the partial pressure ratio
of that component. If one considers the equation of state for component i on mass basis, piV  mi RiT ,
T
pi 
mi Ri then the gas constant of the mixture becomes,
since p 
V


R
 x R (kJ/kgK)
i
(7.14)
i
Since cp, cv, and R values of an ideal gas mixture can be evaluated by the corresponding values of
its constituents, as illustrated in the following examples, processes related to ideal gas mixtures can
be reduced to a process of a single gas.
Example 7.4 As shown in Figure 7.5, 0.1kg of combustion gases in an engine cylinder initially at p1  150 bar, V1  0.0022 m3
has the following volumetric analysis: 5%CO, %12CO2, 8.5%O2 and 74.5%N2. In accord with the relation pv1.3  Constant ,
the gas expands reversibly through a volume ratio of 7:1 in the cylinder. Determine,
a. the work done by the gas,
b. the heat transfer through the engine cylinder wall,
c. the entropy change of the gas mixture
Solution:
The following table calculates the molar mass, the mass fractions of each component, the mixture gas constant and the
constant pressure specific heat by considering the mixture composition.
Component
yi
Mi
yi M i
xi
Ri
xi Ri
c pi
xi c pi
CO
0.05
28
1.40
0.046
0.297
0.0136
1.127
0.0518
CO2
0.12
44
5.28
0.174
0.189
0.0328
1.15
0.2001
O2
0.085
32
2.72
0.089
0.26
0.0231
1.031
0.0917
N2
0.745
28
20.86
0.689
0.297
0.205
1.098
0.7565
∑
a.
30.26
0.274
1.1
With respect to the results on the table above, R  0.274 kJ/kgK, c p  1.1 kJ/kgK, and the work done W12 
n
Due to polytropic expansion,
W12  48.65 kJ.
p1V1  p2V2 .
n 1
1.3
p2  v1 
1
       0.0796 , p2  11.95 bar, V2  0.0154 m3 and the work is
p1  v2 
7
286
b.
THERMODYNAMICS
Apply 1st law to the piston-cylinder device as, Q12  U 2  U1   W12 where, U 2  U1   mcv T2  T1  and
cv  c p  R  0.826 kJ/kgK, T1 
p1V1 15000  0.0022

 1204 K, and T2  671.6 K. Hence the internal energy
mR
0.1  0.274
change becomes, U 2  U1   43.97 kJ, and the heat to be supplied is, Q12  4.68 kJ.
c. The entropy change of the mixture is S  m s2  s1  , since the mixture properties are known,
s2  s1   cv ln T2 / T1   R ln v2 / v1 
or
s2  s1   0.826  ln 0.557   0.274  ln 7 / 1  0.0498 ,
and
S  0.00498 kJ/K.
Example 7.5 Combustion gases of a fuel have the following volumetric analysis: 9.87%CO2, 4.00%CO, 14.2%H2O,
0.9%H2, 71.03%N2. As illustrated in Figure 7.6, the combustion products at 400C enter to the economizer of a boiler with
a mass flow rate of 2.1kg/s. If the gas temperature drops to 150C at the economizer exit, determine the mass flow rate of
water so that its temperature will be increased by 70C.
Solution:
With respect to mixture composition, the following table determines the molar mass, the mass fractions of each component,
and the constant pressure specific heat of the mixture at the average bulk temperature.
Component
yi
Mi
yi M i
xi
c pi
xi c pi
CO2
CO
H2O
H2
N2
∑
0.0987
0.040
0.142
0.009
0.7103
44
28
18
2
28
4.342
1.12
2.556
0.018
19.88
27.924
0.155
0.04
0.0915
0.00064
0.712
1.02
1.065
1.985
14.52
1.056
0.1581
0.0426
0.1816
0.0093
0.752
1.143
For steady state conditions, with negligible kinetic and potential energy changes, if we apply the 1st law to the economizer, the following result is obtained: m mc pm Tin  Tex   m wc pwT , or m w 
m w  7.38 tons of water can be heated from 20C to 90C per hour.
2.1  1.143  250 
4.18  70
 2.05 kg/s. Hence,
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 287
Example 7.6 “Entropy increase in mixing of ideal gases.” As in Figure 7.2, let us consider two different gases, nA
moles of gas A at pressure p, and temperature T, be mixed with nB moles of gas B at the same pressure and temperature in
an adiabatic chamber of constant volume. Formulate the entropy increase for this process.
Solution:
0
0
If we apply the 1st law to the entire chamber of Figure 7.2, dU   Q   W  0 and dU  ncv dT , hence the temperature
is kept constant for adiabatic mixing. Similarly, the final pressure is p, but the components assume partial pressures after
the mixing, p A  y A p and pB  yB p . The entropy change of the system is Sm  S A  S B . For isothermal process,


p 
p 
the entropy change of each component becomes, S A  n A   ln A  and S B  nB   ln B  . Finally, the entropy
p
p 



n  y A ln
l yA 
change due to mixing of two ideal gases is Sm
l
B ln
 yB  .
In accord with the results of Example 7.6, the entropy increase by adiabatic mixing of k components at the same pressure and temperature is,
S m   n
k
 y ln y
i
(7.15)
i
i 1
Due to mixing of different gases, the entropy of the system increases and the system exergy decreases.
The irreversibility of the mixing process or the amount of exergy destroyed can be calculated by,
I12   destroyed  T0 S m
(7.16)
Example 7.7 An ideal mixture of 2kmol Helium (He) and 1kmol of Hydrogen (H2) contained in a piston-cylinder device
initially at a state of 2bar, 40C is compressed isentropically to a pressure of 10bar. Determine,
a.
the final temperature of the mixture and the amount of work done,
b.
the entropy change of each component.
Solution:
c pi (kJ/kmolK)
yi
Mi
He
0.67
4
20.77
13.92
12.456
8.345
H2
0.33
2
28.96
9.76
20.646
6.813
∑
a.
cvi (kJ/kmolK)
23.48
yi cvi
15.158
Since the mixture system is compressed without changing its entropy, S  S He  S H 2  0 or
 n c ln  TT   n ln pp
i pi
2
2
1
1
 0 . Hence, the temperature ratio becomes
help of the table above, T2  313  10 
0.354
W12   n
b.
yi c pi
Component
 y c T  T  or W
i vi
2
1
12
T2  p2 


T1  p1 
/
 y c 
i pi
and with the
 707.2 K. As a result of 1st law, the work done by the mixture is,
 3  15.158  707.2  313  17925.85 kJ.


The entropy change of Helium is, S He  nHe c p ln T2   ln p2  . Since the content of each component
T
p
1
1  He

does not change, the ratio of partial pressures is the same as the ratio of total pressures. Thus,
707.2
10 

S He  3   20.77  ln
 8.314  ln   4.427 kJ/K and the entropy change of Hydrogen gas will be
313
1

+4.427kJ/K so that the sum is assured to be zero.
288
THERMODYNAMICS
7.2.2
Real gas mixtures
Experimental observations reveal that the addition of volumes occupied by each component of
a mixture at the temperature and pressure of the mixture might not result with the mixture volume.
Let us consider the equation of state with compressibility factor for component i and apply Dalton’s
rule to that component as, piVm  ni Z i Tm . Obviously, the addition of partial pressures should yield
the mixture pressure, pmVm 
 n Z T
i
i
m
. Then the gas mixture might be assumed to be a single
component gas with a compressibility factor,
Zm 
 y Z 
i
(7.17)
i
This result does not work for gas mixtures. Because, the above result is correct only if the influence of similar molecules exist. Since a gas mixture consists of dissimilar molecules, the resulting
molecular effect is totally different and the use of Eq. (7.17) yields large deviations in predicting p-V-T
behavior of real gas mixtures. Another approach, however, is to assume the mixture as a pseudo-pure
substance.
Principle 26: Kay’s rule: In predicting p-V-T behavior of real gas mixtures, the
mixture is assumed to be a pseudo-pure substance having critical pressure and
temperature evaluated with respect to mole fractions of the constituents of the
mixture.
In accord with this principle, the gas mixture is a substance with the following critical values,
pcm 
 y p 
i
Tcm 
ci
 y T 
i ci
(7.18)
Together with the critical values evaluated by Eq. (7.18), the generalized compressibility chart
can be used for studying p-V-T behavior of the gas mixture. Similar mixing rules can be generated
for studying p-V-T behavior of gases by analytical methods. For instance, Van der Waals equation
of state, Eq. (2.25), might be appropriate if the constants “a” and “b” of the equation are evaluated
by the following rule: am 
yi ai , bm 
yi bi . However, Van der Waals originally proposed the


following rule in evaluating the equation constants as,
am 
 y
i
ai

2
bm 
yb
i i
(7.19)
Example 7.8 An insulated rigid tank having 200L of volume contains a mixture of 85 % methane and 15 % ethane gases
on mole basis at 3.5MPa, and 40C. Accidentally, the valve on the tank opens and the pressure inside quickly drops to 2MPa
before the valve is closed. For atmospheric pressure of 1bar, calculate the mass that escapes from the tank by assuming that
Kay’s rule is applicable.
Figure 7.7 A storage tank of a binary mixture
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 289
Solution:
Apply the 1st law to a control volume including the tank and the escaped gas mixture: u2  u1  p0 v2  v1  and the enthalpy
change is, h2  h1  p0v2  p0v1  p2v2  p1v1 . The enthalpy change can be expressed with respect to deviations as,
Tcm
h1*  h1
h*  h
 Tcm 2 2 
Tcm
Tcm
yc
i pi
 p0

 1 Z m 2T2   p1  p0 v1 . Assuming the final temperature
p
 2

T2  T1   
as T2  300 K , this relation may be solved by trial and error method. In addition to the compressibility chart, the enthalpy
correction chart has to be used.
pci (bar)
Component
yi
Methane
0.85
Ethane
0.15
Tci (K)
yi pci (bar)
yiTci (K)
c pi (kJ/kmolK)
yi c pi
46.4
191
39.44
162.35
20.8
17.68
48.8
305
7.32
45.75
888.6
133.29
46.76
208.1
∑
pcm  46.76 bar, Tcm  208.1 K; pr1  0.748 , Tr1  1.5 , Z m1  0.945 , and
Tr 2  1.44 , Z m 2  0.965 , and
150.97
h1*  h1
 0.25 .Similarly, at pr 2  0.42 ,
Tcm
h2*  h2
0.945  8.314  313
 0.2 , v1 
 0.702 m3/kmol. Substituting these values into the
Tcm
3500
above relation and solving for T2 results as, T2  312.4 K. There is no need to repeat the calculations.
Use Eq. (7.5) for the molar mass of the mixture, M m  0.85  16  0.15  30  18.1 kg/kmol, the number of initial
3500  0.2
2000  0.2
 0.284 , n2 
 0.159 . The amount of mass escaped,
and final moles; n1 
0.945  8.314  313
0.965  8.314  312.4
m  M m n1  n2   2.26 kg.
Example 7.9 A piston cylinder apparatus contains 0.362kg of the same mixture as in Example 7.8 and initially occupies a
volume of 2L at 40C. The mixture is compressed isothermally until the volume is reduced by 50-percent. Assume that the
mixture obeys Van der Waals equation of state, and calculate,
a. the initial pressure in the cylinder,
b the work done.
Solution:
a.
To reduce the mixture to a single gas behavior Eq. (7.19) has to be used. Hence,
2
am  0.85  232.4  0.15  557.1   272.19 and bm  0.85  0.0427  0.15  0.065  0.046 .


Since the molar mass is M m  18.1 kg, the mole number, and the initial molar volume respectively are,
n
0.362
0.002
 0.02, v1 
 0.1 m3/kmol. Then the pressure is calculated by Van der Waals equation as,
18.1
0.02
p1 
8.314  313 272.19

 20971 kPa.
0.1  0.046
0.12
2
b.
the work is w12 
v2  bm
 pdv T ln v  b
1
1
w12  8.314  313  ln
W112  4051
4051 0.02
m
 1 1
 am    or
 v2 v1 
0.05  0.046
1 
 1
 272.19 

  4051 and the total work is
0.1  0.046
 0.05 0.1 
81.02 kJ.
290
THERMODYNAMICS
7.2.3
Orsat’s apparatus and gas mixture analysis
Quoting the analysis of gas mixtures by volume is the most convenient method for practical determinations and usually Orsat’s apparatus is used for the analysis of combustion gases. As shown in
Figure 7.8, Orsat’s apparatus consists of a levelling bottle, a burette, and three absorption pipettes.
The gas sample is usually at atmospheric pressure. The temperature and thus the density of gas are
held constant by water jacket around the burette. The pipettes are interconnected and contain different
chemicals to absorb CO2, CO, and O2. For instance, pipette 1 contains KOH for absorption of CO2.
Each pipette also contains a number of small tubes for increasing the absorption surface area. The
constituents are chemically absorbed one by one and the remainder volume of the sample is measured
after each absorption process. The difference in volume gives the partial volume occupied by the
constituent in that mixture. Since the temperature of the sample gas is reduced below the saturation
temperature of water vapor present, Orsat’s apparatus gives analysis of dry products of combustion.
The dry product analysis, on the other hand, may be used to calculate the air-fuel ratio of combustion
processes.
Example 7.10 A sample of flue gases from a coal fired boiler is analyzed by an Orsat’s apparatus and the following volumetric analysis is provided: 11.4% CO2, 0.5% CO, 6.5% O2 and 81.6% N2. The gravimetric composition of coal is given
as; carbon 80%, hydrogen 5%, oxygen 4% and non-combustibles are 11%. Determine,
a.
the mass of dry flue gases per kg of coal,
b.
the total mass of flue gases per kg of coal,
c.
the mass of excess air per kg of coal.
Solution:
a.
The first step in determining the mass of flue gases is to convert the volumetric analysis into gravimetric ratios as
presented in the table below.
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 291
Gas
Mole fraction,
(y%)
Molecular mass
M(kg)
Mass in the mixture
x  yM (kg)
Mass fraction
(x%)
CO2
CO
N2
O2
Sum
0.114
0.005
0.816
0.065
44
28
28
32
5.016
0.14
22.84
2.08
30.076
16.67
0.465
75.94
6.91
3
3
Amount of carbon per kg of flue gas= carbon in (CO2) + carbon in (CO) = 11  0.1667  7  0.00465  0.0474
kg carbon/kg gas.
The stoichiometric coefficients are calculated as following:
C  O2  CO2



3 kg(C )  8 kg(O2 )  11 kg(CO2 ) 
C  O  CO



3 kg(C )  4 kg(O2 )  7 kg(CO) 
2 H 2  O2  2 H 2O



1 kg ( H 2 )  8 kg(O2 )  9 kg( H 2O) 
Coal gravimetric analysis indicates that 1 kg of coal contains 0.8kg of carbon. Thus the mass of dry flue gas per kg of
0.8
 16.877 kg/kg-coal.
coal is, mg dry 
0.0474
b.
Together with the stoichiometric equation between hydrogen (H2) and oxygen (O2), the amount of vapor created by
0.05kg of hydrogen is: 9  0.05  0.45 kg-water/kg-coal. Hence the total mass of flue gas per kg of coal becomes
mg  16.877  0.45  17.32 kg/kg-coal.
c. Considering the above table and allowing oxygen for unburned carbon monoxide, the mass of excess oxygen per
kg of flue gas is,
4
mo2  0.0691   0.00465  0.0664 kg/kg-gas, or per kg of coal, mo2  0.0664  16.877  1.12 kg/kg-coal. The amount
7
of excess air containing the required amount of oxygen is mair  1.12 
7.3
100
 4.87 kg-air/kg-coal.
23
Moist Air and its Psychrometric Properties
As indicated in Figure 7.9, atmospheric air is a mixture of dry air, water vapor, and pollutants.
Even though the composition of dry air is fairly constant, the content of water vapor and pollutants in
atmospheric air may vary with respect to the location in the atmosphere. However, the pollutants can
be filtered out by a process and then what is left is a mixture of dry air and certain amount of water
vapor. The combination of dry air and water vapor is called moist air.
As illustrated in Table 7.2, the composition of dry air is fairly constant. Hence, in air conditioning
processes, the amount of water vapor may be changed from zero value to a maximum depending on
the temperature and the pressure of the mixture. On the basis of real gas behavior, it is quite difficult
292
THERMODYNAMICS
to estimate the exact thermodynamic property values of moist air. Up to 300 kPa of pressure, however,
moist air can be treated as an ideal gas mixture with accuracy sufficient for engineering calculations.
For such cases then, Dalton’s rule of additive pressures yield the total barometric pressure of moist
air as,
p  pa  pv
(7.20)
where pa  ma RaT / V and pv  mv RvT / V are the partial pressures of dry air and water vapor
respectively.
In analyzing the moist air properties, the following terms are frequently used and need to be
defined.
Definitions: Saturated air is a mixture which contains the maximum amount of water vapor at a given
temperature and pressure. Dry bulb temperature (DBT) is the temperature of the moist air measured
by any temperature measuring instrument. Saturated vapor pressure ( pvs ) is the partial pressure of
water vapor that corresponds to the saturation pressure at the given temperature of the mixture.
For moist air temperatures in the range of 0°C and 100°C, ASHRAE (American Society of Heating Refrigerating and Air conditioning Engineers) suggests the following regression equation for
saturated vapor pressures,
ln  pvs  
a1
 a0  a1T  a2T 2  a3T 3  a4 ln T 
T
(7.21)
where T is in K, and the regression coefficients respectively are, a1  5800.2 , a0  5.516 ,
a1  0.04864 , a2  0.0000417 , a3  0.0000000144 , and a4  6.545 .
Definition: Relative humidity () is the ratio of mole fraction of water vapor in moist air to the mole
fraction of water vapor in saturated air both at the same temperature and pressure.
Since, nv  pvV / T and nvs  pvsV / T then, the relative humidity may be expressed as,

nv
p
 v
nvs pvs
(7.22)
Definition: Humidity ratio (  ) is the amount of water vapor per kilogram of dry air and is also
called specific humidity (   mv / ma ).
The mass of dry air and water vapor respectively are, ma  paV / RaT , mv  pvV / RvT where
Ra  0.287 kJ/kgK, and Rv  0.4618 kJ/kgK. Then, the humidity ratio becomes,

pvV / RvT
pv
pv
 0.622
or   0.622
paV / RaT
pa
p  pv
(7.23)
As indicated by this relation, specific humidity is a function of both the total pressure and the dry
bulb temperature (DBT) of the mixture.
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 293
Definition: If the unsaturated moist air is cooled at constant pressure then Dew point temperature
(DPT) is the temperature at which the moisture starts to condense.
As illustrated in Figure 7.10, when unsaturated moist air at state (1) is cooled down, the partial
pressure of vapor remains constant until state (2) is reached. The temperature at state (2) is called the
dew point temperature ( T2  DPT ) of state (1) and is the saturation temperature corresponding to the
partial pressure of vapor. Hence the dew point temperature can also be obtained from steam tables.
Since the mixture at state (2) is saturated, some of water vapor condenses as the cooling process continues. As indicated by state (3), the mixture is always at saturated state, but due to condensation, the
partial pressure of vapor decreases. With respect to relative humidity () and DBT of the mixture,
the dew point temperature may be approximated by the following equation.
DPT 
4030 DBT  235 
4030  DBT  235 ln  
 235
(7.24)
where, DBT and DPT are both in 0C.
Example 7.11 The barometric pressure of atmospheric air on a hot and humid day is 755mmHg. The dry bulb temperature
and the relative humidity respectively are 410C, and 91%. Determine,
a.
the partial pressure of water vapor
b.
the specific humidity
c.
the dew point temperature.
Solution:
a.
Equation (7.22) yields pv   pvs , pvs  7.82 kPa at 410C, and pv  7.12 kPa is the partial pressure of water
vapor.
b.
Use Eq. (7.23) as W  0.622
7.12
 0.048 kg-water/kg-air.
99.34  7.12
294
c.
THERMODYNAMICS
By steam tables, the saturation temperature at pv  7.12 kPa is 38.90C. The use of Eq. (7.24) results as following,
DPT 
7.3.1
4030 41  235 
4030  41  235 ln 0.91
 235  39.2 0C which deviates by 0.7-percent from the table reading.
Mass, Energy and Entropy Balances for Moist Air
As shown in Figure 7.11, processing of moist air at steady state conditions, the mass, energy and
the entropy balances may be expressed as,
Continuity: m a1  m a 2  m a
m v1  m v 2  m f 2
(7.25)
Energy:
q  Q / m a  ha 2  ha1   2 hv 2  1hv1  1  2 h f 2
(7.26)
Entropy:
q / Tboundary  sa 2  sa1   2 sv 2  1sv1  1  2 s f 2
(7.27)
In above relations, if condensation takes place, 1  2 otherwise, 1  2 . For constant volume
processes, and for occurrence of condensation, the energy equation reduces to

Q12  ma cva T2  T1   mv 2uv 2  mv1uv1  m2 f u2 f

(7.28)
Even though, the above relations are written on an unmixed component basis, it can also be defined on premixed mixture basis.
Definition: The enthalpy ( h ) of most air is the sum of enthalpy of dry air and water vapor ( h  ha   hv ).
This definition of enthalpy is especially useful for psychrometric chart applications. Since the energy
values are always based on the same reference state, the enthalpy of dry air and water vapor are both
zero at 00C. Then, together with approximate values of cpa and cpv, the mixture enthalpy becomes,

h  c paT   h fg  c pvT
 or
h  1.005T   2501  1.88T 
(7.29)
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 295
Example 7.12 A piston-cylinder apparatus contains two kilograms of air and water vapor mixture at 200C, 100 kPa, and
80 % relative humidity. The mixture is compressed at constant temperature to 200 kPa. Determine,
a.
the final relative humidity and humidity ratio
b.
the mass of liquid water condensed
c.
the amount of work done.
d.
the amount of heat transfer
Solution:
a.
The humidity ratio at state (1), pv1  1 pvs  0.8  2.34  1.872 kPa, and use of Eq. (7.23) yields,
1.872
 0.0118 . Let us consider an intermediate state (2) at which the condensation just starts
100  1.872
and find out the total pressure of the mixture at that state. For 1  2 , and pv 2  pvs  2.34 kPa, Eq. (7.23) yields,
1  0.622
0.0118  0.622
2.34
, p2  125.68 kPa. Since p3  p2 , some of water vapor will condensate as the pressure
p2  2.34
reaches 200 kPa at 200C.
The volume occupied by the mixture at state (1),
respectively are ma 
3  0
0.622
622
b.
2.34
200 2.34
pa1V1 pv1V1

 2 or V1  1.693 m3. The values of ma and mv1
RaT1
RvT1
98.128  1.693
 1.975 kg, mv1  2  1.975  0.025 kg. At state (3),
0.287  293
0.00736
The amount of condensation is m f 3  mv1  mv3 and mv3  0.00736  1.975  0.0145 kg, then
m f 3  0.025  0.0145  0.0105 kg.
c.
pa 2  123.34 kPa, pa 3  197.66 kPa, V2  1.351 m3, and V3 
pa 2
V2  0.843 m3. The work done by dry air is
pa 3
 pa1 
  1.975  0.287  293  ln(98.128 / 197.66) or W13 a  116.301 kJ. The work of water
 pa 3 
W13 a  ma RaT ln 
vapor is calculated at two steps, W12 v  0.025  0.462  293  ln(1.872 / 2.34)  0.755 kJ , W23 v  pvs V3  V2 
W23 v  2.34  0.843  1.351  1.188
kJ. The total work then is
W13  116.301  (0.755)  (1.188)  –118.244 kJ.
d.
Apply the 1st law between states (1) and (3), Q13  W13  m f 3h fg or
Q13
118.244

0.0105 x 2454.1
144.012 kJ
296
THERMODYNAMICS
Example 7.13 One means of condensing steam during an emergency blow down of a nuclear reactor is to use of containment
vessel. The vessel is insulated and has a total volume of 30m3 with liquid water initially occupying 1m3. The initial state inside
the vessel is 350C, and 100kPa. For a short period of time, 55 kg of water from reactor enters the vessel at an average state
of pi  700 kPa, xi  0.5 . Find the temperature and pressure inside the vessel at the end of this water entrance period.
Figure 7.13 Containment vessel for emergency blow down
Solution:
At initial state, pv1  pvs  5.62 kPa, pa1  94.38 kPa, and the volume occupied by saturated air, V1  29 m3. Hence,
ma 
94.38  29
5.62
 0.037 , mv1  0.037  30.96  1.145 kg, mw1  1000  1.14  1001.14 kg,
 30.96 kg, 1  0.622
0.287  308
94.38
1.14
 0.001 , u1  146.67  0.001  2276.73  148.94 kJ/kg, and the enthalpy of inlet water is hi  1730.37 kJ/kg.
1001.14
After the water inlet, the final total mass of water in containment becomes, mw2  mw1   mi  1056.14 kg, and air is still
x1 
saturated at state (2) for which Pa 2  pa1 
1st law for the containment
T2
. The mass of vapor is
T1
mv 2  0.622ma 
psv 2 T1
(a)

pa1 T2
hi mi  mv 2uv 2  mw2  mv 2 u f 2  ma cva T2  T1   mw1u1 (b)
Apply trial and error method to solve equations (a) and (b) simultaneously. Assume T2=530C, and by Eq. (a) mv 2  2.65 kg
,and u f 2  209 kJ/kg, uv 2  2449 kJ/kg and substitute these values into Eq. (b) to find a new temperature, T2=52.50C.
7.3.2 Adiabatic saturation and thermodynamic wet bulb temperature (WBT)
In general, the evaporative cooling processes occur without any external heat transfer and are
adiabatic. As shown in Figure 7.14, if an evaporative humidifier is long enough so that the air at the
exit is saturated then the device is called adiabatic saturator and the temperature of the outlet stream
is the thermodynamic wet-bulb temperature of moist air at state 1.
Figure 7.14 Schematic of adiabatic saturator
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 297
At steady state conditions, considering that m f 2  m v 3  m v1 the energy balance on adiabatic
saturator yields,




m a ha1  ha 3   m v1 hg1  h f 2  m v 3 h f 2  hg 3  0
(7.30)
Introducing the parameters, 1  m v1 / m a and 3  m v 3 / m a , rearrange the above relation as,
1 
ha3  ha1   3 hg 3  h f 2 
hg1  h f 2
or
1 
c pa T2  T1   3 h fg 2
hg1  h f 2
(7.31)
where, T3  T2 and h fg 2  hg 3  h f 2 . The wet bulb temperature of moist air at state 1 is indicated
by temperature T2. Hence, knowing the temperatures DBT(T1) and WBT(T2), Eq. (7.31) calculates the
humidity ratio of state 1. The WBT temperature like DBT is a property of moist air and is independent
of path. The thermodynamic wet-bulb temperature (T2) is always less than the DBT (T1) but greater
than the dew point temperature (DPT).
7.3.3
Psychrometer
In accord with Gibbs phase rule, three independent properties are required for measuring the
psychrometric state of moist air. Two of these parameters are usually the barometric pressure and
the dry bulb temperature (DBT). Due to inaccuracies involved in measurement of humidity ratio, the
third parameter is again depicted to be a temperature and is the wet bulb temperature (WBT) of moist
air. The sling psychrometer, as shown in Figure 7.15a, is widely used for measurements involving
room air applications.
The sling psychrometer consists of two thermometers and is fitted in a frame with a handle for
swinging in air as shown in Figure 7.15b. Readings are taken when both thermometers indicate the
steady state values.
298
THERMODYNAMICS
Figure 7.16 illustrates graphically the thermodynamic properties of moist air and is readily available for standard barometric pressure of 101.32 kPa at sea level and for temperatures between 0°C and
50°C. ASHRAE has also developed psychrometric charts for other temperatures and pressures such
as a chart for low temperature (-40°C to 10°C) and a chart for high temperatures (100°C to 120°C).
In solving moist air problems by a chart, first the barometric pressure that the system works has to
be known. Then, as shown in Figure 7.17a, the state of moist air is located by two independent properties like DBT1 and 1 , and the unknown properties like WBT1, 1 , h1 etc. can be read on chart. Figure
7.17b shows the directions of simple processes like sensible heating, cooling, and humidifying.
Figure 7.16 The psychrometric chart at sea level, p=760mmHg
Figure 7.17 (a) Determination of properties of moist air on a psychrometric chart,
and (b) process lines
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 299
Example 7.14 The DBT and WBT temperatures of moist air at sea level respectively are 350C and 280C. Determine the
followings by using both a chart and related equations,
a.
the humidity ratio,
b.
the partial pressure of water vapor,
c.
the relative humidity,
d.
the moist air density,
e.
the enthalpy
Solution:
a.
After locating the given state on a psychrometric chart, 1 is 20.8 g/kg-air. Consider Eq. (7.31), to calculate the
3.78
humidity ratio at state (1), first the humidity ratio at WBT has to be determined, 3  0.622
 0.024
101.3  3.78
and h fg 2  2552.6  117.4  2435.2 kJ/kg, hg1  h f 2  2565.3  117.4  2447.9 kJ/kg. Substitution of these values
into Eq. (7.31) yields, 1  21 g/kg-air.
b.
The chart yields the partial pressure of vapor as, pv1  3.22 kPa. In computing, consider Eq. (7.23) for 1  0.021
kg-v/kg-air and determine pv1 as pv1  3.3 kPa.
c.
The value of relative humidity by chart is 1  0.57 and the use Eq. (7.22) for numerical calculations results as,
1  pv1 /
d.
vs1
0.58 .
From chart, the moist air specific volume is v1  0.903 m3/kg-air, and the density becomes 1  1.107 kg/m3. To calculate the density, consider ma / V1  p1 / Ra  1Rv1 T1  and after substitution of numerical values 1  1.108 kg/m3.
e.
7.4
The enthalpy value by chart reading is h1  89.2 kJ/kg-air, and by Eq. (7.29) it can be calculated as h1  89.08 kJ/
kg.
Air Conditioning Processes
Air conditioning processes involve altering the temperature and the humidity of moist air so that
comfort conditions (DBT temperature in the range between 18°C and 22°C, with 40 to 50 % relative
humidity) are provided in dwellings. Since most of the air conditioning processes can be modeled as
steady flow processes, together with mass and energy balance equations (Eqs. (7.25) and (7.26)), the
psychrometric chart is an important tool in determining the final state and/or the heat load through
an air conditioning system. As shown in Figure 7.18a, whenever there is no moisture addition or
removal from air, the process is simply represented as a horizontal line on the chart. The following
fundamental processes are carried out in air conditioning applications.
Figure 7.18 Air conditioning processes: (a) sensible heating or cooling, and (b) humidification
300
THERMODYNAMICS
7.4.1
Sensible heating or cooling
For sensible heating, the horizontal line moves right, line (1-2) in Figure 7.18a, and for sensible
cooling, the process line is to the left line (3-4). These two processes are also called constant specific
humidity processes (ω0=constant). The amount of heat added or removed by such a process is,
Q  m a h2  h1 
7.4.2
(7.32)
Humidification evaporative cooling
As moist air is heated the relative humidity drops, and to alleviate the conditions, the moist air
can be humidified. As shown in Figure 7.18b, the humidification of moist air can be done by the
following two methods:
Steam injection into air stream. If water vapor is injected to moist air for increasing the humidity
ratio, the air temperature is essentially kept constant. Due to small amount of steam injection, even if
the steam temperature is higher than the air temperature, the temperature is not affected appreciably
to indicate a temperature change. Such a process (line 1-2 in Fig. 7.18b) is also called constant DBT
process and the amount of steam injected is,
m s  m a 2  1 
(7.24)
Spraying water particles. The injected water particles are in liquid phase and the required heat
of vaporization for evaporation comes from moist air. As a result, the temperature of air stream decreases, but the wet bulb temperature is kept constant. Such process (line 1-3 in Fig. 7.18b) is also
called constant WBT process. A similar constant WBT process is called evaporative cooling process.
You probably notice that on a hot and dry summer day, the environment feels a lot cooler when water
is sprayed into the atmosphere in a café. This is because water absorbs heat from air as it evaporates.
The enthalpy of the mixture is constant ( WBT1  WBT3 ) but the DBT drops as on line 1-3 in Figure
7.18b for evaporative cooling.
Example 7.15 Saturated air at 5°C with a volumetric flow rate 150m3/min is required to be supplied to a conference room
where the temperature must be at 20°C with relative humidity 50%. As shown in Figure 7.19, air is first heated and then
water at 10°C is sprayed to provide the required humidity. Determine,
a.
the temperature at which air must be heated,
b.
the mass rate of hot water flowing through the heater for 7°C of temperature increase,
c.
the mass required to be sprayed per hour.
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 301
Solution:
a.
As shown on the schematic diagram, Figure 7.20, to determine the air temperature at the heater exit, the lines A
and B have to be crossed. The intersection is the exit state of air, and T2=26°C.
b.
The required heat input by the exchanger is Q  m a h2  h1   m wcT where air mass flow rate is m a  V1 / v1 and
by psychrometric chart, v1  0.795 m3/kg and m a  150 / (60  0.795) , m a  3.145 kg/s. Hence the heater capacity
is, Q  3.145  41  20   66.045 kW, and the hot water flow rate becomes, m w  66.045 / (4.18 7)
c.
2.26 kg/s.
The mass flow rate of sprayed water is calculated by m sw  m a 3  1   3.145  8.3  6  103 kg/s or
m sw  26.04 kg/h.
7.4.3
Dehumidification
Dehumidification by cooling is a process by which the humidity ratio of moist air is reduced by
flowing through a cooling coil. As shown in Figure 7.21, the moist air is dehumidified after reaching
saturated conditions. Once the air becomes saturated, it remains saturated even if the temperature is
further decreased ( 2  1 ). After dehumidification, to reduce the relative humidity to a desired level,
the moist air has to be reheated ( 3  2 ) as by line (2-3) in Figure 7.21. The collected condensate
leaves the channel at the same temperature as air at the cooler exit. The heat capacity rates of the
cooling coil and the heater are,
Q c  m a h2  h1   1  2 h f 2  





Qh  m a h3  h2 

(7.25)
302
THERMODYNAMICS
Figure 7.21 Dehumidification of air
Example 7.16 An air conditioning system for an industrial process has to be designed so that 250 m3/min of outside air at
350C and 60% relative humidity has to be processed to 250C with 55% relative humidity. The dew temperature on the coil
surface (T6) is known to be 120C. Show the process on psychrometric chart and calculate,
a.
the mass of water removed per hour,
b.
the cooling capacity,
c.
the temperature at the cooler exit,
d.
the heating capacity,
e.
the percent error in cooling capacity if air at the cooler exit is assumed to be saturated.
Solution:
a.
As shown in Figure 7.22, since the dew temperature on the surface
is 120C, the exit state of air has to be on line (1-6). The interaction
of lines (1-6) and (5-3) yields the coil exit (2) which indicates
that air may not be fully saturated at the exit. Hence by chart, the
thermodynamic values are v1  0.91 m3/kg, 1  0.023 , h1  95
kJ/kg, 2  0.0115 , h2  45 kJ/kg, T2  15 0C, h3  55 kJ/kg,
h6  39 kJ/kg, h f 2  62.9 kJ/kg. Then the mass flow rate of air
is, m a  250 / (60  0.91)  4.578 kg/s, and Eq. (7.24) yields,
m f 2  4.578 (0.023 0.0115) 3600 189.36 kg/h
b.
Referring to Eq. (7.25), the cooling capacity is
Q c  4.578  45  95   0.023  0.0115  62.9   225.588 kW.
c.
In accord with the chart in Figure 7.22, the air temperature at the
cooler exit is T2  15°C.
d.
The heat capacity rate is Q h  4.578  55  45   45.78 kW
e.
If air is assumed to be fully saturated at the cooler exit, then h f 5  50.41 kJ/kg, and the cooling capacity becomes
Q c  4.578  44  95   0.023  0.0115  50.41  230.824 kW. The percent error in estimating the capacity
would be err %
230.824
230 824 225.588
225 588  / 225.588
225 588 100
2.32% .
7.4.4 Adiabatic Mixing of Air Streams
Adiabatic mixing of two air streams is another method for changing the humidity ratio of moist air.
In large dwellings such as hospitals, schools, or in process plants, air needs to be mixed with outside
air to keep it fresh. Depending upon the state of air streams mixed, the process may take place with
or without condensate of moisture.
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 303
Mixing without condensation. As shown in Figure 7.23, if two air streams are not at saturated
state then during the mixing process, no condensation will take place.
Figure 7.23 Mixing of unsaturated air streams
For an adiabatic mixer as shown in Figure 7.23, consider a control volume surrounding the mixer
and apply mass and energy balance equations to yield the mixture conditions as following,
1m 1  2 m 2  3 m 1  m 2 


h1m 1  h2 m 2  h3 m 1  m 2  
(7.26)
Then, the mass ratio of two mixing streams is as follows,
m 1 2  3 h2  h3


m 2 3  1 h3  h1
(7.27)
The above equations reveal that the final humidity ratio and the enthalpy are weighted average of
inlet humidity ratios and enthalpies. Therefore, the final temperature of the mixture is the weighted
average of inlet temperatures and the state of the mixture will be on a straight line connecting the two
inlet states. If the ratio m 1 / m 2 has changed by keeping the inlet states fixed (the points 1&2 in Figure
7.23), the mixing state will still be on the same line connecting the two points but the final state be
moved to points like 3 or 3 depending on the magnitude of the mass ratio.
Mixing with condensation. When very cold air stream is mixed with very warm and humid air,
the state of the resulting mixture might be in two-phase region. As shown in Figure 7.24, since the
mixture cannot hold the liquid droplets, a certain amount of water vapor will condensate and the
resulting air stream will be saturated air at state 4. Such a process hardly occurs in air conditioning
systems, but when occurs, due to release of latent heat of condensation in this adiabatic system, an
increase in temperature of the mixture will take place (T4>T3).
304
THERMODYNAMICS
Figure 7.24 Mixing of cold and hot air streams
Example 7.17 An air conditioning system mixes adiabatically 250 m3/min of outside fresh air at T1=100C, 1  80%
with 125m3/min of indoor air at T2=380C and 2  40%
. Assuming that mixing occurs at 1bar,
a.
determine the dry bulb temperature and the
relative humidity at the mixer outlet,
b.
decide which process to be applied after mixing
to get air at T4=250C and 4  50% .
Solution:
a.
In accord with Figure 7.23, V1  2.08
m 3 / s , 1  0.0176 k g / k g - a i r , T1  38 0 C ,
v1  0.91 m 3 /kg, and m 1  2.08 / 0.91  2.28
3
kg/s. Similarly, psychrometric properties of cold air are: V2  4.167 m /s, 2  0.007 kg/kg-air,
T2  10 0C, v2  0.813 m3/kg, and m 2  4.167 / 0.813  5.125 kg/s. Since none of mixing streams is on saturated
state, no condensation will take place, and Eq. (7.27) is applicable as,
2.285 0.007  3

 0.4458 . Solving
5.125 3  0.0176

  0.0102
kg-v/kg-air. Together with a psychrometric chart in Figure 7.25, knowing that the
for 3 results as; 3
mixture state is on the line connecting the two inlet states, the mixture temperature and relative humidity respectively are T3  20 °C, 3  68% .
b.
7.5
As shown on chart in Figure 7.25, if air at state 3 is heated, the horizontal line crosses the constant relative humidity
curve ( 4  50% ) at temperature T4  25°C.
Cooling Tower Basics
The large capacity refrigeration plants, air conditioning systems, and the power plants all generate
large amount of heat that is rejected at the condenser. To keep the size of the condenser at reasonable limits, however, water having high thermal properties is usually preferred as cooling medium.
Because of limited water resources, water that is circulated through the condenser is generally reused
by cooling it through a cooling tower. A cooling tower cools down water by contacting it with air
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 305
and evaporating some of water. As shown in Figure 7.26, the classification of cooling towers may be
done with respect to the draft type, the flow geometry, and the mode of heat transfer.
Understanding the advantages and limitations of cooling towers as classified in Figure 7.26 is
of vital importance to the project engineer. Figures 7.27a and b show a typical natural draft tower
which operates on the basis of air density variation (chimney effect) and is more effective in regions
of high relative humidity. In mechanical draft towers, Figure 7.27c, the air flow is caused by single
or multiple fans. Fans can also compensate changes in atmospheric and load conditions. Hence, the
thermal performance of mechanical draft towers is less affected by psychrometric changes in air.
Depending upon the air flow direction relative to the fan, the mechanical draft towers can be either
forced or induced draft type.
In counter flow towers, Figure 7.27c, air moves vertically upward through the fill and is in opposite direction to the downward motion of water. Due to high pressure losses of air at inlet and
outlet plenums, the counter flow configuration is generally appropriate for large capacity cooling
towers. The enclosed nature of counter flow tower, however, restricts the exposure of water to sun
and retards the growth of algae. The particular tower in Figure 7.27c may also be categorized as an
induced draft type tower.
In cross flow towers, Figure 7.27d, air flows horizontally through fill and crosses the downward
fall of water. This type of tower is especially useful in regions where prevailing wind directions occur.
The tower can be sited so that the air inlet section facing the prevailing wind direction, the power
needed for air circulation can be reduced.
All the towers types indicated above are evaporative type towers in which the cooling effect
is generated by evaporation of water droplets when air and water are brought into direct contact.
However, in sensible (dry) type towers no direct contact between water and air takes place. Water
is cooled by sensible heat transfer. Due to need of large heat transfer surface area, the use of such
towers is seldom.
306
THERMODYNAMICS
Figure 7.27 Cooling tower illustrations: (a) and (b) natural draft type (c) counter flow type
(d) cross flow type
7.5.1
Calculation of Air Mass Flow Rate
In energy and mass transfer from water to the unsaturated air, there are two driving forces for the
transfer; i. the difference in dry-bulb temperatures, ii. the difference in vapor pressures between the
water surface and the air. Hence, as shown in Figure 7.28, some portion of water is lost to air and has
to be made up by make-up water.
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 307
Figure 7.28 Cooling tower performance
The amount of water lost to air is, m w1  m w 2  m a a 2  a1  , and the water mass flow rate at
the tower exit becomes,
m w 2  m w1  m a a 2  a1 
(7.28)
The energy balance along the control volume on the tower yields, m w1hw1  m w 2 hw 2  m a ha 2  ha1 
,
and substitution of m w 2 in Eq. 7.28 into energy balance equation and rearranging results with the air
mass flow rate as following,


hw1  hw2 
m a  m w1 

 ha 2  ha1   a 2  a1 hw 2 
(7.29)
The conditions within the cooling tower are typically such that the emerging air is very close to
100% relative humidity and is assumed to be saturated air at the tower exit. On the other hand, the
minimum temperature that can be attained by water at the tower exit is thermodynamically limited
to the WBT of the incoming air. However, such a temperature is only obtainable for infinitely tall
cooling tower and a reasonable achievable limit for exit water temperature is within 5K to 10K above
the WBT of the entering air.
Example 7.18 Warm water at 40°C from a power plant condenser enters the cooling tower at a flow rate of 400tons per
hour. The atmospheric air at 95kPa, 25°C, and 70% relative humidity is circulated through the tower and leaves the tower
at 35°C as saturated air. For a wet bulb temperature WBTa1=21°C of entering air estimate,
308
THERMODYNAMICS
a.
the mass flow rate of air,
b.
the make-up water in ton per hour.
Solution:
a.
For WBTa1=21°C at the air entrance, the cold water temperature at the outlet may be taken to be Tw2  WBTa1  5 ,
Tw2  26 °C. The humidity ratio of air at the inlet and outlet respectively are,
a1  0.622
0.7  3.16
5.62
 0.0238 kg-a/kg-water, a 2  0.622
 0.039 kg-a/kg-water. The enthalpy
95  0.7  3.16
95  5.62
difference for water, hw1  hw2  167.5  109.07  58.43 kJ/kg, and for air is
ha 2  ha1  c pa T2  T1   a 2 hg 2  a1hg1  1.005  35  25   0.039  2565.3  0.0238  2547.2  49.47 kJ/kg.
By Eq. (7.29), the needed mass flow rate of air is m a  111.11 
b.
58.43
 135.74 kg/s.
49.47  0.039  0.024  109.07
By Eq. (7.28), the amount of make-up water needed, m mu  135  0.039  0.0238   2.06 kg/s, or
m mu  7.43 tons per hour.
7.5.2
Cooling tower design and performance
As shown in Figure 7.28, representing the total surface area by dA that includes the surface area
of water drops as well as baffles and other fill materials, the rate of heat removed from water is,
 Q  hdA Tw  Ta   4.18m w dT  m a dha
(7.30)
Assuming that water is at WBT of incoming air, then Tw  Ta  hi  ha  / c pm , and Eq. (7.30)
reduces to
A

0
hdA
 4.18m w
c pm

Te
Ti
dT
hi  ha
or
exit
hA
1
 4.18m w Te  Ti 
c pm
inlet hi  ha

(7.31)
where Ti and Te are temperatures of water entering and leaving the tower, h is the convective heat
transfer coefficient, and hi is the enthalpy saturated air at water temperature. The value of hA / c pm
can be numerically determined by using a stepwise integration method. If air and water flow rates are
kept constant, the magnitude of hA / c pm essentially remains constant for a particular tower and is used
for predicting the tower performance at various water inlet temperatures. Cooling tower manufacturers
often treat the term hA / c pm as NTU (Number of Transfer Units) of the tower. Hence, the higher the
value of NTU, the better the performance of the tower is.
The tower characteristics can also be represented graphically as shown in Figure 7.29. In this figure,
water enters the tower at Tin, and exits at Tex. The corresponding saturated air enthalpies respectively
are hi ,in and hi ,ex . Similarly the air enthalpies at the inlet and exit are ha ,in and ha ,ex .
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 309
Figure 7.29 Graphical presentation of cooling tower performance
As shown in Figure 7.29, two terms are regularly used in performance analysis of cooling towers
and are defined as following: 1.Range is the temperature change of water through the tower, and 2.
Approach indicates the difference between the WBT of entering air and exit temperature of water.
A tower with a smaller approach and a larger range is always preferable and provides a higher value
of NTU.
Example 7.19 Warm water at 38°C from a condenser enters a counter flow cooling tower at a flow rate of 108 tons per
hour. The atmospheric air at 100kPa, DBT1=25°C, and WBT1=22°C ( 1 =70%) is circulated through the tower and leaves


the tower at 35°C as saturated air. Calculate NTU  hA / c pm parameter of the tower.
Solution:
As indicated in Figure 7.29, the approach for the condenser is assumed to be 8K, then the water temperature at the condenser
exit is Twe  22  8  30o C . By the given data and by the psychrometric chart, the numerical values of the parameters in
Eq. 7.29 are a1  0.0148 , a 2  0.042 , ha1  63 kJ/kg, ha 2  146 kJ/kg and water enthalpy values are, hw1  159.2 kJ/kg,
hw2  125.79 kJ/kg. Hence, the required air mass flow rate becomes m a  30 
33.41
 12.59 kg/s.
83  0.042  0.0148  125.79
Figure 7.30 Numerical analysis of a counter flow tower
310
THERMODYNAMICS
As shown in Figure 7.30, let us divide the tower into five sections (n=5) with identical water temperature drop at each
section as, T  38  30  / 5  1.6o C . The air enthalpy change of air at each section is ha (i )  ha (i 1)  4.18  30  1.6 / 12.59
or ha (i )  ha (i 1)  15.93 kJ/kg, ha 0  63 kJ/kg.
Section 1: ha (1)  ha (0)  15.93 or ha (1)  78.93 , ham(1) 
Twm(1) 
78.93  63
 70.96 kJ/kg, Tw(1)  31.60 C
2
30  31.6
 30.80 C , and the corresponding saturated air enthalpy is, hsam(1)  115 kJ/kg, and the mean differ2

ence at section 1 becomes, hsa (1)  ha (1)
m  44.23 kJ/kg.
Section 2: Similar calculations can be carried out for this section. The values of related parameters are ha (2)  94.86
kJ/kg, ham(2)  86.31 kJ/kg, Tw(2)  33.20 C , Twm(2)  32.40 C , and hsam(2)  122 kJ/kg. The corresponding enthalpy dif-

ference is hsa (2)  ha (2)
m  35.69 kJ/kg.
Similar algorithm may be applied to the rest of the sections, and the enthalpy of air at the tower exit is determined to be
ha (5)  142.65 kJ/kg which is 2.39% off the real value. Better estimates, however, can be obtained by increasing the number
of sections (n) of the tower. The summary of related calculations is presented in Table 7.3.
Table 7.3 NTU calculation for a counter flow cooling tower
Section
Mean water
temperature Twm(i )
1
30.8
2
3
4
5
Mean enthalpy
of air ham(i )
Mean enthalpy
difference hsa (i )  ha (i )

hsa(i)  ha(i) m
1
m
70.77
44.23
0.0226
32.4
86.31
35.69
0.02801
34.0
101.65
32.15
0.0311
35.6
117.39
28.61
0.0349
37.2
153.77
25.23
0.0396
 h
sa (i )
 ha (i )
m =0.1562
0
Since the temperature drop of water at each section is T  1.6 C , Eq. (7.31) yields,
hA
 4.18  30  1.6  0.1562  31.33 kW/(kJ/kg) or NTU  31.33 .
c pm
7.6
Homogenous and Ideal Binary Solutions
A binary solution is a two-component, two-phase system with one phase is vapor phase and the
other is condensed (liquid) phase. To understand the properties of such solutions is especially important for analyzing the vapor absorption refrigeration systems.
According to Gibb’s phase rule, the thermodynamic state of a binary solution (in liquid or in
vapor phase) can be fixed by three independent properties. In addition to pressure and temperature,
the composition of the mixture which is identified by mass fractions of components A and B is taken
as the third parameter. The mass fractions are expressed as,
xA 
mA
mB
, xB 
and xB  1  x A
mA  mB
mA  mB
(7.32)
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 311
Where, mA and mB represents masses of components A and B in liquid or in vapor phase.
Homogeneity in a binary solution is only obtainable if the components are miscible. Temperature
might affect the miscibility of binary solutions. However, Refrigerant-absorbent type solutions are
miscible and homogeneous under all conditions both in liquid and vapor phases.
Figure 7.31 shows the temperature-pressure-concentration diagram for Lithium Bromide-water
solution. This diagram is also called equilibrium chart for aqueous (LiBr) solutions.
Figure 7.31 The equilibrium chart for (LiBr) solution
In the above plot, solution temperature is the abscissa of the diagram, the saturation temperature
of water corresponding to vapor pressures is shown as the ordinate. The chart applies to saturated
conditions where the solution is in equilibrium with water vapor. In determining the vapor phase partial
pressure of the solute, the solution concentration and the temperature have to be specified. In accord
with the chart, (LiBr) solution with 54-percent of (LiBr) concentration at 800C develops water vapor
pressure of 10.92 kPa. The same pressure can be recorded at 45-percent of LiBr solution at 650C of
solution temperature. For an isothermal process of solution (vertical line on the chart), increasing the
concentration (states 1 to 2) results with a decrease in vapor pressure. For a constant concentration
process, however, increasing the solution temperature (states 1 to 3) increases the vapor pressure.
Example 7.20 Dilute LiBr-water solution at 300C with water vapor pressure at 1.23 kPa enters a heat exchanger in Figure
7.32 with a mass flow rate of 0.5kg/s. The solution is heated to 1000C by steam flow through the exchanger tubes. Some of
water evaporates, and water vapor pressure increases to 7.38kPa. The concentrated solution leaves the exchanger at state
2. Determine,
a.
the solution concentration at the exchanger inlet and outlet,
b.
the mass flow rate of concentrated solution at the exchanger outlet.
c.
the mass flow rate of water vapor at state 3.
312
THERMODYNAMICS
Solution:
a.
From chart in Figure 7.31, at p1  1.23 kPa, T1  30o C , the concentration at the exchanger inlet is, x1  0.49 .
Similarly, at the exchanger outlet, p2  7.38 kPa, and T2  100o C , then the concentration at the equilibrium is
x2  0.66 .
b.
Since LiBr is highly non-volatile substance, the mass balance requires that x1m 1  x2 m 2 , or 0.49  0.5  0.66  m 2 ,
m 2  0.371 kg/s.
c.
Considering the overall mass balance on the solution side, we may state that m 1  m 2  m 3 , and the mass flow rate
of water vapor becomes, m 3  0.5  0.371  0.129 kg/s.
7.6.1
Ideal Solution
A binary solution is called an ideal solution if it satisfies the following constraints:
1. Upon mixing of two components, there should be no change in volume ( vmix  0 ), and solution volume is the sum of volumes of its constituents. Hence the specific volume of the solution at a
particular state is,
v  x Av A  xB vB  x Av A  (1  x A )vB
(7.33)
2. Upon mixing of two components, there should be no heat generation, or absorption ( hmix  0 ).
Then the specific enthalpy of solution at a particular state is,
h  x A hA  xB hB  x A hA  (1  x A )hB
3. The components of the solution should obey the Raoult’s law.
(7.34)
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 313
Principle 27: Raoult’s law: The vapor pressure of component A can be expressed in
terms of the saturation pressure of pure component A ( p As ) at the same temperature
as that of binary system is as following: lim y 1  p A  y A p As  , where y A is the vapor
phase mole fraction of component A in the solution.
A
Ideal solutions should obey the Raoult’s law over the entire range of 0  y A  1 . However, most
solutions obey the law in the limit y A  1 . Hence the partial vapor pressures of components A and B
at the temperature of the solution are given as,
p A  y A p As and pB  yB pBs
(7.35)
As illustrated in Figure 7.33a, for positive deviation from Raoult’s law, the actual equilibrium
pressure will be higher than that predicted by Raoult’s law, or the equilibrium temperature at a given
concentration and pressure will be lower. For refrigerant-absorbent type binary solutions, because
of exothermic process of absorption, the enthalpy of mixing ( hmix  0 ) is negative as in Figure
7.33b.
4. The vapor phase of the solution should obey Dalton’s law of additive pressures. Hence, the
partial vapor pressures of components A and B at temperature T can be calculated as,
p A  y A ptotal and pB  yB ptotal
(7.36)
Since Dalton’s law of additive pressures holds,
ptotal  p A  pB
(7.37)
For refrigerant-absorbent type of binary solutions, one component is non-volatile with respect to
other. For instance, consider Lithium Bromide-water solution, Lithium Bromide salt is non-volatile
314
THERMODYNAMICS
component, and in vapor phase, yB  0 . Thus, together with Eq. (7.36), Eq. (7.37) is reduced to the
following,
ptotal  y A p As
(7.38)
Example 7.21 Let us consider water-lithium bromide solution with 50% mole fraction of (LiBr) at 49°C. Estimate the
total pressure on the (LiBr) solution, and compare with the actual measurement of 0.9kPa.
Solution:
As explained above, (LiBr) being a highly non-volatile
salt, its existence in vapor phase is almost untraceable. In
accord with Raoult’s law, Eq. (7.38), the total pressure
becomes, p  1  yB  pw or p  1  0.5  3.1.69  1.58
kPa.
With respect to actual pressure measurement, the
pressure of mixture is pmix  0.9  1.58  0.68 kPa,
and the actual equilibrium temperature should be less
than the predicted value.
There are three different situations for
which the ideal solution model might be
useful: 1. If there is no any experimental data
for a particular solute-solvent combination,
then we may resort to the model to estimate
the properties of the solution, 2. Considering
the simplicity of mathematical analysis, the
model may be assumed to be satisfactory for
preliminary calculations, 3. For rare cases
like mixing of isotopes, the predictions of the
model might be very accurate.
7.6.2
Enthalpy diagram for
binary solutions
To predict the enthalpy of a solution at
specified temperature and concentration is
vital for energy analysis of systems especially
used in absorption refrigeration field. If such
data of enthalpy for a particular solution is
given, then the ideal solution model may
not be a factor for analysis. Figure 7.34 illustrates the specific enthalpy, temperature,
and the mass fraction data for water-lithium
bromide solution. In the chart, the enthalpy
of liquid water at 0°C is taken to be 0 kJ/kg,
and lithium bromide salt is at 25°C.
Example 7.22 Using the graph in Figure 7.34, evaluate, the missing properties of lithium bromide-water solution.
a.
Solution temperature and (LiBr) concentration respectively are T1  80o C , x1  0.4 , h1  ?
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 315
b.
Solution temperature and enthalpy respectively are, T2  100o C , h2  300 kJ/kg, x2  ?
Solution:
a.
From Figure 7.34, the enthalpy of (LiBr) solution at the given state is h1  186 kJ/kg.
b.
For a (LiBr) solution at 100°C, with enthalpy value of h2  300 kJ/kg, Figure 7.34 helps us to determine the
concentration as, x2  0.22 .
Example 7.23 Consider the heat exchanger in Example 7.20, where the (LiBr) solution with a mass flow rate of 0.5 kg/s
inlets the exchanger at ( T1  30o C , x1  0.49 ) and exits at ( T2  100o C , x2  0.66 ). Steam at 1bar, enters the exchanger
at T4  150o C and leaves at as saturated vapor. Evaluate the mass flow rate of steam.
Solution:
Referring to Figure 7.35, the temperatures, concentrations, mass flow rates, and the enthalpies of both solution and the steam
at various states are represented in the table below. All fluids are at 1 bar pressure.
State no.
1
2
3
4
5
Temp. (0C)
30
100
100
150
100
Concentration (%)
Mass flow rate (kg/s)
0.49
0.66
-
0.5
0.371
0.129
-
Enthalpy (kJ/kg)
60
260
2676.1
2776.4
2675.5
With respect to control volume around the heat exchanger, the energy balance yields, m s h4  h5   m 2 h2  m 1h1  m 3h3 .
Substituting the tabulated values into energy equation, the steam flow rate is m s 
0.371  260  0.5  60  0.129  2676.1
 4.08 kg/s
2776.4  2675.5
References
1.
Y. V. C. Rao, Engineering Thermodynamics Through Examples, Universities Press, ISBN-81-7371-4231, 2003.
2.
J. P. Holman, Heat Transfer, 6th Edition, McGraw-Hill Book Company, ISBN-0-07-029620-0, 1986.
3.
B. E. Poling, J. M. Pravnitz, J. P. O’connell, Gases and Liquids, 5th Edition, McGraw-Hill Publishing, ISBN 0-07149999-7, 2001.
4.
H. D. B. Jenkins, Chemical Thermodynamics at a Glance, Blackwell Publishing, ISBN 978-1-4051-3997-7, 2008.
5.
“Virial Coefficients of Pure Gases and Mixtures”, Edited by M. Frenkel, and K. N. Marsh, Springer-Verlag, ISBN
3-540-44340-1, 2002.
6.
W. P. Jones, Air Conditioning Engineering, 5th Edition, Butterworth-Heinemann, ISBN 0-7506-5074-5, 2001.
7.
R. W. Serth, Process Heat Transfer, Principles and Applications, Elsevier Science, ISBN 0-1237-3588-2, 2007.
8.
“Power Plant Engineering”, Edited by Black and Veatch, Springer Science, ISBN 0-412-06401-4, 1996.
316
THERMODYNAMICS
Problems
7.8
Tank A contains 2.5kg of methane (CH4) at 200
kPa, 15°C, and tank B contains 4.2kg of O2 at 600
kPa, -15°C. The valve connecting these two tanks is
opened and the gases mix adiabatically. Determine
the mixture pressure and temperature.
7.9
A storage tank of 120 m3 containing a mixture of
acetylene (C2H2), propane (C3H8) and butane (C4H10)
is initially at 100 kPa, 300K. The partial pressure of
each component is measured and determined to be
20 kPa for C2H2 and 55 kPa for C3H8. Find the mass
of each component.
7.10
A mixture of 2.2kg of O2 and 2.5 kg of argon (Ar)
in Figure 7.37 is contained in an insulated pistoncylinder arrangement at 140 kPa, 300K. The piston
compresses the mixture reversibly to one-third of
its initial volume. Determine,
a. the final pressure and the temperature,
b. the work done on the mixture.
Closed systems
7.1
7.2
A tank contains 2.5kg of N2 and 5kg of O2. Determine,
a. the average molar mass,
b. the apparent gas constant,
c. the composition in terms of mole fractions,
d. the average molar mass if the gas mixture contained 5kg of N2 and 2.5kg of O2.
A tank contains 4kmol of N2 and 6kmol of CO2
gases at 37°C, 10 MPa. Assuming ideal gas mixture,
determine,
a. the average molar mass,
b. the composition in terms of mass fractions,
c. the tank volume.
7.3
A gas mixture consists of 1kmol of H2, 3.76kmol of
N2 and 5.24kmol of CO. Determine,
a. the mass of each gas,
b. the apparent gas constant.
c. Replace H2 by O2 and repeat (a) and (b) above.
7.4
A rigid tank contains 4.5kmol of O2 and 6.2kmol of
CO2 gases at 17°C, 100 kPa. Determine the volume
of the tank.
7.5
Consider an ideal gas mixture of 1kg mass at a pressure of 115 kPa, and consisting of 15%H2, 48%O2,
and 37%CO by mass. Determine,
a. the molar mass of the gas mixture,
b. the volume percentage of each gas,
c. the partial pressure of each gas,
d. the specific heat at constant pressure.
7.6
An insulated rigid tank is divided into two compartments by a partition. One compartment contains 7kg
of O2 at 45°C and 100kPa, and the other contains
5kg N2 at 25°C and 190 kPa. The partition is then
removed and the two gases are allowed to mix.
After equilibrium is reached, determine the mixture
temperature, and the pressure.
7.7
Tank A in Figure 7.36 contains 3.2kg of N2 at 27°C,
250 kPa, and Tank B contains 1.5kg of O2 at 27°C,
450kPa. Both tanks are rigid and connected by a
valve. After opening the valve, the two gases are
allowed to mix. If the final temperature is 27°C,
determine,
a. the volume of each tank,
b. the final pressure of the mixture.
Figure 7.36 Schematic of Problems 7.7,
and 7.8
Figure 7.37 Gas mixture in a piston-cylinder
arrangement
7.11
An ideal gas mixture of methane (CH4) and ethylene
(C2H4), each one is 50% by mass, initially is at 450
kPa, 330K in a piston-cylinder apparatus, and occupies
a volume of 1.15 m3. The mixture is compressed by
a reversible and polytropic process to a final state of
430K and 0.03m3. Determine,
a. the final pressure and the polytropic exponent,
b. the work and the heat transfer of the process,
c. the entropy change of the mixture.
7.12
Consider a rigid and adiabatic tank with two compartments separated by a membrane.
a. Compartment A contains 2kmol of N2 and B has
2kmol of O2. Both compartments are at 800 kPa,
300K. The membrane separating the compartments breaks up and the gases mix and finally
come to a state at 800 kPa, 300K. Determine the
entropy generation of this process.
b. What would be the entropy generation of mixing,
if we replace the gases with CO2 and CO at the
same conditions and mole numbers?
c. What would be the entropy generation of mixing, if there were 2kmol of the same gas in both
compartments at the same conditions?
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 317
7.13
Air having a mass of 3.5kg at 100 kPa, 10°C is mixed
with 6kg of nitrogen (N2) at 100 kPa, 110°C. The
mixing takes place at constant pressure. Assuming
that air consists of 21%O2 and 79%N2 by volume,
determine,
a. the final temperature of the mixture,
b. the entropy change of the system.
7.14
An insulated rigid tank is divided into two compartments by a membrane. Compartment A contains 70
kg of O2 and compartment B contains 140 kg of CO2.
Both gases are initially at 27°C and 120 kPa. The
membrane breaks up and the gases mix. Assuming
that both gases behave like ideal gas, determine,
a. the entropy generation,
b. the exergy destruction.
c. What would be the exergy destruction if there
were 140 kg of O2 in compartment B instead.
7.15
As shown in Figure 7.38, a rigid and insulated tank
is divided into three compartments each of which
is 0.1m3 in volume. The end compartments contain
argon (Ar) at 6bar, 35°C, and helium (He) at 2.5bar,
110°C, and the compartment at the center is evacuated. After removing the partitions, gases mix and
attain an equilibrium state. Determine,
a. the final pressure and the temperature,
b. the partial pressure of each gas at the equilibrium
state,
c. the entropy change of the system.
Figure 7.38 Schematic of three-zone
insulated tank
7.16
Rigid tank A contains 1.2 kg of argon (Ar) at 37°C,
100 kPa and is connected by valve to rigid tank B
containing 0.6kg of O2 at 117°C, 500 kPa. The valve
is opened and the gases mix achieving an equilibrium
temperature of 77°C. Assuming surroundings is at
37°C, determine,
a. the volume of each tank in m3,
b. the final pressure in kPa,
c. the heat transfer from or to gases in kJ,
d. the entropy change of each gas, and the entropy
generation of the process.
7.17
A well insulated container with a volume of 0.2 m3 is
divided internally into two equal parts by a rigid and
adiabatic partition. As can be seen on Figure 7.39,
the compartments are connected by a pipe and valve
system. Compartment A contains nitrogen at 27°C,
2bar, and B also contains nitrogen but at 57°C, 10bar.
Both gases are ideal with constant specific heats.
a. The valve is opened and the pressure rapidly
equalizes on both sides and then the valve is
closed. No conduction of heat occurs. Determine
the pressure and temperature on both sides of
the tank, and evaluate the entropy generation.
b. The valve is left open and eventually the temperatures become equal on both sides. Determine
the final pressure and temperature for this case,
and evaluate the entropy generation.
c. Repeat parts (a) and (b) if O2 is substituted for
nitrogen in compartment B.
Figure 7.39 A schematic of Problem 7.17
7.18
An insulated vertical cylinder with a frictionless
piston having a cross-sectional area of 0.2 m2 is
initially 1.2 m in height and contains methane at 120
kPa, 300K. As shown in Figure 7.40, the cylinder
also contains 10 L of capsule containing air at 600
kPa, 300K. The capsule accidently breaks and two
gases mix in a constant pressure process. Take air
as an ideal gas and Assume that the surroundings is
at 300K and 100 kPa, determine,
a. the final temperature,
b. the final height of the cylinder,
c. the work done,
d. the entropy generation of the process,
e. the exergy destruction.
Figure 7.40 Schematic of methane storage tank
318
7.19
7.20
THERMODYNAMICS
One kmol mixture of CO2 and ethane (C2H6) occupies a volume of 0.5 m3 at a temperature of 125°C.
The mole fraction of CO2 is 0.25. Using Kay’s rule,
determine,
a. the mixture pressure,
b. the volume occupied by the mixture if the pressure increased by 10-percent.
Steady flow systems
7.23
The camping stove in Figure 7.41 uses a mixture
of methane and propane as fuel, and the mixture
is prepared in the ratio of 3 parts of propane per 1
part of methane on mass basis. The storage tank of
the stove is 0.03 m3 in volume and contains 1.2 kg
of the mixture. It is required to determine the tank
pressure when it is exposed to hot summer sun rays
and its internal temperature becomes 100°C. Hint:
Apply Kay’s rule.
As shown in Figure 7.42, CO2 gas at 87°C is mixed
with N2 at 27°C in an insulated mixing chamber.
Both flows are at 100 kPa and the mass ratio of CO2
to N2 is 5:2. Determine,
a. the exit temperature of the mixture,
b. the entropy generation rate for 2kg/s mass flow
rate of the mixture.
Figure 7.42 Mixing of gases
7.24
A flow of 1.5 kg/s argon at 17°C is mixed with the
flow stream of 2.5 kg/s carbon dioxide gas at 1227°C
in a mixing chamber. Both streams are at 200 kPa
of pressure and the heat loss through the chamber
is 3.2 kJ per kg of the mixture. Determine,
a. the exit temperature,
b. the entropy generation rate for 17°C of surroundings temperature.
7.25
Figure 7.41 A typical camping stove
A mixture of 45% of argon (Ar) and 55% of H2 by
volume has been proposed to be used as the working
fluid in a closed type Brayton cycle. The mixture is
compressed isentropically from (37°C, 0.4 MPa) to
1.2 MPa. Determine,
a. the final temperature of the mixture,
7.21
A 3kg mixture of 45% argon and 55% nitrogen by
mass is in a tank at 2.5 MPa and 180K. Determine
the volume of the tank,
a. by ideal gas model
b. by using the Kay’s rule.
7.22
The gas in an engine cylinder has a volumetric
analysis of 12%CO2, 13.5%O2, and 74.5%N2. The
temperature of the mixture at the beginning of expansion process is 850°C and the gas mixture expands
through a volume ratio of 8:1 in accord with the
relation pV 1.2  Constant . If the cylinder contains
0.005kg of mixture, determine the work done and
the heat transfer of the expansion process.
b. the power needed if the mixture mass flow
rate is 2 kg/s.
7.26
A mixture of 60% N2 and 40% O2 on mole basis flows
through a long and insulated pipe 20 cm in diameter.
At the pipe inlet, the mixture is at 3.2 MPa, -75°C
and the velocity is 45 m/s. If the pressure drop for
the entire pipe is 0.2 MPa, determine,
a. the temperature and the velocity at the pipe
exit,
b. the exergy destruction rate at the surroundings
temperature of 27°C.
Hint: Apply Kay’s rule and the critical values of both components are given as, N2 :
pc  3.39 MPa, Tc  126 K , O2: pc  5 MPa ,
Tc  155 K .
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 319
7.27
A mixture of O2 and N2 with the same mole numbers
enters a compressor at 15 bar, 220K with a mass flow
rate of 2kg/s. The mixture exits the compressor at 55
bar, 400K. Neglecting the changes in ke and pe and
assuming an adiabatic compressor, determine the
required power and the rate of entropy generation,
a. if ideal gas model is used,
b. if Kay’s rule is applied.
7.28
Air (79% N2 and 21% O2 by volume) is compressed
isothermally at 210°C from 3.5 MPa to 7 MPa in a
steady-flow compressor at a rate of 5.1 kg/s. Assume
no irreversibilities occur during the compression,
and determine the power input to the compressor
by treating air as a mixture of
7.30
A steady stream of equimolar mixture of O2 and N2
is heated from -25°C to 0°C at a constant pressure
of 10 MPa. Assuming that the gas mixture obeys
Kay’s rule, determine the heat capacity rate of the
exchanger to be used for heating the mixture with a
mass flow rate 1.5 kg/s.
7.31
A gas mixture consisting of 25% helium, 75%
hydrogen by volume is compressed isentropically
from 37°C, 350 kPa, to 1100 kPa. Determine the
final temperature and the specific work required if
the compression process is carried out by
a. a piston cylinder apparatus,
b. a compressor.
7.32
a. ideal gases,
b. real gases obeying Kay’s rule.
a. the required minimum power to drive the separation unit
b. the rate of entropy generation due to separation
process.
Hint: The separation unit is not insulated and
might be heat transfer between the unit and the
environment.
Figure 7.43 Flow of gas mixture through a
compressor
7.33
7.29
A gas mixture consisting of 25% CO2, 45% O2 and
30% N2 by mass initially is at 120kPa, 20°C. The
mixture inlets a heat exchanger at a mass flow rate
1.2kg/s, and exits at 45°C, and 100kPa. As shown in
Figure 7.44, heating of the mixture is accomplished
by the condensation of saturated steam at 100°C. Assume an adiabatic heat exchanger and determine,
a. the heat capacity rate of the exchanger,
b. the amount of steam condensation per hour,
c. the entropy generation rate.
Figure 7.44 Heating of gas mixture through a
heat exchanger
One of the natural resources for helium is natural
gas which is composed of 0.1% He and 99.9% CH4
on mole basis. To store pure helium, natural gas is
processed through a separation unit. The gas enters
the separation unit at 130 kPa, 17°C with a volumetric
flow rate of 80 m3/s. Pure helium leaves the unit at
100 kPa, 27°C, and pure methane exits at 130kPa,
37°C. Take the surroundings temperature at 27°C,
and determine
An oxyacetylene torch as shown in Figure 7.45 mixes
oxygen and acetylene in a ratio of 5:1 respectively
on a volume basis. Both oxygen and acetylene are
drawn from separate tanks each of which is at 1.5
bar, 20°C, and flow reversibly through the torch to 1
bar, and 190°C. The torch uses 0.15 kg/s of oxygen.
Assume invariable tank conditions, and ideal gas
mixture in the torch, and determine for a welding
period of 5 minutes,
a. the lost work
b. the amount of heat transfer.
Figure 7.45 Oxyacetylene torch and it is use in
metal cutting
320
7.34
THERMODYNAMICS
Combustion gases at the exhaust stroke of an engine
have a volumetric analysis of 12%CO2, 11.5%O2,
and 76.5%N2 at 530°C, 1.8 bar. As shown in Figure
7.46, the gas mixture expands through a turbo charger
to atmospheric pressure of 1 bar and compresses
isentropically atmospheric air at 27°C, 1 bar. Assume mass flow rates through the compressor and
the turbine are the same, and calculate the maximum
air temperature at the compressor outlet.
7.36
Air-water vapor mixture at 100 kPa, 20°C, 30%
relative humidity is contained in a rigid tank of 0.5
m3. The tank is cooled until water droplets just start
to occur on the surface. Determine the temperature
at which the condensation starts.
7.37
A conference room in dimensions of 25mx10mx3.5m
initially contains air at 0.92 bar, 25°C, and relative
humidity of 25%. After a humidification process,
the room temperature drops to 20°C and the relative
humidity becomes 55%. Determine the amount of
water to be added to the room air.
7.38
Air-water vapor mixture is at a state of 35°C dry
bulb temperature, and 15°C wet bulb temperature.
Determine the humidity ratio, and the enthalpy per
kg of dry air from a base of 0°C of dry air and a
pressure of 101.325 kPa.
7.39
A student at a welcome party for senior class is
handed a cold glass of beverage with ice in it. If the
glass surface temperature is +1°C, the moisture just
starts to condense on the glass surface. For a room
temperature of 21°C, the student calculates the relative humidity of the room as 29% at sea level. Is the
student correct? Justify your answer.
7.40
Air-water vapor mixture initially at 105 kPa, 27°C,
and 35% relative humidity occupies a volume of 0.2
m3 in a vertical piston-cylinder apparatus. The piston
moves downward and the content is compressed
isothermally. Determine the pressure at which condensation starts.
7.41
In cleaning eyeglasses, we exhale heavily our breath
on the glass surface, and wipe out the fog occurring
on the surface. Assume that the air in the lungs is
at 36°C, and 80% relative humidity. Determine the
maximum temperature of glass surface that will stop
causing formation of fog on it.
Figure 7.46 Schematic of turbo charger
Moist air and psychrometry
7.35
7.42
A room contains moist air at 27°C, 95kPa, and 21%
of relative humidity. Determine,
a. the vapor pressure,
b. the humidity ratio,
c. the dew point temperature,
d. the mass of water vapor in 150kg of dry air,
e. the relative humidity if the mixture temperature
is decreased by 10°C at a constant vapor pressure.
Fill the blanks of the following table by using ASHRAE psychrometric chart for sea level.
DBT
(°C)
WBT
(°C)
29
16
Dew point
(°C)
Humidity ratio
(kg-v/kg-air)
Enthalpy
(kJ/kg-air)
24
Specific volume
(m3/kg-air)
40
30
21
60
0.0114
45
7.43
Relative humidity
(%)
Air stream of 225 m3/min at sea level is cooled from
the condition, DBT  27o C , WBT  20o C , until it
becomes saturated at a temperature of 10o C . By
using ASHRAE psychrometric chart, determine,
50
7.44
The products of combustion are flowing through a
heat exchanger with 13%CO2, 12%H2O, and 75%N2
on volume basis at a rate of 0.5kg/s and 100 kPa,
85°C.
a. the moisture removed in kg per hour,
a. Determine the dew point temperature.
b. the sensible heat and the total heat removed in kW,
b. The mixture is cooled 5°C below the dew point
temperature. Calculate the time required to collect 15kg of liquid water.
c. the sensible heat factor (SHF) of the system as the
ratio of sensible heat to total heat removed.
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 321
7.45
An air conditioning unit takes the outside air at 27°C,
85% relative humidity cools down to 17°C before
transporting it into the room. As shown in Figure
7.47, the cooling process is accomplished by running
cold water through the tubes of a fin-and-tube heat
exchanger. Determine,
a. the humidity ratio and the enthalpy at the exchanger inlet,
b. the mass of water removed per kg of kg of dry
air,
7.47
Moist air stream enters a refrigeration coil at
DBT  32o C , WBT  24o C at a flow rate of 50
m3/min. The dew point temperature of the coil is
12o C . If the refrigeration capacity of the coil is 3.2
tons, determine the exit temperature of air stream.
Hint: 1ton=3.516 kW.
7.48
Damp egg cartons designed for carrying whole eggs
initially at 35°C and the moisture content is 50% of
the dry box mass. The boxes, as shown in Figure 7.49,
enter a drier and the moisture content is reduced to
5%. Air at atmospheric pressure enters and leaves
the drier at specified conditions in the figure.
a. Determine the mass flow rate of air on mass
basis,
b. In providing drier inlet conditions to air, the
atmospheric air at 20°C, 60% relative humidity is heated prior to entering into the furnace.
Find the heat capacity of the heat exchanger to
be used.
7.49
Atmospheric air at 32°C and 55% relative humidity
is to be conditioned to 21°C and 35%relative humidity by a cooling coil section, a condensate removal
section, and a reheat section. For air mass flow rate
of 250 kg/h, sketch the processes schematically and
determine,
a. the mass of water removed per hour
b. the heat transfer rate in kW.
c. the relative humidity, and enthalpy of air entring
the house,
d. the amount of heat removed per kg of dry air.
Figure 7.47 Cooling process by a fin-and-tube
heat exchanger
7.46
Moist air stream enters the evaporator of an air conditioning system at 35°C, 80% relative humidity, with
volumetric flow rate of 40 m3/min. The air stream is
cooled down to 20°C, 20% relative humidity at sea
level. As shown in Figure 7.48, refrigerant R134a,
flowing through the tubes of the evaporator, enters
the evaporator at 400 kPa, 15% quality and leaves
as saturated vapor. Determine,
a. the rate of heat transfer at the evaporator in
kW,
b. the amount of water condensing per second,
c. the mass flow rate of refrigerant in kg/s.
Figure 7.49 Egg box drying system
7.50
Figure 7.48 The evaporator of an air conditioning system
A stream of return air with a volume flow rate of 16
m3/min at 38°C dry-bulb temperature and 32°C wet
bulb temperature is first dehumidified by removing
75 g of water per minute and then is mixed with a
conditioned second stream. The second stream taken
outside of the building at the following conditions:
DBT  10o C ,   40% , and V  50 m3 /min . This
stream is first heated to 17°C and flows through a
spray section where 280g of steam at 150°C, 3bar
is adiabatically injected per minute. The resulting
322
THERMODYNAMICS
second stream mixes adiabatically with the return air
stream. Finally, the combined stream is distributed to
various rooms of the building. Assume all processes
take place at constant pressure of 101.3kPa, sketch
the system schematically and show the processes on
a psychrometric chart. Determine,
a. the volume flow rate of air at the tower inlet,
b. the mass flow rate of makeup water in kg/s.
7.54
A cross flow type cooling tower is used to cool water
from 37°C to 24°C at a water flow rate of 100 tons
per hour. The air enters the tower at 16°C with relative humidity of 50% and leaves at the top at 32°C
as saturated air. Determine,
a. the mass flow rate of air in kg/s,
b. the mass of makeup water required per hour.
7.55
A counter flow type and pilot-scale cooling tower at
a volumetric flow rate of 20m3/min and relative humidity of 40% at 20°C. Air leaves with DBT  34o C ,
WBT  33o C at atmospheric pressure of 101.3kPa.
As shown in Figure 7.50, the height of cooling
tower is 2 m.
a. the temperature of return air after the dehumidification process,
b. the amount of heat rejected during the dehumidification process,
c. the amount of heat input to the outside stream
to bring its temperature to 17°C,
d. the temperature of outside stream after humidification,
e. the temperature and the relative humidity after final
mixing of return air and outside air streams.
7.51
To maintain proper conditions in a surgery room,
air stream of 35m3/min is to be supplied at 22°C,
and 25% relative humidity. As shown in Figure
7.50, outside air at DBT  22o C , WBT  17o C is
first compressed isentropically and then cooled to
22°C in a heat exchanger. Hence, the liquid water
is separated from the air stream, and the humidity
is dropped to a desired level. Finally, air stream is
throttled to 100 kPa pressure. The power consumed
by the compressor is 4kW. Determine,
a. the pressure at state 2,
b. the amount of liquid condenses,
c. the amount of heat removed at the heat exchanger.
Figure 7.51 Counter flow type cooling tower
Figure 7.50 Dehumidification by compression
and cooling
7.52
7.53
Air at 30°C and 80% relative humidity is to be processed in a steady flow air conditioner system and
delivered at 20°C, 50% relative humidity. Design
a suitable system for conditioning 10m3/min of air
stream. Specify any energy transfers, power requirements, temperatures, intermediate system states, and
sketch the system.
A counter flow type cooling tower is to be used
for cooling 75kg/s of water from 40°C to 25°C by
atmospheric air at 98kPa pressure. Air enters the
tower at 20°C, 60% relative humidity, and exits as
saturated air at 33°C. Neglecting the power input to
the fans, determine,
a. Perform a mass balance on water side of this system
and calculate how much liquid water to be added
to compansate for evaporation in kg/min.
b. Calculate the number of transfer units for this
cooling tower by dividing the tower into four
sections with identical water temperature drop at
each section.
7.56
Consider a cross flow cooling tower as shown
in Figure 7.27d operating with water flow rate of
48kg/s and air flow rate of 42kg/s has a value of
hc A / c pm  45 kW/(kJ/kg) . The enthalpy of entering air is 75 kJ/kg, and the temperature of water at
the inlet is 35°C. Predict the water outlet temperature
when the tower is divided into 8 sections.
7.57
To increase the mass concentration of (LiBr) in a
stream of dilute LiBr-H2O solution at mass flow rate
of 0.4kg/s, the solution inlets a heat exchanger at 35°C,
40% concentration, and exits at 80°C. In heating the
solution, geothermal water at 160°C with a mass flow
rate of 1.5kg/s is available. Water leaves the exchanger
at 100°C. Determine,
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 323
a. the concentration of the solution at the exchanger
exit,
b. the mass flow rate of water vapor extracted from
the solution.
7.58
As shown in Figure 7.52, two compartments of a
container are separated with an adiabatic partition.
The compartment A contains aqueous solution of
(LiBr) at a mass fraction of 0.5. The compartment B
is filled with water vapor. Initially the valve is open
and both compartments are in equilibrium at 10 kPa,
and 80°C. Cold water circulates through the coil in
B, and water vapor starts condensing.
Figure 7.53 An unsteady filling process
a. the work done,
b. the amount of nitrogen flows into the cylinder,
c. the heat transfer of the process.
d. Show that the process is consistent with the
second law of thermodynamics.
Figure 7.52 Behavior of binary solutions
on a process
a. Determine the temperature at which water vapor
starts condensing in B.
b. With respect to equilibrium chart in Figure 7.31,
to maintain the same pressure at both compartments, what will happen to the concentration,
and the temperature of the solution. Will these
properties increase or decrease? Explain.
c. Calculate the mass of solution required so that
1.5 kg of water vapor is transferred from A to
B and the mass concentration of the solution is
increased by 10-percent.
7.60
A gas mixture consisting of 25% propane (C3H8),
45% butane (C4H10) and 30% methane (CH4) by
mass initially is at 3 bar, 27°C in a tank of 0.5 m3.
Due to corrosive environment at 27°C, the discharge
valve on the tank starts leaking and the gas mixture
escapes isothermally and slowly. Calculate the tank
pressure when the half of the tank content leaks.
7.61
A rigid 400 L of tank initially contains air-water
vapor mixture at 150 kPa, 40°C and 10% relative
humidity. A steam supply line at 600 kPa, 200°C is
connected to the tank by a valve. The valve opens, the
steam flows into the tank and the relative humidity
increases. When the relative humidity reaches 90%,
the valve is closed. Because of heat transfer, the tank
temperature remains constant at 40°C. Determine,
d. Neglect the temperature changes during the transfer
of 1.5 kg of water vapor, and calculate the heat
tranferred to solution by heating process in A,
and the heat removed by cooling process in B.
e. Why the heat supplied and the heat removed are
not identical? Explain the difference.
a. the mass of water vapor entering the tank,
b. the final pressure inside the tank,
c. the heat transferred during the process.
Unsteady systems
7.59
As shown in Figure 7.53, vertically positioned pistoncylinder apparatus initially contains 25L of argon at
105kPa, 17°C. The nitrogen line at 300kPa, 27°C is
connected to the cylinder by a valve. Due to stops, the
maximum volume of the cylinder is 40L. The valve
opens nitrogen flows into the cylinder and mixes
with argon. When the valve is closed, the pressure
and temperature inside the cylinder respectively is
250kPa, 35°C. Assume surroundings at 17°C, and
determine,
7.62
As shown in Figure 7.54, a rigid tank of 50 L is
initially half filled with liquid water and the other
half is a mixture of air and water vapor at 300 bar.
Liquid water is evacuated slowly by a valve at the
bottom of the tank until the last liquid drop disappears.
The evacuation process takes place isothermally at
200°C. Determine,
a. the final pressure in the tank,
b. theamount of liquid water taken out of the
tank.
324
THERMODYNAMICS
Figure 7.54 An isothermal discharging process
i.
There are cases for which the DBT and
the WBT temperatures of air-water vapor mixture
might be identical.
j.
When the enthalpy of air is equal to
the enthalpy of saturated air at a wetted surface
temperature, then there is no net heat transfer
between air and the wetted surface.
k.
In sensible heating process of air, DBT,
WBT, and dew point temperatures all increase.
l.
In sensible cooling process of air, DBT,
and WBT temperatures decrease but the dew point
temperature remains constant.
m.
For an unsaturated air, water vapor in
the mixture is at superheated state.
n.
For air conditioning systems operating
in very humid climates, the sensible heating is
always low compared to latent heating.
o.
In heating and humidification processes,
the relative humidity of air increases.
p.
For an ideal solution, there is neither
expansion nor contraction upon mixing.
q.
An ideal binary solution obeys Raoult’s law
in liquid phase and Dalton’s law in gas phase.
True and False
7.63
Answer the following questions with T for true and
F for false.
a.
The sum of mole fractions of ideal gas
and also real gas mixtures is always unity.
b.
The apparent gas constant for a mixture
is always larger than the largest gas constant in
the mixture.
c.
The apparent molar mass of a mixture
of two gases is determined by simply taking the
arithmetic average of the molar masses of the
individual gases.
r.
For a binary solution, a negative deviation from Raoult’s law means that the actual
equilibrium temperature is more than the one
predicted by Raoult’s law.
d.
Taking equal portion of components A
and B at the same pressure and temperature and
mixing them will lead to the largest entropy of
mixing.
s.
For a binary solution, a positive deviation from Raoult’s law indicates that the mixing
process is endothermic.
e.
An equimolar mixture of X2 and Y2 also
yields the same mass fractions.
f.
The bathroom mirror often fogs up when
the surface temperature is higher than the dew
point temperature of the bathroom air.
g.
The same gas at the same pressure and
temperature is contained in two identical and
separate compartments. As a result of mixing of
these gases, entropy is generated.
h.
The molar volume ( v  V / n ) of an
1
1
ideal gas mixture is given by 
where
v
vi
vi  V / ni .

Check Test 7
Choose the correct answer:
1.
A rigid and insulated tank is divided into two compartments by a partition. Initially one compartment
contains 0.04m3 of steam at 157°C, 1bar, and the
other compartment contains 0.12m3 of methane at
27°C, 1bar. The partition is removed and both gases
are allowed to mix. For ideal gas behavior of both
gases, the estimated partial pressure of steam in kPa,
and the mixture temperature in K are,
a. 18.6, 322,
b. 19.6, 322,
c. 18.6, 332,
d. 19.6, 342.
CHAPTER 7 GAS MIXTURES & PSYCHROMETRY 325
2.
An ideal gas mixture of hydrogen (H2) and argon
(Ar) with identical mass fractions enter an adiabatic
turbine at 1000K, 10bar at a rate of 0.5kg/s. The
mixture isentropically expands through the turbine
and develops power in kW as,
a. 1509.2,
b.
1609.2,
c. 1709.2,
d.
1809.2.
6.
7.
3.
An automobile exhaust gas analysis shows that the gas
composition on mole fraction basis is as following,
N2
0.808
CO2
0.100
O2
0.002
CO
0.065
H2
0.025
The specific volume of the exhaust gas in m3/kg is
4.
a. 0.72,
b.
0.82,
c. 0.92,
d.
1.02.
In a test of a new furnace, first 1.5m3/s methane is
adiabatically mixed with 2.5m3/s of propane gas.
Then, the mixture is preheated to 170°C by passing
through an adiabatic heat exchanger which uses
steam at 200°C for heating the gas mixture. Steam
inlets the exchanger as saturated steam and exits
as saturated liquid. As shown in Figure 7.55, both
gases are at 1bar, 25°C before entering the mixing
chamber. The mass of steam in kg/s, and the rate of
total entropy generated in kW/K respectively are,
8.
A conference room in dimensions of 12mx15mx4m
contains air at 100kPa, 25°C and 60% relative humidity. Estimate the amount of water in that room
in kg.
a. 5.91,
b.
7.91,
c. 9.91,
d.
11.91.
Outside air at 10°C, 40% relative humidity is mixed
with inside air of a factory building at 35°C, 70%
relative humidity taken from the ceiling. The volume
flow rates of inside and outside air streams respectively
are 30m3/min and 50m3/min. The mixing takes place
at 100kPa pressure. Estimate the relative humidity
of the mixed stream in percent (%).
a. 75,
b. 80,
c. 85,
d. 90.
If water is sprayed into air at relative humidity of
35% in an adiabatic and constant pressure environment, the following happens,
a. the temperature decreases,
b. the enthalpy decreases,
c. humidity ratio decreases,
d. relative humidity decreases.
9.
A stream of moist air is subjected to two processes in
sequence. First the stream is cooled then humidified
by steam injection into the stream.
a. the enthalpy of air decreases,
a. 0.72, 2.755,
b.
0.72, 3.775,
b. the enthalpy of air remains constant,
c. 0.72, 1.755,
d.
0.92, 1.75.
c. the enthalpy of air first decreases then increases,
d. the enthalpy of air increases.
10.
During an isothermal expansion of 0.5kg of moist
air the following property does not change.
a. the relative humidity,
b. the dry bulb temperature,
Figure 7.55 Pre-heating of gas mixture
c. the dew point temperature,
d. the specific enthalpy.
5.
In a highly explosive environment, a pneumatic motor
has to produce 5hp of power by utilizing a mixture
of 75% argon and 25% helium in mass fractions.
The mixture inlets the motor at 10bar, 37°C and
expands by a reversible isothermal process to 1bar.
The closest value of the required mass flow rate in
kg/h is,
a. 25.6,
b.
26.6,
c. 27.6,
d.
28.6.
11.
The following temperature data are taken from a
forced draft counter flow cooling tower through
which water flows at a rate of 5kg/s.
Inlet
DBT(°C) WBT(°C)
Air
Water
30
39
25
Exit
DBT(°C)
35
32
WBT(°C)
35
The air flow rate in kg/s, and the percentage of entering water vaporizes by passing through the tower
respectively are,
12.
a. 2.66, 1.02,
b.
2.46, 1.02,
c. 2.56, 1.25,
d.
2.66, 1.20.
Dilute (Li-Br)-water solution inlets a heat exchanger
at 70°C, 59% concentration with a mass flow rate
of 0.45kg/s. The solution is heated by hot water
at 115°C, and exits the exchanger at 95°C, 64%
concentration. The mass flow rate of water vapor
produced in kg/h is,
13.
14.
In a heat exchanger, lithium-bromide water solution
flowing at a rate of 0.35kg/s is heated from 50°C,
55% concentration at the inlet to 75°C, 65% concentration at the outlet. Heating is accomplished by
hot water which inlets the exchanger at 115°C and
exits at 105°C. The estimated mass flow rate of hot
water in kg/s is,
a. 1.52,
b.
2.52,
c. 3.52,
d.
4.02.
An important property of binary solutions used
for vapor absorption refrigeration systems is to
exhibit
a. a negative deviation from Raoult’s law upon
mixing,
a. 121.56,
b. 122.56,
b. producing a large heat of mixing,
c. 125.56,
d. 126.56.
c. a positive deviation from Raoult’s law upon
mixing,
d. a large temperature difference between the boiling points of components.
C
H
8
A
P
T
E
R
Power Producing Systems
8.1
General Considerations for Power Cycles
Two important applications of the outcomes of thermodynamics are: 1. The conversion
of heat energy into work, and 2. The transfer of heat from low temperature medium to high
temperature.
To accomplish these duties in a continuous manner, we need systems that operate
on thermodynamic cycles. Referring to the energy conservation principle for cycles as,
 Q   W , and if  W  0 , then that particular cycle is called power cycle and a power
cycle is used to convert heat energy into work (heat engine). The heat in turn is obtained
by means of a combustion process. If the combustion process is part of the cycle, then the
system is called internal combustion system. Air is the working fluid of these systems and
changes its composition due to combustion.



Figure 8.1 The power cycles commonly in use for industrial applications
If the combustion is not a part of the cycle, then the system is named as external combustion system. Since the combustion is external to the cycle, the composition of the working
fluid does not change throughout the cycle. The classification of power producing systems
which are in common use is presented in Figure 8.1. In rating the internal combustion engines
on a power scale from the least power producing one to the most, three types of engines
327
328
THERMODYNAMICS
have found general acceptance in industry. These are respectively SI engine, CI engine with positive
displacement and the rotary type CI engines.
The cycles related to the engines presented above are very complex to analyze. In internal combustion engines, for instance, the air-fuel mixture is usually mixed with the residual exhaust gas remaining
from the previous cycle, and this resulting mixture is actually compressed. Due to friction, pressure
drop takes place in the intake and the exhaust systems. Heat at a certain rate is usually lost through
the pipes connecting the main components of the engine, and non-equilibrium conditions may exist
within the processes of the system. To make the analysis more manageable, however, the following
model has found a general acceptance in literature.
8.1.1
Ideal cycle
To study the effects of major parameters on the engine performance, certain minor complexities
are ignored. Hence, Ideal cycle is a mathematical model for performing analysis on cyclic engines.
Principle 27: An ideal cycle is an internally reversible cycle. For an ideal cycle, a. the working fluid does not cause any pressure drop due to flow through pipes and heat exchangers, b.
pipes connecting the components of the cycle are well insulated, and the transfer of heat is
neglected, c. the expansion and the compression processes are considered to be reversible. If
there is no heat transfer, then those processes become isentropic, d. Changes in kinetic and
potential energies due to flow through the devices are neglected.
Figure 8.2 Conversion of an actual cycle into an ideal one
A comparison of an actual cycle with the ideal correspondence is presented in Figure 8.2. Consistent with the ideal cycle definition, no pressure difference is required for the intake and exhaust
periods, and no time is needed for completion of combustion process. The heat addition and extraction processes take place instantly in the ideal cycle. However, an ideal cycle is not a reversible
cycle. Heat transfer between the working fluid and the low or high temperature medium takes place
at a finite temperature difference and this causes irreversibility. Therefore, the efficiency of an ideal
cycle is always less than the efficiency of a reversible cycle operating between the same high and
low temperature reservoirs.
CHAPTER 8 POWER PRODUCING SYSTEMS 329
8.1.2
Main components of reciprocating engines
Except Wankel rotor, the engines operating with SI and CI principles are generally reciprocating
type in their construction. Therefore, the following terms and abbreviations are defined and used for
the components of these engines. The basic components are illustrated in Figure 8.3.
Definitions: Spark ignition (SI): The combustion process of the cycle is started by a spark plug.
Compression ignition (CI): The combustion process starts when air-fuel mixture self ignites due to
high temperature in the chamber.
Four-stroke cycle: The cycle is completed by four piston movements over two engine revolutions.
Two-stroke cycle: The cycle is completed by two piston movements over one engine revolutions.
Top Dead Center (TDC): The position of the piston when it is at the furthest point away from the
crankshaft.
Bottom Dead Center (BDC): The position of the piston when it is at the closest point to the crankshaft.
Bore (D): The diameter of the cylinder, and due to small clearance, the diameter of the piston face.
Stroke (H): The distance between the (TDC) and the (BDC) of the cylinder.
Stroke volume (Vst): The volume that is swept by the piston as travels through one stroke.




For one cylinder, Vst  H  D 2 / 4 , for N cylinders, Vst  H  D 2 / 4 N
(8.1)
Clearance volume (Vc): The volume left in the combustion chamber when the piston is at (TDC).
Fuel-Air ratio (FA): The ratio of mass of fuel input to the mass of air, FA  m f / m a . The ideal or
stoichiometric FA for many hydrocarbon fuels is around 1:15. Gasoline fueled SI engines usually
have FA input in the range of 1:10 to 1:18. However, FA input of CI engines ranges between 1:18
and 1:50.
Figure 8.3 Components of a four-stroke cycle SI engine
Clearance volume (Vc): The volume left in the combustion chamber when the piston is at (TDC).
Compression ratio (r): The ratio of the maximum cylinder volume to the minimum.
330
THERMODYNAMICS
r
Vmax Vc  Vst
V

 1  st
Vmin
Vc
Vc
(8.2)
Typical values of compression ratio are r  8 to 12 for SI engines, and r  12 to 25 for CI engines.
8.1.3
Mean effective pressure, torque, and power
Mean effective pressure (pm): Being independent of engine size and speed, this parameter is usually used for comparing engines for power output. As shown in Figure 8.4, the two shaded areas are
identical and represent the net work output of the cycle. Hence, the mean effective pressure (pm) is
such a fictitious pressure that the multiplication by stroke volume yields the cyclic net work output
and expressed as following,
Wnet  pmVst
(8.3)
Typical values of pm for SI engines are 1000 kPa to 1400 kPa, and for CI engines are 800 kPa
to 1000 kPa.
Torque ( T ): Torque indicates the engine’s ability to do work, and for a four-stroke engine, is related
to the net work per cycle as,
T  pmVst / 4
CI engines generally have greater torque than SI engines.
(8.4)
CHAPTER 8 POWER PRODUCING SYSTEMS 331
Power (W ): For an engine of N cylinders running at a speed of n revolutions per second, the power
developed is,
W  2 nT  pmVst n / 2  N
or
W  m kg/cycle wnet kJ/kg n cycles/s 
(8.5)
Engines producing power in the range of 2 kW to 5 kW are commercially used for lawn mowers,
chain saws, and for snow blowers. Automobile engines mostly produce power in the range between
50 kW and 200 kW. With respect to Eq. (8.5), more power may be generated by increasing the stroke
volume. Increased volume, however, results with a bulky engine size which is an unfavorable approach
in design. Hence, modern engines are smaller in size but run at high speed.
Example 8.1 An eight cylinder four-stroke cycle SI engine operates at 2200 rpm. Determine,
a.
the number of cycles per minute,
b.
the number of cycles per revolution,
c.
the degrees of engine rotation for each ignition to take place.
Solution:
a.
 2200 
The number of cycles per minute is ncycle  
  8  4400 cycles/min.
 2 
b.
If we divide the number of cycles per minute by the revolutions per minute, the result would be the number of
 2200 

cycles per revolution. Hence, ncycle

  8 / 2200  4 cycles/rev.
 2 
c.
Since one ignition takes place per cycle and four cycles are completed in one revolution, then the angle of rotation
for each ignition is ignition  360 / 4  900.
Example 8.2 An automobile has 2.4 liter SI four-cylinder engine operating on a four-stroke cycle at 3800rpm. At this
speed, the torque output of the engine is 220 Nm, and air enters the cylinder at 90 kPa, 50°C. The engine is square (D=H),
and the compression ratio is 8. Determine,
a.
the cylinder bore and the stroke length,
b.
the clearance volume of one cylinder,
c.
the power produced,
d.
the mean effective pressure,
e.
the specific work output of the engine.
Solution:
a.
  D2 
Since the engine is square, V  4Vst  4 
  D  0.0024 or D = H = 0.091m.
 4 


b.
By Eq. (8.2) r  1 
c.
Equation (8.5) yields the power developed as, W  2  3.14  (3800 / 60)  220  87.5 kW.
d.
Eqs. (8.3) and (8.4) provide the mean effective pressure of the cycle as, pm  4 T / Vst  4  3.14  220 / 0.0006  4605.33
kPa.
e.
The amount of work done per cycle is Wcycle  4 T  4  3.14  220 / 1000  2.76 kJ/cycle. Besides, the amount
Vst
 8 or Vc  Vst / 7  0.0006 / 7  0.000085m3.
Vc
of air that inlets the cylinder per cycle is mcycle  p1V1 / RT1  90  0.006  0.000085  / (0.287  323)  0.00066
kg/cycle. Thus, the engine specific work is w  Wcycle / mcycle  2.76 / 0.00066  4181.81kJ/kg.
332
THERMODYNAMICS
8.2
Four-stroke SI Engine Cycle
As illustrated in Figure 8.5, this basic cycle is fairly standard for all SI engines and consists
of the following processes,
Stroke 1. The piston travels downward with the intake valve at open position and the exhaust
valve closed. Because of pressure differential, air passing through the intake system mixes with
fuel vapor at a desired level and fills the cylinder.
Stroke 2. The intake valve closes when the piston reaches BDC, and the air-fuel mixture is compressed when the piston travels upward. Close to TDC, the spark plug is ignited, and the combustion
starts. This, in turn, raises the pressure and temperature in the cylinder to a very high level.
Stroke 3. Combustion gases at high pressure causes the piston to move away from TDC, and the
pressure and temperature in the cylinder drops. The exhaust valve opens before the piston reaches
the BDC for exhaust blow down.
Stroke 4. The piston moves upward and pushes the remaining exhaust gases in the cylinder after
the exhaust blow down. However, the gas trapped in the clearance volume will be left for the
next cycle.
These engines provide peak pressure values between 1030 kPa and 2060 kPa. They employ
carburetor, gas mixing system, or fuel injection system. A wide variation in speed and power is obtainable for SI engines. They are commonly used in automobiles. The indicator diagram for a typical
four-stroke SI engine is illustrated in Figure 8.6a. As shown in the figure, at point A, the intake valve
opens and the exhaust valve closes. The intake valve closes at B, and combustion starts and ends at
points C and D respectively. The exhaust blow down takes place at E.
CHAPTER 8 POWER PRODUCING SYSTEMS 333
To reduce the complexity of the actual cycle of SI or CI engines, the real cycle is further approximated with an air-standard cycle.
Principle 28: In addition to being an ideal cycle, an air-standard cycle differs from the actual cycle
by the following: 1. The working fluid is air for the entire cycle and the property values of air are
used in the analysis. 2. The combustion process is replaced by heat addition process, and the combustion chamber works as a heat exchanger. 3. The open exhaust process is replaced by a heat rejection
process that restores the air to its initial state. 4. The real open cycle is changed into closed cycle by
assuming that the air after the exhaust process is fed back into the intake of the engine.
The air-standard cycle for four-stroke SI engines is Otto cycle. As shown in Figure 8.6b, Otto cycle
starts with the piston at TDC (point 5). As the piston moves downward, the pressure in the cylinder
is kept constant (5-1). Then, the air trapped in the cylinder is compressed isentropically (1-2). The
334
THERMODYNAMICS
compression stroke is followed by a constant volume heat input process at TDC (2-3). The air at peak
pressure and temperature isentropically pushes the piston downward (3-4). The exhaust blow down is
accomplished by constant volume heat rejection process at BDC (4-1). Finally, as the piston travels
from BDC to TDC, the exhaust gases are expelled from the cylinder at constant pressure (1-5). Figure
8.6c also illustrates Otto cycle on T-s coordinates. It is quite common to find Otto cycle shown with
processes 5-1 and 1-5 left off the figure. The reasoning for that is simply these two processes cancel
each other thermodynamically and not needed for the cycle analysis.
Thermal efficiency of Otto cycle. The thermal efficiency of Otto cycle, as determined by Eq. (4.48),
may be expressed as following,
o 
wnet
q
 1  out
qin
qin
(8.6)
Where, with respect to Figure 8.6b,
qin  u3  u2
and
qout  u4  u1
(8.7)
In both heat addition and extraction processes of the cycle, the temperature of air changes drastically
and causes appreciable change in specific heat values. If such a change in specific heats is taken into
consideration, then the air tables have to be utilized in the analysis. Otherwise, for constant specific
heats, qin  cv T3  T2  , qout  cv T4  T1  and the thermal efficiency in terms of cycle temperatures
becomes,
o  1 
T T / T  1
T4  T1
1 1 4 1
T3  T2
T2 T3 / T2  1
(8.8)
For an isentropic process of an ideal gas with constant specific heats, the following relations hold,
T1 / T2  v2 / v1 
k 1
 v3 / v4 
k 1
 T4 / T3 and Eq. (8.8) reduces to
o  1 
T1
1
1
1
 1  k 1
k 1
T2
r
v1 / v2 
(8.9)
This oversimplified model of the working fluid with constant specific heats concludes that the
efficiency goes up as the compression ratio (r) increases. This result is also true for real SI engines.
Then, the next logical question is that to what values the compression ratio can be increased for the
purpose of enhancing the efficiency. In real engines, the phenomenon called knocking occurs when
the compression ratio exceeds a certain limit, and the temperature in the cylinder self ignites the fuel
without spark. Even though, the knocking might be avoided by increasing the octane number of the
fuel, the compression ratios of SI engines generally are not greater than 12.
Example 8.3 A six-cylinder, 3.0-liter SI automobile engine operating on a four stroke air standard Otto cycle at 3200
rpm. The engine has a compression ratio 8.5:1, and a stroke-to-bore ratio H/D=1.02. Air at the start of compression process
is at 98 kPa, 60°C, and 1075 kJ/kg of heat is generated during the combustion process. Assuming variable specific heats,
determine,
a.
the peak temperature and the pressure,
b.
the net specific work output of the engine,
c.
the mean effective pressure,
d.
the thermal efficiency,
e.
the power produced by the engine.
CHAPTER 8 POWER PRODUCING SYSTEMS 335
Solution:
a.
Air Tables (A24) and Figure (8.6b) will be used in solving the problem. For T1  333K , the reduced volume and the
internal energy of air is Vr1  479, u1  237.6 kJ/kg. Since the compression process is isentropic,
V1 Vr1

 8.5 ,
V2 Vr 2
and Vr 2  56.35 , by air Table, T2  755 K , u2  555 kJ/kg. Since, qin  u3  u2 , u3  555  1075  1630 kJ/kg. By
air table, the peak temperature is T3  1950 K .
 v  T2 
T
755
The state equation yields the peak pressure as following, p2  p1  1 
 1888.6 kPa, p3  3 p2
   98  8.5 
v
T
333
T
2
 2  1 
1950
or p3  1888.6 
 4877.8 kPa.
755
V
V
1
or Vr 4  8.5  3.022  25.68 and the air table yields
b. Since the expansion process is isentropic, 3  r 3 
V4 Vr 4 8.5
u4  754.3 kJ/kg, qout  754.3  237.6  516.7 kJ/kg. The net specific work is wnet  1075  516.7  558.3 kJ/kg.
c.
To calculate the effective mean pressure, first the volume at state 1 has to be found by v1 
m3/kg. Then, the mean pressure is pm 
0.287  333
 0.975
98
wnet
558.3

 648.96 kPa.
1 
 1

v1 1   0.975  1 

 r
 8.5 
558.3
 51.9%
1075
d.
By Eq. (8.6), the thermal efficiency of the cycle is o 
e.
As indicated by Eq. (8.5), the power produced is W  m kg/cycle wnet kJ/kg n cycles/s  . On the other hand, Vst  0.003
m3 and V1 
0.003
 0.003402 m3, the amount of mass per cycle is m kg/cycle   0.003402 / 0.975  0.003489
1  1 / 8.5
kg/cycle. The number of cycles per second is n  3200 / (2  60)  26.67 cycles/s. Hence the power developed is,
W  0.003489  558.3  26.67  51.95 kW.
Example 8.4 The automobile engine in Example 9.2 consumes fuel with a heating value of 41,100 kJ/kg. The FA ratio for
the engine is 1:18, and the combustion is assumed to be ideal. Assuming constant specific heats (k=1.4), determine,
a.
the thermal efficiency,
b.
the specific heat input,
c.
the specific net work,
d.
the power produced by the engine.
Solution:
1
a.
The thermal efficiency by Eq. (7.9) becomes o  1 
b.
To calculate the specific heat input, first the mass of fuel per cycle has to be determined. Since, Vst  0.0024 m3,
81.4 1
 56.4%
Vc  0.00034 m3 and V1  0.00274 m3, then the mass of air-fuel mixture per cycle is m 
kg/cycle. The mass of fuel becomes, m f 
p1V1 90  0.00274

 0.00266
RT1
0.287  323
1
 0.00266  1.4  104 kg fuel/cycle. The heat input per cycle is
19
q f  m f H u  1.4 x104  4.11  104  5.754 kJ/cycle. The specific heat input then is qin 
kJ/kg
qf
m

5.754
 2163.15
0.00266
336
THERMODYNAMICS
c.
Since the thermal efficiency is known, wnet  o qin  0.564  2163.15  1221.58 kJ/kg
d.
By Eq. (8.5), the power output becomes, W  0.00266  1221.58  31.67  102.89 kW
8.3
Four-stroke CI Engine Cycle
To prevent knocking at high compression ratios r  12  , first only air is compressed, and then
the fuel is pulverized directly into the combustion chamber. After mixing with hot air, the fuel evaporates and combustion starts. Such a cycle is called CI engine cycle. Constructively, CI cycle can be
performed either in reciprocating or in rotary type engines. The illustration in Figure 8.7 is about
four-stroke CI cycle for reciprocating engines.
Stroke 1. Same as the intake stroke of SI engine, except that no fuel is added to the incoming air.
Stroke 2. Only air is compressed. Because of high compression ratio, the air temperature assumes
high enough values to self-ignite the fuel at the end of compression stroke.
Stroke 3. As combustion continues, the piston moves towards BDC. The pressure in the cylinder is
almost kept constant until the fuel injection is completed.
Stroke 4. As the piston travels towards TDC, the exhaust gases in the cylinder are discharged.
These engines produce high thermal efficiency followed with low specific fuel consumption, and
provide peak pressures between 2760 kPa and 4830 kPa. Over the years CI engines have become the
premier transportation system for intermediate power needs.
The indicator diagram for four-stroke CI engine is shown in Figure 8.8a. Due to time requirement
for the completion of combustion, the combustion process still continues when the expansion starts.
This, in turn, keeps the cylinder pressure at peak levels well past TDC. Hence, the combustion is well
approximated as constant pressure heat input in air standard CI cycle which is called Diesel cycle. As
shown in Figure 8.8b, except the constant pressure heat input, Diesel cycle resembles Otto cycle and
the occurrence of exhaust blow down is exactly the same for Diesel cycle. The temperature-entropy
diagram of Diesel cycle is illustrated in Figure 8.8c. The increase in volume during fuel injection
period is an important parameter for cycle analysis and is called Cutoff ratio.
CHAPTER 8 POWER PRODUCING SYSTEMS 337
Definition: Cutoff ratio rc  : is the volume change during combustion (see Figure 8.8b) and expressed as,
rc 
v3
v2
(8.10)
Thermal efficiency of Diesel cycle. Referring to Figure 8.8b, the amount of heat transferred to the
working fluid (air) at constant pressure is qin  h3  h2 . Because of exhaust blow down, the amount of
338
THERMODYNAMICS
heat rejected is qout  u4  u1 . Rewriting Eq. (8.6) for constant specific heats, the thermal efficiency
may be expressed by cycle temperatures as,
d  1 
where, T3 / T4  v4 / v3 
k 1
cv T4  T1 
c p T3  T2 
1
T1 T4 / T1  1
(8.11)
kT2 T3 / T2  1
 v1 / v2 v2 / v3 
k 1


 T2 / T1  1 / rck 1 . Further manipulation of
this result with ideal gas equation yields,
T4 T3 k 1
 rc
T1 T2
and
T3 v3
 rc
T2 v2
(8.12)
Substitution of Eq. (8.12) into Eq. (8.11) provides the Diesel cycle efficiency as a function of
compression ratio (r) and cutoff ratio (rc) as following,
1 rck  1 1
d  1 
k rc  1 r k 1
(8.13)
This equation is valid for constant specific heat of air and states that Diesel efficiency can be increased by greater compression ratio, r, and by smaller cutoff ratio, rc. In fact, for rc  1 , Diesel and
Otto efficiencies become identical. However, for rc  1 , the term, rck  1 / k rc  1, is always greater
than unity, and Diesel efficiency is always lower than that of Otto cycle having the same compression


ratio. For instance, if r  10 , and rc  2 , Otto and Diesel efficiencies respectively are o  60.2% ,
 d  53.4% . In practice, however, Diesel cycle always runs at higher compression ratios. Hence, for
r  20 and rc  2 , Diesel efficiency would be  d  64.7%  60.2% . Thus, because of higher compression ratios, Diesel cycle operates at higher efficiencies than Otto cycle.
Example 8.5 A CI engine operating on air-standard Diesel cycle has cylinder conditions at the start of compression process
as 60°C and 120 kPa. Light diesel fuel with a heating value, Hu=31,000 kJ/kg-fuel, is used at a fuel-air ratio of FA=1:20.
Compression ratio is 19, and combustion is ideal. For variable specific heats, calculate,
a.
the peak cycle temperature,
b.
the cut-off ratio,
c.
the heat lost to the environment,
d.
the thermal efficiency.
Solution:
a.
The mass of fuel used per kilogram of mixture is m f  1 / 20  1  0.0476 kg-fuel/kg. Then the amount of heat
input per kilogram of the mixture becomes qin  m f H u  0.0476  31000  1476.19 kJ/kg. For isentropic compression, v1 / v2  19  vr1 / vr 2 and by air tables, vr 2  478.5 / 19  25.18, the temperature and the enthalpy at the end
of compression are T2  1000 K , h2  1046.04 kJ/kg respectively. Referring to Figure 8.8b, since qin  h3  h2 ,
the enthalpy at the end of combustion is h3  1476.19  1046.04  2522.23 kJ/kg. The air tables yield the peak
temperature as T3  2224 K.
b.
Due to ideal gas behavior, the cut-off ratio may be expressed as, rc 
v3 T3 2224


 2.224 .
v2 T2 1000
CHAPTER 8 POWER PRODUCING SYSTEMS 339
c.
The amount of heat lost to the environment is qout  u4  u1 . Due to isentropic expansion,
v3 / v4  vr 3 / vr 4  v3 / v2 v2 / v4  or vr 4  1.94  19 / 2.22  16.6 , and by air tables, u4  881 kJ/kg, T4  1142 K.
Thus the heat lost is qout  881  237.7  643.3 kJ/kg.
d.
The thermal efficiency of the cycle by Eq. (8.6) is d  1 
8.3.1
643.3
 56.4%
1476.19
Dual cycle
Comparing the efficiencies given by Eqs. (8.9) and (8.13), we may state that higher compression
ratios with constant volume combustion would result with higher efficiencies. Thus, an ideal engine
would be compression ignition type but would operate on Otto cycle principle.
In modern CI engines, some portion of fuel is injected before the piston reaches TDC, and the
remaining injected right at TDC. Hence, a portion of combustion takes place at constant volume, and
the rest is completed by constant pressure process. Hence the pressure in the cylinder is kept high into
the expansion stroke. Because of completing the combustion by a dual process of constant volume
followed by constant pressure, the air-standard cycle for analyzing high speed modern CI engines is
called Dual cycle. Dual cycle p-V and T-s diagrams are shown in Figure 8.9. For the same reasoning
explained in Otto cycle, the processes 6-1 and 1-6 in Figure 8.9a cancel each other, and are left off
the figure. Hence, in Dual cycle, after an isentropic compression (1-2), heat addition at constant volume (2-3), and at constant pressure (3-4) are followed. Isentropic expansion (4-5) leads to a constant
volume heat rejection (5-1).
Thermal efficiency of Dual cycle. Referring to Figure 8.9a, the amount of heat transferred to the
working fluid (air) in both processes (2-3) and (3-4) is qin  q23  q34  u3  u2   h4  h3  . Due to
exhaust blow down, the heat rejected at constant volume is qout  u5  u1 . For constant specific heats,
Eq. (8.6) yields the thermal efficiency in terms of cycle temperatures as,
340
THERMODYNAMICS
 du  1 
cv T5  T1 
cv T3  T2   c p T4  T3 
1
T1 T5 / T1  1
T2 T3 / T2  1  kT3 T4 / T3  1
(8.14)
Introducing the following parameters: 1. Compression ratio, r  v1 / v2 , 2. Cut-off ratio,
rc  v4 / v3 , and 3. Pressure ratio, rp  p3 / p2 , and considering that T2  T1r k 1 , T3  T2 rp  T1r k 1rp ,
T4  T3 rc  T1r k 1rp rc , and T5  T4 rc / r 
k 1
 T1rp rck , Eq. (8.14) can be rearranged. Hence, the Dual
cycle thermal efficiency becomes,
 du  1 
rp rck  1
1

(8.15)

r k 1 krp rc  1  rp  1
It can be deduced from this relation that Dual cycle represents the Diesel for rp  1 , and the Otto
for rc  1 . However, for rp  1 , the term rp rck  1 /  krp rc  1  rp  1  assumes smaller values


and the efficiency increases. In fact, Dual cycle efficiency lies between that of the Otto cycle and the
Diesel having the same compression ratio.




Example 8.6 The CI engine of a small truck operating on air-standard Dual cycle with a compression ratio of r  20 has
cylinder conditions at the start of compression process as 50°C and 98 kPa. An ideal combustion of light diesel fuel with a
heating value of Hu=31,000 kJ/kg-fuel, is completed at a fuel-air ratio of FA=1:18. Due to structural limitations, however,
the maximum allowable pressure in the cylinder is 10 MPa. Assume constant specific heats at cv  0.788 kJ/kgK, and
c p  1.075 kJ/kgK, and determine the efficiency of the cycle.
Solution:
With respect to the given data of the problem, the pressure ratio, rp , and the cut-off ratio, rc , of the cycle have to be calculated. Since, p2  p1r k , then the pressure ratio is rp   p3 / p1 r  k  (10000 / 98)  201.364  1.715 . Referring to Figure
1.364 1
8.9a, the temperatures at state 2 and 3 respectively are T2  323  20 
 961.12 K, T3  1.715  961.12  1648.32
K. The portion of heat input for constant volume process becomes, qi1  cv T3  T2   0.788  1648.32  961.12   541.51
kJ/kg. Moreover, the fuel mass per kilogram of mixture is m f  1 / 19  0.0526 kg f / kg and the total heat input by the
combustion process would be qin  0.0526  31000  1631.57 kJ/kg. Then the portion of heat input at constant pressure process is qi 2  1631.57  541.51  1090.06 kJ/kg. Since the heat input at constant pressure may be calculated by
qi 2  c p T4  T3   c pT3 rc  1 , then the cut-off ratio becomes rc  1 
values into Eq. (8.15) results as, du  1 
8.3.2
1
200.364
1090.06
 1.615 .Substituting r , rp , rc
1.075  1648.32
1.715  1.6151.364  1
 50.3%.
1.364  1.615  1  1.715  1
Comparison of Otto, Diesel and Dual cycles
In this section, a comparison of these three cycles is made for the important factors of the cycle
that are compression ratio, peak pressure, heat rejection, and the net work output. Comparisons are
made for the following cases and for each case, the inlet conditions are assumed to be fixed.
CHAPTER 8 POWER PRODUCING SYSTEMS 341
Case 1: Same compression ratio and heat rejection. As shown in Figure 8.10a, the area defined
as A(a14b) represents qout and is the same for each type of cycle. However, due to same compression ratio, qin differs from cycle to cycle. In fact, on (T-s) diagram qin Diesel  A(a123''4b) ,
qin Dual  A(a1253'4b) , and qin Otto  A(a1234b) , and comparison of these area values shows that
qin Diesel  qin Dual  qin Otto . Under these conditions, together with Eq. (9.6), we may be state that
 d   du  o .
Case 2: Same compression ratio and heat addition. Since the heat inputs for all cycles have the same value,
from Figure 8.10b, it can be seen that area values are identical, A(a1234b)  A(a1253'4' c)  A(a123''4'' d ) .
In addition, the initial state and the compression ratio of all cycles are identical. Under these conditions,
the heat rejection of Otto cycle which corresponds to the area value of A(a14b) is the smallest in Figure
8.6b. In fact, the area comparison shows that qout Otto  qout Dual  qout Diesel . Consequently, the Otto
cycle has the highest work output, and with respect to Eq. (8.6), we may state that  d   du  o .
Figure 8.10 Efficiency comparison on the basis of same compression ratio
342
THERMODYNAMICS
Case 3: Same peak pressure, peak temperature and heat rejection. In applications, these three cycles
do not operate on the same compression ratio. CI engines operating on the Dual or Diesel cycle have
much higher compression ratio than SI engines operating on the Otto cycle. Hence, another case would
be to fix the peak pressure and the amount of heat rejection and then compare the cycles.
Referring to Figure 8.11, the heat input for each cycle is represented as qin Diesel  A(a12''34b)
qin Dual  A(a12'534b) and qin Otto
,
 A(a1234b) and comparison of these area values shows that
qin Diesel  qin Dual  qin Otto . With respect to Eq. (8.6), Diesel cycle becomes the most efficient
cycle and the efficiency distribution among cycles is as follows,  d   du  o . With respect to the
(T-s) behavior of these cycles as in Figure 8.11, we may state the most efficient engine as follows:
the most efficient engine would be a high compression ratio CI engine that completes the combustion
as close to constant volume process as possible.
8.4
Gas Turbine Engine
Gas turbine engines are the most reliable systems to meet the high power demand varying in the
range between 106 and 109 Watts. We already know that the relationship between the power produced
 net , in which m represents the mass flow
and the specific work of an engine is given as, Wnet  mw
rate of the working fluid. In reciprocating engines, the four steps of a cycle; intake, compression and
combustion, expansion, and exhaust, do occur in the same cylinder but at different times. Besides, the
mass flow rate of the working fluid, m , has very small values. For instance, 3L engine, in Example
8.3, runs at a speed of 3200 rpm and provides a typical mass flow rate of m  0.0928 kg/s for air which
has to be increased by at least twenty times for receiving power at mega-watt level. Increase in mass
flow rate, however, makes the system bulky and non-producible in size.
Due to high power-to-weight ratio, gas turbines are particularly suited for propulsion applications.
The absence of reciprocating and rubbing members make these systems to run at higher speeds with
less lubricating-oil consumption.
CHAPTER 8 POWER PRODUCING SYSTEMS 343
Figure 8.12 Basic principles of a gas turbine cycle and p-V representation
As shown in Figure 8.12a, the main difference of gas turbines from the reciprocating engines is
that the four steps of the cycle occur at the same time but in different sections. Hence the power is
produced in a continuous manner. Air is sucked into the compressor at atmospheric conditions and
is compressed to state 2 in Figure 8.12b. Heat is added to air by burning the injected fuel in the combustor ( qin ). The combustion is essentially a constant pressure process, but due to flow enlargement,
contraction and frictional effects in the combustor, pressure drop occurs. The combustion gases at high
pressure and temperature enters the gas turbine at state 3, and expands through the turbine. Because
of hot gases leaving the turbine at state 4, certain amount of heat is rejected to the surroundings ( qout
). To provide the above indicated cycle, a gas turbine engine consists of the following sections:
Inlet section. Clean and undisturbed inlet airflow extends the engine life by preventing erosion,
corrosion, and foreign object damage. The air inlet duct assembly is designed for this purpose and
provides clean and unrestricted airflow for the system. As shown in Figure 8.13a, the inlet duct section is produced as a separate item.
Compressor. With respect to the air flow path, the compression of air is accomplished by one
of two basic types of compressors which are named as axial flow and radial flow compressors. As
shown in Figure 8.13a, the axial flow compressor attains higher pressure ratios than the centrifugal
344
THERMODYNAMICS
one and provides more air flow rate for the same frontal area. Because of its simplicity, the centrifugal
compressors, shown in Figure 8.13b, are usually favored for smaller engines.
As illustrated in Figure 8.14, the compressor rotor blades convert mechanical energy into kinetic
energy of air. The stator vanes slow down the airflow by means of their divergent duct shape. Hence,
the kinetic energy is converted into enthalpy of air by increasing the air pressure at the exit of stator
vanes. The vanes are positioned at an angle such that the exiting air is directed into the rotor blades
of the next stage, and the process of increasing the pressure is repeated.
Figure 8.14 A compressor stage and an increase in enthalpy of air
The efficiency of a compressor is primarily determined by the smoothness of the flow. Losses due
to friction and turbulence will be minimized, if the air flows smoothly through the compressor.
Combustor. The combustion section has the task of controlling the burning of a large amount of
fuel. Thus the combustor has to release the heat in a manner that a stable stream of uniformly heated
gas is provided for all operating conditions of the turbine. As illustrated in Figure 8.15, various devices are equipped into the chamber for metering the air flow distribution and stabilizing the flame.
The air-fuel ratio of a combustion chamber may vary in the range of 45:1 to 120:1 which indicates
that the mass added to the working fluid by fuel injection is usually negligible. Actually the amount
of fuel injected into the air stream is governed by the temperature rise required. Since the material of
turbine blades and nozzles can stand temperatures in the range of 900°C to 1700°C, and since the air
entering the chamber has already been heated to a temperature between 200°C and 500°C by the work
done during the compression process, the temperature rise in the combustor is limited to a range of
700°C to 1200°C. In fact, knowing this limit of temperature rise helps us in determining the optimum
compression ratio for maximum cyclic efficiency.
Figure 8.15 Cross-sectional view and air flow distribution in a combustion chamber
CHAPTER 8 POWER PRODUCING SYSTEMS 345
Turbine. As shown in Figure 8.16, the turbine converts the gaseous energy into mechanical energy
by expanding the hot, high pressure gases to a lower temperature and pressure.
The number of stages employed in a turbine depends upon the power required, the rotational
speed, and the turbine diameter. As the air stream enters the turbine section from the combustor, it
is accelerated by the stationary vanes of the first stage. As illustrated in Figure 8.17, the stator vanes
also called nozzles form convergent ducts that converts the gaseous enthalpy into kinetic energy and
the flow is accelerated.
346
THERMODYNAMICS
In addition to acceleration, the stator vanes change the direction of the flow and direct it into the
rotor blades at an optimum angle. Then, the rotor blades extract the fluid kinetic energy. The velocity, the temperature and the pressure of the gas decrease as it leaves the stage. All the gas must flow
across the blades to achieve the maximum efficiency in the turbine. To ensure this, as illustrated in
Figure 8.16, the first three stages of the turbine blades have tip shrouds to minimize the gas leakage
around the blade tips. Besides, a smooth flow of gas through the blades and vanes affects the turbine
efficiency.
An ideal gas turbine engine would perform the processes that make up the Brayton cycle. As it is
represented on p-V and T-s diagrams in Figure 8.18, air standard Brayton cycle consists of the following processes: (1-2): Isentropic compression of air from lower pressure p1 to higher pressure p2. The
temperature of air increases but there is no heat flow. (2-3): Due to heat addition at constant pressure,
both the volume and temperature of air increase from values (V2, T2) respectively to (V3, T3). Since air
behaves like an ideal gas with constant specific heats, the amount of heat added is,
 p T3  T2 
qin  mc
(8.16)
(3-4): The air at high pressure and temperature expands isentropically from p2 to p1, and the
temperature falls from T3 to T4. (4-1): Heat is rejected from the working fluid by exhaust into the
atmosphere. Then, the volume and the temperature decrease from values of (V4, T4) respectively to
(V1, T1), but the pressure remains constant at p1. The amount of heat rejected is,
 p T4  T1 
qout  mc
Figure 8.18 p-V and T-s representation of Ideal Brayton cycle
(8.17)
CHAPTER 8 POWER PRODUCING SYSTEMS 347
Thermal efficiency of Brayton cycle. Starting with the basic definition of cyclic efficiency, the
thermal efficiency of air standard Brayton cycle may be expressed as following,
 1
T T / T  1
qout
T T
1 4 1 1 1 4 1
qin
T3  T2
T2 T3 / T2  1
(8.18)
Because of isentropic compression and expansion of air with constant specific heats, the following
k 1/ k
k 1/ k
relations hold for temperature ratios; T1 / T2   p1 / p2 
  p4 / p3 
 T4 / T3 . Furthermore,
let us represent the cycle pressure ratio as, rp   p2 / p1  , then the ideal Brayton cycle efficiency
becomes,
 1
T1
1
 1  ( k 1)/ k
T2
rp
(8.19)
It is evident from Figure 8.19 that increasing the compressor pressure ratio increases the thermal
efficiency. The pressure ratios used in gas turbine engines vary in the range between 5 and 20.
As illustrated in Figure 8.20, the actual Brayton cycle differs from the ideal counterpart by the
following two facts: 1. Compression and the expansion processes are not isentropic. With respect to
isentropic efficiencies defined by Eqs. (6.29) and (6.33), the actual work input to compressor is more,
and the actual work output of the turbine is less than the ideal counterparts. 2. A pressure drop in the
range of 2% to 5% of the compression ratio takes place during the combustion and also in the heat
rejection processes. This fact even causes more reduction in the turbine work output.
Figure 8.19 Efficiency of an ideal Brayton cycle
Figure 8.20 Comparison of ideal and actual
Brayton cycles
Definition: In a gas turbine engine the ratio of the work consumed by the compressor, wc , to the
turbine work, wt , is called back work ratio ( rbw ) and expressed as,
rbw 
wc
wt
(8.20)
348
THERMODYNAMICS
In simple gas turbine engines, consumption of more than 50-percent of turbine work by the
compressor is common, and the situation becomes even worse if the isentropic efficiencies of the
compressor and the turbine are low.
Example 8.7 The gas turbine engine in Figure 8.21 produces 15MW of power and has a compressor pressure ratio of 6.0.
The air temperature at the turbine inlet is 1000K, and the ambient conditions are 310K and 1bar. Determine the back work
ratio, the mass flow rate of air, and the thermal efficiency for
a.
air standard Brayton cycle,
b.
actual Brayton cycle having compressor and turbine efficiencies of 86% and 89% respectively. Due to fractional
pressure drop, a total of 5-percent of the pressure increase in the compressor is lost in the combustor and in the
exhaust sections.
Solution:
a.
Together with air tables, the reduced pressure at state 2 is pr 2  rp pr1  6  1.55  9.3 , and T2  515 K, h2  518.3
kJ/kg. Similarly, at state 3, the temperature, enthalpy and reduced pressure are T3  1000 K, h3  1046.04 kJ/
114
kg, pr 3  114 . Then, pr 4 
 19 and the corresponding temperature and enthalpy values are T4  625 K ,
6
h4  633.1 kJ/kg.
wc  518.3  310.24  208.06 kJ/kg, wt  1046.04  633.1  412.94 kJ/kg and rbw  208.06 / 412.94  0.503.The
net work is
wnet  wt  wc  204.88 kJ/kg, and the mass flow rate becomes m 
Wnet
 73.21 kg/s. The heat input rate at the
wnet
combustor is qin  h3  h2  1046.04  518.3  527.74 kJ/kg, and the thermal efficiency is  
b.
204.88
 38.8%.
527.74
The pressure drops in the combustor and in the exhaust system are assumed to be identical; pc  pex  0.025  600  15
kPa. Besides, the enthalpy and the temperature at state 2 respectively are h2 
518.3  310.24
 310.24  552.17 kJ/kg,
0.86
and T2  547.4 K . At state 3, the reduced pressure is the same, pr 3  114 , but the pressure ratio of the turbine is not identical with the compressor, p3  585 kPa, p4  115 kPa. The reduced pressure ratio of the
turbine is
pr 3 585

 5.087 , and pr 4  22.41 , h4 s  665.34 kJ/kg. Hence, the enthalpy at state 4 is
pr 4 115
CHAPTER 8 POWER PRODUCING SYSTEMS 349
h4  1046.04  0.89  (1046.04  665.34)  707.22 kJ/kg. From air tables, the temperature of gases at the exit becomes
T4  695 K . Therefore, the compressor work input, turbine work output, and the back work ratio respectively are
wc  552.17  310.24  241.93 kJ/kg, wt  1046.04  707.22  338.82 kJ/kg, rbw  241.93 / 338.82  0.714.Since
15000
the net work output is wnet  96.89 kJ/kg, the mass flow rate of the actual engine becomes m 
 154.81 kg/s.
96.89
Similarly, for qin  1046.04  552.17  493.87 kJ/kg, the thermal efficiency of the engine is  
8.5
96.89
 19.61%
493.87
.
Improving the Thermal Efficiency of Gas Turbine Engines
As studied in Example 8.7, the thermal efficiency of an actual gas turbine is very sensitive to
variations in compressor and turbine efficiencies. To end up with a higher thermal efficiency, the
isentropic efficiencies of both the turbine and the compressor must be as high as possible. With the
advent of computers, the computer aided design of these machines made possible to minimize the
exergy losses. Besides, the following parameters have a direct effect on the engine efficiency: 1. The
pressure ratio, rp, 2. The temperature at the turbine inlet, T3, 3. The turbine exit temperature, T4, 4.
The back work ratio, rbw, 5. The compressor inlet temperature, T1.
8.5.1
Compressor pressure ratio and turbine inlet temperature
For the same amount of fuel consumption, as the pressure ratio increases, the turbine inlet temperature also increases. Hence, these two parameters, the compressor pressure ratio, rp, and the turbine
inlet temperature, T3, are inter-related, and here we are going to study the combine effect.
Figure 8.22 Effect of pressure ratio (rpB>rpA) on thermal efficiency for fixed T3, and T1
Due to limited ability of the material for the turbine blades to withstand the high thermal and rotational stresses, there is an upper bound for the turbine inlet temperature. Figure 8.22 demonstrates that as
the pressure ratio increases (rpB>rpA) the amount of heat supplied and rejected both decrease for a fixed
turbine inlet temperature. However, the decrease in heat supply is less than the heat rejected, and this
causes an increase in the engine thermal efficiency. In fact, there is an optimum pressure ratio, rpo, for
which the thermal efficiency attains a maximum value at a specified turbine inlet temperature, T3.
Let us express the net work output of the engine with respect to pressure ratio as,
350
THERMODYNAMICS
k 1 / k
c pT1  rp   1




c
1 k / k 
wnet  wt  wc  c pT3t 1  rp

(8.21)
For the maximum value of wnet the derivative of Eq. (8.21) has to be zero, dwnet / drp  0 . Hence
for constant values of T1 and T3, taking the derivative of Eq. (8.21) with respect to rp and equating
to zero yields,

T 
rpo  ct 3 
T1 

k /2k 1
(8.22)
This relation indicates that the higher inlet temperature allows for increased pressure ratio and
improves the thermal efficiency. If we exceed the temperature limit for blade durability, then the
blades become brittle and take the shape shown in Figure 8.23a even for a short operational period
of the engine.
Figure 8.23 Turbine blade cooling
Increasing the gas stream temperature beyond the safe limit and keeping the blade surface temperature as low as possible can be accomplished by a blade cooling scheme. Figures 8.23b and 8.23c
illustrate the application of this scheme to the first stage blades of a gas turbine engine.
Example 8.8 Due to application of blade cooling scheme to the first stage of the gas turbine engine in Example 8.7, the
inlet temperature of gas stream, T3, is raised to 1904K. At ambient conditions of 1bar, 310K, determine the thermal efficiency
and the back work ratio for optimum pressure ratio. Assume constant specific heats of air, and take k=1.4.
Solution:
1.4/(20.4)
1904 

As given by Eq. (8.22), the optimum pressure ratio for maximum efficiency is rpo   0.86  0.89 
 15.0 .
310 

0.285
Then, by Eq. (8.19), the thermal efficiency becomes b  1  15
 0.5378 . Since the specific heats of air are assumed
to be constant, the back work ratio may be expressed as,
CHAPTER 8 POWER PRODUCING SYSTEMS 351
k 1/ k
rbbw 
wc  T1  rppo
 
wt  T3  ct
 33100  150.285

 0.46

 1904  0.86 0.89
With respect to these results, the engine efficiency and the back work ratio values are incomparably convenient than the
values at rp=6.0. However, the designer has to realize that the compressors and the turbines become more expensive with
increasing the pressure ratio.
8.5.2
Turbine Exit Temperature
One of the major parameters that results with low engine thermal efficiency is the high temperature of the exhaust gases leaving the turbine. As shown in Example 8.7, the temperature of exhaust
gases is often considerably higher than the temperature of air leaving the compressor. In such a case,
instead of throwing a large amount of heat into the atmosphere, some portion of that energy can be
saved by pre-heating the air entering the combustion chamber. Hence, this process of heat recovery
is called regeneration and the cycle is named as Regenerative Brayton Cyle. A regenerative Brayton
cycle is illustrated in Figure 8.24a, and the related T-s diagram is in Figure 8.24b. It is very clear
from this figure that less fuel will be consumed in the combustor for the same net work output. However, regeneration is only applicable when the gas temperature at the turbine outlet is higher than the
compressor exit temperature. Otherwise, the heat will flow in the reverse direction and resulting with
lower thermal efficiency.
Gas turbine regenerators are usually constructed as shell-and-tube heat exchangers with high
pressure air from the compressor flowing through the tubes, and the low pressure exhaust gas in the
shell side.
The effectiveness,  , of the regenerator measures how well the available temperature potential
of exhaust gases is used for heating the compressor discharged air and is defined as following,

q
qmax

h3  h2
h5  h2
(8.23)
where, q  m a h3  h2  . If we assume that the mass flow rates of compressed air and the exhaust
gas are approximately the same m a  m g , then the maximum temperature of air attainable at the
exchanger exit will be identical with the inlet temperature of the hot exhaust gas T3  T5  . Thus the


352
THERMODYNAMICS
maximum heat flow rate becomes qmax  m a h5  h2  . A greater amount of fuel will be saved by an
exchanger having a higher effectiveness. Higher effectiveness, however, requires a larger heat transfer
surface area that causes a larger pressure drop. The pressure drop especially on the high pressure side
of the regenerator is an important parameter and is usually kept below 2-percent of the compressor
discharge pressure. The effectiveness of the most regenerators in use is generally below 0.85.
Let us formulate the thermal efficiency of air standard Brayton cycle with regeneration for which
the effectiveness is taken to be unity (   1 ). For constant specific heats of air, the thermal efficiency
may be expressed as,
 T6 
  1
qo
T6  T1
T1  T1 
  1  1
 1
qi
T4  T3
T4  T3 
1  
 T4 
(8.24)
Since   1 , we realize that T6  T2 , and T5  T3 . In addition, the temperature ratios in Eq. (8.24)
k 1
k
and
p
may be expressed as, T6 / T1  T2 / T1  r
1 k
k
p
as, T3 / T4  T5 / T4  r
. Substituting these tem-
perature ratios into Eq. (8.24) and rearranging yields,
T1 kk1
  1  rp
T4
(8.25)
As shown in Figure 8.24b, the term T1 / T4 indicates the temperature ratio of the cyclic minimum
to the maximum temperature. This parameter has to be as low as possible for high thermal efficiency.
Similarly, as verified by Figure 8.25, regeneration becomes more effective at lower pressure ratios.
Figure 8.25 Effect of Tmin/Tmax on thermal efficiency of air standard regenerative
gas turbine engine for   1
Example 8.9 Consider the regenerative gas turbine engine in Figure 8.24a that has a compressor pressure ratio of 5.0,
and the isentropic efficiencies of compressor and the turbine respectively are 0.85, and 0.9. The regenerator effectiveness
is 0.72. For ambient temperature at 300K, the turbine inlet temperature is 1300K. Determine,
a.
the back work ratio and the net work output,
b.
the gas temperatures at the turbine outlet, and at the regenerator exit,
CHAPTER 8 POWER PRODUCING SYSTEMS 353
c.
the thermal efficiency of the engine.
d.
compare the efficiency with the efficiency of the corresponding engine without regenerator.
Solution:
a.
Using the air tables (A24), h1  300.19 kJ/kg, pr1  1.386 , and for pr 2 / pr1  5.0 , pr 2  6.93 , the isentropic enthalpy at state 2 is h2 s  477.24 kJ/kg. Through the compressor efficiency, the enthalpy at state 2 is
477.24  300.19
 300.19  508.48 kJ/kg. Similarly, at state 4, the enthalpy is h4  1395.97 , pr 4  330.9 and for
0.85
pr 4 / pr 5  5 , pr 5  66.18 , the isentropic enthalpy at state 5 is h5 s  888.27 kJ/kg. By using the turbine efficiency, the
h2 
enthalpy at state 5 may be calculated as h5  1395.97  0.9  1395.97  888.27   939.04 kJ/kg. Considering the definition
of regenerator effectiveness (Eq. (8.23)), the enthalpy at state 3 is h3  508.48  0.72  939.04  508.48   818.48 kJ/
kg. Hence, the turbine work output and the compressor work input respectively are wt  1395.97  939.04  456.93
kJ/kg, wc  508.48  300.19  208.29 kJ/kg. The back work ratio becomes, rbw  208.29 / 456.93  0.455 , and the
net work is wnet  wt  wc  248.64 kJ/kg.
b.
By air tables, the gas temperature at the turbine outlet is T5  906.2 °C. For the same mass flow rate at both sides of
the regenerator, the energy balance requires that h3  h2  h5  h6 , or 818.48  508.48  939.04  h6 and h6  626.04
kj/kg, T6  618 K.
c.
248.64
64 / 577
577.49
49
qin  1395.97  818.48  577.49 kj/kg, and   248
d.
For the similar engine with no regenerator, qin  h4  h2  1395.97  508.48  887.49 kj/kg, and qout  h5  h1  638.85
43.05% .
638.85
 28.01% .
887.49
This result shows that the effect of regenerator on the engine efficiency is predominantly strong for gas turbine engines
operating at low compression ratios.
kj/kg. Then the engine efficiency is   1 
8.5.3
Back work ratio
The back work ratio of a gas turbine engine has to be decreased as much as possible for increasing the engine thermal efficiency. In accord with Eq. (8.20), the back work ratio can be reduced by
decreasing the work consumed by the compressor, and increasing the work produced by the turbine.
The work required to compress a gas between two specified pressures can be decreased by carrying
out the compression in stages and cooling the gas in between the stages. Similarly, the work output
of a turbine operating between two pressure levels can be increased by expanding the gas in stages
and reheating it in between the stages.
Principle 29: The steady-flow compression or expansion work is proportional to
2
the specific volume of the fluid,
wrev    vdp . Consequently, the specific volume
1
of the working fluid should be as low as possible for a compression process, and as
high as possible for an expansion process through a turbine.
Working fluid is usually cooled down to its inlet temperature through the inter cooler of the compressor. Besides, due to avoiding excessive temperatures, the amount of air drawn into the combustor
is four times larger than the amount needed for a stoichiometric combustion. As shown in Figure 8.26a,
this excess oxygen may be used for burning of additional fuel sprayed into the reheat combustor.
Due to application of inter-cooling and reheating, the working fluid leaves the compressor at a
lower temperature and the turbine at a higher temperature. This makes regeneration more attractive
in these systems.
354
THERMODYNAMICS
Figure 8.26 Gas turbine engine with intercooling, reheating and regeneration
Therefore, in addition to inter-cooling and reheating, these systems also have regenerator, and
as in Figure 8.26a, air leaving the compressor may be heated to a higher temperature before entering
the combustion chamber.
Figure 8.26b is the T-s diagram of an ideal Brayton cycle with inter-cooling and reheating. The
regenerator is also assumed to be an ideal heat exchanger with   1 . The air enters the first stage of
the compressor (state 1), and is compressed isentropically to an intermediate pressure (state 2), and
is then cooled to inlet temperature by an inter-cooler (state 3). In the second stage, it is compressed
to the final pressure (state 4). Air enters the regenerator where it is heated to the turbine exhaust
temperature for   1 and at constant pressure (state 5). After the combustor (state 6), the gas enters
and expands isentropic ally through the first stage of the turbine (state 7). The gas is then reheated at
constant pressure where it enters the second stage of the turbine (state 8). At the turbine exit (state
9), the gas enters the regenerator and is cooled at constant pressure (state 10). The cycle is completed
by purging the exhaust gases into the atmosphere (state1).
As shown in Figure 8.27b, for two-stage compression and expansion systems, air is usually cooled
to the inlet temperature of the first compressor (T3=T1) and the gas is reheated to the inlet temperature
of the first stage (T6=T8). In addition, let us assume that the efficiency of the compressors respectively
are c1 , and c 2 . Then the total work consumed by the compressors is,
k 1
 

1
 1  p2  k
wc  c pT1 
 1 



 c 2
c1  p1 

 
k 1


k


p

 4
 1 


 p2 


 
(8.26)
Since the overall compression ratio is defined as, rp  p4 / p1  rp1rp 2 , then the above relation may
be expressed in terms of rp1 as following,
k 1



k 1
k


r

1
 1 


p
wc  c pT1   rp1  k  1 
   1 


 c 2   rp1 

 c1 



(8.27)
CHAPTER 8 POWER PRODUCING SYSTEMS 355
For minimum work consumption, the derivative of Eq. (8.27) with respect to pressure ratio rp1
has to be zero. After mathematical manipulations, the optimal pressure ratios may be formulated as,
k
   2 k 1
rpo1   c1 
rp
 c 2 
k
and
rpo 2
   2 k 1
rp
  c2 
 c1 
(8.28)
In accord with this result, the compressor with higher efficiency should have the higher pressure
ratio. For a special case of identical efficiencies, both compressor stages share the same pressure ratio,
rpo1  rpo 2 and the intermediate pressure p2 becomes the geometric average of the initial and the final
pressures as following,
rpo1  rpo 2  rp
p2  p3 
and
p1 p4
(8.29)
Referring to the basic definition of cyclic efficiency, the thermal efficiency of this system may
be determined as following,

w
q
net
 1
in
q
q
out
in
 1
qout1  qout 2
qin1  qin 2
(8.30)
Example 8.10 A two-shaft gas turbine engine in Figure 8.27 produces 6 MW of power and operates with two compressor
stages having an overall pressure ratio of 16:1. Air inlets each compressor at 300 K, and the high pressure turbine derives the
compressors with an isentropic efficiency of 80-percent for each one. With an isentropic efficiency of 86-percent for each
turbine, the high pressure turbine derives the compressors and the low pressure turbine generates the power. The exhaust
gases at the exit of low pressure turbine flow through a regenerator having an effectiveness of 0.75.
Neglect the mass of fuel, and all pressure losses, and assume that c pa  1.005 kJ/kgK, ka  1.4 for air, and that c pg  1.15
kJ/kgK, k g  1.33 for combustion gases. Determine,
a.
the temperature of the gas at the turbine inlet,
b.
the back work ratio,
c.
the mass flow rate of air,
d.
the thermal efficiency of the cycle.
Figure 8.27 A two-shaft gas turbine engine with intercooling, reheating and regeneration
356
THERMODYNAMICS
Solution:
a.
The pressure ratio for each compressor is 4 ( rp1  rp 2  4 ) and the temperature ratio for isentropic compression is
T T / T  1
T2 s
 (4)0.285  1.4845 . Then the temperature at the compressor exit is T2  T1  1 2 s 1
 481.68 K  T4 .
T1
c
The total work consumed by the compressors is wc  2  1.005  481.68  300   365.18 kJ/kg. This work has to
 T
be supplied by the high pressure turbine and calculated as, wt1  t c pg T6 1  7 s
T6

0.248

T7 s  1 
 
 0.709 ,
 where
T6  4 

and substitution yields the gas temperature at the turbine inlet as, 365.18  1.15  0.86  T6  1  0.709  or
T6  1268.9 K. The gas temperature at the turbine exit may be calculated by the turbine isentropic efficiency as,
T7  T6  t T6  T7 s   951.34  T9 .
b.
Since the turbines are identical, each one produces the same work, wt1  wt 2  365.18 kJ/kg. Hence the back work
ratio becomes rbw 
365.18
 0.5
2 x365.18
Wnet
6000

 16.43 kg/s.
wnet 365.18
c.
The mass flow rate for the required power production is m 
d.
To find out the heat input at the first combustion chamber (CC1), the temperature at state 5 has to be calculated by the effectiveness of the exchanger, T5  T4   T9  T4   833.92 K. Hence, the heat input at CC1 is
qin1  h6  h5  1.15  1268.9  1.005  833.92  621.15 kJ/kg. Similarly, for the reheat combustor (CC2),
qin 2  h8  h7  1.15  1268.9  951.34   365.19 kJ/kg. Thus the thermal efficiency is

365.18
 37%
621.15 365.19
In the above example, a case for two-stage compression and two-stage expansion has been studied. If more and more
stages are added to the system, the thermal efficiency will increase and approach to Carnot efficiency. However, the contribution of each additional stage to thermal efficiency is not identical, and in fact, becomes less and less. Therefore, the use
of more than two or three stages may not be economically justified.
8.5.4
Temperature of Air at the Compressor Inlet
In case of engine running at
constant speed, the compressor
pumps a constant volume of air
into the engine. Due to different
air density under different atmospheric temperatures, mass flow
rate changes, and affects the power
generated. On a cold day, the density of air is high, and more mass
enters the compressor, and more
power is developed. In fact, as
shown in Figure 8.28, the thermal
efficiency of a simple gas turbine
engine assumes higher values at
T1 / T3  0.25 . On a hot day, however, the air density would be less
and a decrease in output shaft power
will take place.
a
Figure 8.28 Effect inlet temperature ratio on thermal
efficiency of a simple gas turbine engine for c  t  0.85 .
CHAPTER 8 POWER PRODUCING SYSTEMS 357
Let us assume that the temperature at the turbine inlet is fixed at T3  1000 K, referring to Figure
8.28, increasing the atmospheric temperature from 250K to 310K causes a drop of 21.8-percent in the
thermal efficiency of a simple gas turbine engine working at pressure ratio of rp  8.0 .
Several methods like inlet fogging, or injection of water droplets are employed to make the gas
turbine engines insensitive to the variation of the ambient temperature.
8.6
The Jet Engine
The jet engine is a gas turbine engine used in aircraft. The gas turbine of a jet engine produces
high pressure hot gases but has zero net work output. As shown in Figure 8.29a, a nozzle attached to
the turbine exit converts the thermal energy of the hot and high pressure gas into a high kinetic energy
exhaust stream that produces a forward thrust on the engine.
Figure 8.29 The jet engine and T-s representation
Since the high power, light weight, and small volume are the most inevitable design parameters of
these engines, the configuration and the design of jet engines differ significantly from those stationary
gas turbines. To start with, we have not considered the effect of kinetic energy change in stationary
gas turbine analysis. In jet engines, however, due to considerable change in gas speeds, the kinetic
energy term cannot be neglected. Therefore, the analysis can be done in terms of stagnation (total)
temperature or stagnation (total) pressure where kinetic energy is taking into consideration.
The stagnation conditions, as given by Eq. (6.25), are obtained by decelerating the flow isentropically to zero velocity. Hence, at the engine inlet, the stagnation (total) values of temperature and
pressure are,
358
THERMODYNAMICS
T00  T0  V02 / 2c p
and
k / k 1
p00  p0 T00 / T0 
(8.31)
For an adiabatic inlet duct, the stagnation enthalpy will be constant, h00  h01 . However, due to
effect of friction, and turbulence, a drop of 5 to 10-percent in total pressure may take place between
states (0) and (1). For an isentropic flow of air through the inlet section, we may simply state that
T00  T01 , p00  p01 .
For jet engines, the compressor pressure ratio is expressed as the ratio of total pressures as following,
rc  p02 / p01
(8.32)
Then, the isentropic and actual stagnation temperatures at the compressor exit are,
k 1
T02 s  T01rc k
and
T02  T01 
T02 s  T01 
(8.33)
c
The pressure losses in the combustion chamber may be neglected and the combustion is assumed
to take place at constant total pressure, p03  p02 . Neglecting the mass of fuel injected into the combustor, and considering the fact that the power produced by the turbine is absorbed by the compressor,
the following relation holds,
c pa T02  T01   c pg T03 1  T04 / T03 
(8.34)
Similar to stationary gas turbine engines, the durability limit of blade material determines the
stagnation temperature at the turbine inlet, T03 . Hence, for a predetermined value of T03 , the isentropic
discharge temperature, T04s , and the stagnation pressure, p04 , at the turbine exit may be calculated as,
T04 s  T03 
T03  T04 
t
and
k / k 1
rt  p03 / p04  T03 / T04 s 
(8.35)
If we compare Eqs. (8.32) and (8.35), we may state that the total pressure ratio of the turbine is
significantly lower than the compressor pressure ratio ( rt  rc ).
In terms of flight conditions, the nozzle flow is mostly at chocked flow conditions. Then, for
convergent nozzles as explained in Chapter 6, the flow rate attains maximum at this condition, and
the Mach number is unity at the throat. Since, the throat also represents the nozzle exit, by Eq. (6.26),
the temperature at the nozzle exit becomes,
T04  T05  T5  V52 / 2c p  T5 k  1 / 2
(8.36)
Similarly, the pressure of exhaust gases at the nozzle exit is,
k / k 1
p5 / p04  T5 / T04 
(8.37)
Example 8.11 An aircraft flies at a speed of 200m/s at an altitude for which the atmospheric temperature and pressure
respectively are 260K, and 60 kPa and uses a gas turbine engine schematically shown in Figure 8.29b for producing trust.
Assume that no losses occur at the engine intake, and the mass flow rate of air through the engine is 80kg/s. The engine
operates at a compressor pressure ratio of 8, with the turbine inlet temperature at 1200K. The compressor and turbine efficiencies are 0.9 and 0.88 respectively. At the turbine exit, combustion gases flow through an isentropic and convergent
nozzle. Take k  1.4 , and c p  1.005 kJ/kgK throughout the engine, and determine,
CHAPTER 8 POWER PRODUCING SYSTEMS 359
a.
b.
c.
d.
the turbine expansion ratio,
the velocity at the nozzle exit,
the nozzle exit area.
check if the nozzle operates under chocked conditions.
Solution:
a.
40000


The stagnation temperature and pressure at the intake duct are T00  260  1 
 279.9 K,
2  1005  260 

p00  60  279.9 / 260 
3.5
 77.67 kPa. Since there is no losses through the intake duct, then the total pressure is
T

kept constant, p00  p01 , and T00  T01 . The work consumed by the compressor is wc  c pT01  02  1 where
T
 01

T / T  1 , and T02s  8 0.285  1.808 . Substituting backwards yields, T02  1.897 , T  530.97
T02
 1  02 s 01

02
T01
T01
c
T01
K, wc  252.32 kJ/kg, p02  8 x77.67  621.36 kPa, and T03  1200 K.
By Eq. (8.34),
1200  1  0.79 
T04
252.32
 913.6 K.
1
 0.79 , T04  948 K, and by Eq. (8.35), T04 s  1200 
0.88
T03
1.005  1200
 1200 
With respect to isentropic stagnation temperature at state 4, the turbine expansion ratio becomes, rt  

 913.6 
b.
3.5
 3.896
.
For chocked flow conditions, by Eq. (8.36), the temperature at the nozzle exit is T5  948  2 / 1  1.4   790 K,
and the gas velocity at the exit becomes, V5  2  1005  948  790   563.5 m/s.
c.
Since p03  p02  621.36 kPa, and rt  p03 / p04 then p04  621.36 / 3.9  159.32 kPa. By Eq. (8.37), the
pressure at the nozzle exit is p5  159.32  790 / 948 
3.5
 84.17 kPa. The specific volume at state 5 is
0.287  790
 2.693 m3/kg. Considering the mass flow rate of air and the continuity equation, the nozzle exit
84.165

mv
80  2.693
area becomes A5  5 
 0.382 m2.
563.5
V5
v5 
d.
p04 159.32

 2.655 . However, as calculated previously, the
p0
60
p04
p
p
nozzle critical pressure ratio,
 1.892 , is smaller than the applied pressure ratio, 04  04 and the nozzle
p5
p5
p0
The applied pressure ratio for the present nozzle is
flow has to be at chocked conditions.
8.7
Stirling Engine
360
THERMODYNAMICS
If the combustion process of a cycle is not part of that cycle, the composition of the working fluid
will not be altered throughout the cycle, and the engine working with this principle is called the external combustion engine. The Stirling engine is a typical Example for external combustion engines.
Due to complexity of its construction for performing the cycle processes and high manufacturing cost,
however, Stirling engines are generally used for small scale power needs varying in the range 0.5 kW
to 5 kW. Figure 8.30 illustrates some of the applications of Stirling engine.
As shown in Figure 8.31a, Stirling engine generally uses air as the working fluid, but helium is
also used at some particular applications. The cycle is composed of two isothermal and two isochoric
processes. There are two pistons in two different cylinders A and B. As in Figure 8.31a, one of the
pistons is called power piston (P), and the other is displaced piston (D). At state 1 in Figure 8.31b, the
displacer D is at BDC and piston P is at TDC, and the air in cylinder A is at temperature T1.
Isothermal compression process. (1-2) As the crankshaft rotates clockwise direction between 1
and 2, D moves from BDC to TDC compressing the air isothermally at temperature T1, and heat is
rejected ( qout ).
Isochoric heating process. (2-3) As piston P continues moving downward, the cold fluid in cylinder
A is pushed through the regenerator and is heated to temperature T3. Since the sum of volumes above
the piston P and D in both cylinders is constant, the heating process takes place at constant volume.
Isothermal expansion process. (3-4) Air in cylinder B is at temperature T3, and both pistons move
downward as the crankshaft rotates from 3 to 4. Heat is added by high temperature reservoir at T3 to
keep the temperature constant ( qin ) at T3=T4.
Isochoric cooling process. (4-1) Displacer D continues moving downward, and the piston P moves
upward. Hence, air flows from hot cylinder B to cylinder A by passing through the regenerator, and is
cooled down to temperature T1. Again this cooling process takes place at constant volume.
Figure 8.31 Schematic of Stirling engine and p-V, T-s representation
CHAPTER 8 POWER PRODUCING SYSTEMS 361
For an ideal regenerator, it can be shown that the heat absorbed by air from the regenerator during
process (2-3) is identical to the heat given by air to the regenerator at process (4-1). Thus, the exchange
of heat with external sources takes place during isothermal processes.
The thermal efficiency of the Stirling engine is determined by considering the fundamental efficiency relation as,   1  qout / qin . For an ideal cycle with an ideal regenerator, qin  T3 s4  s3  and
qout  T1 s1  s2  . To calculate the entropy change, we may use Eq. (5.30) as, s4  s3  R ln v4 / v3 
and s1  s2  R ln v1 / v2  . As indicated in Figure 8.31b, v1  v4 and v2  v3 , then the volume ratios
become identical as, v1 / v2  v4 / v3 . Hence, the entropy change of both isothermal processes is identical, s1  s2   s4  s3  , and the preceding relations yield,
 1
T1
T3
(8.38)
The ideal Stirling cycle has the same efficiency as the Carnot cycle. However, to accomplish isothermal compression and expansion process in a machine running at a certain speed is considerably
hard and is almost impossible.
Example 8.12 Ideal Stirling engine, as shown in Figure 8.31, operates with 0.1kg of air as a working fluid between temperatures of 10000C and 300C. The highest power piston pressure is p3  3000 kPa, and the lowest pressure of the displacer
piston is p1  500 kPa. Determine,
a.
the power piston minimum pressure, p4
b.
the displacer piston maximum pressure, p2 .
c.
Show that the heat supplied and rejected through the regenerator during isochoric compression and expansion
processes are identical.
d.
Calculate the heat to be supplied and rejected by the cycle in kJ/cycle,
e.
the net work per cycle and the thermal efficiency of the engine.
Solution:
a.
Since the process (4-1) is a constant volume process, applying the ideal gas equation results as,
or p4  500 
b.
c.
1273
 2100.66 kPa.
303
Similarly, for displacer piston, we consider the process (2-3), and apply the gas equation as,
p2  3000 
p4V4 p1V1
,

T4
T1
p3V3 p2V2
, then

T3
T2
303
 714.1 kPa
1273
To raise the temperature of working fluid from T2 to T3 during process (2-3), the amount of heat supplied by the
regenerator is
Q23  U 3  U 2  mcvm T3  T2   0.1  0.8  970   77.6 kJ/cycle. Similarly, the amount of heat rejected during
the process (4-1) through the regenerator is Q41  U1  U 4  mcvm T1  T4   0.1  0.8  970   77.6 kJ/cycle.
The specific heat of air is determined at the arithmetic mean temperature, and the results show that the amount of
heat rejected by hot air is totally absorbed by cold air at the regenerator.
362
THERMODYNAMICS
d.
Since the process (3-4) is isothermal, the work done and the heat supplied are identical for an ideal gas,
W34  Q34  Qin  p3V3 ln  p3 / p4   0.1  0.287  1273  ln 3000 / 2100.66   13.02 kJ/cycle. Similarly, the heat
rejected at low temperature reservoir is, W12  Q12  Qout  p1V1 ln  p1 / p2   0.1  0.287  303  ln 500 / 714.1  3.1
kJ/cycle.
e.
The net work obtained at each cycle is calculated as, Wnet  W12  W34  3.1  13.02   9.92 kJ/cycle. Since
the thermal efficiency of the engine is defined as,  
into efficiency relation yields,  
8.8
Wnet
, substitution of numerical values of Wnet and Qin
Qin
9.92
 76.2% . The same result could also be obtained by Eq. (8.38).
13.02
A Simple Rankine-Cycle Power Plant
Today, much of the electricity consumed in the United States is produced by steam power plants.
Closely related to the essential parameters of plant establishment criteria, these plants are powered
by various fuels. Depending upon the type of fuel consumed, however, a steam power plant may be
classified as coal plant (Figure 8.32), natural gas plant, or as nuclear plant (Figure 8.33). All of these
plants, however, operate under the same basic cycle called Rankine cycle. If there is no any particular
reason for alternative fluids, water is the working fluid of the cycle and is nontoxic and nonreactive
fluid. Besides, due to its low cost, availability, and high energy absorption properties, the use of water
in electricity production will continue for many years and provide the necessary energy for the United
States and for the world economies. The most prominent feature of a simple Rankine cycle is that it
complies with the largest power demand on the power scale. In fact, it is the only cycle which may
produce power at Giga-Watt level by a single unit. As shown in Figure 8.32a, a simple Rankine cycle
consists of the following four components: 1.Boiler. The combustion of fossil fuel (coal or natural gas)
produces hot combustion gases that transfer heat to water passing through boiler tubes. As shown in
Figure 8.34, first, the temperature of feed water is increased to saturation temperature, then evaporated
to form saturated vapor. Mostly, further rising its temperature, the feed water becomes superheated
at the boiler exit. If we neglect the pressure losses, the process (4-1) or (4-1´) in Figures 8.32b, c, and
d represents the constant pressure heat addition in the boiler and calculated as,
qin  h1  h4
(8.39)
The rate of fuel consumed and the heat transferred to steam may be related through the boiler
efficiency as,
Q in  m f H uboiler  m s qin  m s h1  h4 
(8.40)
where, H u kJ/kg-fuel, is the fuel heating value, and ηboiler, describes the portion of the fuel energy
transferred to water and is defined as,
boiler 
m s h1  h4 
m f H u
(8.41)
CHAPTER 8 POWER PRODUCING SYSTEMS 363
364
THERMODYNAMICS
2. Turbine. High pressure and temperature steam at the boiler exit flows to steam turbine where part of
its energy is converted to shaft work and transmitted to an electrical generator. Unless stated, turbine
is essentially an adiabatic device and the work is represented by process (1-2) or (1´-2´) in Figures
8.32b, c, and d and calculated as,
wt  h1  h2  t h1  h2 s 
(8.42)
3. Condenser. The low energy steam flowing out of turbine condenses and becomes saturated liquid
at the condenser exit. The heat rejection process in the condenser is isobaric, and is represented by
process (2-3) or (2´-3) in Figures 8.32b, c, and d. The specific energy rejected at the condenser is,
qout  h2  h3
(8.43)
As shown in Figure 8.32a, the heat rejected in the condenser is transferred to a separate cooling
water loop which in turn delivers this energy to atmosphere by cooling towers. The mass flow rate of
the cooling water is calculated by the following energy balance equation,
m w c pw T  m s h2  h3 
(8.44)
4. Feed water pump. The low pressure of condensate is raised to boiler pressure by a feed water
pump. The process is indicated by line (3-4) in Figures 8.32b, c, and d. Since the working fluid is
at liquid state at the pump inlet, the required work is incomparably smaller than the work needed to
transport vapor at the same pressure differential. Referring to Eq. (6.20), the adiabatic pump work
may be calculated as,
CHAPTER 8 POWER PRODUCING SYSTEMS 365
w p  h4  h3 
h4 s  h3 v  p4  p3 

p
p
(8.45)
The ideal Rankine cycle is the one with the isentropic efficiencies of unity for both the turbine and
the pump. As shown in Figure 8.35a, the ideal Rankine cycle is depicted with area (123461) and is not
Carnot cycle. In fact, Carnot cycle is not a realistic model for Rankine cycle. As defined by area (12561)
in Figure 9.35a, to complete Carnot cycle, the liquid –vapor mixture at state (5) has be compressed
isentropically to saturated liquid state. First, it is almost impossible to device a system which provides
an appropriate quality so that after the compression process water is at saturated liquid state. Secondly,
there is no such a pump that compresses two-phase flow. As an alternative system, if we run the Carnot
cycle as shown in Figure 8.35b, the problem of pumping two-phase flow is eliminated. However, another problem of transferring heat isothermally at decreasing pressure between states (6) and (1) arises.
Moreover, the process (6-1) takes place above the critical point and is unpredictable at what phase the
water will be during this process. Hence, Carnot is not an ideal cycle for Rankine cycle.
Figure 8.35 Unrealistic features of Carnot and comparison with ideal Rankine cycle
Similar to the other cycles as explained in this Chapter, Eq. (8.6) may be used for calculating
the energy based thermal efficiency of Rankine cycle. In determining the overall efficiency of the
plant, however, fuel heat energy released by combustion has to be compared with the electrical
power at the generator outlet as following,
o 
W g
W g Wnet Q in



  g Rb
Q fuel Wnet Q in Q fuel
(8.46)
Example 8.13 As shown in Figure 8.36, a coal fired steam power plant running with a simple Rankine cycle produces
250MW of power at the generator outlet and provides the following data at the specified states.
State no.
Location
Pressure (bar)
Temp.(0C)
Quality(%)
Velocity(m/s)
Enthalpy(kJ/kg)
1
Turbine inlet
50
400
-
-
3195.7
2
Turbine exit
0.25
-
0.9
250
2383.6
3
Condenser exit
0.20
-
0.0
-
251.4
4
Pump exit
60
-
-
-
-
5
Boiler exit
55
450
-
-
3309
366
THERMODYNAMICS
The heating value of coal used at the plant is
H u  29800 kJ/kg-coal. The boiler and the generator runs
with b  78% and  g  95% efficiencies respectively. The
isentropic efficiency of the pump is  p  70% . Determine,
a.
the mass flow rate of water circulating through the
cycle,
b.
the back work ratio of the cycle,
c.
the rate of heat loss between the boiler exit and turbine
inlet,
d.
the thermal efficiency of the cycle,
e.
the overall efficiency of the system,
f.
the coal consumption rate of the plant,
g.
the condenser capacity and the cooling water mass
flow rate for a temperature change of 100C.
Solution:
a.


V2 
2502 
The turbine specific work is, wt  h1   h2  2   3195.7   2383.6 
  780.85 kJ/kg. Referring to the isen

2 
2000 


tropic efficiency of the pump, the pump specific work is w p  h4  h3 
v  p4  p3 
p

0.001  6000  20 
0.7
 8.54
kJ/kg, and the enthalpy at state 4 becomes, h4  251.4  8.54  259.94 kJ/kg. Hence, wnet  wt  w p  772.31 kJ/
kg, the net power at the turbine outlet is Wnet 
flow rate is m s 
W g
g

250000
 263157.89 kW. The corresponding steam mass
0.95
Wnet 263157.89

 340.74 kg/s
wnet
772.31
wp
The back work ratio is rbw 
c.
The heat loss through the pipe line connecting the boiler and the turbine is
Q
m h h   340.74 3309 3195.7  38605.8 kW.
loss
d.
s
5
wt

8.54
 1.09% and is much lower than the values for gas turbine engines.
772.31
b.
1
The boiler capacity is Q in  340.74 3309  259.94   1038936.7 kW. Then the thermal efficiency of the cycle is
R 
Wnnet 263157.89

 25.3% . Without any heat loss through the connection pipes, the heat input in the boiler
Q iin
1038136.7
263157.89
is Q in  340.74 3195.7  259.94   1000330.86 kW and the cyclic efficiency becomes  R 
 26.3% .
1000330.86
e.
Referring to Eq. (9.45), the overall efficiency of the system is o 
kW. Hence, o 
f.
250000
 18.77% .
1331970.13
The coal consumption rate is calculated by m f 
hour.
W g
Q
1038936.7
, where Q fuel  in 
 1331970.13

Q fuel
b
0.78
Q fuel
Hu

1331970.13
 44.69 kg/s or 160.9 tons of coal per
29800
CHAPTER 8 POWER PRODUCING SYSTEMS 367
g.
The condenser heat capacity is Q con  m s h2  h3   340.74  2383.6  251.4   726525.8 kW. The energy balance
at the condenser yields the mass flow rate of cooling water as, m w 
tons of water per hour.
Q con
726525.8

 17381 kg/s or 62571.6
4.18 x10
c p T
8.9 Improving the Thermal Efficiency of Rankine Cycle
In regard to forecasts in energy needs, the worldwide demand for power will increase significantly
over the next two decades, and a considerable portion of this demand will be covered by power plants.
To save primary energy resources by reducing the fuel consumption, the thermal efficiency of these
plants has to be increased. Rankine cycle can be considered as operating between two fixed pressure
levels, the boiler pressure and the pressure in the condenser. The turbine provides a controlled pressure
drop between these two pressure limits, and the pump increases the pressure back to its boiler pressure.
However, unlike gas turbine engines, the back work ratio in Rankine cycle has so small value that it is
not considered as a parameter for affecting the thermal efficiency. The following parameters, however,
have direct effect on the plant efficiency: 1. The condenser pressure, p2 2. The boiler pressure, p1 ,
3. The temperature at the turbine inlet, T1 , 4. The rate of heat rejected at the condenser, Q cond .
As shown in Figure 8.37a, let us consider an ideal Rankine cycle for which the boiler pressure is kept
constant but the condenser pressure is reduced. Comparison of these two cycles indicates that the net work
of the cycle with the lower condenser pressure has increased by the blue area (22´3´4´432), and the heat
rejected has decreased to orange area indicated by (2´66´3´2´). Hence, with respect to Eq. (8.6), the efficiency
of the system has to be increased. In other words, lower the condenser pressure, better the thermal efficiency
of the plant. However, there is a limit to the condenser pressure. To condense the steam flowing from the
turbine outlet, the heat has to be transferred from steam to cooling water. Thus, the saturation temperature
inside the condenser has to be greater than the atmospheric temperature. Then, for a maximum of 30°C of
atmospheric temperature, the condenser pressure has to have a value between 5 kPa and 10 kPa. As the
maximum value of atmospheric temperature drops, the thermal efficiency gets better.
As shown in Figure 8.37b, if we increase the boiler pressure by keeping the condenser pressure
constant, the net work increases by blue area (A55´1´A+34´5´3) but decreases by the amount of red
area (122´A1). As a result, there might be some increase in the net work, but the amount of heat rejected definitely decreases by green area (266´2´2).
368
THERMODYNAMICS
As illustrated in Figure 8.38, for an ideal Rankine cycle with fixed condenser pressure at 0.2
bars and saturated vapor at the turbine inlet, the overall effect is to increase the plant efficiency
as the boiler pressure increases. If pressure in the boiler is increased, the temperature at which
heat is added during evaporation also increases. Due to increase of average temperature of the
working fluid, the cycle efficiency will increase. Although it is an advantage to increase the
boiler pressure as much as possible, there are practical limitations as boiler components have to
be sufficiently robust to withstand the high pressures. Besides, increase in thermal efficiency is
not linearly proportional with the increase in boiler pressure. In Figure 8.38, increase in thermal
efficiency is 8.7-percent for raising the pressure from 50bar to 100bar. However, the efficiency
increase drops to 2.9-percent by increasing the pressure from 100 bars to 150 bars. Thus, the
generally accepted maximum economical pressure and temperature for circulating boilers is at
about 160bar and 560°C.
As discussed below, several modifications can be done on Rankine cycle for further improvement
in thermal efficiency and for better specific net work output.
8.9.1
Rankine cycle with reheat
As shown in Figure 8.39, the specific volume of steam from high pressure turbine increases by
reheating it in the boiler, and provides more work by expanding through the low pressure turbine.
The reheat process tends to increase the average temperature at which heat is added. Depending
on the cycle operating conditions, however, the thermal efficiency may increase or decrease. The
major benefit of reheat is to increase the net work of the cycle so that smaller components may be
used for a specified power output. Besides, reheating ensures that the exhaust steam quality of low
pressure turbine is not below the safety limit of 85-percent.
CHAPTER 8 POWER PRODUCING SYSTEMS 369
There is an optimum pressure of reheat for which the cyclic efficiency is kept at maximum level.
The optimum values of reheat pressure lie in the range 20% to 30% of the initial pressure of steam.
Since the steam is reheated in the boiler, the thermal efficiency has to be defined as following,

wnet
qout
1
qin1  qin 2
qin1  qin 2
(8.47)
Example 8.14 As in Figure 8.39, steam at a pressure of 20bar, 320°C is expanded through a first stage of the turbine to a
pressure of 5bar. It is then reheated at constant pressure to the initial temperature of 320°C and expanded through the second
stage to 10 kPa of condenser pressure. Assume ideal cycle,
a.
Draw the cycle on T-s diagram.
b.
Determine the percent increase in specific net work output obtained by reheating.
c.
Calculate the thermal efficiency of the system with and without reheating.
d.
For steam flow rate of 100kg/s, estimate the extra power delivered due to reheating.
Solution:
a.
370
b.
THERMODYNAMICS
State no.
Pressure (bar)
Temp.(0C)
Enthalpy(kJ/kg)
Entropy(kJ/kgK)
Quality(%)
1
20
320
3069.5
6.845
-
2
5
155
2757
6.845
-
3
5
320
3105.6
7.53
-
4
0.1
45.8
2386.8
7.53
0.917
5
0.1
45.8
191.83
0.649
0.0
6
20
-
193.82
0.649
-
2´
0.1
45.8
2168.3
6.845
0.826
The total specific work of turbines is, wt  h1  h2   h3  h4   3069.5  2757   3105.6  2386.8   1031.3
kJ/kg. Since the pump work is w p  v  p6  p5   0.001  2000  10   1.99 kJ/kg, then the net work output of reheating becomes, wnet  1031.3  1.99  1029.31 kJ/kg.. Without reheating unit, the net work is
wnet  3069.5  2168.3  1.99  899.21 kJ/kg. Hence by reheating the specific net work of the same plant has
been increased by 14.46-percent.
c.
The heat input at the boiler for reheating case is
qin  h1  h6   h3  h2   3069.5  193.82   3105.6  2757   3224.28 kJ/kg. With respect to Eq. (9.47), the
thermal efficiency of the reheat plant is   31.9% .
For the cycle without reheating, the heat input is, qin  h1  h6   3069.5  193.82   2875.68 kJ/kg, and thermal efficiency becomes   31.27% . The change in efficiency is   reheat    31.9  31.27  0.63% .
c.
Increase in specific net work is wnet  1029.31  899.21  130.1 kJ/kg. Due to reheating, the extra power developed by the plant is, Wnet  m s wnet  100 x130.1  13.01 MW.
Comments. Even though the efficiency of the system is slightly improved, the major enhancement is done on the net power developed. As numerically determined, 13.01 MW of extra
power is gained by the implication of reheating. This is why reheating is a preferred method in
large power plants.
8.9.2
Rankine Cycle With Regeneration
As illustrated in Figure 8.35, the heat addition and rejection processes of Carnot cycle take place
at a single high temperature and single low temperature respectively and the exergy of the system is
conserved. This causes a significant efficiency advantage over the Rankine cycle. Because, in Rankine
cycle, a part of the heat addition, the process 4-6 in Figure 8.34, is completed over a wide range of
water temperatures, and results with large exergy destructions. To reduce exergy losses during heat
addition and increase efficiency, the average temperature of water that receives heat in the boiler has
to be increased.
Definition: Preheating the water before entering the boiler by steam extracted from the main stream
expanding in the turbine is called regeneration.
CHAPTER 8 POWER PRODUCING SYSTEMS 371
Even though regeneration reduces the amount of steam flowing into the condenser, the power
generated by the same steam flow rate also lessens. Hence no actual alleviation in efficiency takes
place due to energy monitoring. Increase in efficiency is solely due to increase of feed water temperature at the boiler inlet.
Figure 8.41 illustrates a simple regenerative Rankine cycle with a single feed water heater (FWH)
which transfers heat to feed water incoming from the condenser. The particular heater in Figure 8.41a
is an open type heater and the extracted steam at state 2 in Figure 8.41b is directly mixed with the feed
water at sub-cooled conditions at state 5. Since the mixing takes place at steam extraction pressure,
the pressure of condensed water at state 4 is increased to mixing pressure by pump A. The amount of
extracted steam is such that the mixture becomes saturated liquid at the outlet of the heater (state 6 in
Figure 8.41b). Finally, the pressure of feed water is raised to boiler pressure by pump B.
A feed water heater is basically a heat exchanger and the types shown in Figure 8.42 are used in
regenerative systems.
372
THERMODYNAMICS
Open feed water heaters are simple, inexpensive, and provide saturated liquid at the outlet. For
regenerative systems furnished with open feed water heaters, the number of pumps needed is one
more than the number of heaters ( n p  nheater  1 ). The amount of steam extracted may be determined
by the energy balance around the feed water heater in Figure 8.41a as following,
 6
m 2 h2  m  m 2 h5  mh
Representing the extracted amount as the fraction of steam flowing into turbine and rearranging
yields
h h
m
y 2  6 5
(8.48)

m h2  h5
Hence, feed water at the heater outlet is at the saturation temperature of extracted steam pressure.
Closed feed water heaters are generally shell and tube type and feed water passes through the
tubes while steam flows on the shell side. The steam condensates and either passes through a trap
to a lower pressure heater (drains cascaded backward), or is pumped by a condensate pump into the
main feed water line (drains pumped forward). The steam traps are designed in such a way that only
allows condensed liquid to flow through by reducing the pressure but they are leak free to superheated
or saturated steam.
As necessitated by open type heaters, there is no need for a pump at the inlet of each heater.
However, closed type feed water heaters are more expensive, and as shown in Figure 8.43, are more
complex due to internal tubing network. For all types of heaters, the optimum pressure at which the
steam should be extracted is determined as following: 1. Single heater, the feed water temperature at
the heater outlet should the arithmetic average of boiler saturation temperature and condenser temperature. 2. Several heaters, the temperature difference between the boiler and the condenser should
be divided as equally as possible among heaters.
Figure 8.43 Closed type feed water heater and installation types
CHAPTER 8 POWER PRODUCING SYSTEMS 373
Example 8.15 As shown in Figure 8.41, consider an ideal regenerative cycle with a single open type feed water heater.
Steam enters the turbine at 40bar and 400°C, and some of steam is extracted at an appropriate pressure for regeneration.
The rest of steam expands to condenser pressure of 10 kPa. Determine,
a. the appropriate pressure for steam extraction,
b. percent of steam extracted,
c. efficiency of the system
Solution:
a.
The saturation temperatures at 40bar and at 0.1bar respectively are 250.40C, and 45.80C. Then the mean temperature
becomes Tm  250.4  45.8  / 2  148.1o C , and the steam extraction pressure should have a value between 4bar
and 4.5bar. Let the pressure at state 2 be p2  4 bar.
b.
c.
For p2  4 bar and s2  s1  6.769 kJ/kgK, x2  0.97 , the enthalpy at state 2 is h2  604.7  0.97  2133.8  2674.48
kJ/kg. For isentropic pump, the outlet and inlet enthalpy difference is h5  191.83  0.001  400  10  or h5  192.22
kJ/kg. By Eq. (8.47), the fraction of extracted steam becomes y  604
604.74
74 192
192.22
22  / 2674
2674.48
48 192
192.22
22  0.166
The steam quality and the enthalpy at the condenser inlet respectively are x3  6.769  0.649  / 7.5  0.816 , and
h3  2144.3 kJ/kg. The heat rejected at the condenser is qout  1  y h3  h4   0.834  2144.3  191.8   1628.38
kJ/kg. Similarly, the enthalpy of feed water at the boiler inlet is calculated as, h7  604.74  0.001  4000  400 
or h7  608.34 kJ/kg, and the heat input at the boiler is qin  h1  h7  3213.6  608.3  2605.26 kJ/kg. Hence the
thermal efficiency of the cycle becomes,   1
1628.38
2605.26
37.5%
Essentially we may convert an ideal Rankine cycle to Carnot cycle by using very large number of
heaters. Due to economical considerations, however, this is not only impractical but also impossible.
You may add a heater only if it saves more fuel than its cost and maintenance. Modern power plants
generally use up to 8 feed water heaters. The power plant shown in Figure 8.44 includes reheating
and also regeneration. Two heaters one open and the other closed type are used for regeneration. In
Figure 8.44, the condensed steam (state 12) and the feed water at the outlet of close feed water heater
(state 10) are usually taken to be at the saturation temperature of steam extraction pressure at that
particular heater.
374
THERMODYNAMICS
Example 8.16 As presented in Figure 8.44, a steam power plant operates on ideal Rankine cycle with reheating and uses
two feed water heaters for regeneration. One of the heaters is open type and the other is closed type with drains pumped
forward. Steam enters the turbine at 160 bar, and 560°C, and expands to condenser pressure of 0.1bar. At the exit of high
pressure turbine, some steam is extracted for closed feed water heater at 30bar, and the remaining is reheated to 560°C.
Steam for open type feed water heater is bled from low pressure turbine at 4 bar. A cross drum type boiler with a steam
capacity rate of 50kg/s is used in the plant.
a.
Draw T-s diagram of the cycle.
b.
Determine the fraction of steam extracted at each bled point.
c.
Calculate the thermal efficiency and the net power developed by the plant.
Solution:
a.
State no.
1
2
3
4
5
6
8
12
Pressure (bar)
160
30
30
4.0
0.1
0.1
4.0
30
Temp.(0C)
560
300
560
45.8
45.8
143.6
233.9
Enthalpy(kJ/kg)
3465.4
2993
3591.6
2964
2345
191.8
604.7
1008.4
Entropy(kJ/kgK)
6.513
6.513
7.398
7.398
7.398
0.649
1.77
2.79
Quality(%)
0.899
0.0
0.0
0.0
Figure 8.45 Tabulation of state properties and T-s representation of the cycle
b.
Energy balance around the closed feed water heater yields, y1h2  1  y1 h9  h12  1  y1 h10 , where
h9  604.7  0.001  16000  400   620.3 kJ/kg, and h10  h12 . Substituting the enthalpy values results with the
steam extraction ratio at state 2 as, y1  0.413 and 1  y1   0.587 .
Similarly, for the open type heater, the energy balance yields, h4 y2  h7 0.587  y2   0.587 h8 , wherev
h7  191.8  0.001  400  10   192.19 , and the steam extraction fraction at state 4 is, y2  0.087 .
c.
To determine the thermal efficiency, first we have to evaluate the enthalpy of feed water at the boiler entrance
(state 11), the enthalpy at state 13 is h13  1008.4  0.001  16000  3000   1021.4 kJ/kg. The energy balance at the mixing box yields, h11  y1h13  h10 1  y1  . Substitution of enthalpy values yields h11  1013.7
kJ/kg. Hence, heat input at the boiler is qin1  h1  h11  3465.4  1013.7  2451.7 kJ/kg. In addition, heat
input due to reheating is, qin 2  1  y1 h3  h2   351.37 kJ/kg. The heat rejected at the condenser is calculated as, qout  1  y1  y2 h5  h6   1076.6 kJ/kg. Hence, the thermal efficiency of the plant becomes,
q
1076.6
  1 out 1
61.59% .
qin
2451.7 351.37
The specific net work output of the plant is wnet  0.6159  2451.7  351.37   1726.41 kJ/kg and the power
output is Wnnet  50 1726.41 86.32 MW.
CHAPTER 8 POWER PRODUCING SYSTEMS 375
8.10
Cogeneration
Cogeneration means the combined production
of electricity and heat in an energy conversion
facility. Because of this feature, cogeneration
plants are also called Combined Heat and Power
(CHP) systems. As described by Example 8.17,
cogeneration makes sense only when there is a
continuous demand for large amount of heat at
power plant location. Hence, the energy transferred
Figure 8.46 A cogeneration plant
to water in the boiler is utilized as process heat
as well as electric power.
All continuous processes used in industrial plants such as petrochemicals, hydrocarbon refineries,
food processing units, dairy plants, pharmaceuticals require uninterrupted energy input in the form of
power and heat. Hence, it is appropriate to define a utilization factor εu for a cogeneration plants as,
u 
W
 Q
 Q
net

in
process
1
Q out
Q
(8.49)
in
where Q out represents the heat rejected in the condenser. Notice that the utilization factor of cogeneration plant with an ideal cycle is 100-percent. Actual cogeneration plants have utilization factors
as high as 80-percent.
Example 8.17 Let us consider the plant in Figure 9.46 that requires 2.5 MW of electrical energy and 3.0 MW of heat energy
for its processes. Analyze the methods available to solve the energy needs of the system and determine which method is
more economical.
Solution:
a.
If we solve the electrical energy and heat requirement of the system by separate plants, then a centralized power
plant meets the electrical energy requirement by consuming 6MW of fuel energy. Similarly, as shown in Figure
8.47a, the need for steam can be met by a coal fired steam boiler. Considering the losses of these two systems
(4.5MW), the need for total fuel energy becomes 10MW of conventional generation. However, if we use a cogeneration plant, as shown in Figure 8.47b, to supply the same amount of power and heat, the total fuel energy needed
may be reduced to 7MW.
Figure 8.47 System improvement and energy saving by cogeneration
The cogeneration system configuration used in various plants is classified in the following
sections.
376
THERMODYNAMICS
8.10.1
Steam turbine based cogeneration
Depending upon the energy needs of the facility, steam might be extracted at several pressures as
it flows through the turbine. The steam turbines most widely used for this cogeneration configuration
are two-fold: (1) the back pressure turbine type, (2) extraction-condensing turbine type. The selection
of steam turbine for a particular cogeneration application depends on the required heat to power ratio.
If we define, ξ, representing this ratio as,

Q process
W
(8.50)
net
For ξ values in the range 4.0    15.0 , usually back pressure type steam turbines are used.
However, for ξ values,1.0    8.0 , extraction-condensing steam turbines are preferred. As shown
in Figure 8.48a, at times of high demand for process heat, some steam leaving the boiler is throttled
and then routed to the process heater. The extraction fractions are adjusted so that steam leaves the
process heater as saturated liquid.
Figure 8.48 Typical steam turbine based cogeneration
CHAPTER 8 POWER PRODUCING SYSTEMS 377
Example 8.18 As shown in Figure 8.49, an extraction-condensing cogeneration plant is modified with regeneration. Steam
at 4MPa, and 500°C enters the high pressure turbine and expands to 0.5MPa. At this pressure, 50-percent of steam is extracted and the remainder expands to 20kPa. Part of the extracted steam is used for 20MW capacity process heater and the
rest feeds the open type feed water heater at 0.5MPa. After leaving the process heater and the FWH, the saturated liquid is
pumped to the boiler.
a.
Draw the T-s diagram of the cogeneration plant.
b.
Determine the mass flow rate of steam through the boiler.
c.
Calculate the net power developed, and the utilization factor of the plant.
State no.
Pressure (kPa)
Temp.(0C)
Enthalpy(kJ/kg)
Entropy(kJ/kgK)
Quality(%)
1
4000
500
3445.3
7.09
-
2
500
200
2855.4
7.09
-
3
20
60.06
2336.1
7.09
0.884
4
20
60.06
251.4
0.832
0.0
5
500
-
251.88
-
-
6
500
151.8
640.2
-
-
7
4000
-
643.7
-
-
Solution:
a.
T-s diagram of the plant is indicated above.
b.
With respect to energy balance on FWH, 0.5  y f h6  y f h2  0.5h5 , or y f  0.5 


y p  0.5  0.088  0.412 . The mass flow rate for process heater is m p 
mass flow rate through the boiler becomes, m 
9.02
 21.89kg/s.
0.412
640.2  251.88
 0.088 ,
2855.4  640.2
20000
 9.02 kg/s, and the
2855.4  640.2
378
c.
THERMODYNAMICS
Wt  21.89 3445.3  2855.41  0.5  2855.4  2336.1  18596 kW, W p  21.89 0.5  0.48   3.5  81.86
kW. Wnet  18514.14 kW. The boiler capacity is Q in  21.89  3445.3  643.7   61327 kW. Hence the utilization factor of the plant is  u 
8.10.2
20 18.514
 0.628 .
61.327
Gas Turbine Based Cogeneration
As illustrated in Figure 8.50, a gas turbine based cogeneration system provides recovery of heat
in exhaust flue gases by a boiler for steam production or by a heat exchanger to generate hot water for
district heating purposes. If the heat output of the system is less than that required by the consumer,
it is possible to have supplementary firing by mixing additional fuel to oxygen rich exhaust gas to
boost the thermal capacity.
Gas turbine based cogeneration plants is ideal for chemical industries where the demand of steam
is high and fairly constant in comparison to that of steam turbine based cogeneration.
8.10.3
Combined steam and gas turbine based cogeneration
In comparing the maximum temperatures of Brayton and Rankine cycles, a typical gas turbine
cycle operates at considerably higher temperature than a steam turbine cycle. Hence, a gas turbine
cycle has a greater potential for higher thermal efficiencies. However, the flue gases leaving the
turbine are at very high temperature, and the waste heat has to be recovered. As shown in Figure
CHAPTER 8 POWER PRODUCING SYSTEMS 379
8.51, the exhaust gases from the gas turbine produce high pressure steam in a waste heat boiler.
Depending upon the plant needs, the high pressure steam expands through a back-pressure or
extraction-condensing steam turbine to generate additional power, and some portion of steam might
be used as process steam.
When the ratio of electrical power to thermal load is high, the cogeneration plant based on combined
cycle principle provides better results than the gas turbine cogeneration plant. In combined cycles,
about 66-percent of power is generated in gas turbine and the rest 34-percent in steam turbine, and
the plant utilization factor achieves values above 90-percent.
Figure 8.51 Combined gas-steam turbine cogeneration plant
Example 8.18 The combined gas-steam power plant in Figure 8.51 works with a pressure ratio of rp  9 in gas turbine
part and air enters the compressor at 290 K, and the turbine at 1500 K. The combustion gases leaving the gas turbine
are utilized in a waste heat recovery boiler and steam at 10 MPa and 500°C is produced. The combustion gases leave
the boiler at 250°C. After expanding in a high pressure turbine to 1 MPa, 60-percent of steam is utilized for 1.5 MW of
process heat and the rest is reheated to 500°C at the combustor before expanding in low pressure turbine to 10 kPa. Assume ideal combustion, and all compression and expansion processes are isentropic. Determine,
a. the mass flow rate of air,
b. for fuel heating value of 40,000 kJ/kg, the rate of fuel consumed,
c. the thermal efficiency of the overall plant,
d. the plant utilization factor.
380
THERMODYNAMICS
Steam Turbine
Gas Turbine
State no.
Pressure (kPa)
Temp.(0C)
1
100
17
2
900
3
Enthalpy(kJ/kg)
Entropy(kJ/kgK)
Quality(%)
290.163
-
-
267
544.3
-
-
900
1227
1635.9
-
-
4
100
601
903.2
-
-
5
10000
500
3373.7
6.596
-
6
1000
185
2790
6.596
-
7
1000
180
-
0.0
9
1000
658.94
10
10000
667.94
11
1000
12
10
2460.17
13
10
191.8
500
762.8
3478.5
7.762
0.948
Solution:
a.
The mass flow rate for process heater is m p 
the boiler becomes, m s 
1500
 0.739 kg/s, and the mass flow rate of steam through
2790  762.8
9.02
 1.233 kg/s. Flue gases leave the boiler at 2500C with the enthalpy of 526.63kJ/kg.
0.6
Energy balance around the boiler yields the mass flow rate of air as, m a 
b.
1.233
3373.7
667.94 
903.2 526.63
 8.859 kg/s
Energy balance around the combustor gives us the heat input of the system as,
qin  8.859  1635.9  544.3  0.494  3478.5  2790   10010.6 kW. Hence the fuel consumption rate be10010.6
 0.25 kg/s
40000
The net power of the gas turbine is
comes, m f 
c.
Wnet gas  m a h3  h4   h2  h1   42359.56 k W .
The
power consumed by pumps 1 and 2 respectively are W p1  0.494  0.0011000  10   0.489 kW and
W  1.233  0.001  10000  1000   11.097 kW. Then, the net power developed by steam turbine is,
p2
Wnet steam  1.233  3373.7  2790  0.494  3478.5  2460.17   0.489  11.097 o r Wnet steam  1211.214
kW. The total power of the plant is
W
net
 5450.77 kW, and with respect to Eq. (8.30), the plant efficiency is
  5450
5450.77
77 / 10010
10010.66 54.45%
d.
The amount of heat rejected at the condenser is qout  m 12 h12  h13   0.494  2460.17  191.8   1120.57 kW,
and by Eq. (8.49), the plant utilization factor becomes  u  1
8.11
1120.57
10010.6
88.8% .
Organic Rankine Cycle
Organic Rankine Cycle (ORC) is gaining increasing interest as cost effective and sustainable energy systems in recent years. ORC power plants convert thermal energy of relatively low temperature
sources with temperatures in the range of 1000C to 3500C to electricity.
CHAPTER 8 POWER PRODUCING SYSTEMS 381
Figure 8.52 Comparison of power generation and efficiency of different cycles
Besides industrial waste heat, heat sources such as solar and geothermal energy as well as biomass
can be utilized by ORC systems. Because of small temperature range, only a simple Rankine cycle can
be used. As shown in Figure 8.52, due to low temperature of the heat source, the cycle efficiency is
rather low and varies in the range of 8-percent and 16-percent. Because of free fuel source, however,
efficiency is not a critical parameter.
When appropriate working fluid and operating conditions are selected, the power produced by an
ORC system varies in the range of 0.1MW to 1MW. Organic fluids such as Pentane and Freon family
fluids are preferred in ORC systems. As can be seen by Figure 8.53, working fluids are classified with
respect to the slope of the saturation curve in a T-s diagram as following: (1) wet (negative slope), (2)
isentropic (vertical), and (3) dry (positive slope).
Figure 8.52 Comparison of power generation and efficiency of different cycles
Referring to Figure 8.53, to avoid occurrence of liquid droplets that impinge and cause erosion on
the turbine blades during the expansion, isentropic or dry fluids are selected for ORC systems. Besides,
the specific volume of such fluids (refrigerants) at low temperature is incomparably less than that of
steam, and the desired turbine size can be much smaller and less expensive. High density and latent heat
are desirable properties for working fluids by that high turbine specific work output can be provided.
382
THERMODYNAMICS
An ORC cycle with regeneration used as geothermal power plant is shown in Figure 8.54a. The
working fluid of the cycle is isopentane. The geothermal fluid (usually liquid water) enters the vaporizer at state (w1). If the fluid pressure is sufficiently high, there is no need for a gas extraction system,
and the non-condensable gases will not separate from the fluid. Geothermal fluid cools down in the
vaporizer at constant pressure, and then is re-injected into the well at state (w2). ORC fluid enters the
vaporizer at state (2), and becomes saturated or superheated at the vaporizer exit (state 3) and enters
the turbine. At the turbine exit (state 4), ORC vapor enters the regenerator where the condensed ORC
fluid is preheated prior to the vaporizer entry. Moreover, the cooled vapor enters the condenser at
state 5 and becomes saturated liquid at the condenser exit (state 6).
Figure 8.54 Schematic of ORC power plant with regeneration and T-s diagram representation
Air cooled condensers are generally used for ORC plants with the cooling air entering the condenser at state (a1) and leaving at state (a2). Figure 9.54b represents the T-s diagram of ORC power
cycle with regeneration. Because of regeneration, the heat extracted between states 4 and 5 must be
the same with the heat given between states 1 and 2.
References
1.
Heywood, J. B., Internal Combustion Engine Fundamentals, McGrawHill Publications, ISBN-0-07-028637-X, 1988.
2.
Giampaolo, T., Gas Turbine Handbook-Principles and Practice, 4th edition, CRC Press,ISBN-10: 0-88173-613-9, 2009.
3.
Kiameh, P., Power Generation Handbook-Selection, Applications, Operation, and Maintenance, McGrawHill Publications, ISBN: 00 71396047, 2002.
4.
Leyzerowich, A. S., Steam Turbines for Modern Fossil-Fuel Power Plants, The Fairmont Press Inc., ISBN: 0-88171548-5, 2008.
5.
Barclay, F. J., Combined Power and Process-An Exergy Approach, Professional Engineering Publishing Ltd., ISBN:
1-86058-129-3, 1998.
6.
Turns, S. R., An Introduction to Combustion Concepts and Applications, 2nd Edition, McGrawHill, ISBN: 0-07-230096-5,
2000.
7.
Mollenhauer, K., and Tschoeke, H., Handbook of Diesel Engines, Springer-Verlag, ISBN: 978-3-540-89082-9, 2010.
8.
Oates, G. C., Aerothermodynamics of Gas Turbine and Rocket Propulsion, 3rd edition, American Institute of Aeronautics
and Astronautics, Inc., ISBN: 1-56347-241-4, 1997
CHAPTER 8 POWER PRODUCING SYSTEMS 383
Problems
SI and CI engine cycles, dual cycle
Hint: Assume variable specific heats.
8.5
8.1
A four-cylinder 3.2-liter SI engine operates on a four
stroke cycle at 3000 rpm. At this condition, 1kJ of
indicated work is produced in each cylinder at each
cycle. Determine,
a. the net work output,
a. the mean effective pressure,
b. the torque produced by the engine at the given
speed.
8.2
8.3
An eight cylinder four stroke SI engine operates at
1000 rpm. Determine,
a. the engine rotation in degrees for each ignition,
b. the number of power strokes per revolution, and
per second.
A pickup truck as shown in Figure 8.55 is furnished
with a 4.2 L SI engine operating at 2500 rpm. The
compression ratio of the engine is 10.2. The bore and
the stroke are related as, H  0.95 D . Determine,
a. the stroke length,
b. the clearance volume of a single cylinder,
c. the average piston speed.
A four-cylinder gasoline engine has a displacement
volume of 2.2 L and operates with a compression ratio
of 12:1. Air at the inlet is at 17°C, 100 kPa, and 1600
kJ/kg of heat is added by the combustion process.
Assuming variable specific heats, determine,
b. the thermal efficiency,
c. the mean effective pressure of the cycle.
8.6
An ideal Otto cycle having a compression ratio of
10 operates with a maximum temperature of 2100K.
For atmospheric air at 10°C, 100 kPa, determine,
a. the specific work output, the specific heat input,
and the specific heat removed during the cooling
process,
b. the thermal efficiency of the cycle.
8.7
A 3L, four cylinder SI engine as shown in Figure
8.56, operates on air standard Otto cycle. At the start
of the compression process, air is at 60°C, and 98
kPa, and the compression ratio of the cycle is 9:1.
a. For a combustion efficiency of 95-percent (
c  q / qideal
), determine the specific heat
input if the engine uses gasoline with air-fuel
ratio of 15.5. Take the heating value of the fuel
as 50,000 kJ/kg.
b. Calculate the temperatures and pressures at all
states of the cycle, and also evaluate,
c. the specific work output and the thermal efficiency of the cycle,
d. the torque produced at a speed of 2600 rpm.
Figure 8.55 A pickup truck with SI engine
8.4
An engine operates on the Otto cycle and has a compression ratio of 8.2:1. Fresh air enters the engine at
27°C, and 100kPa. During the combustion process,
heat added at a rate of 750 kJ/kg to air of 0.015 kg
in the cylinder. Draw the cycle on p-V diagram, and
determine,
a. the temperature and pressure at the end of combustion process,
b. the pressure and temperature after expansion,
c. the mean effective pressure, and the cycle efficiency,
d. the power produced at a speed of 1500 rpm by
a single cylinder.
Figure 8.56 A four-stroke SI engine
384
8.8
THERMODYNAMICS
A 5L SI engine as shown in Figure 8.57 is mounted
on a hydraulic dynamometer provides an output of
60 HP at 4000 rpm. Water absorbs the energy output
of the engine as it flows through the dynamometer
at a rate of 1.5 L/s. Water inlets the dynamometer at
10°C. For a dynamometer efficiency of 95-percent,
evaluate
8.11
A construction vehicle as shown in Figure 8.58 has
a Diesel engine with eight cylinders of bore 12 cm
bore, and 20 cm stroke and operates on a four-stroke
cycle. The engine with a compression ratio of 18:1
delivers 250 HP at 1000 rpm. Determine,
a. the engine total displacement volume,
b. the mean effective pressure,
c. the torque at 1000rpm.
a. the temperature of water at the dynamometer
exit,
b. the torque output of the engine,
c. the mean effective pressure at this condition.
Figure 8.58 A construction vehicle equipped
with a diesel engine
Figure 8.57 Structural view of an SI engine
8.9
An engine operates on an Otto cycle with a compression ratio 10.5:1. The volume, pressure and
the temperature at the start of compression process
respectively are 0.003 m3, 115 kPa, and 60°C. The
peak temperature of the cycle is 927°C. Evaluate,
a. the temperature at the remaining two states,
b. the pressure at the end of combustion process,
c. the specific heat added by the combustion process,
d. the specific heat removed from the engine,
e. the specific compression and expansion work
of the cycle,
f. the mean effective pressure and the thermal
efficiency of the cycle.
8.10
8.12
An Otto engine with a turbo charger operates with
a compression ratio of 9:1, and the following information is supplied: 1. Temperature and pressure of
air prior to turbo charging compression respectively
are 21°C, 1 bar. 2. The pressure after turbo charging
compression is 1.35 bar. 3. The heat addition during
the combustion process is 800 kJ/kg. 4. The volume
after the compression is 0.0015 m3. Determine,
a. the mass of air in the cylinder,
b. the pressure and temperature at the end state of
each process,
c. the specific compression and expansion work,
d. the thermal efficiency of the engine.
A single cylinder and four-stroke CI engine with a
bore of 10 cm, and 14 cm stroke operates at 1000
rpm and develops a torque of 90 Nm. The engine
consumes fuel at a rate of 0.00005 kg/s. Take the
diesel fuel heating value as 4x105 kJ/kg and calculate,
a. the mean effective pressure,
b. the thermal efficiency of the cycle.
8.13
For an ideal Diesel cycle with a compression ratio
of 20 and a cutoff ratio of 2, air at the beginning of
the compression process is at 105 kPa and 40°C.
Assume variable specific heats and determine,
a. the temperature and pressure at the end state of the
compression and the combustion processes,
b. the amount of specific heat added,
c. the amount of specific heat removed,
d. the thermal efficiency of the cycle.
8.14
Air inlets a Diesel engine, having a compression ratio
of 18:1, at a condition of 22°C, 100 kPa. To avoid
damaging of the engine block, the highest temperature
of the cycle is limited to 1350°C. Assume variable
specific heats, and determine,
a.
b.
c.
d.
the net specific work output,
the mean effective pressure,
the thermal efficiency of the cycle,
the thermal efficiency of Carnot cycle operating
between the same temperature levels.
CHAPTER 8 POWER PRODUCING SYSTEMS 385
8.15
As shown in Figure 8.59, a CI engine for a small
truck operates on an air-standard Diesel cycle. Due
to structural limitations, the maximum allowable
pressure in the cycle is 95 bar, and at the start of the
compression process, the pressure and the temperature
in the cylinder are 95 kPa, 45°C. Light diesel fuel
with a heating value of 4x105 kJ/kg at an air-fuel
ratio of AF=20:1 is used in the cycle. Determine,
Figure 8.60 Use of diesel engine in heavy
duty vehicles
8.18
Figure 8.59 Structural view of a
diesel engine
a. the compression ratio of the engine,
b. the specific heat added by the combustion process,
and the cutoff ratio,
c. the peak cycle temperature,
d. the thermal efficiency of the cycle.
Air is at 27°C, 1 bar at the start of compression
process in a dual cycle. The compression ratio is 16,
and the heat addition at constant volume and constant
pressure are 310 kJ/kg, and 554 kJ/kg respectively.
The mass of air in the cylinder is 0.018 kg. Show
the cycle on p-V and T-s diagrams and determine,
a. the maximum temperature and the maximum
pressure of the cycle,
b. the specific work output and the mean effective
pressure of the cycle,
c. the thermal efficiency of the cycle
8.19
A 3.2 L four-cylinder CI engine consumes light
diesel fuel (heating value of fuel, H u  4 x105 kJ/
8.16
kg) at an air-fuel ratio of AF  19 :1 , and operates
on an air standard dual cycle. The cylinder conditions
at the start of compression are 47°C, 1bar, and the
compression ratio of the cycle is 15. Half of the fuel
may be considered to be burned at constant volume
and the other half at constant pressure. Determine,
A diesel engine with a compression ratio of 20
receives air at 17°C and 1 bar. The amount of heat
added during the combustion process is 1560 kJ/kg.
The mass of air contained in the cylinder is 0.01 kg.
Assume variable specific heats, and determine,
a. the temperature and pressure at each state of the
cycle,
b. the pressure ratio and the cutoff ratio,
c. the total amount of specific heat supplied,
d. the net work produced,
e. the thermal efficiency of the cycle.
a. the maximum temperature of the cycle,
b. the amount of specific heat added, and removed,
c. the net work produced,
d. the mean effective pressure,
e. the efficiency of the cycle.
8.20
8.17
An interstate truck, as shown in Figure 8.60, equipped
with eight-cylinder four-stroke cycle Diesel engine
with 110 mm bore and 125 mm stroke has a compression ratio of 19 and produces 422 HP at 3000
rpm. Determine,
a. the engine displacement volume,
b. the clearance volume of one cylinder,
c. the mean effective pressure,
d. the torque produced at this speed.
Air at the start of compression in a dual cycle is at
1 bar, 17°C, and the compression ratio is 10:1. Heat
added at constant volume process is 120 kJ/kg and
the maximum temperature of the cycle is 1927°C.
The amount of air in the cylinder is 0.012 kg. Determine,
a. the maximum pressure of the cycle,
b. the specific heat supplied at constant pressure
process,
c. the heat removed, and the net work produced,
d. the thermal efficiency of the cycle.
386
8.21
THERMODYNAMICS
a. the net work of the cycle,
b. the thermal efficiency of the cycle,
c. the mean effective pressure.
8.22
a. the power output of the cycle,
b. the rate of heat input,
c. the thermal efficiency of the cycle.
8.26
An air standard cycle with constant specific heats
is enclosed in a piston-cylinder arrangement and
is composed of the following processes: (1-2)
Isentropic compression with a compression ratio
of 6, (2-3) Constant pressure heat addition, (3-1)
Constant volume heat rejection. Air at the beginning of compression process is at 100 kPa, 27°C,
and 951 kJ/kg of heat is added during the constant
pressure process. Sketch the p-V and T-s diagrams
of the cycle and evaluate,
a. the net work of the cycle,
b. the back work ratio of the cycle.
Gas turbine engines, regeneration, intercooling, reheating, and stirling cycle
8.23
of 15, and a maximum temperature of 777°C. The
volumetric flow rate of helium at the compressor
inlet is 350 m3/min. Assume constant specific heats
evaluated at 300K, and calculate,
A proposed air standard piston-cylinder arrangement
cycle consists of the following processes: (1-2) Isentropic compression process, (2-3) Constant volume
heat addition, (3-4) Isentropic expansion, (4-1)
Constant pressure heat rejection. The compression
ratio of the cycle is 9:1, and the heat added at constant
volume process is 1250 kJ/kg. For air at 95 kPa, 17°C
at the beginning of the compression process, draw
the cycle on T-s diagram and evaluate,
A simple ideal Brayton cycle with air as the working
fluid has a pressure ratio of 10. The air enters the
compressor at 17°C and the turbine at 850°C. Accounting for the variation of the specific heats with
temperature, evaluate,
A large stationary power plant uses simple Brayton
cycle and produces a power of 102 MW. The minimum and the maximum temperatures in the cycle
respectively are 300K, and 1700K. The compressor
compression ratio is 14:1, and air inlets the compressor
at a pressure of 95 kPa. The isentropic efficiencies
of both turbine and the compressor are identical at
88-percent. Evaluate,
a. the power output of the turbine,
b. the back work ratio,
c. the thermal efficiency of the cycle.
8.27
A gas turbine engine at San Francisco Power Station
takes in 110,000kg/h of filtered outside air at 27°C
and compresses it to 6.516bar. Due to combustion,
30MW of heat is added to the air, and the turbine
exhausts to atmospheric pressure of 1bar. Assuming
both the compressor and the turbine are 75-percent
efficient,
a. draw the T-s diagram of the cycle,
b. evaluate the net power output and the efficiency
of the cycle.
a. the air temperature at the compressor and at the
turbine exits,
b. the net work output,
c. the thermal efficiency of the cycle.
8.24
A simple ideal Brayton cycle produces 15 MW with
an inlet state of air at 17°C, 100 kPa, and a compression ratio of 18:1. The amount of heat added in the
combustion process is 946 kJ/kg. Assuming variable
specific heats, determine,
a. the highest temperature of the cycle,
b. the mass flow rate of air.
8.25
A closed and simple ideal Brayton cycle uses helium
as the working fluid. It operates between 0.8 bar, and
15°C at the compressor inlet, has a pressure ratio
Figure 8.61 A view of a gas turbine
power station
CHAPTER 8 POWER PRODUCING SYSTEMS 387
8.28
An air standard cycle is executed in a closed system
and is composed of the following processes: (1-2)
Isentropic compression from (95 kPa, 17°C) to 10
bar, (2-3) Constant pressure heat addition in amount
of 2500 kJ/kg, (3-4) Constant volume heat rejection
to 95 kPa, (4-1) Constant pressure heat rejection
to initial state. Assume constant specific heats at
17°C, draw the cycle on p-V and T-s diagrams, and
determine,
c. the temperature of air at the inlet of the combustion chamber, T3,
d. the thermal efficiency of the cycle.
a. the maximum temperature of the cycle,
b. the thermal efficiency of the cycle.
c. Compare the efficiency of this cycle with the efficiency of air standard Brayton cycle operating
between the same pressure limits and supplied
with the same amount of heat.
8.29
In a regenerative Brayton cycle, as in Figure 8.62,
air inlets at a condition of 7°C, 1.01 bar, and is
compressed through a pressure ratio of 5:1. The heat
exchanger which causes 0.15 bar of pressure drop
heats up the air through 75% of the maximum range
possible at given conditions, and after the combustion
process the maximum cycle temperature becomes
800°C. Isentropic efficiencies of both turbine and
the compressor respectively are 0.88, and 0.82.
Determine the thermal efficiency of the plant.
Figure 8.63 A regenerative gas turbine
used as automobile engine
8.31
Consider a simple Brayton cycle with air entering
the compressor at 1.02 bar, 15°C, and leaving at a
pressure of 6.5bar. The maximum cycle temperature
is 950°C. Isentropic efficiencies of the compressor
and the turbine respectively are 80% and 85%. The
pressure loss due to piping between the compressor
and turbine is 25 kPa. Evaluate,
a. the pressure and temperature at each end state
of the cycle,
b. the compressor and turbine specific work, and
the efficiency of the cycle.
c. To enhance the thermal efficiency, an ideal
regenerator is incorporated into the cycle. Calculate the new efficiency and percent change in
the efficiency.
Figure 8.62 A regenerative gas turbine engine
8.32
8.30
The gas turbine engine shown in Figure 8.63 is used
as an automotive engine. The first turbine (T1) produces enough power to run the compressor, and the
second turbine (T2) generates 200 kW of power to
drive the car wheels. The compressor pressure ratio
is 8, and intake air is at 1bar, 300K. The isentropic
efficiencies of the compressor and the turbines are
identical at 82%, and the regenerator effectiveness
is 80%. The maximum temperature of the cycle is
1700K. Neglect the pressure drops due to flow in
connecting pipes, and determine
a. the pressure at the inlet of second turbine
(T2),
b. the mass flow rate of air through the engine
A regenerative Brayton cycle has a pressure ratio of
10:1, and air circulates with a mass flow rate of 0.2
kg/s through the cycle. The condition of air at the
compressor inlet is 1 bar, -5°C. The temperatures of
air at the inlets of combustion chamber and the turbine
respectively are 327°C, 827°C. The air pressure at
the turbine exit is 1.1 bar. The isentropic efficiencies
of the compressor and the turbine are 82% and 89%
respectively. For 85% of exchanger effectiveness,
determine,
a. the air temperatures at the compressor and turbine
exits,
b. the rate of heat added at the combustion chamber,
c. the back work ratio, and the net work output,
d. the thermal efficiency of the cycle.
388
8.33
8.34
THERMODYNAMICS
Regeneration in Brayton cycle is useful only when
the turbine exhaust temperature is greater than the
compressor exhaust temperature. As the compression ratio increases, however, the difference in these
temperatures decreases and the effectiveness of
regeneration vanishes. Determine an expression for
the pressure ratio of an ideal regenerative Brayton
cycle in terms of Tmin , Tmax , and k for which effectiveness of regeneration becomes zero. Assume
constant specific heats, and evaluate the pressure
ratio for Tmin  17 oC , Tmax  1627 oC and k  1.4
.
a. the maximum air temperature,
b. the power required by the first and the second
compressors,
c. the back work ratio, and the net power developed,
d. the thermal efficiency of the cycle.
8.36
As shown in Figure 8.64, air enters the turbine of
an ideal gas turbine engine at 1600 kPa, 1400K and
expands to 100 kPa in two stages. Air is reheated to
1400K at a constant pressure of 400 kPa between the
stages. Determine the following per kg of atmospheric
air at 27°C flowing through the engine,
a.
b.
c.
d.
the work developed by each turbine stage,
the work consumed by the compressor,
the heat added at both combustors,
the percent increase in the net work as compared
to a single turbine with no reheat.
a. the back work ratio, and the net specific work
output,
b. the mass flow rate of air,
c. the thermal efficiency of the cycle.
8.37
Figure 8.64 A gas turbine engine with reheat
8.35
An engine operates on a reheat and inter-cooling Brayton cycle. The low and the high pressure compressors
provide compression ratios 2:1 and 4:1 respectively.
Air enters the engine at 27°C, 1bar, and is cooled down
to 27°C before entering to high pressure compressor.
Air flowing with a mass flow rate of 0.1 kg/s into
the combustor is heated by the amount of 1050 kJ/
kg, and expands to 200 kPa through the first turbine.
Air is reheated to the same temperature as the first
turbine inlet temperature, and then expands in the
second turbine to 100 kPa pressure. All compressors
and turbines operate with an isentropic efficiency of
85%. Draw the cycle schematically, and show on
T-s diagram. Determine,
A large gas turbine plant producing 10 MW of power
comprises low pressure (LP) and high pressure (HP)
compressors and turbines, and air inlets the plant at
7°C, 95 kPa. Air is compressed to 300 kPa in the
LP stage and then inter-cooled down to 27°C with
a pressure loss of 10 kPa. The air pressure is arisen
to 1200 kPa after the HP compressor. Air flowing
through a regenerator with an effectiveness of 0.65
is heated by hot gases from LP turbine and then exits
the combustor at a temperature of 817°C. Air leaves
the HP turbine at a pressure of 350 kPa, and then
reheated to 767°C with a pressure loss of 10 kPa.
Isentropic efficiencies of all turbines and compressors are identical at 82%. Sketch the system, and
calculate,
As shown in Figure 8.65, a turboprop engine consists of a diffuser, compressor, combustor, turbine
and nozzle. The turbine drives both the propeller
and the compressor. Air inlets the diffuser with a
volumetric flow rate of 90m3/s at 40kPa, 250K, and
has a velocity of 200 m/s. Air is compressed by a
pressure ratio of 9:1, and the turbine inlet temperature
is 1200K. The outlet pressure of the turbine is 55
kPa. Assume that both the diffuser and the nozzle
are isentropic, but the compressor and the turbine
operate with isentropic efficiency of 82%. Except
the diffuser inlet and the nozzle exit, neglect kinetic
energy effects, and determine,
a. the power delivered to the propeller,
b. the air velocity at the nozzle exit.
8.38
The pressure and temperature of air entering the
compressor of an ideal jet engine are 75 kPa, 260K.
The pressure ratio across the compressor is 15:1, and
the turbine inlet temperature is 1550K. After the
turbine, air enters the nozzle and expands 95 kPa.
Evaluate,
a. the pressure at the nozzle inlet,
b. the velocity of air at the nozzle exit.
CHAPTER 8 POWER PRODUCING SYSTEMS 389
8.42
b. the net work output and the efficiency of the
engine.
Figure 8.65 A cutaway view of a
turboprop engine
8.39
Repeat problem 8.38 by assuming that the compressor and the turbine work with isentropic efficiencies
of 85% and 88% respectively. Also assume that the
isentropic efficiency of nozzle is 95%.
8.40
An afterburner in a jet engine is usually used to
increase the speed of air at the nozzle exit due to
energy of combustion. As shown in Figure 8.66, the
state of air after the turbine into the nozzle is 250
kPa, 820K, and the pressure at the nozzle exit is 90
kPa. Suppose now the afterburner is turned on, then
it adds 400 kJ/kg of heat to air with rise in pressure
for the same specific volume (v1  v3 ) . Determine the
nozzle exit velocity before and after the afterburner
is turned on. Note that the nozzle exit pressure is
always at 90kpa.
Consider an ideal air standard Stirling cycle with an
ideal regenerator. The working fluid is air, and the
compression ratio is 10. The minimum temperature
and pressure of the cycle respectively are 1 bar, 27°C,
and air assumes a maximum temperature of 927°C.
Evaluate,
a. the work and heat transfer for each of the four
processes of the cycle,
Rankine cycle, reheat and regenerative systems,
cogeneration, combined cycles
8.43
As shown in Figure 8.67, consider a steam turbine
power plant operating with a conventional Rankine
cycle. Steam is fed to a turbine at a pressure of 8
MPa, and a temperature of 550°C. Exhaust from the
turbine enters a condenser at 10kPa, and exits the
condenser as saturated liquid which is then pumped
to the boiler by raising its pressure.
a. Assume an ideal cycle and determine the thermal
efficiency,
b. Suppose isentropic efficiencies of the turbine and
pump respectively are t  0.82 and  p  0.78 .
Determine the thermal efficiency for this case.
c. For a practical Rankine cycle, in addition to turbine and pump isentropic efficiencies, boiler and
generator efficiencies are specified respectively
as b  0.75 , and  g  0.92 . Determine the
overall efficiency of a practical cycle operating
at conditions given above.
d. What is the steam circulation rate through the
practical cycle for 85 MW of net power received
at the generator outlet?
e. Determine the rate at which heat is discarded to
the environment in the condenser of the practical
cycle.
H  28000
f. Coal with a heating value of u
kJ/
kg is used as the fuel of the boiler. Evaluate the
rate of coal as tons per hour consumed in the
plant.
Figure 8.66 Effect of afterburner on a
jet engine
8.41
A Stirling cycle operates with 0.5 kg of helium as
the working fluid between the temperatures of 700°C
and 27°C. The highest and the lowest pressures of
the cycle are 30 bar, and 5 bar respectively. Determine,
a. the amount of heat added,
b. the net work output,
c. the thermal efficiency of the cycle.
Figure 8.67 A power plant operating with a
simple Rankine cycle
390
8.44
THERMODYNAMICS
Water is the working fluid in a Rankine cycle. Superheated water vapor at 60 bar, 540°C enters the
turbine with a mass flow rate of 7.5 kg/s and exits at
20 kPa. Leaving the condenser as saturated liquid at
20 kPa enters the feed water pump. The isentropic
efficiencies of the turbine and the pump are t  0.87
and  p  0.80 . Cooling water enters the condenser
at 20°C and exits at 28°C. Neglecting any pressure
loss throughout the cycle, show the cycle on a T-s
diagram and determine,
a. the net power developed,
b. the thermal efficiency of the cycle,
c. the mass flow rate of cooling water through the
condenser.
8.47
Water is the working fluid of a Rankine power cycle
shown in Figure 8.69. Turbine is non-adiabatic and
loses heat at a rate of 2500kW. Assume no pressure
and heat losses on the connecting pipes, and sketch
accurately T-s diagram of the cycle. Determine,
a. the net power developed by the cycle,
b. the thermal efficiency of the cycle,
c. the mass flow rate of cooling water through the
condenser.
8.45
Water is the working fluid of a Rankine power cycle
and leaves the boiler at 90 bar, 480°C. Due to pressure
loss and heat transfer effects in the pipe-line between
the boiler and the turbine, the state of steam at the
turbine inlet becomes 85bar, 440°C. At the exit of
adiabatic turbine with isentropic efficiency of 85%,
the pressure is reduced to 20 kPa, and water leaves
the condenser 16 kPa, 40°C. The feed water pump
with 80% isentropic efficiency increases the water
pressure to 95 bar before entering the boiler. For
water mass flow rate of 16.5 kg/s, calculate,
a. the net power developed,
b. the thermal efficiency of the cycle,
c. the rate of heat loss at the pipe line connecting
the boiler and the turbine.
8.46
Figure 8.69
a. the heat capacity of the boiler,
As shown in Figure 8.68, a conventional Rankine
power cycle with water as the working fluid has a
turbine and a pump operating with identical isentropic
efficiencies as t   p  0.85 . Steam enters the turbine
at 80 bar, 520°C, and exits to condenser pressure of
6 kPa. The heating capacity of the steam generator
is 500 MW. In the condenser, the cooling water is
available at 17°C, and we wish to limit the temperature
increase of water by 10°C. Determine,
b. the power output of the turbine,
c. the thermal efficiency of the cycle,
d. the thermal efficiency of Carnot cycle operating
between the same temperature limits.
e. The temperature rise of cooling water through
the condenser is designed to be 12oC. Calculate
the mass flow rate of cooling water.
f.
Evaluate the rate of entropy production for the
turbine, condenser, pump, and the connecting
pipes.
g. Using the results of part (f), place components in
rank order by beginning with the one contributing
the most inefficiencies.
8.48
Figure 8.68 A conventional Rankine cycle
A simple Rankine cycle operates with a condenser
pressure of 30 kPa, and the temperature of steam
at the turbine inlet is 500°C. To prevent corrosion
of turbine blades at last stages by liquid water, the
boiler pressure is set so that water at the turbine exit
is saturated vapor at the condenser pressure.
CHAPTER 8 POWER PRODUCING SYSTEMS 391
a. For isentropic turbine and pump, evaluate the
boiler pressure, the net specific work output,
and the thermal efficiency of the cycle.
a. the vapor temperature at the inlet of each turbine,
b. the thermal efficiency of the cycle.
b. Repeat the same problem for a turbine with an
isentropic efficiency of 88%. Assume the pump
to be isentropic.
Note: The solution in b requires iteration. First guess
the boiler pressure, then evaluate the conditions at
the turbine exit. If you get saturated vapor at the
turbine exit, your guess is correct. Otherwise, you
should refine your guess.
8.49
8.51
a. the vapor pressure at the reheater stage and at
the boiler stage,
As an alternative energy source, a solar powered
Rankine cycle could run on R-134a as a working fluid.
Turbine takes saturated vapor at 80°C. The turbine
and the pump have 80% isentropic efficiencies.
a. In Summer, air cooled condenser operates at
37oC, and solar energy is available at 800W/
m2. Determine solar collector area required to
generate 2 kW of net power.
b. the specific net work output,
c. the specific total heat input, and the thermal
efficiency of the cycle.
8.52
b. In Winter, the air cooled condenser operates at
10oC, and the solar energy available reduces to
500 W/m2. Determine the percent change in the
solar collector area for producing 2 kW of net
power.
8.50
Consider a steam power plant that operates on the
ideal reheat cycle as shown in Figure 8.70. The plant
maintains the boiler at 30 bar, and the reheat section
is at 7 bar, and the condenser is at 10 kPa. The vapor
quality at the exit of each turbine is required to be
90-percent. Determine,
Figure 8.70 A typical reheat Rankine cycle
A reheater is added to the Rankine cycle of Problem
8.48. The condenser pressure is at 40 kPa as in the
problem 8.48, and the temperature of steam at the
inlet of each high and low pressure turbine is 550°C.
Both turbines are isentropic, and water at the exit of
each turbine is saturated vapor. Determine,
Steam at 320 bar, 520°C enters the first stage of a
supercritical reheat cycle plant having two turbine
stages. The steam exiting the first stage turbine at
40 bar is reheated at constant pressure to 520°C.
The isentropic efficiency of each turbine stage and
the pump is 82%. Steam at a pressure of 6 kPa exits
the second stage and enters the condenser. The net
power developed by the plant is 120 MW. Draw the
T-s diagram of the cycle, and determine,
a. the total heat rate transferred to the working
fluid by the steam generator,
b. the thermal efficiency of the cycle.
8.53
Water is the working fluid of an ideal Rankine cycle
with reheat. Steam at 100bar, 600°C enters the first
stage of the turbine and is reheated to 600°C before
entering the second stage. The condenser pressure is
at 10kPa. Plot the variation of thermal efficiency of
the cycle with respect to reheat pressure for pressures
of 5bar, 10bar, 20bar, 30bar, 40bar, and 50bar.
8.54
In a regenerative Rankine cycle, steam leaves the
boiler at a pressure of 15 bar, and temperature of
400°C, and the condenser pressure is 50 kPa. An
open feed water heater (FWH) is used in the cycle
operates at a pressure of 3 bar. The turbine and the
pumps can be considered as isentropic, and water
is saturated at the exit of FWH. Determine,
392
THERMODYNAMICS
a. the fraction of extraction steam flow,
b. the specific net work output,
c. the thermal efficiency of the cycle.
8.55
The power plant in Figure 8.71 has a condenser
temperature of 45°C, and the maximum pressure
and temperature respectively are 60 bar, 800°C.
Extraction steam at 10 bar is mixed with feed water
in FWH such that the exit of FWH is saturated liquid into the second pump. The turbine and pumps
work with isentropic efficiencies of 88% and 75%
respectively. Draw the T-s diagram of the system
accurately, and determine,
a. the fraction of extraction steam flow,
Figure 8.72 A regenerative Rankine cycle
with closed FWH
b. the mass flow rate of steam for 1 MW of power
output,
c. the power required by each pump.
8.57
Reconsider the regenerative Rankine cycle in Problem
8.53, the cycle operating conditions are exactly the
same as given in the problem. However, the steam
extraction pressure for open feed water heater varies
in the range of 2 bar and 10 bar with increments of
2 bar. The turbine and the pumps can be considered
isentropic, and water is saturated at the exit of FWH
at all cases. Plot the variation of,
a. the net specific work,
b. the thermal efficiency of the cycle, as a function
of steam extraction pressure.
Figure 8.71 A regenerative Rankine cycle
with open FWH
8.56
As shown in Figure 8.72, a regenerative Rankine
cycle with closed feed water heater operates between
the pressure limits of 160 bar in the boiler, and 1bar
in the condenser. Steam flow rate through the boiler
is 25 kg/s, and the temperature at the turbine inlet
is 793K. Steam is extracted from the turbine at 40
bar, and is condensed to a saturated liquid to heat
the liquid water coming from the feed water pump
at the boiler pressure. Assume that the FWH exhibits
ideal behavior, and the turbine, and pump operate
isentropically.
a. Determine the heat added in the boiler, and
compare this result with the heat addition to the
same system without a FWH.
b. Calculate the power of the turbine, and the
thermal efficiency of the cycle.
8.58
A cogeneration power plant operates on ideal Rankine cycle using reheater and regenerative feed water
heater. Steam is supplied to HPT at 100 bar, 600°C,
the condenser pressure is 0.1 bar. At 30 bar of exit
pressure, some steam is extracted from the turbine
for closed feed water heater, and is completely condensed at the heater exit. Then, as shown in Figure
8.73, the condensed water is pumped to 100 bar
before mixing with feed water of the main line. The
remaining steam is reheated to 30 bar, 600°C before
entering to LPT. A process heater with 4.32  107
kJ/h capacity is supplied with extracted steam flowing
through LPT at 5bar. The return of process heater is
condensed water at the same pressure, and then is
pumped back to the main line. The plant is designed
so that the LPT turbine produces 20 MW of power.
Show the cycle on T-s diagram and determine,
CHAPTER 8 POWER PRODUCING SYSTEMS 393
b. Reanalyze the whole system for the same operating
conditions and constraints but the air temperature
at the compressor inlet has increased from 290K
to 310K. Determine the drop in power output for
the same mass flow rate of air as calculated in
a. Assume the same isentropic efficiencies and
regenerative effectiveness.
a. the fraction of steam extracted for feed water
heater,
b. the net power output of the system,
c. the thermal efficiency of the cycle.
Figure 8.73 A schematic of Rankine cycle
cogeneration plant
8.59
A proposed combined cycle power plant is shown
in Figure 8.74. A gas turbine engine takes air at a
state of100 kPa, 290K and the gas temperature at the
turbine inlet is 1500K. The gas expands to a temperature of 780K in the turbine. At the turbine exit,
the gas flow splits and enters two separate counter
flow heat exchangers. The gas is cooled down to the
same temperature of 470K by water in each of the
heat exchangers ( T5  T6  470 K ). Then the gas is
exhausted through the stack.
Steam exiting the boiler is at 90 bar, 500°C. The
high pressure turbine (HPT) exhausts to a pressure
of 10 bar, and then steam is reheated to 400°C. The
exit of low pressure turbine (LPT) is at 10 kPa. The
cooling water through the condenser experiences a
maximum temperature rise of 15°C. All the turbines,
compressors, and pumps of this combined cycle have
the same isentropic efficiency of 87%. All the heat
exchangers operate with effectiveness of 77%, and
no pressure drop occurs on the connecting pipes. The
electric generators are 100% efficient. The desired
output power of the gas turbine cycle generator is
40 MW.
a. Determine the pressure ratio in the gas turbine
cycle, the net power of the combined cycle,
the overall thermal efficiency of the system,
and the cooling water flow rate through the
condenser.
Figure 8.74 A schematic of gas turbine based
cogeneration plant
8.60
In a combined gas-steam power plant as shown in
Figure 8.51, air enters the compressor of 85% isentropic efficiency at 1bar, 21°C, and is compressed to
14 bar. The rate of heat addition in the combustion
chamber is 50 MW, and then air enters a turbine of
87% efficiency at 1250°C. The turbine exhaust passes
through a heat exchanger which generates steam for
the Rankine cycle. The air exits the heat exchanger
at 200°C, 1bar, and steam is produced at 120 bar,
450°C. The condenser pressure for the system is 0.1
bar. The isentropic efficiencies of Rankine turbine
and the pump can be considered to be the same at
82%. Determine,
a. the power delivered by each of the turbines,
b. the overall efficiency of the plant.
8.61
As shown in Figure 8.75, the output of the cogeneration plant on the campus of Dartmouth College
is divided as follows: 30% electricity is produced,
40% of the output energy is used for heating, and
30% of energy is waste heat and lost to the outdoors
by fumes through the chimney.
394
THERMODYNAMICS
a. Determine the cost of heating in December that can be met by this form of cogeneration plant if the electricity
consumption of the campus is 2500 kWh for that month. Assume that gasoline yields 25 MJ of heat energy per
liter and the cost of gasoline at the pump is $0.87 per liter.
b. Compare also the cost of purchasing the electricity from the grid at $0.17 per kWh with the results of (a).
Figure 8.75 A schematic of cogeneration plant of Dartmouth College
8.62
A proposal is made to use a geothermal supply of hot water at 15 bar, 170°C to operate a steam turbine. Water is first
throttled into a flash chamber in which liquid and vapor phases are separated at 5bar pressure. The liquid is discarded
while saturated vapor flows into the turbine and exits at 8 kPa of pressure. The turbine isentropic efficiency is 85%,
and water at the turbine exit is reinjected back to the source. Draw the system schematically, and determine the mass
flow rate of geothermal hot water required for producing 1 MW of power at the turbine shaft.
Figure 8.76 Geothermal hot water application of ORC with pentane
8.63
The geothermal supply of hot water in problem 8.62 is to be utilized by running a Rankine cycle with isopentane
as the working fluid. As shown in Figure 8.76, pentane enters the turbine as saturated vapor at 1000 kPa, and exits
at 100kPa. Pentane is then condensed at 100 kPa and leaves the condenser as saturated liquid at 28°C. Assume that
both the pump and the turbine share the same isentropic efficiency of 82%, and the data of pentane at the specified
states are given by the table below.
CHAPTER 8 POWER PRODUCING SYSTEMS 395
State
no.
Location
Pressure
(kPa)
Temperature
(°C)
Specific vol.
(m3/kg)
Entropy
(kJ/kgK)
Enthalpy
(kJ/kg)
1
Turbine inlet
1000
115.5
0.0366
4.92
438.7
2s
Isentropic turbine exit
100
55
0.36
4.92
351
3
Condenser exit
100
28
0.00162
0.0
0.0
4
Pump exit
1000
-
-
-
-
Determine,
a. the mass flow rate of geothermal hot water for producing 1 MW of power at the turbine shaft,
b. the thermal efficiency of the cycle,
c. the mass flow rate of air flowing through the condenser, if cooling air inlets the condenser at 15oC and experiences 10°C of temperature rise,
d. the frontal area of the condenser if the air velocity at the inlet is 20m/s.
True and False
8.64
Answer the following questions with T for true and
F for false.
a.
In an ideal Otto cycle, cycle efficiency
depends on the temperature ratio during compression.
b.
Engine knock takes place when the combustion of air-fuel mixture is out of control.
c.
d.
e.
As a car gets older, the moving parts
become worn, its compression ratio changes,
and the mean effective pressure increases.
Without the knocking problem, the
thermal efficiency of Otto engine is greater
than the efficiency of Diesel engine at the same
compression ratio.
Otto engine in comparison to Diesel
engine is always preferable in producing a large
amount of power.
f.
In an ideal Diesel engine, all of the
processes are internally reversible.
g.
In an ideal Diesel cycle, cycle efficiency
depends on the compression ratio only.
h.
Dual cycle is a compromise between
the Otto and Diesel cycles.
i.
Since the combustor of a gas turbine
engine is an open system, combustion takes
place at constant pressure.
j.
Even if a gas undergoes a constant pressure cooling in the exhaust outside the engine,
it is still within the system boundary.
k.
The thermal efficiency of a Brayton
cycle increases as the gas temperature at the
turbine inlet increases.
l.
In a Brayton cycle, work done by the
turbine is always equal to the sum of the work
consumed by the compressor and the work
output.
m.
The change in kinetic and potential
energies is usually considered insignificant for
combustion turbines.
n.
The pressure ratio of gas turbine engine depends upon the number of stages of the
turbine.
396
THERMODYNAMICS
o.
Check Test 8
To allow higher gas turbine inlet temperatures, steam is used for cooling the blades.
Choose the correct answer:
p.
The mass flow rate of gases through the
turbine is greater than that through the compressor. The difference represents the air leaking
through the turbine casing.
q.
Brayton cycle efficiency is unaffected
by the climatic conditions.
r.
The climatic conditions may appreciably
change Rankine cycle efficiency.
s.
t.
A Rankine cycle comprises two constant pressure processes, and two isothermal
processes.
1.
2.
A regenerative Rankine cycle thermal
efficiency is greater than simple Rankine cycle
efficiency only when steam is extracted at a
particular pressure.
u.
In a regenerative Rankine cycle, thermal
efficiency increases with the increase of number
of feed water heaters.
v.
With the increase of pressure ratio, the
thermal efficiency of a simple gas turbine with
fixed turbine inlet temperature first increases
then decreases.
In an ideal Otto cycle, gas temperature
after compression is higher than after expansion.
x.
In a two stage gas turbine engine with
intercooling and reheating, work ratio improves
but thermal efficiency decreases.
a. 0.45, 620,
b. 0.35, 820,
c. 0.35, 620,
d. 0.45, 820.
The conditions at the beginning of compression in an air standard Diesel cycle are given as
p1  2 bar , T1  380 K . The compression ratio is 19
and the specific heat addition is 850 kJ/kg. Assume
variable specific heats, and then the cutoff ratio and
the net work output of the cycle in kJ/kg respectively
are,
a. 1.85, 650,
c. 1.65, 650,
3.
b. 1.85, 500,
d. 1.65, 500.
Air in an ideal Diesel cycle is compressed from
3.2L to 0.20 L, and then expands to 0.4 L during
the constant pressure heat addition process. Under
air standard conditions with k = 1.4, the thermal
efficiency of the cycle is,
a. 0.58,
c. 0.66,
4.
w.
In an actual internal combustion engine the compression and expansion processes are not isentropic, and
there is always some heat transfer from the cylinder
gas to the cylinder wall. Consider a modification of
the air-standard Otto cycle in which the isentropic
compression and expansion processes having n = 1.28.
The compression ratio is 8.5 for the modified cycle,
and the condition at the beginning of compression
is p1  1 bar , T1  300 K . Assume constant specific
heats, and then the thermal efficiency and the mean
effective pressure in (kPa) for a maximum cycle
temperature of 2000K respectively are:
b. 0.62,
d. 0.70.
An air-standard dual cycle has a compression ratio of
15 and a cutoff ratio of 1.2. At the start of compression process, p1  0.92 bar , T1  290 K . The pressure
increases by a factor of 2.5 during the constant volume heat addition process. Assume variable specific
heats, and for 0.01 kg of air in the cylinder, the total
amount of heat added in kJ per cycle, and the thermal
efficiency of the cycle respectively are,
a. 15.6, 0.58,
c. 19.6, 0.72,
b. 5.6, 0.65,
d. 10.6, 0.79.
CHAPTER 8 POWER PRODUCING SYSTEMS 397
5.
The primary reason for replacing the exhaust process
of Otto and Diesel cycles with a constant volume
process is:
a.
b.
c.
d.
6.
no heat transfer of the actual exhaust process,
no work transfer of the actual exhaust process,
restoring air to its initial conditions,
satisfying the second law of thermodynamics
for the cycle.
In a gas turbine power plant, air enters the compressor at 12°C, 1.05bar, and is compressed through a
pressure ratio of 5:1. Air flowing through a regenerator with 75% of effectiveness and then through the
combustor, it is heated to 910°C while its pressure
drops by 0.2bar. After expanding in the turbine, the
regenerator cools the air. Due to flow of air through
the regenerator, again a pressure drop of 0.2bar takes
place. The isentropic efficiency of the compressor and
the turbine respectively are 0.80, 0.85. For variable
specific heats, the thermal efficiency of the plant
becomes,
a. 45%,
c. 35%,
7.
10.
An ideal Stirling engine operates with 0.01kg of
carbon dioxide as the working fluid between temperatures of 1027°C and 27°C. The highest and the
lowest pressures during the cycle are 20bar and
1.5bar respectively. If the engine runs at a speed of
200rpm, then the power developed in kW is:
a. 3.5,
b. 4.5,
c. 5.5,
d. 6.5.
A simple ideal Rankine cycle operates between the
pressure limits of 4MPa, and 70kPa with 600°C of
steam temperature at the turbine inlet. If the pump
work is disregarded, then the thermal efficiency of
the cycle becomes,
a. 20%,
b. 24%,
c. 28%,
d. 32%.
b. 40%,
d. 30%.
A turbojet engine inlets 50kg/s of air at 0.35bar,
-45°C, while travelling with a velocity of 200m/s.
The pressure ratio of the compressor is 9:1. Assuming that the compressor and turbine operating with
85% isentropic efficiency, and the nozzle is 95%
efficient, if the hot gases enter the turbine at 1100°C,
the power developed in MW by the engine is:
a. 4.5,
c. 6.5,
8.
9.
b. 5.5,
d. 7.5.
An simple Brayton cycle is modified to incorporate
multi-stage compression and expansion with intercooling and reheating. As a result of this modification,
11.
Consider a 500MW ideal reheat Rankine cycle where
steam enters the high pressure turbine (HPT) at 50bar,
500°C, and expands to 10bar. It is then reheated to
500°C before entering the low pressure turbine (LPT)
and expands to 0.1bar condenser pressure. The mass
flow rate of water through the boiler in kg/s, the
thermal efficiency, and percent moisture content in
steam entering the condenser respectively are,
a. 312, 35%, 15
b. 322, 40%, 5,
c. 332, 40%, 5,
d. 322, 45%, 15.
a. compressor work: increases
turbine work:
increases
efficiency:
increases
b. compressor work: increases
turbine work:
increases
efficiency:
increases
c. compressor work: decreases
turbine work:
increases
efficiency:
increases
d. compressor work: increases
turbine work:
increases
efficiency:
increases
Figure 8.77 Schematic of reheat
Rankine cycle
398
12.
THERMODYNAMICS
The demand for energy from an industrial plant at
steady state conditions is 50MW of process heat
at 120°C, and a maximum of 40MW of power to
drive the electrical generators. Steam is available
from the boiler at 60bar, 550°C, and the condenser
pressure is 0.05bar. The process heat is extracted
from the turbine at an appropriate pressure, and
the condensed steam returns to the feed water line
at 120°C. Neglect the feed pump work, assume a
linear relationship between enthalpy and entropy,
and take the isentropic efficiency of the turbine as
95%. Then the steam flow rate in kg/s through the
boiler becomes:
a. 30.6,
c. 46.6,
14.
13.
A steam plant uses a simple ideal Rankine cycle
with one regenerative feed water heater. The boiler
produces steam at 80bar 500°C. For a condenser
pressure of 0.1bar, the pressure (kPa) at which steam
has to be extracted, and the efficiency of the plant
respectively are,
a. 600, 36%,
c. 800, 48%,
b. 800, 42%,
d. 600, 54%.
b. 38.6,
d. 54.6.
Consider the ammonia Rankine cycle power plant as in Figure 8.78. The plant is designed to operate in a location where
geothermal hot water at 170°C is available. Ammonia at the turbine inlet is at 20bar, 120°C, and expands to 12bar of
condenser pressure. Both the turbine and the pump share the same isentropic efficiency of 90%, and the geothermal
water exits the vaporizer at 80°C. If you assume that the specific heat of geothermal water is c p  4.2 kJ/kgK , then
for 5MW of net power output, the mass flow rate from the geothermal well in tons per hour is:
a. 832,
b. 932,
c. 732,
Figure 8.78 Energy production by ammonia Rankine cycle
d. 1032.
C
H
9
A
P
T
E
R
Refrigeration Systems
9.1
General Considerations
Refrigeration is part of our life and without it modern life would be impossible. The
need for refrigeration arises from the fact that keeping a space at its low temperature requires withdrawing the amount of heat leaking into the space through a higher temperature
environment. Because of this fact, refrigeration systems are used for food preservation and
processing, for commercial refrigeration (supermarkets, cold stores, and refrigerated transport), industrial refrigeration (chemical processing, petrochemicals etc.) and for domestics
needs (air conditioning, domestic refrigerator etc.).
Refrigeration is a cyclic process by which the transfer of heat from a low temperature
level at the heat source to a high temperature level at the heat sink is accomplished by work
type of energy consumption. Hence for refrigeration cycles, we may state that  W  0 .

Figure 9.1 Heat transfer loops of a commercial refrigeration system
399
400
THERMODYNAMICS
As shown in Figure 9.1, for a typical cold store in supermarkets, the thermal energy moves from
left to right through five loops of heat transfer as it is extracted from the cold space and expelled into
the outdoors. In the leftmost loop of Figure 9.1, (loop1) indoor air of a conditioned room is driven
by a supply air fan through a cooling coil, and (loop2) its heat is transferred to chilled water. Water
flows through the evaporator of the refrigeration system and transfers the heat to condenser water by
the refrigeration cycle (loop3). Water absorbs the heat from the condenser and the heated water is
pumped to the cooling tower (loop4). The fan of the cooling tower derives air across the hot condenser
water, and thus the heat is transferred to the outdoors (loop5).
Depending upon the refrigeration application, the required temperature level of the refrigerated
space might change. Table 9.1 provides typical refrigeration temperatures and the corresponding
areas of application.
Table 9.1 Temperature levels for typical refrigeration applications
Refrigeration temperature(K)
Area of application
73-200
Liquefaction of gases
200
Dry ice (CO2), Freeze drying (e.g. coffee)
255
Freezers, cold stores (commercial refrigeration)
272-278
Food storage (Domestic refrigerators), cold stores,
mine cooling (industrial refrigeration)
291-293
Air conditioning and heat pumps (residential
refrigeration)
To condition the air or a product to the above indicated temperatures, various refrigeration methods
are devised and are in use in practice. Figure 9.2 shows the classification of refrigeration methods that
are most widely used in domestic and industrial applications.
Figure 9.2 Methods of refrigeration for domestic and industrial applications
In selection of a proper refrigeration method, not only the required refrigeration capacity but also
the temperature level has to be considered. Besides, the application environment is also a factor. Even
though more than one refrigeration method may be suitable for a given application, the selection has
to be tested by factors like cost, reliability, size, and unit power.
CHAPTER 9 REFRIGERATION SYSTEMS 401
Among these cooling methods indicated in Figure 9.2, the vapor compression refrigeration cycle
is the most widely used refrigeration cycle in practice. Since the refrigeration capacities are in the
range of 50W and 1MW, and the evaporator temperatures may vary between +5C and -70C, these
systems are suitable for many applications in commercial, domestic, and air conditioning sectors.
9.2
Vapor Compression Refrigeration Cycle
Let us consider a pressurized tank containing liquid refrigerant (R134a) at 5bar, 10C. If the lid of
the tank is instantly removed, then the pressure on refrigerant will drop to 1bar atmospheric pressure.
Since the refrigerant cannot anymore maintain its liquid state at this condition of 1bar, evaporation starts.
The required energy for evaporation is extracted from its environment. The working principle of a Vapor
Compression Refrigeration Cycle (VCRC) is much similar to this and to get pressurized liquid we use
compressors, and instant pressure reduction is provided by expansion valves.
In accord with the definition of an ideal cycle for an ideal VCRC, the compressor is isentropic,
there is no frictional pressure drops through the cycle, and pressures both at the condenser and at
the evaporator are constant. Stray heat losses are ignored, and change in kinetic or potential energy
through a device is negligible. Hence, as shown in Figure 9.3, the four processes of ideal VCRC are:
(1-2) Isentropic compression in a compressor, wcs  h2 s  h1 , refrigerant enters the compressor at
saturated vapor state, (2-3)
Figure 9.3 Ideal refrigeration cycle and (T-s), (p-h) representations
Constant pressure heat rejection in a condenser, qcon  h2 s  h3 , refrigerant at the condenser outlet is at saturated liquid state, (3-4) Throttling in an expansion device, h3  h4 , and (4-1) Constant
pressure heat absorption in an evaporator, qev  h1  h4 . Then the coefficient of performance of ideal
VCRC is,
q
h h
COPideal  ev  1 4
(9.1)
wcs h2 s  h1
In VCRC, there are two pressure zones, the low pressure zone starts at the exit of the expansion
valve and ends at the inlet of the compressor (p4=p1), and the high pressure zone lies between the
exit of compressor and the inlet of the expansion valve (p2=p3). The transfer of heat at the refrigerated
space and at the condenser may be maintained by respective conditions of Tev  Ti , and T0  Tcon .
402
THERMODYNAMICS
Figure 9.4 An actual vapor compression refrigeration cycle
An actual vapor compression refrigeration cycle is depicted in Figure 9.4. In this particular cycle,
the cold space and the surroundings are respectively maintained at Ti=+2C, and T0=+23C, and the
working fluid is R134a. Due to losses, the pressures at the evaporator exit and the compressor inlet
are not the same. In Figure 9.4, similar behavior can be traced between the compressor exit and the
expansion valve inlet. In addition, due to heat transfer, the refrigerant at the compressor inlet is superheated, and is sub-cooled at the expansion valve inlet.
Figure 9.5 Difference between actual and ideal VCRC
CHAPTER 9 REFRIGERATION SYSTEMS 403
Figure 9.5 presents deviations between actual and ideal refrigeration cycles. In numbering the
actual states of the refrigerant, the actual vapor compression cycle in Figure 9.4 is considered. Then
the main deviations are as follows: 1. (a) type difference takes place due to heat transfer at the
connection pipes, 2. (b) type difference is the pressure drop in regions evaporator-compressor and
compressor-condenser, and 3. (c) type difference is the deviation of compression process from isentropic compression. As a result, the actual work consumed and the actual COP of the refrigeration
plant may be calculated as,
wc  h2 a  h1a 
h2 s  h1a
c
COPactual 
qev h1  h4 a

wc h2 a  h1a
(9.2)
Example 9.1 For an ideal refrigeration cycle using R134a, the evaporating and the condensing temperatures respectively
are -10C, and +34C. If the refrigeration capacity is 80 kW, determine,
a.
the volume flow rate of the refrigerant at the inlet of the compressor,
b.
the power required by the compressor,
c.
the COP of the cycle.
Suppose that the actual cycle is as depicted in Figure 9.4, repeat questions a, b, and c for the actual cycle and evaluate
the isentropic efficiency of the compressor.
Solution:
Ideal cycle:
a.
Energy equation for the evaporator is Q ev  m h1  h4  and m 
80
 1,
 0.555kg/s additionally V1  mv
241.3  97.31
V1  0.555  0.099 or V1  3.296 m3 / min .
b.
Wcs  m h2 s  h1  , s1  s2 s  0.925kJ/kgK and h2 s  271.52kJ/kg , Wccs  0.555
c.
the COP of the cycle by Eq. (9.1) is, COPiideal 
271.52
241.3 16.77 kW
80
 4.77
16.77
Actual cycle:
a.
Employing similar analysis, Q ev  m h1  h3a  , m 
V1  0.505 0.099 60
80
 0.505kg/s and
241.3  82.9
2.969m3 / min
b.
Respect to Eq. (9.2), the compressor power is Wc  0.505  294  253  20.705 kW
c.
The cycle COP on the other hand is COPaactual 
d.
80
 3.86 .
20.705
W
The isentropic efficiency of the compressor is c  cs . However, the conditions at the inlet of the actual case
Wc
differ from the ideal one. Thus Wcs  m h2 sa  h1a   0.505  288.1  253  17.725 kW and the compressor efficiency is c 
17.725
 0.85 .
20.705
Example 9.2 Some refrigeration systems use a liquid-to-suction heat exchanger to increase the COP of the system. As
shown in Figure 9.6, this may positively affect the refrigeration capacity but also increases the work consumed by the compressor. Thus, as an overall result, the heat exchanger may provide a negligible advantage. Consider an ideal VCRC with
R22 as the working fluid. The heat exchanger warms up the saturated vapor at -10C coming from evaporator from -10C
to 0C with liquid coming from condenser at +32C. Determine,
a.
the COP without the heat exchanger on the system,
b.
the COP with the heat exchanger on the system.
404
THERMODYNAMICS
Solution:
Cycle without heat exchanger:
Referring to (h-p) diagram, the evaporator specific cooling rate is qev  h1  h4   246.15  84.14  162.01 kJ/kg . Since
the cycle is ideal, s1  s2  0.9424 kJ/kgK and wc  h2  h1  277.5  246.15   31.35 kJ/kg . Then the performance of the
cycle is COP1 
162.01
 5.16
31.35
Cycle with heat exchanger:
The enthalpy at state 7 may be calculated by energy balance on the exchanger as, m h3  h7   m h5  h1 .
As shown in the figure, the refrigerant has the same mass flow rate at both sides of the exchanger.
Thus, h7  84.14  253.21  246.15   77.08 kJ/kg . Referring to (h-p) diagram in Figure 9.6,
qev  h1  h8   246.15  77.08  169.07 kJ/kg , and s5  s6  0.97 kJ/kgK . The work consumed
by the compressor, wc  h6  h5  285.82  253.21  32.615 kJ/kg . Then the performance of the
169.07
cycle is COP2  32.615  5.18 . Percent increase in COP is COP %
5.18 5.16
100%
5.16
0.38%
and is less than 1-percent.
Refrigeration systems absorb especially in summer about 20-percent of electrical generation
capacity of the United States. Considering the huge amount of energy consumed, these systems are
definitely part of our life and are essential for our modern way of living. Without it, a healthy distribution of food to urban areas or expecting an efficient performance on personnel work on a hot summer
day would not be possible.
9.3
Multi-Pressure Refrigeration
A multi pressure system is distinguished from the single pressure system by having more than
one low side pressure. For instance, a refrigeration system used in a diary, one evaporator serves at
-30C to harden the produced ice cream and another operates at -1C to provide the cooling load of
bottled milk storage room. Even though a multi-pressure system may have several low pressure sides,
in this section multi-pressure systems with only two low-side pressures will be studied and several
combinations of evaporators and compressors are analyzed as follows.
CHAPTER 9 REFRIGERATION SYSTEMS 405
9.3.1
Two-Evaporator and One Compressor Systems
In many industrial applications one compressor serves two or more evaporators operating at different temperatures. In Figure 9.7a, a typical application of one compressor and two evaporator system
is illustrated. One evaporator serves to the Heating, Ventilating, and Air Conditioning (HVAC)
system of the installation and operates at -5C. The other evaporator is for frozen food storage unit
and operates at -20C. As shown in Figure 9.7a, since the two evaporators uses the same refrigerant
but operates at different pressures, a pressure reducing valve after the high temperature evaporator is
provided for maintaining -5C evaporation in HVAC system. The (h-p) presentation of the system is
given in Figure 9.7b. The COP of such systems is calculated by the following equation.
COPsys 
Q ev1  Q ev2
Wc
(9.3)
Example 9.3 A refrigeration system as depicted in Figure 9.7 uses R22 as the working fluid and the low temperature evaporator with a capacity of 5ton operates at Tev2=-20C. The higher temperature evaporator having a capacity of 3tons operates
at -5C. The cycle is ideal and the condenser pressure is 16bars. Determine,
a.
the mass flow rate of refrigerant through each evaporator, and the compressor,
b.
the compressor power input,
c.
the condenser capacity.
Solution:
a.
Refrigerant leaves both evaporators as saturated vapor, and the condenser as saturated liquid. Since,
1 ton refrigeration=3.516kW , then the refrigeration capacities of both evaporators are Q ev1  10.548 kW
and Q ev 2 =17.58 kW . Considering the energy equation for both evaporators and Table A6 for R22,
m 1 =
b.
10.548
248.16 96.83
0.069kg/s , m 2 =
17.58
242.09 96.83
0.121kg/s , m =m 1 +m 2 =0.069+0.121=0.19kg/s
First, the enthalpy of the refrigerant at the compressor inlet has to be evaluated. For an ideal mixing with no heat loss,
 1 thus h1  245.21 kJ/kg . For s1  s2  0.97 kJ/kgK , and p2  16bar , the enthalpy at the compressor
m 1h8  m 2 h7  mh
exit is h2  295.0 kJ/kg . Then the compressor power becomes Wc =m h2  h1   0.19  295  245.21  9.46 kW .
c.
One way of determining the condenser capacity is Q con =m h2  h3   0.19  295  96.83  37.65 kW . The other
alternative way is Q con  Q ev1  Q ev2  Wc  10.548  17.58  9.46  37.588 kW . The difference between these two
results is due to table reading error.
406
THERMODYNAMICS
9.3.2
Two-Compressor and One Evaporator Systems
If an evaporator operates at a temperatures as low as -30C or less, then two-stage compression
with inter-cooling is an ideal method to employ. As explained in the following example, the system
with two stage compression requires less power than a single compressor and savings in energy will
justify the cost of extra equipment. Besides reducing the cost of refrigeration, by multi-stage compression with inter-cooling, the discharge temperature of refrigerant at high stage compressor will
be reduced. The lower discharge temperature on the other hand permits better lubrication and longer
life of the compressor. As illustrated in Figure 9.8, inter-cooling is usually accomplished by letting
the liquid refrigerant from condenser into a flash chamber. The COP of such systems is calculated
by the following equation.
COPsys 
Q ev
Wc1  Wc 2
(9.4)
Example 9.4 a. Determine energy savings in kW-h per year by using the refrigeration system as depicted in Example
6.12 instead of a single stage system operating at the same conditions. Assume that both systems are ideal and operate 8400
hours a year. b. Evaluate percent increase in COP of the system
Solution:
a.
Recalling the data of Example 6.12, the refrigerant is R22, evaporator capacity is 200 kW, and works at Tev=-30C,
and the condenser pressure is 1500 kPa. The total power required is W  64.795 kW . If the system had been a
t
single-stage one with the same evaporator and condenser pressures, then the cycle would be as (19510) cycle on
(h-p) diagram of Figure 9.8. Since s9=s2=s1=0.9787 kJ/kgK, and p9=1500 kPa, Table A6 yields h9=292.97 kJ/kg.
qe
200

 1.386kg/s . Then the power required
Additionally, the refrigerant flow rate would be m 1 
h1  h10 237.7  93.5
for single stage compression is Wc  m h9  h1  or Wc  1.386  292.97  237.7   76.604 kW . Thus the annual
energy savings due to two-stage compression is, ES  76.604  64.795  8400 or ES  99,195.6kW-h/year
b.
q
200
The COP of the system with two-stage compression is COPts  ev 
 3.08 . Similarly for single stage
Wt 64.795
200
3.08  2.61
compression, COPs 
 2.61 . Thus the percent increase in COP becomes COP % 
 100 or
76.604
2.61
COP % 18% which is a considerable increase in COP.
CHAPTER 9 REFRIGERATION SYSTEMS 407
In decision making stage for the use of two-stage refrigeration with inter-cooling several factors has to be considered.
Respect to evaporator operating temperature, the boiling pressure of the refrigerant to be selected should not be below 1 bar
and the specific volume should not assume high values. These two are effective criteria in selection of suitable refrigerant.
In addition, due to sudden drop in compressor efficiency, reciprocating compressors are suitable for
stage system,
Tev  30 C if single
Tev  50 C if two-stage system and Tev  70 C if three stage systems are employed.
In applications such as liquefaction of petroleum vapors or manufacturing of dry ice for which
Tev  70 C , it becomes advantageous to separate the low pressure region from the high and built a
cascade system. This arrangement permits selecting a suitable refrigerant for the low pressure region.
As illustrated in Figure 9.9a, a cascade system is consists of two separate single stage refrigeration
system connected by a cascade condenser that operates like evaporator for the higher system and
condenser for the lower. In sizing the lower system compressor, the specific volume of the refrigerant
at the suction line plays an important role. For instance, at Tev  60 C , the specific volume of R22
is 0.537m3/kg, and of R23 is 0.073m3/kg, so R22 occupies 7.35 times the volume of R23 and is not
suitable for the low pressure region cycle.
Figure 9.9b illustrates (h-p) diagram of a cascade system which uses R22 for the higher pressure
system and R23 for the lower. The main disadvantage of a cascade system is the overlap of temperatures at the cascade condenser. This overlap is necessary for the heat transfer between the two cycles
but results with higher energy consumption due to internal irreversibilities caused by the temperature
difference.
It may be demonstrated that for two-stage Carnot Cascade system, the cascade condenser temperature at which COP of the system becomes maximum is determined as following.
Tcac  Tev  Tcon
(9.5)
where Tev and Tcon are the evaporator temperature of the low pressure and the condenser temperature of the high pressure cycles respectively.
Figure 9.9 A typical two-stage cascade refrigeration system and (h-p) presentation
408
THERMODYNAMICS
9.3.3
Two-Compressor and Two-Evaporator Systems
In food processing installations, the manufactured food is usually passed through a shock tunnel at
-40C for fast freeze and then stored at -20C. This type of installation requires multi-stage compression and evaporators operating at different temperatures. Similar examples may be given in chemical
industries, as often different evaporator temperatures are required in various sections of a refrigeration
plant. As shown in Figure 9.10, such industrial needs are usually solved by two-stage systems having evaporators, if necessary, at the intermediate pressure. As the liquid refrigerant expands through
the low pressure float valve, flash gas develops. The flash gas, in addition to refrigerant from high
temperature evaporator, is removed by the high stage compressor.
Figure 9.10 Two-evaporator and two-compressor system with flash chamber
and (h-p) representation
The COP of the system in Figure 9.10 is calculated by,
COPsys 
Q ev1  Q ev2
Wc1  Wc 2
(9.6)
Example 9.5 In a food processing unit, tuna fish first shocked at -32C and then is stored at -18C by the ammonia refrigeration system shown in Figure 9.10 for which the condenser operates at +40C. The cooling loads of the low and the high
pressure evaporators are estimated to be 250 kW and 150 kW respectively. Assuming isentropic compression, determine,
a.
the percent of evaporation of refrigerant entering the flash tank at 5, m 3 f / m 5l ,
b.
the mass flow rates of refrigerant compressed by the low pressure and by the high pressure compressors,
c.
the power consumed by each compressor,
d.
the COP of the system.
Solution:
a. The numbers in Figure 9.10 only represent the thermodynamic states of the refrigerant, but the mass flow rate at
the float valve is m 5l and the gas discharge from flash tank is m 3 f . Considering the energy balance around the
flash chamber,
m 5l h5  m 7 h2  m 3 f h3  m 7 h7 , and the mass balance around the chamber yields, m 7  m 5l  m 3 f . The enthalpy
values at states 2, 3, 5, and 7 are known. Thus,
m 3 f
m 5l

h2
h2
h5
h3
h7 1475.1 371.35 97.68

 0.625
h7 1475.1 1420.45 97.68
CHAPTER 9 REFRIGERATION SYSTEMS 409
b.
The mass flow rate through low pressure compressor is m 2 
for the high pressure compressor, m 4 
c.
d.
Q eevh
150

 0.1429kg/s .
h3 h5 1420
1420.45
45 371.35
The power of the low pressure compressor is Wcl
for the high pressure compressor, Wch
Q eevl
250

 0.191kg/s , similarly,
h1 h7 1400
1400.81
81 97.68
m 4 h4
m 2 h2
h1   00.191
1911475
1475.11 1400
1400.81
81 14.189 kW , similarly,
h3   00.1429
1429 1737
1737.22 1420
1420.45
45  45.268 kW
Applying Eq. (8.6) for the COP of the system as, COPssys 
250 150
 6.72
14.189 45.268
Comments: Recall that the system is assumed to an ideal cycle. For an actual cycle the COP will assume lower values.
9.4
Refrigeration System Components
As mentioned previously, in addition to control units, driers, and oil separators, typical refrigeration
system consists of the following components: 1.Compressors, 2.Expansion devices, 3.Evaporators,
4.Condensers, and 5.Refrigerants. For efficient operation of the system, a proper matching between
various components is essential. Prior to analyzing the characteristics of component interactions,
the design characteristics and the performance of each component have to be studied.
9.4.1
Refrigeration compressors
A compressor is the heart and the costliest component of the vapor compression refrigeration. To
keep the evaporator pressure at its low level, the compressor constantly draws the refrigerant from
evaporator, and raises the pressure to such level that the refrigerant condenses by rejecting heat to surroundings. As shown in Figure 9.11, the classification chart of compressors is based on two principles
which are the motor-compressor connection principle, and the working principle.
Medium and large capacity refrigeration systems usually use open type compressor which
is externally driven by an electric motor or an engine. The major drawback of this type of compressors is that refrigerant may leak through the seal at which crank shaft extends outside. As
shown in Figure 9.12, in hermetic compressors, the motor and the compressor are enclosed in
410
THERMODYNAMICS
the same housing and the possibility of refrigerant leakage is highly reduced. Some portion of
the refrigeration load of systems with hermetic compressors is used for cooling the motor and
thus the COP of these systems is lower than open compressor based systems. However, hermetic
compressors are usually preferred for small capacity systems like domestic refrigerators, water
chillers, and split type air conditioning units. In some hermetic units, the cylinder head of the
compressor may be removed for replacing the valves and pistons, this type of compressor is
called semi-hermetic compressor.
Reciprocating compressors. It is the most widely used compressor type in refrigeration industry
for cooling capacities in the range of 102 W to 103 kW. As illustrated in Figure 9.13, a reciprocating
compressor is a two-stroke machine. The intake and the discharge processes are completed in one
revolution of the crank shaft, and may be constructed as a single cylinder or multiple of cylinders
up to 16 cylinders.
In determining the actual mass rate of refrigerant flowing into the compressor cylinder, ideal
compressor has to be defined. As explained in Figure 9.14a, an ideal compressor is one in which;
1.There is no clearance volume, 2.No pressure drop during suction and compression period, 3.All
processes (suction, compression, and discharge) are reversible and adiabatic.
CHAPTER 9 REFRIGERATION SYSTEMS 411
In ideal reciprocating compressor, no gas is left in the cylinder at the end of discharge stroke and
suction starts as soon as the piston moves towards BDC. Referring to Eq. (6.15), the power consumed
by the compressor is,

k  pcon
Wcs   vdp  pb vb

k  1  pev
pev

pcon

k 1

 k

 1  h2 s  h1




(9.7)
As in Figure 9.14b, in actual compressors, a clearance volume, Vc, is left between the cylinder
head and the piston to tolerate the valve motion, and the thermal expansion. Hence, the clearance
ratio is defined as,

Vc
Vst
(9.8)
Depending upon the arrangements of valves and the piston speed,  assumes values between
4-percent and 10-percent. Due to clearance, however, some gas will be left in the cylinder at the end
of discharge stroke. This left gas expands as the piston moves towards BDC, and the suction starts
only when the pressure in the cylinder is less than the evaporator pressure. Hence, the volumetric
efficiency of the compressor becomes,
Vb  Va
V  Vc  Vex   vb 
vb
Actual mass intake
v 

 b
 
Mass in stroke volume Vb  Vc
Vb  Vc
 vb 
vb
(9.9)
412
THERMODYNAMICS
Since, Vex  Vc  pcon / pev  , Vst  Vb  Vc , Vb  Vst  Vc , and due to low pressure at the suction
1/ n
stroke, the ideal gas behavior yields, vb / vb  ( pb / pev ) Tev / Tb  . Accordingly, rearranging Eq. (9.9)
results as,

v  1     rp

1
n

b
 
ev
    pp
x
Tev 

Tb 
(9.10)
If the data of refrigerant at the compressor inlet are not known, the term in the right parenthesis
of the above expression may be assumed to be unity. Eq. (9.10) indicates that an increase in pressure
ratio, rp, or in clearance volume ratio, ε, decreases the volumetric efficiency. In fact the limiting value
of pressure ratio for which the volumetric efficiency becomes zero is,
 
rp
max
 1
 1  
 
n
(9.11)
The actual pressure ratio has to be much less than this limiting value. Besides, due to leaks through
cylinder wall, the amount of refrigerant leaving the compressor at the discharge is not the same as
sucked into the cylinder. Since Vleak  Vsuction  Vdischarge , the discharge volumetric efficiency at the
inlet conditions may be defined as,
vd 
Vdischarge
Vst

Vs  Vleak
 v  1   w 
Vst
(9.12)
Where (1   w ) represents percent leak through the wall. As a result, the mass flow rate of refrigerant at the compressor exit is,
m  vd
Vst vd  n   D 2

z 
H
vb
vb  60  4
(9.13)
where, n, is rotational speed (rpm), D and H are the bore and the stroke length of the cylinder, and
z is the number of cylinders. “ vb ” indicates the specific volume at the compressor inlet.
Example 9.6 A reciprocating R134a compressor having 4 cylinders with 100mm of bore, 100mm of stroke length, and
5-percent clearance volume ratio runs at a speed of 3600rpm, and has to be used for a refrigeration system operating between
-18C and +40C evaporator and condenser temperatures respectively. A pressure drop of 0.04bar occurs at the suction
line, and the refrigerant enters the compressor with +6C superheat. The wall leaks estimated to be 10-percent of the stroke
volume. For refrigerant polytrophic index n = 1.15,
a.
Determine the mass flow rate supplied by the compressor.
b.
Assuming saturated vapor at the evaporator exit, evaluate the evaporator capacity in tons of refrigeration.
Solution:
a.
Together with tabulated values of R134a, Eq. (9.10) yields,
1 

 10.16 1.15   1.44  0.04 255 

The discharge volumetric efficiency
v  1  0.05  0.05 


  0.738
  1.44
261 
 1.448 

 
.
CHAPTER 9 REFRIGERATION SYSTEMS 413
by Eq. (9.12) is vd  v  1   w   0.738  0.1  0.638 , and the specific volume at
(-12C, 1.4bar) is vb  0.143m3 /kg . Substituting these values into Eq. (9.13) results as
m 
b.
0.638
 3600  3.14 x0.12
 4
 0.1  0.84 kg/s

0.143
4
 60 
.
Since the refrigeration system is a single pressure system as in Figure 9.5, then the evaporator capacity is,
0.84  236.53  106.19 
Q ev  m h1  h4  
 31.14 ton
3.516
If one considers the flow losses that take place at the compressor valve region, there exists a strong
relation between the volumetric efficiency and the compressor isentropic efficiency. Depending upon
the compressor pressure ratio and the clearance volume, the variation of the ratio of these two efficiencies,   c / vd , is illustrated in Figure 9.15. Since the volumetric efficiency at the discharge
is evaluated by Eq. (9.12), then the compressor isentropic efficiency becomes,
c  vd
(9.14)
Especially in selection of lubrication oil for the compressor or determining the power need, the
outlet temperature of the refrigerant is required. In calculating the actual outlet temperature of the
gas, however, compressor efficiency is needed. The dependence of compressor efficiency on pressure
ratio is not the same as the volumetric efficiency. As presented in Figure 9.15, the upward trend of
curves is indicative of this difference.
Example 9.7 Consider the reciprocating R134a compressor of Example 9.6, and referring to the states indicated in Figure
9.5 evaluate,
414
a.
THERMODYNAMICS
compressor isentropic efficiency,
b.
refrigerant exit temperature,
c.
effective COPef of the system for compressor mechanical efficiency of ηm=0.88.
Solution:
a.
The state of refrigerant at the compressor inlet, p2  1.4bar, T2  12 C and pressure at the exit p3  10.16bar .
The pressure ratio and the clearance volume ratio respectively are rp  7.05,  =0.05 . Thus, by Figure 8.15,
  1.32 , and Eq. (9.14) yields, c  1.32 0.638 0.842
b.
Since compressor isentropic efficiency, respect to Figure 8.5, is defined as, c 
R134a yields, s2  s3s  0.953 kJ/kgK and h3s  283.3 kJ/kg . Hence, h3 
and R134a table yields T3
c.
h3s  h2
and tabulated values of
h3  h2
283.3  241.2
 241.2  291.2 kJ/kg
0.842
60 C
m h3s  h2  0.84  283.3  241.2 
The effective power of the compressor is Wef 

 47.72 kW and COPef of
0.88  0.842
 m  c
the system becomes COPeef 
31.14 3.516
 2.29 .
47.72
Screw compressors. Figure 9.16 illustrates the basic geometry of a screw compressor. The male rotor
is labeled as lobe and the female rotor as flute. As the male rotor turns in an outward direction, a gas
pocket enters into the void created. The trapped gas compressed when the lobes of rotor begin to mesh.
The gas will be discharged at the opposite end where the exhaust port is uncovered. The gas volume
is always reduced to a preset proportion of the inlet volume by the time the outlet port is reached.
This is called built-in volume ratio (rv) which should always correspond to working pressure ratio
for optimum power consumption of the compressor.
As shown in Figure 9.17, for the maximum compressor efficiency, as the working pressure ratio
increases the built-in volume ratio also has to increase. Screw compressors are engineered with no
valves and no rolling elements. In respect to reciprocating compressors, drastic reduction in number
of parts dramatically improves reliability of the compressor and reduces the risk of refrigeration loss.
As is in reciprocating compressors, there is no re-expansion of trapped gas, and more refrigerant can
be compressed for the same size of a compressor. Therefore, screw compressors are commonly used
for water cooled chillers having a cooling capacity of 50 to 500tons of refrigeration.
CHAPTER 9 REFRIGERATION SYSTEMS 415
Figure 9.17 Twin screw compressor efficiency for various built-in volume ratios
Example 9.8 Consider the refrigeration system given in Example 9.6 for which the refrigerant is R134a, and the conditions
at the compressor inlet and outlet respectively are p2  1.4bar, T2  12 C , p3  10.16bar . The reciprocating compressor
has to be replaced by a twin screw compressor with a built-in ratio of 4.6. Determine,
a.
compressor efficiency,
b.
refrigerant exit temperature,
Solution:
a.
For compressor pressure ratio, rp=7.05, and built in volume ratio, rv=4.6, Figure 9.17 yields ηc=0.75.
b.
Referring to isentropic efficiency definition, h3 
T3
h3s  h2
283.3  241.2
 h2 
 241.2  297.3 kJ/kg and
0.75
c
65.4 C .
Scroll and vane type compressors. As in Figure 9.17, scroll compressor compresses the refrigerant
with two inter fitting spiral shaped members. One scroll is fixed, and the other orbits the fixed one
with a circular motion.
Figure 9.17 A scroll set and scroll compression process
At the first orbit, the ends of both scrolls are fully open and low pressure gas flows in. As the
lower scroll completes the second orbit, the gas pocket is sealed off. The third orbit begins with the
crescent-shaped pocket is pushed toward the center decreasing the gas volume increasing the gas
pressure. At the fourth orbit, the discharge port is uncovered and the refrigerant is discharged.
416
THERMODYNAMICS
Due to flat volumetric efficiency curve, a scroll compressor provides more cooling capacity
at extreme conditions. In addition, exhibiting lower noise levels, scroll compressors are gradually
replacing reciprocating compressors in room air conditioning, heat pumps, and in package units for
residential and commercial buildings.
As shown in Figure 9.18, a vane compressor operates on the basis of varying the volume between
an eccentric rotor and a sliding vane as angular position changes. A vane compressor has a small clearance volume and a greater volumetric efficiency than a reciprocating compressor. These compressors
are widely used in small refrigeration systems having cooling capacities less than 3.0 kW.
Dynamic compressors. The most common type is the centrifugal compressor. First, the impeller accelerates the gas particles. Then the gas flows into a diffuser or volute where the flow is decelerated
and some of the kinetic energy is converted into pressure. As in Figure 9.19, the torque imparted
by the impeller is, T   V2t r2  V1t r1 . Considering that the refrigerant enters the impeller at radial
direction, V1t  0 , and assuming that the impeller tip speed and the refrigerant tangential velocities
are identical, the power required by the impeller becomes,
  mV
 22t  m h2 s  h1 
Wcs  mT
(9.15)
In the design of centrifugal compressors, two crucial parameters are the wheel diameter and the
width between the impeller faces. In addition, considering the impeller strength, the tip speed should
not exceed the limiting value of 300 m/s. The following Example illustrates numerically the effect of
refrigerant choice on the impeller size and the tip speed.
CHAPTER 9 REFRIGERATION SYSTEMS 417
Example 9.9 An ideal refrigeration cycle operates between -10C evaporator and +36C condenser saturation temperatures. The rotational speed of the impeller is 3600rpm. Evaluate the tip speed and the impeller diameter for the following
refrigerants.
a.
R22
b.
Ammonia (R717)
Solution:
a.
Together with Eq. (9.15), referring to the tabulated values for R22, s1  s2 s  0.942 kJ/kgK, h2 s  281.5kJ/kg, T2 s  60 C ,
V2t 
h1  h2s 
or V2t  1000  281.5  246.15   188 m/s <300 m/s. Respect to tip speed the impeller di-
ameter becomes,
D2 
b.
60V2t
60  188

 0.99 m
n
3.14  3600
For the ammonia, s1  s2 s  5.46 kJ/kgK, h2 s  1651.2 kJ/kg, T2 s  100 C , and
V2t  1000  1651.2  1430.55   469.73 m/s , which is well above the limiting value of 300 m/s. The impeller
diameter is, D2 
60  469.73
 2.49 m . The size is also impractical for consideration.
3.14  3600
Neither the tip speed nor the impeller diameter is in manageable level for ammonia, and ammonia is not a suitable refrigerant for the system in the example. Even for R22, the impeller size is
still large. To reduce the size to reasonable level, rotor speeds of 10,000rpm are common. For such
a speed, the impeller diameter becomes D2=36cm. A high speed two-stage centrifugal compressor
with magnetic bearings is shown in Figure 9.20. Because of the magnetic bearings, no lubrication
is needed.
Figure 9.20 A high speed two-stage centrifugal compressor
The width of the impeller directly affects the mass flow rate of the refrigerant flowing through
the compressor. Systems with high refrigeration capacity require large width between the faces of the
impeller. Since the centrifugal compressors may handle relatively large flow rates, they are widely
used for systems having refrigeration capacities above 500 tons.
418
THERMODYNAMICS
9.4.2
Expansion devices
The purpose of an expansion device is twofold: 1. To reduce the pressure of the liquid refrigerant, 2. Depending upon the evaporator needs, regulate the flow rate. The two most commonly used
expansion devices in refrigeration industry are the superheat controlled expansion valve, and the
capillary tube.
Thermostatic expansion valves. A schematic of the valve and the refrigerant flow lay out is given
in Figure 9.21. To avoid the flow of liquid droplets into the suction line, these valves are always factory preset so that the refrigerant at the evaporator exit is superheated between 4C and 7C above the
saturation temperature. As the evaporator cooling load increases, however, more refrigerant will be
vaporized and the refrigerant outlet temperature as well as the sensing bulb temperature will increase.
Since the bulb is partially charged with the same liquid refrigerant as used in the system, the higher
temperature will result with higher saturation pressure. To maintain the equality of pressures on the
diaphragm, pb  psp  pev , the spring force increases by pushing the valve downward. This, in turn,
increases the flow area and the refrigerant flow rate. For the reverse case of a drop in evaporator cooling load, the superheat at the evaporator outlet will be less than the pre-set value. The temperature and
the corresponding saturation pressure of the refrigerant in the bulb will drop accordingly. Again, the
equality of pressures on the diaphragm requires the decrease of the spring force by upward motion of
the valve. Thus the flow area and the refrigerant flow rate will be decreased.
Figure 9.21 A schematic of thermostatic expansion valve
Example 9.10 Consider a R134a thermostatic valve operating between -4C evaporator and +40C condenser saturation
temperatures. At nominal operating conditions, sub-cooling of 4C at the condenser outlet and superheating of 4C at the
evaporator exit take place.
a.
Determine the pressure caused by the spring on the diaphragm.
b.
If the evaporator cooling load increases by 10-percent, evaluate the pressure to be applied by the sensing bulb on
the diaphragm just at the beginning of this load change.
Solution:
a.
At nominal operating conditions, the evaporator pressure is pev  2.52bar , and due to 4C of superheat, the sensing bulb pressure is, pb  2.92bar , and the difference between these two pressures, psp  pb  pev  0.4bar , is
balanced by the spring.
b
The nominal cooling load is qev  h1  h4  248  100.25   147.75 kJ/kg . Just at the beginning of the load change,
  h1'  h4  1.1  147.75  162.5 kJ/kg and the enthalpy at the evaporator
the mass flow rate is the same, but qev
exit is h1'  262.7 kJ/kg . Then the temperature at the evaporator exit becomes T1'  12 C . The corresponding bulb
pressure is pb  4.42bar and the valve is pressed down for larger mass flow rate.
CHAPTER 9 REFRIGERATION SYSTEMS 419
If an evaporator coil is equipped with a refrigerant distributor, then a considerable pressure drop
will take place, and for such a case, as shown in Figure 9.22, expansion valves with external equalizer are used.
The advantage of an external equalizer is as following, let us consider the case in Example 9.10,
the pressure at the evaporator inlet is  pev i  2.52bar . Due to refrigerant flow through distributor,
pressure drops by 0.67 bar, the pressure at the evaporator exit is  pev e  1.85bar . Hence, the superheat

to be given by the evaporator becomes T  4  12   16 C and this, in turn, requires unnecessarily
large evaporator surface area. With the external equalizer, however, the evaporator pressure acting on
the diaphragm is the exit value. Since the pressure caused by the spring is 0.4 bar, then the balanced
pressure by the bulb becomes, pb  1.85  0.4  2.25 bar . This time, the amount of superheat requirement is much less than the case without external equalizer and becomes T  4  6.5   10.5 C .
When the sensing bulb is charged with the same refrigerant as used in the cycle, this type of charge
is called straight charge. Respect to refrigerant saturation curve, as in Figure 9.23a, for the same
value of p  pb  pev the amount of superheat caused would be different for different evaporator
temperatures. Since T2  T1 , compressor flooding at high evaporator temperatures is always possible. As shown in Figure 9.23b, this may be prevented by charging the bulb with refrigerant different
from the one used in the cycle. This is called cross charge.
420
THERMODYNAMICS
Capillary tubes. Because of their simplicity and low cost, capillary tubes are used as a throttling
device in small refrigeration systems with a capacity of 3 tons or less. The inside diameter is in
the range between 0.5mm and 2.0 mm and the length varies from 1.0 m to 6.0 meters. It is usually
made of copper. The pressure drop in a capillary tube takes place due to following two factors: 1.
The flow of refrigerant must overcome the frictional resistance, 2. As the refrigerant flows, the
pressure drops, and liquid refrigerant flushes into a mixture of liquid and vapor, and due to drop
in the average density, the refrigerant accelerates and causes additional pressure drop. Hence,
the compressor and the tube assume balanced conditions at the suction and at the discharge, so
that the compressor pumps the same amount as the capillary tube feeds the evaporator. Capillary
tubes are not capable responding to variations in suction pressure, discharge pressure or cooling
load. They are susceptible to clogging by impurities, and the refrigerant charge has to be held
within close limits due to possibility of compressor flooding. However, they are very simple,
contain no moving parts, and are inexpensive devices. During the off-cycle period, capillary
tube allows the high and low side pressures to equalize, and thereby the starting torque required
by the compressor is reduced. They are ideal for hermetic compressor based systems which are
critically charged and factory assembled.
If a capillary tube is to be used for a refrigeration system, the diameter and the length have to
be selected so that the compressor and the capillary tube achieve a balanced point at the desired
evaporator temperature. Both analytical and graphical methods are used in sizing the capillary
tubes. Usually a tube longer than the calculated value is installed, and by cut-and-try method, the
tube is shortened until the desired balanced point is achieved. In graphical procedure, the variation of the maximum mass flow rate through a capillary tube of 1.63 mm diameter and 2.03 m
length is given in Figure 9.24. The flow rate is at the maximum for the given conditions because
at the tube exit the chocked flow conditions occur.
CHAPTER 9 REFRIGERATION SYSTEMS 421
Figure 9.25 also provides the correction factor, φ, for different diameters and lengths. Hence the
mass flow rate at any diameter di and length L is determined by,
m   m *
(9.16)
In this equation, the mass flow rate m * of the standard tube size is first evaluated by Figure 9.24
for the same flow conditions.
Figure 9.25 The correction factor for off-standard capillary tubes
Example 9.11 Consider a capillary tube operating between +36C condenser and -8C evaporator saturation temperatures
has to be selected to throttle 0.01kg/s of R22. At the tube inlet, the refrigerant is at 30C. If a copper tubing having 1.5mm
of diameter is selected, determine the tube length by graphical method.
Solution:
For R22, the condenser pressure is pcon  1389.7 kPa and the sub-cooling at the inlet of the tube is Tsc  36  30  6 C ,
then by Figure 9.24, the mass flow rate for standard tube becomes m *  0.01242 kg/s and the correction factor is
  m / m *  0.805 . The curve for Di=1.5mm in Figure 9.25 yields L=1.95m.
9.4.3
Refrigerant Condensers
A condenser is a major system component through which high pressure and hot gas from the
compressor is cooled and condensed, and is often sub-cooled below the saturation temperature.
Based on the type of the cooling fluid, refrigeration condensers may be classified into the following three categories: 1.Water cooled condensers, 2.Air cooled condensers, and 3.Evaporative
condensers.
422
THERMODYNAMICS
The most common type water cooled condenser is shell-and-tube type. As presented in Figure 9.26,
the hot gas from the compressor enters at the top inlet. To provide effective sub-cooling, water inlets
the copper tubes from the pipe near the bottom, and discharges near the top. In horizontal condensers, to prevent gas bubbles to enter the liquid line, one-sixth of the shell volume is filled with liquid
refrigerant. Usually water cooled condensers are preferred over the air cooled condensers when there
is a long distance between the compressor and the heat rejection location. Conveying water rather
than refrigerant in long pipe lines is always less expensive and more secure.
Air is the external fluid for air cooled condensers. Figure 9.27 presents a typical forced convection
type air cooled condenser. If the working fluid is ammonia, the material of construction is aluminum
fins on stainless steel tubing, but aluminum fins on copper tubing is used for halocarbon refrigerants.
The diameters of copper tubes for air cooled condensers vary in the range from 6.5 mm to 19 mm.
Aluminum fins having fin spacing between 200 and 500 fins per meter enlarge the air side heat transfer
area 10 to 30 times of bare tube area. The number of tubes in the flow direction is called tube rows.
The air cooled condensers have 2 to 8 rows of tubes carrying the refrigerant.
Air cooled condensers are simple in construction. The fouling is small and the maintenance cost
is low. Table 9.2 compares the basic features of water and air cooled condensers.
CHAPTER 9 REFRIGERATION SYSTEMS 423
Table 9.2 Comparison of basic features of condensers
Parameter
Water cooled
Air cooled
5 – 10
5 -25
Volumetric flow rate per ton of
refrigeration (m3/h)
0.4 – 1.2
500 - 1200
Face velocity (m/s)
1.5 – 3
2-6
Heat transfer area per ton of refrigeration (m2)
0.5 – 1.0
10 - 15
Maximum temp. difference
Tcon  Tcoolant (°C)
As shown in the table above, due to low density and specific heat of air, the volumetric flow rates
for the same cooling load is approximately thousand times larger for air cooled condensers. Therefore,
air cooled condensers cost 2 to 3 times more for the same heat rejection rate. However, water cooled
condenser usually require cooling tower for recycling the used water.
In thermodynamic analysis of condensers, Eq. (4.63) may be used in determining the heat load of
the condenser. Completion of a cycle requires that the condenser must reject both the energy absorbed
by the evaporator and the energy input of the compressor.
condenser load=evaporator load+compressor power
(9.17)
Because of the difficulty in determining the exact energy input of the compressor some manufacturers provide ratings in terms of evaporator load by defining a heat rejection factor  as following,
=
Q con Q ev  Wcomp
1

1


COP
Qev
Qev
Figure 9.28 Variation of condenser heat rejection factor
(9.18)
424
THERMODYNAMICS
Figure 9.28 illustrates variation of the heat rejection factor with respect to condenser saturation temperature at two different evaporator temperatures for refrigerants R134a, and R22. For
constant condenser temperature, reducing the evaporator temperature reduces the COP of the
system, and  increases. Similarly, increasing the condenser temperature at constant evaporator
temperature will decrease COP and increase the heat rejection factor, . As illustrated in Figure
9.28, since the heat generated by the electric motor of a hermetic compressor is transferred to
the refrigerant, the heat rejection factor is always larger for a hermetic compressor than an open
type one.
The required surface area of a condenser is calculated by the fundamental equation of heat exchangers as,
 =UAT
Q
con
m
(9.19)
where the overall heat transfer coefficient, U (W/m2K) based on outer surface area and for water
cooled condenser is calculated as following,
1 1  d 0  di ln d 0 / di   d 0
=  

2k w
U 0 h0  di 
 di
 1  d0 
    R fi
 hi  di 
(9.20)
Figure 9.29 Plate fin-tube exchanger characteristic geometry
For air cooled condensers, due to finned surface on the air side, the overall heat transfer coefficient
takes the following form,
 A  d ln d 0 / di   A0  1  A0 
1
1
=
 0  i
      R fi
2k w
U 0 0 h0  Ai 
 Ai  hi  Ai 
(9.21)
where, ηo, is finned surface efficiency and is calculated by,
o  1 
Af
A0
1   
f
(9.22)
CHAPTER 9 REFRIGERATION SYSTEMS 425
As shown in Figure 9.29, depending upon the geometry of the finned surface, the total external
area Ao is determined by,

 d 02  4 s1 s2
Ao  A f  Ato  n1n2
 1 s f  n1n2 d 0 1  ts f

4   d0 d0



(9.23)
where sf represents the number of fins per meter. In addition, the total inside area of tubes per
unit length is
Ai  n1n2 di
(9.24)
The fin efficiency may be determined by transforming the plate fin to an annular fin by defining
an equivalent outer radius as r2  s1s2 /  and r2c  r2  t / 2 . Then Figure 9.30 yields the appropriate value of ηf for the specific fin radius ratio of r2c/r1 of the exchanger.
In water cooled condensers, the coolant (water) flows through the tubes and depending upon the
flow structure, the one of following correlations has to be depicted for calculating the average heat
transfer coefficient of the tube side.
1. For Reynolds numbers, Re  Vdi /  , less than 2300, the flow is laminar, and the average heat
transfer coefficient for the combined entry length with constant wall temperature is proposed by
Seider and Tate as following,
426
THERMODYNAMICS
k
hi  1.86 
 di
1/3
  d i   b 
  Pe  

L   w 

0.14
(9.25)
where, Pe is the Peclet number and is defined as Pe  Re Pr . Equation (9.25) is valid for
1/3
 d   
0.48<Pr<16700 , 0.004<b / w <9.75 , and recommended for  Pe i   b 
L   w 

0.14
 2 . Below this
limit, flow is fully developed and the heat transfer coefficient is calculated by hi  3.66(k / di ) .
2. If the flow is in transition region, 2300 < Re < 104, Gnielinski recommended the following correlation for the average heat transfer coefficient.
hi 
k  f / 2 Re 1000 Pr
di 1  12.7( f / 2)1/2 Pr 2/3  1


(9.26)
where,
f  1.58ln Re 3.28 
2
(9.27)
Properties have to be evaluated at the average bulk temperature. The above relation predicts the
average heat transfer coefficient in the range 2300 < Re < 104, and 0.5 < Pr < 2000 with 10-percent
error.
3. For fully developed turbulent regime, Re > 104, Dittus-Boelter correlation is appropriate for determining the average heat transfer coefficient as,
k
hi  0.023 
 di
 0.8 0.4
 Re Pr

(9.28)
The fluid properties have to be evaluated at the average bulk temperature.
The vapor refrigerant condenses on the outer surface of the horizontal tubes, and forms a film on
the surface. The average heat transfer coefficient due to film-wise condensation is calculated by well
known Nusselt correlation as,


 k 3f  f  f   g gh fg 

h0  0.725 
 nd 0  f Ts  Tw  


0.25
(9.29)
where, n, indicates the average number of tubes per row, and to be determined by the tube
layout. The subscripts “f” and “g” are for saturated liquid and saturated vapor respectively. The
term Tm in equation (9.19) is logarithmic mean temperature difference (LMTD) and expressed
as,
Tm 
Tcon  Ti   Tcon  Te 
 T  T  
ln  con i 
 Tcon  Te 
(9.30)
CHAPTER 9 REFRIGERATION SYSTEMS 427
As shown in Figure 9.31, referring to the temperature distribution of fluids in a typical condenser,
Eq. (9.30) should no longer be valid. Because of gaseous state, the heat transfer coefficient of the
de-superheating section is lower than the condensation coefficient, but the temperature difference is
higher. When multiplied, these two deviations compensate each other. As a conclusion, the use of Eq.
(9.30) provides satisfactory results.
It is good practice to consider the scale formation on the exchanger surface and its effect on the
overall heat transfer coefficient. Depending upon the operating conditions of the condenser, the fouling
values may be found in TEMA (Tabular Exchangers Manufacturer’s Association) standards.
Example 9.12 Consider R134a refrigeration plant having a capacity of 10tons and operates between -8C evaporator and
+40C condenser temperatures. A hermetic type compressor is used in the plant. The tube layout of the condenser is shown
in Figure 9.32, has two water passes and a total of 28 copper tubes with diameters of di/d0=14mm/16mm. Water enters the
condenser tubes at 25C and leaves 30C. Determine the tube length.
Solution:
Respect to the tube layout in Figure 9.32, the total number of rows is
28
11, and the number of tubes per row, n 
 2.54 . The tube wall
11
temperature at the exit approximately is 35C and at the inlet 33C. Then
the average wall temperature is 34C, and T  Ts  Tw  6 C . Together
with these values, substituting the properties of R134a at 40°C into Eq.
(9.29) results as,
 (0.0747)3  1146.7  1096.7  9.81  163020 
h0  0.725 

2.54  0.016  0.0001634  6


0.25
 1552.74
W/m2K Hermetic compressor working between temperatures of -8C
and +40C, Figure 9.28 yields the heat rejection factor as   1.35 and
the condenser capacity is Q con  (10  3.516)  1.35  47.466 kW. Water
3
properties are taken at the average bulk temperature of 28C as c p  4.179 kJ/kgK,   996 kg/m ,   0.00086 kg/ms ,
44.466
 2.271 kg/s and the water velock  0.614 W/mK and Pr  5.85 . Then the mass flow rate of water becomes m w 
4.179  5
m w
2.271
996  1.058  0.014

 1.058 m/s , Re 
 17150  104
ity and the Reynolds number are V 
 wn p Ai 996  14  0.0001538
0.00086
428
THERMODYNAMICS
fully turbulent flow and Eq. (9.28) yields hi  0.023 
(9.20), the overall heat transfer coefficient is,
0.614
0.8
0.4
 17150   5.85   4989.3 W/m2K. Thus, by Eq.
0.014
1
1
 16  0.014ln 16 / 14   16  1
 16 


 
    0.000176
U 0 1552.74  14 
2  390
 14  4989.3  14 
or U 0  928.62 W/m2K. Eq. (9.30) yields the logarithmic mean temperature difference as T  12.33K and Eq. (9.19)
m
47466
provides the heat transfer surface area as, A 
 4.146 m2. Since A  nt  d 0 L  , then the tube length is
928.62  12.33
4.146
L
 2.95 m .
3.14 0.016 28
In air cooled condensers, the coolant (air) flows over the finned tube bank and the correlation to
evaluate the average heat transfer coefficient of the finned surface is given by Kayansayan as following,
h0  0.15c p Gm Re 0.28 Pr 2/3  0.362
(9.31)
Properties have to be evaluated at the average bulk temperature of air. The above relation predicts
the average heat transfer coefficient in the range of 500  Re  3  104 , and 11.2    23.5 with 10percent deviation from the experimental data. In this expression, Gm is the maximum mass flux and
referring to Figure 9.29, the value of Gm, the Reynolds number, Re, and the exchanger fining factor,
ε, are calculated as following,
Gm 
m air
Gm d 0
Total finned surface area Ao

, Re 
and  
Amin
b
Ato
outer tube surface area
(9.32)
where, Ao and Ato are determined by Eq. (9.23).
Condensation in horizontal tubes is complicated by variety of two-phase flow regimes. Due to
increase in the amount of liquid present, the flow pattern changes as condensation proceeds.
As described in Figure 9.33, initially the condensate forms an annular film around the tube periphery while vapor flows in the core region. This flow regime is called annular flow regime, zone
1 in Figure 9.33. As condensation proceeds the condensate begins to accumulate at the bottom of the
tube, and at low flow rates, the condensate flows along the bottom portion with vapor above, stratified flow, zone 2 in Figure 9.33. At high flow rates, flow changes to slug flow, zone 3, finally vapor
pockets flow as bubbles in the flow direction which is called bubbly flow, zone 4. In refrigeration
CHAPTER 9 REFRIGERATION SYSTEMS 429
condensers, if the gas Reynolds number, Rev  4m /  di  g  35000 , then the flow regime zone 2 will
be dominant in the condensation process. Thus Chaddock and Chato’s correlation gives the average
heat transfer coefficient over length of the tube as following,


 k 3f  f  f   g gh*fg 

hi  0.555 
di  f T




0.25
(9.33)
where, the modified enthalpy of evaporation is,
h*fg  h fg  3 / 8 c p , f T
(9.34)
In both relations, T  Tcon  Tw , represents the difference between the condensation temperature
and the surface temperature.
Example 9.13 Reconsider the R134a refrigeration plant in Example 9.12, air cooled condenser is to be designed for
the same load and the same operating conditions. The condenser is plate fin-and-tube type with the following geometric
parameters: s1  40 mm , s2  35 mm , t  0.2 mm , s f  312 m-1,   15.81 , di / d 0  14 / 16 . The number of tubes per
row is, n1  100 , and the number of rows is, n2  4 . Air flowing on the finned surface side inlets at 25C and exits at 30C.
Determine the length of the condenser.
Solution:
Respect to the tube layout in Figure 9.34, H  n1s1  100  40  4000 mm.
Refrigerant mass flow rate,
a single tube, m r1 
Rev 
m r 
Q ev
35.16

 0.255 kg/s and through
h1  h3 393.87  256.41
0.255
 0.00255 kg/s. The vapor Reynolds number is,
100
4m 1
4  0.00255

 18488  35000 , and Chato’s correlation
 di v 3.14  0.014  12.55  106
0.25
 0.07433  1146.7  1096.7  9.81  166390.5 

 1553.23
yields, hi  0.555  


0.014  0.0001634  6


Q con
47.466

 9.436
W/m2K. In addition, the air side mass flow rate is, m a 
c pa Te  Ti  1.006  5
Figure 9.34 Finned tube
geometric parameters
kg/s, the minimum flow area, Amin  100  40  16  1  0.2  0.312  103  2.248 m2/m, and the air side Reynolds
number becomes Re 
Gm d 0 4.196  0.016

 3668.6 . Apply Eq. (9.31) for h0 as,

18.3  106
0.28
h0  0.15  1006  4.196  3668.6 
0.67
 0.707 
0.362
 15.81
Finned surface equivalent fin radius,
 29.5 W/m2K.
r2  40  35 /   21.12 m m ,
r2c / r1  21.22 / 8  2.65 ,
Lc  r2c  r1  13.22 mm, Ap  2.64 mm2 and by Figure 9.30, fin efficiency is  f  0.856 , and the finned surface ef-
ficiency is 0  1 
278.87
1  0.856  . To determine the overall heat transfer coefficient, use Eq. 9.21 as following,
297.70
1
1
1
 297.7  0.014ln(16 / 14)  297.7 
 297.7 
2





  0.000176 or U 0  19.86 W/m K, and
U o 29.5  0.865  17.584 
2  390
 17.584  1553.23  17.584 
430
THERMODYNAMICS
47446
193.75
 193.75 m2, then the exchanger length is L 
 0.65 m.
19.86  12.33
297.7
Calculations are to be repeated for the new tube length L=0.65 m. However the overall effect on heat transfer surface area
might be negligible.
Tm  12.33K . Eq. (9.19) yields Ao 
9.4.4
Refrigerant Evaporators
Classification of refrigerant evaporators is done on the basis of several different parameters. For
instance, evaporators are classified according to 1. The transfer process of being natural or forced
convection, 2. Refrigerant flowing inside or outside of tubes, 3. The evaporator being flooded or direct
expansion type during the evaporation process, and also depending upon the type of fluid cooling,
evaporators are classified as 4. Air cooling or liquid cooling evaporators. In this classification method,
as shown in Figure 9.35, one type of evaporator may accommodate the other types, like air cooling
evaporator in Figure 9.35a, which is usually forced convection type, refrigerant flows through the
tubes, and is direct expansion type.
These evaporators consist of coils placed in a number of rows with various fin arrangements to
increase the heat transfer area. Plate fins accommodating tube rows are used in halocarbon based
refrigerants. However, in ammonia applications, steel tubes with spiral fins are usually used.
Liquid cooling evaporators may be direct expansion type or flooded type. In Figure 9.35b,
shell-and tube type flooded evaporator contains liquid refrigerant on the shell side. As the refrigerant
evaporates from the top, the float allows liquid refrigerant inlet the evaporator.
CHAPTER 9 REFRIGERATION SYSTEMS 431
The type of evaporator as illustrated in Figure 9.36 is used in household refrigerators and is called
roll-bond type or bonded plate evaporators. In construction, two aluminum sheets are embossed in
such a manner that when they are welded the embossed section makes a channel for refrigerant flow
through the entire plate.
The thermal design of evaporators is much more difficult than the design of a condenser. The
complexity arises from the following facts: 1. Pressure drop in evaporators is much more critical and
significantly affects the overall heat transfer coefficient. To reduce the pressure drop effects, multiple
circuits have to be used in large systems. 2. Depending upon operating conditions, the air side surface
of an air cooling evaporator may partly become wet. Therefore the effect of mass transfer on evaporator heat transfer has to be considered. In case of frost formation on the surface, the thermal resistance
of the evaporator will increase respect to time. 3. In case of refrigerant flowing through the tubes,
due to rapid change of flow regime from inlet to outlet, the heat transfer coefficient will be altered
accordingly. An accurate estimate of boiling side heat transfer coefficient becomes a very hard task.
Because of these difficulties, a brief explanation will be given for shell-and-tube type liquid cooling
evaporators with the refrigerant boiling on the shell side. In addition, for air cooling evaporators, flow
boiling inside the tube will be studied.
Since the heat transfer coefficients in nucleate pool boiling regime are stable and high in magnitude,
flooded evaporators usually operate in this regime. Rohsenow presented the following correlation for
the heat flux of liquid boiling in nucleate pool boiling regime.

 g  f  g
q  h fg  f 






0.5
 c pf T

 Csf h fg Pr n





3
(9.35)
where, the coefficient Csf and the exponent n depend on the surface-liquid combinations and is
Csf  0.013 , n  1.7 for halocarbon-copper combination. T is the temperature difference between
the surface and the fluid, and all fluid properties are evaluated at the saturation temperature. As can
 
be noticed by Eq. (9.35), q  h fg
2
, and knowing that h fg decreases with the increase in pressure,
the heat transfer coefficient will increase as the refrigerant pressurized.
In air cooling evaporators, forced convection boiling takes place inside the tubes, and the flow
regimes of such a boiling process is explained in Figure 9.37. Since the wall temperature exceeds the
saturation temperature of the liquid, the vaporization starts as in the sub-cooled flow boiling regime.
As the fluid further flows in the downstream direction, the vapor quality, x, increases, and due to
density difference, the average velocity in the pipe also increases, and sequentially bubbly, slug, annular and mist flow regimes take place.
432
THERMODYNAMICS
For saturated flow boiling region in smooth tubes, the following correlation has been developed
by Kandlikar,

 f
h 
 1.136 
 g
hl 


Where, G 




0.45
x
0.72
1  x 
0.08

 q
f Fr   667.2  ev
 Gh fg







0.7
1  x 0.8 Gsf
(9.36)
Q
m R
, is the mass flow rate of refrigerant per unit cross sectional area, and qev  ev , is
A0
Ac
the heat flux on the outer surface of the evaporator. In addition, Fr, is the liquid phase Froude number,

and is defined as Fr  G /  f
 / gd . The coefficient G
2
i
sf
is a function of surface-liquid combina-
tion, and is 1.63 for R134a, 1.1 for R22, and 1.0 for ammonia (R717) with stainless steel tubing. The
stratification parameter, f(Fr), is unity for vertical tubes and for horizontal tubes with Fr  0.04 . If
Fr  0.04 for horizontal tubes, then f becomes f Fr   2.63Fr 0.3 . Eq. (9.36) is applicable only if
the average vapor quality is in the following range, 0  x  0.8 , and all properties of the refrigerant
has to be evaluated at the saturation temperature. The liquid phase heat transfer coefficient, hl , in Eq.
(9.36) has to be evaluated for turbulent regime and the flow conditions should satisfy the turbulent
flow conditions in tubes.
Example 9.14 Reconsider the R134a refrigeration plant of Example 9.12, air cooling evaporator is to be designed for the
same load and the same operating conditions. The evaporator is plate fin-and-tube type with the following geometric parameters: s1  40mm , s2  35mm , t  0.2mm , s f  312 1/m,   15.81 , di / d 0  14 / 16 , the number of tubes per row,
n1  25 , and the number of rows, n2  4 . As shown in Figure 9.38, the refrigerant flowing through the expansion valve is
distributed into four circuits. Hence each row consisting of 25 tubes makes one circuit. Air flowing on the finned surface
side inlets at +5C and exits at 0C. Find the length of the evaporator.
Solution:
The mass flux of the refrigerant, G 

m R
0.255

 416.33 kg/m 2s, the Froude number of the refrigerant,
n2 Ai 4  0.000153
 / gdi  416.33 / 1146.7 2 / 9.81 0.014  0.959 and f ( Fr )  1.0 . The exchanger length is assumed
to be L=1m, then At 0  n1n 2  d 0 1  ts f  4.71 m /m and A 0 = At 0  74.47 m /m and the heat flux at the surface,
Fr  G /  f
2
2
2
CHAPTER 9 REFRIGERATION SYSTEMS 433
qev 
Q ev 35160

 472.14 W/m2 then
A0
74.47
qev
472.14

 6.956 x106 substituting these parameters into Eq.
Gh fg 416.33 x163020
(9.36) and assuming that the average vapor quality is x=0.55 yields,
h
 1146.7 
 1.136  

hl
 50 
0.45
 0.550.72  1  0.55 
0.08
 1.0  667.2  0.000006956 
0.7
The Reynolds number of the flow inside the tube, Re f 
The fluid Prandtl number, Pr f 
hl  0.023 
c p
kf

 1  0.55 
0.8
 1.63
h
 2.976 .
hl
Gdi 416.33  0.014

 35759.25 , the flow is turbulent.
f
0.000163
0.0001634  1320
 2.887 , and Dittus-Bolter equation yields,
0.0747
0.0747
0.8
0.4
 35759.25   2.887  or hl  823.77 W/m2K. Thus, the boiling side heat transfer coefficient
0.014
becomes hi  823.77  2.976  2451.56 W/m2K.
The air side bulk temperature is Tb  3 C and the mass flow rate, m a 
Q ev
35.16

 6.99 kg/s, the minic pa Ti  Te  1.006  5
mum flow area, Amin  25  40  16  1  0.2  0.312  103  0.562 m2/m, Gm 
Reynolds number is Re 
6.99
 12.437 kg/m2s and the air side
0.562
Gm d 0 12.437  0.016

 11610.47 . Then Eq. (9.31) yields,

17.14  106
0.28
h0  0.15  1006  12.437  11610.47 
0.67
 0.71
0.362
 15.81
 63.11 W/m2K. The evaporator has exactly the
same geometry and the same materials of Example 9.13, therefore,  f  0.856 , 0  0.865 and assuming clean air side surface,
R f 0  0.0 . Hence, Eq. (8.21) yields,
1
1
1
 74.47  0.014ln(16 / 14)  74.47 




  2451.5 , U 0  39.59
U 0 63.11  0.865  4.396 
2  390
4.396


W/m2K, Tm  10.3 C . The fundamental equation of heat exchangers yields the outer surface area as, Ao 
m2. Then, respect to Figure 9.34, the length L is, L 
35160
 86.22
39.59 x10.3
86.22
 1.16 m. Again, the calculations are to be repeated for the new
74.47
tube length L=1.16m. However the overall effect on heat transfer surface area might be negligible.
9.4.5
Wilson’s plot
The idea of Wilson’s plot is to determine the individual thermal characteristics of a heat exchanger
by conducting a number of experiments. For instance, by varying the flow rate and measuring the inlet
and outlet temperatures of water in a water cooled condenser, m w c pw Te  Ti   U 0 A0 Tm , the overall
heat transfer coefficient can be determined for specified condensing conditions of the refrigerant. Since
1
the only term changing in Eq. (9.20) is d 0 / d i  , it can be expressed in the following form,
hi
C
1
 C1  2
U0
hi
(9.37)
Assuming fully turbulent flow through condenser tubes, Eq. (9.28) may be reduced to hi  C3Vi 0.8 .
Substituting this result into (9.37) and rearranging yields,
C4
1
 C1  0.8
U0
Vi
(9.38)
434
THERMODYNAMICS
where, C1 
1  d 0  di ln d 0 / di   d 0
 

h0  di 
2k w
 di

 R fi

(9.39)
In Figure 9.39, as Vi assumes very large values, 1 / Vi 0.8 approaches zero, and the y-intercept
represents the experimental value of C1 which is also expressed by Eq. (9.39). For a new condenser,
however, Rfi=0, and the tube thermal resistance can be estimated, then for known C1 Eq. (9.39) yields
the condensation side heat transfer coefficient h0.
Figure 9.39 Wilson’s plot for a water cooled condenser
Example 9.15 A heat exchanger manufacturer measures the following data on a shell-and-tube ammonia condenser.
Average velocity of water inside tubes Vi(m/s)
1.00
0.50
Overall heat transfer coefficient U0(W/m2K)
2099
1381
Water flows inside of steel tubes having 46 mm inside and 50 mm outside diameter and the tube thermal conductivity
is 55 W/mK. Determine,
a.
the heat transfer coefficient of the condensing side,
b.
the overall heat transfer coefficient for water velocity of 0.35 m/s.
Solution:
a.
Since the overall heat transfer coefficient is expressed by Eq. (9.38), the constants of the equation may be evalu1
C4
1
C4
and
. Solving for the constants, C1  1.423  104
 C1  0.8
 C1 
ated by the given data as,
2099
1381
1
0.50.8
and C4  3.34  104 . Considering that C1 is evaluated for a new condenser with R fi  0 in Eq. (9.39), and
1
 50  0.046  ln 50 / 46 
 1.423  104   
or h0  9578.6 W/m2K.
h0
2  55
 46 
b.
Substituting C1 and C2 values into Eq. (9.38),
9.4.6
1
3.34  104
, then U o  1092 W/m2K
 1.423  104 
U0
0.350.8
Refrigerants
As explained by the classification chart in Figure 9.40, the refrigerants used for refrigeration
processes are classified into 1. Primary refrigerants, and 2. Secondary refrigerants.
CHAPTER 9 REFRIGERATION SYSTEMS 435
Secondary refrigerants are antifreezes or brines which are liquids used for absorbing heat from
the cold space and rejecting it to the refrigeration system at the evaporator. Secondary refrigerants do
not change phase but transports cold energy from low temperature medium. For carrying cold energy
in the temperature range of -10C to -30C, commonly used secondary refrigerants are brines and are
essentially water solutions of ethylene glycol, propylene glycol, or calcium chloride.
Primary refrigerants are used as working fluid in vapor compression cycle, and are in the form
of either compounds or mixtures. The refrigerants in the form of compounds can be either natural
compounds or synthetic compounds. Organic compounds like hydrocarbons (CH4, C2H6, C3H8, etc.)
and inorganic compounds like (Ammonia NH3, Carbon-dioxide CO2, Sulfur-dioxide SO2, and water
H2O) are called natural compounds. The natural compounds have been used since the early ages of
refrigeration. A revolution in refrigeration technology came about with the invention of synthetic compounds in early 1930s. Synthetic compounds are chemical combination of carbon, hydrogen, chlorine,
and fluorine and in general format expressed by the following chemical formula, Cm H n Fp Clq and
commonly indicated as CFCs, HCFCs, or HFCs. Refrigerants in the form of mixtures can be either
Azeotropic or Zeotropic. Azeotropi mixture has two components, and cannot be separated into its
constituents by evaporation or condensation. An azeotropic mixture offers different properties from
its constituents. Similarly, zeotropic mixtures contain three different components.
As indicated in classification chart, a large number of refrigerants are developed over the years for
a wide variety of applications. To distinguish one refrigerant from the other, a numbering system is
adopted by ASHRAE, and all refrigerants are designated by letter R followed by a unique number.
In developing the numbering system for synthetic and for organic refrigerants, the chemical formula is taken into consideration, and for general molecular formula of Cm H n Fp Clq the refrigeration
number is designated as following,
(9.40)
436
THERMODYNAMICS
Example 9.16 Determine the refrigeration numbers of the following compounds: a. CHClF2, b. C2H2F4, and c. Propane,
CH3CH2CH3.
Solution:
a.
For CHClF2, X  1  1  0 , Y  1  1  2 , and Z  2 . If X is zero, it is indicated in the numbering system and the
refrigerant is R22.
b.
For C2H2F4, X  2  1  1 , Y  2  1  3 , and Z  4 . The refrigerant is R134a where “a” stands for an isomer of
the same chemical composition.
c.
For propane, CH3CH2CH3, X  3  1  2 , Y  8  1  9 , and Z  0 . Propane is indicated by R290 as a refrigerant.
The designation of inorganic refrigerants starts with number 7 and the numbers following 7 represent its molecular weight. For instance, Ammonia, molecular weight is 17, it is an inorganic refrigerant
and the refrigeration number is R717. Similarly, CO2, molecular weight is 44, and the refrigeration
number is R744. Azeotropic refrigerants are designated by 500 series like R502, which is a mixture
of 48.8% R22 and 51.2% R115. R502 offers better advantages than R22, especially behaves better
with oil. Zeotropic mixtures (non-azeotrops) are 400 series refrigerants and contain three synthetic
compounds. For instance, R404A, is a mixture of 44%R125, 52% R143a, and 4% R134a. Especially
at low temperature applications, R404A exhibits low temperature discharge and avoids the need for
inter-stage cooling.
Refrigerant selection. In addition to the temperature application ranges; low temperature (-40C to
-25C), medium temperature (-25C to -5C), and high temperature (-5C to +10C), due to several
environmental issues like “ozone layer depletion” and “Global Warming Potential” and their relation
to refrigerants in use, the selection of an appropriate refrigerant for a particular application has become
a difficult task. In fact, once a refrigerant is selected, to replace it for some unavoidable reason is an
expensive procedure and may require several changes in the system design. Therefore, the following
constraints have to be considered in selecting a refrigerant: 1. Thermodynamic constraints, 2. Physical
property constraints, 3. Environmental and safety constraints, 4. Economic constraints.
1. Thermodynamic constraints: A refrigerant should have high latent heat of vaporization, high
density at the compressor inlet, and low condenser pressure to allow for light weight and small size
compressor construction. These thermodynamic requirements are somewhat contradictory. The relation between saturation pressure and temperature as given by Clasius-Clapeyron equation explains
this contradiction as following,
h fg
dp

dT Tv fg
Where, v fg  vg  v f , vg  v f and vg 
(9.41)
RT
. Substituting this result into Eq. (9.41) and integratp
ing between evaporator and condenser pressures yields,
 p  h fg  1
1 

ln  con  


 pev  R  Tev Tcon 
(9.42)
At a particular evaporator and condenser temperature, this relation indicates that high value in hfg
cannot be obtainable without an increase in pressure ratio. Hence, these two contradicting properties
have to be balanced. Similarly, the critical values of the refrigerant must be outside of the working
range. In fact, high critical temperature yields higher COP, but the vapor pressure will be low and will
cause low volumetric capacity. Again these two contradicting facts must be balanced for a particular
application.
CHAPTER 9 REFRIGERATION SYSTEMS 437
2. Constraints on physical properties: Appreciable sub-cooling requires the specific heat of the liquid
phase to be small but the vapor phase specific heat should be large for small degree of super heating. In both phases, the thermal conductivity should be high for high heat transfer rates. In addition,
refrigerants should have smaller viscosity values for less frictional pressure drops.
3. Constraints related to environment and safety: Refrigerants should be non-corrosive, non-toxic,
and non-flammable. Depending upon a particular application, toxicity of a refrigerant might change.
For instance, some CFC and HCFC compounds are non-toxic when mixed with air. However, in
contact with a flame or a heating wire, they become very toxic. Hence, decision should be made on
a particular application basis. Respect to non-flammability of the refrigerants, ASHRAE divides the
refrigerants into six groups as, A1, A2, A3, B1, B2, and B3. In this grouping, A1 is the least hazardous
group and contains refrigerants like R22, R134a, R744, while B3 is the most hazardous and contains
refrigerants like R1140.
Refrigerants should be chemically stable as long as they are in the system, and also should be
miscible for the return of oil into the compressor. For instance, ammonia is not a miscible refrigerant,
and systems using ammonia should have oil separators. High dielectric strength especially is a must
for refrigerants used in hermetic compressors.
Ozone Depletion Potential (ODP) of a refrigerant basically exists due to presence of chlorine or
bromine in the molecular structure. Since ODP should be zero, refrigerants like CFCs and HCFCs
cannot be used under new regulations. In addition, a refrigerant should have low value of Global
Warming Potential (GWP). For instance, ODP=0 for R134a but GWP value is high and has to be
reconstructed under new regulations.
4. Economic constraints: Refrigerants should be inexpensive and easily available products.
9.5
Heat Pump
A heat pump works in reverse order with respect to the refrigeration cycle. Heat pump extracts
heat from a heat source at lower temperature, and supplies heat to air or water at a higher temperature.
As shown in Figure 9.41, a heat pump contains all the main components (compressor, evaporator,
condenser, expansion valve, fans and filters) of a refrigeration cycle.
Figure 9.41 Modes of a heat pump
In addition, to use the system for both heating and cooling purposes, the heat pump cycle is equipped
with an apparatus called reversing valve. The reversing valve arranges the direction of refrigerant flow
438
THERMODYNAMICS
by manipulating between four points of the system which are the compressor discharge, the compressor
outlet, the evaporator outlet, and the condenser inlet. Therefore, the valve is usually called as four-way
reversing valve. The functions of the indoors and the outdoors coils can be reversed by the activation
of the valve, and the condenser is changed to evaporator and the evaporator turns to condenser.
T-s and p-h representation of heat pump cycle is given in Figure 9.42, and with respect to Figure
9.42b, the coefficient of performance of the heating effect in a heat pump system is,
COPhp 
Q23
Wc
(9.43)
In accord with the type of heat source from which heat is absorbed by the refrigerant, the heat
pump systems are classified as following: 1. Air sourced heat pumps, 2. Water sourced heat pumps.
For air sourced heat pumps, surroundings air acts as a heat source from which heat extracted during the heating, and as a heat sink to which heat is rejected during cooling. One of the fundamental
problems of these pumps is the formation of frost on the outdoor coil during the heating mode of
operation in cold weather. To melt the accumulated frost, the reverse cycle switches heating mode of
operation to cooling mode by which the outdoor coil changes from evaporator to condenser and the
hot gas flows through. After the frost melts, the heat pump switches back to the normal heating mode
again. This process is known as defrosting.
Water source heat pumps may be either ground water type, or surface water type or a combination
of these two. Ground water heat pumps use well water as a heat source during heating and as a heat
sink during cooling, and are suitable for low-rise residents or commercial buildings.
9.6
Vapor Absorption Refrigeration (VAR)
In vapor compression refrigeration, we use high grade of energy like work (electricity) to raise the
refrigerant vapor pressure from evaporator to condenser pressure. In vapor absorption refrigeration,
however, we use low grade energy like waste heat at a temperature of 100°C to 200°C in compressing the refrigerant. To accomplish this, we replace the compressor by a Thermal Compressor. As
shown in Figure 9.43, a thermal compressor consists of an absorber, a pump, a generator and a throttling device. Besides, an absorption refrigeration system uses a binary solution which contains the
refrigerant and the absorbent. The need for a binary solution and how the thermal compressor works
is explained in the following experiments.
CHAPTER 9 REFRIGERATION SYSTEMS 439
Experiment 1: Let us consider two vessels A and B connected with a pipe and a valve as shown in
Figure 9.44a, and initially the entire system is in equilibrium with surroundings. Vessel A contains
pure water and the corresponding vapor pressure at 35°C is 5.63 kPa, and lithium bromide solution
(LiBr) with 45-percent concentration is in vessel B at the same temperature. Referring to Section 7.6,
since LiBr is highly non-volatile, the pressure on the solution is the vapor pressure of water, and with
respect to Figure 7.31, the solution pressure is 2.4 kPa.
If we open the valve on the connecting pipe, due to pressure difference, water vapor will flow from
vessel A to B and absorbed by the solution in vessel B. The absorption process is exothermic. To keep
the solution temperature constant, however, the released heat of absorption ( Qa ) has to be transferred
to surroundings. Suppose also that by some means the concentration of solution in vessel B is kept
constant. Then, because of evaporation of water in vessel A, the pressure will be reduced to 2.4 kPa.
Coexistence of both liquid and vapor phases at 2.4 kPa in vessel A requires that the water temperature
should drop to 20.5°C which in turn causes a refrigeration effect of ( Qe ) on the surroundings.
Figure 9.44 Schematic of thermal compressor working principle
Experiment 2: Let us consider the reverse process of experiment 1 as shown schematically in Figure
9.44b. Suppose that solution in vessel B absorbs enough water vapor, so that the pressure increases
to 5.63 kPa and the solution is diluted. Let us apply heat ( Qg ) to dilute LiBr solution in B for regeneration. Due to heat addition, water evaporates and flows into vessel A where it is condensed by heat
440
THERMODYNAMICS
rejection ( Qc ) into the surroundings. Transferring the amount of water absorbed during refrigeration
process to vessel A, the solution in vessel B returns to its initial concentration and system also restores
itself back to its initial conditions.
These two experiments described above explain the basic principles of absorption refrigeration
system. To get continuous refrigeration effect, however, the system has to be modified as shown in
Figure 9.43. Essentially, together with a solution pump and an expansion valve, four basic components
of absorption refrigeration system are the generator and the condenser on the high pressure side, and
the absorber and the evaporator on the low pressure side.
9.6.1
Thermal analysis of components
In performing analysis for equipment sizing and evaluating the performance of vapor absorption refrigeration system, the following assumptions are considered: 1. Refrigerant is pure water. 2.
Except the flow restrictors and the solution pump, no pressure change takes place within a part of the
system. 3. In Figure 9.45, at states 2, 4, 5, 7 only saturated liquid and at states 1 and 6 only saturated
vapor exists. 4. The throttling process at the flow restrictor is assumed to be adiabatic. 5. The solution
pump is isentropic.
Generator and condenser sections. Starting with the high pressure side of the cycle, the purpose
of the generator is to deliver the refrigerant vapor to condenser by separating it from LiBr solution.
As shown in Figure 9.45a, steam or hot water flows through the tubes of the generator, the water in
the solution evaporates and the absorbent solution exits the generator at a higher concentration.
Since the temperature of cooling water flowing through the condenser tubes is less than the temperature of steam in generator tubes, the pressure of condenser section is less than the pressure in the
generator. This pressure difference causes the flow of water vapor into the condenser section.
Heat input to the generator is determined by the following energy balance on the generator,
Q g  m 4 h4  m 6 h6  m 3 h3
(9.44)
CHAPTER 9 REFRIGERATION SYSTEMS 441
Besides, considering the concentrations at the inlet and exit of the generator, the mass flow rates
are related as following,
m 3 x3  m 4 x4
and
m 6  m 3  m 4
(9.45)
Water vapor surrounding the condenser tubes condenses as heat is transferred to the cooling water
at the following rate,
Q c  m 6 h6  h7 
(9.46)
High pressure liquid water flows through an expansion device into the evaporator section of the
cycle. As an expansion device usually an orifice type restriction is used for maintaining the pressure
difference between the condenser and the evaporator.
Evaporator and absorber sections. Principally the evaporator and the absorber are contained inside
the same shell. Due to pressure drop at the expansion valve, water (refrigerant) evaporates and cools
down warm water returning from the chilled-water system. As shown in Figure 9.45b, evaporator
pump continuously circulates and sprays the refrigerant over the tubes for better heat transfer.
Referring to Figure 9.45b, at steady state working conditions, the mass balance on the evaporator
gives m 1  m 8 . In addition, the energy balance for the evaporator is,
Q e  m 1h1  m 8 h8
(9.47)
The resulting refrigerant vapor (water vapor) in the evaporator is drawn into the low pressure
absorber where it is absorbed by LiBr-water solution. The released heat due to absorption is rejected
to the cooling water circulating through the absorber tubes. The absorber spray pump in Figure 9.45b
mixes the concentrated solution returning from the generator with dilute solution and delivers the
mixed solution to the absorber spray. In accord with equilibrium chart, the mixing process is a necessity. If concentrated solution were sprayed directly on the absorber tubes, increase in temperature
would cause the solution to crystallize. To avoid this possibility, the concentration is reduced by
mixing with dilute solution.
With respect to Figure 9.45b, the mass balance around the absorber yields,
m 2  m 1  m 5
and
m 2 x2  m 5 x5
(9.48)
Similarly the energy balance yields the heat rate to be withdrawn from the absorber as,
Q a  m 1h1  m 5 h5  m 2 h2
(9.49)
The COP of absorption refrigeration systems is defined as,
COP 
Qe
Qg  W p

Qe
Qg
(9.50)
The work input of the solution pump ( W p ) is usually negligible compared to the generator heat
input ( Q g ). Comparing the COP values of absorption refrigeration system and vapor compression
refrigeration system working at the same refrigeration and heat rejection conditions, the COP of
442
THERMODYNAMICS
absorption refrigeration system is lower than the COP of vapor compression refrigeration system.
However, comparing only the COP values of two systems is not justifiable. Because, high grade
mechanical energy is more expensive than low grade thermal energy and comparing exergetic efficiencies instead would be more meaningful. Both systems provide exergetic efficiencies at the same
order of magnitude.
The complete system and the representation of the cycle on the equilibrium chart are shown in
Figure 9.46. The heat exchanger in the system transfers heat between the two streams of solutions. It
heats the dilute solution from the absorber and cools the strong solution returning from the generator
to the absorber. The heat exchanger helps the system to increase COP.
Example 9.17 As shown in Figure 9.46, consider a water-LiBr absorption refrigeration system with a cooling capacity of
100kW. Evaporator and condenser temperatures of the system respectively are 6°C and 45°C. Dilute LiBr solution leaves
the absorber at 35°C with 55% concentration. The mass flow rate delivered by the solution pump is 0.53kg/s. The solution
passes through the heat exchanger where it is heated to 65°C. In generator, the solution absorbs the heat from steam, the
refrigerant begins to boil and separate from the solution. At the generator exit, the solution is at 90°C with 60% concentration. The concentrated solution passes through the heat exchanger and is cooled to 55°C (state 7). Determine,
a.
the ratio of mass flow rates of refrigerant (water) to solution, m 1 / m 2
b.
the mass flow rate of solution leaving the generator,
c.
the temperature of the strong solution leaving the exchanger,
d.
the heat capacities of absorber, generator, and the condenser,
e.
the power consumed by the solution pump, and the COP of the system.
CHAPTER 9 REFRIGERATION SYSTEMS 443
Solution:
State no.
Pressure (kPa)
Temp.(°C)
Enthalpy(kJ/kg)
Mass flow rate( m kg/s)
Concentration(x%)
2
0.934
35
83
0.53
55
3
9.66
35
83
0.53
55
4
9.66
65
145
0.53
55
5
9.66
90
212
-
60
7
9.66
-
-
-
60
a.
The table above gives the properties of the solution at specified states. The energy balance for the evaporator yields,
100
Q e  m 1 h1  h9  , and for 100 kW of evaporator capacity, m 1 
 0.043 kg/s. Hence, the ratio of
2512  188.45
mass flow rates is m 1 / m 2  0.043 / 0.53  0.081.
b.
With respect to Figure 9.46a, mass conservation of LiBr requires that x3m3  x5m5 . Besides, the following relations for mass balance can be expressed as, m 3  m 5  m 8 and m 8  m 1 . Hence, the flow rate of strong solution at
the generator exit is m 5  0.53  0.043  0.487 kg/s.
c.
The energy balance for the heat exchanger yields, m 3 h4  h3   m 5 h5  h7  and the enthalpy at state 7 is
h7  212 
0.53  145  83
0.487
 144.5 kJ/kg. Hence, by using the equilibrium diagram, the temperature of the strong
solution at the exchanger exit becomes T7
d.
55o C .
Energy balance around the absorber yields, Q a  m 7 h7  m 1h1  m3h3 or
Q a  0.487 144.5 0.043 2512 0.53 83 134.39 kW. Similarly, for the generator, the energy balance requires that Q g  m 5h5  m 8h8  m4 h4 , or Q g
0.487 x 212 0.043 x 2583.2 0.53 x145 137.47 kW. Similarly,
the energy balance for the condenser is Q c  m 8 h8  h10  , or Q c  0.043
e.
2853.2
188.45  114.58 kW.
The density of LiBr solution for concentration in the range ( 0.2  x  0.6 ) and for temperature range of
( 0o C  T  200o C ) is expressed as,  x  1145.36  470.84 x  1374.79 x 2  0.33  0.571x T  273 . At the solution pump inlet, x2  0.55 and T2  35o C , then the solution density is   1620 kg/m3. The pump is isentropic,
9.66  0.934   2.85 W. The pump work is negligibly small, and the
and the required pump work is W p  0.53 
1620
COP of the system is COP 
100
 0.7274 .
137.47
Example 9.18 a. Drive a relationship for maximum COP of a vapor absorption refrigeration system at which the generator, and the evaporator temperatures are Tg and Te respectively. The heat rejection at the absorber and the condenser takes
place at the same environmental temperature of To .
b.
Determine COP max of the system in Example 9.17 by applying the findings in part (a) and compare with the
actual COP value.
444
THERMODYNAMICS
Solution:
In Figure 9.47, the vapor absorption system is represented as a combination of heat engine and a refrigerator. If we apply
first law of thermodynamics stated for cycles as,
ance as,
  Q   W to this system, we end up with the following energy balQ e  Q g  Q a  c  W p
In accord with the second law, the total entropy change of the system and the environment has to be positive or zero at the
limit.
Stotal  Ssys  So  0
Since the system completes a cycle and works at steady state conditions, Ssys  0 , then the rate of entropy change of
environment becomes
Q g Q a  c
Q

0
So   e 
Te Tg
To
Figure 9.47 Reversible absorption refrigeration cycle as a combination
heat engine and refrigerator
Expressing Q a  c with respect to Q g and Q e by using the energy equation and substituting into So expression yields,
 Tg  To
Q g 
 Tg


 T T
  Q e  o e

 Te

 
  W p

Q e  Te / To
COP


Neglecting the pump work, COP of the system becomes,
Q g  1  Te / To
  To
 1 
  Tg




 T   T /T 
For a reversible cycle then COP assumes the maximum value as COP max  1  o   e o 
 Tg   1  Te / To 




 
Heat Engine Refrigerator
A shown in Figure 9.47, a reversible absorption cycle can be considered to be a combined system of reversible heat
engine and a reversible refrigerator and the system COP becomes the product of engine efficiency and the refrigerator COP.
As can be deduced from COP max expression, the COP of the system increases as the generator ( Tg ) and evaporator ( Te )
temperatures increase and the surroundings temperature ( To ) decreases.
CHAPTER 9 REFRIGERATION SYSTEMS 445
b.
In Example 9.17, the component temperatures are Tg  363K , Te  279 K , and To  300 K . At this condition,
79 / 300 
 300  279
the maximum value of COP is COP max  1


 363  1 279 / 300 
actual COP .
9.6.2
2.305 and is three times larger than the
Refrigerant-Absorbent Pairs
The following properties are desirable for refrigerant-absorbent type of binary solutions:
1. The solution should have high thermal conducti210231vity and low viscosity for high performance.
2. There should be neither crystallization nor solidification of the absorbent for the predicted
working conditions of the system.
3. A large difference in boiling temperatures between the refrigerant and the absorbent is desirable. Thus, only pure refrigerant may circulate through the evaporator or the condenser of the
system.
4. Heat of mixing must be as small as possible.
5. The refrigerant should exhibit solubility as high as possible.
6. The solution should be chemically stable, non-corrosive, and inexpensive.
Providing the above indicated features, the most commonly used refrigerant-absorbent pairs are
presented in Table 9.3.
Table 9.3 Commonly used refrigerant-absorbent pairs
Refrigerant
Absorbent
Absorber state
Ammonia
Water
Liquid
Ammonia
Lithium nitrate
Solid
Ammonia
Calcium chloride
Solid
Water
Lithium bromide
Solid
Methylene Chloride
Dimethyl ether
Liquid
9.6.3
Crystallization
The equilibrium chart (Figure 7.31) and the enthalpy chart (Figure 7.34) both contain crystallization zone where LiBr salt solidifies and blocks the pipes and valves. When a hot and strong solution
is cooled to low temperatures in a heat exchanger, crystallization may take place. To avoid this to
happen, decrease in condenser pressure due to low cooling water temperature has to be prevented.
In industrial applications, the condenser pressure is maintained at certain level irrespective of the
cooling water temperature by the following methods: 1. Depending on the condenser pressure, the
cooling water flow rate is regulated. As the condenser pressure increases due to vapor accumulation,
the flow rate should be increased. To prevent the drop of condenser pressure below a certain value,
the flow rate should be lessened. 2. Additives are used for preventing the crystallization.
Keeping condenser pressure at certain level is against the performance of the system, but has to
be done for proper operation of the system.
446
THERMODYNAMICS
9.7
Miscellaneous Refrigeration Methods
Referring to Figure 9.2 for classification of refrigeration methods, the most commonly applied
methods of refrigeration in miscellaneous group include: 1. Ejector refrigeration, 2. Peltier cooling,
3. Magneto-electric refrigeration, and 4. Evaporative cooling.
9.7.1
Ejector refrigeration
In ejector refrigeration, ejector or jet pump is a thermally driven system that can be used for cooling applications as shown in Figure 9.48. When ejector is used instead of an expansion valve, the
high pressure primary fluid from the condenser flows through the ejector, and accelerates through
the nozzle. The reduction in pressure induces the low pressure vapor refrigerant from the evaporator
at state 7, known as the secondary fluid.
The two fluids mix at the mixing section, and enter the diffuser section where the pressure recovery occurs. The mixed fluid flows to gas-liquid separator, and the liquid portion passes through the
expansion valve before entering the evaporator. The pressure of the remaining vapor is raised to the
condenser pressure by the compressor of the cycle. In comparison with conventional systems, the ejector provides the following advantages: 1. Ejector utilizes the energy of high pressure liquid otherwise
dissipated in the expansion valve, 2. Ejector works as a pump and reduces the compressor work.
9.7.2
Thermo-Electric Cooling
As shown in Figure 9.49a, thermoelectric cooling is created by passing electric current through
the circuit built by two different wires (usually bismuth-telluride material is used). The intensity of
the current is proportional to the temperature difference between the two ends of the circuit and this is
known as Peltier effect. A thermoelectric module is manufactured by placing ceramic wafers between p
and n poles of bismuth-telluride material. The p and n poles make a couple and a module may contain
several hundreds of these couples. In Figure 9.49b, depending upon the intensity of current flowing
through the module, the upper and the lower plates assume different temperatures.
CHAPTER 9 REFRIGERATION SYSTEMS 447
Figure 9.49c shows the assembly of a simple thermoelectric refrigeration system. To reduce the
effect of thermal contact resistance, the module in the figure is installed through mechanical clamping, epoxy bonding, or solder bonding. Thermoelectric cooling is used in portable refrigerators, space
applications, micro-processors, cameras, laser devices. Thermoelectric devices are light in weight,
small, quite, and inexpensive, and may function in environments that are too severe, too sensitive,
and too small for other refrigeration methods.
9.7.3
Evaporative cooling
Evaporative coolers utilize the evaporation of water to cool the air. As shown in Figure 9.50, the
operation principle is the same as that used in cooling towers. The warm outdoor air is brought into
contact with water to cool it to a temperature close to the wet bulb temperature of air. The cooled air
is used for human comfort and as well as for certain processes in textile industries. Large commercial
systems employ cellulose filled pads over which water is sprayed. Due to low cost of operation, this
type of cooling is especially attractive for comfort cooling in dry regions.
448
THERMODYNAMICS
9.7.4
Magnetic refrigeration
There are two main reasons why magnetic refrigeration is important in today’s refrigeration
technology. First, even though a magnetic refrigerator would cost more than today’s refrigerators, it
would conserve above 20% of energy that the current expansion-compression refrigerators consume.
Hence, the operating costs are drastically reduced. Second, there exist serious concerns about the
ecological impact of the conventional refrigeration systems, and a magnetic refrigerator is totally
environmentally friendly technology.
Magnetic refrigeration relies upon the temperature change of some materials when exposed to
a changing magnetic field. These materials are called magneto caloric effect (MCE) materials, and
Gadolinium (Gd) exhibits one of the largest known magneto caloric effect. At room temperature
refrigeration, however, Gadolinium based alloys, GdSiGe alloys, are more efficient. As explained in
Figure 9.51a, the magneto caloric effect (MCE) results from coupling of a system of magnetic moments
with an external magnetic field, and the outcome is the cooling or heating of the material. Referring
to Figure 9.51a, a magnetic refrigeration cycle is composed of the following steps: 1. The sample of
magneto caloric material (MCM) is at ambient temperature, and is magnetically in disorder state, 2.
The MCM sample is moved into a magnetic field, and this process causes the sample to be heated up,
3. The sample MCM is cooled down by a convective cooling process, 4. The temperature is reduced to
ambient temperature and the sample is at magnetically ordered form, 5. The sample is removed from
applied magnetic field, its temperature drops and thus cooling effect results. The sample temperature
rises to ambient temperature and becomes magnetically disordered.
Figure 9.51 Magnetic refrigeration
CHAPTER 9 REFRIGERATION SYSTEMS 449
As shown in Figure 9.51b, in producing continuous refrigeration effect, the element Gd, in powdered form, is stuffed in pockets inside a ring shaped regenerator. The regenerator rotates and the
powdered (MCE) material is placed in and out of a gap where a powerful magnetic field exists. Hence,
a continuous refrigeration effect is created by consuming work for rotating the regenerator disc.
References
1.
A.R. Trott, and T.C. Welch, Refrigeration and Air-Conditioning, 3rd edition, Butterworth-Heinemann, ISBN-0-75064219-X, 2000.
2.
S. Kakaç, A.E. Bergles, F. Mayinger, and H. Yüncü, Heat Transfer Enhancement of Heat Exchangers, Nato Scientific
Affairs Division, Kluwer Academic Publisher, ISBN 0-7923-5637-3, 1998.
3.
S.K. Wang, Handbook of Air Conditioning and Refrigeration, 2nd edition, McGrawHill Publications, ISBN: 0-07068167-8, 2001
4.
T. Kuppa, Heat Exchanger Design Handbook, Marcel Dekker Inc., ISBN: 0-8247-9787-6, 2000.
5.
R. Miller, and M.R. Miller, Air Conditioning and Refrigeration, McGrawHill Publications, ISBN: 0-07-146788-2,
2006.
6.
S.G. Kandlikar, and M.S.V.K. Dhir, Handbook of Phase Change: Boiling and Condensation, Taylor and Francis, ISBN:
1-56032-634-4, 1999.
7.
P.K. Bansal, and A.S. Rupasinghe, An Empirical Model for Sizing Capillary Tubes, International Journal of Refrigeration, Vol.19, no:8, pp. 497-505, 1996.
450
THERMODYNAMICS
Problems
Refrigeration cycle fundamentals
9.1
A new plastic molding facility has a large amount of
waste heat available at 150oC. The local temperature
of the surroundings is 22oC. As an entry level engineer, you are to investigate possible energy savings
by determining,
a. the thermal efficiency of a Carnot engine,
b. the coefficient of performance of a Carnot heat
pump,
c. the coefficient of performance of a Carnot refrigerator operating between these temperatures.
9.2
A large commercial refrigerator has a coefficient of
performance of 3.7, and is driven by a 10 HP electric
motor. Determine the refrigeration capacity of this
refrigeration unit in tons of refrigeration.
9.3
9.4
For an ideal vapor-compression refrigeration cycle
using R-22, the evaporator temperature is -10oC, and
the condensing temperature is 30oC. Assuming that
the refrigerant leaves the evaporator as saturated
vapor, and exits the condenser as saturated liquid,
sketch the cycle on pressure-enthalpy diagram and
evaluate,
a. the specific compression work,
b. the heat absorbed at the evaporator,
c. the heat rejected by the condenser,
d. the coefficient of performance of the unit.
A refrigeration unit is to be designed for a meat
market and uses R-22 refrigerant to maintain meat
at 0oC while operating in an environment of 37oC.
The refrigerant leaves the evaporator as saturated
vapor, and exits the condenser as saturated liquid.
Assume that the compressor is isentropic, and the
temperature of 1000 kg of meat initially at surroundings temperature has to be dropped to 0oC in 1 hour.
Take c p  3.34 kJ/kgK for meat, and determine,
a. the minimum power required for reducing the
meat temperature to 0oC,
b. the power needed for keeping the storage room
at 0oC, if the total heat loss from the room is
12kW.
c. Compare these two values of power and decide
what power must be supplied to the refrigeration
unit.
9.5
A vapor-compression refrigeration unit as shown in
Figure 9.52 is used to maintain the room temperature
at 17oC when the ambient temperature is at 37oC.
Saturated vapor enters the adiabatic compressor at
0oC. The isentropic efficiency of the compressor is
82%. The refrigerant leaves the condenser at 50oC
as saturated liquid. Assume that no pressure drop
occurs in the evaporator and the condenser or in the
connecting pipes. If the following refrigerants are to
be used as the working fluid of the system,
a. determine the power input for reducing the
ambient air from 37oC to 17oC with 15 m3/min
of volumetric flow rate at the unit exit,
b. select the suitable working fluid for this particular
application.
Refrigerants to be used: 1. Refrigerant R22, 2. Refrigerant R134a, 3. Refrigerant R717 (Ammonia)
9.6
Determine the variation of COP of the refrigeration unit
in Problem 9.5 for evaporator temperatures varying in
the range of -10oC and +10oC with 5oC of increments
by using three different working fluids as R22, R134a,
and R717 in the system. Tabulate the results.
9.7
A vapor-compression refrigeration system uses R-22
and the compressor isentropic efficiency is 88 %
when the evaporator and the condenser temperatures
respectively are -20°C and +25°C. The refrigerant
enters the compressor as saturated vapor, and it is
saturated liquid at the condenser exit. For refrigerant
mass flow rate of 7.5kg/min, determine,
a. the volumetric flow rate of refrigerant at the
compressor inlet,
b. the refrigeration capacity in tons of refrigeration,
c. the coefficient of performance (COP) of this
unit.
CHAPTER 9 REFRIGERATION SYSTEMS 451
9.8
A refrigeration unit, as in Example 9.2, includes a
liquid-to-suction exchanger. The refrigerant vapor
from the evaporator is heated from -12°C to +4°C
with the liquid flowing from the condenser at 40°C.
Assume that the compressors have 82% of isentropic
efficiency for both cases indicated below. Evaluate,
a. the COP of the unit without the heat exchanger,
b. the COP of the unit with the heat exchanger,
c. the refrigeration capacity of the unit without the
heat exchanger for refrigerant flow rate of 20L/s
at the compressor inlet,
d. the power needed to run the compressor in (c).
e. With the same compressor power in (d), determine the refrigeration capacity of the unit with
the heat exchanger.
9.9
As shown in Figure 9.53, an ideal vapor-compression
refrigeration unit with ammonia as the working
fluid, the evaporator temperature is at -30°C, and
the condenser operates at 10bar pressure. The mass
flow rate of ammonia is 1.2 kg/min. Determine,
a. the refrigerating capacity in tons of refrigeration,
b. the coefficient of performance of the unit.
c. Consider the case of keeping all conditions except
the condenser pressure of the cycle constant. The
condenser pressure is varied between 10bar and
20bar with 5bar increments. Calculate the variation of COP with respect to condenser pressure,
and graph the results in COP-p diagram.
e. the rate of exergy destruction in the condenser
as a percentage of the power input.
Figure 9.54 Actual cycle with pressure drop and
(R-s) representationv
9.11
9.10
As shown in Figure 9.54, in a refrigeration unit with
a capacity of 15 tons, the refrigerant R134a leaves the
evaporator as saturated vapor at 4 bar, and enters the
adiabatic compressor at 3.2 bar, 10°C and exits at 12
bar, 60°C. Water enters to water cooled condenser
at 17°C and exits at 27°C. The pressure and the
temperature at the condenser exit respectively are
10 bar, 35°C. Assume To  17o C , and calculate,
a. the mass flow rate of refrigerant,
b. the power input for the compressor in kW,
c. the mass flow rate of the cooling water,
d. the coefficient of performance of the system,
A manufacturer of vapor-compression refrigeration
machines tests its units by directing the condenser
heat to the evaporator and adding additional cooling
necessary to fully cool the unit. As shown in Figure
9.55, the refrigerant is R717 with the mass flow
rate of 600 kg/min. The isentropic efficiency of the
compressor is 85%. A 60% ethylene glycol water
solution with density of 1100 kg/m3, and specific
heat of 3 kJ/kg K is used in experiments as condenser
cooling fluid. The ethylene glycol solution enters
the condenser at 0°C and exits at 25°C. The states
of the refrigerant in completing the cycle are given
in the above table. Determine,
a. the cooling capacity of the unit in tons of refrigeration,
452
THERMODYNAMICS
b. the mass flow rate of ethylene-glycol solution
c. the cooler capacity and the temperature of the solution at the cooler exit.
Figure 9.55 A schematic of a test set-up for refrigeration systems
State no.
1
2s
3
4
9.12
Location
Compressor inlet
Compressor exit
Condenser exit
Evaporator inlet
Pressure (bar)
2
7
7
2
Temp.(°C)
-5
80
13.8
-18.86
Quality(%)
1.0
0.0
-
Entropy(kJ/kgK)
5.722
5.722
0.939
-
Enthalpy(kJ/kg)
1452
1625
244.69
244.69
Compare the refrigeration effect, the compressor work, and the exergetic efficiency for refrigerants R22 and R404A
for an ideal refrigeration unit having +5°C evaporator and 55°C condenser temperature. The thermodynamic properties of R404A are given in a table at the bottom of the page:
State no.
1
2s
3
4
Location
Compressor inlet
Compressor exit
Condenser exit
Evaporator inlet
Pressure (bar)
7
26
26
7
Temp.(°C)
+5
60
55
-
Quality(%)
1.0
0.0
-
Entropy(kJ/kgK)
1.603
1.603
1.28
-
Enthalpy(kJ/kg)
367.62
390
287
287
Multi pressure refrigeration
9.13
An ideal two-stage compression refrigeration system as
shown in Figure 9.56, with evaporator capacity of 20 tons
of refrigeration at -32°C is equipped with intercooler operating at 5bar. The saturated vapor exiting the intercooling
tank is compressed to 16bar of condenser pressure. R717
(ammonia) is the working fluid of the system. Determine,
a. if the system has a single compressor, the power required
for the same evaporator and condenser conditions,
b. the total power required by the two-compressor system,
c. the COP of the two-compressor system.
d. Including the saturation curve, graph the h-p diagram
of the cycle.
Figure 9.56 A two-compressor refrigeration system with intercooling
CHAPTER 9 REFRIGERATION SYSTEMS 453
9.14
Consider the two-stage vapor-compression unit as illustrated in Figure 9.56, and assume that R134a is the working
fluid. The system produces 3200 tons of refrigeration with the evaporator, intercooler and the condenser operating at
pressures of 1bar, 6bar, and 20bar respectively. Additionally, the isentropic efficiencies of both compressors are 90%.
The thermodynamic properties of the refrigerant at various states of the cycle are provided by the table below,
State no.
Location
Pressure (bar)
hf (kJ/kg)
hg (kJ/kg)
sf (kJ/kgK)
s (kJ/kgK)
1
2s
3,7
4s
5
Compressor inlet
Compressor exit
Flash chamber exit
Compressor exit
Condenser exit
1
6
6
20
20
79.48
150
231.35
259.19
281
0.29
-
0.939
0.939
0.909
0.909
Enthalpy
(kJ/kg)
Evaluate,
a. the mass flow rate of the refrigerant flowing through each compressor in kg/s,
b. the total power required to run the system in kW,
c. the COP of the system.
9.15
A two-stage and ideal vapor-compression refrigeration
unit as in Figure 9.56, operates with R134a, and the
pressures of the evaporator, intercooler, and of the
condenser are 1.2bar, 7bar, and 24bar respectively.
If the refrigeration load is estimated to be 10 tons,
calculate,
a. the ratio of mass flow rates, r  m 3 / m 1 ,
b. the COP of the system,
c. the heat removal rate at the condenser.
9.16
A frozen food manufacturer needs 25tons of refrigeration for storing frozen food at -18°C of evaporator
temperature. As shown in Figure 9.57, the same
refrigeration system is also used for air conditioning
of the office buildings with another evaporator operating at +5oC with a capacity of 8tons. The condenser
operates at 60oC, and the isentropic efficiency of the
compressor is 80%. The refrigerant of the cycle is
R22. Including the saturation curve, graph the h-p
diagram of the system and determine,
a. the enthalpy of the refrigerant at state 1,
b. the power of the compressor,
c. the mass flow rate of condenser cooling water,
if the water temperature rise is 10°C,
d. the COP of the system.
Figure 9.57 A two-evaporator
refrigeration system
9.17
Consider a two-stage refrigeration system that operates with -36oC evaporator and +60°C condenser
temperature. The cycle consists of two ideal vapor
compression cycles connected by a heat exchanger
that serves as the condenser for the low pressure
cycle and the evaporator for the high pressure. The
evaporating and the condensing temperatures of
both fluids in the exchanger are 0°C. Assume that
R22 is the working fluid for the two cycles, and the
heat capacity of low pressure evaporator is 55 tons
of refrigeration.
a. Evaluate the power required and the COP for a
single stage system having the same evaporator and condenser conditions of the cascade
system.
b. Calculate the mass ratio of the two refrigerant
flows.
c. Determine and compare the total power input,
and the COP values of the cascade system with
the corresponding values for the single stage
system.
9.18
A vapor compression refrigeration system with
a cooling capacity of 100 kW uses three-stage
compression with intercooling, and R134a is the
working fluid. As shown in Figure 9.58, the system
is equipped with two intercoolers and two mixing
chambers, and all the compressors of the system
are isentropic. Assume that the refrigerant is saturated liquid at the inlet of each expansion valve,
and is saturated vapor at the evaporator outlet. The
thermodynamic properties of refrigerant at various
states are given by the table above.
Determine,
a. the mass flow ratios y13  m 13 / m 10 , and
y14  m 14 / m 8 supplied to mixing chambers,
b. the mass flow rates for each compressor,
c. enthalpy and entropy values of the refrigerant
at the mixing chamber outlet (states 3 and 5),
d. power supplied to each compressor in kW,
454
THERMODYNAMICS
e. the heat capacity of the condenser,
f. COP of the system.
State no.
Pressure (bar)
hf (kJ/kg)
hg (kJ/kg)
Quality(%)
Entropy(kJ/kgK)
Enthalpy(kJ/kg)
1
1
16.29
231.4
1.0
0.939
-
2
5
71.31
256.1
-
0.939
264.2
4
9
99.54
266.2
-
6
12
115.2
271
-
-
Figure 9.58 A three-compressor refrigeration system
9.19
9.20
A new commercial refrigerator-freezer combination unit is designed as a dual evaporator system.
The freezer compartment is to be at -32°C and the
refrigerator compartment is at +4°C. The system
uses R134a which is saturated liquid at 40°C at the
condenser outlet. The cooling capacity of both the
refrigerator and the freezer is the same and is 0.5kW
for each. The compressor isentropic efficiency is
87%. Draw the system schematically, and show the
cycle on h-p diagram. Determine,
a. the required mass flow rate through each evaporator, and the compressor.
b. the enthalpy of the refrigerant at the compressor
inlet,
c. the power to be supplied to the compressor,
d. the coefficient of performance of this design.
A cascade and ideal vapor-compression refrigeration
system as in Figure 9.59 uses propane for low pressure
and ammonia for high pressure cycles. The evaporator with a refrigeration capacity of 12 tons operates
at -40oC, and the condenser temperature is +50°C.
Assume that the heat is transferred at the cascade
condenser by 8°C temperature difference between
the refrigerants, and the evaporation temperature
of ammonia is 0°C. Neglect the heat exchanger
effect, graph the h-p diagram of the system, and
determine,
a. the total power required by the cascade system,
b. the COP of the cascade system.
c. Suppose that the cascade system is replaced with
a single stage ammonia system operating at the
same evaporator and condenser temperatures as
given above, determine the power required by
this new system,
d. evaluate the COP value of this single stage ammonia system and compare with the COP of the
cascade system.
CHAPTER 9 REFRIGERATION SYSTEMS 455
9.21
A single stage ideal vapor-compression cycle with
evaporator capacity of 50 tons operates between
-26°C evaporator and +50oC condenser temperatures, and uses R22 as the refrigerant. This system
is modified to a two-stage system as shown in
Figure 9.60 with the intercooling pressure at 6bar.
Assume that the refrigerant is always at saturated
state at the exit of evaporator, condenser, and the
intercooler. Graph the modified two-stage cycle on
h-p diagram, and evaluate,
a. the power needed by the original single stage
system,
b. the total power to be supplied to the revised
two-stage system, and the COP of the system.
Figure 9.61 A semi-hermetic six-cylinder refrigeration compressor
9.23
Figure 9.60 An alternative two-stage refrigeration system with intercooling
Refrigeration components
9.22
The volumetric efficiency, v of a reciprocating com-
n
pressor may be approximated as, v  1     r1/
p ,
where  is the ratio of the clearance volume to the
piston displacement volume, rp is the compressor
pressure ratio, and n is the polytropic exponent.
As shown in Figure 9.62, assume that   0.08 ,
Vleak / Vst  0.05 , and the cylinder stroke volume
is Vst  50 cm3 . The compressor is to be used with
ammonia, n = 1.25, and operates at a speed of
1200rpm. For evaporator temperature varying in
the range between -20°C and +10°C with 10°C of
increments, and the condenser temperature being
kept constant at +60°C, then determine and plot the
variation of the following parameters with respect
to evaporator temperature,
a. the refrigeration capacity of the evaporator,
b. the compressor power input,
A semi-hermetic reciprocating compressor as shown
in Figure 9.61 is used with refrigerant R22 and has
6 cylinders with 80mm of bore, 120mm of stroke
length. The clearance volume is 3-percent of the
displacement volume and the polytropic exponent is
1.16. The wall leaks during the compression process
are estimated to be 5% of the stroke volume. Assume
that +4°C of superheat and 2 kPa of pressure drop
occur at the suction line of the compressor. If the
compressor is to be used for a refrigeration system
operating between -18°C of evaporator and +60°C
of condenser temperature and if the compressor runs
at a speed of 1000 rpm, then determine,

a. the volumetric efficiency of the compressor, v
b. the discharged volumetric efficiency, vd
c. the mass flow rate of refrigerant at the compressor exit,
d. the evaporator capacity in tons of refrigeration,
e. the isentropic efficiency of the compressor,
f. the exit temperature of the refrigerant.
Figure 9.62 A schematic view of a refrigeration compressor
456
9.24
9.25
9.26
THERMODYNAMICS
condenser pressure at 12bar. The refrigeration load
of the chiller is 50 tons. Assume that the compressor
with an isentropic efficiency of 80% operates at a
speed of 3000rpm. Determine the impeller wheel
diameter for the following refrigerants:
Analyze the effect of condenser temperature on the
COP of the refrigeration cycle if the following reciprocating compressor with R22 refrigerant is used for
the system (the polytropic exponent, n  1.13 ). The
compressor characteristics are: d  50mm , H  40mm
,   0.035 , Vleak / Vst  0.06 , n  1000rpm , and
the number of cylinders is, z  4 . The evaporator
temperature is kept constant at -18°C and the condenser
temperature varies in the range between +32°C and
+56°C with 8°C of temperature increments.
A single cylinder reciprocating compressor is to
be designed for a domestic refrigerator with 100W
of cooling capacity. The refrigerator operates at
evaporator temperature of -18°C and the condensing temperature is +56°C, and R134a is used as the
refrigerant of the system (the polytropic exponent,
n  1.15 ). The compressor characteristics are as
follows:   0.045 , Vleak / Vst  0.04 , and the
rotational speed is n  2900rpm . Assume that the
bore and the stroke are identical, and determine,
a. the cylinder diameter,
b. the compressor power input,
c. the refrigerant temperature at the compressor
exit,
d. the COP of the system.
a. R134a
b. Ammonia.
c. Comparing diameters, and the tip speeds, decide
which refrigerant is more appropriate for the
given system.
d. Calculate the impeller width at the compressor
exit for the selected refrigerant.
9.28
A method for reducing the impeller diameter to reasonable limits in refrigeration systems is to increase
the number of stages. As shown in Figure 9.64, the
compression might be isentropic and executed in
three stages with 3600rpm rotational speed. Ammonia is compressed from saturated vapor at 2 bar
to condenser pressure at 8 bar. The refrigeration load
of the chiller is 50 tons. If the all impeller wheels
are at the same diameter, determine,
a. the wheel diameter,
b. the impeller width for 1 m/s velocity at each
impeller exit.
Calculate the power required by two compressors in
an ammonia system which serves a 45 ton of evaporator operating at -25°C. The system uses two-stage
compression with intercooling, and the compressors
are identical with 88% of isentropic efficiency. The
condensing temperature is 40°C.
9.29
As shown in Figure 9.65, a thermostatic expansion valve in R134a refrigeration system supplies
refrigerant to an evaporator coil and is set to +4°C
superheat in order to open the valve at an evaporator
temperature of -8°C.
a. Calculate the pressure difference between the
opposite sides of the diaphragm to open the
valve.
9.27
A centrifugal compressor as in Figure 9.63 is to
be used for chilling water, and the refrigerant has
to be compressed from saturated vapor at 2 bar to
b. Suppose that the same valve is used in a refrigeration system with the evaporator operating
at 4bar pressure. The pressure drop across the
evaporator is 0.4 bar. Calculate the superheat
required to open the same valve.
CHAPTER 9 REFRIGERATION SYSTEMS 457
9.34
A capillary tube with a diameter of 1mm and 2.1m
of length is to be used for R22 refrigeration system
operating between 0°C of evaporator, and +45°C of
condenser temperatures. Determine the maximum
refrigeration capacity of the refrigeration system.
9.35
An ammonia refrigeration plant with a capacity of 50
tons of refrigeration operates between -18°C of evaporator and +40°C of condenser temperatures, and the
isentropic efficiency of the compressor is 88-percent.
As shown in Figure 9.66, the condenser of the system
is a shell-and-tube type exchanger with two tube passes,
and contains a total of 50 copper tubes with diameters
of 16mm/18mm. The inlet and outlet temperatures of
cooling water respectively are 20°C and 30°C. Determine the tube length of the condenser.
Figure 9.64 A three-stage cenrifugal
refrigeration compressor
9.30
A thermostatic expansion valve is used in R22 based
refrigeration system operating at 0°C of evaporator
temperature. The adjustable spring of the valve is
set to require 40 kPa of additional pressure to move
the needle.
a. Ignoring the pressure drop in the evaporator,
determine the degree of superheat at the evaporator exit.
b. If the evaporator pressure drop is 10kPa, what
will be the degree of superheat at the evaporator
outlet?
9.31
An expansion valve manufactured for ammonia refrigeration system provides 10 tons of refrigeration when the
pressure difference across the valve is 11.5bar which
corresponds to -16°C of evaporating temperature and
+35°C of condensing temperature. At these conditions,
the valve causes 6°C of superheat at the evaporator
outlet. If the same valve with the same spring force is
used for an evaporator operating at -4°C, the pressure
drop across the valve is measured to be 10 bar. Evaluate
the refrigeration capacity at these conditions.
9.32
9.33
A refrigerator that operates with an evaporator temperature of -10°C and a condenser temperature of
+30°C uses R134a as the working fluid. Saturated
liquid refrigerant from the condenser flows through
the expansion valve into the evaporator. The refrigerant at the evaporator outlet has 5K of superheat.
a. For a cooling rate of 6 kW, determine the mass
flow rate of refrigerant through the valve and
the spring pressure on the valve diaphragm.
b. Suppose that the evaporator load suddenly increases
to 7 kW, evaluate the initial pressure difference
on the diaphragm caused by this change.
A capillary tube is to be designed for a freezer operating
between -12°C of evaporator and +40°C of condenser
temperatures. Refrigerant R134a with a mass flow
rate 0.005 kg/s enters the tube at 35°C. If the tube
diameter is selected to be 2mm, then determine the
tube length by using the graphical method.
Figure 9.66 Tube layout for two-pass water
cooled condenser
9.36
A fin-and-tube type air cooled condenser operates at
constant refrigerant temperature of 50°C. Refrigerant
R22 enters the condenser as saturated vapor and exits
as saturated liquid. Air enters the condenser at 35°C
with a volumetric flow rate of 2 m3/s. The air side
surface area is 45 m2 and the overall heat transfer
coefficient is 28 W/m2K. Determine,
a. the temperature of air at the condenser exit,
b. the heat rejection rate in kW,
c. the mass flow rate of the refrigerant.
9.37
As shown in Figure 9.67, a fin-and-tube type heat exchanger with s1  50 mm , s2  40 mm , t  0.2 mm ,
s f  125 m-1,   22.29 , di / d 0  14 / 16 , and
 f  0.8 . The number of tubes per row, n1  30 , and
the number of rows is, n2  4 . The heat exchanger
is made of copper tubes and aluminum fins, and is to
be used as air cooled condenser of R22 refrigeration
plant. The plant with a refrigeration capacity of 40
tons operates between -10°C of evaporator and +32°C
of condenser temperatures, and uses hermetic type
compressor. Air inlets the condenser on the finned
side at 20°C and exits at 25°C. Determine,
a. the cooling capacity of the condenser,
b. the surface area and the length of the condenser
tubes.
458
THERMODYNAMICS
constant temperature of +4°C. For a fin efficiency
of 85%, calculate,
a. the UA value of the evaporator.
b. the required mass flow rate of the refrigerant.
9.40
Figure 9.67 Air cooled condenser
9.38
A water cooled R22 condenser as shown in Figure
9.68 has four tube passes and a total of 44 copper
tubes with diameter of 20mm/16mm is installed.
The tube layout is presented in Figure 9.68. If the
refrigeration system operates between -4°C of
evaporator and +40°C of condenser temperatures
and uses an open type compressor, the condenser
capacity is determined to be 100 kW. For cooling
water inlet and outlet temperatures at 20°C and 30°C
respectively, calculate,
a. the evaporator cooling capacity in kW,
b. the refrigerant flow rate in kg/s,
c. the mass flow rate of water in kg/s,
d. the total heat transfer coefficient of the condenser
in W/m2K,
e. the condenser tube length in m.
A fin-and-tube type heat exchanger, as in Figure 9.67,
has the following geometric properties: s1  50 mm ,
s2  40 mm , t  0.2 mm , s f  125 m-1,   22.29 ,
di / d 0  14 / 16 . The number of tubes per row is,
n1  30 , and the number of rows is, n2  4 .the
exchanger is made of copper tubes and aluminum
fins, and is to be used as an evaporator for R22 refrigeration plant with 40 tons of capacity. The plant
operates between -18°C of evaporator and +40°C of
condenser temperatures. The refrigerant R134a is
distributed into four circuits at the evaporator inlet,
so that each row of 30 tubes makes one circuit. The
air temperature is dropped from +5°C to -5°C by
flowing through the finned surface. Determine,
a. the refrigerant mass flow rate,
b. the volumetric flow rate of air at the evaporator
inlet in m3/min,
c. the required evaporator length for a fin efficiency
of  f  0.82 .
9.41
Experimental data for a water cooled shell-andtube type R22 condenser yields the following correlation for the overall heat transfer coefficient:
104 / U o  1  4 / Vi0.8 , where Vi is in m/s, and U o
is in W/m2K. Water flows inside of copper tubes
( k  290 W/mK) at a velocity of 0.25m/s, and the
inlet and outlet temperatures respectively are 25°C,
and 35°C. The condenser is equipped with 80 tubes
in 3m of length with inside and outside diameters at
35 mm and 40 mm. The refrigeration system operates
between 0°C of evaporating and 45°C of condensing
temperatures. Determine,
a.
b.
c.
d.
Figure 9.68 Tube layout for four-pass water
cooled condenser
9.39
A direct-expansion and fin-and-tube type evaporator
has the following properties: 1. Refrigerant side: surface area, Ai  20 m 2 , and heat transfer coefficient,
hi  1100 W/m 2 K , 2. Air side: the bare tube surface
area, Ato  18.5 m 2 , and the total area of the finned
surface, Ao  180 m 2 and the heat transfer coefficient,
ho  50 W/m 2 K . Air inlets the evaporator at 20°C
and exits at 12°C. The evaporator operates at a
the heat rejection rate of the condenser in kW,
the water flow rate in kg/s,
the evaporator cooling rate in kW,
the mass flow rate of the refrigerant in kg/s.
9.42
Provide the refrigerant R number of the following
fluids:
a. CClF3,
b. CHF3,
c. CO2,
e. CH2ClF.
d. H2O,
9.43
Determine the chemical formula of the following
refrigerants and specify whether the refrigerant is a
halogenates or inorganic compound. If it is a mixture, specify whether it is azeotropic or zeotropic
mixture:
a.
d.
R290,
R502,
b. R11,
e. R404A
c. R744,
f. R600
CHAPTER 9 REFRIGERATION SYSTEMS 459
Heat pumps
Figure 9.69 Air-to-air heat pump for heating of
a dwelling
9.44
As shown in Figure 9.69, an air-to-air heat pump
with R22 as the working fluid maintains a building
at 22°C when the outside temperature is at +5°C and
supplies heat at a rate of 50 kW. The evaporator and
the condenser pressures respectively are 3 bar, and
14bar. A hermetic type reciprocating compressor
is used for the heat pump and furnished with the
following characteristics:
1. The cylinder diameter and stroke: D  100 mm ,
H  80 mm , 2. The clearance volume:   0.02 , 3.
Percent leakage: Vleak / Vst  0.05 4. The number
of cylinders: z  4 , 5. The isentropic efficiency:
c  85% . Determine,
a. the power input to the compressor,
b. the coefficient of performance, COP, and the
exergetic efficiency of the heat pump,
c. the volumetric efficiency of the compressor
used,
d. the rotational speed of the compressor.
9.45
The heat pump in Figure 9.70 uses ground water as
a thermal source. Water enters the evaporator of the
pump at 15°C and leaves at 7°C. Refrigerant R134a
at a condition of 3.2 bar, and +4°C (state1) inlets the
compressor and exits at 12bar, 60°C of temperature
(state2). Pressure loss due to flow in the condenser is
negligible, but the refrigerant exits the condenser as
subcooled liquid at 34°C. After the expansion valve,
the refrigerant pressure drops to 3.2 bar. As shown
in Figure 9.70, the indoor air circulates through the
condenser, and the inlet, and outlet temperatures of
air respectively are 18°C and 36°C. For 75 m3/min of
air circulation through the condenser, determine,
a. the heating capacity of the condenser,
b. the compressor power input,
c. the isentropic and the exergetic efficiencies of
the compressor,
d. the coefficient of performance, COP, and the
exergetic efficiency of the heat pump,
e. the mass flow rate of ground water through the
evaporator.
9.46
Figure 9.71 presents the performance plots of a typical heat pump. Let us consider a heat pump with 4.2
tons of heating capacity. Develop linear equations
for the following cases:
a. the heating capacity Q (kJ/h) as a function of
hp
outdoor air temperature, To (°C),
b. the power input Whp (kW) as a function of
outdoor air temperature, To (°C),
c. the heating COP as a function of outdoor air
temperature, To (°C).
d. For a house with UA  50 W/K that is maintained
at Ti  20o C , find the heat pump balance point
outdoor temperature.
Figure 9.71a Heat pump characteristics, heat
supplied.
Figure 9.71b Heat pump characteristics,
power consumed.
460
9.47
THERMODYNAMICS
Using the performance curves shown in Figure 9.71
develop linear equations for a heat pump with 3.8
tons of refrigeration capacity to predict the following
parameters:
a. the cooling capacity Q r (kJ/h) as a function of
outdoor air temperature, To (°C),
b. the power input Wr (kW) as a function of outdoor
air temperature To (°C),
c. If the house requires 2100 kW-h of seasonal
heating, determine the energy saving by using
domestic water instead of air as a heat source.
Vapor absorption refrigeration (VAR)
9.49
The operating temperatures of a vapor absorption
refrigeration system are as follows: 1. Condenser:
85°C, 2. Absorber: 42°C, 3. Evaporator: 5°C. The
refrigeration capacity of the system is 33 tons, and
the heat input to the system is 180 kW. The work
consumed by the solution pump is neglected.
a. Determine the system COP and the heat rejected
at the condenser.
b. An engineer claims that by design improvement
in components of the system, the heat input to
the system may be reduced to 150 kW for the
same operating conditions and the refrigeration
load. Is this true?
9.50
Consider an ideal vapor absorption refrigeration
system that receives heat from a solar collector at a
temperature of 65°C, performs refrigeration at 10°C,
and rejects heat to surroundings at 37°C. Determine
the COP of this system.
9.51
For the LiBr-H2O absorption system as in Figure 9.73,
the solution temperature leaving the heat exchanger
and entering the generator is 45°C. The operating
temperatures of absorber and the generator respectively
are 30°C and 105°C. Moreover, the evaporator and
the condenser operate respectively at temperatures
5°C and 35°C. The flow rate of the solution pump
is 0.4 kg/s. Sketch the system schematically, and
clearly indicate the location of the regenerative heat
exchanger, and determine,
c. the cooling COP as a function of outdoor air
temperature To (°C).
d. It is known that placing the outdoor condensing
unit in the shade could improve the cooling COP
of the system. Assume that the average outdoor
temperature during air conditioning season is
35°C in the sun and 30°C in the shade. If a house
requires 15  106 kJ/year of cooling, and electricity costs $0.10/kW-h, calculate the savings
from locating the condenser in the shade.
9.48
An engineer suggests the use of domestic water
instead of outside air as a heat source for heat pump
heating of a house. Domestic water enters the house
at a higher temperature than the outside air by which
the performance of the heat pump may be increased.
As shown in Figure 9.72, the pump withdraws Q w
from the inlet water at Tw  20o C and decreases its
temperature to surroundings temperature, To  4o C .
The house with UA  65 W/K is maintained at
Ti  20o C , and the heat pump with R22 as the working fluid supplies the necessary heat, and operates
o
between Tsource  8 C of evaporating and + 30o C of
condensing temperatures. If the isentropic efficiency
of the compressor is 82%, determine,
Figure 9.73 LiBr-H2O absorption system for
air conditioning applications
Figure 9.72 Use of domestic water in house
heating
a. the COP value for the following heat sources:
1. Domestic water, and 2. Surroundings air,
b. the required mass flow rate of domestic water
for the given conditions.
a. the flow rates each of the following fluids: 1.
The refrigerant, 2. The weak solution, 3. The
strong solution,
b. the heat transfer rates at the generator and at the
absorber,
c. the refrigeration capacity of the system in tons and
the heat rate rejected at the condenser in kW,
d. the COP of the system.
CHAPTER 9 REFRIGERATION SYSTEMS 461
9.52
9.54
A Diesel engine develops 400 kW of power with
a thermal efficiency of 36-percent. Almost all of
the heat rejected by the engine is transferred to the
generator of an absorption refrigeration system with
COP = 0.82.
a. Derive a relation for the evaporator capacity in
terms of the engine efficiency, the engine power
output, and the COP of the system.
b. Determine the evaporator capacity of the refrigeration system in tons for the above indicated
diesel engine.
9.53
Consider a single effect
LiBr-H2O absorption refrigeration system as in
Figure 9.46. The concentrated solution from the
o
generator at T5  100 C ,
x5  0.65 enters the exchanger of Figure 9.74 and
heats up the solution from
the absorber at T3  30o C
, x3  0.5 . The mass flow
rate of the dilute solution is m 3  0.5 kg/s . Determine
the maximum temperature of the solution at the
generator inlet so that crystallization starts in the
system.
As shown in Figure 9.75, a single effect vapor absorption refrigeration system with LiBr-H2O solution operates at
+4°C of evaporating and 40°C of condensing temperature. The thermodynamic properties of the solution along the
cycle and the mass flow rates are provided by the table below. The refrigerant is assumed to be at saturated vapor
state at the evaporator exit, and at saturated liquid state at the condenser outlet. Determine,
a. the refrigeration capacity in tons,
b. the heat capacity rates of the condenser, the generator, and the absorber,
c. the heat capacity rate and the effectiveness of the solution heat exchanger,
d. the power required by the solution pump,
e. the COP of the system.
State no.
Pressure (kPa)
Temp.(°C)
Enthalpy (kJ/kg)
Mass flow rate(kg/s)
Concentration(x%)
1
0.81
33
85.8
0.05
57
2
7.35
33
85.8
0.05
57
3
7.35
63
147
0.05
57
4
7.35
90
5
7.35
6
0.81
-
-
0.045
7
7.35
77
2645
0.005
0.0
8
7.35
40
167.5
0.005
0.0
9
0.81
167.5
0.005
0.0
10
0.81
2508.7
0.005
0.0
0.045
0.045
4
Figure 9.75 Schematic of LiBr-H2O vapor absorption system
462
THERMODYNAMICS
True and False
9.55
l.
An external equalizer to thermostatic
expansion valve is required when a large temperature lift occurs in the evaporator.
Answer the following questions with T for true and
F for false.
a.
b.
c.
d.
A method of increasing the refrigeration
capacity of a compressor is that while decreasing the evaporator temperature, the condenser
temperature is to be increased.
m.
The cross charging of a thermostatic
valve prevents compressor flooding at high
pressure drops across evaporator.
n.
For a specified refrigeration load, as
the evaporator temperature decreases, the size
of the compressor has to be increased.
In refrigeration systems using capillary
tube, pressure equalization takes place when the
compressor stops.
o.
Because of high index of compression,
ammonia compressors are usually equipped with
cooling water jackets.
The mass flow rate through a capillary
tube assumes a minimum value when the chocked
flow conditions exist.
p.
Receivers are usually used in large
installations for balancing the excess mass
requirement of refrigerant.
Air cooled condensers are simple in
construction and cost less for the same heat
rejection rate.
q.
If a shell-and-tube type heat exchanger
is used as a flooded evaporator, it is usually
equipped with a float valve as an expansion
device.
r.
In using fin-and-tube type evaporators
for low temperature applications, the fin spacing
should be kept larger.
s.
A refrigerant with high critical temperature yields high COP and low volumetric
capacity.
t.
The chemical formula of R22 is
e.
The main differences between actual
and ideal refrigeration cycles are essentially
collected in three groups.
f.
Multi-evaporator systems are usually
needed when refrigeration at different temperatures is required.
g.
The refrigerant used for multi-stage
systems should have boiling pressure above 1
bar, and high specific volume at the evaporator
conditions.
h.
The main advantage of cascade systems
is the freedom of using different refrigerants in
each cycle, but causes overlap of temperatures
at the cascade condenser.
i.
The COP of a refrigeration system becomes higher if a hermetic compressor instead
of open type is used in the system.
j.
In reciprocating compressors, the clearance volume decreases the volumetric efficiency,
and thus reduces the power consumption.
k.
The screw compressors are commonly
appropriate for installations with large mass flow
rate of refrigerant.
CCl3F.
u.
If a refrigerant of zeatropic mixture
leaks, it’s composition changes.
v.
A vapor absorption refrigeration system
with LiBr-H2O is commonly used for frozen food
storage applications.
w.
In LiBr-H2O absorption systems, crystallization is possible to occur when the condenser
pressure falls.
x.
In refrigerant-absorbent pairs, a large
difference in boiling temperatures of the components is desirable.
CHAPTER 9 REFRIGERATION SYSTEMS 463
Check Test 9
Choose the correct answer:
1.
A vapor-compression refrigeration unit uses R22 and works between temperature limits of -18°C, and 27°C. The
isentropic efficiency of the compressor is 79%, and states of R22 in the cycle are given by the following table:
State no.
Location
Pressure (bar)
Temp.(°C)
Quality(%)
Entropy(kJ/kgK)
Enthalpy(kJ/kg)
1
Compressor inlet
2.64
-18
1.0
0.9559
242.92
2s
Compressor exit
12.0
-
-
0.9559
3
Condenser exit
30.2
0.0
0.302
81.9
4
Evaporator inlet
-
-
-
81.9
2.64
For a refrigerant mass flow rate of 5.0kg/min, the cooling capacity of the unit in kW, the actual temperature of
refrigerant at the compressor exit in oC are respectively given as,
a. 13.42, 68,
b. 13.42, 58,
c. 10.42, 68,
3.
d. 11.42, 64.
Ammonia enters the compressor of an actual refrigerator at 150kPa, -20°C at a rate of 0.06kg/s and leaves
at 10bar. The isentropic efficiency of the compressor
is 85%, and the refrigerant is cooled in the condenser
to 24°C. The power input in kW, and the cooling load
of the system in tons are respectively expressed as,
a. 29.9, 19.3,
b. 9.9, 25,
c. 19.9, 19.3,
d. 19.9, 9.9.
2.
If the surroundings temperature is at 300K, the
actual coefficient of performance (COP), and the
exergetic efficiency of the above refrigeration unit
are accordingly determined to be,
a. 3.4, 0.75,
b. 3.0, 0.70,
c. 3.4, 0.65,
d. 3.4, 0.6.
4.
As shown in Figure 9.76, a cascade refrigeration system without a regenerative heat exchanger at the evaporator
outlet with R22 in both loops is used to produce 40 tons of refrigeration with -40°C of evaporating and +36°C of
condensing temperatures. The refrigerant, in loop B, evaporates at -12°C, and in loop A, condenses at -4°C. The
isentropic efficiencies of both compressors are identical and equal to 80%. Considering the data given by the table
above, the mass flow rates of refrigerants in loops A and B respectively are,
a. 0.73, 1.31,
b. 0.63, 1.41,
c. 0.63, 1.11,
d. 0.73, 1.11.
State no.
Location
Pressure (bar)
Temp.(°C)
Quality(%)
Entropy(kJ/kgK)
Enthalpy(kJ/kg)
233.27
Loop A
1
Compressor inlet
1.052
-40
1.0
1.0005
2s
Compressor exit
4.36
-
-
1.0005
3
Condenser exit
-4
0.0
0.16
40.46
4
Evaporator inlet
1.052
-
-
-
40.46
5
Compressor inlet
3.3
-12
1.0
0.9457
245.36
6s
Compressor exit
13.89
-
-
0.9457
7
Condenser exit
36
0.0
0.326
89.29
8
Evaporator