THERMODYNAMICS Principles & Applications
Prof. Dr. Nuri KAYANSAYAN DOKUZ EYLUL UNIVERSITY Me c ha nic a l Engineering Department
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: 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 Bu baskının bütün hakları Nobel Akademik Yayıncılık Eğitim Danışmanlık Tic. Ltd. Şti.ne aittir. Yayınevinin yazılı izni olmaksızın, kitabın tümünün veya bir kısmının elektronik, mekanik ya da fotokopi yoluyla basımı, yayımı, çoğaltımı ve dağıtımı yapılamaz.
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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
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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
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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)
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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"
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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
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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
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Isentropic compressibility Volume expansivity Emissivity and exchanger effectiveness Efficiency Boiler efficiency Isentropic efficiency
J
Exergetic efficiency Isothermal compressibility
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Angular displacement Viscosity Joule-Thomson coefficient Kinematic viscosity Density Surface tension and Stefan-Boltzmann constant Exergy rate Specific exergy
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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.
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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.
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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 ) 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 3.32 106 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 99.63C 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, cpcv. 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 (p0), “ 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
272Tc2 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, 1312.8, 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 103T 6.607 106 T 2 2.002 109 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.
For an ideal gas, cvo c po R .
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.
CHAPTER 2 THERMODYNAMIC PROPERTIES OF SYSTEMS 61 p.
q.
At the critical state, the saturated liquid and the saturated vapor specific volumes intercept.
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 0C 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 38C 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. 58.8C,
b.
58.5C,
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/kgC, b.
2.97 kJ/kgC,
c. 3.97 kJ/kg°C, d.
14.
0.97 kJ/kg°C.
10 kg of steam at 600C 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 30C, 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 30C, 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 250C. 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 250C, 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 * 2V / 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 25C, 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 50C,
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, 50C) 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 , V2 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,
1V1 2V2 3V3
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 V3 8.653 m3 /min . Considering the definition of volumetric flow rate, the velocity of air at the exchanger exit is: V3
V3 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 +5C. As the valve is opened, the saturated vapor flows through a heat exchanger and exits the exchanger at 500kPa in pressure, 50C 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 +5C, 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, 40C 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 (2p / )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
V2 0.1 , m a 0.1176 kg/s and m liq 2 0.1176 (0.029 0.015) 1.64 103 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
2p
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 2p /
(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, Vo / Vw 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,
Vo 12 x1.066 ,
or
Vo 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 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 (mT0. 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 27C. Work is done against the air by running the fan and the final temperature becomes 527C. 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 227C and 27C supplies work to a reversible refrigeration system that cools down a space at -50C and rejects heat to an environment / Q . at 27C. Evaluate Q 1 2 Solution: T In regard to Eq. (5.16), the work output of the reversible heat engine is, Whe Q1 1 0 , and the amount of work consumed T1 T0 1 T Q T by the reversible refrigeration is, Whe Q1 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 22C, the required heat supply for an outside temperature of 2C 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 -15C 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 20C 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 1C temperature difference (2000kJ/hC). 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 10Sint / 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 227C and the other at 77C. The rate of heat conduction through the rod is 500 kJ/h. For surroundings at a temperature of 27C, 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 Q1 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 97C 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 27C 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 = 25C. 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=25C. 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 0C 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 25C, 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 = 0C, 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 20C 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 20C.
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 Wrev X W . Hence the maximum power is Wrrev = 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, 200C) 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, 50C), and is heated to a temperature 10C 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=180C, T 3=50C, T 4=170C, 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 25C 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, 20C) at a flow rate of 2.5 kg/s is mixed with steam (200 kPa, 300C) 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 25C. Water leaves the chamber at (200 kPa, 60C). Determine, a.
The mass flow rate of steam,
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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, 747C), and exits at (70 kPa, 500C). For a surroundings temperature of 17C, 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, 600C) with a velocity of 80m/s and exits at (50 kPa, 100C) and the speed at the exit is 140 m/s. For ambient conditions of 100 kPa, 25C, 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.
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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 Qin 1 0 Q e T i T0 0 m s h1 h2 T0 s1 s2 or I 4464.528 kW .
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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=17C, 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)
Wt 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
Wt / 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 Wc in Eq. (5.44), the exergy equation reduces to the following form, m 2 1 Wc 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 Wc / m
(5.54)
Example 5.30: Determine the exergy efficiency of the adiabatic compressor explained in Example (4.26). Surroundings at T0=27C, p0=100kpa. Solution: Referring to Figure 4.43, the compressor intakes the air at atmospheric conditions, and compresses it to (500 kPa, 227C). 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, Wc 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
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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 140C at a rate of 24 kg /min and leaves the exchanger at 0.5 MPa and 60C. 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, 25C. The air pressure at the exit is 110 kPa. If surroundings is at T0=25C, 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 33 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 = 87.65C
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)
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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 27C, 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 Q1 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
Wnet W net T Wrev rev Q1 1 0 T1
(5.57)
Similarly the exergy supplied to the refrigeration system in Figure 5.38b is Wc , 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 Wc 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 We 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, Qin 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 62C. 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, Qin Wel Ql 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 = Qin 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, Wel 1 kW, and energy out, Qin 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 (+55C) and the surroundings, b. heat interaction of 550 kJ between a reservoir at (-55C) and the surroundings, c. heat interaction of 550 kJ between the two reservoirs (+55C) and (-55C). 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 10C 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 20C at 100 kPa, and the other (B) 1 kg of water at 500C and 20 MPa. The surroundings is air at 100 kPa, and 20C. 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 =20C, p 2 =100kPa, T2=100C b. air, p1=1bar, T1=27C, p2=3bar, T2=377C c. carbon dioxide, p1=150kPa, T1=30C, p2=300kPa, T2=300C 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 40C 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, 22C. 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 400C 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 180C 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 600C 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 27C. 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 260C 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 20C. 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, 400C 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 550C until its temperature reaches 400C, 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 5C. 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 20C to 60C by flowing through a heat exchanger. The hot fluid of the exchanger being geothermal water (cp=4.31 kJ/kgK) is at 165C 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 75C is losing heat to the surroundings at 5C by natural convection. The rate of heat loss is calculated by Q=UAT 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 90C and exits at 60C 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, 240C in a piston-cylinder device expands to 300 kPa, 160C by doing work and releases 3 kJ of heat to surroundings at 100 kPa, 27C. 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 40C. 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.
An evacuated space has a negative exergy
i.
The exergy change of a flowing stream is determined by dh T0 ds .
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 maximum work obtainable from a
Choose the correct answer: 1.
a.
2.
non-flow system is h1 h0 T0 s1 s0 m.
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 )
Wnet (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.
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 vp and the above equation for shaft work can be modified as, wshaft u vp 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 vp 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, vp 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 = 22C, 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 350C and submerged into a tank containing 200L of water at 22C. For an ambient temperature at 22C, 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=24.713C. 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 2500T0 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 5000T T kJ/h. h
h
0
a.
Determine the minimum driving power for the heat pump for Tr=-12C, Th=21C, and T0=0C.
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, 350C) 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, 25C) to (5 bar, 210C). 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, Wrev 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.
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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, -15C 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 V2 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 V1 / 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 -30C, and in chemical industries, the liquefaction of certain gases might require -150C 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 -30C, 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, qe 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 Wc1 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,
Wc 2 m 3 h4 h3 1.466 273.9 252 32.095 kW . The total power consumed is Wt Wc1 Wc 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 cT vp
(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: vp ke wshaft
- cT vp 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 ( sgen 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 1M 2 . Substituting these two ratios into the gas equation, k 1M 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 M0. 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, pbT3).
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, V1 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: V2 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 1.4 104 kg fuel/cycle. The heat input per cycle is 19
q f m f H u 1.4 x104 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) 201.364 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, Wnet 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 ( qin ). 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 ( qout ). 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 qin 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 qout 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 qout T T 1 4 1 1 1 4 1 qin 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
Wnet 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 pT3t 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 ct 3 T1
k /2k 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/(20.4)
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 ct
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 kk1 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
Wnet 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 ( qout ). 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 ( qin ) 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 uboiler 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 Wnet Q in g Rb Q fuel Wnet 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 Wnet flow rate is m s
W g g
250000 263157.89 kW. The corresponding steam mass 0.95
Wnet 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
Wnnet 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, Wnet 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 Wnnet 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 Wt 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. Wnet 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.
Wnet 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 0.0011000 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
Wnet steam 1.233 3373.7 2790 0.494 3478.5 2460.17 0.489 11.097 o r Wnet 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 qout 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 +5C and -70C, 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, 10C. 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=+2C, and T0=+23C, 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 -10C, and +34C. 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 V1 mv 241.3 97.31
V1 0.555 0.099 or V1 3.296 m3 / min . b.
Wcs m h2 s h1 , s1 s2 s 0.925kJ/kgK and h2 s 271.52kJ/kg , Wccs 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 V1 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 Wc 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 Wcs 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 -10C coming from evaporator from -10C to 0C with liquid coming from condenser at +32C. 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 -30C to harden the produced ice cream and another operates at -1C 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 -5C. The other evaporator is for frozen food storage unit and operates at -20C. 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 -5C 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 Wc
(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=-20C. The higher temperature evaporator having a capacity of 3tons operates at -5C. 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 Wc =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 Wc 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 -30C 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 Wc1 Wc 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=-30C, 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. qe 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 Wc m h9 h1 or Wc 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 Wt 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 -40C for fast freeze and then stored at -20C. 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 Wc1 Wc 2
(9.6)
Example 9.5 In a food processing unit, tuna fish first shocked at -32C and then is stored at -18C by the ammonia refrigeration system shown in Figure 9.10 for which the condenser operates at +40C. 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 Wcl for the high pressure compressor, Wch
Q eevl 250 0.191kg/s , similarly, h1 h7 1400 1400.81 81 97.68
m 4 h4
m 2 h2
h1 00.191 1911475 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
Vst 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 -18C and +40C evaporator and condenser temperatures respectively. A pressure drop of 0.04bar occurs at the suction line, and the refrigerant enters the compressor with +6C 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
(-12C, 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 Wef 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 Wcs 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 -10C evaporator and +36C 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
